In previous videos it has happened that I had difficulties in reading what you wrote on the blackboard, I thought it was because of the chalk colors or the illumination of the room. This video is perfectly clear. Please keep this same setting while filming, it works very well.
Absolutely beautiful. Certainly nothing I would be comfortable trying on my own at my current level of math. But using so many different results and techniques... I just love it!
Informative. It's straightforward to generalize from 2 infinite sums presented here to N such sums. E.g., N=3 would correspond to inf't sum over indices (m,n,k) of 1/[mnk(m+n+k)], & so on. The general sol'n is Gamma(N+1)xZeta(N+1), where "Gamma" denotes the familiar Gamma function, which is just the factorial over the natural numbers. Note that it works even for N=1, which is just the inf't sum over n of 1/n^2. Also, recall that the Zeta funct'n has a simple form over even natural numbers. E.g., for N=3 the triple sum is just (pi^4)/15, & for N=5 it's (8/63).(pi^6), etc.
Thanks for pointing this out. This is exactly how I started it then got stuck on sum(Hn/n^2) and completely forgot that I solved it some time ago when Michael brought it up. So used the triple integral method which happens to be the same approach here, and honestly, it easier than going thru the harmonic series.
13:20 these two integrals inside are exactly same thing. \int_0^1\frac{dy}{1-y z}=\int_0^1\frac{dx}{1-x z} , the only change is a private integration variable. No need to consider them separately.
I think the change of order is more like Tonelli's theorem (it's like Fubini's Theorem but for positive functions, and it works with combinations of sums/integrals and integration w.r.t. sigma-finite measures, or whatever) Dominated Convergence Theorem helps you deal with +/- signs and functions that vary with the index but within a shared integrable absolute ceiling. I guess Monotone Convergence Theorem can also be applied, because adding positive terms to a sum will give you a pointwise-increasing total as n increases.
We can write 1/(m*n*(m+n)) as f(m,n)-f(m,m+n)-f(m+n,n) where f(x,y)=(x+y)/(x^2y^2). If we create a Stern-Brocot-like tree, where instead of fractions we put pairs, and (m,n) as a parent leads to (m,m+n) and (m+n,n), eventually we hit every pair of coprime numbers. If we add up 1/(m*n*(m+n)) level by level on this tree, we'll get f(1,1) minus the lowest-level leaves, so the sum is at most f(1,1)=2. It's exactly 2, however. If we go down any branch, there will eventually be a point where one of the two numbers will be greater than n. Let us add up f(a,b) where (a,b) is one of these points: that is, either a>n>=b and a-bn>=a and b-a
Very neat and clear. These multiple summation/integration tricks were really cool (did the transformations on my own up 'til that point, then was "watching & learning"). Like videos like that
This video is fascinating. Forty years ago, the Dominated Convergence Theorem was not commonly taught at the undergraduate level because it is established using concepts from measure theory. Are measure theory and the Dominated Convergence Theorem currently taught to undergraduates?
Nice ,but we directly see Harmonics here S=Sigma from m=1 to infinity 1/(m^2) sigma from n=1 to infinity (1/n -1/(m+n)) =sigma from m=1 to infinity H_m/m^2 =2zeta(3) So easy
What is the definition of a double infinite sum? What is the order of limits? Is it lim(sum(lim(sum(...))))? Or maybe lim(lim(sum(sum(...))? Perhaps the limit is multivariable? Is there even a difference between these?
lim lim sum sum =?= lim sum lim sum There are a few theorems allowing you to interchange the middle lim and sum (in other words lim sum = sum lim). I'm pretty sure dominated convergence is one of them. So if the sum is absolutely convergent (sum |f| exists) the above statement is correct. Also sum_0^infty sum_0^infty is defined as lim sum lim sum.
Hi Michael, W.r.t. the double-sum that you solved in this video, here is a slight variation of it (I thought I would share with you): Find Sum(n=1,inf) Sum(m=1,inf) 1/(m*n*(m+n+2)) The answer is 7/4 Interesting how this one has a closed-form.
