Recent content by Stumped1

How would I apply the change of variable technique to solve the integral $$\int x\frac{2x}{1000^2}e^{-(x/1000)^2}dx$$ w/ out the $$x$$ I used $$u=(x/1000)^2$$ , and $$du=2x/1000^2$$ Now, I am calculating $$E(x)$$, and now sure how to deal w/ the extra $$x$$. Thanks for any help!

expand $$e^{\frac{z}{z-2}}$$ in a Laurent series about $$z=2$$ I cannot start this. my attempt so far has been $$e^\frac{z}{z-2}=1 + \frac{z}{z-2} + \frac{z^2}{(z-2)^2 2!} + \frac{z^3}{(z-2)^3 3!}$$ This is unlike the other problems I have worked. Seems I need to manipulate this equation...

If this is not closed, continue w/ my approach?

Seems that it is since it will be one rotation of the cycloid, since it is parameterized by $$0 \leq\theta \leq 2\pi$$

$$\int(z^2+1)^2dz$$ Evaluate this over the cycloid$$x=a(\theta-sin\theta)$$ and $$y=a(1-cos\theta)$$ for $$\theta =0$$ to $$\theta = 2\pi$$ Am I on the right track, or do I need to approach this a different way? for $$z^2$$ we have $$(x+iy)^2$$, so $$x^2-y^2 + i2xy$$ for the real part...

Thanks for all the help on this. Now that I have a proof, I am still curious about the way first mentioned. Using $$\frac{1}{n^{1+i}}=\frac{1}{n^i n}$$ and $$n^i=e^{ilogn}$$ we have $$\sum \frac{1}{e^{ilogn}n}$$ Since $$n $$is ever increasing, and $$e^z=e^{ilogn}$$ is periodic, is this...

That is correct. Thanks for looking at this!

prove that $$tan^{-1}z=z-z^3/3 + z^5/5 -z^7/7 + ...$$ for $$|z|<1$$ I know of a proof for this that takes the derivative, does long division, then integrates. I would like a proof of this using the known Maclaurin series for e^z, cosz, or sinz.Is there a way to do this using these? Thanks...

The question asks to prove that $$\sum_{n=1}^{\infty} \frac{1}{n^{1+i}}$$ diverges. I am having trouble with this. using the ratio test $$\lim_{n\to\infty}\left|\frac{1}{(n+1)^{1+i}}\cdot\frac{n^{1+i}}{1}\right|$$ How can I simplify this further to find the limit? Or is there another...