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APPENDIX

Bessel Functions

It is important to be able to carry out controlled computational experiments to evaluate computing procedures. For this purpose it is necessary to have reliable numerical data. The various tables of Bessel Functions, e.g.,

BAAS, Bessel Functions, I (1937), II (1952); The Airy Integral (1946), Cambridge Univ. Press.

RSMT, Bessel Functions, III (1960), IV (1964), Cambridge Univ. Press. Harvard, Annals of Computation Lab., 2-14, 1945-1951, Harvard

Univ. Press. NBS, Tables of Bessel functions of fractional order, I (1948), II (1949),

Columbia Univ. Press.

are ideal for this purpose, supplementing the less extensive tables and graphs in

E. Jahnke, F. Erode, F. LOsch, Tables of higher functions, McGrawHill, 1960.

M. Abramowitz-I. A. Stegun, Handbook of mathematical functions, NBS Applied Math. Series, 55, 1964.

We therefore give an account of some aspects of the theory of Bessel Functions which are relevant in the present context. The definitive treatise is

G. N. Watson, Bessel Functions, Cambridge Univ. Press, 1922-1941.

We shall confine our attention largely to In{x)-the Bessel function of the first kind of integral order n. Two definitions are:

(1) 1 12.... 11 .... In(x) = 21T cos (-x sin 0 +nO) dO = 1T cos (nO -x sin O)dO

(2) 00 (-1)' G)n+2r

In(x) = L, ,-r=O r .(n + r).

Our first objective is to establish the equiValence of (1), (2) and we begin by finding some properties of the In's. We indicate by subscripts on the numbers of our formulas the definitions on which they are based.

Using the fact that cos A -cos B = 2 sin !(A + B) sin !(B - A) we obtain

142 Appendix

from (1), with n ~ 1:

I n - 1(X)-Jn +1(X) = ! r'7l" sin (-x sin O+nO) sin OdO

= 2J~(x)

assuming it permissible to differentiate (1) with respect to x, i.e., to interchange the operations d/ dx and I ... dO on the right.

We note that when n = 1 we have

1 r2'71" J1(x) = 2'IT.Io cos (0 - x sin 0) dO

= ! 1'71" cos 0 cos (x sin 0) dO + ! 1'71" sin 0 sin (x sin 0) dO.

The first integral vanishes because we can write it as

1 L'7I" 1 1'71"/2 1 1'71" - =- +- =11 +12; 'IT 'IT 'IT ",/2

if we change the variable from 0 to </J = 'IT - 0 in 12 we see that 12 = -11•

Thus

J1(x) = ! L'7I" sin 0 sin (x sin 0) dO.

Assuming that differentiation under the sign of integration is permissible we find

and so we have

Jo(x) = ! 1'71" -sin (-x sin O)(-sin 0) dO

= ~1 1'71" sin 0 sin (x sin 0) dO

1. THE DIFFERENTIAL EQUATION

We first note that the series (2) is convergent for all x: the ratio of the r+1-st term to the r-th is -x2 /(4r(n + r)) and this tends to zero as r~oo for

Bessel Functions 143

any x, n. We can therefore differentiate term by term to find

, _ co (-1Y{n+2r) (~)n+2r-l In(x)- r~o r!(n+r)!2 2 '

" _ co (-1Y{n+2r)(n+2r-1) (x)n+2r-2 In(x) - L '( )'22 -2 .

r=O r. n+r .

Consider the coefficient of xn+2r in x2J~(x)+xJ~(x)+(x2-n2)Jn(x): it is (-i)'/[r!(n + r)!2n+2r] times

(n +2r)(n +2r-1)+(n+2r)-4r(n + r)- n2

which is zero. This establishes (52). It is easy to check that for general values of v

(2') G)V co (-1)' (x)2r J (x)= - -

v r~o r!r(r+ 1 + v) 2

also satisfies the equation

(8)

Indeed, if v is not an integer, Lv(x) also satisfies this equation and J", Lv form a pair of independent solutions to (8). We note from (2') that, for n =0, 1,2, ...

( x)-n co (-1)' (X)2r Ln(x)= 2: r~or!r(r+1-n) 2:

and the coefficients of the terms with index 0, 1, ... , n -ion the right vanish because of the behavior of the r(t) when t = 1- n, 2- n, . .. ,0. Hence, putting s = r - n,

( x)-n co ( -1)' (x)2r Ln(x)= 2: r~nr!r(r+1-n) 2:

{~)-n co (-1t+. (x)2'+2n = \2 .~o r(s+ 1 + n)r(s + 1) 2:

( x)n co (-1)' (X) 2.

=(-1t 2: .~os!r(s+l+n) 2: = (_1)RJn (X).

Therefore, in the case when v is an integer we have to seek elsewhere for another solution of (8). This leads to the introduction of the Bessel functions of the second kind - we shall not be concerned with these here.

144 Appendix

2. MIxED RECURRENCE RELATIONS

These are simple consequences of (2). We deal only with the first relation. The coefficient of x2v+2r-1 on the right is

1 (-1)' 1 1 (-1)' 1 --'---'---. _. 2r+2v =_. . -. (r+ v)

2v r!f(r+l+v) 22r 2v- 1 r!(r+v)f(r+v) 22r

which after cancelling the factor (r + v), is exactly the coefficient of x2v+2r-1 on the left.

H we differentiate out the left hand sides of (92 ) and multiply across by x-v and -XV we obtain

v -J~(x)+- J)x) = Jv+1(x).

x

From these, by adding and subtracting, we obtain

2v - Jv(x) = Jv- 1(X)+Jv+1(x); 2J~(x) = Jv- 1 x)-Jv+1(x). x

We now show how to derive the first part of (1l~ on the basis of the original definition (1). [The second part of (112) is just (31),]

Elementary trigonometry gives, if x -F 0,

cos «n + 1)8- x sin 8)+cos «n -1)8- x sin 8)

= 2 cos 8 cos (n8 - x sin 8)

= 2n cos (n8 - x sin 8) -~ (n - x cos 8) cos (n8 - x sin 8). x x

H we integrate across with respect to 8 between 0,21T we find

2n 2 L2mr 21TJn +1(X) + 21TJn _ 1(X) =- 21TJn (X)-- cos 'P d8

x x

where 'P = n8 - x sin 8. The last integral vanishes and so

We are now able to identify (1), (2). We do this by checking the results first for n = 0, and then using the fact the recurrence relations are satisfied on the basis of (1), (2).


Since the series for cos t is convergent for all t we can expand the integrand in (1) as a power series and integrate using the fact that for all even r

smrOdO= .-1(/2)'71" • (r-1)(r-3) ... (1) 'IT

o r(r-2) . .. (2) 2

to find

1 r'7l" . 1 r'7l" 00 (-1)'x2r sin2r 0 - cos (x sm 0) dO =- L (2 )' dO 'IT 'IT r=O r .

00 [( _1)'x2r 1 r'7l" . ] 00 (-1)' (x)2r = L (2 ),.- sm2r OdO = L -( ,)2 -2 .

r=O r. 'IT r=O r.

This completes the identification for the case n = o. We now note that a special case of the second relation of (92) is

J6(x) = -J1(x). Combining this with (41) and the identification already established for n = 0, we establish identification for n = 1.

To complete our program we use (31) and (112) to proceed to the case n + 1 from nand n-1.

3. SOLUTION OF THE DIFFERENTIAL EQUATION y"+ AXY = 0, y(O) = 0

We write J = J 1/3(~A 1/2X3/2) and shall show that y = x 1/2 J is the solution. Differentiating and substituting we find

y" + AXY = (AX3/2_h-3/2)J +~A 112 J' + AX 3/2 J".

By definition J1/3(x) satisfies

X2 !1/3(X) + xJi/3(x) + (x2-b)J1/ix) = o. We now replace x by ~A 1I2X3/2 in this equation and solve for h 3/2 J" getting

AX3/2 J" = -~A 1/2 J' - (h3/2 -ix-3/2)J

which when substituted in the expression for Y" + AXY cancels out everything.

4. SoLUTION OF THE DIFFERENTIAL EQUATION y' =X2_y2, yeO) = 1

It has been found convenient to introduce the modified Bessel function. For n integral

(12)

satisfies the differential equation

(13)

146 Appendix

The properties of the I's can be obtained by translation of those of the 1's just as we obtain properties of the hyperbolic functions from those of the trigonometric ones, or by a parallel development. Some care has to be taken in the determination of the multiplier on the right in (12) when we deal with cases when n is not an integer.

We show that the complete solution of the differential equation u"x 2 u =0 is

(14)

and

Differentiating we find, omitting the argument 1X2 of the I's,

u' =1X-1/2(CIIl/4 + ... )+ XlI2 . X(clI~/4 + ... )

u" = -iX-1/2(CIIl/4 + .. . )+hl/2(clI~/4 + ... )

+~xlI2(cII~/4 + ... )+ XS/2(clI1I4 + ... )

so that

u" - x2u = -iX-3/2[(1 + 4X4)(clIlI4 + .. . )-8x2(clI~/4 + ... )-4x4(clIlI4 + ... )].

Now IlIit) satisfies

t2y"+ ty'-<l6+t2)y = 0

and replacing t by 1X2, this becomes

The same equation is satisfied by LlI4 and so we see that u" - x2 u = O. We have already noted that when v is not an integer, l±v are independent solutions of (8) and I±v are consequently independent solutions of (13). This establishes our assertion about (14).

The equation y' = X2 - y2 is of the Riccati type and can be transformed to the linear equation u" - x2 u = 0 by the substitution y = u' /u. From what we have just seen the solution is

Cl(Xl/2IlIih2»' +CiXlI2L l/4(1X2)' y=

CIXlI2Il/i1X2) + C2Xl/2 LlIix2)

where the c's will be determined by the initial condition y(O) = 1. We now use the following transforms of (72):

Bessel Functions

in the case when n = ±~ to get

C1X 3/2 L3/4(~X2) + C2X3/2 13/4(~x2) y=

Cl X 1/2 Il /ih2) + C2X 1/2 L 1/4(h2)'

For x near zero Iv(x)~(x/2t/f(v+1) so that

y ~2clf(3/4)/c2f(1/4).

Hence (Cl/C2) = f(1/4)/f(3/4) and the solution is

rmI_3/i~x2) + 2f(~)I3/4(~X2) y = X f(~)Il/4(h2)+2f(~)Il/4(h2) .

147

From the NBS tables of Bessel functions of fractional order, II we find

(1) = (3.625609908)(0.9800767696) + 2(1.22541 6702)(0.39858 50517) y (3.625609908)(0.81967 59660)+2(1.225416702)(1.25197 01940)

1.848449429 0 75001 5703 2.464547638' .

5. RELATIONS OF JACOBI AND HANSEN

We shall now establish the fact that

Jo(z) + 2J2(z) + 2Jiz) + ... = 1,

which was mentioned in Chapter 10. We do this on the basis of the series definition (2).

For s"?m the coefficient of (iz)2s in J2m (z) is (-l)m-s[((s-m)!)

«s+m)!)]-l. Hence the coefficient of (~Z)2S in Jo(z)+2I:~lJ2m(Z) is

-1 -- + + -1 s __ s[ 1 2 2 ] ( ) s!s! (s-l)!(s+l)! ... ( ) 0!(2s)!

= ~;:;; [(~S)-2C ~ 1) + ... +(_1)S2(~S)]. The expression [ ... ] can be readily identified with the expansion of (-W [1-1]2S by replacing, for r = 1, 2, ... , s,

2( 2S) b (2S) (2S) s-r y s-r + s+r .

Hence all the coefficients in Jo(z)+2I:~lJ2m(Z) vanish, except the first, which is 1. This is what we want, but so far our work has been formal.

148 Appendix

To justify our manipulation, which is essentially the transformation of a sum by rows of a double series into one by columns, it will be enough to establish absolute convergence of the double series. This is easy, for changing signs in the argument of the preceding paragraph shows that the double series is dominated by

" 11 1)2S 1 ( )2s _ " IzI2s _ I I ~ (zZ (2s)! 1+1 - ~(2s)!-cosh Z <00.

We shall next establish the following relation of P. A. Hansen, which can also be used for normalization in backward recurrences:

J~(x)+2Ji(x)+2Pz(x)+ . .. = 1.

We note that for x real and r = 1, 2, ... this gives the inequalities

We begin by determining the coefficient of (-1)S(!xfs in Pn(x). It is

(-1)n Stn 1 r=O r!(r + n)!(s - n - r)!(s - r)!

( -1)n s-n (S) ( S )

= (S!)2 r~o r s-n-r

= (-1)n (.!.)2( 2s ) s! s-n

by a standard result, ascribed to Vandermonde. [To get a combination of (s - n) objects from 2s objects we take r from one fixed set of s objects and (s - n - r) from the remaining set of s objects, and do this for r = 0, 1, ... , s - n.]

To get the coefficient of (-1)S (!x )25 in the sum

J~+2Ji+ ...

we have to add, and we get, as before,

6. BOUNDS ON DERIVATIVES OF Jo(X)

It is easy to show that for x real, r = 0, 1, 2, ...

s;e 1 s=1


In fact from the representation (1), if we can differentiate under the integral SIgn,

(-1)' f" 1~2r)(x) = -:;;:- .10 sin2n 0 cos (x sin 0) dO

and

(_1)'+1 f" l~r+1)(X) = 7T .10 Sin2r+1 0 sin (x sin 0) dO.

The stated result follows since the integrands do not exceed 1 in absolute value.

7. ZEROS OF BESSEL FUNCTIONS

We have made use on several occasions of the zeros of Bessel functions. We shall state some of the results: (15) Iv (x) has an infinite number of real zeros for any real v and, if v> -1, all the zeros are real. (16) Except possibly for x = 0, the zeros are simple. (17) Denoting by iv,I> iv,2' ... the positive zeros of lv(x), arranged in ascending order, for v> -1 we have

0<iv,1 <iv+1,1 <iv,2<iv+1,2<iv,3<'"

i.e., the zeros of IV+1 are interlaced with those of Iv. (18) Denoting by iv, i~ the least positive zeros of lv, l~ we have, when v>O,

iv>v,i~>v.

We shall establish part of (15) and (16), (17), (18). Bessel and Lommel showed that if -!<1I:5! then lv(x) has the following signs in intervals m7T, (m+!)7T:

+ - + -

6 !~ ~ ~11' 211' ~11' 311' rlT 4~ • +. • -. .

2n7T (2n + 1)7T(2n + 2)7T

It follows that lv(x), -!<V:5!, has an odd number of zeros in G7T,7T), (~7T, 27T), .. .. We shall establish the sign pattern when v = O. We have

2 r<1I2)" 10(X)=;.Io cos (x sint) dt.

Suppose x = (m +!0)7T where 0:5 0:51. Then if we put x sin t =!7TU, dt = du/.J(2m+0)2- u2 we get

lo(x) =~ f2m+1I cos (!7TU) du . 7T Jo J(2m+0)2- u2

150 Appendix

If we write

J 2m+8 = L2 14 12m 12m+

8 + + ... + +

o m-2 m

= -VI +V2+" . + (-l)mvm + (-l)mv:"

then it is easy to show (draw a graph of the integrand) that

and so sign Io(x) = (-l)m.

The step from - ~:s; v:s;~ to general v is carried out as follows: Rolle's theorem and the recurrence relations

show that between any consecutive pair of zeros of x-VIv(x) there is at least one zero of x-vI v+1(x) and between any consecutive pair of zeros of xv+IIv+1(x) there is at least one zero of xv+IIv(x). Thus we can move up and down from - ~:s; v :s;~.

To establish (16) we use the differential equation

Z2Y"+zy'+(Z2_ V2)y =0

to conclude that if there was a double zero at a point x'" 0 then y"(O) and all higher derivatives would vanish there and so y would be identically zero.

To establish (17) we use, ~ before, the recurrence relations. To establish (18) we proceed as follows: When v> 0 the power series of Iv and I~ show that they are positive for

small positive x. The differential equation, written in the form

shows that if x < v and Iv > 0 then xJ~ is positive and increasing and so Iv is increasing. That is, so long as 0 < x < v both Iv and xJ~ are positive and increasing and so cannot vanish.

8. THE Amy INTEGRAL

[Detailed references to the sources of this discussion will be found at the end of this section.]

This is defined as

(19) 1 Loo Ai(x) = 'Tf' cos (lt 3 + xt) dt


and satisfies the differential equation

(20) y"=xy.

A direct proof of this fact, according to Hardy, "is not a particularly simple matter". It, together with

(21) Bi(x) = ! r {exp (_~t3+ xt)+sin (~t3+ xt)} dt,

has been tabulated extensively [British Association, Math. Tables, Partvolume B, 1946]. The pair Ai(x), Bi(x) are an independent pair of solutions of the equation (20) and were expressed in terms of the Bessel functions I±1/3' J±1/3 with argument ~X3/2 by Wirtinger and Nicholson. (See (25) below.)

First we have to discuss the convergence of (19). This follows from the fact which can be proved by integration by parts, that,

LX> Q(x)eiP(x) dx

where P, Q are polynomials, P having real coefficients, is convergent if and only if q ~ p - 2 where q = degree Q, p = degree P. We shall discuss the problem directly by a method which we use again.

Since 1 = (t2+ x - x)/t2 we have

I(b) = f" cos Gt3 + xt) dt = I~ (t2 + x) cos (~t3 + xt) dt/t2

-x fO cos(~t3+xt)dt/f=I1+12' say.

We apply the Second Mean Value TheoreIl}. to 11 to get, for some B~b:

so that

152 Appendix

Clearly

Thus

and so the integral (19) is convergent and indeed uniformly convergent in any interval [-X, Xl We shall use the uniformity at an essential stage later.

The next step is to differentiate (19) formally to get

(22) -1T-1 1"0 t sin (1t3 + xt) dt

and then to show that this integral is uniformly convergent with respect to x, so that we can apply a standard theorem on differentiation under the integral sign. [See e.g., Titchmarsh, 59.]

Since

we have

where

11 = 1= {(t2+ x) sin (1t3+ xt)}t- 1 dt,

I2=-x i= {(f+x)sin(1t3+xt)}t-3 dt,

13 = x31= {sin Gt3+ xt)}t-3 dt.

The Second Mean Value Theorem can be applied to the first two integrals giving

11 = b-1 sin GBi+ xB1)

12 = -xb-3 sin GBi + xB2)

where b :5B1> b :5B2. The third integral is estimated trivially as

1131 :5~x3b-2.


Hence

and so the integral I(b) is convergent and indeed uniformly convergent with respect to x in any interval [-X, X]. It follows, from the theorem cited, that

Ai'(x) = -1T-1 100

t sin (~t3+ xt) dt.

If we differentiate formally again we are led to

tOO t2 sin (~t3 + xt) dt:

but this integrand does not converge. This follows from the fact stated above (p. 151), or it can be shown directly as follows. If we change the variable from t to y = ~t3 + xt then dy = (f + x) dt and since for large t, y ~ ~t3 the integral becomes approximately

foo sin y dy

which is divergent. To get the result we require we follow Stolz and Gibson in using a

technique of de la Vallee Poussin. We describe it formally in a general case and then deal with the special case rigorously.

Suppose that

F(x) = r f(x, t) dt

and that fx(x, t) is continuous. Then

f(X, t)- f(O, t) = IX fx(x, t) dx.

If we integrate across with respect to t between 0, T and let T ~ 00 we get

(23) F(X)- F(O) = ;~ [IT dt IX fAx, t) dX]

= ~ [IX dx IT fx(x, t) dt]

The easy case is when we can interchange limT->oo and S . .. dx to get

F(X)- F(O) = IX dx 100 fx(x, t) dt

154 Appendix

which when differentiated gives

F'(x) = r fx(x, t) dt.