Basically the power of Harmonics is immense, it dominates logarithmic integrals and even such sums .We can escape those manipulations once we command harmonics
Did the sum over three indices and got 3*2*zeta(4). I bet there is a general formula for any number of indices: t!Zeta(t+1) where t is the number of indices
You used twice the dominated convergence theorem, while I think in both cases it should have been easier by monotone convergence theorem. For using DCT, what would be the dominating function, in both case?
Counting this Sum other way gives sigma(H_n/n^2) from 1 to infinity where H_n is harmonic number from 1 to infininty implies zeta(3)=1/2(sigma(H_n/n^2)) isnt it interesting🤔🤔🤔
Yes, this is an example of a relation among so-called multiple zeta values. It corresponds to zeta(3) = zeta(2,1). See here for another proof: ruclips.net/video/_HHwnVArDm4/видео.html
Michael I'm glad you're teaching at a slower pace now, in the past I've had to slow you down to .85x in order to follow without proof by intimidation lol I do have a question, u never explained why we're evaluating this as an integral to begin with. And why are we using those specific tools? Some of this stuff I can gather on my own but it'd be nice to hear the logic explained!
you are right. There's small potential difficulty that the argument doesn't work when xz=1. Even though it doesn't work it doesn't matter, since whatever is the value of the function at a single point (or more generally on a null subset) - it can be even infinity, it doesn't change the integral.
TL;DR - There is no closed form expression involving pi, exp, trig, inverse functions, roots, and arithmetic, but there is a much nicer infinite single-sum form.
This is incredibly counter-intuitive to me. You start with a discrete function on a rational domain with a lower bound of 1, apply a bunch of contiinuous transformations with real domains and end up with an analytically continued function with a complex domain that excludes {1}.
Let n be fixed, then 1/mn(m+n)=1/n[1/m(m+n)]=1/n^2[1/m - 1/(m+n)] Sum 1/mn(m+n) for m=1 to inf =1/n^2 [Sum(1/m-1/(m+n)) for m=1 to inf] =1/n^2 [1/1-1/(n+1)+1/2-1/(n+2)+...+1/n-1/(n+n)+1/(n+1)-1/(n+n+1)+...], this is telescoping series =1/n^2[1/1+1/2+1/3+...+1/n] =Hn/n^2 And it converges to 2Zeta(3)
Hi Michael, Hope you get to see this comment. I can tell you have a passion for solving tough questions regarding limits, sums, and integrals (they are my favorite, too) Below I am providing one tough question that has a closed sum, thought it might interest you enough to create a video on it. Sorry I couldn't provide a picture of it (e.g. in LaTeX), I know that might have expressed the question better... anyway, here it is: Find Limit(n->inf) Integral(x=0,inf) [Sum(m=1,inf) (m+x)/((m^n+x^n)^n)] dx The answer is 3/2 If you would like the steps to get to the answer, let me know, I can e-mail it to you. Thanks, and keep up the great work. Your videos are awesome. Cheers!
Can anyone tell me the solution of this functional equation or its source please :) The problem: Let α,β ∈ Q* . Find all functions f: R ->R Such that: f([x+y]/α)=[f(x)+f(y)]/β
Solution: If α≠β, then f(x)=0 only. [f(0)=0, f(1/α)=C, f(1)=C/β, f(2)=2C/β, f(2/α)=2C, ... by induction f(k)=kC/β and f(n/α)=nC. Use α=p/q for either C=0 or contradiction with α≠β. You can choose f(x/α)=C, f(x)=C/β for arbitrary x and repeat this argument. So f(x)=0. ] If α=β, then let B is a base of Q in R. I.e. B is a uncountable set, such that every real number can be written a FINITE linear combination r = q_1*b_1+...+q_n*b_n with rational coefficients. Then for every element b∈B choose arbitrary coefficients c. This give you all possible solutions f(r)=f(q_1*b_1+...+q_n*b_n)=q_1*c_1+...+q_n*c_n. Example choice b_1=1, c_1=1, and for all other b∈B/{1} choice c=0: f(x)=x for rational and f(x)=0 for irrational.
