1-1/4+1/9-1/16+... (zeta function & eta function)
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- เผยแพร่เมื่อ 7 ก.พ. 2025
- Eta function is the alternating version of the zeta function. We will figure out a connection between the Eta function and the Zeta function so we can evaluate 1-1/4+1/9-1/16+...
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Check out the sum of reciprocals of squares,
pi^2/6 by Max Z, • Proof by intuition don...
pi^2/6 by Dr. P, • Video
Next time find the nontrivial roots of the zeta function pls.
Nxn908xxx Yes we all want bprp to be a millionaire.
Lol
If u prove the Riemann hypothesis by disproving it with an example, u wont get a million dollars. U get the million if u have a generell prove for it.
@Mushushu My thought process is: if you find all of the nontrivial zeroes, and your definition of find includes having an exact value, then those exact values will either all have a real part of 1/2 or not, and you will already know this if your definition of finding includes knowing the exact values. Therefore, this would be sufficient to solve the Riemann Hypothesis I believe.
I have find one
The Riemann "theta" function.. lol
Also: me when physics teacher thinks I don't write time units 4:23
The annoying thing is that we have a product expansion for zeta(s) which works when re(s)>1: which doesn't work in the critical strip (0
As most of us know, the best way to solve an equation that has 0 as the subject is to factorise the other side
Very nice video! It is so astounding and cool how you can do algebraic manipulation of the zeta function minus the eta function in order to come up with a nice simplification of it. Keep up the good work making math videos!
Allen Minch thanks!!!!
Hi blackpenredpen, I’m glad you like my comment. I want to tell you that I am a 9th grade student and that I really like math; I am currently doing precalculus, and I can surely say I enjoy algebra 1 and 2. A year ago when I was doing algebra 2, I think I had a sense that precalculus is largely algebra 2 review, but I didn’t really have much exposure to calculus or the involvement of algebra 1 and 2 in calculus. I originally found your channel just as a cool math channel last summer, and I certainly have enjoyed your videos that don’t involve any calculus. But I have watched many calculus videos of yours as well, and they have given me a great foundation for doing calculus next year - they have made it very clear to me that algebra plays a huge role in calculus and have given me much more exposure to calculus than I had a year ago. At this point I definitely enjoy your calculus videos as well and have gotten a sense from them that I think I really will like calculus when I do it next year. So I am quite happy to enjoy interesting math videos on your channel and to have had great opportunities to be exposed to calculus through your channel. Keep up the good work in making these cool math videos!
You can use the Fourier series of x^2 to solve the alternating reciprocal of the squares problem in the beginning.
I was just today learning beta function and gamma function, though I need to study more about them. Now, I am learn zeta function and eta function.
Why so complicated? Take that zeta function at s=2, then multiply everything by 2*2^(-2). Then subtract that result with the original summation, and we are done
Dude you have an awesome channel & Im learning so much man. The logic behind some of the results we come to accept & the reasoning often times is beautifully simple, it just takes a lot of knowledge & experience to get things right. For example, many of your vids explore questions such as the i-th root of i and you demonstrate that using our familiar mathematical tools plus a bit of creativity allows us to tackle these strange problems. Well theyre actually not so strange problems, they are rather standard curiosities that arise naturally when new tools or concepts arrive & so at some point you wonder what this or that would evaluate to in order to test this new object & see what other discoveries & connections can be made.
I spent a long time practicing how to draw zeta after my math love was rekindled by this channel and similar ones. Not only did I want to draw it, but I wanted to make sure that, when written quickly, it couldn't be mistaken for anything else.
I also practiced how to write s and 5 where they can't get mixed up. 5 always had a flat top with a sharp corner, while the top of s is always curved. I wonder if o vs sigma vs 0 is going to show up, too!
Now practice the greek letter xi (ξ). There’s a meme where a person writes a random squiggle instead of the proper symbol.
