You can also calculate ζ(4) using Fourier series of f(x) = π^2 - x^2 on [-π, π], and then using Parseval's theorem which relates sum of squares of coefficients with the square of the original function. Parseval might be a cool idea for a future video.
Exactly. It is incorrect to give values for x in the Fourier form. Let's say I am a student of Mr. Penn. And I say "Mr. Penn, pi=0". -How, so? - Well, if I make Fourier series for f(x) =x and make x=pi, I get pi=sum(zeroes). So pi=0. -But why you make x=pi? -Because is exactly what you teach me in this video. Nothing more and nothing less. 😁😂
I paused at 4:20 and did the calculation myself. The difference was I took x=0 instead of x=pi at the evaluation step. It still works and I got the same answer in the end! However, there was some extra computation... Choosing x=0 turns the cos(nx) into 1 but it doesn't eliminate the (-1)^n . It's easy to fix by splitting the sum into pairs of adjacent odd and even values of n. This led me to a general formula for the alternating version, Sum(n = 1 to inf) of (-1)^n / n^(2k) splitting and re-indexing means it's equal to: Sum(n = 1 to inf) of ( -1/(2n-1)^(2k) + 1/(2n)^(2k) ) In other words, negative of the odd powers, plus the even powers. Set S = zeta(2k) E = just the even powers Factor out the 1/2^(2k) and you just get E = S / 2^(2k) Then the odd powers are S - E So the alternating version = -(S-E) + E = 2E - S = S ( 2 / 2^(2k) - 1) = S * (2^(1-2k) - 1) In other words AlternatingZeta(2k) = Zeta(2k) * (-1 + 1/2^(2k-1)) For example, The Basel problem k=1 is zeta(2) is pi^2 over 6 The alternating version (-1)^n / n^2 Sums to: pi^2 / 6 times (-1 + 1/2^1) or -pi^2 over 12 The alternating version of zeta(4) is -7/8 times pi^4 over 90 Because -1 + 1/8
There's a beautiful proof using Eisenstein series for this, you can find guided exercises for the calculation in Diamond and Shurman's modular forms book. Of course it would be amazing to get a video on the topic too. Btw this result is not just a curiosity, but useful for calculating q-series of modular forms and it even connects Bernoulli numbers and L-functions to algebraic K-theory
The easiest way I found to find zeta(2n) without some sort of formula are the polygamma functions and their reflection formulae. It does require you to take many derivatives though.
It's also possible to obtain ζ(4) Euler's way by "expanding" the 5th order coefficient of the product sin(x)=x*(1-x^2/pi^2)*(1-x^2/(4*pi^2))*... wich is also x^5/120 (taylor serie)
Another way: we can equate the coefficient of x^4 in the traditional method of finding zeta(2) using the idea: roots of sin x/x. So now, zeta(4)=(zeta(2))^2-2*. So, zeta(4)=pi^4(1/36-1/60)= pi^4/90.
I belive in Introduction to the Analysis of the Infinities, Euler works out at least up to ζ(12). He uses a different method relating coefficients of polynomials (and by limits, taylor series) with symmetric functions of the roots.
I would love to see a video on the connection between the Riemann zeta function and the Euler-Mascheroni constant gamma. I've been fascinated with the Riemann zeta function and the Riemann hypothesis for years, and a certain set of videos by 3Blue1Brown over the last decade has only deepened that obsession, but I don't have the depth of formal math education I would like to have.
By using repeated integration by parts, this can be generalized to the Fourier series of x^(2n) on [-pi, pi] and it yields a recursive formula for zeta(2n) in terms of all previous zeta(2k) values. If we define z_n to be the coefficient of pi^(2n) in zeta(2n), we get a recursive formula for a sequence of rational numbers related to the Bernoulli numbers.
It would be interesting to focus on the quick resolution of that bn integral, just by assessing the parity of the functions. Would you put a focus on this topic in one of your next videos and why it is so? Thanks!
It's not very much to elaborate on though: For an odd function f(x), we have f(x) = -f(-x). Hence, if you integrate over a symmetric interval around the origin, such as [-π,π], it is the same thing as integrating f(x)-f(x) over [0,π], which obviously is equal to zero.