@@lubosdostal8523 thanks I'm sorry I forgot to write that α≠β But I have a few questions I think when you plug x=y=0 we will get That βf(0)=2f(0) I think we should discuss two cases when f(0)=0 & β=2 And one more question what is c and what did you plug to get f(1) & f(1/α) I'm so sorry for asking a lot of questions 😅
@@GG-tw8rw Ok, you are right, the case β=2 has to be done separately... Its similar to the case α=β. The case β≠2: Sorry for previous typoes. Set f(α)=C and try to insert possible x,y: 1 = (α + 0) / α => f(1) = (f(α) + 0)/β = C/β 2 = (α + α) / α = (2α + 0) / α => f(2) = (f(α) + f(α))/β = 2C/β = (f(2α) + f(0))/β = f(2α)/β => f(2α) = 2C 3 = (2α + α) / α = (3α + 0) / α => ... ... 0 = (x + (-x)) / α => ... ... For α=p/q then: ? = f(p) = f(qα) = ? ... => C=0 and f(x)=0 for x ∈ Q ... Start again with arbitrary irrational r ... f(r)=D to show that D must also be 0. And finish it by yourself, you can do it 😅
Hey @Michael Penn , i'm a big fan of your channel , i was wondering if you could solve a problem from a math contest in my school , the problem reads : a , b , c are three positive real numbers , prove that 2a/(3a²+b²+2ac) + 2b/(3b²+c²+2ab) 2c/(3c²+a²+2bc) ≤3(a+b+c)
v o i d the inequality doesn't appear to hold. For a=b=c the left hand side becomes 1/a, and the right hand side becomes 9a. That doesn't hold for all real positive a. Just pick a small enough number.
do you actually mean 2a/(3a²+b²+2ac) + 2b/(3b²+c²+2ab) 2c/(3c²+a²+2bc) ≤3/(a+b+c)? if so, you just need to notice that 2a/(3a²+b²+2ac)≤1/(a+b+c) by plain calculation and that's the end of story
For completeness, if you meant to divide by (a+b+c), the steps would be as follows. We know squares are nonnegative, so (a-b)²≥0, a²-2ab+b²≥0, a²+b²≥2ab, 3a²+b²+2ac≥2a²+2ab+2ac, (added 2a²+2ac) 3a²+b²+2ac≥2a(a+b+c), 1/(a+b+c)≥2a/(3a²+b²+2ac), where we can divide because the variables are all positive. We can repeat the steps for the other two terms and add the inequalities to finish the proof.
Thank you for the video! By the way, right now I'm struggling with an infinite sum of square powers of x, |x| < 1 (1 + x + x^4 + x^9 + x^16 + ...) Do you think there is a way to represent it as some kind of integral or connect it to another sum? Any thoughts appreciated)
Thanks 4 u :) U does not speak as fast as usualy eng speakers speak, so it is easier 4 people from other coutries to understand😁 sry 4 my english)) Привет из России!😘 Can you explain the proof of Stirling's approximation?
The double sum of a triple integral isn’t how I expected to be spending my Saturday evening.
Alas, here we are.
A year later, this is how I spent my Saturday Morning!
Another year lated I am spending my sunday morning
In previous videos it has happened that I had difficulties in reading what you wrote on the blackboard, I thought it was because of the chalk colors or the illumination of the room. This video is perfectly clear. Please keep this same setting while filming, it works very well.
Absolutely beautiful. Certainly nothing I would be comfortable trying on my own at my current level of math. But using so many different results and techniques... I just love it!