I always write the random squiggle^^
GreenMeansGO I actually don't find xi to be hard. It's a backwards three with a loop on top and a small hook on the bottom.
Both of those two top series can be produced from this known result from Fourier series: sum of cos(k x) / k^2, k goes from 1 to infinity = (pi^2)/6 - (pi x)/2 + (x^2)/4.
"This is like 'n', but.... not." xD
ZETA - ETA= Z... Right? Anyone? Help!
Chris Sándor Kacsó You tried :D
Actually it is
ETA( Z-1)
Incorrect
C'est toujours agréable de vous suivre dans vos démonstrations
I actually found something like this myself but didn't find anything on the internet about it.
What I found was that for a sum like
1+1/2^s-1/3^s+1/4^s+1/5^s-1/6^s+... (here the minus sign repeats once every three terms)
The sum is equal to (1-2/3^s)zeta(s)
when the minus sign repeats every n-terms, the sum is equal to:
S=(1-2/n^s)zeta(s)
So for 1-1/2^2+1/3^2-1/4^2+1/5^2 ... = (1-1/2^(2-1))zeta(2)=1/2 pi^2/6=pi^2/12
What I found funny the first time I saw this was when I plugged in s=1 en n=2, I got
1-1/2+1/3-1/4+1/5-1/6+1/7-... = 0*inf using the above formula, but this turns out to be ln(2) as bprp showed in a recent video. This shows again that 0*inf can be anything you want.
Do you have a proof? I'd be quite interested c:
I know which function's better! (yay me)
ζeta lolll nice
Thank you! :D
@@eta3323 Shouldn't it be ζήτα?
@@he2he Zeta eta tau alpha?
I keep thinking, instead of using (and reusing) letters of various alphabets to represent variables and constants, we should learn how to write the actual name of the variable or constant _in Chinese characters._
That way nobody can mistakenly reach the conclusion that E = m(a^2+b^2). (not that anyone would make that particular mistake, but the point remains) It would also eliminate the common post-equation explanation "where E is energy, m is mass, a is acceleration, and b is the length of one of two perpendicular sides of a triangle" because that's already in the equation.
Apologies to the entire Sinosphere for my attempted example:
能量=的块质*光速^2
or maybe
能量=质块*光速^2
The point of using letters is to reduce the amount of time you spend writing down what you think, it's like not proving that ax + b = 0 x = -b/a if a 0 each time you use it. We know it, but the key is always to know why, what it represents. Not understanding what the equation means, is like not knowing it at all, and once it's known, it can surely be simplified by using letters to represent the things that are implied. It's how I like to see it ;) I might have made one or two mistakes btw, I'm still learning English !
If you subtract the eta function from the zeta function, you get a complex number!!! (zeta - eta = z)
eta(s)=(1-2^(1-s))*zeta (s)
Set: s=1
eta (1)=(1-2^(1-1))*zeta (1)
eta (1)=(1-1)*zeta (1)
eta(1)=0
Then: (1/1)-(1/2)+(1/3)-(1/4)+•••=0
Then: 1+1/3+1/5+1/7+•••=1/2+1/4+1/6+•••
Friend: What are you doing?
Me: Calculating values of zeta function for 2s.
Friend: Wow, such a complex math, isn´t it?
Me: Nah, it isn´t even real.
Hahaha
10:20
quick maffs
You also need to remember it works only if zeta of s is finite. For example for s=1 it doesn't work.
If s=1 it will be Zeta(1) - Eta(1)=Zeta(1). I think it is right, because we can see it in full equation (2*1/2+2*1/4+...)
But Eta(1) don’t equals 0 (??). Mystic?
Виктор Поляк It is right because both sides are equal to infinity. So no matter what eta(1) is, infinity -eta(1) is infinity so value of eta(1) is lost here. That's why you can't just say what is infinity-infinity, it could be anything. However eta (1) as defined be the series itself, is well defined, unique number. Besides this method is supposed to work the other way around usually, to approximate zeta using eta, since eta converges faster.
