bio | website | |
---|---|---|
location | University of California, San Diego | |
age | 24 | |
visits | member for | 1 year, 5 months |
seen | 5 hours ago | |
stats | profile views | 169 |
I am a graduate student at University of California, San Diego under the advice of Jim Agler.
Aug 19 |
comment |
Is there an algebraic number that cannot be expressed using only elementary functions?
In your post on stackexchange you allowed $i,$ and other elemenary complex analysis: do you still want to do this? It seems if you include integration, it would be possible to express any algebraic number this way. Anyway for an algebraic number $a$ with minimum polynomial $p,$ $a = \int_{\gamma} \frac{zp'(z)}{p(z)} dz,$ where $\gamma$ is a circle with rational radius and center, containing $a$ and no other roots of $p.$ |
Aug 7 |
comment |
Invertibility of random Vandermonde matrix
Yeah, you definitely need to use non-atomicity. I think you need something a bit stronger, since if the points lie in certain varieties, you may run into trouble. |
Aug 7 |
answered | Invertibility of random Vandermonde matrix |
Aug 4 |
comment |
What summations of elementary trig functions are known to have (elementary) closed forms?
If you mean $\Sigma^{\infty}_{k=0} \tan{k},$ I think that sum does not converge. Can you be more explicit about what sums you're talking about? |
Aug 3 |
comment |
largest eigenvalue of a symmetric matrix
I mean $I$ to be the N by N identity matrix. |
Aug 3 |
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largest eigenvalue of a symmetric matrix
Then it would seem that taking $\Phi = -I$ and $P = I,$ where I, would imply that the phenomenon is indeed a coincidence. Although, with some extra special structure, the phenomenon may hold. |
Aug 3 |
comment |
largest eigenvalue of a symmetric matrix
What is $\Phi?$ |
Jul 31 |
awarded | Enlightened |
Jul 31 |
awarded | Nice Answer |
Jul 31 |
awarded | Yearling |
Jul 30 |
comment |
Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform
$\cos x$ isn't in $L^2.$ |
Jul 30 |
awarded | Critic |
Jul 29 |
revised |
Homogenous polynomially convex hull of $[0,1]^n$
grammatical fix |
Jul 29 |
asked | Homogenous polynomially convex hull of $[0,1]^n$ |
Jul 29 |
comment |
Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform
I guess the simplest answer would be that the function $\|v\|_S = (\sum_{k\in S} \langle v,f(x-k) \rangle^{2})^{1/2}$ defines a norm on $V$ which is finite dimensional, and all norms on a finite dimensional space are equivalent in the sense that they induce the same topology. |
Jul 29 |
comment |
Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform
Sorry, I meant to write $f(x-m).$ |
Jul 29 |
comment |
Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform
Bartgol, he's talking about the functions $f(x-k) \in L^2,$ not the values of $f$ at the integers which wouldn't be well defined anyway. |
Jul 29 |
comment |
Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform
Well suppose you had a linear dependence, i.e. there $\sum^n_{k=-l} a_kf(x-k) =0,$ where $a_{-l}$ and $a_n$ are non-zero. By applying the shift, we get that $\sum^n_{k=-l} a_kf(x-(k+t)) =0.$ That is, for any $m > n$ or $m< -l$ we can write $f(m)$ in terms of a recurrence relation. |
Jul 29 |
answered | Prove that …, f(x-2), f(x-1), f(x), f(x+1), f(x+2),… is algebraically linearly independent without the Fourier transform |
Jan 11 |
awarded | Teacher |