I have known this series as a higher mathematics problems but as soon as you switched the order of summation I got goosebumps just seeing how easily it can be solved
you're kidding, that's nuts dude. i love this. one of my favorite things is when a problem produces a completely nonintuitive result that you really have to dig to understand.
Its more interesting to get 1 actually... Since most infinite series are oscillatory in nature... To take a random infinite series and end up with a whole number is much more fascinating.
It's actually much easier for the last part to use the sum convergence test, since you just proved that the sum of zeta(s) - 1 converges that must mean the limit of individual terms goes to 0 and so zeta(s) goes to 1.
You can also try this for just even values of s in the Riemann Zeta function. And if you have done this, you can just subtract the value from 1 and you get the sum for odd values of s. I have done this in the past, try it, its really worth it!! (Btw: Michael Penn also did some great videos about sums of the zeta function using generating functions)
@@laxminarayanbhandari855 I mean first you sum zeta(2)-1+zeta(4)-1+zeta(6)-1… and then you take the value (S) this sum converges to and calculate 1-S and then what you get is zeta(3)-1+zeta(5)-1+zeta(7)-1…
@@User-gt1lu according to Wolfram Alpha, the second sum is divergent. I checked till 300 terms, but the sum doesn't go above 1/4, which is expected value according to your method. Don't know what is going on.
I tutor math. Have an honors precalc student that got this exact question (without mention of the Reimann zeta function) on a worksheet. I recognized it as the infinite sum of zetas immediately, and I'm thinking "how the heck is this kid, who JUST got introduced to sequences and series supposed to do this!? I'm not even sure how to do this!" Thing is, it was presented as zeta(2)-1 expanded on one line, then zeta(3)-1 expanded on the second line, etc... and, if you add VERTICALLY, the first term of zeta(2)-1 = 1/4, the first term of zeta(3)-1 = 1/8, the first term of zeta(4)-1 = 1/16... THEY'RE INFINITE GEOMETRIC SEQUENCES! Each of which we can take the sum of using good ole (1-r)^-1 So, yeah, solved it that way, as an nice alternative that doesn't require you to know anything about the Reimann zeta function.
started watching your videos a few years ago. I love it, but I am just also surprised that you still exist on youtube. Did not realize that this many people actually like to watch someone talking about college math.
A lot of these constants show up in the polygamma functions and after some expermination, I just realized the relationship, and it kind of makes sense because you can calculate finite sums of a harmonic series with the digamma function ψn(1) = ζ(n) Γ(n+1) where ψn is the nth polygamma function The missing constant at n = 1 is the euler mascheroni constant
Hello! I was messing around in Wolfram Alpha with the harmonic series, but instead of the usual sum from n=1 to inf 1/n, I decided to plug in i, the imaginary unit into the numerator, and see what would happen if I played around with it. The series which I wanted to share with you is the following: sum from n=1 to inf (i)^((n+k)*pi/2)/(n). o if k=0, it converges; o if |k|>=1, it diverges; o if 0
Do you think that you could perhaps do a STEP 3 question in the near future. They are quite more fun than the STEP 2 that you attempted earlier. It is fine if you have other plans through.
Oh my god I only seen your 100 world record video for the first time 30 minutes ago.. your hair has swapped places on your face 🤯🤯🤯 HOPE YOU ARE STILL WELL
Do you realize that, having shown that the series whose general term is z(s)-1 converges, you have also proven, as a byproduct, that z(s)-1 tends to zero for s tending to infinity (the necessary condition for convergence) and therefore z(s) tends to 1 in the same limit for s?
I wonder if it would be possible to do this kind of thing but instead of a series of zeta functions it would be a product, since z(∞)=1 it might converge
What about the odd numbers?! Is it possible, if I subtract the upper limit (I mean when s approaches to the infinity ♾️) and even numbers to obtain the odd numbers? (it means: 1 - Zeta(even)= Zeta(odd) )
@@user-wu8yq1rb9t they are two totally different sums. Try writing zeta(even) and zeta(odd) in their summation form and then expanding them like here: ζ(2)=1/1^2+1/2^2+1/3^2.......... You will automatically realise why I am saying this.
I have a question, when you use the geometric series you use the 1/1-r formula that is derived when the series goes from 0 to infinity, but this starts at 2 ¿How's that not a mistake? Thanks for your attention
He uses a slightly modified version, in general as long as the sum is infinite and geometric with ratio r the formula 'a/(1-r)' works where 'a' is the first term in the series. This formula works for any indexing which can be easily shown. The reason we can have such a formula is that starting from a different index in a geometric series is equivalent to multiplying every term by a constant.
