I might pick up Serre, Local Fields to help me cope with the first chapter of CF, which left me a little scarred with all the bilinear forms and determinants
I can use that pleasant $efg=(L:K)$ formula (and its extension to non-local fields) just fine, but I couldn't reproduce a damn thing from the first chapter, nor the ideas in the proofs
Certainly not as comprehensive, but Pete L Clark (he posts on math.se and mathoverflow a lot) has written a pretty good set of notes on valuation theory
@PedroTamaroff I disagree. The question meant to ask for a function to use in image processing. It's no big deal, though: OP had their answer, albeit in a comment.
Let $\chi$ be any nonprincipal charachter $\mod k$ Prove that for all integers $a<b$ we have $$ \left| \sum_{n=a}^b \chi (n) \right| \leq \frac{1}{2} \phi(k) \,.$$
@N3buchadnezzar by geometry it suffices to show that no more than half the values are on any semicircle, for which it suffices to show there is an m with X(m)=-1 (using symmetry), which follows from 2|phi(k)
Suppose there are only $\varphi(k)/2$ nonzero elements inside the interval $a,b$.
Then obviously $$\left|\sum_a^b\chi(n)\right|\leqslant \frac 1 2\varphi(k)$$
Now suppose there are more than $\varphi(k)/2$ nonzero elements in $a,b$.
Then shift by a full period, use that $$\sum_{1\leqslant n\leqslant k} \chi(n)=0$$ to get the other part, which has less than $\varphi(k)/2$ has before.
@PedroTamaroff Nah, the reason $0 \rightarrow G \rightarrow H \rightarrow 0 \implies G \cong H$ is good is when you have, say, a long exact sequence and you can prove that every third term is trivial. Then you've got a pleasant isomorphism.
I have to leave by 8:40 to get to class on time, and I usually leave early so that I'm not rushed for time. But I don't wanna go yet so I'm procrastinating until I have to.
@PedroTamaroff I think I put it the wrong way since it gives a negative result $$\sum_{n=1}^{\infty} \left(\frac{H_{2n-1}}{2n-1}-\frac{H_{2n}}{2n}\right)$$
If $\chi$ is a real value characher $\mod k$ then the sum $$ S = \sum_{n=1}^k n \chi(n) $$ is an integer. I am looking to prove that if $(a,k)=1$ then $a \chi(a) S \equiv S \pmod{k}$
Ok. How to use integration by parts on $\displaystyle\int\limits_{\Omega} a_{ij}\frac{\partial^2 u}{\partial x_i \partial x_j}\, dx$, where $x=(x_1,\ldots, x_n)$ and $\Omega \subset \mathbb{R}^n, a_{ij}, u \in C^2(\Omega)$
robbie i have some problem i think its small but it dont see it
what is $1 - P(X > u, Y >u)$ if $X$ and $Y$ are independent, i was just thinking it is $1-(1 -P(X\leq u , Y \leq u)) = 1 -f_X(u)f_Y(u) $ what am I doing wrong
Most of you are aware that $f'+\alphaf$ can be rewritten as $(exp(\alpha x)f(x))'/exp(\alpha x). This is a useful trick for solving some real analysis problems.
Can anything similar be told about $f'' + \alpha f$ ?
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