However we are interested in the case when this inversion is not possible and we proceed as follows. Break up the inner integral in (23) into

IT fx(x, t) dt = (f)(X, T)+ iT I{I(x, t) dt

where

while

IT I{I(x, t) dt

is uniformly convergent with respect to x so that

~~ iX dx iT I{I(x, t) dt = iX dx IX> I{I(x, t) dt.

In these circumstances

F(X) - F(O) = LX dx Loo I{I(x, t) dt

and, as before

F'(x) = Loo I{I(x, t) dt.

In the special case we have f(x, t) = t sin (it3 + xt) and we want to find appropriate (f), 1/1 so that

IT t2 cosGt3 +xt)dt=(f)(x, T)+ iT I/I(x,t)dt.

We write

so that

IT t2 cos (it 3 +xt) dt= iT (t2 +x) cos (it3 +xt) dt

-x iT cos (it3 +xt) dt = sin (iT3 +xT)-x iT cos (it3 +xt) dt.

Bessel Functions

Now as T --+ 00,

Ix • (lT3 T)d cos(iT3)-cos(iT3+x1) 0 SID 3 + IL IL = --+

o T

and the uniform convergence of the integral

IT cos (it 3 + xt) dt

has already been established. We are therefore able to conclude that

d~ [-1T-1 r t sin Gt3 +xt) dt] = +1T-1X r cos Gt3 +xt) dt

i.e.

Ai"(x) = xAi(x)

as required.

REFERENCES

155

G. H. HARDy, On certain definite integrals considered by Airy and Stokes, Quart. J. Math. 41 (1910), 226-240 = Coll. Papers 4 (1969), 460-474.

O. STOLZ, Grundzuge der Differential- und Integral-Rechnung, B3 (1899), Leipzig.

G. A. GmsoN, Advanced Calculus, Macmillan, London 1931,439, 452. C. J. de la VALLEE POUSSIN, Etude des integrales a limites infinies pour

lesquelles la function sous Ie signe est continue, Ann. Soc. Scient. de Bruxelles 16B (1892), 150-180.

E. C. TrrcHMARsH, Theory of functions, 1939. Oxford.

9. REPRESENTATION OF Ai(x) AS A POWER SERIES

Let

Next, differentiating n times the equation y" - xy = 0 we get:

(24) y(n+2) _ xy(n) - ny(n-l) = 0

and then putting x = 0 we get

y(n+2)(0) - ny(n-l)(O) = O.

156 Appendix

We therefore find a Maclaurin series for Ai(x) of the form

ao(l+ ;, x3+ ... ) Ai(x) = ;

-a1(x+ 41 x4+ ... )

ao = 3-2/3/[(2/3) = 0.35503 where

al = r 1/ 3/f(1/3) = 0.25882'

A real variable derivation of the expressions for ao and al follows. Write

where n > 0, a> 0, b > O. Put a = r cos 8, b = r sin 8, rx = y, ern = u, srn = v, to get

u(8)== U = 100 e-YOOS6yn-l cos (y sin 8)dy,

v(8) == v = 100

e-YOOS6yn-l sin (y sin 8) dy.

We differentiate formally to get

du = roo e-YOOS6yn sin 8 cos (y sin 8) dy - roo e-YOOS6yn cos 8 sin (y sin8) dy ~t t .

Both these integrals are uniformly convergent with respect to 8 in -7T/2 < -8o!5:.8 !5:.80 < 7T/2 because they are dominated by

1"" r(n +1) e-COS6oYyR dy = . o (cos 8ot+1

Hence the differentiation is legitimate. We can write

du = roo yR ~ {e-Y oos6 sin (y sin 8)} dy d8 t dy

so that, integrating by parts,

du = -n roo yn-l . e-YCOS6 sin(y sin 8) dy = -nv d8 t .

In the same way

dv d8=nu


and, combining, we get

so that

u(O) = A cos nO + B sin nO.

When 0=0

v (0) = 0,

so that

u(O) = r(n) cos nO, v(O) = r(n) sin nO.

It is clear that c, s are continuous functions of a at a = 0 and we can deduce from

that

foo -ax -1 cos b d r(n) cos e xn X x=-- nO o sin rn sin

n7T

foo b r(n) cos-2 cos x dx = ___ _ o x 1- n bn '

fooSinbX _ r(n)sinT ~dx- bn ,

o X

O<l-n<l

0<1-n<2.

10. REpRESENTATION OF Ai(x) IN TERMS OF BESSEL FUNCfIONS

From the power series representations of J±1/3(X) and I±l/ix) we easily check that if x > 0, ~ = ~X3/2, then

(25) Ai(x) =iXl/2{Ll/i~) - 11/i~)},

Ai(-x) =ix1/2{J -1/3(~)+ J1/3(~)}.

Solutions to Selected Problems

Chapterl

1.3. Solution Take 8> 0 and choose a so that 8(1 + a) = 1. This implies a> 0 and so

Hence for any 8> 0, 8n < 8 if n> no(8) = (a8 tl. Thus 8n -+ O. In case 8 -+ 0 we observe that 18nl = 181n.

1.5. Solution Differentiating we find

~ [lOg (l_1.)X] =~ [x log (1-1.)] = 10g(1-1.) +_1 . ~. x dx x dx x x 1-1. x2

x

1 2 =-2+-3+ ... >0.

2x 3x

We use the logarithmic series and the binomial series (in the geometric case) at *.

Similarly

d [ (l)X-l] (1) 1 1 1 - log 1-- =log 1-- +-= ----- ... <0. dx x x X 2X2 3x3

1.6. Solution See Problem 1.13 below for the value of M(1,0.2). The results of

Gauss in the case ao =./2, bo = 1 are:

ao = 1.414213562373095048802 at = 1.20710 6781186547524401 a2 = 1.19815 6948094634295559 a3 = 1.198140234793877209083 a4 = 1.198140234735592207441

bo = 1.000000000000000 00000 0 bt = 1.18920711500272106671 7 b2 = 1.1981235214 03120122607 b3 = 1.198140234677307205798

b4 = 1.198140234735592207439

162 Solutions to Selected Problems

Note that the observed values of a,. - bn are much smaller than the estimated values of a,. - bn •

1.7. Solution

M(6000, 1200) = 6000 M(l, 0.2) = 3.12480 9828 x 103 •

1.B. Solution It is well known that if A, G, H are the arithmetic, geometric and

harmonic means of two positive numbers then A > G > H with equalities if and only if the given numbers are equal. This implies the monotony statements. Boundedness is obvious. Hence a,.-+a, bn -+(3. Since a,.+1 = !(a,. + bn ), we have a = !(a + (3) so that (3 = a = I, say. It is also clear that

so that a,.bn = aobo and, passing to the limit, [2 = aobo. Since I> 0, it follows that I is the geometric mean of ao, boo

To discuss the rate of convergence we note that

a,.+1 - bn +1 = !(a,. + bn ) - [2a,.b.J(a,. + bn )]

= ![(a,. + bn )2-4a,.bn ]/(a,. + bn )

= (a,. - bn )2/(4a,.+I).

Thus the convergence is of the same type as that of the Gaussian algorithm.

Observe that this algorithm suggests a method for finding square roots. If we want to find IN then we start with

bo=N/xo.

A moment's thought reveals that this is really the familiar algorithm

a,.+1 = ~( a,. + ~) which will be discussed in Chapter 3.

If we take ao =./2, bo = 1, then 1= 2114 = 1.18920711.

1.9. Solution Let ao = aol, (30 = bOI and form the Gaussian sequences for ao, (30. We

assert that ~ = a;;\ (3n = b;;l. This is true for n = O. Assume that it is true for n = r then

Chapter 1

and

Hence lim a,. = lim bn = [M(ao1, bo1}r1.

1.10. Solution [We use numerical values from the NBS Handbook.]

a} Borchardt

ao=0.2, (30 = 1, cos 6 =0.2, sin 6 = ../0.96 = 0.97979 590.

From NBS Handbook p. 169:

cos 1.369 = 0.20042 95292, sin 1.369 = 0.9797081218.

If

0.2= cos (1.369+e) = cos 1.369- e sin 1.369-!e2 cos 1.369+ ...

then a first approximation to e is

= 0.0004295292 . 0 00043 8 e1 sin 1.369 .. .

We now compute

cos (1.36943 8) = cos 1.369 = 0.2004295292

- e1 sin 1.369 - 0.0004291122

-!eicos1.369- 191

= 0.20000 03979.

The next approximation is given by

= 0.000000 3979 = 0 00000 041 e2 sin 1.36943 8' .

We take

6 = 1.36943 841

and then

I = (sin 6/6) = 0.715472776.

(30 = 1, cosh t =..J2, sinh t= 1.

From NBS Handbook p. 22: t = 0.881373587 and

I = sinh t = 1.13459266. t

163


b) Carlson.

bo =0.2:

ao=..fi, bo= 1:

1- (Q.96 _ ~0.48 _ I 0.48 - V2lOg5 - log 5 - V1.60943 79124

= 0.546114246.

I = V 210: J2 = VIO~ 2 = VO.6931~ 71806

= 1.20112240.

[The logarithms are taken from the NBS Handbook, p. 2.]

1.11. Solution The direct treatment is the following. Take the case when Ao = cos2 8,

Bo = 1. Then induction gives

A..+l = cos 8 cos (2-18) ... cos (Tn8) cos2 (2-n- 18),

Bn+1 = cos 8 cos (2-18) ... cos (2-n8).

As before

2n+1 sin (2-n 8)Bn +1 = sin 28

and so

lim Bn = sin 28/[lim 2n+1 sin (2-n8)] = sin 28/28.

1.12. Solution

and

Since (a,J f3n) = cos 6/2n we have

f(n) = 2n arc cos (a,J~n) = 8

g(n) = 2n(~~-a~l/2 = 2n cos (2-16) ... (cos 2-n8)[1-cos2 2-n8]112

= 2n~n sin 2-n6

= sin 8

so that !(n), g(n) are independent of n. Next we note that

f(n) arcsin «~~-a~)112/~n) g(n) = ~n«~~-a~)112/~n)

Since ·an ~ l·and ~n ~ l we have

so that I = g(O)/f(O).

Chapter 1 165

1.14. Solution Since K' depends on k through k 2 some tables have m = k 2 as the

argument. Then

K'(m) = 1'71"/2 (1-(1-m)sin20t1/2 dO

= r {(1-t2)(1-(1-m)f)}-1/2dt.

Then k = 0.2 corresponds to m = 0.04 and we find from NBS Handbook, p. 608

K' = 3.01611 24925

and

1.57079 63268 M(1, 0.2) = 3.0161124925 0.52080 1638.

1.15. Solution What is involved here is essentially the relation

lim 1'71"/2 In (0) dO = r/\imfn(t~) d6 when fn(6)

= (a~ cos2 0 + b~ sin2 Ot1/2.

Although this relation is valid in the present case it is not in general: e.g., if for n = 0, 1, 2, ...

gn(O)=O for 6=0, andfor 2-n - 1'77":50:52-1'77"

gn (2-n - 2 '7T) = 2n +2 '7T- 1

gn linear in 0, 2-n - 2 '77" and in 2-n - 2 '77", 2-n - 1 '77".

See diagram. In this case

t'IT/2Iim gn(O) dO = O.

For further remarks on similar problems see Chapter 5. Since

it is clear that

-1<{. (1I)<b-1 an - n 11 - n •


Also

b~t-----==;;;;;;o;o-...,

M-l~ a~l~V ____ ----l

O'--------!--ln 9 2

It is also clear (see diagram) that for any n

Suppose the common value of the integrals S;;/2 In (6) d6 was different from S;;/2 d61M and that they differ by, say, Eo. This supposition leads to a contradiction in the following way:

Since a..l M, bn t M we have Ib;;-1-a;;-11'IT12«a.. -bJ'lT12b~. Since a,. - bn ~ 0 we can find an no such that for n;a.: no,

(a,. -bnhT/2b~<IEol.

The last two displayed formulae are in contradiction. Thus we have established the relation required.

1.16. Solution a) If A is the common limit of {a,.} and {gn} then A1JX is the common

limit of the Borchardt sequences with initial values (1 + x)/2JX, 1. The limit in this case is sinh tIt where cosh t = (1 + x)/2JX. It is clear that sinh t = v'(cosh2 t-l) = (1-x)/2JX and that

exp t = cosh t + sinh t = 1!JX, i.e., x = exp (-2t)

so that - 2t = log x. Thus

A i-x -2 Jx= 2Jx 'logx'

i.e.,

log ..fi =! log 2 = 0.346573 5903.

x-l A=-

log x

x-l or logx=--.

A

Chapter 1 167

The sequence {a,. - g,.} converges to zero geometrically, with approximate ratio 4 whereas the sequence {:l(a,. +2gn )} converges to its limit geometrically with approximate ratio 16.

b) If B is the common limit of {a,.} and {gn} then BJ../(1 + x2 ) is the common limit of the Borchardt sequences with initial values l/.J(1 + x2), 1. Hence

so that

B

B= x . arctan x' I.e.,

x

x arctan x = B'

arctan 1 =i1T = 0.78539816.

[B. C. Carlson, Math. Comp. 26 (1972), 543-549 gives techniques to accelerate the convergence of these sequences. See also his paper, Amer. Math. Monthly 79 (1972), 615-618.]

1.17. Solution Clearly

These give

Hence

a,.+1 - bn+1 _ !(a,. - bn ) = a,. - bn

log (a,.+1/bn+l) -! log (a,./bn) log (a,./bnr

a,. -bn ao-bo log (a,./bn) log (ao/bor

Now, by the Mean Value Theorem,

log (a,./bn) = log a,. -log bn = (a,. - bn)/c,., for some c,. between a,., bn •

Hence

Let n~ao and we have


1.18. Solution Clearly x..+l lies between x.., Yn and Yn+l lies between x.., x..+l and so

also between x.., Yn. Further

Hence if x.. ~ Yn then x..+1 s.: Yn+l. It follows that the odd terms and the even terms of each sequence are monotonic and bounded and therefore have limits. The relation established above ensures the equality of all the limits and that as n ~oo

The argument used in Problem 1.13 shows that

100

(t+ X~)-3/4(t+ y~)-1I2 dt = Loo (s + Ft3/4(S + F)-1/2 ds

= Loo (s + 12)-5/4 ds

= [-4(s + F)-1I4J;' = 41-1/2.

Changing the variable from T to T by T = (1 + n-1I4 shows that

100 (T+1)-3/4y--1I2 dT=4 r (1-T4)-1I2 dT.

Hence 1(1,0) =A -2 where A = J~ (1_T4)-1I2 dT = 1.31102877714605987.

1.19. Solution We find:

cos qJ = [1- (1 + k') sin2 8](1- k2 sin2 (J}-I12

(1-k~ sin2 cp)1I2=[1-(1-k') sin2 8](1-Psin2 8)-112.

Differentiating the expression for sin cp gives

cos qJ(dqJ/d8)

= (1 + k')[(cos2 8 -sin2 8)(1- k 2 sin2 8) + k 2 sin2 8 cos2 8](1- P sin2 8)-3/2.

Combining these three relations gives

dqJ(1- k~ sin2 qJ )-112 = (1 + k') d8(1- P sin2 8)-112.

Chapter 1 169

As 0 increases from 0 to !?T the numerator in sin <p increases from 0 to !(1 + k') and then decreases to zero while the denominator decreases from 1 to k'. Hence <p increases from 0 to ?T. Since

{1' d<p(I- ki sin2 <p t 1l2 = 211'/2 d<p(I- ki sin2 <p )-112

the result follows. This result is due to J. Landen (1719-1790). From it we can recover

Theorem 1 by taking P=I-(b~/a~) which gives k'2=bMa~ and k1 = (ao-bo)/(ao+bo) so that

i.e.,

r 1'/2 rl'/ 2 Jo dO(a~ cos2 (I + b~ sin2 0)-112 =.10 d<p(ai cos2 <p + bi sin2 <p )-112.

Repeating this indefinitely we find

1.20. Solution This problem derives from J. Schwab, Elemens de geometrie, [ere

partie, Geometrie plane. Nancy, 1813. I am indebted to I. J. Schoenberg for sending me copies of parts of

Schwab's book and for suggesting that I record the fact that the so-called "Borchardt" - Algorithm was known to Schwab before Borchardt (1817-1880) was born. One sector of a regular n-gon is indicated below. It is clear that

an = (2n)-1 cot (?TIn),

bn = (2n)-1 cosec (?TIn).

The relations (1) follow by elementary trigonometry.


The convergence of the sequences (2) follow from our discussion of "Borchardt's Algorithm" and the limit is

(2n)-1 cosec (1T/n){[sin (1T/n)]J( 1T/n)} = 1/(21T).

It is clear "geometrically" that the common limit is the radius of a circle with circumference 1.

Schwab gives 7D values in the case of polygons of 6, 12,24, ... ,6144 sides. With our normalization

a6144 = 0.159154931, b6144 = 0.15915 4950

while

(21T)-1 = 0.159154943.

In the case n = 2, a2 = 0, b2 = 0.25. Compare also Problem 9.7.

Chapter 2

2.1. Solution Let f be a positive decreasing function defined for all x ~ 1. Then the

sequence {sn}, where Sn = L~=l f(r), and the integral j f(x) dx converge or diverge together. The "infinite integral" .17 f(x) dx is said to converge or diverge according as limx.-oo F(X) exists or not, where

F(X) = IX f(x) dx.

Since f(r)~f(x) when r:5x:5r+1 we have, for n~2,

e(n) = f(l) + f(2) + ... + f(n -1) - r f(x) dx = ~t: 1'+1 (f(r) - f(x» dx ~ O.

Chapter 2 171

Rewriting this as

f(1)+f(2)+ .·· +f(n - 1);::: r f(x)dx

and passing to the limit, in the convergence case we find

f f(r) 2! f oo f(x) dx. 1 1

Moreover, e(n+1)-e(n)=S~+I(f(n)-f(x))dX2!O, so that the sequence {e(n)} is an increasing one. Since

e(n) = f(1) - J: 1'+1 (f(x) - f(r + 1)) dx - I~1 f(x) dx

we see that e(n)::5f(1). Hence en--,;l, 0::5I::5f(1). See diagram.

y

o 2 3 n n+1

The sum of the shaded areas to the right of the ordinate through 1 is manifestly positive; translating them to the left as indicated, the sum is manifestly less than f(1).

It follows in the case of convergence, that

rf(n)::5f(l)+fOO

f(x)dx. 1 1


We have, therefore, obtained both upper and lower bounds for the infinite series in terms of the infinite integral.

2.2. Solution We discuss rn-1/ 2 • From graphical considerations, as in Integral Test,

we have

Also,

SN-l:51N X- 1/2dx=2N1/2-2, i.e., ~:52N-l/2_1.

These two relations show that SN = (J(Nl/2 ) exactly. [Compare Problem 2.5 below.]

The first of these relations gives

SN> 10 if 2(N + 1)1/2> 12, i.e., if (N + 1) > 36, i.e., if N> 35.

Actually S32=9.94, S33= 10.11 so that the least number of terms required is 33.

The solution to the first part is 20, that to the third is 12367 and that to the fourth is approximately 1.6 x 104321.

For further examples of this type see: G. H. Hardy, Orders of infinity, Cambridge, 1924. R. P. Boas, Jr., and J. W. Wrench, Jr., Partial sums of the harmonic

series, Amer. Math. Monthly 78 (1971) 864-870. R. P. Boas, Jr., Growth of partial sums of divergent series, Math.

Compo 31 (1977), 257-264. R. P. Boas, Jr., Partial sums of infinite series and how they grow, Amer.

Math. Monthly 84 (1977), 237-258.

