Sum of 1/n^2+1

[@drpeyam]

If you want to know more, make sure to also look at blackpenredpen’s videos on Fourier series, they’re an awesome supplement to this video!

[@AsadAli-cr3ve]

Good ,carry on👏👏👏👏

[@franciscojavierorozcobeiza9378]

Dr. Peyam's Show I’ve watched of bprp and he should be expressed his alternating serie in terms of hyperbolic fncs as you... Awesome thank you for upload videos

[@AndDiracisHisProphet]

Fouryay

[@blackpenredpen]

AndDiracisHisProphet yayyayyayyay

[@yaaryany]

@@blackpenredpen The god himself has commented ..

[@imaginaryunit5056]

Ohhh, it's awesome Let let me make a notice: We can sinh(2pi)/sinh^2(pi) invert to 2sinh(pi)*cosh(pi)/sinh^2(pi) and get 2cosh(pi)/sinh(pi) = 2cth(pi) The total is 1/2(pi*cth(pi)+1)

[@scottgoodson8295]

It's been noticed by other people, but the sum from -∞ to ∞ can be quite succinctly expressed as π*coth(π), which is arguably more amazing as it almost looks like it has a relationship with arctan (they are inverses in quite different ways to tanh and tan, respectively). Just a cool note!

[@Czeckie]

word is that charles darwin noted "A mathematician is a blind man in a dark room looking for a black cat which isn’t there" when he saw dr. Peyam struggling to keep the lights on.

[@zachary7067]

1:57 "𝜋 m x" I see what you did there :)

[@triedtested8797]

what did they do there? I was confused af ;-;

[@zachary7067]

@@triedtested8797 Whenever he used m instead of n, it sounded like he was saying his name, Peyam (with an extra x at the end).

[@timothyaugustine7093]

Lol pi am xoxo

[@whythosenames]

4:39

[@wduandy]

1:17 "WTF" hahahahahaha

[@ffggddss]

Very nice! Thanks!! Even using Parseval's Theorem, which I've always been fond of! Suggestion: When you get a thing like sinh(2π)/sinh²π , use the double angle formula to simplify further - ½π sinh(2π)/sinh²π = ½π·2sinh(π) cosh(π)/sinh²π = π/tanh(π) ½π sinh(2)/sinh²(1) = ½π·2sinh(1) cosh(1)/sinh²(1) = π/tanh(1) Fred

[@drpeyam]

That’s even prettier, thank you!

[@ffggddss]

@@drpeyam You're welcome; your derivation is also quite pretty! FYI, I see that "Imaginary Unit" arrived at much the same thing about a month ago. And BTW, tanh(π) = 0.99627208...; very close to 1. So ½(π/tanh(π) + 1) ≈ ½(π + 1) is a very good approximation. The exact sum is 2.07667404746858..., and fiddling with the partial sums, I find that it takes about a dozen terms to get up to 2. The approximation is ½(π + 1) = 2.07079632 Fred

[@minagawargious]

I absolutely love your enthusiasm.

[@Rundas69420]

When the light went out, the math-ghost appeared and told you to fix your video description since there it says from 0 to 1 which would be 3/2 and therefore trivial. Also mathematicians don't kill 2 birds with one stone but rather 2 functions with one identity. xD

[@がつかくん]

in japan,at a.m.4 but i don't care It was fantastic！ thank u

[@curtiswfranks]

Math literally slaughters birds!

[@GreenMeansGOF]

7:48 I thought my phone locked. XD

[@drpeyam]

Hahaha

[@ffggddss]

Yeah, two power outages in one video! You guys should be more attentive in paying your electric bills!! ;-) Fred

[@maxsch.6555]

Same XD

[@Shad0wWarr10r]

1/(n^2+1)

[@emanuelmartinez3585]

I love to evaluate infinite series ☺️...parseval identity is kind of Magic

[@drpeyam]

Me too!!! 😄

[@fotnite_]

Just had this as a homework problem, except we were using a theorem that allows you to calculate this via Residue Theorem

[@IS-eb9lf]

If you do that with e^ax, than you can get formula for 1/(n^2 + a^2), it will be (pia*coth(pia)-1)/2a^2 The fun fact: everything works with complex numbers, so 1/(a^2-n^2) converges to the (pia*cot(pia)-1)/2a^2

[@biswadeepghosh5568]

Great work.

