who are you???

Where can I suggest integrals to Michael?😀

@@i.am.jihoonk I’m just a memer

@@ld1ego_733 Post in the comments, but Michael can miss it so you may have to repost. I know some people who kept posting the same suggestion for several days and eventually got the video.

@Good Place to Stop is right. You can also email me -- i won't list my email here, but it is easy enough to find online. The one bad thing with this channel getting bigger is that I tend to miss lots of nice suggestions, especially when I am busy with other things. I promise it is never due to a lack of my interest or gratitude though. Thanks eveyone!!

You did a video on this topic, I think The Bose integral

Thanks, I'll never tell my secret!! j/k: I hand place all of the characters using canva and use a color pallet generator to make sure the colors work together.

@@angelmendez-rivera351 Ofcouse I said he did a video about this "topic" as a whole

@@MichaelPennMath you only use canva for the thumbnails?

I too deal with this Trauma inflicted by Michael's mistake

Or use 2n-1

@skulliam4 That would be more elegant and, since he usually excludes zero from the naturals, a bit more consistent.

@@MAULIKPATELnamste jajajjsjsjjajajjs

It is important correct the mistake in the next step by making another small mistake "in the opposite direction" :)

ya that is what i was thinking too and i wonder if he will pull out polylog suddenly some how

To add, that "special" version is sometimes called the Fermi Integral or Fermi-Dirac integral, and it is related to the behavior of fermions in the same way that the Bose integral is related to bosons. And writing this comment is the first time I realized that bosons are named after Bose and fermions are named after Fermi.

You are correct, but in the next step he rewrites that sum as (sum of all n's) - (sum of all even n's) so it gets to the right answer

That's also how I did it. I agree: simpler.

I think the same method used in this video ought to work.

Could u plz post them in the comment section, so that we can try :)

We can't expand 1/(1+e^x) as a geometric series because e^x>1 for x>0. However we have x/(1+e^x) = xe^(-x)/(1+e^(-x)) and now we can expand 1/(1+e^(-x)) as a geometric series to get x/(1+e^x) = sum_{n=0}^{\infty} (-1)^n x e^{-(n+1)x} dx, which we can integrate termwise because of uniform convergence. A quick integration by parts gives int_0^{\infty} x e^{-(n+1)x} dx = 1/(n+1)^2, so we get the same result at the end.

He does the same thing, except he doesn’t plug it into that formula but derives it.

It's not magical. He shows where the property comes from

I just spotted the same error. He makes another error later that cancels this one out.

Yeah you are right :) I was just about to comment on the same thing.

Or make it 2n-1 at 12:04

Actually, it would have changed the result, but he makes another mistake later that cancels out this mistake.

Exactly i realised too ( we re right ????)

n must be evaluated from 0 to i finity in the first part to get (1)

Well, Things clear up at 12:48 in the video, while splitting up the term.

@@MAULIKPATELnamste Doens't change the fact that you are still missing a +1 constant term, by neglecting the (2n+1) = 1 (when n=0) term, which evaluates to 1. So basically, the final answer should be 1 + pi^2 / 12, unless I am missing something.

@@Hvidbergen The correct answer is pi^2/12. What you're missing is that he went on to make another mistake that canceled out this mistake. It was on the very next step/line.

I don't see how it diverges, |sin(n)| averages 0.6 and the zeta function converges for all n>1.

@@paul_w try it yourself with software

@@stenzenneznets I did and looking at the partial sums it seems convergent to around 3.3 . But obviously this is no proof, it could diverge very slowly. But again I don't see how it would diverge (cf previous post).

I can prove it diverges, but I world like to see how Michael manages to do it

At 12:48 onwards, the summations clear up automatically.

@@MAULIKPATELnamste yeah, thanks noticed that. I guess his error was more of a mis-use of the equals sign

@@tomatrix7525 He just made a small mistake writing the sum. No need to get complicated.

