3,943 reputation
1635
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location Besançon, France
age 34
visits member for 1 year, 10 months
seen 22 hours ago

I am a statistician at INSEE, the french national institute of statistics.

Mathematics is a hobby for me, and MSE helps me not to forget too much what I learned in university (MS in applied mathematics at Lyon 1). I like especially real and complex analysis, trigonometry, numerical analysis, finite groups theory, number theory and combinatorics. I am not allergic to other fields.


Jul
26
comment To prove $\cos(A+B) = \cos A \cos B - \sin A \sin B$
@BCLC In the $xy$ plane, the cross product is always colinear to the $z$ axis, so it's fully described by its component on this axis. And this component is... a $2x2$ determinant.
Jul
26
comment Finding $f(x)$.
@ClaudeLeibovici I think it's when you differentiate I.
Jul
26
comment Solutions to functional equation $f(f(x))=x$
@user2992539 Only works if $a$ is an odd integer.
Jun
24
comment Horizontal axis for reference angles
Yet another convention: you could perfectly have x>0 and y>0 in any quadrant, not necessarily the usual one. Screen images are an example where this convention is not followed, since usually the low $y$ are at the top of the screen.
Jun
23
comment Derivative in calculus $f(t)= 7\sinh(\ln t)$
If you want the derivative of this, it's of course $(2t(2t)-2(t^2-1))/(4t^2)$ (derivative of a quotient $u/v$ is $(u'v-uv')/v^2$).
Jun
23
comment equilateral triangle area geometry
You may also find this interesting: en.wikipedia.org/wiki/Koch_snowflake
Jun
23
comment how to solve this question of polynomials
What's the meaning of your equation (3), and the following conclusion (4)? (especially the notation $p(1,1/3)$ is rather obscure) Actually, from $p(-1)=0$ you get $(-1)h+k=0$, and from $p(1/3)=4$, $(1/3)h+k=4$.
Jun
15
comment Trigonometry without sine and cosine
Since the angles involved are not "easy" (like 45° or 60°), you will inevitably need trig functions to compute exactly the height. Hence, without trig functions, either you must do a careful drawing, or you look up in Wikipedia or elsewhere the actual height of Eiffel tower...
Jun
14
comment Measuring diaognals without Sine Law
Is there a reason not to use law of cosines? The solution is very easy with it.
May
28
comment Pearson's Chi Squared / Cochran–Mantel–Haenszel test analog to N-way ANOVA
You may try also on stats.stackexchange.com
May
28
comment Dilogarithm in closed form
At least you can tell the result is a real number (plug the dilog series expansion in your expression). Also, there may be a clever Fourier series to use.
May
27
comment Fascinating Lampshade Geometry
Nevertheless, the parametric approach is very instructive, and one can rather easily rediscover all basic results of perspective, with vanishing line, etc. You simply start with a point (the eye, equivalent of the light source here), a plane (the wall), and compute projection of lines, circles...
May
27
comment Fascinating Lampshade Geometry
@TedShifrin No, I knew the answer was incomplete, but I thought suggesting the idea was enough ;-)
May
27
comment Fascinating Lampshade Geometry
@TedShifrin Actually, it does. Section of a cone by a plane parallel to the axis must be a hyperbola (or two intersecting lines, but here the light source is not on the wall)
May
27
comment Fascinating Lampshade Geometry
Nices answer, but you could have guessed: the conic projection of a circle is a conic, and here it's obviously unbounded.
May
27
comment Differential equation $y''=e^y $
+1 for the multiplication trick, but you should explain the integral, WA is not an answer (it could be wrong, WA too has bugs)
May
26
comment Proof Strategy - Prove that each eigenvalue of $A^{2}$ is real and is less than or equal to zero - 2011 8C
Notice that if $k$ is an eigenvalue of $A$, with eigenvector $v$, then $Av=kv$, thus $A^2v=Akv=kAv=k^2v$. And eigenvalues of an antisymmetric matrix are pure imaginary. Regarding the logic behind your exercise, you may have a look at quadratic forms.
May
26
comment zeros of a polynomial
You may use Routh-Hurwitz theorem. Also, even if it's not needed, Rouché's theorem applied to $z^6$ and $6z+10$ shows there is no root with $|z|<1$, so all roots lie in the annulus $1<|z|<2$.
May
26
comment Convert a recursive sequence formula to explicit
There are several ways to do what you want, but the most similar to your method is writing $b_n=a_{n+1}$, this will give you a system of two first order recurrence equations, one of them being $a_n=b_{n-1}$.
May
26
comment Integrating $\frac{1}{(ax^2+bx+c)^n}$ two ways
These are not memory tricks, they are basic integration methods. If you don't expect to know the basics, no need to dig deeper.