Indeed, by reindexing the sum with s=s'+2 then s' starts from 0 and the exponent becomes (s'+2) so you get (1/n)^s'.(1/n)^2. The latter factor is a constant (independent of s') in the numerator.
Please stop inverting sigmas without telling a word as if it was okay. I know you will not check fubini theorem each time vecause it is boring but just tell a little word that you can do this because we have a friendly sequence
Just to let you know mate, when you say things like 'calculus 2', that is largely meaningless to most of the world, being an American concept. What does it mean? Perhaps you could be more inclusive to everyone else - just a suggestion, love everything you do.
Thank you for the video 😉
Thank you for the nice problem. I actually have not thought about it before.
Another method to solve it😜 same day fresh! czcams.com/video/LUTeBIXzKGM/video.html
@@blackpenredpen hello sir..... I am biggest fan of your 🙏🙏and I want to become a mathematician and I have a wish that i will talk with you.........
Sir one youtuber solved Riemann hypothesis this is not a joke really this is link his video czcams.com/video/8y7fMmdAcb8/video.html
Nice shirt! We should do a t-shirt trade....
Behind every failed mathematician is a nice T-shirt.
Another method to solve this problem😆 czcams.com/video/LUTeBIXzKGM/video.html
Micheaaaaaall
@@sonkim6876 ?
Sure. Send me an email.
I have known this series as a higher mathematics problems but as soon as you switched the order of summation I got goosebumps just seeing how easily it can be solved
I have a quick question about the non-trivial zeros of the Riemann Zeta Function...
We’re gonna need more colors of pens 😳
I’m just gonna assume it’s always 1/2 + iC
😂😂
you're kidding, that's nuts dude. i love this. one of my favorite things is when a problem produces a completely nonintuitive result that you really have to dig to understand.
Thank you for all the effort you put into making all these videos. I've started to really like maths in general thanks to you :)
I am glad to hear. Thank you.
switching order of summation is a discrete version of fubini's theorem
what is fubini theorem? does it allow switching integrals?
@@dudono1744 with some conditions.
Its more interesting to get 1 actually... Since most infinite series are oscillatory in nature... To take a random infinite series and end up with a whole number is much more fascinating.
It's actually much easier for the last part to use the sum convergence test, since you just proved that the sum of zeta(s) - 1 converges that must mean the limit of individual terms goes to 0 and so zeta(s) goes to 1.
True!
seems easier to just notice that 1/n^infinity is 0 for n>= 2
@@dudono1744 Swapping that limit with the sum require some justification, that justification is actually equivalent to the sum argument.
I absolutely love this channel ❤️. The knowledge is appreciable . Respect from India🇮🇳
What an interesting video, very well explained! Videos like these inspire me to share my own maths content!
You can also try this for just even values of s in the Riemann Zeta function. And if you have done this, you can just subtract the value from 1 and you get the sum for odd values of s. I have done this in the past, try it, its really worth it!!
(Btw: Michael Penn also did some great videos about sums of the zeta function using generating functions)
What exactly do you mean by your first paragraph?
@@laxminarayanbhandari855 I mean first you sum zeta(2)-1+zeta(4)-1+zeta(6)-1… and then you take the value (S) this sum converges to and calculate 1-S and then what you get is zeta(3)-1+zeta(5)-1+zeta(7)-1…
@@User-gt1lu according to Wolfram Alpha, the second sum is divergent. I checked till 300 terms, but the sum doesn't go above 1/4, which is expected value according to your method. Don't know what is going on.
I saw Penn’s videos about generating functions, he even used that method for some trivial proofs. That guy is awesome
The Riemann zeta function videos I made are still some of my favorites!
Your maths are 🔥 as always. But also your beard is on point - almost starting to look like a video game final boss!
😆 thanks!!
I tutor math. Have an honors precalc student that got this exact question (without mention of the Reimann zeta function) on a worksheet. I recognized it as the infinite sum of zetas immediately, and I'm thinking "how the heck is this kid, who JUST got introduced to sequences and series supposed to do this!? I'm not even sure how to do this!" Thing is, it was presented as zeta(2)-1 expanded on one line, then zeta(3)-1 expanded on the second line, etc... and, if you add VERTICALLY, the first term of zeta(2)-1 = 1/4, the first term of zeta(3)-1 = 1/8, the first term of zeta(4)-1 = 1/16... THEY'RE INFINITE GEOMETRIC SEQUENCES! Each of which we can take the sum of using good ole (1-r)^-1 So, yeah, solved it that way, as an nice alternative that doesn't require you to know anything about the Reimann zeta function.
