NUMERICAL MATHEMATICS & COMPUTING 6th Edition · Example 2 Example Approximate e8 using Taylor...

NUMERICAL MATHEMATICS & COMPUTING6th Edition

Ward Cheney/David Kincaid c©

UT Austin

Engage Learning: Thomson-Brooks/Colewww.engage.com

www.ma.utexas.edu/CNA/NMC6

September 16, 2011

Ward Cheney/David Kincaid c© (UT Austin[10pt] Engage Learning: Thomson-Brooks/Cole www.engage.com[10pt] www.ma.utexas.edu/CNA/NMC6)NUMERICAL MATHEMATICS & COMPUTING 6th EditionSeptember 16, 2011 1 / 13

1.2 Review of Taylor Series

ex = 1 + x +x2

2!+

x3

3!+ · · · =

∞∑k=0

xk

k!(|x | <∞)

sin x = x − x3

3!+

x5

5!− · · · =

∞∑k=0

(−1)kx2k+1

(2k + 1)!(|x | <∞)

cos x = 1− x2

2!+

x4

4!− · · · =

∞∑k=0

(−1)kx2k

(2k)!(|x | <∞)

1

1− x= 1 + x + x2 + x3 + · · · =

∞∑k=0

xk (|x | < 1)

ln(1 + x) = x − x2

2+

x3

3− · · · =

∞∑k=1

(−1)k−1xk

k(−1 < x ≤ 1)


Example 1

Series are often used to compute good approximate values ofcomplicated functions at specific points.

Example

Approximate ln(1.1) using five terms in Taylor Series.

Take x = 0.1 in the first five terms of the series for ln(1 + x)

ln(1.1) ≈ 0.1− 0.01

2+

0.001

3− 0.0001

4+

0.00001

5= 0.09531 03333 . . .

This value is correct to six decimal places of accuracy!.


Example 2

Example

Approximate e8 using Taylor Series.

The result is

e8 = 1 + 8 +64

2+

512

6+

4096

24+

32768

120≈ 570.06666 5

We find e2 = 7.38905 6, e4 = 54.59815 00, e8 = 2980.95798 7.

Many terms are needed to compute e8 with reasonable precision!


RULE OF THUMB

Rule

A Taylor series converges rapidly near the point of expansion and slowly(or not at all) at more remote points.


Partial Sums of Taylor Series

A graphical depiction of the phenomenon can be obtained by usingthe Taylor series

sin x = x − x3

6+

x5

120+ · · ·

and some of the partial-sum functions

S1 = x

S3 = x − x3

6

S5 = x − x3

6+

x5

120


Figure

Figure: Approximations to sin x


Formal Taylor Series of f at the point c

These series are examples of the following general series:

Theorem

f (x) ∼ f (c) + f ′(c)(x − c) +f ′′(c)

2!(x − c)2 +

f ′′′(c)

3!(x − c)3 + · · ·

f (x) ∼∞∑k=0

f (k)(c)

k!(x − c)k


Maclaurin series

The special case c = 0

f (x) ∼ f (0) + f ′(0)x +f ′′(0)

2!x2 +

f ′′′(0)

3!x3 + · · ·

f (x) ∼∞∑k=0

f (k)(0)

k!xk


Example 3

Example

What is the Taylor series of the function

f (x) = 3x5 − 2x4 + 15x3 + 13x2 − 12x − 5

at the point c = 2?

f (x) = 3x5 − 2x4 + 15x3 + 13x2 − 12x − 5 ⇒ f (2) = 207

f ′(x) = 15x4 − 8x3 + 45x2 + 26x − 12 ⇒ f ′(2) = 396

f ′′(x) = 60x3 − 24x2 + 90x + 26 ⇒ f ′′(2) = 590

f ′′′(x) = 180x2 − 48x + 90 ⇒ f ′′′(2) = 714

f (4)(x) = 360x − 48 ⇒ f (4)(2) = 672

f (5)(x) = 360 ⇒ f (5)(2) = 360

f (k)(x) = 0 ⇒ f (k)(2) = 0


Example 3 (cont.)