2.3. Solution All the series converge. We can use results obtained in the solution of

Problem 2.1. For instance, in the case of r n-101/l(lO, we have

100 = rcc X-101/100 dx:5 f n-101/100:51 + rcc X-10l/1OO dx = 1 +[ -100x-1I1oor J1 1 J1

Similarly cc

10:5L n-11/ 1O :511, 1

=101.

cc

1/9:5L n-1°:510/9, 1

Chapter 2

Actually 00 00

L n-lO = 1.00099 4575, L n-2 = (1/6)'1T 2 = 1.64493 40668. 1 1

2.4. Solution In the notation of Problem 2.2 we have

1 fa(n) =-e(n)+--n-a

I-a

and, letting n~oo, we find

limfa(n)=-I+-11 . -a

When a = 1 we have

{log n -(1 + 1/2+ ... + l/n)}~-'Y = -0.5772156649.

Generally, lim fa (n) = -'(a), , being the Riemann Zeta-function. Here are the first 4 values of fl/2(n).

n

1

2

3

4

1

2h- (1 + ~) = 1.12132034

2J3-(1+ ~+ ~)=1.1796446

4- (1 + J"z+ ~+~) = 1.21554295

The limit is -,(1/2) = 1.46035 45088. See T. M. Apostol, Calculus II, p. 618-9, Exercises 15.23 #3.

2.5. Solution

173

This result states that the remainder after n terms of a series I ,-i-a is exactly of order a. We use again facts established in the solution to Problem 2.1.

Let f(x) = (n -1 + x)-l-a. Then f(x) decreases as x increases. Further, as X~OO.

Ix [(n-l+X)-a]x n-a

F(X)= 1 f(x)dx= (-a) 1 ~~,


so that I~ f(r) is convergent and lies between h f(x) dx and n-1- a + h f(x) dx. Thus

-::5 L f(r) = L r-1 - a +-= n-a -+-n-a 00 00 n-a (1 1)

a r=1 r=n a a n

so that we can take

2.6. Solution The first series is obtained by expanding the integrand in the lemniscate

integral S~ (1- x4t1l2 dx by the binomial theorem and integrating term-byterm.

From Stirling's Formula [po 111] we find that the general term an is approximately (l/47T)n-3/2• It follows that the error after n-terms is 0'(n-1I2) so that direct summation is unfeasible.

The second series is also obtained from the lemniscate integral by writing the integrand as (1- X2)-1I2(1 + x2tl/2 and expanding the second factor by the binomial theorem and the integrating term-by-term using the fact that

11 2n(1_ 2)-1I2d =(2n-1)!!~ ott t (2n)!! 2·

Again using Stirling's Formula, we find that bn is approximately (2n)-I. The series I (-1tbn is an alternating one ({bn } is monotone) and the error in such a series is less in absolute value than the first term omitted.

See Problem 9.10.

2.7. Solution Denote the limit by C. Then we are concerned with the ultimate

behavior of the ratio (a,. - C)/(a,.+1 - C). We may assume a,. = 1 and bn = x = 1 - y, where y > 0 is small. Then

C 2 = 1-x2 = 2y_y2 = 1-(y/2) 2 log X-I 2y +y2+(2y3/3)+... 1 +(y/2)+(y3/3)+ .. .

= 1-y +(y2/6)+ ... .

Hence

C = 1- (y/2) - (y2/24) + ....

Since

an +l = J(1 + x)/2 = J1- y/2 = 1-(y/4)-(y2/32)+ ...

Chapter 2 175

we have

a.. - C (y/2) + (y2/24) + .... 1/2 a..+1- C (y/4) + (y2/96) + .... .

2.8. Solution This follows by induction from the fact that if f(x) = o(h(x)) and

g(x)=o(h(x)) then f(x)+g(x)=o(h(x)) and the fact that xr=o(eX ) as x~oo.

Alternatively, if Pr(x)=aO+a1x+ ... +a.xr, note that if x~1

and use the fact that if f(x) = o(h(x)) then Af(x) = o(h(x)), for any constant A.

2.9. Solution

where

n

f(x)= L fk)(a)(x-a)k/k!+E,.(x) k=O

Mlx-al n +1

1E,.(x)l< (n+1)! .

This can be written as

or as

2.11. Solution Since

d {( t)n} te t (' t)n-l

dt et 1-~ =--; 1-~

if t increases from 0 to x (where x:::; n) we have

176 SolutiQns to Selected Problems

Integrating across with respect to t between 0, x give

Dividing across by eX gives

This is result required: if n ~ x then

Chapter 3

3.2. Solution If {x,,} converges to a limit l;t: 0 then I =!(l + Nil) so that 1= ±IN.

Graphical considerations suggest that there will be convergence to IN if Xo is positive and to -IN if Xo is negative. It will be enough to discuss the first case. Further, Bn+1 = !B~X~1 so that convergence is quadratic. If Xo < IN then x1>JNand we find X1>X2> ... >IN. The sequence {x,,}n;;,,1 is therefore a bounded decreasing one and its limit is necessarily IN.

3.3. Solution Deal with this as in Problem 3.4 below, or transform it algebraically

into Problem 3.4. Quadratic convergence to N 1/ 2 takes place for appropriately chosen Yo since

3.4. Solution For further details see W. G. Hwang and John Todd, J. Approx. Theory

9 (1973), 299-306. See diagram opposite. When 0 < Xo < (3N)1/2 the sequence converges quadratically to N. When

xo> (SN)1/2 the sequence oscillates infinitely. There is an increasing sequence 13r with 13-1 = (3N)1I2 which converges to y = (SN)1I2 and is such that when 13r<XO<f3r+1 the sequence {x,,} converges to (-1YN1/2. For Xo= 0, 13-10 130, ... the sequence converges to O. For Xo = Y the sequence oscillates: x" = (-1)ny. The behavior for negative Xo is obtained by symmetry.

Chapter 3 177

3.5. Solution If we use the relation (8) of §3.2 the only division we need to do is by 6

which can be done mentally. We have

X1-m= -(xo-·!N'f(xo+2JN)/(2N)

and so if we choose Xo so that Ixo-.J31 < 10~8 we have

IX1-J3I:::; 1O-16(3J3+ 10-8)/6< 10-16 .

Preliminary calculations e.g., the use of the above relation with Xo = 1.732, which is the usual approximation to J3 which one remembers, suggests a choice of

Xo = 1.73205 081.

We compute (exactly)

x~ = 3.00000 00084 21656 1

[xo-1O-8]Z= 3-0.00000 002621936

which guarantees xo-J3 < 10-8 • We can write the recurrence relation in the


form

[ N-X~] Xl=XO 1+---zN

and complete our calculation as follows: From our computation of x~ we find

(3- x~)/6 = -0.00000 00014 03609 3

and multiplying by Xo we get

xo(3-x~)/6=-0.00000 00024 31122 62.

Adding Xo to the last line gives

J3 = 1.73205 08075 68877 38.

O. Emersleben gives 29(353) in place of the last two figures 38 so that our result appears just good to 15D. NBS Handbook (p. 2) gives 29(35).

3.6. Solution We find, by elementary algebra, when x,.+1 =~X(x,.)-!Y(x,.) that

x,.+l-vW=~B<x,. + NX~1)_JN]-M{2x!,(3x~-N)}-JN]

3 1 = 4x,. (x,. _IN)2 2(3x~-N) (x,. -JNf(2x,. +JM

= (x,. - IN)3(5x,. + 3JN)/4x,. (3x~ - N)

so that this sequence has cubic convergence to m. In a similar way, when x,.+1 = !Y(x,.)+!Z(x,.) we obtain

x,.+1- vW=![{2x!'(3x2- N)}-JN] + ![x,. (3N - x~/2N}-JN]

(x,. -IN)2(2x,. +IN) (x,. _IN)2(x,. +2JN) 2(3x~-N) 4N

= -(x,. -IN)3(3x~+9x,. vW+4N)/(4N(3x~-N),

again showing cubic convergence to /N. These examples illustrate how cubically convergent schemes can be

obtained by combining two quadratic schemes. Similarly, by combining these two cubically convergent schemes we can get one with quartic convergence.

Chapter 3 179

3.7. Solution Draw the graph of y = x[ 4- Nx 3 ]/3. It is obvious geometrically that if

0<XO<4113N-1I3 the sequence will converge quadratically to N-1/3. To prove this we note the identity

x,,+1- N-1I3 = _(3N)-1[x,. - N-1/3]2[X~+2xN-1I3+3N-2/3].

If xo=O or xo=4113N-1/3 there will be convergence to zero. If xo<O or Xo > 41/3 N-l {3 there will be divergence to -00.

[1.2345678]-113 = 0.93216 97773.

3.B. Solution If lim Xp = I exists then [(I, l) = I and this gives In = a, I = a lIn. Observe

that [(x, y) = [(y, x). Since

(1)

it follows that if x = I, [(x, y) = I for any y. It follows that if Xo or Xl is I then X2 = X3 = ... = I. We have to discuss separately the cases when xo:5 Xl < I, Xo < I < Xl> I < xo:5 Xl. For instance, take the case when l < xo:5 Xl. It is clear from (1) that if x> I then [(x, y) increases with y, X being fixed: since [(x, l) = I we have [(x, y) > l if y> ,. Hence X2> ,. It also follows from (1) that X2 < Xo. Proceeding we see that {x,.} is a decreasing sequence, bounded below by ,. It must converge to I.

For further details see Amer. Math. Monthly, 55 (1948) Problem E 790, p. 367.

3.9. Solution If 1= a lln then

[(x, y)-l = (x-l)(y -l)g(x, y)

where g(x, y) ~ 0 for x = I, y = I. Let 0 = 1.618 be the positive root of the quadratic q2 - q -1 = O. Then 0 is the order of convergence. If

xp -l=O'(e r.)

then the above equation shows that Tp+2 = Tp+1 + Tp and the solution to this Fibonacci difference equation (d. Chapter 10) is Tp = AOP + BO-P ~ AOP.

3.10. Solution In case i) show that Tn decreases steadily to ../N. In case ii) show that T2n decreases steadily to ../N - this means that we

can obtain upper and lower bounds for ../N. For further information on this problem, and extensions to deal with

the solution of quadratics and the determination of cube and other roots, see


A. N. Khovanskii. The application of continued fractions and their generalizations to problems in approximation theory, Noordhoff, 1963.

3.11. Solution This problem is fully discussed theoretically by K. E. Hirst, J. London

Math. Soc. {2} 6 (1972), 56-60 and earlier in E. T. Whittaker and G. Robinson, The Calculus of Obseroations, Blackie, London {4}, 1944, p. 79.

3.13. Solution Suppose xo> O. Then x.. is clearly positive for all nand

x..+l- x.. = 2x..(N - x~)/(3x~+ N) ~O according as x.. ~.JN. Also

x..+l -./N= (x.. -JF:ry3/(3x2 + N) ~O according as x.. ~./N. Consequently if 0 < xo.JN we have x.. t and x.. :5.JN while if .IN < Xo we have x..! and x.. ::::.IN. Convergence takes place for any xo> 0 and the limit satisfies

(31 2 + N)l = x3 + 3Nl i.e., P = Nl i.e., 1= 0, ±.IN and so 1=.JN. We have x..+l-.JN=(x.._JN)3/(3x~+N) so that convergence is cubic.

The behavior of this sequence can be illustrated graphically in the usual way. Observe that if y = (x 3 + 3Nx )/(3x2 + N) then y - 3x as x - 0 and y - xl3 as x~±oo. Also y' = 3[(x2 - N)/(3x 2 + N)Y::::O for all x and y' =0 for x = ±N1I2•

y Y'x

y-3x

y·tx

2 3 x

3.16. Solution This is not entirely trivial. See e.g., J. L. Blue, ACM Trans. Math.

Software 4 (1978), 15-23.

4.1. Solution -0.882027570.

4.2. Solution

Chapter 4

Chapter 4

0.67434714 ±i 1.1978487. the remaining root is -2.

4.3. Solution If q(x)=qoxn- 2+ ... +qn-3X+qn-2 and r(x)=qn-1X+qn then

i = 0, 1, 2, ... , n - 2.

4.4. Solution

181

We take the case of a double root. Then f'(~) = 0 but f'(~) '¢ O. We find

Xn+1 - ~ = !(Xn -~) + O'(Xn - ~f

and we have linear convergence, not quadratic. We can, however, restore the quadratic convergence by changing the Newton formula to

in the case of a double root and to

in the case of a root of multiplicity r.

4.5. Solution We show that the conditions (8), (9), (10), (11) are satisfied. First

(8) H(-I) = 26>0 and H( -1-~6)= -6-~63<0.

Next,

(9) H'(x)=9x 2-3>0 if Ixl>l/.J3. Then

(10) H"(x) = 18x <0 if x <0.

Finally the tangent to the curve y = H(x) at (-1, 28) has equation

y-28 x+l =6, i.e., y =6x+6+26

and cuts the x-axis at -1- 6/3, so that (11) is satisfied.


Thus the Newton process converges. For instance taking 0 = 1 we begin:

xo=-1

Xl = -(4/3) = -1.3333

X2 = -(148/117) = -1.2650

= -1.2406

lim x" = -1.2600

See Problem 3.4 and reference there given.

4.6. Solution Suppose we have

f(x) = (x -a)q(x)+ r.

Then f(a)=r and since f'(x)=q(x)+(x-a)q'(x) we have f'(a)=q(a). We have to show that q(x)=foxn-l+ ... +fn_l and that r=fn- We

multiply to get

(x -a)q(x) = foxn +(fl -afo)xn- 1+ . .. + (fn-l -afn-2)X +(fn -afn-l)- fn

and then the recurrence relations defining the t give

(x -a)q(x) = aoxn +aIXn- 1 + ... +an-1x +~ - fn

i.e.,

(x - a)q(x)+ fn = f(x).

4.7. Solution 0.77091 70, -0.3854585 ± 1.56388 45i

4.8. Solution See M. Davies and B. Dawson, Numer. Math. 24 (1975), 133-135,

George H. Brown, Jr., Amer. Math. Monthly 84 (1977), 726-727. We may assume the zero at ~ = 0 and we use Maclaurin series for

convenience instead of the Taylor series used on page 49. We assume f4}(X) is bounded in the neighborhood of the origin. Then

f(x) = xf'(O) + x2 f'(0)/2! + x 3 f"'(0)/3! +0'(x4 )

f'(x) = f'(0) + xr(O) + x2 f"'(0)/2! + 0'(x3 )

r(x) = r(O) + xf"'(O) + 0'(x2)

Chapter 5 183

and the correction term is, suppressing the arguments,

x [21' + xf' +lx2f'" + O(x3)][f' + xf' +!x2f'" + O(x3)]

2[f' + xf' +!x2f'" + O(X3 )]2- x[f' +!xf' +ix2f'" + O(x3)][f' + xf'" + O(x2)]

X [2/,2 + 3x/,f' + x2(f12+1/'n + O(x3 )]

21'2 + 3xf'f' + X2(~f'I2+ f'f'~ + O(x3 ) •

The new estimate is therefore

showing that convergence is cubic. If 1'(0) = 0 then the new estimate is of the form

~x +O(X2)

showing that convergence is linear.

4.9. Solution

4.10. Solution

4.4934094579, 7.7252518369.

a = 1, b =~, c = U.

f(a)=-l, f(b)=!

f'(x) = -4x ~O in [a, b]

f"(x) = -4::;;0 in [a, b]

The tangent to the curve at x = b has equation y = 2x +~ and cuts the x-axis at x = -1, between a and b.

Thus all the conditions (8) (9) (10) (11) are satisfied. If xo=-0.8 then Xl =-0.7125<~=-2-l/2=-0.7071 while if Xo=

-0.6>~ then Xl = -0.7142<~.

Chapter 5

5.4 Solution Suppose lim M,. = O. Take any E > 0 and then no = no( E) such that

Mn ::;; E if n ~ no; this implies that Irn (x)\::;; E for all x and all n ~ no. Hence no(E, x)::;; no and so no(E) is finite.

On the other hand, suppose no( E) < 00 for all E > O. If Mn ~ 0 is false there is an E > 0 such that for every N (however large)

there is an n" > N such that M ... > s. Choose N = no(E). Then \rn(x)\ < E for n > N and so M.... ::;; E, a contradiction.


5.5. Solution The solutions are 1; x; x 2 +[x(1-x)]/n and x 3 +[3x2(1-x)]/n+

[x(1- x)(2- x)]/n2 • We shall establish the result in the cases k = 1 and k = 2. In the case k = 1 we have

L (;)xr(1- x)n-r (;) = L (; = Dxr(1- xt-r

= x L (n -1)xr- 1(1_ x)(n-l)-(r-l) r-1

= x[x + (1- x)]n-l

=x.

The general term in the case k = 2 which is

( n) xr(1- x)n-r ~ r n2

can be split into two by cancelling out an r and an n as in the case k = 1 and then writing r=(r-1)+1. We find

L (;) xr(1_x)n-l (;Y = n: 1 L (;=~) xr(1-xt-r

+1. L (n -1) xr(1- xt-r n r-1

n-1 (n-2) = -- x2 L xr- 2(1- x)n-r n r-2

as stated.

5.6. Solution

n-1 x =--x2 +-

n n

2 x(1-x) =x +--'--

n

x (n-1) +- L x r - 1(1- xt-r

n r-1

ql(X) is the Bernstein polynomial, adjusted to the interval [-1,1] and obtained in Problem 5.8 below.

qz(x) is (approximately) the polynomial of best approximation, obtained by Remez.

q3(X) is the truncated Legendre expansion. qix) is obtained by truncating the Chebyshev expansion obtained in

Problem 5.11 below.

Chapter 5 185

In the evaluation of lie; (x )11 we can confine our attention to O:s; x :s; 1. The value of \iel(X)\\ is shown to be 0.375 in Problem 5.8. It is less easy to find the other norms exactly and they can be estimated as

max \e;(x)\, x =0(0.01)1

which gives

l\eix)1\ = 0.067621, l\eix)l\=

Remez gives Ile2(x)II+0.067621 and this is assumed positively at x = ±1, x9±0.28 and negatively at 0 and at x9±0.78 exemplifying the equalripple behavior characteristic of the polynomial of best approximation.

5.7. Solution

a) S =J292{ -!+(1-eei6 +e2e2i6 - ••• )}

= J292{ -!+ (1 + eeilltl}

=J2{!-!e2}{1 +2e cos 0 +e2}-1

2 3+cos 0

1 1+x·

b) e,.(x) = {1!X -1Tn (X)} 2 J2 {!(1- e2 ) - (-l)e"[cos n6 + e cos (n -1)6]

3+cosO 1+e2+2ecos6

+(-1)" en COS nO} (1-e 2 )

(-l)"J2e n e"-l cos n6 = 2e(3+cos 0) [cos nO+e cos (n -1)6]-(-1)n 4

(-1)"e"-1 = J2 [4 cos nO+4e cos (n-1)6-3J2cos nO

4 2(3+cosO) - J2 cos 0 COS nO]

= (-1)n-1e" x {COS (n + 1)6 + 2e cos nO + e2 COS (n -1)6}. 4 1 + 2e COS 0 + e2


The result we need can be established by elementary trigonometry. For it is easy to verify that

where

cos (n + 1)6 + 2e cos n6 + e2 cos (n -1)6 1+2ecos6+e2 cos (n6+c/»

cosc/>=3cos6+1 3 + cos 6 '

• ..I.. _2.J2sin 6 sm 'fI- 3 + cos 6·

It is now clear that

max 1(1 +xtl -1Tn(x)1 =kn.

