Philippe Malot
Reputation
1,365
Next privilege 2,000 Rep.
 Feb20 comment Suppose $f:\mathbb{R} \to \mathbb{R}$ is such that $f(x) \leq 0$ and $f''(x) \geq 0, \forall x.$ Prove $f$ is constant. Proof idea: $f$ is convex so it lies above its tangents. Their slopes can't be non-zero or else $f$ would have a $+\infty$ limit at $-\infty$ or $+\infty$. Feb17 answered Evaluate $P(A^C\cup \!\,B^C)$ Jan8 revised Finding radius of convergence for taylor series of $\frac{z}{e^z-1}$ added 23 characters in body Jan8 comment Finding radius of convergence for taylor series of $\frac{z}{e^z-1}$ Good remark! I'll modify my answer. Jan7 comment Finding radius of convergence for taylor series of $\frac{z}{e^z-1}$ Can you solve the equation $e^z-1=0$? Jan7 answered Finding radius of convergence for taylor series of $\frac{z}{e^z-1}$ Jan7 comment Finding radius of convergence for taylor series of $\frac{z}{e^z-1}$ $0$ is clearly a false singularity of this meromorphic function on $\mathbb{C}$. It can obviously be extended as an holomorphic function on a neighborhood of $0$ by setting $f(0)=1$. Dec9 awarded Caucus Oct7 revised Proof the bijectivity of the exponential function mod $2 \pi i$ Incorrect tag. Improved LaTeX. Oct7 suggested approved edit on Proof the bijectivity of the exponential function mod $2 \pi i$ Sep30 awarded Explainer Sep18 comment Prove $\sin(x)< x$ when $x>0$ using LMVT Or you can juste choose $x$ in $(0,+\infty)$ from the beginning. Sep8 awarded Yearling Sep4 revised How to prove property of complex powers Improved LaTeX, changed tag Sep4 suggested approved edit on How to prove property of complex powers Aug28 awarded Announcer Aug2 comment Video lectures of algebraic geometry (Hartshorne, Shafarevich, … ) Harpreet Bedi has nice videos here. Jul4 revised If $Du=0$ a.e. , does $u=c$ a.e.? Added LaTeX, improved the message. Jul4 suggested approved edit on If $Du=0$ a.e. , does $u=c$ a.e.? Jun10 comment Show that $|z+1|\le|z+1|^2 +|z|$ for all $z \in \mathbb{C}$ I think it's badly written : "If $1\leqslant |z+1|$, then $|z+1|\leqslant|z+1|^2$ hence $|z+1|\leqslant |z+1|^2+|z|$, now I must show that the inequality is also true when $|z+1|<1$, but this is where I'm stuck."