copper.hat
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 Oct 27 comment Problem 23 chapter 5 PMA Rudin (point b) (This is about Point 2.*) I think it might be easier to note that $f(x) < x$ for $x \in (\beta,\gamma)$ and so from Point 2, you have $x_{n+1} \le x_n$ for all $n$ (and $x_n \in [\beta, \gamma)$). So you know that $x_n \downarrow x^*$, and $f(x^*) = x^*$. Hence you must have $x^* = \beta$. Oct 27 answered Proof by induction help!!! Oct 27 comment prove trig equivalence :-) ${}{}{}{}{}{}$ Oct 27 comment prove trig equivalence You didn't try very hard. It is just multiplication. Oct 27 comment prove trig equivalence Bingo! ${}{}{}{}$ Oct 27 comment prove trig equivalence There are many ways to prove this. Writing $\sin,\cos$ as exponentials will certainly work. Have you tried? Oct 27 comment Question about optimization? No. Take $g(x) = (x-2)^2$ and $f(x) = x$, then you cannot write $\nabla f(2) + \lambda \nabla g(2) = 0$ for any $\lambda$. You need some regularity conditions so the multiplier of the cost gradient is non zero. Oct 27 comment Question about optimization? Generally Kuhn Tucker only appears when you have some form of constraint. Did you mean to ask something else? It would be a Kuhn Tucker point, but only in a vacuous sense, since $\nabla f(x) = 0$. Oct 27 answered Prove that if $f(x)= \sum^{\infty}_{n=0} a_n x^n$ has infinite radius of convergence Oct 27 comment Prove that if $f(x)= \sum^{\infty}_{n=0} a_n x^n$ has infinite radius of convergence An infinite radius of convergence doesn't mean that that ratio test holds. Take $1+0+{x^2\over2}+0+{x^4\over4}+...$ as an example. Oct 27 comment Example to the statement that $a_{n+1} - a_n \rightarrow 0$ as $n \rightarrow \infty$ does not imply that sequence $a_n$ converges. @MPW: The difference converges to 1... Oct 27 comment Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$ I have no idea what you are doing with the $k$ above. Look at Math's answer below. The key to that answer is the fact that ${7 \pi \over 12} = {\pi \over 3} + {\pi \over 4}$. Oct 27 answered Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$ Oct 27 comment Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$ The statement is still false, you are missing a square. Oct 27 comment Prove: $(1+i\sqrt{3})(1+i)(\cos\phi+i\sin\phi)=2\sqrt{2}\left(\cos\left(\frac{7\pi}{12}+\phi\right)+i\sin\left(\frac{7\pi}{12}+\phi\right)\right)$ The statement $(1+i\sqrt{3})(1+i)=8$ is clearly false. Oct 27 comment Numerical approximation of the Jacobian matrix What does the Jacobian of $g$ have to do with the ODE? (I'm trying to figure out what you are asking.) Oct 27 comment Nullified terms of this polynomial? @gsamaras: I have no idea. I try to ignore downvotes (not very successfully). Oct 27 comment Numerical approximation of the Jacobian matrix You need to provide a little more information. The Jacobian of what? Oct 27 comment Understanding the actual meaning of “square root”. There are two solutions to $x^2 = y$ (for $y > 0$), by convention we define $\sqrt{}$ to be the non negative solution. Oct 27 answered Existence of partial derivative and their continuity in a neighbourhood of a point where it is differentiable