11.9: Power Series
( \newcommand{\kernel}{\mathrm{null}\,}\)
Recall that we were able to analyze all geometric series "simultaneously'' to discover that \sum_{n=0}^\infty kx^n = {k\over 1-x}, \nonumber if |x| < 1, and that the series diverges when |x|\ge 1. At the time, we thought of x as an unspecified constant, but we could just as well think of it as a variable, in which case the series \sum_{n=0}^\infty kx^n \nonumber is a function, namely, the function k/(1-x), as long as |x| < 1. While k/(1-x) is a reasonably easy function to deal with, the more complicated \sum kx^n does have its attractions: it appears to be an infinite version of one of the simplest function types---a polynomial. This leads naturally to the questions: Do other functions have representations as series? Is there an advantage to viewing them in this way?
The geometric series has a special feature that makes it unlike a typical polynomial---the coefficients of the powers of x are the same, namely k . We will need to allow more general coefficients if we are to get anything other than the geometric series.
A power series has the form \sum_{n=0}^\infty a_nx^n, with the understanding that a_n may depend on n but not on x.
\sum_{n=1}^{\infty} x^n \over n is a power series. We can investigate convergence using the ratio test:
\lim_{n\to\infty} {|x|^{n+1}\over n+1}{n\over |x|^n} =\lim_{n\to\infty} |x|{n\over n+1} =|x|. \nonumber
Thus when |x| < 1 the series converges and when |x|>1 it diverges, leaving only two values in doubt. When x=1 the series is the harmonic series and diverges; when x=-1 it is the alternating harmonic series (actually the negative of the usual alternating harmonic series) and converges. Thus, we may think of
\sum_{n=1}^\infty {x^n\over n} \nonumber
as a function from the interval [-1,1]) to the real numbers.
A bit of thought reveals that the ratio test applied to a power series will always have the same nice form. In general, we will compute
\lim_{n\to\infty} {|a_{n+1}||x|^{n+1}\over |a_n||x|^n} =\lim_{n\to\infty} |x|{|a_{n+1}|\over |a_n|} = |x|\lim_{n\to\infty} {|a_{n+1}|\over |a_n|} =L|x|, \nonumber
assuming that \lim |a_{n+1}|/|a_n| exists. Then the series converges if L|x| < 1, that is, if |x| < 1/L, and diverges if |x|>1/L. Only the two values x=\pm1/L require further investigation. Thus the series will definitely define a function on the interval (-1/L,1/L), and perhaps will extend to one or both endpoints as well. Two special cases deserve mention: if L=0 the limit is 0 no matter what value x takes, so the series converges for all x and the function is defined for all real numbers. If L=\infty, then no matter what value x takes the limit is infinite and the series converges only when x=0. The value 1/L is called the radius of convergence of the series, and the interval on which the series converges is the interval of convergence.
Consider again the geometric series, \sum_{n=0}^\infty x^n={1\over 1-x}. Whatever benefits there might be in using the series form of this function are only available to us when x is between -1 and 1. Frequently we can address this shortcoming by modifying the power series slightly. Consider this series:
\sum_{n=0}^\infty {(x+2)^n\over 3^n}= \sum_{n=0}^\infty \left({x+2\over 3}\right)^n={1\over 1-{x+2\over 3}}= {3\over 1-x}, \nonumber
because this is just a geometric series with x replaced by (x+2)/3. Multiplying both sides by 1/3 gives
\sum_{n=0}^\infty {(x+2)^n\over 3^{n+1}}={1\over 1-x}, \nonumber
the same function as before. For what values of x does this series converge? Since it is a geometric series, we know that it converges when
\eqalign{ |x+2|/3& < 1\cr |x+2|& < 3\cr -3 < x+2 & < 3\cr -5 < x& < 1.\cr } \nonumber
So we have a series representation for 1/(1-x) that works on a larger interval than before, at the expense of a somewhat more complicated series. The endpoints of the interval of convergence now are -5 and 1, but note that they can be more compactly described as -2\pm3. We say that 3 is the radius of convergence, and we now say that the series is centered at -2.
A power series centered at a has the form \sum_{n=0}^\infty a_n(x-a)^n, with the understanding that a_n may depend on n but not on x.