Convergence of Series3 + Series Solution

8/3/2019 Convergence of Series3 + Series Solution

http://slidepdf.com/reader/full/convergence-of-series3-series-solution 1/24

Ms. Deepali Gupta. JIIT, NOIDA

Alternating Series

A series in which the terms are alternately positive and negative is

an alternating series.

∞+−+− ...............uuuu 4321

∞+−

++−+−+

......n

)1(.............

4

1

3

1

2

11

1n

Lebnitz’s Rule for convergence of an alternating series.

if converges..........uuuuu(-1)seriesThe 4321

1n

n

1n +−+−=∑∞

=

+




• un′sare all positive.

•un≥ un+1 for all n ≥ N, for some integer N.

u1

s1

u2

s2 s3s4

u4

u3

nn

lim u 0→∞

• =




Test the convergence of the series

..........5

1

14

1

13

1

12

1

1)11( −⎟ ⎠

⎞

⎜⎝

⎛

++⎟ ⎠

⎞

⎜⎝

⎛

+−⎟ ⎠

⎞

⎜⎝

⎛

++⎟ ⎠

⎞

⎜⎝

⎛

+−+

011n

nlimulimn

nn

≠=+

=∞→∞→

.......5

6

4

5

3

4

2

32 +−+−

⎟

⎠

⎞⎜

⎝

⎛ +−+−+−+−+− .....

4

1

3

1

2

11.....)11111(

Series is oscillatory.




Absolute Convergence : A series ∑unconverges absolutely if the

corresponding series of absolute values ∑|un| converges.

.........8

1

4

1

2

11 +−+−

.........8

1

4

1

2

11 ++++

The geometric series converges absolutely because the corresponding

series of absolute values converges.

Absolute convergence of series implies ordinary convergence.




.........5

1

4

1

3

1

2

1

1 −+−+−

The series is convergent by LebnitzTest

.divergentis.........5

1

4

1

3

1

2

11 ++++

Conditional Convergence : A series that converges but does notconverge absolutely.

.convergentabsolutelyisn

nsin

1n2

∑

∞

=




Pow er Ser ies

The power series method is the basic method for solving linear

differential equations with variable coefficients. It gives solutions in theform of power series.

..............)xx(a)xx(aa)xx(a2

020100m

m

0m+−+−+=−∑

∞

=

where a0,a1,a2….are constants coefficients, x0 is the center of the series.

..........!3

x

!2

x

x1e.g.e

32x

++++= .........!5

x

!3

x

xxsin

53

−+−=

....xa...........xaxaaxan

n

2

210

0m

m

m ++++=∑

∞

=

A power series about x = 0 is a series of the form

A power series about x = x0 is a series of the form




The power series is said to converge at a point x if the limit

n

0

m

1n

nm

)xx(alim −∑=

∞→

exists, and in this case the sum of the series is the value of this limit.

The series always converges at the point x = x0.

For e.g.∑

∞

=

++++=0n

32n .......x!3x!2x1x!n

x)1n(x!n

x)!1n(

u

un

1n

n

1n +=+

=+

+ ∞→∞→ nas

The series diverges for all values of x except x = 0.




∑∞

=

++++=0n

32n

.......

!3

x

!2

xx1

!n

x

x1n

1

x)!1n(

x!n

u

un

1n

n

1n

+=

+=

++

∞→→ nas0

The series converges for all values of x.

∑∞

=

++++=0n

32n .......xxx1x

The series converges for |x| < 1 and diverges for |x| > 1.

This means that to each series of this kind there correspond a positive real

number R, called the radius of convergence, with the property that the

series converges if |x| < R and diverges if | x | > R.




∑

∞

=

+++=0n

2

210

n

n

.......xaxaaxa

Lxa

alim

xa

xalim

n

1n

nn

n

1n

1n

n== +

∞→

++

∞→

The series converges for L < 1 and diverges for L > 1.

n

1n

n a

alim

1

R+

∞→

= nn

nalim

1

R∞→

=

if this limit exist. If R = ∞ the power series converges for all values of x.

If R = 0 the power series converges only at x = 0.




Example

Find the radius of convergence of the series

∑∞

=

+−+−=−

0n3

9

2

63n3

n

n

.......8

x

8

x

8

x1x

8

)1(

This is a series in powers of t = x3 .

8

1

8

8

a

alim

1n

n

n

1n

n

==+

+

∞→

Thus R = 8. Hence the series converges for |t| < 8, i.e. |x| < 2.




Suppose that the power series converges for |x| < R with its sum

given by f(x):k 2

k 0 1 2

k 0

f (x) a x a a x a x .........∞

=

= = + + +∑

Then f(x) is continuous and has derivatives of all orders.

Also the series can be differentiated term wise

k 1 2k 1 2 3

k 1

f (x) ka x a 2a x 3a x .........

∞

−

=′ = = + + +∑

k 2

k 2 3

k 2

f (x) k(k 1)a x 2a 3.2a x .........∞

−

=

′′ = − = + +∑




Analytic Function: A function f(x) is called analytic at a point x = x0 if it can

be represented by a power series in powers of x –x0 i.e f(x) = a0 + a1(x-x0)+a2(x-x0)

2+ ….. with radius of convergence R > 0.

A rational function is analytic except at those values of x at which its

denominator is zero.

3xand2at xexcept

everywhereanalyticis65xx

x:functionRational

2

==

+−

points.allatanalyticarecosx,xsin,efunctionspolynomialAllx

log(1+x) is not analytic at x = -1.




Ordinary Point:: A point x = x0 is called an ordinary point of the equationy’’+ P(x)y’ + Q(x)y = 0 (1) if both the functions P(x) and Q(x) are

analytic at x = x0.

