Another digestable approach to the bassel problem perfectly explained. Good job

I love all Apostol books! I learn most of my first math topics thanks to him! Great video.

I just liked the video and subscribed: I absolutely love this kind of derivations! Excellent job 👍

10:22 The top equation suggests that u=1-sin^2(theta)=cos^2(theta) This contradicts the second equation u=1-cos^2(theta) Which both need to be satisfied by the same substitution to do the substitution also cos(2theta)=cos^2(theta)-sin^2(theta)=2cos^2-1=1-2sin^2 not 1-sin^2(theta) Also also if u=cos(2theta) du=-2sin(2theta) Some of this is kind of pedantic but this stuff is important.

Absolutely amazing!

Apostol's approach is nice. If we want the simplest way to solve the Basel problem , it is the Fourier series.

If you look at what Mengoli actually wrote, he did not really use the alternating series directly. Instead, he considered two sequences of numbers, one consisting of the sums over 1/k with k going from n to 2n-1, and the other consisting of the sums over 1/k with k going from n+1 to 2n. The first he called the hyperlogarithms, the second the hypologarithms; and then he argued that the hyperlogarithms are descending, whereas the hypologarithms are ascending, and that the hyperlogarithms are always greater than the hypologarithms. Hence the two sequences converge to a common limit, and he argued that this limit is the logarithm of 2 (or actually, of the "double ratio"). The partial sums of the alternating series over 1/k consist of the hyperlogarithms and hypologarithms by turns, so his argument amounts to providing that this alternating series converges to ln(2). But he never actually talked about the alternating series.

I actually love this video.❤

as some may have noted, 10:40 should just've been u = cosθ, since tan(½θ) = ±√[(1-cosθ)/(1+cosθ)] though yes i comprehend the motivation, to annihilate the term tan⁻¹, but what you obtained was 1-sin²θ=cos²θ-1 yet the actual is √[(1-u)/(1+u)] = (1-cosθ)/sinθ again, since (1-cosθ)/sinθ = tan½θ Still a very well done vid. I remember stumbling upon this approach for the first time back when i was still self-learning calc using stewart's, in which he used the sub x=(u-v)/√2 and y=(u+v)/√2

Thank you Sir! This is a wonderful presentation of a wonderful mathematical gem! Gems like this show, beyond a shadow of a doubt, that Science in general and Mathematics in special are ...wayyy better than... sex! Even though chalk and chalkboard reign supreme, your electronic presentation of the problem is not only innovative but outright revolutionary as the formulas (in very pretty font!) are introduced gradually as they would be written by hand, allowing the information to be more readily absorbed by the mind. The speed of human understanding is the speed of speaking and writing as it is very well explained in a You Tube video by Professor Patrick Winston.

Imho this is a very good proof because of the following generalization used for prooving the famous Apery theorem concerning ζ(3): "... ζ(3) was named Apéry's constant after the French mathematician Roger Apéry, who proved in 1978 that it is an irrational number...The original proof is complex and hard... Beukers's simplified irrationality proof involves approximating the integrand of the known triple integral for ζ(3)..."

Sehr gut!

The alternating harmonic series can be made to have a value of -ln(3) if one merely rearranges the series to have blocks of 1 positive followed by 36 negatives

For second integral, using h for hypothenuse, we have sqrt(1-u) = h sin T and sqrt(1+u) = h cos T. After squaring we get two expressions for u, u = 1 - h^2 [sin T]^2 = h^2 [cosT]^2 -1. So this gives h^2 = 2! Therefore, u = 1 - 2 [sin T]^2 and u = 2 [cos T]^2 - 1. Finally, adding we get the correct expression: 2u = 2 ([cos T]^2 - [sin T]^2) simplifying to u = cos (2T).

This is a nice proof indeed!

Nice. What was the motivation for the U,V substitution?

Beautiful video! I just got a question, at minute 10:38 there is the relationship "cos(2a)=1-sin²(a)", which, at least to my knowledge, is not valid. cos(2a) should be "cos²(a)-sin²(a)", which can be written as "1 - 2sin²(a)". Was this the relationship you meant to use? (I apologize for any spelling mistakes, english is not my mother tongue)

At 2:16 is it really valid to drag the sum out of the integral? The problem is, in order to be integrated element-wise, a functional series has to converge uniformly over the entire region of integration (including edges). In the presented case, however, the given series don't converge at (1, 1) even pointwise, not to mention uniform convergence. Am I missing something?

At 10:51 you dropped 2 from the argument. du = -2*sin(2*theta) dtheta

What software was used to produce this video please?

at 11:30, arccos of 1/2 is pi/3 and not pi/6, right??

At 10:54, shouldn't it be du = -2sin(2theta)dtheta ? The argument 2theta has to be included right?

6:00 result is right but the explanation is wrong. We can multiply by 2 not only because of the symmetry of the Integration area but also because of the symmetry of the integrand with respect to reflections across the u-axis!

5:20 i kinda didn't understood

9:10 You said 62 but it says 72. Doesn't really matter just wanted to comment

you lost me with the substitutions in the second integral. I’m not sure what all went wrong as you got to the right answer, but I don’t understand how you justified some of the steps. One example: if u=cos(2θ) then du=-2sin(2θ)dθ you neglected the 2θ again I can’t argue with the correct answer but some of the steps along the way are a little questionable. It seems to all work out in the end but I would like a more rigorous proof.

If u = 1 - sin²(θ), then du = -2sin(θ)cos(θ) = -sin(2θ) Also, cos(2θ) = 1 - 2sin²(θ) Also also, if you have u = cos(2θ), then du = -2 sin(2θ), but this isn't even the correct u, as cos(2θ) != 1 - sin²(θ)

arctan(tan(x)) is only x if x is between -π/2 and π/2, you didn't show that. Also tangent is with a soft g, why are you saying tanghent?

Tom Apostol

You can be before you solve; you can't be before you're going to solve since you a.) already announced that you're going to solve and b.) always were about to. OK?

Jesus! Since when cos(2t)=1-(sin(t))^2? LOL! you just need a substitution of u=cos(2t) for the second integral. Luckily the bullshit in the middle did not affect the final result.

This is cute, but does a rise to the level of being worthwhile to mention Apostol's heritage??

I believe tangent is pronounced "ˈtanjənt".