504 reputation
39
bio website
location
age
visits member for 2 years, 7 months
seen 4 hours ago

6h
comment Does $\sum_{n=0}^\infty (-1)^n (e-(1+\frac{1}{n})^n)$ converge absolutely, conditionally, or diverge?
If not $\sum_{k=0}$ then where to start the summation?
6h
comment Does $\sum_{n=0}^\infty (-1)^n (e-(1+\frac{1}{n})^n)$ converge absolutely, conditionally, or diverge?
Can you elaborate? I still don't see why it's obvious that the sum is > 1 / 2n...
7h
comment Does $\sum_{n=0}^\infty (-1)^n (e-(1+\frac{1}{n})^n)$ converge absolutely, conditionally, or diverge?
How do you know that $e - (1+\frac{1}{n})^n > \frac{1}{2n}$?
Oct
21
comment Deduce the Bolzano-Weierstrass Theorem from the Heine-Borel Theorem
but doesn't the open cover (that's not finite) also contain only finitely many $x_n$ because it's composed of the intervals $I_x$ that themselves only contain finitely many $x_n$?
Oct
18
comment Prove if $a$ is a nonnegative real number and $n$ is a positive integer, there exists a $b \geq 0$ such that $b^n = a$
So, if I wanted to add in another step for clarity, I could have added $b^n + \sum_{k = 0}^{n-1}(-1)^k {n \choose k}b^k\frac{1}{m^{n-k}} > b^n - \sum_{k = 0}^{n-1} {n \choose k}b^k\frac{1}{m^{n-k}}$, right? Thank you for the explanation, this is very clear.
Oct
18
comment Prove if $a$ is a nonnegative real number and $n$ is a positive integer, there exists a $b \geq 0$ such that $b^n = a$
How does the negative sign appear in front of the summation with $\delta$ in the second to last expression? Can't the original summation be positive or negative depending on $k$ and $(-1)^k$?
Oct
16
comment Evaluate $\int_{\partial C} \frac{dz}{(z-a)(z-b)}$ where $\partial C$ is the boundary of a rectangle ($a$ and $b$ are not on $\partial C$)
Thanks for this response. After thinking about your answer for a bit, what if $a = b$? Then it seems $I_a$ and $I_b$ are undefined?
Oct
15
comment Prove: Convergent sequences are bounded
Helpful even two years later!
Oct
10
comment Show $f(z) = \frac{z}{e^z-1}$ is analytic in the neighborhood of the origin and find the first 4 terms in its power series representation
So, I should be able to show that $z$ is analytic and $1/(e^z-1)$ is analytic, thus their product is analytic everywhere except at $z=0$?
Oct
10
comment Show $f(z) = \frac{z}{e^z-1}$ is analytic in the neighborhood of the origin and find the first 4 terms in its power series representation
the derivative of $e^z$ is $e^z$ so it is $1$...
Oct
10
comment Show $f(z) = \frac{z}{e^z-1}$ is analytic in the neighborhood of the origin and find the first 4 terms in its power series representation
How would one show this?
Oct
10
comment Show $f(z) = \frac{z}{e^z-1}$ is analytic in the neighborhood of the origin and find the first 4 terms in its power series representation
I'm not sure if that helps. Then I can get $\frac{x+iy}{e^x \cos y + i e^x \sin y - 1}$ but I'm still not able to separate out the imaginary from the real components?
Oct
8
comment Derive branch cuts for $\log(\sqrt{1-z^2} + iz)$ as $(-\infty,-1)$ and $(1,\infty)$?
Thanks, Daniel for your help with this question and my previous one. What is the reasoning for not needing a branch cut for the logarithm? (also: can you recommend a book that covers these concepts in more detail at an introductory level?)
Oct
8
comment Derive branch cuts for $\log(\sqrt{1-z^2} + iz)$ as $(-\infty,-1)$ and $(1,\infty)$?
I did this to show that the branch cuts for the square root term are $(-\infty,-1)$ and $(1,\infty)$.
Oct
7
comment How do we define the branch cuts for $\sin^{-1}z = \frac{1}{i} \log(\sqrt{1-z^2} + iz)$ as $(-\infty,-1)$ and $(1,\infty)$?
I'm a little lost at "Look at how $\sin$ maps the strip $|\operatorname{Re}z|<\frac{\pi}{2}...$". Could you elaborate on that part?
Oct
4
comment Prove that $\sum_0^\infty \frac{z^n}{1-z^{2n}}$ converges for all $\left|z\right|<1$ and $\left|z\right|>1$.
I did! Thank you very much. It was helpful to see how you deal with the modulus values.
Oct
3
comment Prove that $\sum_0^\infty \frac{z^n}{1-z^{2n}}$ converges for all $\left|z\right|<1$ and $\left|z\right|>1$.
graydad, thanks. But I'm still a little confused. Wouldn't the $-1$ in the denominator $|z|^n - 1$ imply that $\frac{1}{|z|^n - 1} > \frac{1}{|z|^n}$?
Oct
2
comment Prove that $\sum_0^\infty \frac{z^n}{1-z^{2n}}$ converges for all $\left|z\right|<1$ and $\left|z\right|>1$.
thanks. why does it follow that $\frac{\left|z\right|^n+1}{(1+|z|^n)(|z|^n-1)} < \frac{2}{|z|^n}$?
Oct
2
comment Prove that $\sum_0^\infty \frac{z^n}{1-z^{2n}}$ converges for all $\left|z\right|<1$ and $\left|z\right|>1$.
@DanielFischer does your first hint follow from the reverse triangle inequality?
Oct
2
comment Prove that $\sum_0^\infty \frac{z^n}{1-z^{2n}}$ converges for all $\left|z\right|<1$ and $\left|z\right|>1$.
@DanielFischer thanks. I suspect the second hint is for Case 2, which would allow us to revert to Case 1, once I can solve it.