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.
11:06 Also the -2sin(2theta) dtheta seemed to be replaced with -2dtheta without reason. Also bounds should be half the arccos of 1 and 2 if cos(2*theta)=1 &1/2 since 2*theta=arccos(1) & arccos (1/2) and so theta=1/2*arccos(1),arccos(1/2)
@@thepathintegrator No, the equations at 10:22 just don't make sense. The top equation says u = 1 - sin^2(theta) which equals cos^2(theta). The bottom equation says u = cos^2(theta) - 1. So u = u - 1?
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.
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?
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
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)..."
@@thepathintegrator It is remarkable that the duble integral can be calculated and this leads to the solution of Basel's problem and in the same time the value of the corresponding triple integral it is not known
Yes one might think it cannot be that hard, just do it like the double Integral but no. The zeta function indeed is a very weird thing which makes it so interesting. What is so special about the uneven numbers?
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)
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
@@thepathintegrator The alternating harmonic series is conditionally convergent, it is a well known fact from Real Analysis that If a series is conditionally convergent, then: 1) re-arrangements of the terms do not necessarily sum to the same values 2) for any real number r, there exist re-arrangements such that the series sums to r 3) there exist re-arrangements such that the series diverges to positive/negative infinity
@@thepathintegrator I read about it once, regarding the exact value of -ln(3) being the rearrangement of 1 positive term followed by 36 negative terms There is an old theorem from Riemann that says there is a _simple_ rearrangement of the alternating harmonic series to get any exact value of ln(sqrt(r)), where r is a rational number. So r = 1/9 shows, via this theorem, that -ln3 is a simple rearrangement of the alternating harmonic series
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.
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).
Most interestingly I couldn’t find a real motivation for this change of coordinates. I’m guessing that Tom Apostol was playing around with the integral and found this to be the best way. Let me know if you (anyone who reads this) finds something about it! :)
@@birdbeakbeardneck3617 yep, thats how I thought about it too: 2xy is sort of dual to x^2 + y^2 with respect to u = x + y, v = x - y, and since generally it is a good idea to take advantages of symmetries, this substitution would be (and of course, is) a sensible step in this direction
@@thepathintegrator It is what works; asking what the motivation is is like asking "why is he even doing math?" This is how real math is done; people experiment and find stuff. Then, once you do all this hard work, you only keep the single line of reasoning that provides the elegant answer, thus making people think math is done via single line of reasoning found via intuition alone People will then say "Wow, what intuition motivated such an answer?!?!" Whilst they are totally blind to the actual process.
Yes, this video contains a few minor errors, as others have noted before. If u = cos(2ϴ) (what had been defined before), then the bounds of the integral at the upper right at 11:30 should be ϴ₁ = arccos(u₁)/2 = arccos(1/2) / 2 = π/3/2 = π/6 ϴ₂ = arccos(u₂)/2 = arccos(1) / 2 = 0 And at 10:45 bottom right this should of course read cos(2ϴ) = 1 - 2sin²(ϴ) so that 1 - u = 2sin²(ϴ) 1 + u = 2cos²(ϴ) and du/dϴ = -2sin(2ϴ), rather than -2sin(ϴ) at 10:50.
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!
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.
I’m sorry I lost you at the second integral. You’re right that I dropped the 2. Concerning the substitution I tried to give an explanation of why Tom Apostol back then chose it, but it didn’t quite work out unfortunately. So just stay maybe with the fact that he chose cos(2theta) to be the best substitution and then you get the solution
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)
@@thepathintegrator Well, your end result is correct, so no, the integral value doesn't change. I haven't worked it out, since I was doing this in my head. I need to work this out on paper to see what happens
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.
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?
Another digestable approach to the bassel problem perfectly explained. Good job
Thank you so much I’m happy you liked it ☺️🍀
What is bassel, please?
*Basel, as in the place Basel in Switzerland. ;)
... and a great demonstration that 1 = 0 at 10:26.
@@chrisl442lmao I just noticed
I love all Apostol books! I learn most of my first math topics thanks to him! Great video.
How amazing 🤩 he was a very good teacher. Thanks for your appreciation:)
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.
11:06
Also the -2sin(2theta) dtheta seemed to be replaced with -2dtheta without reason. Also bounds should be half the arccos of 1 and 2 if cos(2*theta)=1 &1/2 since 2*theta=arccos(1) & arccos (1/2) and so theta=1/2*arccos(1),arccos(1/2)
You mean not using the same substitution variable both times?
In fact the substitution u=cos(2θ) may be pulled directly from the half angle formula for tangent, i.e. tan(θ/2)=sqrt((1-cosθ)/(1+cosθ))
@@thepathintegrator No, the equations at 10:22 just don't make sense. The top equation says u = 1 - sin^2(theta) which equals cos^2(theta). The bottom equation says u = cos^2(theta) - 1. So u = u - 1?
Indeed, well noted, though it can easily be resolved. Also we should all agree that otherwise this video is a job well done.
I just liked the video and subscribed: I absolutely love this kind of derivations! Excellent job 👍
Thank you so much and welcome on board fellow path integrator :)
Apostol's approach is nice. If we want the simplest way to solve the Basel problem , it is the Fourier series.
Yes but that would require knowledge of Fourier analysis 🧐
Absolutely amazing!
Thank you so much 😊
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.
Oh wow didn’t know about the history so much thanks for the input 🙏 super interesting
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?
