yeah; ...really

the first thing he did made no sense to me...

​@@seanmcghee2373 lol it makes sense to me but how he figured that out is beyond me. It would have taken a lot longer for me to get there

@@lukealadeen7836 It's probably a classic trick.

What is the reflection formula?? Is thst a commonly known thing?

@@leif1075 It's a relation between f(a-x) and f(x). e.g. Euler's reflection formula Γ(1-z)Γ(z)=π/sin(πz) or even/odd functions for a=0.

​@@leif1075in this case, the identity is Li₂(x) + Li₂(1-x) = π²/6 - ln(x)ln(1-x)

Let S = -int_0^{1/2} ln(1-t)/t dt integration by parts gives = -ln(t)ln(1-t)|_0^{1/2} - int_0^{1/2} ln(t)/(1-t) dt = -ln(2)^2 - int_0^{1/2} ln(t)/(1-t) dt Swapping t with 1-t gives = -ln(2)^2 - int_{1/2}^1 ln(1-t)/t dt Adding S to both sides 2S = -ln(2)^2 - int_0^1 ln(1-t)/t dt Then using the fact that sum x^n/n^2 = - int_0^x ln(1-t)/t dt the result follows.

Well, we had to pull out the -1/2 times log(2)^2 at some point. Noticing that the sum we want to calculate is equal to a part of the product of the other two known sums given, it should be possible to assail this problem by taking this product of two infinite sums and subtracting all the parts that don't belong to the infinite sum we're interested in, but that seems more laborious than the method in the video (and is probably not a method of a nature different to the method used here as integrals are just infinite sums, too).

Let f(x) = Σ x^n/n^2. Then f'(x) = -1/x ln(x). If A is the integral of f from 0 to 1/2, and B is the integral from 1/2 to 1, you get B - A = ln(2)^2 by integrating by parts and changing variables, and B + A = f(1) = pi^2 / 6. Therefore f(1/2) = A = 1/2 (pi^2 / 6 - ln(2)^2).

@@Technopolo I think there is a typo here - as previously noted by @Spaghetti (and later more laboriously by me) f'(x) = -1/x ln(1-x) which is essential to the solution you outline involving IBP and the change of variable.

You can even improve that benchmark by realizing that 1/2 is the largest term of the (1/2)^n series. Then we already know that the result has to be less than (pi^2)/12

You know what they say about magicians... ain't no magic trick without misdirection

It would be good to see this approach using properties of the dilogarithm. The playlist for the dilogarithm is a short one.

Just a mistake

@ I was wondering the same thing. Usually Prof. Penn clears mistakes up either with a wrist-erase or post-production.

I was wondering the same. I assumed it had to be 1/2 though, but it was slightly confusing.

Just classic Michael Penn mistake

@@davidcroft95 i probably make more mistakes than all of KZfaq combined rofl 🤣 i really should brush off my math rusty-nessz

Notice that he's splitting the integral in two, leaving the integral of ln(1/2)/1-y and then, the integral of -ln(y)/1-y. So ln(1/2) in the first one can simply be factored out, leaving 1/x with x = 1-y.

That's the second order polylogarithm evaluated at 1/a. I think the only "nice" value is at a=2 though (other than a=1, which is zeta(2), or the Basel problem), based on some testing with WolframAlpha.

@@1s3k3b5 actually theres more

@Noam .Menashe [Edit - Hmm , just realised this is similar to @Spaghetti earlier comment , anyway ...] Actually, assuming that by y you mean the f(x) defined by Michael to generalise the infinite sum, where f(1/2) is the quantity to be found then, ignoring Li2(x) etc, I think this differentiation approach is nicer than Michael's nested integrals. It also loses the infinite sums sooner. Recall we are given that f(1)=π^2/6, and obviously f(0)=0. As you derived, (I just differentiated the series twice and used the geometric series sum) xf''+f'=1/(1-x) but LHS is (x f')' so integrating twice gives f(x)=Integral[- ln(1-z)/z,{z,0,x}] - an integral Michael also produces. Recall though that we already know that f(1) is the Basel sum, which he had to demonstrate. This integral is not trivial but it has an interesting integration by parts: particularly useful for x=1/2. Integral[ ln(1-z)/z] = ln(1-z)ln(z)+Integral[ln(z)/(1-z)]. So, f(1/2)=-[ln(2)]^2-Integral[ln(z)/(1-z),{z,0,1/2}]. Now time for Michael's change of variable (r=1-z) in that last integral, switching to Integral[- ln(1-r)/r,{r,1/2,1}], but this is just f(1)-f(1/2). So 2 f(1/2)= -[ln(2)]^2+π^2/6 and result follows. Apologies for the clunky rendering of the equations.

The idea is to think of the term as F(b)-F(a) (antiderivative) then write this as an integral of f(x) from a to b.