Informative. It's straightforward to generalize from 2 infinite sums presented here to N such sums. E.g., N=3 would correspond to inf't sum over indices (m,n,k) of 1/[mnk(m+n+k)], & so on. The general sol'n is Gamma(N+1)xZeta(N+1), where "Gamma" denotes the familiar Gamma function, which is just the factorial over the natural numbers. Note that it works even for N=1, which is just the inf't sum over n of 1/n^2. Also, recall that the Zeta funct'n has a simple form over even natural numbers. E.g., for N=3 the triple sum is just (pi^4)/15, & for N=5 it's (8/63).(pi^6), etc.
Michael, I enjoy watching, learning about 10% of what you say, and learning how ignorant of high level math for the 90%. Fantastic!
You can do it in another way with partial fractions, and get to the form Hn/n^2 which is 2zeta(3) (proved by another video of yours)
Do you have the link of that video please ? I can't find it =(
@@zza7195 ruclips.net/video/5OPLW8wH_Po/видео.html
@@nombreusering7979 thk
Thank you obamasphere
Thanks for pointing this out.
This is exactly how I started it then got stuck on sum(Hn/n^2) and completely forgot that I solved it some time ago when Michael brought it up. So used the triple integral method which happens to be the same approach here, and honestly, it easier than going thru the harmonic series.
13:20 these two integrals inside are exactly same thing. \int_0^1\frac{dy}{1-y z}=\int_0^1\frac{dx}{1-x z} , the only change is a private integration variable. No need to consider them separately.
This is a great math channel - I don't know how I (or Google's algorithm) missed it for six months (!)
Same
Third tool is especially useful. And I am also expecting some overkill series as well!
Its a substituted version of one of the lemmas of Gamma function basically evaluated at 2
Can you do a video about burnside’s lemma? And it’s applications to math olympiad it could be a part of your overkill series!
I think the change of order is more like Tonelli's theorem (it's like Fubini's Theorem but for positive functions, and it works with combinations of sums/integrals and integration w.r.t. sigma-finite measures, or whatever)
Dominated Convergence Theorem helps you deal with +/- signs and functions that vary with the index but within a shared integrable absolute ceiling. I guess Monotone Convergence Theorem can also be applied, because adding positive terms to a sum will give you a pointwise-increasing total as n increases.
We can write 1/(m*n*(m+n)) as f(m,n)-f(m,m+n)-f(m+n,n) where f(x,y)=(x+y)/(x^2y^2). If we create a Stern-Brocot-like tree, where instead of fractions we put pairs, and (m,n) as a parent leads to (m,m+n) and (m+n,n), eventually we hit every pair of coprime numbers. If we add up 1/(m*n*(m+n)) level by level on this tree, we'll get f(1,1) minus the lowest-level leaves, so the sum is at most f(1,1)=2.
It's exactly 2, however. If we go down any branch, there will eventually be a point where one of the two numbers will be greater than n. Let us add up f(a,b) where (a,b) is one of these points: that is, either a>n>=b and a-bn>=a and b-a
very nice
Please start lectures on discrete mathematics!
Very neat and clear. These multiple summation/integration tricks were really cool (did the transformations on my own up 'til that point, then was "watching & learning"). Like videos like that
A closer formula to this series for simply supported plate by the navier method
Such a nice sum!
This video is fascinating. Forty years ago, the Dominated Convergence Theorem was not commonly taught at the undergraduate level because it is established using concepts from measure theory. Are measure theory and the Dominated Convergence Theorem currently taught to undergraduates?
Still waiting for that "And that's a good place to stop"
I think that was an incredible use of simple tools.
elegant solution
Indeed, very nice and very nice presentation. I wonder what happens if we have something similar, like triple sum.
Love your taps that cut the video so seemlessly
"And that's a good place to stop".