Therefore, we have

f (x) = 207 + 396(x − 2) + 295(x − 2)2

+ 119(x − 2)3 + 28(x − 2)4 + 3(x − 2)5


Maclaurin Series Illustration

We show how well we can approximate a function f (x) at a pointx = a by taking only a few terms of the Maclaurin series.With only the first term, the function is assumed to be a constant:

f (a) ≈ f (0)

With two terms, the slope at 0 is taken into account by way of thestraight line from f (0) to f (a); namely,

f (a) ≈ f (0) + f ′(0)a

With three terms, the curivature due to f ′′(0) comes into play and weobtain a parabolic curve:

f (a) ≈ f (0) + f ′(0)a +1

2f ′′(0)a2

Each additional term improves the accuracy in the approximation forf (a).In Figure 1.3, we show these partial sums in the Maclaurin serieswhen f (x) = ex and x = a = 1.


Figure 1.3

0 0.2 0.4 0.6 0.8 1

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

f0=1 1

f1=1+x

2f2=1+x+x2/2

2.5

f=ex

2.7183

Figure: Approximations to ex


Complete Horner’s Algorithm

An application of Horner’s algorithm is that of finding the Taylorexpansion of a polynomial about any point.

Let p(x) be a given polynomial of degree n with coefficients ak .

Suppose that we desire the coefficients ck in the equation

p(x) = anxn + an−1xn−1 + · · ·+ a0

= cn(x − r)n + cn−1(x − r)n−1 + · · ·+ c1(x − r) + c0

Taylor’s Theorem asserts that

ck = p(k)(r)/k!

but we seek a more efficient algorithm.


Complete Horner’s Algorithm (cont.)

Notice that p(r) = c0, so this coefficient is obtained by applyingHorner’s algorithm to the polynomial p with the point r .

The algorithm also yields the polynomial

q(x) =p(x)− p(r)

x − r= cn(x − r)n−1 + cn−1(x − r)n−2 + · · ·+ c1

This shows that the second coefficient, c1, can be obtained byapplying Horner’s algorithm to the polynomial q with point r sincec1 = q(r).

This process is repeated until all coefficients ck are found.


Complete Horner’s Algorithm (cont.)

The pseudocode for executing it is arranged so that the coefficientsck overwrite the input coefficients ak .

integer n, k , j ; real r ; real array (ai )0:nfor k = 0 to n − 1 do

for j = n − 1 to k doaj ← aj + raj+1

end forend for

This procedure can be used in carrying out Newton’s method forfinding roots of a polynomial, which we discuss later.

Moreover, it can be done in complex arithmetic to handle polynomialswith complex roots or coefficients.


Example 4

Example

Using the complete Horner’s algorithm, find the Taylor expansion of thepolynomial

p(x) = x4 − 4x3 + 7x2 − 5x + 2

about the point r = 3.


Example 4 (cont.)

The work can be arranged as follows:

1 −4 7 −5 23 ) 3 −3 12 21

1 −1 4 7 233 6 30

1 2 10 373 15

1 5 253

1 8

The calculation shows that

p(x) = (x − 3)4 + 8(x − 3)3 + 25(x − 3)2 + 37(x − 3) + 23


Taylor’s Theorem in Terms of (x − c)

Theorem

Taylor’s Theorem for f (x) If the function f possesses continuousderivatives of orders 0, 1, 2, . . . , (n + 1) in a closed interval I = [a, b], thenfor any c and x in I ,

f (x) =n∑

k=0

f (k)(c)

k!(x − c)k + En+1

where the error term En+1 can be given in the form

En+1 =f (n+1)(ξ)

(n + 1)!(x − c)n+1

Here ξ is a point that lies between c and x and depends on both.


Example 5

Example

Derive the Taylor series for ex at c = 0, and prove that it converges to ex

by using Taylor’s Theorem.

If f (x) = ex , then f (k)(x) = ex for k ≥ 0.Therefore, f (k)(c) = f (k)(0) = e0 = 1 for all k .We have

ex =n∑

k=0

xk

k!+

eξ

(n + 1)!xn+1

limn→∞

∣∣∣∣ eξ

(n + 1)!xn+1

∣∣∣∣ ≤ limn→∞

es

(n + 1)!sn+1 = 0

if |x | ≤ s.If we take the limit as n→∞ on both sides of ex above.

ex = limn→∞

n∑k=0

xk

k!=∞∑k=0

xk

k!