Me who only just learnt derivatives of trigonometric functions: *Ah yes, a fun video for me!*
Such a good explanation I didn’t need calc 2 to understand
This comment should have more likes.
started watching your videos a few years ago. I love it, but I am just also surprised that you still exist on youtube. Did not realize that this many people actually like to watch someone talking about college math.
Finally, an approximation for 1
This video pops into my recommendations just after I learn how to solve the exact same thing for my tomorrow's test !
Getting my Ed from CZcams Uni. Thank you oh so blessed one. 🐑🔥🔥🔥
I have discovered same thing with the sum but with Euclid's series not by these substitutions and that was kind of interesting too!!!
Equal to 1.... my god, its been a while since my mind has been blown that hard.
I love the painting on ur wall.
this is truly amazing
Thanks!
1:50 "Don't go yet" hahaha
😆
function zeta(n){
let zeta=0
for(let i=1 ; i
A lot of these constants show up in the polygamma functions and after some expermination, I just realized the relationship, and it kind of makes sense because you can calculate finite sums of a harmonic series with the digamma function
ψn(1) = ζ(n) Γ(n+1)
where ψn is the nth polygamma function
The missing constant at n = 1 is the euler mascheroni constant
oily macaroni
Hello!
I was messing around in Wolfram Alpha with the harmonic series, but instead of the usual sum from n=1 to inf 1/n, I decided to plug in i, the imaginary unit into the numerator, and see what would happen if I played around with it. The series which I wanted to share with you is the following: sum from n=1 to inf (i)^((n+k)*pi/2)/(n).
o if k=0, it converges;
o if |k|>=1, it diverges;
o if 0
This is amazingly cool!
Also what would be the product of the zeta function of all the integers from 2?? :)
That’s a cool question! I just checked on WFA and got about 2.29. Not sure how to get that tho.
Well, I checked on MSE. No closed form exists. The product can only be approximated.
so it's a whole new constant ?
Do you think that you could perhaps do a STEP 3 question in the near future. They are quite more fun than the STEP 2 that you attempted earlier. It is fine if you have other plans through.
Check out J Pi math. He’s done a few STEP problems on his channel.
@@blackpenredpen thank you.
Beautiful piece of math
Oh my god I only seen your 100 world record video for the first time 30 minutes ago.. your hair has swapped places on your face 🤯🤯🤯 HOPE YOU ARE STILL WELL
Please upload some tricks of integration and different forms of integration 🙏🙏
Beautiful!
I want you teach algebra now, especially probability and permutations
Nice video...👍☺️
Finally!
Thank you so much
Brilliant stuff my man
Nice sir ...
Pretty cool !!!
Sum in the Thumbnail starts at k=1 but the calculation is k=2!!
Just fixed. Thanks.
Dude 2!! = 2
You did a good job.
Do you realize that, having shown that the series whose general term is z(s)-1 converges, you have also proven, as a byproduct, that z(s)-1 tends to zero for s tending to infinity (the necessary condition for convergence) and therefore z(s) tends to 1 in the same limit for s?
Yes because 2:47
13:19 quick note: The inequality is true for s ∈ ℝ . For complex number it's not always true.
the values of zeta on N\{1} is just a partition of 1 ,cool
Maaaaaan this result is huge :OO
1 is a small number
I agree, cool question but I did not get satisfy with the answer
6:46 actually it should be "n is bigger than 1" i suppose
Yes.
Daamn just 1, that's so cool lol
Is the infinity then multiple of pi?
Looks like cool, but can't see what it means.
I have precalc and that’s it, but ima still watch the video
Awesome.
Sir please solve my doubt. Is w(x. lnx) =ln(x)? Here w is inverse of x.e^x. Lambert function
6:36 I don't see the thing about switching order of summation in the description.
@blackpenredpen Your channel name used to have your birthdate in it, right? I thought you looked familiar.
No. I never put my bday in my channel name before. This channel has always been “blackpenredpen”
Very Cooooooooooooooooooooool!
