$\text { Use the power series } \frac { 1 } { 1 + x } = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } x ^ { n } \text { to determine a power series centered at } 0$ for the function $g ( x ) = \frac { 1 } { 11 x ^ { 2 } + 1 }$.
A) $\sum _ { n = 0 } ^ { \infty } ( - 11 ) ^ { n } x ^ { n }$
B)$\sum _ { n = 0 } ^ { \infty } ( 11 ) ^ { n } x ^ { 2 n }$
C) $\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } x ^ { 2 n }$
D)$\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } x ^ { n }$
E)$\sum _ { n = 0 } ^ { \infty } ( - 11 ) ^ { n } x ^ { 2 n }$

Question

Quizplus · Accepted Answer

To find the power series for $g(x) = \frac{1}{11x^2 + 1}$, substitute $x^2$ for $x$ and $-11$ for the coefficient in the series $\frac{1}{1 + x} = \sum_{n=0}^{\infty} (-1)^n x^n$. This gives $\sum_{n=0}^{\infty} (-11)^n x^{2n}$.

For the Function g(x)=111x2+1g ( x ) = \frac { 1 } { 11 x ^ { 2 } + 1 }g(x)=11x2+11​

For the Function $g ( x ) = \frac { 1 } { 11 x ^ { 2 } + 1 }$