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bio website cse.iitb.ac.in/~aruniyer
location India
age 30
visits member for 2 years, 6 months
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4h
answered Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$
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
25
revised Prove that $E[Y_1 + Y_2\mid X=x] $…
Corrections and general facelift
Sep
25
suggested suggested edit on Prove that $E[Y_1 + Y_2\mid X=x] $…
Sep
24
comment Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
@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
answered Is vector W in the span of V1,V2,V3
Sep
24
revised Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
deleted 124 characters in body
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
revised Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
added 362 characters in body
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
answered Subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$ for the norm $\Vert x\Vert_M=\Vert Mx\Vert$
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 Maximum Likelihood Estimation for function with several variables
Set v1 = v2 = 1. If you use arbitrarily large values (> 1) for v3 and v4, the function value will keep on increasing. Hence, your function is unbounded above.
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
accepted Positive Semi Definite Matrix
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!