So cool
I'm a bit curious to know why, in the last step (before the ad break) the 2 was inserted in one denominator but not the other?
yes, I too
I hit the 25 th like😁😁😁
Do you know something about the news of the proof of the Riemann Hypothesis? As a phycisist it just go beyond me . Not my major
I wonder if it would be possible to do this kind of thing but instead of a series of zeta functions it would be a product, since z(∞)=1 it might converge
It does conv as I checked on Wolframalpha. Not sure how tho.
@@blackpenredpen I remember somewhere that a product converges iff the sum of the logs of the terms converges, maybe there's something there
Good tutor bro... Make it more deep as close as Chinese students levels...
I need to√-a! When (a) is positive integer
Kind of a boring answer. What would the corresponding alternating series be equal to?
My stupid ass trying to solve the Riemann hypothesis it is not solvable
Teacher: what is the sum of (zeta(n) -1) from n=2 to infinity?
Student: I don't know. Let me guess a number. Shall I start at 1?
Teacher: ...
Hahaha!
“You just need calculus 2 to understand this video” well shit
Lol, I answered that question in quora a few weeks ago.
Kirby rules!
Nice ;D
How about an infinite series of an infinite series of an infinite series... infinite number of times?!?! o.o
When u realize his channel's name is black"PEN"red"PEN" but he uses red and black "MARKERS"...XD
Understand nothing
But still like it 👍👍
now do zeta(-1)
;)
you mean sum of zeta (s) from -1 to - infinity?
What about the odd numbers?!
Is it possible, if I subtract the upper limit (I mean when s approaches to the infinity ♾️) and even numbers to obtain the odd numbers? (it means: 1 - Zeta(even)= Zeta(odd) )
Nope. It doesn't work like that.
@@laxminarayanbhandari855
Thank you for your attention.
But why?
@@user-wu8yq1rb9t they are two totally different sums. Try writing zeta(even) and zeta(odd) in their summation form and then expanding them like here:
ζ(2)=1/1^2+1/2^2+1/3^2..........
You will automatically realise why I am saying this.
@@laxminarayanbhandari855
Maybe the infinity involved odd numbers too.
But if I consider the infinity as odd+even, we can subtract
@@laxminarayanbhandari855
I'll try it
Thank you so much
So did you get the answer of Reimann hypothesis?? If yes then I guess you will get 1 million dollars for the 7 millenium problems.
Pls reply 🙏🙏
Sir, would you please give me a pdf of the total syllabus ofCalculus 1 and calculus 2
Depends on the school. You can get a rough idea from a Google search.
the openstax ones are decent. try them.
why that painting ? i mean "the scream"
Telescoping is tricky
His beard is the reason
I FIND CALCULUS INTERESTING
You should change your channel name to blackpenredpenbluepen!
zeta ( 3 ) = apery's constant
鬍子越來越長了
I have a question, when you use the geometric series you use the 1/1-r formula that is derived when the series goes from 0 to infinity, but this starts at 2 ¿How's that not a mistake?
Thanks for your attention
He uses a slightly modified version, in general as long as the sum is infinite and geometric with ratio r the formula 'a/(1-r)' works where 'a' is the first term in the series. This formula works for any indexing which can be easily shown. The reason we can have such a formula is that starting from a different index in a geometric series is equivalent to multiplying every term by a constant.
Indeed, by reindexing the sum with s=s'+2 then s' starts from 0 and the exponent becomes (s'+2) so you get (1/n)^s'.(1/n)^2. The latter factor is a constant (independent of s') in the numerator.
Please stop inverting sigmas without telling a word as if it was okay. I know you will not check fubini theorem each time vecause it is boring but just tell a little word that you can do this because we have a friendly sequence
We want you to prove, Papa Riemann was true!
Ok, now *solve for all complex zeros that have a real part of 1/2*
the sum is 0 bruh
Just to let you know mate, when you say things like 'calculus 2', that is largely meaningless to most of the world, being an American concept. What does it mean? Perhaps you could be more inclusive to everyone else - just a suggestion, love everything you do.
So in this entire video, you are assuming that s∈ℝ
9:20 - 9:50
wat
@@eduardvalentin830 ok
its clear now
how can we make n=0? n=0! lol
All that work for zero
Prove : trillion of a root 2 = million of 5 times of 7
I think only aliens can solve this question...
challenge: zeta(i)
Well, no closed form exists, due to the fact that no closed form exists for Gamma(s+it), t≠0.