I need help with the following:
Find the power series representation of the function defined by:
F(x) = the integral from 0 to x of cos (t^{1/2}/2) dt
Thanks!!!
By the second fundamental theorem of calculus, we have $\displaystyle \displaystyle \begin{align*} F'(x) = \cos{\left( \frac{x^{\frac{1}{2}}}{2} \right)} \end{align*}$.
If you find the power series for $\displaystyle \displaystyle \begin{align*} F'(x) \end{align*}$ and integrate it, you will have the power series you need.
We should know that the Taylor series for cosine is
$\displaystyle \cos(y)=\sum_{n=0}^{\infty}\frac{(-1)^{n}y^{2n}}{(2n)!}$
Now let $\displaystyle t=\frac{\sqrt{t}}{2}$
This gives
$\displaystyle \cos\left( \frac{\sqrt{t}}{2}\right)=\sum_{n=0}^{\infty}\frac {(-1)^{n}t^{n}}{4^n(2n)!}$
Now just integrate both sides from zero to x to get
$\displaystyle \int_{0}^{x}\cos\left( \frac{\sqrt{t}}{2}\right)dt=\int_{0}^{x}\sum_{n=0} ^{\infty}\frac{(-1)^{n}t^{n}}{4^n(2n)!} dt$
$\displaystyle F(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{4^n(2n)!(n+1)}$