But if you apply this method for partial sums you can easily prove that eta (1) is log 2 and I think it's actually the easiest and most elegant proof.
How did *you* know what I was learning again?!
really?
Yah!
Technically... the product formula of the zeta function!
Oon Han make more videos lol
I just uploaded one yesterday
If you factoris the eta function in two parts you get the Gandi's series multiplite by the Zeta function.You want to prove that this is 1/2 times the Zeta function that means that the gandi's series should be 1/2 in our example, but the Gandi's series, which you can easily proof, does not converges. Gandi's series is an divergent series.(by the way the Geometrical series is a good way to proof that the Gandi's series does not converg)
wow.really enjoyed it
Yay, a bprp video! My day is made!
Hi there is a simple and easiest way to find the zeta function and i personally find it true with same formula and if (s) = 1/2 then also the result will be there
That's the easiest proof that zeta1 diverges. Eta(1)=-ln2 and when u apply the formula u will get 0*zeta1=-ln2
Yes except Eta(1) = ln(2). It's not negative.
Sorry I were wrong
Literally did that summation in my head.
Literally
Can you do a video on the value of the eta function at one? When you use the formula connecting the zeta and eta functions, you get an indeterminate form. But after using L'Hopitals rule, the new sum is hard to evaluate.
Good job .. 🤗🤗
I got that answer because I watched 3blue1brown and he proved that 1^2+3^2+5^2.... equals pi^2/8. Using this, I know the answer is pi^2/8-(pi^2/6-pi^2/8) which is pi^2/4-pi^2/6 which is pi^2/12.
And if you sum them? Do you get pi^2 / 4 then? Because then all the unevenpowers doubled and the even powers cansel out.
Classical scholar here: it's helpful to think of lower-case ZETA as a cursive form of capital ZETA, which is exactly like Roman Z.
Pretty cool, I love it
Wao sir.. Your explanation is absolutely awesome. 🤣👌👌👌👍👍👍👍 & love u 2 sir.
This was great! :)
Amazing!
At 6:39 if you have same base can’t you add the powers. Like 4=2*2 take one of the 2 and add their exponent or power in this case s. So 2 to the power s+s. When at 16 then 4*4 take one 4 and add their power?
How can I find the convergence of eta function for s?
We call z(s) the function zeta(s)*2^(1-s)
z(s) + eta(s) = zeta(s)
We proved that you can add functions by lexical-additivity
For all s in reals,
Pi(s) + Id(s) = Pi(s) + s(s) = Psi(s)
Psi(x) is the second Chebyshev function:
Psi(x) = sum_(p^k
Wow, my favorite, zeta! Is this vidio sign of vidios about primes? :)
vidiot!
While Shifting equations, number of terms of each equations are getting changed.
Infinites aren't single number.
Even if we consider Infinites to be compressed single values, then those properties should be evident in real values too, as Infinites are part of totality of the reality.
These types of mistakes leads us to wrong values for infinite series like: 1+2+..=(-1/12).
Please Refer:
www.quora.com/Does-infinity-contain-itself/answer/Sanoy-Samuel?ch=10&share=aca2f541&srid=p7ti4
awesome vid
Brilliant
Can you also make a video how to integrate?
Albert Einstein integrate....what?
blackpenredpen integration i don't know how that works
why would you use zeta(s) instead of zeta(x) if you only insert real values
Proof that Γ(n).Γ(1-n)=π.cosec(nπ)=π/sin(nπ)
Very cool.
Looks good. Now find a way to use this to solve the Riemann Hypothesis and you’ll be a millionaire!
So cool
Gr8...
Hey bprp I just wanted to ask you do you teach at UC Berkeley by any chance?