The steps in this solution are to a certain extent standard methods of solving analysis problems, but to realise what method is needed at a particular step would take time no matter what level of mathematician you are. It just takes lots of practice. So to answer your question, I would expect he came up with this solution himself as he is a professional mathematician, but it would have taken him some time to figure out the steps of the solution

@@gearoidmccarthy8408 That is something I considered, but I'd like to hear from Michael himself. Some of these problems seem comparable in complexity to the Basel Problem. They ought to take on the order of days or weeks, but he is able to post most every day.

power series of ln(1-x) is -x^n/n so - * - = +

I lost track of how many minus signs I lost track of...

@@krishdewani4949 thx

From the fact that u have to plug in X=1/2 in order to get back to the original expression

There are two places where the Prof mistakenly wrote the integral from 0 to 1.

I think it's correct at 1:14 but the next step should be from 0 to 1/2 and at 2:15 it's right again

@@mavenfromheaven1190 ah yes, i just assumed that 0 to 1 was correct in the first place without checking xD

What about it did you like?

Using the Lambert function W, you have x=e^(W(ln y)). Maybe there are better ways.

Partial fractions separates polynomials, 2^n is exponential. If you try to use partial fractions, it won't work because the assumption that 1/(2^n * n^2) = A/2^n + B/n^2 is wrong. You end up getting 1 = A*n^2 + B*2^n. There are no A and B you can choose to make that statement true for all n. You can change the numerators to something else, but it still won't work as exponentials will never cancel polynomials.

Oh yes, a failed to realise this. Thanks

Just 2 mistakes. All of them should be 1/2

What he did is ok, but I know it can be confusing. Just don't care about the variable; think of one integral as evaluating an area below a function from 0 to 1/2, and the other integral the area from 1/2 to 1. How you call the variable that parts the integral doesn't matter here. So the sum of the 2 integrals is like "ok, let's get the total area" which is the sum of the other 2 areas.

@@Ligatmarping I was just thinking of an integral as the reverse of a derivative. I forgot that it is also the area beneath a curve. Thanks! That helped.

It’s integral evaluation short-hand.

@@NotoriousSRG but then why is the integral sign (S) used together with |?

Taking the time to input series into Wolfram Alpha and dutifully report. We all gained immense insight into the solution process due to your contribution.

@@MyOneFiftiethOfADollar I can sum your cynicism as your dollars diminish from x= 1/50 to zero.

A "tool" which is usually useful is to try to solve something changing those limits carelessly and, if it turns out to go to the solution, verify then that the change is ok, which happens in a lot of cases.

@@Ligatmarping your quote symbols enveloping tool is noted and appreciated. I love this channel and learn a lot, but sooo much of the stuff is not in the spirit of discovery, BUT more how to use unmotivated tricks after one already knows the solution.

@@MyOneFiftiethOfADollar first sorry if my english isn't accurate... I hope to express the following properly. I'm a math teacher at the university of Buenos Aires and I get that questioning a lot when showing solutions. "Where does that come from?". It is hard to draw the line of what is a trick and what is not. I agree that just seeing the solution does not motivate the trick, but a lot of the techniques to solve thing when investigating a topic, come from a "seemingly useless trick" which you saw in the solution of a particular case. What I do prefer, is to let someone "find" the trick or at least think about the problem some time, so that you afterwards kinda appreciate what's the idea on the trick and which step it helps to solve. I usually speak normally in english but this is something a little complicated to express, so sorry if I didn't manage to show my thoughts about it.

true

I don't see any problems other than a lot of unused variables and redundant code

They are integrals of the same function, and share a limit of integration (1/2). Thinking of the integral as the area under a curve, it's like saying two areas under the curve that border each other. So, we can add those areas together and descirbe them as one big area, all in one piece (aka as just one big integral).

For any integrable f(x), the integral from a to b of f(x) + the integral from b to c of f(x) is equal to the integral from a to c of f(x). This follows immediately from the linearity of integration, or you can get it from FTOC. From FTOC: since f(x) is integrable, there exists an antiderivative F(x) such that F'(x) = f(x). The first integral is defined as F(b) - F(a) by FTOC; similarly, the second integral is defined as F(c) - F(b) by FTOC. Adding the two together gives F(b) - F(a) + F(c) - F(b). Canceling the + and - terms of F(b) yields F(c) - F(a). Again, by FTOC, this is nothing more than the integral from a to c of F'(x), and since F'(x) = f(x), this is the integral from a to c of f(x), as was to be shown.

Ah, thanks for the replies. I understand it now. It has been quite some time since I did any integration 😅

First, let u=pi/x. The limit is then lim(u -> 0, pi / (u * tan(pi/2 - u))). Recall that tan((pi/2)-x) = cot(x). So, we have lim(u -> 0, pi * tan(u)/u) = pi * lim(u -> 0, sin(u)/u) / lim(u -> 0, cos(u)) = pi * 1 * 1 = pi What gives it away is knowing trig identities and the sinx/x or tanx/x limit.

There’s a link in the description to a form to suggest problems. Have you tried that?

No. An exact solution is better than an approximation

do you mean the third centered formula on the second page?

It's a case of Li2(X)+Li2(1-x).