# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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3

I

EXAMPLE

4

A function f that has a Taylor series expansion about x = x0

f (n)(x )

f (x) = rr°-(x x0)n >

n n!

n=0

with a radius of convergence p > 0, is said to be analytic at x = x0. According to statements 6 and 7, if f and g are analytic at x0, then f ± g, f ¦ g, and f /g [provided that g(x0) = 0] are analytic at x = x0.

Shift of Index of Summation. The index of summation in an infinite series is a dummy parameter just as the integration variable in a definite integral is a dummy variable. Thus it is immaterial which letter is used for the index of summation. For example,

2nxn ^ 2JxJ

n! *^ 1 !

n=0 * j=0 J *

Just as we make changes of the variable of integration in a definite integral, we find it convenient to make changes of summation indices in calculating series solutions of differential equations. We illustrate by several examples how to shift the summation index.

TO

Write anxn as a series whose first term corresponds to n = 0 rather than n = 2.

n=2

Let m = n 2; then n = m + 2 and n = 2 corresponds to m = 0. Hence

TO TO

T.anxn =E a+2*m+2. (0

n=2 m=0

By writing out the first few terms of each of these series, you can verify that they contain precisely the same terms. Finally, in the series on the right side of Eq. (1), we can replace the dummy index m by n, obtaining

TOTO

J2anx" = J2 an+2x"+2- (2)

n=2 n=0

In effect, we have shifted the index upward by 2, and compensated by starting to count at a level 2 lower than originally.

Write the series

TO

Y^(n + 2)(n + 1)an (x x0)n2 (3)

n2

as a series whose generic term involves (x x0)n rather than (x x0)n2.

Again, we shift the index by 2 so that n is replaced by n + 2 and start counting 2 lower. We obtain

]T(n + 4)(n + 3)an+2(x x0)n. (4)

n=0

You can readily verify that the terms in the series (3) and (4) are exactly the same.

236

Chapter 5. Series Solutions of Second Order Linear Equations

EXAMPLE

5

EXAMPLE

6

Write the expression

x2Y (r + n)anxr+n1 (5)

n0

as a series whose generic term involves xr+n.

First take the x2 inside the summation, obtaining

+ n)anxr +n+1. (6)

n=0

Next, shift the index down by 1 and start counting 1 higher. Thus

J2(r + n)anxr+n+1 = ? (r + n 1)anj xr+n. (7)

n=0 n=1

Again, you can easily verify that the two series in Eq. (7) are identical, and that both are exactly the same as the expression (5).

Assume that

CO CO

T.na,x-' = ? a,xn (8)

n=1 n=0

for all x, and determine what this implies about the coefficients an.

We want to use statement 10 to equate corresponding coefficients in the two series. In order to do this, we must first rewrite Eq. (8) so that the series display the same power of x in their generic terms. For instance, in the series on the left side of Eq. (8), we can replace n by n + 1 and start counting 1 lower. Thus Eq. (8) becomes

J2(n + 1)an+1xn = J2 anx ¦ (9)

n=0 n=0

According to statement 10 we conclude that

(n + 1)an+1 = an ¦ n = 0, 1, 2, 3,...

or

a

an+1 = -+Γ ¦ n = 0 1 2 3-- (10)

Hence, choosing successive values of n in Eq. (10), we have

a1 a0 a2 a0

a = an ¦ a = ¦ a = = ¦

1 °9 2 2 2 ¦ 3 3 3! ¦

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