12,307 reputation
22180
bio website mathproblems123.wordpress.com
location Chambery, France
age 26
visits member for 3 years, 2 months
seen 2 hours ago

PHD - interests: PDE, Free boundaries, Shape Optimization


Dec
11
comment Simple Functional Equation $\frac{f(a)-f(b)}{a-b}\cdot(-a)+f(a)=-ab$
@NickKidman: You're right :) You don't need to verify in order not to lose points. What I wanted to say is that even at high levels verification is considered important. It doesn't mater that it is a trivial computation. You might find that your answer is wrong.
Dec
11
comment Differentiability at a point on a compact set implies difference quotients are bounded
Not necessarily. If the ratio is bounded then $f(x)=f(x_0)+g(x)(x-x_0)$ where $g$ is a bounded function. Pick a bounded continuous, non-differentiable function and you have found a counterexample.
Dec
10
comment Proving $(0,1) $ is not countable
@Inquest: Yes, but how do you prove that $\Bbb{R}$ is uncountable then?
Dec
8
comment Trigonometric equation: $2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$
@dona12: then write it now. You can edit your question.
Dec
8
comment Trigonometric equation: $2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$
Try to make the title about the question, not about begging for help.
Dec
5
comment Korean Math Olympiad 2005 (trapezoid & tangent circles)
Ok, with respect to my comment, if the circles of diameters $AD,BC$ are tangent then the circles $O_1,O_2,O_3,O_4$ all pass through the same tangency point, so there is no non-degenerate circle which can be tangent to all of them.
Dec
5
comment Korean Math Olympiad 2005 (trapezoid & tangent circles)
This problem cannot be correct in the presented form. Consider an isosceles trapezoid such that the circles of diameters $AD$ and $BC$ are tangent. Then there exists a circle which is tangent to all of the $O_1,O_2,O_3,O_4$ and the quadrilateral is not a parallelogram.
Dec
2
comment The series $\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2}$ and the series $\sum_{k=1}^{\infty} \frac{|\pi -n/k|}{k}$
@johnmangual: I think the actual question is in the end where he compares the two series. He explicitly says that he chooses $k(n)$ such that $n/k(n) \to \pi$.
Nov
30
comment Convex hull of infinite points
There exists the convex hull of every set of a vector space. en.wikipedia.org/wiki/Convex_hull
Nov
30
comment Dimension of diagonal matrices
How many free variables do you have when you impose the matrix to be diagonal?
Nov
27
comment If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then to prove $x=\sqrt{3}+\sqrt{2}$
It $x^3=a$ has a unique solution for the two $a$ that you find as a solution of $a^2-18\sqrt{3}a+1=0$ so in fact you get a solution for every $a$. In the end you have two solutions.
Nov
26
comment How find this determinant $\det(\cos^4{(i-j)})_{n\times n}$
In fact, for $n\geq 6$ the result is zero.
Nov
26
comment How find this determinant $\det(\cos^4{(i-j)})_{n\times n}$
If you replace $4$ by $1$ or $2$ the result is zero. In this case, some numerical computations show that $\det(A)\neq 0$.
Nov
22
comment Proof that a ODE system admits symmetric solutions
in your system if $(x_1,x_2,x_3)$ is a solution then $(-x_1,-x_2,x_3)$ is also a solution.
Nov
21
comment Does $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } } $ diverge?
I guess there are tons of questions on proofs for the divergence of harmonic series. math.stackexchange.com/questions/163921/harmonic-series math.stackexchange.com/questions/172504/… etc.
Nov
21
comment Rational translates of the unit circle cover the plane
Indeed, it follows that $(x-a)^2+(y-b)^2=1$
Nov
21
comment Rational translates of the unit circle cover the plane
Nice, quick argument.
Nov
21
comment Condition for an additive function to be continuous
There is another error in the proof. You cannot extend directly the boundedness from one subinterval to the whole $(0,1)$. if $(x,y)$ is on the arc where the circle is bounded and $p\neq q$ are two rational such that $(px,qy)$ is also on the unit circle then you can find explicitly $x$ and $y$ in terms of $p,q$ and it turns out that $x^2$ and $y^2$ are both rationals, which excludes most of the remaining points... Anyway, the idea to go into 2 dimensions is interesting, and it might lead to something towards the solution.
Nov
20
comment Condition for an additive function to be continuous
You say that $F$ is additive in each variable. How do you deduce that $F(Q)-F(R)=F(Q-R)$? This seems to work only if $QR$ is parallel to one of the coordinate axes.
Nov
19
comment Need help badly: n-dimensional Lebesgue measure of a hyperplane is zero.
@user100463: So after making people work to write a good answer for your question you want the bounty for yourself?