The fact that the error e,. (x) has n + 2 extrema, alternately ±len implies, because of the fundamental equal-ripple theorem of Chebyshev, that 1Tn(x) is the polynomial of degree n of best approximation to (1 + X)-l

in [0,1]. Note that it is obtained by taking the truncated Chebyshev expansion of (1 + X)-l and dividing the term involving Tn(x) by (1-e 2).

c) tl consists of the first two terms of the series in a) and the maximum error is

5 -W98 * 0.0502.

t2 is 1Tl and the maximum error is ie * 0.0429 which is assumed, with

alternating signs at 0, J2-l. When n = 6, the error in using 1Tn{x) is ie6 =l= 6.4 x 10-6 ; if we did not

modify T6 the error is

and this is in absolute value less than

d) B l =l·l·(l-x)+!·l·x=l-tx.

max 1(1 +X)-l-Bli =~-h*0.0858.

e) 6, 8, 10, 13, 18, 23, 33, 53, 110.

Chapter 5

5.8. Solution Since

B4 = (1- x4)f(0) +4x(1- x)3f(1/4) +6x2(1- x)2f(1/2)

+4x3(1- x)f(3/4) + x4f(1)

we have, using x = (1/2)(1 + y), 1-x = (1/2)(1- y),

so that

16B! = (1- y)4f(0) +4(1- y2)(1- y)2f(1/4) +6(1- y2ff(1/2)

+ 4(1 + Y )2(1- y2)f(3/4) + (1 + y)4f(1)

16B! = [[(0) -4f(1/4) + 6f(1/2) -4f(3/4) + f(1)]y4

+ [-4f(0) + 8f(1/4) - 8[(3/4) + 4[(l)]y3

+[6[(0) -12f(1/2) +6[(I)]y2

+ [-4[(0) - 8[(1/4) + 8[(3/4) + 4f(l)]y

+ [[(0) + 4[(1/4) + 6[(1/2) + 4[(3/4) + [(1)].

187

In the special case of Iyl we have [(1/2) = 0, [(1/4) = [(3/4) = 1/2, [(0) = [(1) = 1 which gives

B~(IYI, y)= _h4+~y2+i.

Consider the difference e * (y ) = 1 y 1-B ~ in the range [0, 1]. Its derivative is

h 3 -h + 1 =~(y -1)2(y +2)

which is positive in [0,1]. The error e*(O) =-3/8, increasing to e*(1)=0.

5.9. Solution We deal with r2,1 first. Equating coefficients of 1, x, x2, x3 in

we find

ao= (30 ! a1 = (31 + (30

o =~(30+ (31 + (32 .

0= t(3o + ~(31 + (32

The last two equations are homogeneous in (30, (31) (32 and therefore have a non-trivial solution:


Substituting in the first two we get

and so we find

r2,l(x) = (6+2x)/(6-4x+ x2),

The general case is handled in the same way. The IL equations arising from tV +1 , ••• ,t"'+v involve the IL + 1 coefficients 130,"" 13", and we can find a non-trivial solution. Putting these in the equations arising from 1, t, ... ,tV gives the <xo, ••• ,<Xv directly.

Observe that if XX divides D so that fJO=fJl= ... =fJX-l=O then XX

divides N for <Xo = ... = <XX-l = O. Thus we can divide out by XX and normalize with the constant term in the denominator unity.

Observe also that the rational function NID is necessarily unique. To see this suppose that NI D and NI D are IL, v entries in the Pade table for f(x). Then Df-N and Dt-N each involve powers of X larger than IL+V. Hence E = ND - ND involves powers of x larger than IL + v but E is of degree IL+V at most: hence E=O and so (NID) = (NrD).

We can therefore assume NID irreducible and, say, D(O) = 1. This will give uniqueness.

In some cases the general Pade fraction can be written down explicitly. see e.g., Y. L. Luke (The special functions and their approximations, I, II Academic Press, New York, 1969) who gives, e.g., information on (1 + Z-1)1I2, (1 + Z-1)1/3, Z 10 (1 + Z-l), Z-l arctan z.

1, 1+x, x2

1+x+Z'

1 2+x 6+4x+x2

1-x ,

2-x ,

6-2x ,

2 6+2x 12+6x+x2

2-2x+X2' 6-4x+X2' 12-6x+x2'

5.11. Solution a) Suppose lcos 01 = tao + L a.. cos nO. If we multiply across by cos mO

and integrate between -7T, 7T we get, on the right, 7Ta",. Similarly, on the left, we get

1m = t: COS mOlcos 01 dO = 2 fT COS mOlcos 01 dO

1,"/2 I'" = 2 COS mO COS 0 dO - 2 COS mO COS 0 dO. ,"/2

Chapter 5

Changing the variable to cp = 7T - 6 in the last integral gives

C'II"/2 1m =2 ~ [1 +(-lr] cos m6 cos 6d6.

If m is odd, 1m = O. If m = 0, 10 = 4 S~/2'11" cos 6 d6 = 4. If m = 2n

12n = 2 {'II"/2 [cos (2n + 1)6 + cos (2n -1)6] d6

= 2[(sin (2n + 1)7T/2)(2n + 1)-1 + (sin (2n -1)7T/2)(2n -1)-1]

= 2( -1)n[(2n + 1)-1- (2n -1)-1]

= 4(-I)n+l(4n2-1tl.

This gives the result required.

189

b) To establish the second result we use an integral representation for the Bessel function In(z). [See Appendix]. Suppose sin @7T cos 6) =~ao+ L ~ cos n6. We have to evaluate

In = t: cos m6 sin (~7T cos 6) d6.

By putting 6 = 7T - '" we find 1m = -( -lr 1m so that 1m = 0 if m is even. Since the cosine is even, 1m = 2 So COS m6 sin (~7T cos 6) d6. Writing this integral as SO/2 + S:/2 and changing the variable to '" = 7T - 6 in the second part, we find when m is odd, that 1m = 4 So/2 cos m6 sin (~7T cos 6) d6. It can be shown that if

(1) In(z) = (27T)-1 t: cos (n6 - Z sin 6) d6

then, if n is odd,

(2) 2 C'II"/2

In(z)=;;-(-I)(n-l)/2 k cos nTJ sin(z cos TJ) dTJ.

Hence, if m is odd, 1m =27T(-I)(m-1)/2Im(~7T)'as required. To establish (2) we use the addition formula for the cosine in (1) to

get

In (z) = ! I'll" COS n6 cos (z sin 6) d6 + ! {'II" sin n6 sin (z sin 6) d6.

Putting 6 = 7T - cp shows that the first integral vanishes for n odd so that in this case

In(z) = ! r sin n6 sin (z sin 6) d6, n odd.


Writing j1T - j1T/2 + j1T 0- 0 1T/2 and changing variable in the second integral to 1/1 = 7T - 0 gives

2 I1T!2 I n (z) = 7T sin nO sin (z sin 0) dO.

If we now put 0 =17T-TJ we obtain (2).

5.14. Solution The last approximation to Jo(x) is due to E. E. Allen (see NBS

Handbook, p. 369) and it is asserted that for -3:5x:53,.!e,.(x)!:55xlO-s.

5.15. Solution For x = 0, every term is zero and so the sum is zero. For x;e 0 the series

is a geometric one with first term x2 and common ratio (1 + X2)-1 < 1; it is therefore convergent with sum

x 2J(1-(1 + X2)-1) = 1 + x 2 •

The sum function is clearly discontinuous at x = 0 and so convergence cannot be uniform in any interval including this point.

Consider rn(x). For x=O, rn(O)=O, For x;eO, rn(x)=(1+x 2 )-n and !rn(x)!<s if (1+x2)">s-1 i.e., if no (s,x»logs-lJlog(1+x2)). As x~O, noes, x) is clearly unbounded for log (1 + x 2) ~ log 1 = O. This establishes non-uniform convergence in the sense of Definition 3.

Using the second approach we observe that

and so {Mn} is certainly not a null sequence.

5.16. Solution Since Sn (0) == 0, lim Sn (0) = O. For x;e 0, by Problem 2.8, lim Sn (x) = O.

Hence s(x)=limsn(x) is identically zero in [0,1]. Hence gs(x) dx=O. On the other hand, integrating by parts,

r sn(x)dx= r (nx)e-nx(ndx) = r te-tdt

= [-te-t];j+ r e-t dt

= -en + 1)e-n + 1 ~ 1, as n ~OO.

We show that there is non-uniform convergence. It is more convenient to use the Mn - definition. Since Sn (x) - S (x) = Sn (x) we need to

Chapter 6 191

evaluate

Since sn(x)=n2[I-nx]e-nX, sn(x) increases from 0, at x=O to n/e at x = n-t, and decreases to .n2/en at x = 1. Clearly Mn = n/e and this is certainly not a null sequence.

Chapter 6

6.3. Solution Try, for instance, Xn = 2(1!2t + 3(1!3t + 4(1/4t.

6.4. Solution See John Todd and S. E. Warschawski, On the solution of the

Lichtenstein - Gerschgorin integral equation in conformal mapping, II, NBS Applied Math. Series 42 (1955), 31-44, esp. 41.

6.5. Solution See J. Liang and John Todd, The Stieltjes Constants. J. Research Nat.

Bureau Stand. 76B (1972), 161-178. The first sum is 0.1598689037, the second 0.0653725926 and the third 0.0094139502.

6.6. Solution

L a..xn = L xn(1 + ;1)nao = (1- x(1 + ;1))-lao

= (1- x)-I(I- x(l- x )-1;1)-1 ao

=1~xL(l~Xr;1nao 6.7. Solution

By induction, beginning with ;1°ao = (-I)O[aoJ. If ;1nao is in the given form we have

;1 n+l ao = ;1(;1 nao) = ;1 nal -;1 nao

= (_l)n+l [ao-(~) al+" . + (-Ita..

- a l + ... + (-It+l (~) a.. + (-I t +1an+1]

_ n+l [ (n + 1) )n (n + 1) -(-1) ao- 1 a1+· .. +(-1 1 a.. +(-It+1an+l] .


Integrating tenn by tenn,

1= r (l-x)ndx= r (1-(~)X+ ... +(-1)nxn)dx

[ ( n)x2 xn]1 = x- 1 "2+ ... +(-1)n n+1 0

= 1_(n)!+ +(_1)n_1_ 1 2 ... n+1

where a,,=(n+1)-1. On the other hand, I=[-(1-x)n+1/(n+1)~=

1/(n + 1). Hence the Euler transfonn of the logarithmic series is

L (-1)n2-n- 1/(n + 1).

That this has the same sum as the original follows from the relation

log(1-x)-1=x+(x2/2)+(x3/3)+ ... , Ixl<1

by putting x = 1/2.

6.S. Solution Observe that

anU, =(-1)" [r!1-(~) r!2 +(~) r!3 - ... J

= (-1)n r xr(l-x)ndx

= (-1)n r!n! (n+r+1)!

Hence the Euler transfonn of the tail is

6.9. Solution We use the result of Problem 6.7 above to compute amf(O) when

f(s) = n-S • In fact

amf(O)=(-1r [1- (7) n-1+ (~) n-2- .. . + (-1)mn-m J =( -1)m[1-n-1r. The Euler transfonn of I (-n)-S (which has sum n/(n+1) when n>1)

Chapter 6 193

is therefore

(1/2) f (-1)"2-8(-1)8(1- n-1)8 = (1/2) f (n -1)8, 8=0 8=0 2n

a geometric series, again with sum n/(n + 1). Thus from a geometric progression with common ratio - n -1 we obtain

one with common ratio (n -1)/(2n) so that we get acceleration of rate of convergence when n = 2, retention when n = 3 and deceleration when n = 4.

6.10. Solution We consider using a head of r terms

1 1 r 1 1 1--+-+ +(-1) - --3 5 ... 2r-1

and applying the Euler transform to the tail

(-1)' [2r~1- .. .].

The n-th difference of the sequence (2r+ 1t\ (2r+3t\ . .. is

lln(Ur+l) = (-It [2r~ 1-(;) 2r~3 + ... +(-It 2r+;n+ 1]

= (-l)n r x2r(1- x2)n dx

i"'/2 = (-1)" sin2r 6 COS2 n+1 6 d6

= (-lt2n n!(2r-1)!! (2n +2r+ I)!!'

These results are easily established using the reduction formulas for

1112",

l(m, n) = sinm x cosn x dx.

With this notation we have

(1) (m + n)l(m, n) = (m -l)l(m -2, n) = (n -l)l(m, n -2),

l(m, n) = l(n, m), 1",12

1(0, 1) = cos xdx = 1.

r=O.


We establish (1) by integrating by parts: we find

1",/2

I(m, n) = (sinm x cos x) cosn- 1 X dx

[Sinm +1 x ] ",/2 1"'/2 Sinm+l = 1 cosn-1 X + -- (n -1) cosn-2 X sin x dx

m+ 0 m+1

so that

(m + 1)I(m, n) = (n -1) 1"'/2 sinm+2 x cosn-2 X dx

f. ",/2

= (n -1) 0 (1-cos2 x) sinm x cosn-2 X dx

= (n -1)I(m, n - 2) - (n -l)I(m, n).

Hence

(m +n)I(m, n) = (n -1)I(m, n -2).

We therefore obtain, if we start the transform from the beginning,

Observe that this series is much more rapidly convergent than the Gregory series. In fact

1 n!2n • n! f7r -n-l

u..=2· (2n+1)!-V;;·2 .

We can check that the new series actually has the proper sum, e.g., as follows. Consider the Maclaurin series for

f(x) = (arcsin x)/.J1- x2 •

This can be found by differentiating to get

so that

(1-x2)f'-xf-1 =0.

If we differentiate this n times by Leibniz' theorem we get

Chapter 6

If we put x = 0 then

This gives

22 22.42 f( )- +- 3+ __ 5+ x - x 3! x 3!5! x ....

Putting x = 7T/.J2 we find

~/ fI=~[1+!+1.2+ ... ] 4 'J2: J2 3 3.5

which is the result required.

6.11. Solution

VO-V1 +V2+'" = (1- E)-lvo

= (1/2)(1- m)-lvo

= (1/2)[vo+ Mvo+ Wvo+ ... ].

6.12. Solution

195

We take so=O, Sn =I~=l (-ly-1a" for n~l so that (-1)n- 1a.. = Sn - Sn-1 for n ~ 1. By definition Sn = I~=l 2-V ( -1)v-1av-1a1'

We use induction to prove that Sn = 2-n I~=o (~)ST' This is trivial for n = 1 since Sl = (1/2)a b Sl = a1•

Assume the result established for a particular value, n. Then

Sn+1 = Sn + 2-(n+1)(-1)" anal

= Sn +T(n+l) nf (_ly-l ( n ) a" from Problem 6.7 v=l v-1

= 2-(n+1) [2 i (n) Sv + nf (Sv - Sv-1) x ( : 1)] v=o v v=l V

by induction hypothesis

This completes the induction proof. The fact that Sn ~ I implies Sn ~ I now follows from a general theorem

of Toeplitz. [See e.g., K. Knopp, Theory of Infinite Series, p. 72]. We now


give a direct proof in the present special case. Because (0) + (~) + ... + (;:) =

(1 + 1t = 2n we have

and therefore it will be sufficient to establish the result when {s,.} is a null sequence. For convenience put a",r = 2-n(~) so that 1 ~ an,r> 0, L~=o an,r = 1.

Notice that for r fixed an,r is a polynomial in n (of degree r) divided by an exponential 2n = en10g2 - hence an,r ~ 0 as n ~ 00. [Compare Problem 2.8].

Given any e > 0, we can find no = no( e) such that ISn I < ~e if n ~ no. Then

ISnl < lan,oso+ ... + an,nos...,1 + lan,no+1sno+l + ... + an,ns,.1

< lan,oso + ... + an,nos...,1 + ~e

We can now choose nl~nO such that lan,oso+" .+an,noSnaI<~e. This is because, no being fixed, {an,oso + ... + an,nos..J is a null sequence. Hence, if n ~ nl> we have 15..1 < e i.e., {Sn} is a null sequence. This completes the proof.

6.13. Solution

r=4:

1 1 -2

1 1 2: '3

1 1 -6 -4

1 '3

1 4

1 12

111 1-2:+"3-4= 112/192

192 log., 2 = 133.08

Chapter 6

6.15. Solution

11 dx i1 00 (-l)n -1 r= [1-xr+x2r - ... ]dx=L-1-·

o +X 0 +m

Those integrals can be evaluated explicitly, e.g.,

00 (-l)n 11 dx [1 x+1 1 2X-1]1 --= -~= -log +-arctan--~ 1+3n 0 1+x3 3 v'(x2-x+1) J3 J3 0

1 7r =-log2+ ~=0.83564885.

3 3,,3

For the other integrals see Problem 9.13. The results are

6.16. Solution

f (-It =0.86697299, o 1+4n

00 (-l)n L -1 5 =0.88831357, o + n

00 (-It L-1 6 =0.90377177. o + n

197

We have seen in Chapter 2 that if En = Xn - N-1 then En+1 = -NB~. Hence Bn+2=-NE~+1 =-N(-NB~2=-N3B~. Clearly

This gives

N2B~(1-~e~f = Bn+2- en (1 +2Nen - N3B~r

The second term on the right is O(E~) and swamps the first which is O'(e~). Hence En+2=0(E~~0. However

En+2=1+ l+O'(E~ Bn+2 NBn(l +O'(En))

= +1 +0(1) NBn

--+-00.


Chapter 7

7.1. Solution In the usual notation [(x) = eXEt(x) and we find from NBS Handbook,

pp.242,243

7.2. Solution We have

[(20) = 0.04771 8545

[(10) = 0.09156 33339, [(5) = 0.17042 2176

U(x) = iOOCOjidt_ rCOS,frdt= .Ji7T[1-2CiX)]

and, similarly, V(x) =.J( 7T/2)7T[1- 2S2(x)]. From Watson, Bessel Functions, p. 744, we find

C2(5) = 0.32845 7, so that U(5) = 0.42999 5

S2(5) = 0.46594 2 V(5) = 0.08537 1

C2(10) = 0.43696 4 U(10) = 0.15800 8

S2(10) = 0.60843 6 V(10) = -0.27180 9

Ci15) = 0.56933 5 U(15) = -0.17379 7

Si15) = 0.57580 3 V(15) = -0.19001 0

Ci20) = 0.580389 U(20) = -0.20150 5

S2(20) = 0.46164 6 V(20) = 0.961392

The required values of the trigonometrical functions can be found in NBS Handbook, p. 175. From p. 3 we have

.J!7T= 1.2533141373.

7.3. Solution Computation for x = 3

1.0000000 -0.05555 56

0.00925 93 -0.00257 20

0.0010002 -0.0005001

0.00030 56 -0.0002207

0.00018 34 -0.00017 37 ~ least term

0.0001834

Chapter 7

Sum of first nine terms 0.9519037; error bound 0.00017 37. Correct value of g(3): 0.951814, actual error 0.00009 O. Correct value of erfc 3: 2.20905 x 10-5 •

199

X = 4: error estimate 1.6 x 10-7 , g(4) = 0.971304, erfc4 = 1.54173 x 10-8 .

x=5:

x=6:

g(5) = 0.98109 4, erfc 5 = 1.53746 X 10-12•

g(6)=0.986651, erfc6= 2.15197 x 10-17•

The correct values of erfc x were obtained from pp. 242, 295 - of National Bureau of Standards Applied Math. Series #41, Tables of the error function and its derivatives. Washington, D.C. 1954. From these the correct values of g can then be obtained from

g(x) = ';;xex2 erfc x

using, e.g., the tables of exp x in the NBS Handbook, p. 138.

7.4. Solution For more details about Problems 7.4-6 see E. T. Goodwin and J.

Staton, Quart. J. Mech. Appl. Math. 1 (1948), 319-326; a brief version of the table in this paper is in NBS Handbook, p. 1004.