@@colonelburak2906 It's also important to note that the integral must exist, aka it doesn't diverge. For example the integral [-1; 1] of 1/x dx is not equal to 0 because it actually diverges, even though 1/x is odd.
@@xavierwainwright8799 True indeed. One should always be careful to check the domains of functions, especially when integrating and differentiating. Once I asked my students to differentiate log(-sin^2(x)) for x in R, just to see whether they checked if the domain is non-empty in the first place... (They didn't.)
It doesn't fail, the problem is that it just gives you the same information as x^4; it gives an equation with zeta(2) and zeta(4). It's essentially because the opposite terms cancel out when you do the definite integrals, as x^3 is an odd function. So no polynomials give you information about the odd values of zeta.
@@stanleydodds9 nj wildburger (not sure of spelling) in a video explains that ζ(3) having a non-closed form was published by an French Engineer, not a Mathematician, LOL☺A tempting conjecture would be for all positive integers n > 3, ζ(n) lacking a closed form.
@@ojas3464 are you talking about Apery's proof that zeta(3) is irrational? I don't know of anything about a "closed form". I'm pretty sure it's unknown if it has a closed form.
@@stanleydodds9 Thanks for your time replying. I seem to have confused between the irrationality of ζ(3) and existence / nonexistence of a closed form for ζ(3)
@@ojas3464 Actually it's an open question whether a closed form/solution for zeta(3), and really for any odd number can be found. I remember reading about it in Axler's "Measure, Integration and Real Analysis" in the Fourier Analysis chapter.
@@zeravam I'm aware there are no exact values known so far, but afaik it's not proven yet if it is really impossible to calculate them in a closed form. Maybe it's possible though, just not solved by anyone yet. As far as I'm up to date, it's still an open question.
If you use x=0 you do not get an extra factor of (-1)^n that you can use to cancel out the (-1)^n that's already there. That means you end up with an expression involving sum (-1)^n / n^4, which isn't what you want.
Choosing x = 0 is not wrong, but it does require more work. [More specifically, the same trick that is used to quickly find the sum of (-1)^n/n^2 if you know the sum of 1/n^2.]
👍Solutions to S L Loney, Analytical trigonometry compares coefficients in Taylor Expansion, with Product Expansion. Ahlfors (not sure of the author) uses Formal Power Series. Would be nice to have algorithm comparisons on execution difficulties for solving a given problem☺
The thing is, thinking about pi this way (in terms of circles) is quite limiting on what you can "intuitively" understand. It's better to think about pi as being a constant that fundamentally comes from the roots and period of sine, cosine, and exp. In analysis, sine and cosine are not defined by circles, angles, etc.; it's the other way around. sine and cosine come from the exp function, and are special because of (and defined by) their power series and/or their derivatives.
Very good Michael. Why stop there! ζ(8) = ∑ 1/n^8 = π^8/9450 which gives π ~ 3.14153 adding first two terms only which is correct to 4dp Ok I cheated I looked it up in Wolfram.😏
Is ok to give the value of x, one of the interval extremety? If we make the Fourier for f(x)=x, we get: x=2*sum [(-1)^(n+1)*(1/n)*sin(nx)] sin(n*pi) is always 0. So if we make x=pi we get pi=2*sum(0)=0 So, pi=0?! In your case is just a luck that for X=pi is the correct result. I think is more correct using Fourier and Parseval Theorem.
The periodic extension of f(x) = x on [-pi, pi] is not continuous at the odd multiples of pi, and then it is known that the Fourier series converges to the average of the two endpoints, I.e (pi + (-pi))/2 = 0. So fortunately math is not broken… The periodic extension of g(x) = x^4 on [-pi, pi] is continuous (since g is even) so the fact that equality holds for x = pi is not just a coincidence. Also, you could have instead plugged in x = 0, which would also give the same result but requiring a little bit more work. There is no “more correct” way of evaluating zeta(4); the argument in the video is completely valid.
@@joelganesh8920 Yes it is. Parseval's theorem for the function f(x)=x^2. There no longer exists any ambiguity related to the convergence of the associated Fourier series. Ok, is not a coincidence that is ok for X=pi, but this must be explained making the extension of the Fourier series on reunion of intervals [k*pi, (k+1)*pi], where the series for x^4 is continuous in every k*pi.