Crazy result, really unexpected
such a fucking fresh channel loving it man
Nice ,but we directly see Harmonics here
S=Sigma from m=1 to infinity 1/(m^2) sigma from n=1 to infinity (1/n -1/(m+n))
=sigma from m=1 to infinity H_m/m^2
=2zeta(3)
So easy
What is the definition of a double infinite sum? What is the order of limits? Is it lim(sum(lim(sum(...))))? Or maybe lim(lim(sum(sum(...))? Perhaps the limit is multivariable? Is there even a difference between these?
lim lim sum sum =?= lim sum lim sum
There are a few theorems allowing you to interchange the middle lim and sum (in other words lim sum = sum lim). I'm pretty sure dominated convergence is one of them.
So if the sum is absolutely convergent (sum |f| exists) the above statement is correct.
Also sum_0^infty sum_0^infty is defined as lim sum lim sum.
Have you done something to improve the sound? This vid is much easier to listen to than previous ones. I'm glad because I love the quirky content.
Great video, simple and clear. Thank you!!
Hi Michael,
W.r.t. the double-sum that you solved in this video, here is a slight variation of it (I thought I would share with you):
Find
Sum(n=1,inf) Sum(m=1,inf) 1/(m*n*(m+n+2))
The answer is 7/4
Interesting how this one has a closed-form.
Very good method of teaching . My heart is filled with joy .May God bless him .
plz do summation of binomial coefficients also.
Wow
I tried to learn this stuff 45 years ago, to no avail. I'm still confuzzled.
Series integral....=integral series only when there is the uniform convergention. Series x^n doesn't convergent uniformly on (0,1),
Wow so early, you make excellent videos Michael
Basically the power of Harmonics is immense, it dominates logarithmic integrals and even such sums .We can escape those manipulations once we command harmonics
Did the sum over three indices and got 3*2*zeta(4). I bet there is a general formula for any number of indices: t!Zeta(t+1) where t is the number of indices
You used twice the dominated convergence theorem, while I think in both cases it should have been easier by monotone convergence theorem. For using DCT, what would be the dominating function, in both case?
Great teaching style
Counting this Sum other way gives
sigma(H_n/n^2) from 1 to infinity where H_n is harmonic number from 1 to infininty
implies zeta(3)=1/2(sigma(H_n/n^2))
isnt it interesting🤔🤔🤔
Yes, this is an example of a relation among so-called multiple zeta values. It corresponds to zeta(3) = zeta(2,1). See here for another proof: ruclips.net/video/_HHwnVArDm4/видео.html
Cool Apery sum.
Make more such videos
That’s indeed awesome, amazing！
Hi Michael, can you do a video about z-transformation application in determining series like fibonacci's?
I was hopping for PI or e to pop outa no where. Guess not this time then
wikimedia.org/api/rest_v1/media/math/render/svg/b786732ab9ec65d7792e594026e43d1cb659188c
12:20 Missed an edit there, did you? ;)
a nice math channel 👍
Michael I'm glad you're teaching at a slower pace now, in the past I've had to slow you down to .85x in order to follow without proof by intimidation lol
I do have a question, u never explained why we're evaluating this as an integral to begin with. And why are we using those specific tools? Some of this stuff I can gather on my own but it'd be nice to hear the logic explained!
Starting at 6:40 he explains why we are using integration. Basically it is something a little easier to work with, oddly enough.
Very nice!!!
Greetings from Chile. I wonder what's the thought process to change a fraction to an integral to make the problem solvable O_o
When you used the geometric series it should (technically) be noted that the integrals were going from 0 to 1 right? @9:25
you are right. There's small potential difficulty that the argument doesn't work when xz=1. Even though it doesn't work it doesn't matter, since whatever is the value of the function at a single point (or more generally on a null subset) - it can be even infinity, it doesn't change the integral.
Nice touching the board and having the colour appear.
TL;DR - There is no closed form expression involving pi, exp, trig, inverse functions, roots, and arithmetic, but there is a much nicer infinite single-sum form.
Is there any other solution without using integral? if there is, can someone explain it to me?
i looove your videos
At the start first thing came up to mind iz the reiman zeta function
This was a very nice double sum indeed 🍺🍺🍻.
Isn't the Riemann zeta function at 3 equal to apery's constant?