突然定義されたイータ関数
これ何に使うんですか！
趣味？
great
Please, upload for Riemann's hypothesis
Well I did get the correct answer of pi^2/12 but not like he did. I just guessed that it would be half of pi^2/6 because you're subtracting half the reciprocals instead of adding all of them. Was that just way too naive a way of getting the right answer?
Please solve for me the integral of sqrt of sinx dx.
So that means 1 - 2 + 3 - 4 + ... = 0.25, right?
This are s not 5 I love this guy
let S_0 = ( 1/1² + 1/2² + 1/3² + ... ) = (pi²/6) and let S_1 = ( 1/1² - 1/2² + 1/3² - ... )
let S_2 = ( 1/1² + 1/3² + 1/5² + ...) and let S_3 = ( 1/2² + 1/4² + 1/6² + ... )
i first split the sum into positive and negative partial sums:
S_1 = S_2 - S_3
S_3 is equal to a quarter of the original sum:
S_3 = S_0 / (2²) = S_0 / 4 = pi²/6 * (1/4) = pi² / 24
Also, S_2 + S_3 = S_0 ; So, S_2 + S_3 = pi² / 6
All in all, we have S_2 + (pi² / 24) = pi² / 6
So, S_2 = pi² / 8
finally, S_1 = S_2 - S_3 = (pi² / 8) - (pi² / 24) = pi² / 12
EDIT: ALSO, we really need LaTeX or something else for youtube comments.
COOL MATH BOY!!!!!
Hey bprp have you checked that an Indian (from Hyderabad )mathematician has claimed that he solved Riemann Hypothesis .I read that today in my newspaper.
What is the result of zeta(-1)? Is it -1/12 or not?
It is. Numberphile didn't just make that number up. If the real part of s is less than or equal to 1 then zeta(s) is not defined by an infinite sum but by analytic continuation. In other words, zeta(s) is the unique extension of the infinite sum to the entire complex plane (except s = 1) that is continuous, has continuous 1st derivative, continuous 2nd derivative, and continuous n'th derivative for all natural numbers n.
Soooooo cool
blackpenredpen can you please make a video about i factorial (i!) ??
i have a sugestion ,after any vid let us a problem to solve for the follow vid
0:40 I got them.
When you plugging s=1, we have 0 times infinity. It is not ln2.
😀 👍👌
Can you or someone else watching this (Pappa Flammy?) do a vid on what the zeta function actually is?
Second series converges to pi^2/12 I suppose. Huh, I was right.
Fourier series my boi
Pretty cool, isn't it?
Can't do the Zeta either. Worse is only the Xi
I can never write xi successfully...
Even if you writte down xi successfully no one will be able to tell
what is 'zeta either' ?
", too"?
Zeta vs eta, zeta wins😅😅😎😎
Can I learn this if I'm black?
Gotenks of course...lol 😂🤣
Sergio H it just that I have trouble learning math but I have a strong love for it
Jinger McBlabbersnitch :)))) thanks! This is the right teacher for me.
Gotenks bruh...LONG story short: when I came out of high school I didint know how to do BASIC BASIC EASY math. Eventually i went to a community college...and the professor made me fall in love w math even though I was in a pre algebra class....fast forward a year and a half and now I'm taking multivariable calculus and graph theory with linear algebra. I also work as a math tutor at my college! Since then math has been my passion...if I could do it then u can also!!! 😂😇😇😇
Of course it took a lot of work haha tho
Sergio H yes I want to have a master's degree in mathematics
Now show that when Re(s)=1 how the functions and all their derivatives are related.
😍😍
Could we figure out zeta(1) , zeta(3) …… ???
My favourite is ξ(-1)
I found the non trivial zeroes of the zeta function, however, this comment section is far too small to contain my proof.
semi awesomatic haha immitating fermat
If you take eta away from zeta you get z
neat
501th view, 51th like
Late by 57 minutes :(
hi
First....
Second
That''s easy to answer. Eta does not produce any Gundams where as ZETA produces Gundams.
I do not understand anything.
ok.