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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