Write

1 1 [ U]-l 1 n-l (-u)r (-u)n u+x=~ 1+~ =~r~o ~ +xn(u+x)"

Multiplying across by e-t and integrating we see

so that

Since x > 0 and u ~ 0 we have u + x > u and so

Irn(x)l:sx-n ro un- 1e-u2 duo

Hence Irn(x)1 does not exceed the absolute value of the first term omitted. The integrals occurring can be expressed in terms of the Gamma

function (by using the change of variables from u2 to t).

[0 e-u2ur-1 du =!f(!r).


Using the facts that f(s + 1) = sf(s) and that rm = J;. we find

The error being bounded above by the absolute value of the first term omitted we see that, correct to 4 decimals for x ~ 0,

since

.;; {1 I} 1 F(x)=- -+- --2 x 2x3 2X2

I I 1 1 -4 r3(x) <-4:5-X 10 . 2x 2

Note that the optimal value of n is about 2x2. For instance for x = 5, we find f(50/2H= 6.2x 1023, f(49/2) = 1.26 x 1023 (NBS Handbook, p. 272).

It can be shown that the Euler process can be successfully applied to the asymptotic series.

7.5. Solution Integrating by parts we find

foo ue-u2 du 1 1 roo e-u2 du

o u+x 2x 21 (u+xY

Write u/(u+x) in the left hand side of the above relation as 1-[x/(u+x)] and we find

foo ue-u2 du = J;. _ xF(x). o u+x 2

Differentiating the definition of F(x) with respect to x we get

, foo e-u2 du F(x)= -1 (U+X)2'

Combining the three displayed relations gives the result required.

7.6. Solution We want to get a "power series" representation for F(x). There is difficulty because F(x) is only defined for x> 0 and the

differential equation obtained in Problem 7.5 suggests that F(x) behaves like -log x near x = o.

Chapter 7

In order to get more information about F(x) for x -0 we write

F(x) = roo t2tt ) .10 e x+t

and note that for t - 0, et2 -1 + t2 • We therefore suspect that F(x) and

roo dt Fo(x) = .10 (1 + t2)(x + t)

201

should behave similarly for x - O. We can evaluate Fo(x) analytically: using partial fractions we find

and, integrating,

_ 1 [x+t 100 _ 1 1

Fo(x) - (1 + x2) log (1 + t2)1/2 + x arctan t - (1 + x2) [-log x +z1TX].

We therefore have

l oo [ 1] dt F(x)-Fo(x) = e-t2_-1 2-. +t x+t

Letting x --+ 0 we find, subject to justification,

lim {F(x) + log x}= loo [e-t2 - 1 ~t2] ~t.

Changing the variable from t to T = t2 in the last integral gives

1 roo [ 1] d lim{F(x)+logx}="2.1o e-T- 1+T TT,

Easy manipulations of the integrals in the lemma (p. 80) show that

roo [e-T __ 1 ] dT = -'Y, Jo 1 +T T

so that

(1) lim{F(x)+logx}= -!-y.

To get the representation given we multiply across the differential equation by the integrating factor ex2 and rewrite it as

(eX2y + log x)' = .[;ex2 - x- 1(eX2 -1).


Integrating the power series on the right we find

X2 - <X> X2n+ 1 <X> X2n e F+logx=v''lTL '(2 1) L----'--(2 )+const. o n. n+ 1 n. n

By letting x ~ 0, using (1), we see that the constant must be -h and hence we obtained the required result.

We find F(l) = 0.6501.

7.7. Solution The standard definitions of the Fresnel integrals are

C(y) = r cos G'lTf) dt;

1 IX sin t dt 1 IX Six)= ,- -r=- 11/2(t) dt,

'V2'lT 0 'Vt 2 0

1 IX cos t dt 1 IX C2(x)= r;::- ---r=- 11/2(t)dt.

v2'lT 0 vt 2 0

C(x), Sex) and C2(!'lTX 2) , Si!'lTx2) are tabulated in the NBS Handbook. In the present context it is more convenient to have Cix), S2(X). If we

write u = v'(x + t) we find

Using the fact that .ro sin u2 du = .ro cos u2du = v' 'IT/8 we find

(1) [(x) = v'!'lT(cos x -sin x)-2 cos x f'x sin u2 du +2 sin x rx cos u2 duo

If we put t = u 2 in the integrals on the right we get

= v'!'7l{cos x(1-2Six»+sin x(2C2(x) -1)].

Reference to FMRC, p. 461 tells us the Six), C2(x) are tabulated in

Chapter 7

Watson, Bessel Functions. For

In particular we can use

to get

S2(1) = 0.24755 8, sin 1 = 0.84147 1

Ci1) = 0.721706, cos 1 = 0.54030 2,

f(1) = 0.80952 55.

In order to derive the power series for f(x) we rewrite (1) as

- rJi f(x) = v'!17(cos x -sin x)-2.1o sin (u2- x) du

203

and contemplate expanding the integrand as a power series in u2 - x and then integrating. For this purpose we need to find In = S~x (u 2 - x)" duo We have

so that

Also

Hence

so that

= -2nIn -2nxIn - 1

22(2r + 1)(2r)x2 12,+1 = (4r+ 1)(4r+3) 12,-1

I _ -22,+I«2r+ 1)!)X(4'+3)/2

2,+1 - (4r+3)!!


and

f(x)=~!17(cosx-sinx)+(-2) f (-lYI2r+I/(2r+l)! r=O

_ '" (-1Y«2r+ 1)!)24r+3x(4r+3)/2 = ~!17(cos x -sin x) + L (4 3)'

r=O r+ .

Checking the ratio of terms in the last series, or estimating the general term, we see that convergence is reasonable for, e.g., 0:5 x:5 3.

An asymptotic representation of f(x) can be found by repeated integration by parts. Thus

f(x) = [ -(x + tr I/2 cos {1;'+ [' cos t( -!(x + t?/2) dt

= x- I12 + [ -!(x + t)-312 sin t]+ f'" !-~(x + t)-3/2 sin t dt

= x-l/2_ ~:~ [' (x + t)-312 sin t dt.

Generally we find

This will be convenient for say, x;;:: 10. Finally, to get a differential equation satisfied by f(x) we differentiate

under the integral sign to get

, 1 r= sin t dt f (x) = -2.10 (x + t)3/2'

If we integrate by parts we find

f '(x) = ~ r= sin t dt 4.10 (x + t)5/2 .

[ -cos t]= 11= cos tdt 1 1 r= cos tdt f(x)= (X+t)1/2 0-2 0 (x + t)312 = xl/2-2.1o (X+t)3/2·

If we integrate the expression for f"(x) by parts we get

" [3 -2 sin t ] = 3 2 1= cos t dt 1 r= cos t dt f (x)= 4·3 (x + t?/2 0 +4·3"· 0 (X+t)3/2=2.1o (x + t)3/2·

From the last two displayed formulas we find

f"(x) + f(x) = x-1I2,

the required equation which can be used in the intermediate range 3-10.

Chapter 7

We give some values of f(x)

7.B. Solution We have

Also, if x > 0, we have

Hence

so that

Hence

o 1.2533141 1 0.8095255 2 0.6429039 3 0.5468906 4 0.4828942 5 0.436552

10 0.314027 15 0.257368 20 0.223196

x 2n- 1 IG(x)-{ .. . }I=(2n)!/x2~0 as x~oo.

Also

x 2n IG(x)-{ .. . }+O· x-2n l = (2n)!/x ~ 0 as x ~ 00.

The last two relations show that we indeed have

1 2! 4! G(x)----+-

x x 3 x 5 '"

according to the strict letter of our definition.

205


For x = 5, two terms give G(x) to within 10-2 ; for x = 10 five terms give G(x) to within 4x 10-5 and for x = 15 seven terms give G(x) to within 2x 10-7 •

7.9. Solution The only trouble is finding the error in the binomial expansion and from

it to show the true asymptotic character of the series derived formally. Integrating the relation

d - (1 + t)v = v(1 + t)v-l dt

from 0 to x we get

(1+x)V-1=v r (1+t)v-l dt

= v r (1 + x - T)v-l dT,

the last line being obtained by changing the variable from t to T = x - t. Thus we have

(1+x)v=1+v r (1+x-tt- 1 dt.

Integrating by parts gives

(1 +xt = 1 +vx+v(v-1) r (1 +x-t)V-2t dt.

Repeating this operation we find

(1 + xt ={1 + vx+ v(v-1)x2/2! + ... + v(v-1) .. . (v- n + 1)xn/n!}+ rn+l(x)

where

rn +l(x)=v(v-1) ... (v-n) r (1+x-tt-n - 1(t n/n!)dt.

Assuming x>O it is clear that when O::5;t::5;x when 1 +x~(1 +x-t)~ 1 and so, provided n> v -1, the first factor in the integrand is less than 1 and

\ ( )\ <\v(v-1) ... (v-n)\ n+l rn+1 x - (n+1)! x

i.e., the remainder is less in absolute value than the first omitted term.

Chapter 8

In our application we have

where

and where

= (-1)n 1.3 .... . (2n -1) . .r:.. tn 2nn! x2n

I 1-< 1.3 .... . (2n + 1) yn+1 rn+1 2n+1(n + 1)! x2n '

Both infinite integrals exist. Integrating we get

(1) ( ) = f (-1)r 1.3 .... . (2r+ 1) + R g x 1... 2r 2r n+1

n=O X

where

The asymptotic character of (1) is clear since for each n

Chapter 8

8.2. Solution Differentiating the product l(x) by Leibniz' Rule we find

l'(x) = (x - x2)(x - x3) ... (x - Xn-1)(X - Xn)

+(x - Xl) (x - x3) ... (x - Xn-1)(X - Xn)

+ ... +(x - X1)(X - x2)(x - X3) ... (x - Xn-1)'

207

Putting x = ~ all terms vanish except the i-th so that l'(~) = n (~- Xj), the product being over j = 0,1, ... , n, j;i: i.


8.6. Solution We can write down L 3(x) in the general form and simplify it to

L 3(x)=x(2x 2 +x-2). Draw a graph otL3(x) using its values at the nodes and the fact that it has turning points at x = (-1 ± m)/6 with values (19=t= 13m)/54.

The error can be written down directly by noting that it is a polynomial of degree 4 with leading coefficient 1 which vanishes at ±1, 0, 2. Thus

E(x) = x(x2 -1)(x -2).

It is clear that E(x) is symmetric about x =! and

E'(x)=4x3 -6x2 -2x +2=4(x -!)(x2 - x -1).

It follows that E(x) has turning points at !, !(1±v'5) with values 9/16 and -1. At -!, ~ the value is -15/16.

The results just obtained indicate how the error in interpolation varies with the position at which we interpolate - as common sense suggests, it is better to interpolate near the center of a table, than at its ends.

8.7. Solution We have L 1(x) = -x(x -l)(x -2)/6, etc. giving

L1 10 11 12

x = 0.4 -0.064 0.672 0.448 -0.056

x = 0.5 -0.0625 0.5625 0.5625 -0.0625

To find the interpolant 1(0.4) given 1-1> 10' 11> 12 we have just to form the scalar product of these numbers with the numbers in the row labelled x = 0.4 above.

For further information on interpolation, and much good advice on computation, see Interpolation and Allied Tables.

8.8. Solution Since (sin x)" = -sin x and since Isin xl:s 1 the error does not exceed

i x (0.2)2 xl = 0.005. This means that it is not reasonable to use more than two decimals in the table since the round-off error would then be much

Chapter 8 209

smaller than the interpolation error. These conclusions can be checked by actual calculations, for instance from the values

sin 1.4 0.98545

sin 1.5 0.99749

sin 1.6 0.99957

Linear interpolation between 1.4, 1.6 gives 0.99251 with an error of 0.00498.

8.9. Solution If f(x) is a polynomial of degree n at most then

n

f(x)= L f(xJHx). i=O

If we take f(x) = Xi, j = 0,1,2, ... , n in this we find

SO= 1, j= 1, 2, ... , n.

To deal with the case j = n + 1 we note that if

'7Tn(x) = (x - xo)(x - Xl) ••• (X - Xn) = x n+1 + a1xn + ... + a n+1

then for i = 0, 1, 2, ... , n we have '7Tn (X;) = 0,

Multiplying across by HO) and summing from i = 0 to i = n we get

Special case:

,o{x) (x - 2)(x - 3)(x - 4) -6

12(x) (x -l)(x - 2)(x - 4) -2

, 0(0) = 4,

I () (x -l)(x - 3)(x - 4) 1 x 2 '

1 ( ) = (x -l)(x - 2)(x - 3) 3 X 6 '

So = 1, Sl = S2 = S3 = 0, S4 = - 24.

'7T3(X) = (x -l)(x -2)(x-3)(x -4) = x 4 -10x3+35x2-50x +24.


8.10. Solution

then

Lemma. If

r r (n -i)! co-c1+c2-c3 + ... +(-1)c,.=(-1) r!(n-r-l)1"

Proof. The sum required is clearly the coefficient of xr in

(1 + xt[xr - x r- 1+ xr- 2_ . .. +(-1)']

i.e., in

(-1)'[1- x + ... + (-l)'xr][l + x]n

i.e., in

i.e., in

i.e., in

(-1)'[1 + x]n-1

which is the result given. L,.(±1) is clearly +1 since ±1 are always nodes. Hence L,.(±1)~ 1. If n is odd it is clear from symmetry that L,. (0) = O. We deal now with

L 2n (0). We write

k=1,2, ... , n,

and note that

(2n -l)(xk - Xj) = 2(k - j).

Using the explicit form of the Lagrangian interpolant we see that

L 2n (0) = ~ \Xk \ (-It[(2n -1)(2n - 3) ... (3)(1)]2/( -(2n -l)xk) k=l 2(k-l)· 2(k-2) ... (2)(-2)(-4) ... (-2(2n-k»

=(-1)n-1[(2n-3)!!]2 ~ (signxk)(2n-l) 22n- 1 k=l (k -1)!«2n - k)!)(-l)k

n-1 [(2n-3)!!]2 ~ k. (2n-l) =(-1) 22n-1(2n-2)! k~l (-1) SlgnXk· k-l .

Chapter 8 211

Remembering that Xl> ..• , x" are negative and Xn+l' ... , X2n positive and that the coefficients in a binomial series are symmetric we see that the sum in the last expression is just

2[co - Cl + ... + (-It-l cn _l]

where Co, C1o ... are the coefficients in (1 + x?n-l. From the lemma this is

Hence

(2n-2)! 2(-I)n-l ((n -1)!)2·

[ (2n-3)!! J2 L 2n (0) = 2n- l (n -I)! .

Application of Stirling's Theorem gives L2n(0)~(1Tn)-1. Indeed Stirling's Theorem is

so that

It follows that 4 (0) ~ o. It is remarkable that it is only at the points 0, ±1 that we have 4(X) ~ Ix!. For a proof of this result of S. N. Bernstein see, e.g., 1. P. Natanson, Constructive Function Theory (in Russian, 1951, in German translation by K. Bogel, Akademie-Verlag, Berlin, 1955.)

8.11. Solution If f(i) = fb i = 0, 1, ... , n then we can obtain a representation of f(x) as

a Lagrangian polynomial f(x) = L~~o tl;(x) where

Since f(x) is monic and each l;(x) is monic we must have


It follows, from this and the definition of IL, that

n 1 IL L .'( _ .)'~1.

i=O Z. n I.

However

1 1 (n) (1 + It 2n

L i!(n-i)! n! L i = n! nr Hence IL ~ n !/2n but since we may take {; = (-1 Yn !/2n we can have equality and uniqueness, apart from sign.

8.12. Solution We use the facts that 4(~) = 8(i, j). It is obvious that H(x;) = {;, all i.

Clearly

H'(x) = L [24(x)l[(x){I-21[(x;)(x - .x;)} +{li (xW{-21[(x;)}]{;

+ L [24(x)I[(x)(x -:x:;) +{4(x)FJfr

and again

H'(.x;) = [21[(x;) -21[(.x;)]{; + {r = {r.

With E(z) defined by (12) it is clear that E(x) = 0 and E(.x;) = 0 for i = 0,1, ... , n. Hence E'(z) vanishes for (n + 1) values of z distinct from the .x;. But it is clear that E'(x) = 0 for i = 0, 1, ... , n. Thus E'(z) vanishes for 2n + 2 distinct values of z. Repeated applications of Rolle's Theorem shows that E(2n+2)(z) must vanish somewhere in the interval including x and the .x; say at ~. Since H is a polynomial of degree 2n + 1 at most we have

o = f2n+2)(~)_[f(x)-H(x)][(2n +2)!/0 (x - xY]

which gives (11).

8.13. Solution We use induction to prove that

n~i>j

For n = 1,

[ 1 xo] Vl =det 1 Xl =xl-XO·

Suppose we have established the result for n = r. Consider

Chapter 8 213

and expand it in terms of minors of the first row. Clearly V r +1 will be a polynomial of degree at most r+ 1 in x. Also V r+ 1 will vanish for x = Xl> X2, ... ,Xr+l so that

Vr+l = k(x - Xl)(X - X2) ... (x - Xr+l),

where k is a constant. Now k is the coefficient of xr+1 which is also evidently (-ly+1V(xl> x2, ... , x;.). The induction hypothesis gives

k=TI .. (Xi-X.) r~t>J~l J

so that (_1)'+1 = (x - X1)(X - x2) ... (x - xr+1) TI (Xi - Xj).

r::2!i>j

and so

V(xo, Xl> ... , Xr+1) = TI (Xi - Xj). r+l~i>j

This completes the proof. It is clear that 1, x, ... ,xn form a basis for 'Vn and since these are

independent (by an application of the Fundamental Theorem of Algebra), the space has dimension n + 1.

Since, for r = 0,1, ... , n, L,. = I~=o x~li(x) is a polynomial of degree at most n, coinciding with xr at xo, Xl> ... ,Xn it follows that L,. must be identical with xr. In other words

1 1 1

x

xC: x~ x~ 1n(x)

Since the Vandermonde matrix is non-singular, it follows that 'o(x), 11 (x), ... , 1n (x) form a basis for 'Vn"

8.14. Solution This is a table of 1010 Jo(5.495 + 10-3 x). To the zero

io,2 = 5.5200781103

of Jo(x) corresponds a zero of f(x) at

x = 2.50781103.


There is a zero of [(x) between 2 and 3. Using all six points for Lagrangian interpolation to subtabulate indicates a zero between 2.5 and 2.6 since

[(2.5) = -:2 65784

[(2.6) = +3136604.

Estimating the position of the zero by linear interpolation gives 2.5078. We therefore subtabulate again getting

f(2.507) = [(2.508) = f(2.509) =

and observe that the second difference is . Linear (inverse) interpola-tion is permissible and we find the value given.

8.15. Solution For discussions of wide generalizations of parts of this problem see, e.g. P. W. Gaffney, J. lost. Math. Appl. 21 (1978), 211-226. C. A. Micchelli, T. J. Rivlin, S. Winograd, Numer. Math. 26 (1978),

191-200. For instance, if [(0) = 0, [(1) = 1 and 1[,(x)I::::;2 in [0, 1] then the graph

of [(x) must be within the left hand parallelogram. If [(0) = 0, [(1) = 1 and 1f'(x)I::::;2 in [0,1] the graph of [(x) must lie within the lens shaped region on the right.

y-2x-x2

Chapter 8 215

8.16. Solution We want to show that there is a cubic H(x)=a+bx+cx2 +dx3 such

that

(1) H(xJ=t, H'(xJ=ir, i=0,1,

where Xo f= Xl and fo, flo n, f~ are arbitrary. The relations (1) when written out in full give a set of four linear

equations for the unknowns a, b, c, d. The determinant of this system is

[1 Xo x~ x~ 1 '0 1 2xo 3x~

h = det 1 Xl xi xf . o 1 2XI 3xi

We evaluate h by row operations as follows. Take row3 -rowl and divide the third row through by Xl - Xo. We find

Taking row2 -rowl and row3 -rOWI we find, on dividing through the second and third rows by (Xl - xo),

Hence the system is non-singular and a, b, c, d can be found. For information on the use of Hermite interpolation see, e.g., H. E.