I wonder if this method is also working for all zeta(n) where n >= 2. like if we can find the Fourier transform of some polynomial of degree m in [-\pi,\pi], in which the sum of the reciprocals of the n-th power appears, then one can calculate zeta(n) using the same procedure shown in the video.
Yes, basically. If the left and right endpoint had different values (but still continuous in (-pi,+pi)) then it would still converge to the average of those two values. en.m.wikipedia.org/wiki/Convergence_of_Fourier_series
great job！ but it still can't calculate zeta(3) and the series of zeta(4)/[zeta(2)^2] means what ？ 我只是業餘數學愛好者 數學的有趣 就是在於滿足 數學愛好者們對於"calculation"的需求 每個喜歡發表"數學研究成果"的"作者" 都是希望 對數學有興趣的閱覽者 能對該項目"產生興趣" 但語言的隔閡 造成了很多"高等數學內容"失去 很多 原本有興趣的閱覽者
There's a mistake somewhere in calculating a_0, because of the extra factor of 2. Edit: as Star Fox pointed out, my observation was wrong and there are no errors.
And that's a good what? AND THAT'S A GOOD WHAT PROFESSOR?!?
We may never know
I'm very anxious about this, too
"The rest of the outro is left as an exercise for the interested viewer"
Legend has it he didn’t stop and solved the Riemann hypothesis.
@@ericmccormick1639 "That's a good place to start solving all remaining millennium problems onto the next board"
1:45 "Little Twiddle" - thanks! Until now, I did not know the technical term for the "~" symbol. Always a good day when you learn something new!
Sometimes we call this 'tilde' as well. Same goes with latex.
You can also calculate ζ(4) using Fourier series of f(x) = π^2 - x^2 on [-π, π], and then using Parseval's theorem which relates sum of squares of coefficients with the square of the original function. Parseval might be a cool idea for a future video.
also works with f(x)=x² on [-π, π]
Moreover, it is not necessary to know the sum of the series ∑ (1; ∞)1/n^2.
You had my curiosity, but now you have my attention!
Exactly. It is incorrect to give values for x in the Fourier form.
Let's say I am a student of Mr. Penn.
And I say "Mr. Penn, pi=0".
-How, so?
- Well, if I make Fourier series for f(x) =x and make x=pi, I get pi=sum(zeroes). So pi=0.
-But why you make x=pi?
-Because is exactly what you teach me in this video. Nothing more and nothing less. 😁😂
Why would anyone think of the Fourier series at all to.solve this problem..some infinite series that is related yes but why this one?
You know you're a math nerd when you know this value by heart.
Damn I didn't 😞
@@agrajyadav2951 That's ok, just remember it from now on. 😉
I love the animation, it's definitely a nice fresh take! Great vid as always:)
I have used Michael's approach to find a recursive formula for zeta(2*m), where m is any positive integer. Thank you Michael!
I'll never know if that was a good place to stop or not. The agony.
Awesome edits! The resulting pace of the video feels great
I paused at 4:20 and did the calculation myself. The difference was I took x=0 instead of x=pi at the evaluation step. It still works and I got the same answer in the end! However, there was some extra computation...
Choosing x=0 turns the cos(nx) into 1 but it doesn't eliminate the (-1)^n . It's easy to fix by splitting the sum into pairs of adjacent odd and even values of n.
This led me to a general formula for the alternating version,
Sum(n = 1 to inf) of (-1)^n / n^(2k)
splitting and re-indexing means it's equal to:
Sum(n = 1 to inf) of ( -1/(2n-1)^(2k) + 1/(2n)^(2k) )
In other words, negative of the odd powers, plus the even powers.
Set S = zeta(2k)
E = just the even powers
Factor out the 1/2^(2k) and you just get
E = S / 2^(2k)
Then the odd powers are S - E
So the alternating version = -(S-E) + E = 2E - S = S ( 2 / 2^(2k) - 1)
= S * (2^(1-2k) - 1)
In other words
AlternatingZeta(2k) = Zeta(2k) * (-1 + 1/2^(2k-1))
For example,
The Basel problem k=1 is zeta(2) is pi^2 over 6
The alternating version (-1)^n / n^2
Sums to: pi^2 / 6 times (-1 + 1/2^1) or -pi^2 over 12
The alternating version of zeta(4) is -7/8 times pi^4 over 90
Because -1 + 1/8
How in the hell do you people make this seem like common sense 😞
The good place to stop was obviously before "a good place to stop".