The correct answer is pi^2/12. What you're missing is that he went on to make another mistake that canceled out this mistake. It was on the very next step/line.

in theory yes, but the problem is that the function has infinitely many poles (i*pi*(2k+1) for all integers k) and moreover the integral starts at 0, so you would have to find an appropriate contour that you can calculate.

It is not clear what "a_k-1-2a_k+a_k+10 for all k = 1, 2, ... , k-1" means.

@@ricardocavalcanti3343 It means a sub k, 2a sub k and so on.

In case what a_o or 2a_k means, it means a sub 0 and 2a sub k.

@@keshavb3128 But what is the condition on a_k-1-2a_k+a_k+10?

@@ricardocavalcanti3343 that a_0=a_n=0;a_k-1-2a_k+a_k+10 is only true when k=1,2,..., k-1 and are nonnegative.

Quick to state, not to solve! :-) Here is my solution: (1) For F(n) = (2^{4n+2}+1)/65 to be an integer (not necessarily prime), n must be of the form n=3k+1, where k is a nonnegative integer. Indeed, 2^{4(3k+1)+2}+1 = 2^{12k+6}+1 = 64^{2k+1}+1 = (-1)^{2k+1}+1 (mod 65) = 0 (mod 65). (2) Next, we note that 2^{12k+6}+1 = (2^{4k+2}+1)(2^{8k+4}-2^{4k+2}+1) = f(k)g(k). These factors have the following divisibility properties: (2.1) For any nonnegative k, f(k) is a multiple of 5: f(k) = 2^{4k+2}+1 = 4x16^{k}+1 = (-1)x1^{k}+1 (mod 5) = 0 (mod 5). Moreover, if k=3m+1, where m is a nonnegative integer, then f(k) is also a multiple of 65: f(3m+1) = 2^{4(3m+1)+2}+1 = 0 (mod 65) by (1). (2.2) If k=3m or k=3m+2, where m is a nonnegative integer, then g(k) is a multiple of 13. Indeed, g(k) = 2^{8k+4}-2^{4k+2}+1 = 16^{2k+1}-4x16^{k}+1 = 3^{2k+1}-4x3^{k}+1 (mod 13) = (3^{k+1}-1)(3^{k}-1) (mod 13). Therefore, (2.2.a) g(3m) = (3^{3m+1}-1)(3^{3m}-1) (mod 13) = (3x27^{m}-1)(27^{m}-1) (mod 13) = (3x1^{m}-1)(1^{m}-1) (mod 13) = 0 (mod 13); (2.2.b) g(3m+2) = (3^{3m+3}-1)(3^{3m+2}-1) (mod 13) = (27^{m+1}-1)(9x27^{m}-1) (mod 13) = (1^{m+1}-1)(9x1^{m}-1) (mod 13) = 0 (mod 13). Having established the preliminary results above, now we can show that there is no integer n such that F(n) = (2^{4n+2}+1)/65 is a prime number. For this, we have to examine the following cases: (3.1) If n0, it follows from (2.1) and (2.2.a) that F(n) = F(9m+1) = f(3m)g(3m)/65 = 5Mx13N/65 = MN, where M and N are integers greater than 1; therefore, F(n) is not prime in this case; (3.2.d) In a similar way, if k=3m+2 and m>=0, it follows from (2.1) and (2.2.b) that F(n) = F(9m+7) = f(3m+2)g(3m+2)/65 is not prime either; (3.2.e) Finally, if k=3m+1 and m>0, it follows from (2.1) that F(n) = F(9m+4) = f(3m+1)g(3m+1)/65 = 65MxN/65 = MN, where M and N are integers greater than 1; therefore, also in this case F(n) is not prime. These cases exaust all possibilities, thus proving that F(n) is not prime for any integer n.

@@ricardocavalcanti3343 Well done ! That was "the right way" to solve this exercise. (There were actually 2 intermediate questions that I deleted, otherwise it was too simple :-) ).

has been pointed out by others, that it should start at n=0

I’m here 🙋

@@goodplacetostop2973 yooo lol you are always here

@@prithujsarkar2010 As always 👍