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bio website cse.iitb.ac.in/~aruniyer
location India
age 30
visits member for 2 years, 6 months
seen 5 hours ago

Oct
27
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 25 characters in body
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
revised Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$
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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
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 1354 characters in body
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
27
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 367 characters in body
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 171 characters in body
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 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 $\Vert x\Vert_M=\Vert Mx\Vert$
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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.