consider the function f(x)=5 tan^(-1)(x^2). a. differentiate the taylor series centered at 0 for f(x). b…

Explanation:

Step1: Recall the Taylor - series of $\tan^{-1}(u)$

The Taylor - series of $\tan^{-1}(u)$ centered at $u = 0$ is $\tan^{-1}(u)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}u^{2n+1}=u-\frac{u^{3}}{3}+\frac{u^{5}}{5}-\frac{u^{7}}{7}+\cdots$, for $|u|\lt1$. Let $u = x^{2}$, then $f(x)=5\tan^{-1}(x^{2})=5\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}(x^{2})^{2n+1}=5\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{4n + 2}$.

Step2: Differentiate the Taylor - series of $f(x)$

Using the power - rule for differentiation $\frac{d}{dx}(x^{m})=mx^{m - 1}$, we have: [ \begin{align*} f^\prime(x)&=\frac{d}{dx}\left(5\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{4n+2}\right)\ &=5\sum_{n = 0}^{\infty}\frac{(-1)^{n}(4n + 2)}{2n + 1}x^{4n+1}\ &=10\sum_{n = 0}^{\infty}(-1)^{n}x^{4n + 1}=10x-10x^{5}+10x^{9}-10x^{13}+\cdots \end{align*} ] So the answer for part (a) is C.

Step3: Identify the function represented by the differentiated series

We know that the geometric series formula is $\sum_{n = 0}^{\infty}ar^{n}=\frac{a}{1 - r}$, for $|r|\lt1$. The series $10\sum_{n = 0}^{\infty}(-1)^{n}x^{4n+1}=10x\sum_{n = 0}^{\infty}(-x^{4})^{n}$. Here, $a = 1$ and $r=-x^{4}$. So the sum of the series is $\frac{10x}{1 + x^{4}}$.

Step4: Find the interval of convergence of the power series for the derivative

For the geometric series $\sum_{n = 0}^{\infty}ar^{n}=\frac{a}{1 - r}$, the series converges when $|r|\lt1$. In our case, $r=-x^{4}$, so we need $|-x^{4}|\lt1$, which simplifies to $|x|\lt1$. When $x = 1$, the series becomes $10\sum_{n = 0}^{\infty}(-1)^{n}$, which diverges (it is an alternating series that does not converge by the divergence test since $\lim_{n\rightarrow\infty}(-1)^{n}$ does not exist). When $x=-1$, the series becomes $10\sum_{n = 0}^{\infty}(-1)^{n}(-1)^{4n + 1}=10\sum_{n = 0}^{\infty}(-1)^{5n+1}$, which also diverges.

Answer:

a. C. $10x - 10x^{5}+10x^{9}-10x^{13}+\cdots$ b. $\frac{10x}{1 + x^{4}}$ c. $(-1,1)$