Jean-Sébastien
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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 $\sum_i\sum_j a_{ij}=\sum_j\sum_i a_{ij}$ 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