Salzer, J. Res. Nat. Bur. Standards 52 (1956), 211-216.

8.17. Solution From Problem 8.16, Solution it follows that the second derivatives at Xo

in the left and right panels are, if h+ = Xl - XO, h_ = Xo - X-I'

and

-6foh~2 + 6flh~2 - 4nh~l- 2f~h~l.

Equating these gives a linear equation for f~ with coefficient 4h=1 +4h~1 f= O. Hence f~ is determined uniquely.


In the general case, when, fo, f1> ... ,fn and n, f~ are given, quantities f~, ... ,f~-l can be determined uniquely so that all the abutting cubics fit smoothly, because the system of equations is non-singular, having a dominant diagonal.

In the special case n = 0 and the spline is given by

-2x3 -3x2 +1 in [-1,0]

2x3 - 3X2 + 1 in [0, 1].

Observe that there is a jump in the third derivative from -12 to 12 at x = O. These results can also be obtained by using the results of Problem 8.12.

8.18. Solution For simplicity suppose the three points are equally spaced and, without

loss, take them to be Xl = -1, Xo = 0, Xl = 1. Then we have fo = a, f±l = a±b+c so that b=!(fl-f-l) and c=!(f-1-2fo+fl). It is clear that x= -b/(2c) and {= a - b2/(4c). For instance, given f-l = -0.3992, fo = -0.4026, fl = -0.4018 we estimate

x = (13/42) = 0.3095, {=0.4028.

For developments of this method see e.g., H. E. Salzer, Formulas for finding the argument for which a function has a given derivative, MfAC 5 (1951), 213-215.

8.19. Solution This is Newton's interpolation formula. It is, by uniqueness, necessarily

a rearrangement of the Lagrangian expression. It can be obtained formally by writing

and truncating the binomial expansion: if f is a polynomial of degree n, then 0= !J.n+lf = !J.n+lf = ....

To establish the formula we can use induction and the basic recurrence relation between binomial coefficients:

The quartic is

so that q(2.5) = 2884.

Chapter 9 217

The difference table for q(x) is:

0 789 567

1 1356 345 912 123

2 2268 468 200 1380 323

3 3648 791 200 2171 523

4 5819 1314 3485

5 9304

The constant term in the quartic is clearly q(O) = 789 and the leading term must be (200/24)x4 to give a constant fourth difference of 200. If we subtract off these terms and divide through by x we are left with a quadratic whose coefficients are those of the three middle terms of the quartic.

The Newton interpolation formula is:

[(2.5) = 789 + (2.5 x 567) + (2.5 x 1.5)345/2!

+ (2.5 x 1.5 x 0.5) 123/3!

+ (2.5 x 1.5 x 0.5 x (-0.5))200/4!

Truncating this we get successively

789 2206.5, 2853.375, 2891.8125, 2884.

Note that 2206.5 is got by linear extrapolation from q(O), q(1); if we interpolate between q(2), q(3) we get 2958.

Chapter 9

9.1. Solution Write down the fundamental polynomials and integrate them. For

instance

x(x-1)(x-2) I (x) - --61[X 3 - 3x2 + 2x], -1 - (-1)(-2)(-3)

and

12 L1(X)dx=i, -1

(2 lix) dx = i. L1


This gives the so called 'i Rule'

Q = i/( -1) +U(O) +U(1) +U(2).

The fact that the sum of the weights is 3, which is the integral of I(x) = 1 between -1, 2, is a check.

We shall show that if Ie C4 [-1,2] then t: I(x) dx -[U(-1)+~/(0)+U(1)+U(2)] = -iof4)(~)

where -1~~~~. In order that our computations have more symmetry we shall prove, equivalently,

p(h)= 13k I(x) dx - 34h [f(-3h) + 3/(-h) + 3/(h)+/(3h)] = _~h5/(4)(~) -3k

where -3h~~:53h and we assume fe C4 [-3h, 3h]. We differentiate E(h) four times with respect to h obtaining:

i.e.,

~E'(h) = [f-3 - 1-1 - 11 + 13]+ h[f~3 + 1~1 - I~ - I~];

~E"(h) = -2[f~3 - 1~1 + I~ - I~]- h[3f~3 + 1"--1 + n + 3/~];

aE"'(h) = 3[["--3 - f"--l - fl + [3] + h[9f~~ + f~l-fl- 9f3'];

~E(4)(h) =4[f"--'I-fn-27h[f~j +1\4)]- h[f~I + ft4)].

Hence, since f4) is continuous, we have

~E(4)(h) = -8hr)(~l) - 54hr)(~2) - 2hr)(~3) = -64hr)(~4)

where -h~~l~h, -3h~~2~3h, -h~~3~h and so -3h~~4~3h. We also note that E(O) = E'(O) = E"(O) = E"'(O) = O. Hence, integrating

the relation

we get

E"'(h) = -144 r hr)(~it)) dt = -72f4)(~s)h2; integrating again

E"(h) = -24hy4)(~6)' E'(h) = -6h4f4)(~7)'

and, finally,

where -3h~~~3h.

9.2. Solution

Chapter 9 219

We may assume the interpolating quadratic to be of the form q(x) =

[(0) + ax + bx2 • Since q(±1) = [(±1) we have two linear equations for a, b. Actually we only need b since

Q = t: q(x) dx =4[(0)+e~)b =~[2[(-1)-[(0)+2t(1)].

We note that this quadrature is of "open" type; it does not involve the values at the end points and can therefore be used as a "predictor" in the solution of differential equations e.g., in Milne's method.

We also note that it is to be expected that the error incurred in open formulas is larger than that in closed ones.

It does not seem possible to obtain an error estimate in this case by the method used in the previous problem. We use an essentially general method based on a modification of Steffensen's classical account given by D. R. Hayes and L. Rubin (Amer. Math. Monthly, 77 (1970), 1065-1072). Another general method for this type of problem is due to Peano (cf. A. Ghizzetti and A. Ossicini, Quadrature Formulae, Birkhiiuser and Academic Press, 1970); see also B. Wendroff, Theoretical Numerical Analysis, Academic Press, 1966).

Let L(t, x) denote the Lagrangian polynomial based on the nodes -h, 0, h and let 7T(X) = (x + h)x(x - h). Then we put

(1) [(x) - L(f, x) = 7T(x)R(x)

and get

(2) i2h

1-Q = 7T(x)R(x) dx. -2h

This defines R(x) except at the nodes where we define it by continuity. With this convention it is easy to verify that R(x) is continuously differentiable.

If we write

l(x) = rx 7T(t) dt 12h

then l(x)=Hx2(x 2 -2h2)-8h4 ] and l(±2h)=0. Hence, integrating (2) by


parts, we find

r2h 1-Q = [1(t)R(t)f-~h - t2h l(t)R'(t) dt

12h

= - l(t)R'(t) dt. -2h

Now as l(x)::50 in (-2h, 2h) we may apply the Mean Value Theorem to get

12h

1-Q = - R'(~) l(t) dt, -2h

=-R'(~)x -1;:h5•

We shall now show that R'(~)=f4)(C)/24 where -2h::5C::52h, which gives

(3) 1-Q =¥Sh5(4)(~).

Take a fixed x* in [-2h, 2h] and consider

~(x) = f(x)- L(f, x)-'7T(x)[R(x*)+(x - x*)R'(x*)].

Since L, 11' are of degree 3 at most we have

(4) ~(4)(X) = (4)(X)-4!R'(x*).

We shall show that there is a C such that ~(4)(C) = O. We observe that

(5) ~'(x*)=O for any x*, ~"(x*) = 0 if x* is a node.

The first of these results follows from

~'(x) = f'(x) - L'(x)-1T(x)R'(x*)-1T'(X)[R(x*)+ (x - x*)R'(x*)]

using the fact that f'(x)-L'(X)-1T(X)R'(x)=1T'(x)R(x) which comes by differentiating (1).

The second of these follows because differentiating (1) again we get

~"(x) = f'(x)-.L"(x)-21T'(x)R'(x*)-1T"(X)[R(x*)+(x - x*)R'(x*)J

and using f'(x)- L"(x)-21T'(x)R(x) = 1T"(x)R(x) we find

~"(x*) = 1T(x*)R"(x*)

which vanishes if x* is a node. When x* is in general position, ~(x) has (in the special case under

discussion) 4 zeros: x* and the three nodes. Hence ~'(x) has three zeros by Rolle's Theorem, and an additional distinct one at x* by (5). Hence ~"(x) has three zeros.

Chapter 9 221

When x* is a node, ~(x) has three zeros and ~'(x) has two zeros by Rolle's Theorem and an additional distinct one by (5). Hence ~"(x) has two zeros by Rolle's Theorem and an additional distinct one at x* by the second part of (5). Again ~"(x) has three zeros.

In both cases, by Rolle's Theorem, ~(4)(X) has a zero, say at ~. It then follows from (4) that

t<;:~) = R'(x*)

where ~ depends on x*, but x* was arbitrary. This completes the proof of (3).

9.3. Solution Assume that m::5 f'(x)::5 M in [a, b]. By the error estimate for linear

interpolation applied to f(x) in the subinterval [a,., a,.+1] where a,. = a+rh, h =(b-a)/n, we have

f(x) - L(x) = (x - a,.)(x - a,.+1)f'(c,.)/2!

where a,.::5 c,.::5 a,.+1' This is true for r = 0, 1,2, ... , n -1. Since for a,.::5 x::5 a,.+1 we have

we can integrate these ineqUalities between a,., a,.+1 to get

!Mf'+l (x - a,.)(x - a,.+1) dx::5 f'+1 [f(x) - L(x)] dx ~ ~

::5!m f'+l (x - a,.)(x - a,.+1) dx ~

which, after a little algebra, reduces to

-Mh3/12::5 f'+l [f(x)- L(x)] dx ::5-mh3/12. ~

We now sum the last inequalities with respect to r and get

m::5 123 [b [L(x)-f(x)]dx::5M. nh Ja

Now for a fixed h, the middle term above is a constant between m and M; since f'(x) is continuous in [a, b] and bounded there by m, M, it must assume this value (at least once), say at c. Hence

fb [L(x)-f(x)]dx=T-I=nh3 f'(c)=(b-a)3f'(c) a 12 12n2 '


9.4. Solution We give two rather similar proofs.

(1) We verify by integration by parts, that

f uv'" dx = uv"-u'v'+u"v-f u"'vdx

if u, veach have continuous third derivatives. Taking u.=ix(1- X)2, v = f(x) + f(-x) the above relation gives t: f(x) dx = i[f(-1) +4f(0)+f(1)]-r uv", dx.

Since If4)(t)1 ::5M4 and since, from the Mean Value Theorem,

v.'" = f"'(x) - f"(- x) = 2x(4)(~);

we have

IEI::5 2M4l1 x 2(1- xf dx/6 = MJ90.

(2) Let L 2(x) be a quadratic interpolating f(x) at -h, 0, h. Then repeated integration by parts gives for F(x) = f(x)- L 2(x), since F(±h) = F(O) = 0, t: (x + h)3(3x - h)P4)(X) dx + r (x - h)(3x + h)3p4)(X) dx = 72 t: F(x) dx.

Now, L 2(x) being a quadratic, P(4)(X) = ( 4)(x) and so 1P(4)(x)1 ::5M4. Hence

72IEI::52M4i" (h-X)3(3x+h) dx =4M4h s/5.

This gives the result required. For a.derivation of this result under weaker conditions see Anon, Amer. Math. Monthly 76 (1969), 929-930.

9.5. Solution We have M=(4/3)[2f(-1)-f(0)+2f(1)] and the adjusted S=

(2/3)[f(-2)+4f(0)+f(2)] so that

S - M = (2/3)[f(-2)-4f(-1) + 6f(0)-4f(1) + f(2)]

= (2/3)114f(-2)

which is bounded by (2/3)M4.

9.6. Solution It would be enough to do this in the two-point case but the characteris

tic pattern 1 424 1 first turns up in the 3 point case. We have, when

Chapter 9 223

b-a = 1,

T~l) = U(O) +!tG) +U(1)

T~2) = U(O) +um +um +tf(~) +U(1)

so that

= fi[f(O) + 4fm + 2fm + 4f(~) + f(1)].

9.11. Solution Suppose 1Tn(x) = k"xn + .... Then, by orthogonality,

r 1Tn(x)k"xnw(x) dx = r (1Tn(x)fw(x) dx = 1

so that

r 1Tn(x)xnw(x) dx = k;:l.

If there was another orthonormal system {1T~l)(X)} where 1T~l)(X) = k~l)xn + ... we would have

f b 1T~l)(x)xnw(x) dx = (k~l»)-l. a

This implies that

f b 1Tn (X)1T~l)(X)W(x) dx = k,Jk~l) = k~l)/k" a

so that

k~ = (k~1»)2

and, both being positive, this gives kn = k~l). It follows that

(1) r [1Tn(X)-1T~1)(x)]2w(x) dx = 1-2+ 1 = O. a

The result (1) implies that 1Tn(X)=n~l)(x) if we assume that

w(x) is not zero in any subinteroal of [a, b].

This condition is satisfied in all the classical cases. To establish uniqueness we proceed as follows.


If 1Tn (X) ~ 1T~l)(X) then, these being polynomials and so continuous, there will be an interval (c, d) included in (a, b) such that in it

l1Tn(x) -1T~l)(x)l;::: 8 > O.

Hence

a contradiction.

9.12. Solution Let Xl, . .. , Xn be the zeros of 'lTn(x). Let H(x) be the Hermite interpol

ant introduced in Problem 8.12. If we mUltiply the error estimate (11), (p. 91), across by w(x) and integrate between [a, b] we get

f b 1b f2n)(~(x)) n a (f(x)-H(x))w(x)dx= a (2n)! !1(x-xY w(x)dx.

Since H(x) is of degree 2n -1, and H(X;) = !(X;), we have

r H(x)w(x) dx = L AiH(X;) = L AJ(X;).

Thus

1-Q = rb f 2n )(g(x)) [fI (x - :x;f]w(x) dx Ja (2n)! i~l

f(2n)(g) 1b [ n ] = (2n)! a !1 (X - X;? w(x) dx

using the Mean Value Theorem, since the last integrand is positive. In the Chebyshev case we have

1 r'IT = (2n)!22n- 2 cos2 nO dO

1T =---....,. (2n)!22n- 1 •

To deal with the Legendre case we note that it can be verified (e.g., by integration by parts, Apostol II, 179) that if

Pn (x) = [1/(2n(n !»]Dn{(x2 -1)"}

Chapter 9 225

then

Also

Hence

and so the coefficient of f 2n)(e) is

9.13. Solution (a) The indefinite integral is

1 x2-x+1 1 xJ3 -log +-arctan--6 (x+1)2 J3 2-x

when -1 < x < 2. Hence the definite integral is

-log 4 + arctan J3 _log 2 + 1T~ = 0.60459 9788 _ 0.230149060 6 J3 3 3,,3

= 0.374450728.

(b) The indefinite integrals are, respectively:

1 (1 ) J(10+2v'5) 4x+v'S-1 4x-v'S-1 +-log + x + arctan arctan ;

5 10 J(10+2v'5) J(10-2,J5)


The definite integrals, between 0 and 1, are respectively:

../2 8 [log (3 +../2) + Tr] = 0.86697299;

J"S1 3+J"S 11 2 J10+2J5 3+J"S - og --+5 og + arctan -;::::==:::::: 10 3-J"S 10 J10+2.j5

JlO-2.j5 3-J"S + arctan -;::::====

10 JlO-2.j5

J"S TrJ"S = Slog !(J"S + 1)+! log 2+50 J(10+2.j5) = 0.88831357;

IOg(~J3) +~=0.90377177. 236

The general results of which these are special cases are due to A. F. Timofeyev (1933)-see, e.g., formula 2.142 in I. S. Gradshteyn and I. M. Ryzhik, Table of integrals, series and products, Academic Press, 1965. See also I. J. Schwatt, An introduction to the operations with series. Philadelphia, 1924.

9.14. Solution A short table of this integral in which the argument of y is in degrees is

given in NBS Handbook p. 1001. When y =0.5 in radians we find

S(0.5, 0.5) = 0.296657503.

The differences between the given estimates are 9312, 253, 1 in units of the tenth decimal. The ratio 9312/253 is about the fourth power of the ratio of the intervals. We may suspect that Simpson's Rule, or some other of the same accuracy was used. We can use Richardson extrapolation to get a correction of

16 (625 -16) x 9312 = 245

leading to an estimate of 0.2966575002.

9.15. Solution These integrals were evaluated by E. Brixy (ZAMM 20 (1940), 236-

238); unfortunately some of his results seem wrong. It is desirable to draw rough graphs of the integrands and to check that the integrals all exist. Note

Chapter 9 227

that, e.g., the integrand in 16 is of the form 010 at x = 0 and so some minor adjustment in the program may be necessary.

Since JMx)=-J1(x) we have

Is = -r [Jb(x)IJo(x)] dx = [log [JO(X)]-I]~ = log [lIJo(1)] = 0.267621.

Since xl2(x) = J1(x)-xlHx) we have

I = r1 J1(x)-xlHx) d = rl!!:... [1 {_x }] d 6.10 xll(X) x.lo dx og J1(x) x

= [log {J1;x)}I = log [1IJ1(1)]-10g 2 = 0.127718.

Since xlO(x)=2J1(x)-xlz{x) and since, as we have just seen in the discussion of h, [Jz{x)IJ1(x)] = [log {xIJ1(x)}]' we have, on integration by parts,

14 = r (2- x [log {xIJ1(x)}],) dx

= 2-[x 10g{xIJl(x)}~+ r log {xIJ1(x)} dx

= 2+ log J1(1) + log 2+ 17 •

This result can be used as a check on the values of 14 , 17 •

11 = 1.09407,

14 = 1.91448 5,

12 = 2.08773, 13 = 0.08673 4,

17 = 0.04220 4, Is = 0.02795 4.

9.16. Solution If 0:5 8 :5~1T then 0 :5 cos 8:51 and so COS2n-1 82:: cos2n 8 2:: COS2n+1 8

which implies I 2n- 1 2:: I 2n 2:: I 2n+ 1 where

It is easy to find a reduction formula for this integral: in fact

nIn = (n -1)1,.-2, n 2:: 2,

and 10 = ~1T, II = 1. These give

(2n(n!))2

12n+ 1 = (2n + 1)1"


Inserting these values in the inequalities

we get

[2n- 1«n-1)1)]2 (2n)! '1T [2n(n!)]2 ------'-> . ->"---~

(2n-1)! 22n(n!)2 2 (2n+1)!

which gives

1 ~> (2n)!(2n + 1)! '1T> 1 + 2n 24n(n!)4 2-

which is the result required.

9.17. Solution This is established in a way similar to that used to get Wallis' Formula. We write

i1 tn dt In = (1- t4)1/2

and observe that In decreases as n increases. It is easy to show that 13 = ~ for [-~(1- t4)1I2]' = t3(1- t4)-1I2.

We next obtain the reduction formula

In+3 = [n/(n + 2)]In - 1 •

We then use this, and the facts that

to conclude that

and hence that

Use of the reduction formula gives

I 1.5 ..... (4r-3) L 4. 3.7 ..... (4r-1) 0,

I - 3.7 ..... (4r-1) I 4.+2 - 59 ( ) 2 ...... 4r+1

and

1 =!. 2.4 ..... (2r) 4.+3 2 3.5 ..... (2r+ 1)"

Chapter 9 229

We now observe that

1012 = (4r+ 1)14,14r+2 = (4r+ 1)(14r+3)2 X (14,/14r+3) X (14r+~14r+3)'

We let r~oo and the last two factors tend to unity and the first, because of Wallis' Formula, tends to ~1T.

