use term by term differentiation to find the power series for f(x) = 2x/(1 - x^2)^2 centered at a = 0. you…

use term by term differentiation to find the power series for f(x) = 2x/(1 - x^2)^2 centered at a = 0. you may need to manipulate your given series first before differentiating. the power series representation for f(x) is sum(n = 0 to infinity) sqrt(2) x^n help (formulas) (hint: a power series must have all nonnegative integers as the exponents for the variable, so you may need to reindex the power series you find after differentiating.)

use term by term differentiation to find the power series for f(x) = 2x/(1 - x^2)^2 centered at a = 0. you may need to manipulate your given series first before differentiating. the power series representation for f(x) is sum(n = 0 to infinity) sqrt(2) x^n help (formulas) (hint: a power series must have all nonnegative integers as the exponents for the variable, so you may need to reindex the power series you find after differentiating.)

Answer

Explanation:

Step1: Recall geometric - series formula

The geometric series $\frac{1}{1 - t}=\sum_{n = 0}^{\infty}t^{n}$, for $|t|\lt1$. Let $t=x^{2}$, then $\frac{1}{1 - x^{2}}=\sum_{n = 0}^{\infty}(x^{2})^{n}=\sum_{n = 0}^{\infty}x^{2n}$, for $|x|\lt1$.

Step2: Differentiate $\frac{1}{1 - x^{2}}$

Differentiate both sides of $\frac{1}{1 - x^{2}}=\sum_{n = 0}^{\infty}x^{2n}$ with respect to $x$. Using the power - rule for differentiation $(x^{m})^\prime=mx^{m - 1}$, the derivative of the left - hand side: $(\frac{1}{1 - x^{2}})^\prime=\frac{2x}{(1 - x^{2})^{2}}$. The derivative of the right - hand side: $(\sum_{n = 0}^{\infty}x^{2n})^\prime=\sum_{n = 1}^{\infty}2nx^{2n - 1}$.

Answer:

$\sum_{n = 1}^{\infty}2nx^{2n - 1}$