Zeta of 3 is the definition of apery's constant
That editing though
This is incredibly counter-intuitive to me. You start with a discrete function on a rational domain with a lower bound of 1, apply a bunch of contiinuous transformations with real domains and end up with an analytically continued function with a complex domain that excludes {1}.
It doesn't need to be analytically continued to be defined at 3.
@@SugarBeetMC true but picky! :)
Why should it be a problem that the Riemann zeta function is not defined in 1? Zeta(3) doesn't care what happens at 1
Hello my friends. I come from the future and Michael is going to coin a catchy phrase in the end of his videos "This is a good place to stop"
Wonderful!
Amazing!
Let n be fixed, then
1/mn(m+n)=1/n[1/m(m+n)]=1/n^2[1/m - 1/(m+n)]
Sum 1/mn(m+n) for m=1 to inf
=1/n^2 [Sum(1/m-1/(m+n)) for m=1 to inf]
=1/n^2 [1/1-1/(n+1)+1/2-1/(n+2)+...+1/n-1/(n+n)+1/(n+1)-1/(n+n+1)+...], this is telescoping series
=1/n^2[1/1+1/2+1/3+...+1/n]
=Hn/n^2
And it converges to 2Zeta(3)
Is the sum equal to sum of H_n/n^2, where H_n is the nth harmonic number.
I see that someone else noticed the same thing.
Nice one!!
Hi Michael,
Hope you get to see this comment.
I can tell you have a passion for solving tough questions regarding limits, sums, and integrals (they are my favorite, too)
Below I am providing one tough question that has a closed sum, thought it might interest you enough to create a video on it.
Sorry I couldn't provide a picture of it (e.g. in LaTeX), I know that might have expressed the question better... anyway, here it is:
Find
Limit(n->inf) Integral(x=0,inf) [Sum(m=1,inf) (m+x)/((m^n+x^n)^n)] dx
The answer is 3/2
If you would like the steps to get to the answer, let me know, I can e-mail it to you.
Thanks, and keep up the great work. Your videos are awesome.
Cheers!
Can anyone tell me the solution of this functional equation or its source please :)
The problem:
Let α,β ∈ Q* . Find all functions f: R ->R
Such that:
f([x+y]/α)=[f(x)+f(y)]/β
Solution:
If α≠β, then f(x)=0 only. [f(0)=0, f(1/α)=C, f(1)=C/β, f(2)=2C/β, f(2/α)=2C, ... by induction f(k)=kC/β and f(n/α)=nC. Use α=p/q for either C=0 or contradiction with α≠β. You can choose f(x/α)=C, f(x)=C/β for arbitrary x and repeat this argument. So f(x)=0. ]
If α=β, then let B is a base of Q in R. I.e. B is a uncountable set, such that every real number can be written a FINITE linear combination r = q_1*b_1+...+q_n*b_n with rational coefficients. Then for every element b∈B choose arbitrary coefficients c. This give you all possible solutions f(r)=f(q_1*b_1+...+q_n*b_n)=q_1*c_1+...+q_n*c_n.
Example choice b_1=1, c_1=1, and for all other b∈B/{1} choice c=0:
f(x)=x for rational and f(x)=0 for irrational.
@@lubosdostal8523 thanks
I'm sorry I forgot to write that α≠β
But I have a few questions
I think when you plug x=y=0 we will get
That βf(0)=2f(0)
I think we should discuss two cases when f(0)=0 & β=2
And one more question what is c and what did you plug to get f(1) & f(1/α)
I'm so sorry for asking a lot of questions 😅
@@GG-tw8rw Ok, you are right, the case β=2 has to be done separately... Its similar to the case α=β.
The case β≠2: Sorry for previous typoes. Set f(α)=C and try to insert possible x,y:
1 = (α + 0) / α => f(1) = (f(α) + 0)/β = C/β
2 = (α + α) / α = (2α + 0) / α => f(2) = (f(α) + f(α))/β = 2C/β = (f(2α) + f(0))/β = f(2α)/β => f(2α) = 2C
3 = (2α + α) / α = (3α + 0) / α => ...
