find the power - series representation for g centered at 0 by differentiating or integrating the power…

find the power - series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). give the interval of convergence for the resulting series.\ng(x)=\frac{x}{(1 + 9x^{2})^{2}} using f(x)=\frac{1}{1 + 9x^{2}}\ng(x)=sum_{k = 0}^{infty}square

find the power - series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). give the interval of convergence for the resulting series.\ng(x)=\frac{x}{(1 + 9x^{2})^{2}} using f(x)=\frac{1}{1 + 9x^{2}}\ng(x)=sum_{k = 0}^{infty}square

Answer

Explanation:

Step1: Recall geometric - series

The geometric - series $\frac{1}{1 + t}=\sum_{k = 0}^{\infty}(-t)^{k}$, for $|t|\lt1$. Let $t = 9x^{2}$, then $f(x)=\frac{1}{1 + 9x^{2}}=\sum_{k = 0}^{\infty}(-9x^{2})^{k}=\sum_{k = 0}^{\infty}(-1)^{k}9^{k}x^{2k}$, for $|9x^{2}|\lt1$ (i.e., $|x|\lt\frac{1}{3}$).

Step2: Differentiate $f(x)$

Differentiate $f(x)$ with respect to $x$. $f^\prime(x)=\frac{-18x}{(1 + 9x^{2})^{2}}$. And $f^\prime(x)=\sum_{k = 1}^{\infty}(-1)^{k}9^{k}\cdot2k\cdot x^{2k - 1}$.

Step3: Find $g(x)$

Since $g(x)=\frac{x}{(1 + 9x^{2})^{2}}$, then $g(x)=-\frac{1}{18}\sum_{k = 1}^{\infty}(-1)^{k}9^{k}\cdot2k\cdot x^{2k - 1 + 1}=-\frac{1}{18}\sum_{k = 1}^{\infty}(-1)^{k}2k\cdot9^{k}x^{2k}=\sum_{k = 1}^{\infty}(-1)^{k + 1}k\cdot9^{k-1}x^{2k}$. The interval of convergence is $|x|\lt\frac{1}{3}$.

Answer:

$\sum_{k = 1}^{\infty}(-1)^{k + 1}k\cdot9^{k - 1}x^{2k}$, interval of convergence: $(-\frac{1}{3},\frac{1}{3})$