Other Power Series

We now derive some further power series to add to the collection described in 7.1. Starting with the geometric series

$${\frac{{1}}{{1-x}}}$$ (7.1)

we replace x by - x to get

$${\frac{{1}}{{1+x}}}$$

Integrating both sides then gives

log(1 + x) = K + x - $${\frac{{x^2}}{{2}}}$$ + $${\frac{{x^3}}{{3}}}$$ +...+ (- 1)ⁿ$${\frac{{x^{n+1}}}{{n+1}}}$$ +...    for | x| < 1,    

where K is a constant of integration. Evaluating both sides when x = 0 shows that K = 0, and so we get the series

$$\sum_{{n=0}}^{\infty} {\frac{{(-1)^n x^{n+1}}}{{n+1}}}$$ (7.2)

Note:It is easy to get this result directly from the Taylor Series. The next one is not quite so easy.

We return to equation 7.1, and replace x by - x² to get

$${\frac{{1}}{{1+x^2}}}$$ = 1 - x² + x⁴ - x⁶ +...+ (- 1)ⁿx²ⁿ +...    for | x| < 1.    

Again integrating both sides, we have

arctan(x) = K + x - $${\frac{{x^3}}{{3}}}$$ + $${\frac{{x^5}}{{5}}}$$ +...+ (- 1)ⁿ$${\frac{{x^{2n+1}}}{{2n+1}}}$$ +...    for | x| < 1,    

where K is a constant of integration. Again putting x = 0 shows that K = 0, and so

$$\sum_{{n=0}}^{\infty} {\frac{{(-1)^n x^{2n+1}}}{{2n+1}}}$$ (7.3)

Example 7.11   Find the radius of convergence R of the power series

$${\frac{{x^2}}{{2}}}$$ - $${\frac{{x^3}}{{3.2}}}$$ + $${\frac{{x^4}}{{4.3}}}$$ - $${\frac{{x^5}}{{5.4}}}$$ ^... + $${\frac{{(-1)^n x^n}}{{ n(n-1)}}}$$ + ^...

By differentiation or otherwise, find the sum of the series for | x| < R.

[You may assume, without proof, that

$$\int$$log(1 + x) dx = K + x log(1 + x) - x + log(1 + x),

for some constant of integration K.]

Solution. Apply the ratio test to the given power series. Then

$$\left\vert\vphantom{\frac{a_{n+1}}{a_n}}\right.$$$${\frac{{a_{n+1}}}{{a_n}}}$$$$\left.\vphantom{\frac{a_{n+1}}{a_n}}\right\vert$$ = $$\left\vert\vphantom{\frac{x^{n+1} n(n-1)}{x^nn(n+1)}}\right.$$$${\frac{{x^{n+1} n(n-1)}}{{x^nn(n+1)}}}$$$$\left.\vphantom{\frac{x^{n+1} n(n-1)}{x^nn(n+1)}}\right\vert$$$$\to$$| x|    as    n$$\to$$$$\infty$$.

Thus the new series has radius of convergence 1. Denote its sum by f (x), defined for | x| < 1. Inside the circle of convergence, it is permissible to differentiate term - by - term, and thus f'(x) = log(1 + x) for | x| < 1, since they have the same power series. Hence

$$\int \int {\frac{{x + 1 -1}}{{ 1 + x}}}$$ f (x) = (7.4)

= K + x log(1 + x) - x + log(1 + x). (7.5)

Putting x = 0 shows that K = 0 and so f (x) = (1 + x)log(1 + x) - x.

We have now been able to derive a power series representation for a function without working directly from the Taylor series, and doing the differentiations -- which can often prove very awkward. Nevertheless, we have still found the Taylor coefficients.

Proposition 7.12   Let $\sum$a_nxⁿ be a power series, with radius of convergence R > 0, and define

f (x) = $$\sum_{{n=0}}^{\infty}$$a_nxⁿ    for | x| < R.    

Then a_n = $${\frac{{f^{(n)}(0)}}{{n!}}}$$, so the given series is the Taylor (or Maclaurin) series for f

Proof. We can differentiate n times by 7.9 and we still get a series with the same radius of convergence. Also, calculating exactly as in the start of Section 5.6, we see that the derivatives satisfy f⁽ⁿ⁾(0) = n!a_n, giving the uniqueness result. $\qedsymbol$

Example 7.13   Let f (x) = $${\frac{{1}}{{1-x^3}}}$$. Calculate f⁽ⁿ⁾(0).

Solution. We use the binomial theorem to get a power series expansion about 0,

$${\frac{{1}}{{1-x^3}}}$$ = 1 + x³ + x⁶ + x⁹ +...+ x³ⁿ +...    valid for | x| < 1.    

We now read off the various derivatives. Clearly f⁽ⁿ⁾(0) = 0 unless n is a multiple of 3, while f^(3k)(0) = (3k)! by 7.12.