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Nov
16
awarded  Popular Question
Nov
15
awarded  Yearling
Nov
13
awarded  Promoter
Nov
11
asked Algorithm to compute fastest method of collecting $k$ re-spawning items which spawn at $n$ specified points
Oct
8
revised $\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers of $x$ and $y$
added 4 characters in body
Oct
8
answered $\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers of $x$ and $y$
Sep
5
awarded  Nice Question
Aug
13
awarded  Notable Question
Aug
7
comment Countability of Sets
@TomCollinge Look at the second paragraph of that article. Some books don't count finite sets as countable.
Jul
9
answered Calc 2: convergent of divergent sequences
Jul
2
awarded  Curious
Jun
21
awarded  Popular Question
Jun
4
answered Sum of $\mathbf{x}\mathbf{x}'\mathbf{x}\mathbf{x}'$ when $\mathbf{x}$ is a binary vector of length $n$
May
4
comment Let $G$ be a connected graph. If $G$ has no cut vertices, then G has no bridges.
@CharlesNosbig yeah, I would say that is the correct answer. Although if this is homework, it might be cool to mention that it does work for larger graphs.
May
4
revised Let $G$ be a connected graph. If $G$ has no cut vertices, then G has no bridges.
added 15 characters in body
May
4
comment Let $G$ be a connected graph. If $G$ has no cut vertices, then G has no bridges.
It does not. That is a counter example.
May
4
answered Let $G$ be a connected graph. If $G$ has no cut vertices, then G has no bridges.
Feb
5
awarded  Popular Question
Jan
23
comment Integrating a Taylor series term-by-term
Well you actually have 3 limits going on, so maybe its the limit of $z \to \infty$ that is resisting the swap. $$\int_0^\infty \frac{\sin x}{x} dx = \lim_{z\to\infty}\int_{1/z}^z \frac{\sin x}{x} dx = \lim_{z \to \infty}\lim_{n \to \infty}\sum_{k = 0}^n\frac{\sin x_k^*}{x_k^*} \Delta x_k^*.$$ Now using the Taylor series, we have, $$\lim_{z \to \infty}\lim_{n \to \infty}\lim_{m \to \infty}\sum_{k = 0}^n \sum_{j=0}^m \frac{(-1)^j}{(2j + 1)!} (x_k^*)^{2j} \Delta x_k^*.$$ Note: $x_k^*$ depends on $z$.
Jan
23
comment Integrating a Taylor series term-by-term
When converting from the left hand side to the right hand side, you are swapping limits. I am not sure if that is exactly the problem here, but you certainly have to be careful with it. For example, $$\lim_{m\to\infty}\lim_{n\to\infty} \frac{n}{m + n} = \lim_{m \to \infty} 1 = 1,$$ but $$ \lim_{n\to\infty}\lim_{m\to\infty} \frac{n}{m + n} = \lim_{n \to \infty} 0 = 0.$$