DavidP

Cool - mod 11 works indeed. Thought I tried that before... Thank you so much!

How do you prove that 4\cdot 100^k-31=9n^2 has no integral solution. I don't find any mod for a contradiction.

@Ted Shifrin: Ok - I will think about it a little bit more. Thank you very much for your time!

I was not removing the $\cos(n)$, I wanted compare with (asymptotic equivalent sequence) and write (formally) $\sum\limits_{n\geq 1} \frac{\cos(n)}{\log(n)+\cos(n)} \approx \sum\limits_{n\geq 2} \frac{\cos(n)}{\log(n)}-\sum\limits_{n\geq 2} \frac{\cos^2(n)}{\log^2(n)}$ and since on the right hand side the first series converges and the second diverges, so does the left hand side. Is that a valid conclusion?

I was just asking, if my conclusion is right, if I can prove this above. For convergence for the series see for example there: math.stackexchange.com/questions/744125/… The limit I checked with WA. If the conclusion is alright, I will try to prove the limit "by hand".

My thought was to use that $frac{\cos(n)}{\log(n)+\cos(n)}\sim \frac{\cos(n)}{\log(n)}-\frac{\cos^2(n)}{\log^2(n)}$, since $\lim_{n\to\infty} \frac{frac{\cos(n)}{\log(n)+\cos(n)}}{\frac{\cos(n)}{\log(n)}-\frac{\cos^2(n)}{\log^2(n)}}=1$. So we can compare and since $\sum\limits_{n\geq 2} \dfrac{\cos(n)}{\log(n)}$ converges and $\sum\limits_{n\geq 2} \dfrac{\cos^2(n)}{\log^2(n)}$ diverges, the original series diverges also.

Is it possible to approximate $frac{\cos(n)}{\log(n)+\cos(n)}\sim \frac{\cos(n)}{\log(n)}-\frac{\cos^2(n)}{\log^2(n)}$ to conclude the series $\sum\limits_{n\geq 1} \dfrac{\cos(n)}{\ln(n)+\cos(n)}$ diverges, or is this approximation not good enough?

Hi! Is there a clever way to count this? 6 red, 5 blue and 7 yellow balls should be placed in six distinguishable boxes, such that every box contains three balls. Balls with same color are indistinguishable.

Any chance there might be a closed form for $I(a,b)=\int_{-\infty}^{\infty}\frac{e^{-x^2}}{\cosh(a)+\cosh(bx)}\mathrm dx$? Mathematica can't find anything.

You're welcome.

@UnderMathUate See also there: en.wikipedia.org/wiki/Frullani_integral

@TedShifrin Thanks!

Hi! Is the calculation in the comment from JJacquelin wrong, or am I missing something? Shouldn't the inner integral go from theta' to 0. WA gives also negative value. math.stackexchange.com/questions/806709/…

I have also a combinatorics question: Is there a nice double counting argument for the identity 2\cdot\sum_{k=0}^n (n-k)\binom{2n}{k}=n\binom{2n}{n}? The right-hand side counts something like choose a team from n person out of n men and n woman and then choose a captain. Unsure about the left-hand side.

Not really... Next weak we get maybe a stronger Lockdown. Looking forward to summertime. Hope the vaccine helps to get rid of the pandemic problem but now there isn't enough.

Thanks! I have a look. Here is everything fine too, although the situation is not getting any better. I have to teach my classes online this month und maybe longer. It's a little bit frustrating, for example the websites for video-conferences are often not available.

@user21820 Hi! How are you? Hope everything is fine!

@user21820 Hi! Thanks a lot. That seems to work nicely. The definition for gamma function is ∫[0,∞] x^(c−1)·exp(−x) dx, but then the same argument works with the factor c instead of 2c. Right? 

Yes - thanks! Take care too. Bye!

Everything fine? Are you enjoying your weekend?

Thank you!

Yes of course. That is better. I just wanted to see if I am misunderstanding the author or if he is wrong. :)

The first term is finite, being a finite integral. For the second term, when t \geq \epslion we have: t^(x-1)e^(-t)=(t^(x-1)e^(-t/2))e^(-t/2)<=e^(-t/2)

In the case of the second integral, first note that t^q exp(-t/2) \to 0 as t \to \infty for any q in \R. Hence for any x in \R there exits an \epsilon such that 0 \leq t^(x-1)e^(-t/2) \leq 1 for t = \epslion. So we further split the integral at \epsilon.

He splits \int_0^{\infty} t^(x-1)exp(-t) dt in \int_0^1 + \int_1^{\infty} and proves that the first one exists. That is clear. The his text is:

Yes - but isn't this estimation wrong. The first inequality - as he wrote - is for t=epsilon and the he uses this for every t>=epsilon.

@user21820 Hi! I have a short question. I’m reading the book „The Gamma Function“ by J. Bonnar. On page ten he writes there ex. epsilon such that 0<=t^(x-1)e^(-t/2)<=1 for t=epsilon. Then he splits an integral at epsilon in two (1 to epsilon and epsilon to inf) and says the first is finite and for the second one for t >= epsilon he uses t^(x-1)e^(-t)=t^(x-1)e^(-t/2)e^(-t/2)<=e^(-t/2). This ist not really neat, or what do you think?

Yes! See you!

That's a good idea. Thanks!

Thanks! Everything is fine here too. I have a short question if you don't mind. Just did some math for fun and I am unable to check my solution. I calculated sum k=1 to inf int 0 to 1 ln(x)ln(1-x^k)/x dx and got zeta(3)pi^2/6. I tried with Wolfram but that didn't work. Do you have some programm which can check the result?

@user21820 Hi! How are you?

@user21820 Thx for the talk. See you soon. All the best - and stay healthy in this time.

We devided the square in two triangles (half the square) and proved that this triangle is half the rectangle. This is actually a shear, a rotation and another shear. When you have the triangle ABC and the square (over b) ACDE and the rectangle AFGH then we took the triangle ACE and sheared to triangle ABE, rotated with center A (-90°) to AFC and sheared to AFG.

We already proved a^2=pc and b^2=qc with shear-mapping. So the proof for pythagorean's theorem is now not difficult. :)

So lessons are prepared. Tomorrow we will learn ordering fractions at the number line, calculating the intersection from linear functions and the pythagorean theorem.

Me too - and I hope things are going peacefully in the next time. I was a little bit scared when I read how many people have guns in the US and saw pictures from crazy people in front of the houses where they were counting the votes...

Can't imagine our president talking in TV and switched off by press with the comment he is lying all the time. :)

Yeah - excitedly I followed the event in newspaper and internet. And actually I'm quite happy that Biden won.

Good to hear. Here is everything quite well too. I was preparing some stuff for school tomorrow, when I decided to take a little break, drink some coffee and drop in to say "Hi" and see how things are by you before I get back to work. :)

@user21820: Hi! How you doin? Hope everything is fine?!

@user21820: Ok - then enjoy your weekend! See you again soon.

Yes - Thanks. I did some correction today and enjoy now a lazy saturday evening.

@user21820: Hi! How are you? Hope you're good.

But I will try to find a suitable c, after doing some sports. Thanks once again and have a nice weekend. Greetings from Germany :)

Yeah but actually I learned a lot about asymptotic bounds, so I am very happy with this one.

Alright. Thank you so much!

Which line? The 2 / sqrt(n) ?

Ah, know it's clear. I try it.

Or should I now write: n^(1/k) ( (1+1/n)^(1/k) - 1)

Yes, this one i got too. But how do you factor out from n^(1/k)?