[@shiina_mahiru_9067]

Rooms always scared about your math 😂 The fact that the formula involve only pi but not pi^2 also amazed me

[@jarikosonen4079]

Title? Sum of 1/(1+n^2), not sum of 1/n^2+1... but it looks good results if needed somewhere or even if not needed. It looks this is same as 1/2 π coth(π) - 1/2. Not sure which one is simpler though. This looks solid proof even I couldn't follow all the steps. Can you find solution for sum 1 to infinity of 1/n^3 also (and the other odd powers)? This looks like zeta functions expansion... Zeta(-1)=-1/12 or zeta(0)=-1/2 is strange cases. It looks like there would be "overflow" in the positive realnumbers that will change the sign of the zeta function at zeta(1)~+/-infinity point. It looks like for 1/(n^2+b) sum giving: b | 0 | π^2/6 1 | 1/2 (π coth(π) - 1) 2 | 1/4 (sqrt(2) π coth(sqrt(2) π) - 1) 3 | 1/6 (sqrt(3) π coth(sqrt(3) π) - 1) ... b | 1/(2*b) (sqrt(b) π coth(sqrt(b) π) - 1)

[@drpeyam]

Yeah, but nobody is gonna type parentheses when looking for a YT video...

[@shanmugasundaram9688]

The series converges to 2.076674.

[@sundayscrafter1779]

Hi there, there is a little something I do not understand. Isn’t the Fourier series expansion you are using in the beginning something true for functions 2Pi-periodic only? I mean, I have the feeling we have e^2Pi = e^0 with your development.

[@giacomolanza1726]

Fourier series can be applied either to periodic functions, or for functions defined within a closed interval. In the latter case the Fourier series is a good approximation of the starting function only within the interval (outside it it will approximate a periodic repetition of the considered arc of function)

[@ferashamdan4252]

Many thanks for your time sharing this proof. I did enjoy your descriptions.

[@abolfazlmohammadpour1456]

God bless you Dr

[@lunaticluna9071]

I have found a general formula for the integral of 1/(1+|x^n|) from -inf to inf. it is (2pi/n)*csc(pi/n)

[@giacomolanza1726]

Sound interesting - do you have a demonstration?

[@danielmastalerz3640]

Sir, you definitely deserve more subscribers!

[@davidrheault7896]

The solution simplifies to 1/2+1/2*pi*coth(pi) which is approx 2.076674048

[@Contradi]

You are a gem.

[@blackpenredpen]

Andrew H Agree!!!!!

[@darkmask4767]

Let a>0. The sum of 1/(n²+a²), n= 0 to infinity = π*coth(πa)/(2a)+1/(2a²). This can be calculated by finding the Fourier series of e^(ax) on (-π,π) and using Parseval's identity with e^(ax)

[@Enrico-Migliore]

Great demo. Thanks.

[@xCorvus7x]

8:52 Parseval, the hero of legend, knight at the Round Table, keeper of the Holy Grail of Analysis. Too bad that this result is just a special case somebody stumbled upon while doing Fourier analysis and not a result from the general solution for Σ a/(b+n^c) as n goes from 0 to ∞ (or -∞ to ∞) for complex a, b, c.

[@michelkhoury1470]

Very very nice solution 👍 I solved this by using also Parseval's Identity and this is a very very beautiful Identity with Fourier series 😉

[@michelkhoury1470]

But I took f(x)=exp(-x) but the proof is going to be similar since the module of (1-iN) is the same as the module of (1+iN)

[@alephnull4044]

"We're *literally* killing two birds with one stone." Birds were harmed in the making of this video.

[@giannismaris13]

Riemann is proud after this 😅

[@gergodenes6360]

You can take sinh(2pi)/sinh^2(2pi) to be 2*cth(pi), so that simplifies the expression to be (pi*cth(pi)+1)/2, it's a lot more beautiful than with sinh^2 and stuff

[@xxnotmuchxx]

What about sum of 1 / (n^2 + 1/a) from n=1 to infinity?

[@mihaly1027]

I forget how many languages do you speak?

[@drpeyam]

5 :)

[@gorthorki]

Can you do this technique with other functions? I need more closed summation formulas 🤤

[@davidrheault7896]

We can apply this strategy on Zeta function as well to evaluate at integers, such as Zeta(4) = pi^4/90 by using the function f(x) = x^2

[@112BALAGE112]

Excellent as always.

