as power series\ncalculus - calculator allowed\n1. what is the coefficient of x^4 in the taylor series for…

Explanation:

Step1: Use double - angle formula

We know that $\sin^{2}x=\frac{1 - \cos(2x)}{2}$.

Step2: Recall Taylor series of cosine

The Taylor series of $\cos t$ about $t = 0$ is $\cos t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}t^{2n}=1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\frac{t^{6}}{6!}+\cdots$. Let $t = 2x$, then $\cos(2x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}(2x)^{2n}=1-\frac{(2x)^{2}}{2!}+\frac{(2x)^{4}}{4!}-\frac{(2x)^{6}}{6!}+\cdots$.

Step3: Find $\sin^{2}x$ series

$\sin^{2}x=\frac{1 - \cos(2x)}{2}=\frac{1-(1-\frac{(2x)^{2}}{2!}+\frac{(2x)^{4}}{4!}-\frac{(2x)^{6}}{6!}+\cdots)}{2}=\frac{\frac{(2x)^{2}}{2!}-\frac{(2x)^{4}}{4!}+\frac{(2x)^{6}}{6!}-\cdots}{2}=\frac{4x^{2}}{2\times2!}-\frac{16x^{4}}{2\times4!}+\frac{64x^{6}}{2\times6!}-\cdots$.

Step4: Identify coefficient of $x^{4}$

The coefficient of $x^{4}$ in the series of $\sin^{2}x$ is $-\frac{16}{2\times4!}=-\frac{16}{2\times24}=-\frac{1}{3}$.

Answer:

$-\frac{1}{3}$