Singular points: If the point x = x0 is not an ordinary point of thedifferential equation, then it is called a singular point of equation (1).

Regular singular point: A singular point x = x0

is called a RSP of (1) if

both (x-x0)P(x) and (x-x0)2Q(x) are analytic at x = x0.

Irregular singular point: A singular point which is not regular is called an

irregular singular point.




0xyy =+′′ y″+ P(x)y′ + Q(x)y = 0

P(x) = 0, Q(x) = x. At x = 0, both P(x) and Q(x) are analytic,

hence x = 0 is an ordinary point.

0y2y)x1(xy)1x(x 222=+′−−′′−

0y)1x(x

2y

)1x(x

)x1(xy

222222=

−+′

−

−−′′

222222 )1x(x

2)x(Q

)1x(x

)x1(x)x(P

−=

−

−−=




Singular points are x = 0, 1, -1.

2222

)1x()1x(x

2)x(Q

)1x)(x1(x

1)x(P

+−

=

+−

−=

22

2

2 )1x(

2

)x(Qx)1x)(x1(

1

)x(xP −=+−−=

• Since xP(x) and x2Q(x) are analytic at x = 0. Hence x = 0 is a

regular singular point.

22

2

2 )1x(x

2)x(Q)1x(

)1x(x

1)x(P)1x(

+=−

+−=−

• Since (x-1)P(x) and (x-1)2Q(x) are analytic at x = 1. Hence x = 1is a regular singular point.




22

2

)1x(x

2)x(Q)1x(

)1x)(x1(x

1)x(P)1x(

−

=+

+−

−=+

• Since (x + 1)P(x) is not analytic at x = -1. Hence x = -1 is a

irregular singular point.




Solut ion in Ser ies

0yy =+′′

im01m.F.C:lnsoExact 2 ±==+

xsincxcoscy 21 +=

Since P(x) = 0, Q(x) = 1, x = 0 is an ordinary point of the equation.

k 2

k 0 1 2

k 0

Let y a x a a x a x .........∞

=

= = + + +∑k 1 2

k 1 2 3

k 1

y ka x a 2a x 3a x .........∞

−

=

′ = = + + +∑

k 2

k 2 3

k 2y k(k 1)a x 2a 3.2a x .........

∞−

=′′ = − = + +∑






.........x!5

ax

!4

ax

!3

ax

!2

axaay 51403120

10 −++−−+=

.........xaxaay 2

210 +++=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+−+⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+−= .......!5

x

!3

xxa........

!4

x

!2

x1ay

53

1

42

0

xsinaxcosay 10+=

Thus any solution of differential equation about an ordinary point will be of

the form

.........)xx(a)xx(aay

2 +−+−+=02010




Ex am ple on RSP

0yy2yx3 =+′+′′

0yx3

1y

x3

2y =+′+′′

x = 0 is an regular singular point of the differential equation.

m k

k

k 0

Let y a x be the solution of DE.∞

+

=

= ∑

m k 1

k

k 0

y (m k)a x∞

+ −

=

′ = +∑ m k 2

k

k 0

y (m k)(m k 1)a x∞

+ −

=

′′ = + + −∑

where m may be a negative integer, a fraction or even an irrational no.




0yy2yx3 =+′+′′

m k 1 m k 1 m k k k k

k 0 k 0

3(m k)(m k 1)a x 2(m k)a x a x 0∞ ∞

+ − + − +

= =

+ + − + + + =∑ ∑

[ ] m k 1 m k

k k

k 0 k 0

3(m k)(m k 1) 2(m k) a x a x 0∞ ∞

+ − +

= =

+ + − + + + =∑ ∑

[ ]m 10k 0 x : 3m(m 1) 2m a 0−= − + =

0aas0mm3 0

2 ≠=−

31,0m0)1m3(m =⇒=−

The coefficient of lowest degree term xm-1 is obtained by putting k = 0 in

the first summation and equating it to zero. Then the indicial equation is




[ ]m

1 0k 1 x : 3(m 1)m 2(m 1) a a 0= + + + + =

)2m3)(1m(

aa 0

1++

−=

[ ]m 1

2 1k 2 x : 3(m 2)(m 1) 2(m 2) a a 0+= + + + + + =

)5m3)(2m(aa 1

2++

−=)5m3)(2m3)(2m)(1m(

a 0

++++=

[ ] m k 1 m k

k k

k 0 k 0

3(m k)(m k 1) 2(m k) a x a x 0∞ ∞

+ − +

= =

+ + − + + + =∑ ∑




[ ]m 2

3 2k 3 x : 3(m 3)(m 2) 2(m 3) a a 0+

= + + + + + =

)8m3)(3m(

aa 2

3++

−=

)8m3)(5m3)(2m3)(3m)(2m)(1m(

a 0

++++++

−=

0mFor =

01 a2

1a −= 02 a20

1a = 03 a480

1a −=

⎟

⎠

⎞⎜

⎝

⎛ +−+−== ........x

480

1x

20

1x

2

11ay0,mforHence 32

01




3

1mFor =

01 a4

1a −=

02 a56

1a = 03 a

1680

1a −=

⎟ ⎠ ⎞⎜

⎝ ⎛ +−+−== ........x

16801x

561x

411xay,

31mforHence 3231

02

Thus complete solution is

21 ByAyy +=

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +−+−+⎟⎟

⎠

⎞⎜⎜

⎝

⎛ +−+−= ........

1680

x

56

x

4

x1xb........

480

x

20

x

2

x1ay

3231

0

32

0

Date post:	07-Apr-2018
Category:	Documents
Upload:	harshit-agrawal
View:	218 times
Download:	0 times

Convergence of Series3 + Series Solution

Documents