The good place to stop is in your heart
and that's a good
12:51
And tha's a good...-----
I love the new animations throughout the video.
@Michael: Could you please do a video on the connection between the values of zeta(2n) and the Bernoulli numbers?
There's a beautiful proof using Eisenstein series for this, you can find guided exercises for the calculation in Diamond and Shurman's modular forms book. Of course it would be amazing to get a video on the topic too.
Btw this result is not just a curiosity, but useful for calculating q-series of modular forms and it even connects Bernoulli numbers and L-functions to algebraic K-theory
no idea why, but blackpenredpen's picture popping up made me chuckle. great video as always!
The easiest way I found to find zeta(2n) without some sort of formula are the polygamma functions and their reflection formulae. It does require you to take many derivatives though.
polygamma function ??
@@angelmendez-rivera351 digamma, trigamma, etc...
Wonderful.
Now could you please do zeta (3)?
And that's a good
It's also possible to obtain ζ(4) Euler's way by "expanding" the 5th order coefficient of the product sin(x)=x*(1-x^2/pi^2)*(1-x^2/(4*pi^2))*... wich is also x^5/120 (taylor serie)
this method can be used to find higher even numbers of the zeta function.
Another way: we can equate the coefficient of x^4 in the traditional method of finding zeta(2) using the idea: roots of sin x/x. So now, zeta(4)=(zeta(2))^2-2*. So, zeta(4)=pi^4(1/36-1/60)= pi^4/90.
Thanks.
I belive in Introduction to the Analysis of the Infinities, Euler works out at least up to ζ(12). He uses a different method relating coefficients of polynomials (and by limits, taylor series) with symmetric functions of the roots.
I would love to see a video on the connection between the Riemann zeta function and the Euler-Mascheroni constant gamma. I've been fascinated with the Riemann zeta function and the Riemann hypothesis for years, and a certain set of videos by 3Blue1Brown over the last decade has only deepened that obsession, but I don't have the depth of formal math education I would like to have.
By using repeated integration by parts, this can be generalized to the Fourier series of x^(2n) on [-pi, pi] and it yields a recursive formula for zeta(2n) in terms of all previous zeta(2k) values. If we define z_n to be the coefficient of pi^(2n) in zeta(2n), we get a recursive formula for a sequence of rational numbers related to the Bernoulli numbers.
extraordinaire et assez bien présenté.
Merci pour ces belles perles.
It would be interesting to focus on the quick resolution of that bn integral, just by assessing the parity of the functions. Would you put a focus on this topic in one of your next videos and why it is so? Thanks!
It's not very much to elaborate on though: For an odd function f(x), we have f(x) = -f(-x). Hence, if you integrate over a symmetric interval around the origin, such as [-π,π], it is the same thing as integrating f(x)-f(x) over [0,π], which obviously is equal to zero.
@@colonelburak2906 It's also important to note that the integral must exist, aka it doesn't diverge. For example the integral [-1; 1] of 1/x dx is not equal to 0 because it actually diverges, even though 1/x is odd.
Plot an odd function over a symmetric domain, you'll quickly see why the area under the curve over this domain is 0..
@@xavierwainwright8799 True indeed. One should always be careful to check the domains of functions, especially when integrating and differentiating.
Once I asked my students to differentiate log(-sin^2(x)) for x in R, just to see whether they checked if the domain is non-empty in the first place... (They didn't.)
Do the general case with Eisenstein series!
Thank you, professor!
That was fun.
Nice one!
What would be interesting is see where this method fails for f(x)=x^3. A video on that would be very interesting.
It doesn't fail, the problem is that it just gives you the same information as x^4; it gives an equation with zeta(2) and zeta(4).
It's essentially because the opposite terms cancel out when you do the definite integrals, as x^3 is an odd function. So no polynomials give you information about the odd values of zeta.
@@stanleydodds9 nj wildburger (not sure of spelling) in a video explains that ζ(3) having a non-closed form was published by an French Engineer, not a Mathematician, LOL☺A tempting conjecture would be for all positive integers n > 3, ζ(n) lacking a closed form.
