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Feb
26
comment Finding out the coeffcient next to $x^2$ in $(\cdots(x-2)^2-2)^2\cdots-2)^2$.
I don't know why it is, but it seems the recurrence is $a(n)=20a(n-1)-64a(n-2)$, which gives the solution $(4*16^n-4^n)/3.$
Feb
19
comment Prove that $\sum_i\sum_j a_{ij}=\sum_j\sum_i a_{ij}$
See math.stackexchange.com/questions/497096/…
Jan
9
reviewed Approve Prove that $ \frac12 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2} $
Nov
29
awarded  Civic Duty
Oct
27
reviewed No Action Needed Find $a\in\mathbb{N}$ such that $n^4+a$ is not prime $\forall n\in\mathbb{N}$
Oct
23
revised How to change to same units
edited tags
Oct
23
comment Integral of product of two measurable functions
This is essentially a special case of Holder's inequality.
Oct
23
answered Intuition behind product rule of probability
Oct
22
answered Proof of Aristarchus' Inequality
Oct
22
reviewed Approve Polynomial inequality proof
Oct
22
comment Mean value of the image of an exponentiallly distributed time under a smooth curve
$f(t)$ should be $\phi(t)$ in 3.
Oct
22
reviewed Reviewed prove that quadrangle is isosceles trapezoid
Oct
21
comment How to prove that $\int_{-\pi}^{+\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx$ is positive
I dont have a solution, just some computation with maple/ WA
Oct
21
comment How to prove that $\int_{-\pi}^{+\pi}\cos{(2x)}\cos{(3x)}\cos{(4x)}\cdots\cos{(2005x)}dx$ is positive
The integral of $\prod_{k=2}^{n}cos(kx)$ seems to be $0$ for values of $n$ congruent to $-1, 0 \mod 4$ and a positive rational fraction of $\pi$ for $1,2\mod 4$
Oct
21
reviewed Approve $x^{y^z}$: is it $x^{(y^z)}$ or $(x^y)^z$?
Oct
20
comment Power series $ \sum_{r=1}^{n}x^{r}=\:?$
Almost surely this has been answered here before
Oct
20
comment $\int_{-\infty}^{\infty} \frac{1}{2\pi} \exp\{ -\frac{1}{2} ((y-x)^2 + x^2) \} dx$
square completion
Oct
20
answered Are there any open mathematical puzzles?
Oct
19
answered Determine $a<0$ such that $\int_a^0 f(x) dx = f(a)$
Oct
18
comment use combinatorial reasoning to calculate $ \sum{\binom{100}{a}\binom{200}{b}\binom{300}{c}}$
To add some reference, this is sometimes known as Vandermonde's convolution