Given $H\le G$ groups and representations $V\in{\rm Rep}(H)$, $W\in{\rm Rep}(G)$, using the Frobenius formula one can observe $({\rm Ind}_H^GV)\otimes W$ and ${\rm Ind}_H^G(V\otimes {\rm Res}_H^GW)$ have the same characters, hence they are the same representations. However I can't "see" why they are the same. The elements look the same - sums of $g\otimes v\otimes w$s, but $G$ doesn't act the same on them: for $f\in G$, we have
$$f(g\otimes v\otimes w)=fg\otimes v\otimes fw\quad {\rm versus}\quad f(g\otimes v\otimes w)=fg\otimes v\otimes w.$$ The redundancies in these symbols are also different: $$gh\otimes v\otimes w=g\otimes hv\otimes w \quad {\rm versus} \quad gh\otimes v\otimes w=g\otimes hv\otimes hw.$$ So $g\otimes v\otimes w\mapsto g\otimes v\otimes w$ cannot be the isomorphism $({\rm Ind}_H^GV)\otimes W\cong {\rm Ind}_H^G(V\otimes{\rm Res}_H^GW)$, which I assume is natural. So what is?
@r9m Well, I know that if we use $\tan\left(\frac\pi2x\right)-\frac{2x}{\pi(1-x)}$ in place of the $\tan\left(\frac\pi2x\right)$ above, the integral converges...
@r9m this means that we can replace $\tan\left(\frac\pi2x\right)$ by $\frac{2x}{\pi(1-x)}$
@r9m This means your limit should be the same as $$\lim_{n\to\infty}\frac1{\log(n)}\frac2\pi\sum_{k=1}^n\frac{2n+1}{(2n+1-k)k}$$
@Chris'ssis I don't remember... I don't have the result off the top of my head.
@Chris'ssis: I am working on these $$ \sum_{n=0}^j(-1)^{j-n}\binom{2k+1}{j-n}\binom{n+k}{k}^2=\binom{k}{j}^2 $$ $$ \sum_{n=j+k}^{2k}(-1)^n\binom{2k+1}{n+1}\binom{n-j}{k}^2=\binom{k}{j}^2 $$
This question is in continuation to this and motivated to answer this question. If I have an answer to it, then I can prove a special case of this question.
Consider two smooth plane curves $C \equiv (X_C(s),Y_C(s))$ and $S \equiv (X_S(s),Y_S(s))$ represented in arc length parametrization. Curve...
@Nick I just have a relationship between two tables defined by almost the same recurrence and in the limit the ratio between the row sums becomes zero when the recurrence gives the number of divisors of n and n respeectively. Therefore the limit equal zero.
@Mats: Well, you have enough time and free will to do all that. I'm stuck in school doing silly things like proving relations are transitive, symmetric and reflexive.
Studying is easy, you should just exclude the words in the text, and focus on relationships between what could be considered as nouns. But it takes time and effort.
When I studied chemical engineering I often found the need to rewrite lecture notes, handouts and books in order to gain a thorough understanding of the subject I was reading. As much as time permitted I used to draw mindmaps of the reading material combining the symbols on the left in the image ...
@skullpatrol: No, for like the first time in my academical life, I have absolute understanding of the topics to a degree none of my peers nor teachers ever will acquire.
And that's whats putting pressure on me.
If I dare to commit even the slightest blunder in the 3 hours of my test...
@Nick I said a bit too much when I said "studying is easy". It was not until my last year that I became a decent student. And heat transfer was my worst subject, could never really master how to translate a heat transfer problem into an integral. Of course it could have something to do with that I did not attend the lectures in that subject.
What is a value called that is just before the limit of a function?
Confusing. I mean what is the value called that appears just before a singularity? That is what I meant.
@skullpatrol The calculation suggests that the sum of the divisors divided by the odd numbers is 1/2. But that is where it blows up, so no solid justification.
Just so the Mathematics chat denizens are aware, I have just added three answers to the (homework) tag should be deprecated meta-question so that users can unambiguously vote on their preferred option in this matter.
Prove that the sets $S$ and $D$ have the same cardinality, where $S = \{(x,y)\mid-1\leq x \leq 1\text{ and }-1\leq y\leq 1\}$ and $D = \{(x,y)\mid x^2 + y^2 \leq 1\}$.
If you multiply by k and look at the position of the largest term in the triangle, then the index will have a asymptotic given by the LambertW function.
@MatsGranvik Sorry, I can't visualize binomial coefs as elts of the Pascal triangles. I almost always think of it either algebraically or combinatorially.
