I want to show that $\displaystyle \int_{-\infty}^{\infty} \frac{\sinh ax}{\sinh \pi z} \cos bx \ dx = \frac{\sin a}{\cos a + \cosh b} \ ( -\pi < a < \pi) . $
So I let $f \displaystyle (z) = e^{ibz} \frac{\sinh az}{\sinh \pi z} $ and integrated around a rectangular contour with vertices at $R, R...
Find the value of the limit:
$$\lim_{n\to\infty} \sum_{k=0}^n \frac{{k!}^{2} {2}^{k}}{(2k+1)!}$$
I'm trying to find out if this limit can be computed only by using high school
knowledge for solving limits. Thanks.
@Chris'ssis something should be done about this anonymous downvote campaigners ! like making it compulsory to give a reason (by a comment) for downvoting/upvoting ..
@Chris'ssis account just for downvoting is outright childish :| .. I pity them .. (besides I have only dowvoted once .. on china math's question .. that too was an accident when I was accessing M.se from tab :P)
@r9m But think about it, you as a person don't have the courage and the dignity to do that from the original account, but from a fake one. Cowards ... :-)
@Committingtoachallenge I have a seminar on tomorrow and the subject that i got was fixing of triangle. I have no idea what that means i googled it but did not get the answer.
This guy reminds me of a colleague of mine, when he came into campany was totally intrigued by my analysis and calculations and began to do some investigation, he thought I'm helped by someone else, that my work is not really mine, it was hard to him to accept that he with PhD couldn't understand my work or had some terribly difficulties. That was until one day when I gave him a lesson and explained some of my work in details. Since that day he never tried to do anything else but to respect me.
Manufacturing analysis can be terribly hard, it depends on what you analyze, it's hard to imagine the madness there. You don't analyze functions, but very complex processes.
This is a question about pointwise convergence of a Fourier transform of functions of the form $f: \mathbb{R}^N \to \mathbb{R}$, which is potentially a $N$-dimensional generalization to pointwise convergence of $1$ dimensional Fourier transform. This question arose when I am trying to generalize ...
@robjohn Yes! Thank you for the explanation, can you help with some topology ? I would like to prove that if $f:\Bbb{R}\rightarrow \Bbb{R}$ is continuous then the graph is closed.
I am trying to prove that its complement is open, my problem is I would like to justify that if $y$ and $f(x)$ are distinct point in $\Bbb{R}$ then I can found to disjoint open set.
I made it, I just choose an open ball centered at $y$ and $f(x)$ with radius $d(y,f(x))/2$.
@MarcGato If $f\colon X \to Y$ is continuous and $Y$ is a Hausdorff space, then the graph is closed. The proof is pretty much the one you used for $X=Y=\mathbb{R}$, just you can't get the separating neighbourhoods of $y$ and $f(x)$ explicitly in the general setting.
For $Y = \mathbb{R}$ (or similar spaces), you can also use the map $g\colon X\times \mathbb{R} \to \mathbb{R}$ defined by $g(x,y) = f(x)-y$, and see that the graph of $f$ is just $g^{-1}(\{0\})$, which is closed, since $g$ is continuous [prove that].
@DanielFischer great idea to denote $g=f-y$, I was thinking about the diagonal, your method is easier. For the fact that $g$ is continuous it's because g is a linear combination of continuous function, no
@MarcGato Yes, basically. You need to know or show that $(-) \colon \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ is continuous, and a few fundamental properties of the product topology.
I'm sure you can do this easily, but I'm looking for an easy way that only uses series manipulation. Is that possible?
$$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$$
@Committingtoachallenge I also added a supplementary question, that is $$\sum_{n=1}^{ \infty} (-1)^{n+1} \left( \psi^{(0)}\left( \frac{1+n}{2} \right)- \psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n} \right)$$
hehe, that question is for having fun! Math should make people laugh and have fun!
Here is a more advanced version $$\sum_{n=1}^{\infty} \left(\psi^{ (0)}\left(\frac{1+ n}{2} \right)- \psi^{(0)} \left( \frac{n}{2} \right)- \frac{1}{n}-\frac{1}{2n^2}\right)$$
@robjohn we did many such series in the last months, lots of them :-). It's said though there is not a culture of these series ... (the solutions hardly come to the surface as if they were inequalities)
I wanna see the creativity at work, to see something I didn't see before.
@robjohn I mean the inequalities, in general, are hard, and they are not answered that easily. The realm of inequalities is maybe one the toughest areas known.
One needs many many years of learning for being pretty good, but not very good (in my opinion).
(well, out there might be some exceptions)
Oh ...