Example 6

Example

Derive the formal Taylor series for f (x) = ln(1 + x) at c = 0, anddetermine the range of positive x for which the series represents thefunction.

We need f (k)(x) and f (k)(0) for k ≥ 1.

f (x) = ln(1 + x) ⇒ f (0) = 0

f ′(x) = (1 + x)−1 ⇒ f ′(0) = 1

f ′′(x) = −(1 + x)−2 ⇒ f ′′(0) = −1

f ′′′(x) = 2(1 + x)−3 ⇒ f ′′′(0) = 2

f (4)(x) = −6(1 + x)−4 ⇒ f (4)(0) = −6

... ⇒...

f (k)(x) = (−1)k−1(k − 1)!(1 + x)−k⇒ f (k)(0) = (−1)k−1(k − 1)!


Example 6 (con.t)

Hence by Taylor’s Theorem, we obtain

ln(1 + x) =n∑

k=1

(−1)k−1(k − 1)!

k!xk +

(−1)nn!(1 + ξ)−n−1

(n + 1)!xn+1

=n∑

k=1

(−1)k−1xk

k+

(−1)n

n + 1

(1 + ξ

)−n−1xn+1

For the infinite series to represent ln(1 + x), it is necessary andsufficient that the error term converge to zero as n→∞.

Hence, the series represents ln(1 + x) if 0 ≤ x ≤ 1, but not if x > 1.

The series also represents ln(1 + x) for −1 < x < 0, but not ifx ≤ −1.


Mean-Value Theorem

The special case n = 0 in Taylor’s Theorem is known as theMean-Value Theorem.

Theorem

Mean-Value Theorem If f is a continuous function on the closed interval[a, b] and possesses a derivative at each point of the open interval (a, b),then

f (b) = f (a) + (b − a)f ′(ξ)

for some ξ in (a, b).

For some ξ ∈ (a, b)

f ′(ξ) =f (b)− f (a)

b − a

The approximation of derivatives is discussed more fully later.


Taylor’s Theorem in Terms of h

Other forms of Taylor’s Theorem are often useful. These can be obtainedby changing the variables.

Corollary

If the function f possesses continuous derivatives of order0, 1, 2, . . . , (n + 1) in a closed interval I = [a, b], then for any x in I ,

f (x + h) =n∑

k=0

f (k)(x)

k!hk + En+1

where h is any value such that x + h is in I and where

En+1 =f (n+1)(ξ)

(n + 1)!hn+1

for some ξ between x and x + h.


Taylor’s Theorem in Terms of h (cont.)

The error term En+1 depends on h:

As h converges to zero, En+1 converges to zero with essentially thesame rapidity with which hn+1 converges to zero.

To express this qualitative fact, we write

En+1 = O(hn+1)

as h→ 0.

This is called big O notation, and it is shorthand for the inequality

|En+1| ≤ C |h|n+1

where C is a constant.

Roughly speaking, En+1 = O(hn+1) means that the behavior of En+1

is similar to the much simpler expression hn+1.


Taylor’s Theorem in Terms of h (cont.)

For example, we can write out the cases n = 0, 1, 2 as follows:

f (x + h) = f (x) + f ′(ξ1)h

= f (x) +O(h)

f (x + h) = f (x) + f ′(x)h +1

2!f ′′(ξ2)h2

= f (x) + f ′(x)h +O(h2)

f (x + h) = f (x) + f ′(x)h +1

2!f ′′(x)h2 +

1

3!f ′′′(ξ3)h3

= f (x) + f ′(x)h +1

2!f ′′(x)h2 +O(h3)


Example 7

Example

Expand√

1 + h in powers of h. Then compute√

1.00001 and√

0.99999.

Let f (x) = x1/2.

Then f ′(x) =1

2x−1/2, f ′′(x) = −1

4x−3/2, f ′′′(x) =

3

8x−5/2, and so

on.

