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bio website cse.iitb.ac.in/~aruniyer
location India
age 30
visits member for 2 years, 7 months
seen 34 mins ago

1d
comment How to prove $n < \left(1+\frac{1}{\sqrt{n}}\right)^n$
You can lower that limit of 144 even further. Note that $$ 1 + \sqrt{n} + \dfrac{(n-1)}2 + \dfrac{n(n-1)(n-2)}{6} \dfrac1{n^{3/2}} = \frac{n^{3/2}}{6} + \frac{n}{2} + \frac{n^{1/2}}{2} + \frac{1}{3n^{1/2}} + \frac{1}{2}$$. The RHS is greater than (using the first three terms) $$\frac{n}{2} + \frac{n^{1/2}(n + 3)}{6}$$. It is quite easy to show that $$\frac{n^{1/2}(n + 3)}{6} > \frac{n}{2}$$. This gives you the result, without needing calculus or computing for particular values.
Nov
23
comment Prove that $tr(A^-)=\sum_{i=1}^n\lambda_i^{-1}$
math.stackexchange.com/questions/546155/…
Nov
15
comment Jensen’s inequality
@Did Then hopefully the wikipedia link is sufficiently helpful.
Nov
14
comment Eigenvalues of a sum of rank-one matrices?
This is excellent!
Nov
8
comment Counterexample for $\| (\Phi^\top \Xi \Phi)^\dagger \Phi^\top \Xi P \Phi \|_2 \leq 1$
Are they no conditions on phi like norm phi <= 1?
Oct
27
comment Limit of $n(a^{1/n}-1)$ as $n \to \infty$
You mean $(e^x - 1)/x$. Also as n tends to infinity, x tends to 0.
Oct
27
comment Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$
Now, I have rewritten the answer somewhat differently and hopefully, the derivation is a bit more clearer now.
Oct
27
comment Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$
I didn't replace $\cos^{n+1}$ with $\cos^{2k}$. I replaced it with $\cos^{2k+2}$. I just wrote the formula for general even powers. For (a) and (b), I have updated the answer with details on how to prove them, they are quite simple to prove.
Oct
26
comment Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$
Either is fine, if you set n+1 = 2k, then in subsequent steps, you simply use n = 2k - 1.
Oct
14
comment How to find the gradient of norm square
Is your question correct? Since the range of g is R, the norm is pointless, f(x) is simply g(x)^2, so gradient is simply 2g(x)dg/dx
Oct
5
comment Asking for help with an inequality
I haven't checked the entire post carefully, but my knee jerk reaction would be - Cauchy Schwartz? $|\langle x, y \rangle| \leq \|x\|\|y\|$ - en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#L2
Sep
24
comment Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
@AlgebraicPavel Sure. Since I suggested a slightly different approach to the OP earlier, I decided to mark the two approaches separately, so as to not obfuscate anything. Although, it might have been a tad bit unnecessary.
Sep
24
comment Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
@AlgebraicPavel In the first approach, we can show that the $\|MAM^{-1}\|$ is an upper bound to the subordinate norm and then attempt to find an $x$ that realizes this upper bound thereby proving the equality. In the second approach, one doesn't have to find such an x. The equality follows from the definition of the subordinate norm itself.
Sep
24
comment Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
We are not choosing $x = M^{-1}y$. Remember that M is non-singular, therefore, $x \to Mx$ is a bijection.
Sep
24
comment Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
I have updated the answer with a simpler approach, which avoids finding a good x.
Sep
24
comment Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
Yeap. That is what at least I have in my mind.
Sep
24
comment Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
Always start with the definition. Also, if this is a homework, do tag your question as homework.
Sep
24
comment Prove that there is no function $f:\Bbb{R}\to\Bbb{R}$ with $f(0)>0$ such that $\forall x,y\in\Bbb{R}, f(x+y)\geq f(x)+y f(f(x))$
If you put x = 0, then the LHS in the first inequality you gave should be f(0) - yf(f(0))
Sep
15
comment Positive Semi Definite Matrix
True, the general matrix will also continue to be indefinite. @Omnomnomnom has given necessary and sufficiency conditions as well. Thank you so very much to both of you for your answers.
Sep
15
comment Positive Semi Definite Matrix
This is an excellent piece of information! Thank you so very much!