What is the power series expression for f(x)=x^2sin(x). Find the radius of convergence too.
I know that the power series of sinx is (-1)^n (x^(2n+1))/(2n+1)! But what about the x^2? Do you just take the derivative of it?
If $\displaystyle \displaystyle \sin{x} = \sum_{n = 0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$
then $\displaystyle \displaystyle x^2\sin{x} = x^2\sum_{n = 0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} = \sum_{i=0}^{\infty}\frac{(-1)^nx^{2n + 3}}{(2n+1)!}$ (from the Distributive Law).
To check the radius of convergence use the ratio test.
If...
$\displaystyle \displaystyle \sin x= \sum_{n=0}^{\infty} (-1)^{n}\ \frac{x^{2n+1}}{(2n+1)!}$ (1)
... multiplying both therms of (1) by $\displaystyle x^{2}$ You obtain...
$\displaystyle \displaystyle x^{2}\ \sin x= \sum_{n=0}^{\infty} (-1)^{n}\ \frac{x^{2n+3}}{(2n+1)!}$ (2)
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