Now let x = 1 and take n = 2 for illustration

√1 + h = 1 +

1

2h − 1

8h2 +

1

16h3ξ−5/2

Let h = 10−5

√1.00001 ≈ 1 + 0.5× 10−5 − 0.125× 10−10 = 1.00000 49999 87500


Example 7 (cont.)

By substituting −h for h in the series, we obtain

√1− h = 1− 1

2h − 1

8h2 − 1

16h3ξ−5/2

Hence, we have

√0.99999 ≈ 0.99999 49999 87500

Both numerical values are correct to all 15 decimal places shown.


Alternating Series

Principle

If the magnitudes of the terms in an alternating series convergemonotonically to zero, then the error in truncating the series is no largerthan the magnitude of the first omitted term.

This theorem applies only to alternating series!


Alternating Series Theorem

Theorem

If a1 ≥ a2 ≥ · · · ≥ an ≥ · · · 0 for all n and limn→∞ an = 0, then thealternating series

a1 − a2 + a3 − a4 + · · ·

converges; that is,

∞∑k=1

(−1)k−1ak = limn→∞

n∑k=1

(−1)k−1ak = limn→∞

Sn = S

where S is its sum and Sn is the nth partial sum. Moreover, for all n,

|S − Sn| ≤ an+1


Example 8

Example

If the sine series is to be used in computing sin 1 with an error less than1

2× 10−6, how many terms are needed?

We have

sin 1 = 1− 1

3!+

1

5!− 1

7!+ · · ·

If we stop at 1/(2n − 1)!, the error does not exceed the firstneglected term, which is 1/(2n + 1)!.

Thus, we should select n so that

1

(2n + 1)!<

1

2× 10−6

We obtain log(2n + 1)! > log 2 + 6 = 6.3.

With a calculator, we find that log 10! ≈ 6.6.

Hence, if n ≥ 5, the error is acceptable.


Example 9

Example

If the logarithmic series is to be used for computing ln 2 with an error of

less than1

2× 10−6, how many terms are required?

To compute ln 2, we take x = 1 in the series

S = ln 2 ≈ 1− 1

2+

1

3− 1

4+ · · ·+ (−1)n−1

n= Sn

By the Alternating Series Theorem, the error involved when the seriesis truncated with n terms is

|S − Sn| ≤1

n + 1<

1

2× 10−6

Hence, more than two million terms would be needed!

We conclude that this method of computing ln 2 is not practical!


Example 10

Example

It is known thatπ4

90= 1−4 + 2−4 + 3−4 + · · ·

How many terms should we take to compute π4/90 with an error of at

most1

2× 10−6?

A naive approach is to take

1−4 + 2−4 + 3−4 + · · ·+ n−4

n is chosen so that the next term, (n + 1)−4, is less that1

2× 10−6.

This value of n is 37.


Example 10 (cont.)

But this is an erroneous answer because the partial sumS37 =

∑37k=1 k−4 differs from π4/90 by approximately 6× 10−6.

What we should do is to select n so that all the omitted terms add up

to less than1

2× 10−6

∞∑k=n+1

k−4 <1

2× 10−6

By a technique familiar from calculus, we have

∞∑k=n+1

k−4 <

∫ ∞n

x−4 dx =x−3

−3

∣∣∣∣∞n

=1

3n3

Thus, it suffices to select n so that

(3n3)−1 <1

2× 10−6 ⇒ n ≥ 88


Figure

Figure: Illustrating Example 10


Summary 1.2

The Taylor series expansion about c for f (x) is

f (x) =n∑

k=0

f (k)(c)

k!(x − c)k + En+1

with error term

En+1 =f (n+1)(ξ)

(n + 1)!(x − c)n+1

A more useful form for us is the Taylor series expansion for f (x + h)

f (x + h) =n∑

k=0

f (k)(x)

k!hk + En+1

with error term

En+1 =f (n+1)(ξ)

(n + 1)!hn+1 = O(hn+1)


Summary 1.2 (cont.)

An alternating series

S =∞∑k=1

(−1)k−1ak

converges when the terms ak converge downward to zero.

Furthermore, the partial sums

Sn =n∑

k=1

(−1)k−1ak

differ from S by an amount that is bounded by

|S − Sn| ≤ an+1


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