9.18. Solution

1= r X4 dx = (1/5)(b5-a5) = (l/5)(b - a)[a4+a3b +a2b2+ ab3+ b4].

Q = 3~:3a) [a4+3ea3+ br +3(a~2br +b4]

= (b -a) x[27a4+27b4+(16a4+32a3b +26a2b2+8ab3+ b4) 8x27

+(a4+8a3b +24a2b2+32ab3+ 16b4)]

(b-a) =-- [lla4+ 10a3b + 12a2b2+ 10ab3+ llb4].

2x27

Hence the actual error is:

1-Q= -(b-a)[ 4-4 3b+6 2b2-4 b3+b4]= -(b-a)5 270 a a a a 270 .

The estimate is

In the case of N panels, the estimate will be

where M4 is maxasxs" If(x)l. If we take M4 to be 1 then we have to choose N to make

5 x64xN4> e-1

i.e., N>(6480 e)-1/4. E.g., with e = 10-8 we must take

N> (6480)-114 X 102 =€= 102/9 =€= 11.

and so about 34 evaluations of the integrand. Note that if we change M4 to 10 or to 0.1 we only change N by a factor of 10±1I4~ 1.78 or 0.60.


9.19. Solution We have to show that

J = r ~(x)(1-li(x))w(x) dx = o.

Since l;(.xi ) = 1 it follows that (x - x;) divides 1- 4(x), the quotient being a polynomial of Qn-2(X) of degree at most n - 2. Thus

J= r {(x-x;)~(X)}Qn_2(X)W(X) dx

and, as the quantity in braces is a multiple of '1Tn(x), it follows by orthogonality that J = O.

9.20. Solution Since If4)(x)1 $1, the absolute values of the errors in Simpson's Rule

and in the i-Rule are bounded by

h5/90 and 3h5/80,

h being the interval. Suppose we use N panels. Then the total error in the Simpson case is

N. (b-a)5.~ 2N 90

and that in the i-case is

If the error is to be less than e(b - a)5 we must have

N=~1/2880e

and

N = ~1/6480e.

The cost, in terms of evaluations of the function [, is

The basic error estimate in the Gauss-Legendre case is (Problem 9.12)

when we use the interval (-1, 1). For a general interval (a, (3) we have to

Chapter 9 231

write

f: f(x) dx = t:l f(i«(3 -a)t+!(a + (3)). !«(3 -a) dt =!«(3 -a) t:1 F(t) dt

where t=(2x-a-(3)/«(3-a). As pr)(t)=[!«(3-a)]rfr)(x) our error estimate should be mulitplied by [!«(3 - a) fn+l. For n = 2 we find

Returning to our main problem, if we again use N panels, the total error will be

(b -a)5 1 N· N ·4320

and the relevant value of N is

N =~1/4320B.

There are two evaluations per subinterval (at "awkward" abscissas) and so the corresponding cost is

2N=~1/270B.

The relative efficiencies are therefore about

1.36 G-L.; 1.22 S; 1(i).

9.21. Solution See NBS Handbook, p. 492. In particular

f(3) = 1.3875672520-0.9197304101 = 0.4678368419.

9.22. Solution We find

'YlO = 0.62638316;

'Y30 = 0.59378975;

At the first stage

At the second stage

'Y~O = 'YlO - (1/20) = 0.57638 32;

'Y~o = 'Y30 - (1/60) = 0.57712 ~O.


9.23. Solution This is a table of P4(x) = (35x4 -30x2+3)/8; the endings 4,6 are

exactly 375,625. The exact value of the integral is

Simpson: (Milne)3: (3/8)4:

[(7x 5 -10x3 + 3x)/8]A·2 = 0.46728.

0.467352 0.466304 0.467439

We shall examine these results more closely. The errors in the first three methods are estimated as multiples of a mean of the fourth derivative of the integrand i.e., 105. The multiples are obtained in Problems 9.4, 9.2, 9.1. We have to be careful in scaling. The observed errors are respectively

-72x10-6 , 976 x 10-6 , -159x10-6 •

The theoretical errors are

i.e.,

-7X 10-5 ,

in close agreement.

9.24. Solution

3 x!~x 10-5 x 105,

-15.75 x 10-5

The relation given is ascribed to C. Hermite (E. W. Hobson, Plane Trigonometry (1925), 378). Write for r = 1, 2, .•. , n,

2 sin ra cos (nx -rx) = sin (nx +r(a - x» + sin (-nx+r(a + x».

We then use the formula

cos (0 _lcp)-COS (0 +1(2n -l)cp) sinO+sin(0+cp)+ ... +sin(0+(n-1)cp)= 2 2. 1 2

SID2CP

to get

{ }=cos(nx+!(a-x»-cos(oo+!(a-x» ... 2sin!(a-x)

cos (nx-!(a + x»-cos (00 +!(a + x» + 2sin!(a+x)

sin 00.

Using the relation 2 cos A sin B = sin (A + B)-sin (A - B) four times and the fact that

4 sin !(a - x) sin !(a + x) = 2(cos x -cos a)

Chapter 9 233

we find

2(cos x -cos a){ . .. }=sin (nx +a)-sin (n -1)x -sin (n+ 1)a + sin (00 - x)

+ sin (n -1)x -sin (nx - a) -sin (n + 1)a + sin (00 + x)

-2 sin na cos x +2 sin 00 cos a

= sin (nx +a)-sin (nx -a)-2 sin (n + 1)a + sin (00 + x) + sin (oo-x)

-2 sin 00 cos x +2sin 00 cos a

= 2 cos nx sin a -2 sin 00 cos a -2 sin a cos 00

+2sinoocosx

-2sin 00 cos x +2 sin 00 cos a

= 2 sin a (cos nx -cos 00),

which gives the result required. If we take a = 8m = !(2m -1)7r/n, x = 8 we get

cosn8 1 { . 8 8 = -. -8- (_1)m-l + 2 sm (m -1}Om cos 8

cos -cos m sm m

+ 2 sin (m - 2) 8m cos 28 + ... + 2 sin 8m cos (m -1)8}

so that

(_1)m-l ao= . 8 ' sm m

9.25. Solution

2 sin (m - r}Om a,. = r = 1, 2, ... , m -1.

(a) For a popular method using double integrals see T. M. Apostol, Calculus, II, §11.28, Exercise 16.

(b) The following method is given by A. M. Ostrowski, Aufgaben sammlung . .. III, p. 51, 257.

We prove that

(1) [f e-t2 dtT = 11T -r e-x2(1+f2) dt/(1 + t2 )

by observing first that both sides vanish for x = 0 since SA dt/( 1 + t2) = 11T and then noting that the derivatives of the two sides are identical: the derivative of the right hand side is

-( -2x) r e-x 2(l+f2) dt = 2 r e-x2- u2 du, if u = xt

= 2e-x2 LX e-u2 duo


If we now let x~oo in (1) the left hand side has limit [io e-t2 dtf. The integral on the right, since t ~ 0, is bounded by J~ e -x2 dt = e -x2 and this tends to zero as x~oo. Hence the limit of the right hand side is hr.

(c) The following method is due to A. J. Bosch, Nieuw Tijdschr. Wisk. 64 (1977), 210-211.

We use the results of Problem 9.16. Since

12n+1 = (Tr/2(2n + 1))IZ"~

and since 12n+2<12n+l <12n we have

2n+ 1 Tr 1

2n +2 12n <2(2n + 1) lZ"n <12n

which gives

2n+l nTr 2n + 2 (n11J < 2(2n + 1) < (n11J

so that

For O:sx:sl we have

l-x2 <e-x2 «1 +X2)-1

and taking n-th powers and multiplying across by In and integrating we get

.,/r" r (1- x2)n dx <.,/r" r e-nx2 dx = L"" e-t2 dt <.,/r" r (1 + x 2)-n dx.

Changing the variable in the integral on the left by x = sin (J shows that

r nTr lZ"~ Tr 2 1 I

"nI2n+ 1 = 2(2n + 1) • J;;~4· .;; = 2" Tr.

The integral on the right is less than

(where we use a change of variable x = tan 8). Hence the infinite integral

The numerical integration shows up some interesting points. Namely

Chapter 10 235

one can estimate the tail

by the formula of Chapter 7. This tail will certainly be less, e.g., than ~ x 10-6 if x ~ 3.5. We can therefore restrict ourselves to J~.5 e-x2 dx which can be handled by any of the usual formulas.

If we use the trapezoidal rule it is found experimentally that remarkable accuracy is obtained for quite large steps. For further discussion of this see, e.g., E. T. Goodwin, Proc. Cambridge Phil. Soc., 45 (1949),241-245. D. R. Hartree, Numerical Analysis, Oxford, 1952, 111.

Chapter 10

10.1. Solution The general solution is Un = Aan + B{3n where a, {3 are the roots of

x2 -x-1=0

and we obtain A, B from

getting

1=A+B }

1=Aa+B{3

= 1 +v's (1 +v's)n + (-1 +v's)(1-v's)n. Un 2v's 2 2J5 2

The following values can be used as a check

U34 = 9227465 U 35 = 14930352 U36= 24157817 U37 = 39088169 U 38 = 63245986 U 39 = 1023 34155

(c. A. Laisant, L'enseignement math. 21 (1920), 52-56).

10.2. Solution We deal with the general case. Suppose we have obtained

IIn (x) = k"xn + k~xn-l + ....


Then, if

A.. = kn+1/k..,

is a polynomial of degree at most n and can be expressed as a linear combination of 110 , 1110 ... I1n. The coefficient of I1n is

The coefficient of I1n- 1 is

r I1n+1I1n- 1 w dx - A.. ib xl1nl1n- 1 dx

a a

= 0- A.. r I1n(xl1n- 1) dx = -Ank..-1/k.. = -k..+1k..-1/k~. The coefficients of the earlier l1's vanish by orthogonality. Hence we have

The results for special cases can be obtained from NBS Handbook, p. 782 and include

(a) (b)

(c) (d)

(n + 1)Pn+1 = (2n + 1)xPn - nPn-l>

(n + 1)L,. = «2n + 1)-x)L,. - nL..-1o

10.3. Solution The relation is equivalent to

Vn = aUn +b, k(1-k) = ac-b2 +b.

Convergence can only take place to k or 1- k. We may assume that k~! when it is real. The following gives a complete answer:

(1) k not real, Vn ~ 00 monotonically (2) IVol> k, Vn ~ 00 monotonically (3) IVol> k, !::; k ::;~, Vn ~ 1- k and monotonically if !::; k::;1 (4) I vol < k, ~ < k ::; 2, Vn oscillates finitely (5) IVol<k, k>2, Vn ~oo except when Vo belongs to a certain set (cardinal c, measure zero) in which case it oscillates finitely.

Chapter 10 231

For further details see: T. W. Chaundy and Eric Phillips, Quarterly Journal (Oxford), 7 (1936),

74-80.

10.4. Solution. For further details see E. Hansen, Signum Newsletter 3 (1968), #3.

19 = 0.09161 23, 110 = 0.0838771.

10.7. Solution The exact solution to this equation is y = (1- x)-t, with a pole at x = 1. Consider integrating at an interval h = N-1 and write x,. = rh, y, = y(rh)

where

We have

Y,+1 = y, + hy;.

1 1 h 1 h2 y, -=----=--h+-y r+1 y, 1 + hy, y, 1 + hy,'

r = 0, 1, ... , n -1.

Summing we get

1 1 r-l y -=--rh+h2 L --s-=1-rh+hR, say. y, Yo s=O 1 + hys '

We shall now estimate the R" which obviously increase steadily from Ro = 0 to RN • Oearly

hy, 1 1 R,+1-R,=1+hy, =h-1 -r+R,+1 N+1+R,-r

We use the method of the "Integral Test". First, neglecting the positive quantities R, we have

RN< L < log (N+1). N-l 1 iN dx r=O N+1-r N+1-x

Next,

~1 1 f~l ~ RN> > ,~o N +1 + log (N+ 1)-r 1 N +1 + log (N+1)-x

We have therefore shown that

RN = log N +O(log log N)

= log h-1 + O(log log (h-1))

=10 N +10g(N+1) g log (N +1)+2'


and hence

1 { ~loglog(h-l»)} = 1+ . h log h-1 log (h-1)

For further details see M. L. 1. Hautus and G. W. Veltkamp, Nieuw Arch. Wisk. {3}, 17 (1969), 79-80.

10.8. Solution In practice there will be round-off so that, e.g., 100 times 0.001 will not

be 0.1 exactly. A refined program will allow for this and take a small step (backwards or forwards) before printing so as to get the argument 0.1 (or, as near this as the machine allows!).

Library programs will often have an automatic choice of an appropriate interval incorporated.

The solution of y' = x 2 - y2, y(O) = 1 has the value 0.750015703 at x = 1. See Appendix for details.

The results from a Univac computation for the equation

y' = !"'~ + !"'y, y(O) =0

are given below.

x h =0.00025 h =0.00125 h =0.0025 h =0.0125

0 00000000000 00000000000 00000000000 00000000000 0.0125 00054793700 00054084766 00052730731 00034938562 0.0250 00159912174 00159108463 00157549032 00135795112 0.0375 00299888889 00299020417 00297323965 00273099005 0.0500 00469064935 00468145066 00466341119 00440237794

0.0625 00664121481 00663158054 00661263636 00633609674 0.0750 00882794453 00881792703 00879819092 00850825638 0.0875 01123406696 01122370425 01120325726 01090141626 0.1000 01384648727 01383580841 01381471217 01350207833 0.1125 01665459906 01664362708 01662193028 01629937288

0.1250 01964957501 01963832876 01961607102 01928429044 0.1375 02282391333 02281240853 02278962286 02244919744 0.1500 02617113183 02615938188 02613609625 02578751306 0.1625 02968555308 02967356957 02964980810 02929348394 0.1750 03336214844 03334994148 03332572516 03296202153

Chapter 10 239

x h =0.00025 h =0.00125 h =0.0025 h =0.0125

0.1875 03719642145 03718399993 03715934729 03678858123 0.2000 04118431874 04117169062 04114661815 04076907044 0.2125 04532216083 04530933327 04528385574 04489977709 0.2250 04960658706 04959356649 04956769724 04917731319 0.2375 054034511"39 05402130364 05399505479 05359856951

0.2500 05860308623 05858969672 05856307932 05816067873

This is a classical problem studied by Moigno in 1844. Awkwardnesses arise because the expansion of the solution around the origin is of the form:

10.9. Solution The solution is y = eX E 1(x) and the following values are obtained from

NBS Handbook, p. 243:

x= 1 0.596347361 2 0.361328617 3 0.262083740 4 0.20634 5650 5 0.177297535

10 0.0915633339 20 0.04771 85455

10.11. Solution Writing z = y' we replace our scalar differential equation by a vector

differential equation

where [~] is given for x = 0: [;,~~J.

We have

{Y*(1) = yeO) + hz(O)

z*(1) = z(O)+ hf(O, yeO), z(O))

and

{Y**(1) = y(O)+ h[z(O)+ hf(O, yeO), z(O))]

z**(1) = z(O)+ hf(h, y(O)+ hz(O), z(O)+hf(O, yeO), z(O)))

giving

{Y(1) = yeO) + hz(O) +1h2f(0, yeO), z(O))

z(1) = z(O) +1hf(0, yeO), z(O)) +1hf(h, yeO) + hz(O), z(O)+ hf(O, yeO), z(O)))


Clearly y(1), z(1) only depend on y(O), z(O), hand f. We have to evaluate f twice, once with arguments 0, y(O), z(O) (which we use three times), and once with arguments

h, y(O)+ hz(O), z(O) + hf(O, y(O), z(O)).

We are free to change h whenever it is desirable without change in the formulas.

Equation (14) has been studied by P. Painleve - the solution becomes singular, as in the simple case of

y(O)=A

which has a solution y = (A - X2)-1. Thus the problem is one of "polar exploration". The Painleve equation in the case y'(O) = 1 and on the real axis has a series of poles of order 2 of which the first two are at

1.70 ... , 2.97 ....

10.12. Solution Discretizing using a mesh size h = 1, we get a system of three

homogeneous linear equations for y(l), y@, y(~):

(y(0))-2y(1)+y@ = -l~ . lAY (1)

y(1)-2y@+ y(~) = -l~· ~AY(~)

y@-2y(£)+(y(1))= --h. £AY(£)

Note that y(O) = y(1) = O. In order that we should have a nontrivial solution the determinant of

this system must vanish, i.e.,

[(A -128)

det 32 o

This reduces to

64 (A -64)

64 3"

o ] 32 =0. (A _1~8)

A 3_ A 2[11 ;64] + A[128 X364 X 5] _ [64X6;X 128] = O.

Writing A = 64A we get

i.e.,

Chapter 10 241

The last equation has an obvious root A = 1 and the residual equation is

3A2 -8A+2=0

which has roots A = (4±JiO)/3. Thus the roots are

Al = 64(4 - JiO)/3 = 17.8688, A3 = 64(4+ JiO)/3 = 152.794.

The corresponding values of y(~), y@), y(i) are:

[ 2+~J' [~], [2-~J. 2+M -1 -2+M -2+M . 2+M

These have sign patterns (+++), (++-), (+-+). Integrating the differential equation with A = 17, 18, 19 enables us to

get a better estimate for the smallest eigenValue. In order to check our integration subroutines we proceed as follows.

Integrate the differential equation with

A = 18.6624, 18.9225, 19.1844, 19.4481

The corresponding values of the solution at x = 1 should be

0.01066, 0.00122, -0.00816, -0.01747

and we interpolate inversely by the Aitken method

1066 6624 122 9225 9561

-816 11844 581 564 -1747 14481 601 564

giving A = 18.9564 which is the correct result. The motivation of our choice of the four A'S is the following: more

details are given in the Appendix. The solution to

y"-AXY =0, y(O) = 0, y'(O) = 1

is a multiple of

which has the value


when x = 1. We want to choose A so that this vanishes. The zeros of J1/3

have been tabulated in the NBS volume cited in the Appendix and are, to 4D,

2.9026, 6.0327, 9.1705, ....

These correspond to the following values of A:

18.9564, 81.8844, 189.219, ...

(which we can compare with the roots of the cubic). It is Al = 18.9564 with which we are concerned.

The assigned values of A correspond to the following values of ~A 4

~(18.6624)1/2 =~(4.32) = 2.88, 2.90, 2.92, 2.94

The values of J1/3(X), x = 2.88(0.02)2.94 are 0.01066, 0.00122, -0.00816, -0.01747.

10.13. Solution [For a more theoretical discussion of this problem see, e.g., p. 279 of G.

Birkhoff and G. C. Rota, Ordinary differential equations, Ginn, Boston, 1962.]

This is a simple form of the Schrodinger equation. Consider the case when V is even, vanishing outside [-1, 1]. We ask whether there are values of E for which there are nontrivial solutions (Le., I/I¢ 0) which are even, or odd and which are such that I/I(x) ~ 0 as x ~ ±oo.

Since 1/1 is required to be even or odd we can restrict our attention to x ~ O. When x ~ 1 the equation reduces to

I/I"(x) + EI/I(x) = 0

with general solution

l/I(x)=A exp(vCE(x-1»+B exp(J.=E(1-x».

