14,792 reputation
32660
bio website facebook.com/…
location Germany
age 22
visits member for 1 year, 5 months
seen Jun 14 at 17:39

I study physics at the "Technische Universität Darmstadt", but I spend most time doing math. My favorite topics are Topology and Algebra, even though my knowledge there is strictly limited.


Jul
2
awarded  Curious
May
26
comment Show that $f(x)=e^x$
How did you define the Exponentialfunction? because often it is defined that way
May
22
comment Induced Matrix Norm
Because $\|A\|_1 = \sup_{x\in B_1(0)\setminus\{0\}} \frac{\|Ax\|_1}{\|x\|_1} \geq \frac{\|Ax\|_1}{\|x\|_1}$
May
22
comment Induced Matrix Norm
Well if you find such an $x$ you know that $\|A\|_1 \geq C$ and $\|A\|_1 \leq C$.
May
21
answered Convergent subsequence for sin(n)
May
15
comment determinant inequality $\det(A^2+AB+B^2)\geq\det(AB-BA)$
@Clin yes this is your inequality in the case they commute. I just want to know if there is an easy arguement why this should be true
May
15
comment Solve a problem of convergence of integral
Hint: The Surface of the open ball with radius $r$ in dimension $N$ is a constant times $r^{N-1}$, while the measure of the ball is a constant times $r^N$. So I guess you can show that if there is no such sequence the integral of the function can't be finite
May
15
answered the limit of a limit
May
15
comment determinant inequality $\det(A^2+AB+B^2)\geq\det(AB-BA)$
At least if $A$ and $B$ commute and are diagonalizable we have on the lhs $(\lambda_1^2 + \lambda_1 \mu_1+\mu_1^2)\cdot (\lambda_2^2 + \lambda_2 \mu_2 + \mu_2^2)$, where the $\lambda,\mu$ are the eigenvalues, if they are real we are done with arithmetic geometric mean and in the complex case the eigenvalues must be complex conjugate s.t. $\lambda_1=\overline{\lambda_2}$
May
15
comment determinant inequality $\det(A^2+AB+B^2)\geq\det(AB-BA)$
With Mathematica you can surely bruteforce it into an inequality with 8 unknowns. If we assume $A$ and $B$ to commute is there a simple argument why the lhs is positive?
May
14
awarded  matrices
May
13
comment Misunderstanding Cayley's Theorem…?
@user1729 still not injective as the image of $\{0,1\}$ is the same as the image of $\{0\}$
May
13
comment Misunderstanding Cayley's Theorem…?
@user1729 your mapping is not injective as all singletones are mapped on the identity if i am not wrong
May
13
comment Residue method and order of poles
well you have $$(1- (1-sT+\frac{(sT)^2}{2}-\frac{(sT)^3}{3!}\pm \dots)) \cdot s^{-3}$$
May
13
answered Residue method and order of poles
May
13
comment Misunderstanding Cayley's Theorem…?
@sidht because the numbers of Elements in $A^B$ is $|A|^{|B|}$, in the finite case this is quite immediate because you may chose $|A|$ between $|B|$ Elements
May
13
comment Misunderstanding Cayley's Theorem…?
I mean at least for $|G|\geq 4$
May
13
comment Misunderstanding Cayley's Theorem…?
Shouldn't you get something like $|\mathcal{P}(G)|=|2^G|=2^{|G|} \leq |S_G|$ very easy ?
May
13
comment Misunderstanding Cayley's Theorem…?
A permutation on $G$ is just a bijective function from $G\to G$. So if $G$ is not finite by swapping 2 Elements we will have infinite many bijections
Apr
19
awarded  Good Question