...
0 = (x + (-x)) / α => ...
...
For α=p/q then: ? = f(p) = f(qα) = ? ... => C=0 and f(x)=0 for x ∈ Q
...
Start again with arbitrary irrational r ... f(r)=D to show that D must also be 0.
And finish it by yourself, you can do it 😅
@@lubosdostal8523 thank you so much 😍❤
This is why I love math. That is a very nice sum!
Awesome! 😍🔥
Nice video. It's a toughest question.
Hey @Michael Penn , i'm a big fan of your channel , i was wondering if you could solve a problem from a math contest in my school , the problem reads : a , b , c are three positive real numbers , prove that
2a/(3a²+b²+2ac) + 2b/(3b²+c²+2ab) 2c/(3c²+a²+2bc) ≤3(a+b+c)
v o i d the inequality doesn't appear to hold. For a=b=c the left hand side becomes 1/a, and the right hand side becomes 9a. That doesn't hold for all real positive a. Just pick a small enough number.
do you actually mean 2a/(3a²+b²+2ac) + 2b/(3b²+c²+2ab) 2c/(3c²+a²+2bc) ≤3/(a+b+c)? if so, you just need to notice that 2a/(3a²+b²+2ac)≤1/(a+b+c) by plain calculation and that's the end of story
@@looming314 Yeah she mostly meant that.
For completeness, if you meant to divide by (a+b+c), the steps would be as follows. We know squares are nonnegative, so
(a-b)²≥0,
a²-2ab+b²≥0,
a²+b²≥2ab,
3a²+b²+2ac≥2a²+2ab+2ac, (added 2a²+2ac)
3a²+b²+2ac≥2a(a+b+c),
1/(a+b+c)≥2a/(3a²+b²+2ac), where we can divide because the variables are all positive.
We can repeat the steps for the other two terms and add the inequalities to finish the proof.
2a/(3a^2 +b^2+2ac) = 2/(a+b^2/a + 2a + 2c) =< 2/(2b +2a +2c) by the AM-GM inequality.
A me risulta.... Sommatoria m from 1 to infty di (1/m^2)×Hm...È corretto? Non riesco a semplificarlo ulteriormente. GRAZIE
Thank you for the video!
By the way, right now I'm struggling with an infinite sum of square powers of x, |x| < 1 (1 + x + x^4 + x^9 + x^16 + ...)
Do you think there is a way to represent it as some kind of integral or connect it to another sum?
Any thoughts appreciated)
Maybe Google Theta function or Jacobi Theta function
@@angelmendez-rivera351 this is certainly an amazing first step, but I think it would be needed to continue to transform this sum into a closed form
The sort of calculation you'd see in string theory. = )
2 × Apery's constant = sum( lnx/x^2), x=1 to infinity.. 😎😌
I remember this from StackExchange
What!? Not a good place to stop? 🤣
Sick
everyone gangsta till he pulls out the triple integral 😳😳😳😳
Nice one
2.388536
Still looking for a good place to stop.
So about 2.404. My guess had been about 2.
Thanks 4 u :) U does not speak as fast as usualy eng speakers speak, so it is easier 4 people from other coutries to understand😁 sry 4 my english)) Привет из России!😘
Can you explain the proof of Stirling's approximation?
Super
Great as usual. One remark: when watched landscape on my phone the quality is poor.
He's driving me up the wall because he keeps on saying "zee" instead of Z
It is hard on Canadians for sure.
Canadians need to cope lmao
I’m sorry that was mean
You missed the thing " and this is a good place to stop."
11 обезьян поставили дизлайк XD
A me sembra corretto e l'ho fatto in 3 passaggi... Complimemti
1st
No I am
😅
End is anticlimactic. I thought some π type thing will show up.
Nice one
You missed the thing " and this is a good place to stop."