In order that cp(x) ~ 0 as x ~ 00 we must have E < 0, A = O. When 0::S;X:51, the solution depends on the form of V(x). However it

will involve two arbitrary constants, say 1/1(0), 1/1'(0). Now if 1/1 is even, 1/1'(0) = 0 and if 1/1 is odd, 1/1(0) = O. Hence one constant is determined and the solution is of the form

I/I(x) = Ccp(x)

where C is arbitrary and cp involves E. To match the solutions at x = 1 we must have

Ccp(1) = 1/1(1) = B, Ccp'(1) = 1/1'(1) = -r-EB

Chapter 10 243

which can be satisfied if and only if

T(E) == J=Bp(1) + cp'(1) = O.

This is the equation which determines the characteristic values (if any) of E. We now return to the special V(x) of the problem and note that an odd solution of

I/I"(x) + (E + 5)I/I(x) = 0

is

cp(x) = sin.JE + 5x

for which cp(1) = sin.JE + 5, cp'(1) =.JE + 5 cos.JE + 5. The characteristic values are determined by

i.e.,

Hsin .JE+5+.JE+5 cos.JE+5 =0

~E+5 tan.JE+5= - E.

Remembering that E is negative we consider the behavior of the two sides of the last equation in the range -5:5 E :5 0 where they are real. The left side increases from 0 to 00 in (-5, -5 +hr2) and from -00 to tan J5 in (-5 +i1T2, 0). The right side decreases from 0 at E = -5 to -00 at E = O. There is exactly one intersection near E = -1. This can be estimated by Newton's method. If

f(E) = tan.JE + 5 +.J(E + 5)/E

then

f(-1) = tan 2+2,

and the next approximation is

-1-4(2+ tan 2) ~-1 + 0.74016 . -09 1 5 +sec2 2 10.7744 . . 3 .

10.14. Solution In general it is not possible to obtain I/I(x) explicitly as in Problem

10.13. We must proceed as follows. According as to whether we want an even or odd solution we integrate the equation

I/I"(x) = (V(x) - E)I/I(x)


numerically from 0 to 1, beginning with 1/1(0) = 0, 1/1'(0) = 1 or 1/1(0) = 1, 1/1'(0) = 0, for some value of E. We then check whether

T(E) = HI/I(I) + 1/1'(1)

is or is not zero. If it is not zero we try to get a better guess for E and repeat the process until we get T(E) close enough to zero.

Various artifices can be used to determine E, e.g., inverse interpolation as on p. 133.

The result is that there is an even solution corresponding to E ~ 0.337.

10.15. Solution The table is meant to be one of (3x+912)3-38299 99877. There were errors in the entries for arguments 217, 218, 219: the

corrected values are respectively-116 39330; 10389619; 325 03132

10.16. Solution Compare Problem 7.7.

y(1.5) = 0.7127926, y(2.0) = 0.6429039.

10.17. Solution

) _anf(n+(b/a» a . x,. - f(b/a) Xo·

b). x,. = (Ain + Bin+l)/C

where

A = {cio + (a + b )it}xo - ilXl>

and

J. = (_c)1/2n J (2a-1(-c)I/2) n n+(b/a) •

See D. H. Lehmer, Proc. Conference Numerical Mathematics, Winnipeg (1971), 15-30.

10.18. Solution

f(7) =2983

10.19. Solution a) These are the odd prime numbers. b) If Pn is the nth prime number of the form 4r+l so that PI =5,

P2 = 13, P3 = 17, . .. then the sequence is that of the least integer N for

Chapter 10 245

which Pn is a factor of W+1. Thus 22+1=5, 52+1=2.13, 42+1=17, 122+ 1 = 5.29, ...

c) These are the values of n2 - n +41 for n = 0,1,2, ... and are all prime numbers until n = 41.

10.20. Solution

Ai'(l.4) = -0.10850 959

Ai"(l.4) = 1.4Ai(1.4) = 0.11485327

10.21. Solution See preceding solution, or British Assoc. Math. Tables, Part Vol. A.

10.22. Solution

x(O) = 1= e-t4 dt =! 1= e-T . T-3/4 dT = !f(!) = 0.906402479.

x'(O) = r 2t2e-t4dt=~r e-Tl1!4dT=~f(~)=0.612708351. Differentiating we find

y"=x2y+ r t(4t3-3xt)exp(-~x2+2xt2_t4)dt -1= exp (-h2+2xt2- t4) dt.

Integrating by parts the second term on the right we find

{= ... dt = [-t exp {!x 2+2xt2- t4 )Jo+ r exp (-~x2+2xt2- t4 ) dt

The first term on the right vanishes at both limits and the second cancels the third term on the right in the previous display. Hence

The solution is

and for x = -2(1) 8 this has values

0.0078, 0.2059, 0.9064, 0.9869, 0.6702, 0.5241, 0.4487,

0.3637, 0.3363, 0.3143.


10.23. Solution See, British Assoc. Math. Tables, Vol. 2. For recent theoretical work

see papers by E. Hille.

10.24. Solution See, J. L. Synge, On a certain non-linear equation, Proc. Royal Irish

Academy 62A (1961), 17-41. Z. Nehari, On a non-linear differential equation arising in nuclear physics, Proc. Royal Irish Acadymy 62A (1963), 117-135.

10.25. Solution We use an interval h = 0.1 and work to 4D. From the power-series

y(x) = 1 + x + x 2 + (x 3 /3) + (x4/12) + ...

we compute y(0.1), y(0.2), y(O.3) and [(0.1), [(0.2), [(0.3) and we then predict

YP (0.4) = 1 + (4/30)[2.4206 -1.4428 + 3.3994] = 1.5836

We then compute [(0.4) = 1.9836 and check by

Yc (0.4) = 1.2428 + (1.30)[1.4428 + 6.7988 + 8.9836] = 1.5836.

We can accept this value and proceed, or alternatively, try a larger h = 0.2.

x

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

YP 1.0000 1.1103 1.2428 1.3997 1.5836 1.7974

Yc y'=[(x,y)=x+y

1.5836 1.7974

1.0000 1.2103 1.4428 1.6997 1.9836 2.2974

The correct value at x = 1 is 2e - 2 = 3.43656 ... . .

Bibliographical Remarks

RECOMMENDED LITERATURE

T. M. APoSTOL, Calculus, I, II (Wiley, 1967-9). A. M. OsTROWSKI, Vorlesungen uber Differential- and Integralrechnung, 3

vols., Aufgabensammlung zur Infinitesimalrechnung, 3 vols. (Birkhauser, 1965-72).

TESTS AND MONOGRAPHS

F. S. ACTON, Numerical methods that work (Harper and Row, 1970). E. K. BLUM, Numerical analysis and computation; theory and practice

(Addison-Wesley, 1972) C. BREZINSKI, Acceleration de la convergence en analyse numerique

(Springer, 1971). S. D. CONTE and C. DE BOOR, Elementary numerical analysis (McGraw-Hill,

1973). G. DAHLQUIST and A. BJORCK, tr. N. ANDERSON, Numerical methods

(Prentice-Hall, 1976). J. W. DANIEL and R. E. MOORE, Computation and theory in ordinary

differential equations (Freeman, 1970). P. J. DAVIS and P. RABINOWITZ, Methods of numerical integration (Academic

Press, 1975). C.-E. FROBERG, Introduction to numerical analysis (Addison-Wesley, 1965). R. W. HAMMING, Introduction to applied numerical analysis (McGraw-Hill,

1971). H. M. NAUTICAL ALMANAC OFFlCE, Interpolation and allied tables (H. M.

Stationery Office, 1956). P. HENRlCI, Elements of numerical analysis (Wiley, 1964). P. HENRlCI, Discrete variable methods in ordinary differential equations

(Wiley, 1962). P. HENRlCI, Computational analysis with HP 25 pocket calculator (Wiley,

1977).

248 Bibliographical Remarks

F. H. Hn.DEBRAND, Introduction to numerical analysis, 2nd ed. (McGrawHill, 1976).

A. S. HOUSEHOLDER, Principles of numerical analysis (McGraw-Hill, 1953; Dover, 1974).

A. S. HOUSEHOLDER, The numerical treatment of a single non-linear equation (McGraw-Hill, 1970).

E. ISAACSON and H. B. KELLER, Analysis of numerical methods (Wiley, 1966).

L. M. Mn.NE-THOMSON, The calculus of finite differences (Macmillan, 1933). Modem Computing Methods (H. M. Stationery Office, 1961). B. R. MORTON, Numerical approximation (Routledge and Kegan Paul,

1961). B. NOBLE, Numerical methods, 2 vols. (Oliver and Boyd, 1964). A. M. OsTROWSKI, Solution of equations, 3rd ed. (Academic Press, 1973). A. RALsTON and P. RABINOWITZ, A first course in numerical analysis, 2nd ed.

(McGraw-Hill, 1978). T. J. RIvuN, The Chebyshev polynomials (Wiley, 1976). H. RUTISHAUSER, Vorlesungen uber numerische Mathematik, 2 vols.

(Birkhauser, 1976). H. RUTISHAUSER, Numerische Prozeduren - ed. W. Gander, L. Molinari and

H. SvecovR (Birkhauser, 1977). F. ScHEID, Theory and problems of numerical analysis, (Schaum-McGraw

Hill, 1968). L. F. SHAMPINE and R. C. ALLEN, Numerical computing: an introduction

(Saunders, 1973). L. F. SHAMPINE and M. K. GORDON, Computer solution of ordinary differential

equations. The initial value problem (Freeman, 1975). E. STIEFEL, tr. W. C. and C. J. RHEINsolDT, An introduction to numerical

mathematics (Academic Press, 1963). J. STOER, Einfuhrung in die numerische Mathematik, I 2nd ed. (Springer,

1976).

J. STOER and R. BUURSCH, Einfuhrung in die numerische Mathematik, II 2nd ed. (Springer, 1978).

JOHN TODD, ed., Survey of numerical analysis (McGraw-Hill, 1962). JOHN TODD, Numerical analysis, Chapter 7, Part 1 of E. U. Condon-H.

Odishaw, Handbook of Physics, 2nd ed. (McGraw-Hill, 1967). JOHN TODD, Introduction to the constructive theory of functions (Birkhauser,

1963). J. F. TRAUB, Iterative methods for the solution of equations (Prentice-Hall,

1966).

Bibliographical Remarks 249

B. WENDROFF, Theoretical numerical analysis (Academic Press, 1966). E. T. WHfITAKER and G. ROBINSON, The calculus of obseroations, 4th ed.

(Blackie, 1966). J. H. WILKINSON, Rounding errors in algebraic processes (Prentice-Hall,

1963). D. M. YOUNG and R. T. GREGORY, A suroey of numerical mathematics, 2

vols. (Addison-Wesley, 1972-73).

TABULAR MATERIAL

M. ABRAMOWITZ and I. A. STEGUN, ~s. Handbook of mathematical functions, National Bureau of Standards, Applied Math. Series, 55 (U.S. Government Printing Office, 1964).

BARLOW'S TABLES, ed. L. J. Comrie (Spon, 1961). L. J. COMRIE, Chamber's Shorter six-figure mathematical tables, (Chambers,

1950). A. FLETCHER, J. C. P. MILLER, L. ROSENHEAD and L. J. COMRIE, An index of

mathematical tables, 2nd ed., 2 vols. (Addison-Wesley, 1962).

In addition to the literature mentioned above, there are many useful expository articles available, and for up-to-date surveys of special topics, reference can be made to Symposia Proceedings, such as those which appear in the ISNM series.

There is a developing interest in the history of numerical mathematics and computing machines. For the classical material see, respectively. H. H. GOLDSTINE, A history of numerical analysis from the 16th through the

19th century (Spnnger, 1978), B. RANDELL, The origins of digital computers, 2nd ed. (Springer, 1975).

For more recent history see the obituaries of the founders and the periodical Annals of the history of computing, 1979-.

Contents

Vol. 2, Numerical Algebra

Notations and Abbreviations

Preface .

Chapter 1. Manipulation of Vectors and Matrices Chapter 2. Norms of Vectors and Matrices ... Chapter 3. Induced Norms ...... . Chapter 4. The Inversion Problem I: Theoretical Arithmetic Chapter 5. The Inversion Problem II: Practical Computation. Chapter 6. The Characteristic Value Problem - Generalities. Chapter 7. The Power Method, Deflation, Inverse Iteration. . Chapter 8. Characteristic Values. . . . . . . . . . . . .. Chapter 9. Iterative Methods for the Solution of Systems Ax = b Chapter 10. Application: Solution of a Boundary Value Problem. Chapter 11. Application: Least Squares Curve Fitting. . . . . . Chapter 12. Singular Value Decomposition and Pseudo-Inverses

Solutions to Selected Problems

Bibliographical Remarks

Index ........ .

7

9

.13

.16

.19

.29

.44

.53

.65

.71

.83

.99 105

110

. 117

.212

.214

INDEX

M. Abramowitz 141, 249 Acceleration methods 66, 68 A. C. Aitken

algorithm 86, 87, 241 transform 67

G. B. Airy integral 124, 129, 137, 150, ISS, 157,245

Algorithms Borchardt 15, 166, 170 Carlson 16 Gauss 13

E. E. Allen 190 Arithmetic-geometric mean 13, 161 Asymptotic series 75

Bad examples 47, 88, 119, 130 F. L. Bauer 102 S. N. Bernstein 58, 211

polynomials 59, 63, 184, 187 F. W. Bessel 149

functions 64, 65, 115, 116, 119, 189, 227, 241

modified functions 145 Garrett Birkhoff 242 G. D. Birkhoff 91 J. L. Blue 180 R. P. Boas, Jr. 172 C. W. Borchardt 15, 163, 166 A. J. Bosch 234 E. Brixy 226 G. H. Brown, JI. 182

B. C. Carlson 16, 18,21,22,164,167,168 T. W. Chaundy 237 P. L. Chebyshev 186

coefficients 61 polynomials 60, 64, 186

E. B. Christoffel- J. G. Darboux formula 109

C. W. Clenshaw 62 L. Collatz 54, 181 Complementary error function 77 L. J. Comrie 81, 249

M. Davies 182

B. Dawson 182 R. Dedekind 37, 40 Deferred approach 100 Difference equation 118 Differencing 121

Elliptic integral 18, 21, 165 F. Emde 141 R. Emden equation 138 O. Emersleben 178 Error analysis

backward 48 forward 48 interpolation 89, 91 quadrature 98, 99, 105, 109, 113, 218,

229 Error function 77, 117 L. Euler 116

constant 31, 116, 231 method 125 transform 68, 69, 71, 72, 192, 193

Exponential integral 73, 81

R. M. Federova 82 W. Feller 87 Fibonacci 133, 179, 235 Fixed point 36, 38 A. Fletcher 81 FMRC 81, 202 G. E. Forsythe 46, 47 J. B. J. Fourier

coefficients 61 method 50 series 64

L. Fox 83, 202 A. J. Fresnel integrals 76, 202 Fundamental theorem of algebra 84, 213

P. W. Gaffney 214 Gamma function 79 K. F. Gauss arithmetic-geometric mean 161 K. F. Gauss - P. L. Chebyshev quadrature

105, 224 K. F. Gauss - A. M. Legendre quadrature

116,224,225,230

252

Walter Gautschi 121 A. Ghizzetti 219 G. A. Gibson 153, 155 E. T. Goodwin 82, 199, 235 I. S. Gradshteyn 226 J. P. Gram - E. Schmidt process 103 J. A. Greenwood 81

E. Halley 54, 182 E. Hansen 134, 237 P. A. Hansen 148 G. H. Hardy 151, 155, 172 H. O. Hartley 81 D. R. Hartree 138,135,245 C. Hastings 61 M. L. J. Hautus 135, 237 D. R. Hayes 219 P. Henrici 50 C. Hermite 232

interpolation 90, 224 K. Heun method 125, 135, 238, 239 E. Hille 246 K. E. Hirst 180 W. G. Horner's method 52 W. G. Hwang 43, 176

Instability 119, 130 Interpolation

Aitken algorithm 86 errors 89, 91 Hermite 90, 224 inverse 88, 241 Lagrange 84, 210 Newton 96, 216 spline 96, 215

Inverse interpolation 88, 241 Iteration 38

C. G. J. Jacobi relation 147 E. Jahnke 141

E. Kamke 124 A. N. Khovanski 180 K. Knopp 195

J. L. Lagrange interpolation 84 C. A. Laisant 235 E. Landau 24 J. Landen 22, 168 A. V. Lebedev 82 H. Lebesgue constants 91 A. M. Legendre

Index

expansion 184 polynomial 184, 236

D. H. Lehmer 244 G. W. Leibniz' Theorem 129, 207 Lemniscate constants 17, 109, 114 J. Liang 191 L. Lichtenstein - S. A. Gerschgorin equa-

tion 191 Local Taylor series 129 F. LOsch 141 E. C. J. von Lommel 149 Y. L. Luke 188

Mean value theorems first 51, 89, 98, 99, 167, 224 second 151, 152

C. A. Micchelli 214 J. C. P. Miller 81 W. E. Milne quadrature 113

I. P. Natanson 211 NBS Handbook 76, 81, 92, 93, 94, 163,

164, 165, 178, 190, 198, 199, 200, 202, 231, 236, 239

Z. Nehari 246 E. H. Neville 87 I. Newton

process 39, 48 interpolation formula 216

J. W. Nicholson 151, 157

F. W. J. Olver 121 Order of convergence 29 Order symbols 24 Orthogonal polynomials 103

Chebyshev 104, 134, 236 Hermite 134, 236 Laguerre 104, 134, 236 Legendre 104, 134, 236

A. Ossicini 219 A. M. Ostrowski 47, 233

H. Pade fractions 64, 188 P. Painleve 240 J. F. Pfaff 19 Eric Phillips 237 E. Picard method 125 S. K. Picken 62 H. Poincare 75 Practical computation 42 Principal value integral 106

Quadrature Gaussian 103, 115 Lagrangian 98 Milne 113, 219, 232 Romberg 100 ~ Rule 99, 113, 218, 229, 232 Simpson 99, 113, 222, 232 Trapezoidal 97, 221

Predictor - corrector 127

Recurrence relations 33 for reciprocals 33 for square root 37

Remainder Theorem 85 E. Ja. Remez 184 L. F. Richardson 66, 100, 126 T. J. Rivlin 214 G. Robinson 180 M. Rolle's Theorem 90, 212, 220 W. Romberg quadrature 100 L. Rosenhead 81 G. C. Rota 242 L. Rubin 219 H. Rutishauser 102 I. M. Ryzhik 226

D. H. Sadler 3, 83, 202 H. E. Salzer 215, 216 I. J. Schoenberg 19, 169 E. SchrOdinger 136, 242 J. Schwab 22, 169 I. J. Schwatt 226 Second mean value theorem 151, 152 R. M. Sievert integral 115 T. Simpson's rule 102, 113, 222, 230 Spline 96 J. Staton 82, 199 I. A. Stegun 141, 269

Index

E. Stiefel 102 J. Stirling 18

formula 79, 111, 174, 211 G. G. Stokes 56 O. Stolz 153, 155 Stopping rule 42

253

J. C. F. Sturm - J. Liouville problem 124, 131

Subtabulation 95, 214 J. L. Synge 246 Synthetic division 53

Olga Taussky Todd 3 H. C. Thacher 121 Theoretical arithmetic 42 ~ quadrature 99, 116, 229, 230 A. F. Timofeyev 226 E. C. Titchmarsh 152, 155 O. Toeplitz 195

UNIVAC 238

C. J. de la Vallee Poussin 153, 155 A. Vandermonde 95, 148, 213 G. W. Veltkamp 135, 237

H. S. Wall 44, 179 J. Wallis' formula 116, 229 S. E. Warschawski 191 G. N. Watson 141, 198, 203 B. Wendroff 219 E. T. Whittaker 180 K. T. W. Weierstrass' Theorem 58 A. van Wijngaarden 72 J. H. Wilkinson 47 S. Winograd 214 W. Wirtinger 151, 157 J. W. Wrench, Jr. 172

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