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1h
comment How to evaluate this infinite product ? (Fibonacci number)
I was getting close to this, but you were faster. (+1)
2h
comment Stirling approximation note
Yes. That is definitely what I was trying to say. Most cases are not looking at the limit. They are monomials $x^0$ where $x=0$.
2h
revised Stirling approximation note
explicitly include power series
2h
comment Stirling approximation note
@MichaelHardy: That falls under my point 2. However, the problem for most people who think that $0^0$ is not defined is that it is an indeterminate form. Since $x^y$ can take on a lot of different values near $(0,0)$, we could choose any one. However, we choose the one that works in most common cases, such as power series, the binomial theorem, etc.
3h
revised Stirling approximation note
improve the explanation
3h
revised Stirling approximation note
add some explanation about the definition of $0^0$
4h
answered Stirling approximation note
9h
comment If $u_{n+1}\le u_n+u_n^2$ and $\sum u_n$ converges, prove that $\lim\limits_{n\to +\infty}(n\cdot u_n)=0$
You say: "Hence, the sequence $\displaystyle\{n\cdot u_n\}$ either has limit $0$ or has limit $\alpha>0$." Why does the sequence have to have a limit at all? Where did you use the hypothesis that $u_{n+1}\le u_n+u_n^2$? Ian's comment explains why this hypothesis is necessary.
10h
comment Partial Differential Equations $xu_x+yu_y=1$
It doesn't matter whether we specify the question in polar or rectangular coordinates. Using only functions of $\frac yx$ misses solutions that are valid in either case.
10h
comment Partial Differential Equations $xu_x+yu_y=1$
$\arg(1,1)=\frac\pi4$ and $\arg(-1,-1)=-\frac{3\pi}4$. $\frac yx=1$ in both cases, but $\arg(x,y)$ is different. Therefore, $\arg(x,y)$ cannot possibly be a function of $\frac yx$.
11h
comment Integral with complicated exponentials
If the integral were $$\int_0^\tau\frac{x}{(1 + ax^2)^2} e^{-\frac{bx^2}{(1 + ax^2)}}\,\mathrm{d}x$$ it would have a simple closed form. As it is, I can't find a simple closed form.
12h
comment Partial Differential Equations $xu_x+yu_y=1$
here is a plot of $\sin\left(\arctan\left(\frac yx\right)\right)$.
13h
comment Partial Differential Equations $xu_x+yu_y=1$
Note that $\sin\left(\arctan\left(\frac yx\right)\right)$ is not smooth on the line $x=0$. Nor is it equal to $\sin(\theta)$ when $x\lt0$.
13h
comment Partial Differential Equations $xu_x+yu_y=1$
It is more general since functions of $\frac yx$ are $\pi$-periodic functions of $\arg(x,y)$, and we can use any $2\pi$-periodic functions of $\arg(x,y)$. Take for example, $\log(r)+\sin(\theta)$. $\sin(\theta)$ cannot be written as a function of $\frac yx$.
13h
revised Partial Differential Equations $xu_x+yu_y=1$
mention the relation to polar coordinates
13h
revised Partial Differential Equations $xu_x+yu_y=1$
be more specific
13h
answered Partial Differential Equations $xu_x+yu_y=1$
1d
awarded  Nice Answer
1d
answered How to prove $\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$?
2d
comment Computing $\int_{-\infty}^\infty \frac{\sin x}{x} \mathrm{d}x$ with residue calculus
Euler's Formula implies $$\begin{align}\int_{-\infty-i}^{\infty-i}\frac{\sin(z)}z\,\mathrm{d}z &=\frac1{2i}\int_{-\infty-i}^{\infty-i}\frac{e^{iz}-e^{-iz}}z\,\mathrm{d}z\\ &=\frac1{2i}\int_{-\infty-i}^{\infty-i}\frac{e^{iz}}z\,\mathrm{d}z -\frac1{2i}\int_{-\infty-i}^{\infty-i}\frac{e^{-iz}}z\,\mathrm{d}z\end{align}$$ Then we include the large semicircular contours to close each contour; one in the upper half-plane for the first, since $e^{iz}$ decays quickly there, and the lower half-plane for the second, since $e^{-iz}$ decays quickly there.