1. ## Power series

These two problems have stumped me. I doubt they're particularly difficult. I'm just having trouble grasping the concepts.....

2. Originally Posted by osiron23

These two problems have stumped me. I doubt they're particularly difficult. I'm just having trouble grasping the concepts.....
I will help with the first....but...for the second...physics ::shivers::.

a) So first we must find an expression for $\cos\left(x^2\right)$

We first note (I assume you know this) that $\forall{x}\in\mathbb{R}~\cos(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n}}{(2n)!}$

So now if we substitute in $x=\psi^2$ we get

$\forall\psi^2\in\mathbb{R}~\cos\left(\psi^2\right) =\sum_{n=0}^{\infty}\frac{(-1)^n\psi^{4n}}{(2n)!}$

And noting that $x\in\mathbb{R}\implies\psi^2\in\mathbb{R}$ and that since a variable is a variable let us just replace $\psi$ with x[/tex]. So

$\forall{x}\in\mathbb{R}~~\cos\left(x^2\right)=\sum _{n=0}^{\infty}\frac{(-1)^nx^{4n}}{(2n)!}$

b) So since this series converges for all real values we can say it converges on any $[0,\xi]\subset\mathbb{R}$. So now let us consider

\begin{aligned}\int_0^{\xi}\cos\left(x^2\right)dx= &\int_0^{\xi}\sum_{n=0}^{\infty}\frac{(-1)^nx^{4n}}{(2n)!}\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\int_0^{\xi}x^{4n}dx\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n\xi^{4n+1}}{(4n+1)(2n)!}\end{aligned}

And since integrating a power series changes the endpoints of its interval of convergence we are done (because the set of real numbers does not have endpoints)

c) So you must just calculate $\sum_{n=0}^{N}\frac{(-1)^n}{(4n+1)(2n)!}$ where $N$ is given by the first integer that makes $\frac{(-1)^n}{(4n+1)(2n)!}<.001$

Ok...lets do this

a) Just multiply by $\frac{\frac{1}{R^2}}{\frac{1}{R^2}}$

b) Why dont you take a stab at this

Hint: Consider $\left(\frac{1}{1+x}\right)'$

3. ## Thanks

Thanks a lot!!! Is number 5 really a physics problem though? I'm in trouble then.