[@ChefSalad]

There's a much better explanation for the -inx part of the Fourier coefficient formula than complex dot products needing to have a complex conjugate term. Imaging having f(x)=∑C(n)*e^(inx) . Then multiply both sides by e^(−imx)*dx for each m and take the integral. You get ∫f(x)e^(−imx)dx=...+∫C(m−1)*e^−ix*dx+∫C(m)*e^0*dx+∫C(m+1)*e^ix*dx+... The thing to notice is that ∫C(m)e^inx*dx=0 when n≠0 and ∫C(m)*e^inx*dx=2π*C(m) when n=0. This means that 2π*C(m)=∫f(x)*e^(−imx)*dx. The reason that this works is because sin(x) and cos(x) both integrate to zero over one complete period but e^0=1 which is a constant and so that one term doesn′t intergrate to zero. By multiplying all the terms by e^−imx, we make the term we want intergrate to 2πC(n) and all other terms integrate to zero. It has to be negative imx because the term we want is c(n)*e^inx, and when we multiply we get c(n)*e^(inx)*e^(−imx)=c(n)*e^(i(n−m)x). When n=m, then we get c(n)*e^(0) for the term we want but the other terms still have e^iqx and so integrate to zero. In other words, it needs to be negative to cancel out the positive we already have on the term we want.

[@kqp1998gyy]

Beautiful. Thank you 🌷

[@thephysicistcuber175]

4:52 Hi Peyam!

[@Goku17yen]

Hey Peyam, i have a quick question about your recording strategy. Do you connect the lapel mic to your phone and then record it there, and sync the audio separately with your camera, or are you recording the video with the same camera you have the mic in? Thanks so much

[@drpeyam]

In my older videos I recorded the videos without a mic, just with the phone I have, but in my newer videos I record it with a mic, which is plugged in my phone. To this date I still don’t know if it makes a difference 😅

[@Goku17yen]

Dr. Peyam's Show lmao ikr!! I can’t hear a difference so I’m wondering what’s the point of the mic lol, I tried it out with multiple mics as well, so it must not make much of a difference through all brands

[@drpeyam]

Yeah, bprp said the same thing to me! It must be a placebo effect or something 😂

[@Goku17yen]

Dr. Peyam's Show for real, now I feel like I wasted a load of money on a bunch of mics!! Lol

[@blackpenredpen]

Goku17yen That's why I still use the ball.

[@ajayananddwivedi7154]

What to do if there is 2n^2 instead of n^2

[@danielmilyutin9914]

sinh(2x) = 2sinh(x)cosh(x). Thus, final formula can be simplified to ctgh(x).

[@zivssps]

at the "also" part at the end, it's supposed to be 0.5 instead of pi/2.

[@timurpryadilin8830]

Dr. Peyam, can you please proof that there is no general formula for roots of polynomials with degree more or equal to 5. It sounds very interesting

[@drpeyam]

Sadly I can’t, I know no Galois theory at all :(

[@Nerdwithoutglasses]

Sir, is it possible to solve sum (n=0 to inf) 1/(n^4+1) or 1/(n^8+1) this way?

[@Flanlaina]

Converges or diverges?

[@brendanlawlor2214]

Beautiful series compare with Euler sum for squares reciprocals being pi^2/6. Artists have no idea of the beauty of nature provided by math

[@PackSciences]

8:36 Missed the opportunity to use the cardinal hyperbolic sine.

[@quocanhnguyenle4952]

This can be simplified to (π*cothπ+1)/2.

[@lindsaywaterman2010]

Sinh (2pi) = 2* Sinh(pi)* Cosh(pi). When Sinh (2pi) he substituted, a Sinh(pi) in the denominator is cancelled leaving a Coth(pi) with an additional factor of 2. Therefore, this reduces to the classical formula.

[@jesalkotak5595]

Sum of e^(-n)/(1+n^2) = ?

[@djemidjma5274]

Description says from 0 to 1. Fix ples

[@Prasen1729]

At 2:21, not so clearly explained why you put a 2. Though that's another reason to study Fourier coefficients but funny as hell at 7:41, it's not oh my gosh, it's oh my sinh. :-D

[@carlosgiovanardi8197]

Nice video. Very instructive, Peyam Tabrizian. By the way, i have the following doubt: is it possible to obtain a similar formula for other exponents (4, 6, ...) and for other numbers other than 1? For example: Sum of 1/(25+n^4) I appreciate any reference.

[@drpeyam]

I’m not sure how, but I think it’s possible to do that

[@carlosgiovanardi8197]

​@@drpeyam, WA gives results in terms of digamma function 1/1+n^4 www.wolframalpha.com/input/?i=sum+1%2F%281%2Bn%5E4%29 1/1+n^6 www.wolframalpha.com/input/?i=sum+1%2F%281%2Bn%5E6%29 1/1+n^8 www.wolframalpha.com/input/?i=sum+1%2F%281%2Bn%5E8%29 1/25+n^4 www.wolframalpha.com/input/?i=sum+1%2F%2825%2Bn%5E4%29 1/25+n^6 www.wolframalpha.com/input/?i=sum+1%2F%2825%2Bn%5E6%29 1/25+n^8 www.wolframalpha.com/input/?i=sum+1%2F%2825%2Bn%5E8%29 and when the exponent tends to infinity, sum tends to 0.5 or 1.5 depending on the initial term math.stackexchange.com/questions/2070991

[@carlosgiovanardi8197]

dear Peyam, I remind you about this subject. I will be very grateful if you have any further information about this kind of sums involving digamma functions.