@@ojas3464 are you talking about Apery's proof that zeta(3) is irrational? I don't know of anything about a "closed form". I'm pretty sure it's unknown if it has a closed form.
@@stanleydodds9 Thanks for your time replying. I seem to have confused between the irrationality of ζ(3) and existence / nonexistence of a closed form for ζ(3)
@@ojas3464 Actually it's an open question whether a closed form/solution for zeta(3), and really for any odd number can be found. I remember reading about it in Axler's "Measure, Integration and Real Analysis" in the Fourier Analysis chapter.
Much more interesting would be zeta(3): 😉
Thank you, Michael for your Video.
Great! Now, do with zeta(3)!
Are you being sarcastic, right? Sabes que para Z(3) no hay valor exacto, solo aproximado
ζ(-1) is easier
@@zeravam prove it
@@zeravam I'm aware there are no exact values known so far, but afaik it's not proven yet if it is really impossible to calculate them in a closed form. Maybe it's possible though, just not solved by anyone yet. As far as I'm up to date, it's still an open question.
Hi Dr.!
This is so cool!
Thank you! ^.^
I did a talk the day before this came out for a closed form formula of zeta(2n)
The fourier transfor was correct over the -pi to +pi - how did you leap to minus infinity to +infinity required for the zeta ?
At 10:02 why choose x=pi rather than x=0?
Is there something 'wrong' about choosing x=0??
If you use x=0 you do not get an extra factor of (-1)^n that you can use to cancel out the (-1)^n that's already there. That means you end up with an expression involving sum (-1)^n / n^4, which isn't what you want.
Choosing x = 0 is not wrong, but it does require more work. [More specifically, the same trick that is used to quickly find the sum of (-1)^n/n^2 if you know the sum of 1/n^2.]
@@JauntyAngle Thank you 🙂
Outro -> "and" for n -> infinity.
It was, in fact, not a good place to stop.
Love it!!
👍Solutions to S L Loney, Analytical trigonometry compares coefficients in Taylor Expansion, with Product Expansion. Ahlfors (not sure of the author) uses Formal Power Series. Would be nice to have algorithm comparisons on execution difficulties for solving a given problem☺
I watched the video, but I kept going because I wasn't sure where a good place to stop was.
I remember this from Fourier analysis. Why does this fail for odd values again?
Values of ζ(2n) are known and there are Bernoulli numbers and π^{2n}
It seems that you may recover a recursive formula for zeta(2n)?
That was pretty neat! Can ζ(n) be determined this way by recursion for all other even n?
Why is pi showing up everywhere? Where is the circle in this infinite sum?
Go watch 3blue1brown's video on the basel problem (p²/6) it explains this result quite well
The real circle is the friends we made along the way.
@@AymenElassri I know that video (was really nice). But as far as I know, he has not done a video on the value of zeta(4) so far - did he?
The thing is, thinking about pi this way (in terms of circles) is quite limiting on what you can "intuitively" understand. It's better to think about pi as being a constant that fundamentally comes from the roots and period of sine, cosine, and exp. In analysis, sine and cosine are not defined by circles, angles, etc.; it's the other way around. sine and cosine come from the exp function, and are special because of (and defined by) their power series and/or their derivatives.
@@stanleydodds9 Thanks, I get it, but then why sum of reciprocals of integers is related to sines and cosines which are never integers?
12:50
I'm pretty sure he only said, "That's a g"
Yes. It was like a coitus interruptus. XD
We never knew if this was a good place to st...
Very good Michael. Why stop there!
ζ(8) = ∑ 1/n^8 = π^8/9450
which gives π ~ 3.14153 adding first two terms only which is correct to 4dp
Ok I cheated I looked it up in Wolfram.😏
@BiBenBap did exactly this problem in a video yesterday. Excellent coincidence.
12:51 😬🤭
What is a good book on doing Fourier series of functions like these?
That was a nice place to stop. Nice video, thank you.
Absolutely amazing video
incomplete outro 12:51
When he says "that's a good place to stop", he really means it! 🤣
Is ok to give the value of x, one of the interval extremety?
If we make the Fourier for f(x)=x, we get:
x=2*sum [(-1)^(n+1)*(1/n)*sin(nx)]
sin(n*pi) is always 0.
