TenaliRaman
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 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 26 revised Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ added 192 characters in body Oct 25 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 approved 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 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.