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02:00 - 19:0019:00 - 00:00

19:04
@PeterTamaroff I am stuck
@LittleChild Can you find $$\int_{\pi}^\pi t^2\sin ntdt$$ $$\int_{\pi}^\pi t^2\cos ntdt$$?
Yes, by parts
OK, so what's the fuzz?
but this exponential thing seems to be looping
I dont seem to be getting anywhere
:(
@PeterTamaroff That has the question I am talking of. He doesnt explain at all. The font looks like ants
@LittleChild But $e^{-inx}=\cos nx-i\sin nx$
19:13
do I have to substitute ? someone said that let $e^{-inx}$ stay as it is. It will make things easier
in teh comments to the video
@LittleChild Note $\int_a^b (f+g)=\int_a^b f+\int_a^b f$
the other 1 should be g
and both integrals have to exist
what is derivative of $cos(nx)$ ? $-sin(nx) / n$ ??
@Ethan
no its not that
$-n\sin(nx)$
19:23
ok :)
lol
$sin(\frac{nx}{t})$'s derivative would be $\frac{n}{t}.cos(\frac{nx}{t})$ right ?
$$\sum_{n\leq x} \ln(d(d(n))=\ln(2)x\ln(\ln(x))+O(x)$$
$\sum_{d\mid n} 1 = d(n)$
Please use \ for logarithms, sin and other functions..
My eyes are tired of bleeding
@LittleChild Yes, that.
19:28
cool .. beginning to see a pattern here :)
@PeterTamaroff The proof for $\int_0^1 \log \Gamma(x) \,\mathrm{d}x$ is cute
@N3buchadnezzar What are you talking about?
I was just sharing some random thoughts, browsing through some integrals on the site =)
20:13
Thanks for sharing :D
21:00
@robjohn
@PeterTamaroff yes?
@robjohn Can you help me out here? Specially with last sentence, i.e., $$\lim_{n\to\infty}f_n(n+1/2)$$
21:23
@robjohn Hello?
are all the rational curves have genus zero
do*
?
@Theorem What is a rational curve?
@PeterTamaroff : a curve which has a rational parametrization .
@Theorem And how is the genus defined?
@Theorem ?
21:46
@PeterTamaroff I'm having a look at estimating your roots $x_n$.
@AntonioVargas Thanks!
@AntonioVargas I edited the question to add something else. =) This is a lot of fun, indeed!
22:05
@AntonioVargas Are you there?
@PeterTamaroff Yessir.
@AntonioVargas Did you see what I added?
I remember $\sum_{k=0}^{n} n^k/k!$ being estimated on MSE a while ago, so I'm trying to find it.
The maxima seem to go to zero as $1/\sqrt n$.
@AntonioVargas Oh, yes.
@AntonioVargas $$e^{-n}\sum_{k=0}^n \frac{n^k}{k!}\to \frac 1 2$$
I figure some proof out there should be able to be modified to get the alternating case.
It can probably be done from the integral representation using Laplace's method.
22:08
@AntonioVargas Ah! I learned Laplace's method just some days ago.
Wouldn't your integral representation help?
I would bet so.
22:26
Greetings noble minds!
@N3buchadnezzar today my research led me to a mind-blowing result that involves an infinite sum of cosine integrals.
Wop wop
Reminds me of an old putam problem
@PeterTamaroff sorry, afk for a while
@N3buchadnezzar on the other hand there might be a very easy trick around that I simply don't see (and no mind-blowing thing around) :-)
:-)
I found a easy and cute integral today
@N3buchadnezzar Which one?
22:38
Mind you this integral is very simple, but the idea in it is just brilliant
@N3buchadnezzar Minded.
$$
I := \int_{ \psi }^{ \varphi } 2^2 \left( x^2 - 1\right) e^{2x} \mathrm{d}x
$$
Where $\psi$ og $\varphi$ are respectively the smalest and greatest solution of $x(x-1)=1$
@N3buchadnezzar OK, the appearance of $2^2$ already suggests something is going on...
@N3buchadnezzar this statement is annoying to me. Really! (especially the last part with the solutions)
:-)
@Chris'swisesister Bear in mind, English is not my first language ;)
22:43
howdy
@PeterTamaroff It is just there to get rid of some pesky fractions, the clever ting is that one does not have to compute explicit formulas for $\psi$ and $\varphi$, but instead one uses that $\psi(\psi-1)=1$ and $\varphi(\varphi-1)=1$.
I thought that was quite cute blushes
\begin{align*}
\int_\psi^\varphi (x^2-1) e^{2x} \mathrm{d}x
= \frac{1}{4} \Bigl[ \bigl(2\overbrace{x(x-1)}^{1}-1\bigr) e^{2x} \Bigr]_\psi^\varphi
= \frac{1}{4} \Bigl[ e^{2x} \Bigr]_\psi^\varphi = \frac{e^{1+\sqrt{5}} - e^{1-\sqrt{5}}}{4}
\end{align*}
@N3buchadnezzar nice. Here is the cute one I was talking about $$\sum_{k=1}^{\infty} \text{Ci}(2 \pi k)$$
Nice integral.
@Chris'swisesister Where $\operatorname{Ci}(x) = -\int_x^\infty \frac{\cos x}{x} \,\mathrm{d}x$ right?
22:48
Pesky minus sign.
aka 'Cosine Integral'
@Chris'swisesister Maple gives $-0.3860783245$
That's corect.
@N3buchadnezzar this is actually $1/2(1/2-\gamma)$
The inverse symbol generator gave me frac(\gamma*50)+(3)
I suspected something was up :p
@Chris'swisesister Did you use the sum formula for the Ci function, listed on that site?
@N3buchadnezzar yes.
22:56
=)
Did you see the LambertW integral?
@N3buchadnezzar do you mean Lambert W function? I never heard of LambertW integral.
Per say it is not defined as anything special it is just the fact that
$$ \int_0^\infty \operatorname{LambertW}\left( \frac{1}{x^2} \right) \,\mathrm{d}x = \sqrt{2\pi} $$
@PeterTamaroff your roots are too much for me right now, it seems. Szego studied some similar ones a long time ago but his paper is in German so I've never been able to get much out of it.
23:07
@N3buchadnezzar this integral also appears on wiki
@N3buchadnezzar I need to leave now. Maybe you give me some hints on the sum I posted above (next time). :-)
bye
23:22
n00b question: if the statement 2/(x+2) isn't a polynomial. What is it?
@Nick It's called a rational function.
23:34
@AntonioVargas: Thank you, that explains a lot. But I still dont understand why sqrt(x) isn't a polynomial.
@Nick A polynomial only has integer powers. That has a one-half power, $\sqrt{x} = x^{1/2}$, so it isn't a polynomial.
@AntonioVargas: Yes, I know that but why is that? The technical definition of a polynomial includes the statement "A polynomial can have constants, variables and exponents, but never division by a variable.". In x^{1/2}$ , Where is the division by a variable.
@Nick You're right, that's not division. And that's also not the whole definition of a polynomial.
From wikipedia: "A polynomial is either zero or can be written as the sum of a finite number of non-zero terms. Each term consists of the product of a constant (called the coefficient of the term) and a finite number of variables (usually represented by letters), also called indeterminates, raised to whole number powers."
whole number powers... that was the magic word
aww, thank you
np
@N3buchadnezzar Thanks, that was fun.
23:58
@AntonioVargas The other one is interesting too =)
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