So if we make x=pi we get
pi=2*sum(0)=0
So, pi=0?!
In your case is just a luck that for X=pi is the correct result.
I think is more correct using Fourier and Parseval Theorem.
The periodic extension of f(x) = x on [-pi, pi] is not continuous at the odd multiples of pi, and then it is known that the Fourier series converges to the average of the two endpoints, I.e (pi + (-pi))/2 = 0. So fortunately math is not broken… The periodic extension of g(x) = x^4 on [-pi, pi] is continuous (since g is even) so the fact that equality holds for x = pi is not just a coincidence. Also, you could have instead plugged in x = 0, which would also give the same result but requiring a little bit more work. There is no “more correct” way of evaluating zeta(4); the argument in the video is completely valid.
@@joelganesh8920 Yes it is. Parseval's theorem for the function f(x)=x^2. There no longer exists any ambiguity related to the convergence of the associated Fourier series.
Ok, is not a coincidence that is ok for X=pi, but this must be explained making the extension of the Fourier series on reunion of intervals [k*pi, (k+1)*pi], where the series for x^4 is continuous in every k*pi.
What is the reasoning for setting x to pi?
RJ
Is Zeta(p) proportional to pi^p for any integer p>0 ?
how did I get here I can't even do long division
great video it looks like we can do this for all ζ(n) especially for n even but I wonder why it is not going to work for n odd
Try it with x³ for yourself, you'll see where it fails. ;)
any ideas how to find zeta(3) ?
TanQ!
Um, isn't x^4 an even function only when the value of the x-variable itself is even?
No. Apparently you don't know what "even function" actually means? It means that f(-x) = f(x) for all x for which f is defined.
What's the sin or cos of senpai?
;^}
Nice
I wonder if this method is also working for all zeta(n) where n >= 2. like if we can find the Fourier transform of some polynomial of degree m in [-\pi,\pi], in which the sum of the reciprocals of the n-th power appears, then one can calculate zeta(n) using the same procedure shown in the video.
Similar methods should work for even n. Odd n has been very tough -- we have no idea about any closed form for zeta(3).
Q: The Fourier series converges point-wise on [-\pi,\pi], and not just (-\pi, \pi) because (-\pi)^{4}=\pi^{4}?
Yes, basically. If the left and right endpoint had different values (but still continuous in (-pi,+pi)) then it would still converge to the average of those two values.
en.m.wikipedia.org/wiki/Convergence_of_Fourier_series
@@landsgevaer Thanks Dave!
Fire your editor for cutting the ending, mr. Penn 😁
Dear teacher, would you like to solve this problem for me please?
Find all number of factor of N that are square integer such that
N= 5!8!9!
Brilliant video, but maybe credit should have been given to Euler, who first found this closed form.
great job！
but it still can't calculate zeta(3)
and the series of zeta(4)/[zeta(2)^2]
means what ？
我只是業餘數學愛好者
數學的有趣
就是在於滿足 數學愛好者們對於"calculation"的需求
每個喜歡發表"數學研究成果"的"作者"
都是希望 對數學有興趣的閱覽者
能對該項目"產生興趣"
但語言的隔閡 造成了很多"高等數學內容"失去
很多 原本有興趣的閱覽者
Hi,
ok, great!
12:53 : that's a good.... what?
There's a mistake somewhere in calculating a_0, because of the extra factor of 2.
Edit: as Star Fox pointed out, my observation was wrong and there are no errors.
No mistake : check again the result of the integral, it's 2π^4/5. The result he plugged in on the next board is π^4/5, hence dividing a0 by 2.
@@MasterChakra7 ah sorry about that. I did think it was weird that there wasn't any mistake that I could spot yet the result was different.
my 2cents approach, S 1/n4 = (S 1/n2)^2 = (π2/6)^2 = π4/90 obviously 😗 😎
No. A sum of squares of terms is *not* equal to the square of the sum of those terms. Also, the square of π²/6 is *not* π⁴/90.
If you're trying to write the pi symbol squared, you need "^" between it and the
2, for example.
a good what...?
a good train?
a good neutron star?
a good move?
.... I'm so confused right now
I don’t understand why you can approximate x⁴ by this formula?