Question

a) Use the Taylor series given to find a power series for g(x)= arctan (2x) arctan (x) = sum_{infinity}^{1} (-1)^n x^(2n+1)/(2n+1). for -1 le x le 1.

b. Determine the interval of convergence for g(x).
c. Find the coefficient for the x^15 term of the power series for g(x).
d. Would it be reasonable to use this series expansion to approximate g(10)= arctan 20? Why or why not?

Answer 1

\[\sum_{0}^{\infty} (-1)^n x^(2n+1)/(2n+1) for -1 \le x \le 1\]

Answer 2

it should be 1 for the bottom instead of 0

Answer 3

D

Answer 4

what?

Answer 5

So you're given the series for the inverse tangent function with the usual argument \(x\):
\[\large\tan^{-1}x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)} \]
(The bottom index is in fact 0. If it were 1, you'd have the series for \(\tan^{-1}x-x\).)

The Taylor series for any composite function of the form \(\tan^{-1}[f(x)]\) can be found by simply substituting any \(x\) term in the original series with \(f(x)\). In this case, \(f(x)=2x\), so replace all \(x\)'s with \(2x\):
\[\large g(x)=\tan^{-1}2x=\sum_{n=0}^\infty\frac{(-1)^n(2x)^{2n+1}}{(2n+1)}\]
The interval of convergence for a power series is found by applying the ratio test to the seires:
\[{\large\lim_{n\to\infty}}\left|\frac{(-1)^{n+1}(2x)^{2(n+1)+1}}{(2(n+1)+1)}\cdot\frac{2n+1}{(-1)^n(2x)^{2n+1}}\right|\]
If this limit is less than 1, then the series converges.

The \(x^{15}\) term will occur when \(2n+1=15\), or \(n=7\), so plugging in \(n=7\) you have
\[\frac{(-1)^7(2x)^{2(7)+1}}{(2(7)+1)}=\frac{(-1)(2x)^{15}}{15}=\color{red}-\frac{\color{red}{2^{15}}x^{15}}{\color{red}{15}}\]
And lastly, yes. Since \(g(x)=\tan^{-1}2x\), you know that \(g(10)=\tan^{-1}(2\cdot10)=\tan^{-1}20\). With many terms in the series, you'll be able to very accurately approximate the value.