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5.1.1. The series may either converge or diverge when |x x0| = p.
0 . Series _ .
Series converges---------------- <+?--------Series _
diverges absolutely diverges
- p X0 x0 + p
Series may /* converge or diverge
FIGURE 5.1.1 The interval of convergence of a power series.
Determine the radius of convergence of the power series
(x + 1)n
We apply the ratio test:
(x + 1)n+1
(n + 1)2n+1 (x + 1)
|x + 1|
n^? n + 1
|x + 1|
Thus the series converges absolutely for |x + 11 < 2, or 3 < x < 1, and diverges for | x + 1 | > 2. The radius of convergence of the power series is p = 2. Finally, we check the endpoints of the interval of convergence. At x = 1 the series becomes the harmonic series
which diverges. At x = 3 we have
^ (3 + 1)n
= ? (1)n n
which converges, but does not converge absolutely. The series is said to converge conditionally at x = 3. To summarize, the given power series converges for 3 < x < 1, and diverges otherwise. It converges absolutely for 3 < x < 1, and has a radius of convergence 2.
If a (x x0)n and bn(x x0)n converge to f (x) and g(x), respectively,
for |x x0| < p, p > 0, then the following are true for |x x0| < p.
6. The series can be added or subtracted termwise and
f (x ) ± g(x ) = (an ± bn )(x x0)n
Chapter 5. Series Solutions of Second Order Linear Equations
7. The series can be formally multiplied and
f (x )g(x ) =
J2an (x - x0)n
J2bn (x - x0)n
= J2 cn (x - x0)n
where cn = a0bn + a^n_ 1 + + anb0. Further, if g(x0) = 0, the series can be formally divided and
f (x ) g(x )
= dn (x - x0)n ?
In most cases the coefficients dn can be most easily obtained by equating coefficients in the equivalent relation
XX(x - x0)n =
Y,dn (x - x0)n
Y,bn(x - x0)n
= S J2dkbn-k) (x - x0)n ?
Also, in the case of division, the radius of convergence of the resulting power series may be less than p.
8. The function f is continuous and has derivatives of all orders for |x x01 < p. Further, f, f",... can be computed by differentiating the series termwise; that is,
f(x) = a1 + 2a2(x x0) +--------------+ nan (x x0)n1-------+------
= J2 nan (x x0)n1,
n = 1
f'(x) = 2a2 + 6a3(x x0) +-------------+ n(n 1)an (x x0)n2 +----------
= ^2n(n - 1)an(x - x0)n 2,
and so forth, and each of the series converges absolutely for |x x0| < p.
9. The value of an is given by
f (n)(x0) an = n
The series is called the Taylor1 series for the function f about x = x0.
10. If an (x x0)n = bn (x x0)n for each x, then an = bn for n = 0, 1,
2, 3,.... In particular, if Y an(x x0)n = 0 for each x, then a0 = a1 = =
an = ??? = 0.
1Brook Taylor (1685 -1731) was the leading English mathematician in the generation following Newton. In 1715 he published a general statement of the expansion theorem that is named for him, a result that is fundamental in all branches of analysis. He was also one of the founders of the calculus of finite differences, and was the first to recognize the existence of singular solutions of differential equations.
5.1 Review of Power Series
A function f that has a Taylor series expansion about x = x0
to f (n)(x )
f (x) = Y2 TT-0-(x - x0)n >
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 '2^ j !
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.
Write anxn as a series whose first term corresponds to n = 0 rather than n = 2.
Let m = n 2; then n = m + 2 and n = 2 corresponds to m = 0. Hence
Y.axn =E am+2xm+2. (1)
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
J2anx" = J2 an+2x"+2- (2)
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
^(n + 2)(n + 1)an(x - x0)n-2 (3)
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