This is possible because of the Beppo Levi Theorem.
I actually love this video.❤
Thank you so much :)
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
Thanks for the tip and also I appreciate your compliment a lot :) I’m glad you like my work
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)..."
Yes I also found that out during my research it’s absolutely amazing! I’m currently also trying to understand the “simpler” proof by Beukers properly
@@thepathintegrator It is remarkable that the duble integral can be calculated and this leads to the solution of Basel's problem and in the same time the value of the corresponding triple integral it is not known
Yes one might think it cannot be that hard, just do it like the double Integral but no. The zeta function indeed is a very weird thing which makes it so interesting. What is so special about the uneven numbers?
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)
Yes, there's a mistake there
@alebisello9106 thank you so much for your appreciation :) yes you’re right this was a mistake in the video I’m very sorry :/
What software was used to produce this video please?
I am using keynote and notability :)
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
Interesting where did you find out about it?
@@thepathintegrator The alternating harmonic series is conditionally convergent, it is a well known fact from Real Analysis that
If a series is conditionally convergent, then:
1) re-arrangements of the terms do not necessarily sum to the same values
2) for any real number r, there exist re-arrangements such that the series sums to r
3) there exist re-arrangements such that the series diverges to positive/negative infinity
@@thepathintegrator I read about it once, regarding the exact value of -ln(3) being the rearrangement of 1 positive term followed by 36 negative terms
There is an old theorem from Riemann that says there is a _simple_ rearrangement of the alternating harmonic series to get any exact value of ln(sqrt(r)), where r is a rational number.
So r = 1/9 shows, via this theorem, that -ln3 is a simple rearrangement of the alternating harmonic series
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.
Thank you so much for your very kind and heart warming words. I’m happy you like my way of sharing the beauty of maths ☺️
At 10:54, shouldn't it be du = -2sin(2theta)dtheta ? The argument 2theta has to be included right?
Yes true little mistake I’m sorry 😞
@@thepathintegrator its fine bro it happens sometimes, ive had the same mistake many times before
Thanks bro :) man signs and factors drop so easily ^^
Sehr gut!
Vielen Dank! :)
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).
hypotenuse - why "TH"?
Good explanation!
Nice. What was the motivation for the U,V substitution?
Most interestingly I couldn’t find a real motivation for this change of coordinates. I’m guessing that Tom Apostol was playing around with the integral and found this to be the best way. Let me know if you (anyone who reads this) finds something about it! :)
There is always some art involved in solving integrals 😉
I think he wanted to introduce saquares(u+v times u-v) in the denominator which is knoz to hive arctan which gives pi
Pretty cool stuff
@@birdbeakbeardneck3617 yep, thats how I thought about it too: 2xy is sort of dual to x^2 + y^2 with respect to u = x + y, v = x - y, and since generally it is a good idea to take advantages of symmetries, this substitution would be (and of course, is) a sensible step in this direction
@@thepathintegrator It is what works; asking what the motivation is is like asking "why is he even doing math?" This is how real math is done; people experiment and find stuff. Then, once you do all this hard work, you only keep the single line of reasoning that provides the elegant answer, thus making people think math is done via single line of reasoning found via intuition alone
People will then say "Wow, what intuition motivated such an answer?!?!" Whilst they are totally blind to the actual process.
at 11:30, arccos of 1/2 is pi/3 and not pi/6, right??
Yes, this video contains a few minor errors, as others have noted before. If
u = cos(2ϴ) (what had been defined before),
then the bounds of the integral at the upper right at 11:30 should be
ϴ₁ = arccos(u₁)/2 = arccos(1/2) / 2 = π/3/2 = π/6
ϴ₂ = arccos(u₂)/2 = arccos(1) / 2 = 0
And at 10:45 bottom right this should of course read
cos(2ϴ) = 1 - 2sin²(ϴ)
so that
1 - u = 2sin²(ϴ)
1 + u = 2cos²(ϴ)
and
du/dϴ = -2sin(2ϴ), rather than -2sin(ϴ) at 10:50.
This is a nice proof indeed!
I do think so too, Tom Apostol really nailed it with this one ☝️
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!
Yes you’re right thanks for pointing that out!
At 10:51 you dropped 2 from the argument. du = -2*sin(2*theta) dtheta
Yeah true thanks for pointing this out :)
9:10 You said 62 but it says 72. Doesn't really matter just wanted to comment
Thanks man 🙏 luckily it wasn’t written wrongly
5:20 i kinda didn't understood
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.
I’m sorry I lost you at the second integral. You’re right that I dropped the 2. Concerning the substitution I tried to give an explanation of why Tom Apostol back then chose it, but it didn’t quite work out unfortunately. So just stay maybe with the fact that he chose cos(2theta) to be the best substitution and then you get the solution
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?
My English still has room for improvement
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)
Does the integral value change?
@@thepathintegrator Man, if you got the correct result using incorrect calculations, you are not a mathematician, you are a fortune teller.
@@thepathintegrator Well, your end result is correct, so no, the integral value doesn't change. I haven't worked it out, since I was doing this in my head. I need to work this out on paper to see what happens
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.
Lucky yeah
Tom Apostol
Thanks 🙏
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?
I’m sorry I don’t quite understand what you mean
This is cute, but does a rise to the level of being worthwhile to mention Apostol's heritage??
What do you mean?
I believe tangent is pronounced "ˈtanjənt".
Yeah infinite serious😂