@robjohn It is probably backwards, but it still seems strange. If M is huge then i guess it is possible that $\|x - x_\text{min}\| > \|x_\text{mini}\|$.
In order to do so I got a tip that I need to use $$ \| x_\text{min} + t (x - x_\text{min})\|^2 \geq \|x_\text{min}\|^2 $$, but like I said. I do not understand why the inequality holds.
Ah, I see! I understand now why, but unable to explain it, Hmm
@Charlie Well, I'm now better. I still haven't got the proof of perron-frobenius thing. Haven't looked at it actually. I understand the proof now though, still somewhat edgy I think.
Hmm if you had to describe some interaction between euclidean polygons, points, vectors, and lines, what would be your preferred notation for defining these things?
Okay so I need to compute an automorphism on $\mathbb{Z}_2\times\mathbb{Z}_2$ using the fact that if $f\colon G\to H$ is an automorphism and $G=\langle K\rangle$, then $f$ is determined by where it takes members of $K$.
I think that $\mathbb{Z}_2\times\mathbb{Z}_2 = \langle(0,1),(1,0)\rangle$ u...
@PeterTamaroff I tried starting with $z^n$ because the sum of this is know as long as $|z|<1$. Differentiating twice yields. $n(n-1) z^{n-2}$. Now one can use $z=z-2i$, and then...
@N3buchadnezzar Just a derivation. In general, if $$f(x)=\sum_{n=0}^\infty a_n(x-a)^n$$ then $${f^{\left( k \right)}}(x) = \sum\limits_{n = k}^\infty {\frac{{n!}}{{\left( {n - k} \right)!}}{a_n}} {(x - a)^{n - k}}$$
Note that letting $x=a$ one gets the famous $$f^{(k)}(a)=k!a_k$$, that is $$\frac{f^{(k)}(a)}{k!}=a_k$$ which means an absolutely convergent series over some disc $|z-a|<|w|$ coincides with it's Taylor series there.
Am I write if I say that every decreasing sequence is limited from above? Hence the sequece which is decreasing and isn't limited from above doesn't exist?
Say there are three boys, who share 7 bottles filled with milk, 7 empty bottles, and 7 bottles half filled with milk. How can they share such that they each get the same amount of milk? If they do not have to have same amount of bottles each, are there more solutions?
Okay so I need to compute an automorphism on $\mathbb{Z}_2\times\mathbb{Z}_2$ using the fact that if $f\colon G\to H$ is an automorphism and $G=\langle K\rangle$, then $f$ is determined by where it takes members of $K$.
I think that $\mathbb{Z}_2\times\mathbb{Z}_2 = \langle(0,1),(1,0)\rangle$ u...