[@paulfoss5385]

While I wasn't able to work out the closed form, I was able to find the value out to ten decimal places in a spread sheet. The series converges at a glacial pace, but I reasoned that the majority of the difference between the terms in 1/(n^2+1) and 1/(n^2) would occur at small values of n, as 1/(n^2)-1/(n^2+1)=1/(n^4+n^2). So I found the difference between (pi^2)/6 and the series 1/(n^2) and added it to one plus the series 1/(n^2+1). Not an elegant nor efficient method, but eh, it's what I could do. :/

[@drpeyam]

Interesting!

[@houdahbe5538]

can u please help me sum of 1/sqroot of (n^2+in) i=0 to n ????

[@xxnotmuchxx]

Wolframalpha says it is .5 pi coth(pi) - .5

[@gnikola2013]

Parseval himself came to Peyam to see his derivation, that's why the lights went out

[@blackpenredpen]

Kiritsu That must be it!!!

[@mikeh283]

The third time I came here I came for the parseval identity.

[@aneeshsrinivas892]

is there a closed form for ∑1/(n^4+1)

[@robertoxmusica]

Yes: Compute '∑1/(n^4+1) from n=-∞ to ∞' with the Wolfram|Alpha website (www.wolframalpha.com/input/?i=%E2%88%911%2F%28n%5E4%2B1%29+from+n%3D-%E2%88%9E+to+%E2%88%9E) or mobile app (wolframalpha:///?i=%E2%88%911%2F%28n%5E4%2B1%29+from+n%3D-%E2%88%9E+to+%E2%88%9E).

[@dgrandlapinblanc]

Amazing. Thanks.

[@thephysicistcuber175]

Do this with complex analysis.

[@srpenguinbr]

Typo on the tittle, should be 1/(1+n²)

[@drpeyam]

It’s not a typo; people are usually too lazy to put parentheses when searching for a video

[@mjthebest7294]

Angel well no... it would be' divergent in that case :) Because (1 + 1/n²) > 1 for all n :)

[@robertoiturralde2990]

series (n!)/(2^n+1), n=1 to infinity, converge or diverge please help me to solve this.

[@drpeyam]

Ratio test

[@andreamonteroso8586]

this is good!

[@youshouleviathan1662]

Curve Agnesi? Amazing

[@LucaIlarioCarbonini]

@Dr Peyam I've tried to search for a video of yours about Perseval with no success. In case you have not made one yet may I suggest that subject for a video in your KZbin channel?

[@drpeyam]

Sum of 1/n^2 kzbin.info/www/bejne/j37PlnyEl86Boqc

[@LucaIlarioCarbonini]

Pikachu already told me! kzbin.info/www/bejne/aIe8c3p8lLKDf7c

[@LucaIlarioCarbonini]

What I suggest is a proof of Perseval's theorem, I can remember bunches of inequalities waving all around and making me feel that I had to find a job...

[@LucaIlarioCarbonini]

Found! kzbin.info/www/bejne/Z6LOonaKrL6ip7c

[@josephmartos]

1:10 i wanted to say "what the f*uck" is that... and you just wrote it down xD

[@jadegrace1312]

Wait why is this not pi²/6

[@drpeyam]

It’s not the sum of 1/n^2

[@jadegrace1312]

@@drpeyam oh right, I just assumed it was shifted over but that would mean the 1 was inside of the squared. That makes sense

[@koushikbhattacharya3778]

he is happy about something

[@pablojulianjimenezcano4362]

This is crazyyyyy

[@tszhanglau5747]

Probably some kids were playing with the switches...

[@nestorv7627]

ily

[@LouisEmery]

WTF means "what to find"? I realize he is not a native English speaker.

[@abderrahimabderrahim6659]

coming from mr beast

[@mathyland4632]

same! I just used Wolfram alpha at the time, but now that the challenge is over, I'm curious where that answer comes from.

[@-_-rain5432]

1st

[@HowardS185]

Lovely - Just you don't LOVE math!

[@housamkak646]

Oooommmmmmmmmgggggggggggggggggggggggggggggg