This one $$\sum_{n=1}^{\infty} \left(\psi^{ (0)}\left(\frac{1+ n}{2} \right)- \psi^{(0)} \left( \frac{n}{2} \right)- \frac{1}{n}-\frac{1}{2n^2}\right)$$ IS NOT more advanced.
(that's because we can sum separately over $1/(2 n^2)$)
@robjohn I also added the series $$\sum_{n=1}^{\infty} \left( \psi^{(0)}\left(\frac{1+n }{ 2}\right) -\psi^{(0)} \left( \frac{n}{2} \right)-\frac{1}{n} \right)^{2014}$$
@Akshay some context will be nice .. what seminar is it ? Googling spits out a variety of results that starts from elementary concept of geometric congruence, geometric algorithms to stuff that are totally out of my league =P .. It might be helpful if you said a little bit more about the target audience of this seminar :)
@DanielFischer 'cos the answers are insane one liners ! and people either completely ignore or take for granted (without questioning their validity) what they cannot relate to .. :P
@Chris'ssis wive/husband is another example that fits the profile :P
I talk too much .. maybe I need some of @TedShifrin smacks !
@r9m lol :-) Now, I agree that I don't explain things in my answer but still ... to some user things happen in a certain way and with other ones things happen completely differently ..
@ypercube Well, I didn't ask you to upvote me, but keep in mind that in this life you would have never known the answer to that question. Maybe it's not that important. Just for the record.
I'm sure you can do this easily, but I'm looking for an easy way that only uses series manipulation. Is that possible?
$$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$$
where $\psi^{(0)}(x)$ is digamma function
Here is a su...
@Hippalectryon Yeah, but I learned to write things that way.
I usually laugh loud when I attend problems, I have a lot of fun, I'm not used to quarrel, this is weird to me. Mathematics is for fun, it must be like that.
From where can I practice questions related to the following topics:
Arrow's Impossibility Theorem
Nash's Axiomatic Bargaining.
Harsanyi's Axiomatic characterization of Utilitarianism.
?
@r9m do you see any awesome way to prove that $$\int_0^{\infty}\int_0^{\infty} \frac{\sin(y)}{y(x+y)(1+x^2)} \ dx \ dy=\frac{\pi}{2}\left(\gamma-e \operatorname{Ei}(-1)\right)$$?
There is not a single person left in this room who has not tried to conduct a psychoanalytical study of @robjohn on the basis of his mean orange square avatar.
Let us take as an example this question.
Consider whether you think it should be considered off-topic (for being purely mathematical in nature).
Should one's opinion on whether a mathematical question be closed be based on what one finds interesting, the specific written rules of this site, a...
@Chris'ssis LHS reduces to $\displaystyle \frac{\pi}{2}\int_{0}^{1} \frac{1-e^{-x}}{x}\,dx$ .. after that its all Merry go round the mulbury bush Puiseux Series of $Ei$ :D
@Unnamed I was asking if any one has some practice questions related to the topic "Nash Axiomatic Bargaining". I need to practice for my upcoming college exams.
@Chris'ssis -_- of course I started from RHS !! who the hell wants to use pen and paper (in my case chalk and cupboard) after a 8 hour long class ?? :P
@r9m I particularly like the limit from the page 178, that one is crazy awesome. I wonder if I can come up with a nicer proof (although that one is very nice).
@Chris'ssis: He's being portrayed by the guy from Slumdog Millionaire... You know, Prince Zuko who works at the best exotic marigold hotel. He's pretty scrawny, you can definitely beat him up.
@Chris'ssis: He's being portrayed by the guy from Slumdog Millionaire... You know, Prince Zuko who works at the best exotic marigold hotel. He's pretty scrawny, you can definitely beat him up.
I'm sure you can do this easily, but I'm looking for an easy way that only uses series manipulation. Is that possible?
$$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$$
where $\psi^{(0)}(x)$ is digamma function
Here is a su...
@Nick You think you're smart giving those replies, but you also miss another point: one day I hope to be like Ramanujan or far better in terms of integral, series and limits (although it's hard to make such appreciations, it's important to get the idea behind the statement).
@Nick I don't believe in humans that cannot be overpassed in terms of performance. You can believe that all as long as you want to.
Any limit can be broken, anything, impossible is just a word ... If you don't bilieve in yourself highly and if you cannot imagine yourself breaking all those limits you wanna break then you'll never break them.
@Chris'ssis: No, I think I'm delusional, high on prescription medication and pretty much bored and in search of cheap chuckles when I was giving those replies. Don't take it seriously, silly goose.
@Chris'ssis I believe you... and I do believe in you too
No one should fear of the limits posed by the others by their performance, no one needs to live within these limit-walls as if they were condemned. The attitude is important, not the fact that one might fail doing that.