Komisch, in Theorie und Anwendung der unendlichen Reihen von Konrad Knopp wird dieses G nicht erwähnt. Erst im Transkript konnte ich den Namen CATALAN lesen. Sie sprechen ihn soooo schnell aus, dass ich den guten CATALAN ni ht wahrnehmen konnte. Wieso so rasant?
0:00 Form Zero 0:30 sin(x) = f(e^ix) esoteric identity 0:49 multiply top and bottom by e^-ix 1:15 rewrite division as multiplication 1:20 series expansion of recognised component
I wonder why you think your approach of evaluating the real parts of the integrals on both sides of the equation should not be not rigorous. The only thing which is not rigorously proven is the change between summation and integration, which most people including myself as a mathematician usually don‘t want to spend a lot of time on… :-)
Nice. You can modify the integral slightly to eliminate the Catalan constant: Zeta(3)=(2Pi^3/7)Int_0^1/2 x(x-1)csc(Pi x). I derived this by a more circuitous route. You can also show Zeta(3)=(4Pi^3/7)Int_0^1/2 x(1-2x)tan(Pi x)=(4Pi^3/7)Int_0^1/2 x(1-2x)cot(Pi x).
5:00 this step is 100% rigorous, why are you doubting yourself? If z=w then Re(z)=Re(w). Yes you could show the steps one by one by taking the summation k=0..N, THEN let N-->infinity, which would 100% show that the Re part of the first integral is zero. But other than this minor question of the convergence of the first sum, there's no doubt as to the validity of this operation.
It's the structure of the cosine integral....it doesn't look like it goes to zero....maybe I'm just paranoid but I'm bothered by it....nd it's all your fault 😂😂😂...you've sparked my interest in mathematical rigor beyond just plain problem solving
@@zunaidparker Because it's not true in general that an integral evaluates precisely to a real result. i.e: it may actually result in a complex number, and you'd be left with an incomplete solution. Some may argue that it's a real integrand over a defined interval - however, Mathematical rigor simply demands one thing: consistency. It seems like a hand-wavy argument to me, something which a physicist may do to solve an integral (knowing that it actually converges to the desired value). This approach assumes that the integral is purely real, which is inconsistent in general. Atleast that's my two cents.
@@daddy_myers An integral of a real function over a real interval gives a real answer. That is 100% clear, you could go back to the definition of a Riemann integral to see that it is only real multiplication, summation and limits involved, but that feels like an overkill level of fundamentals for where this problem is pitched at.
@@maths_505 each integral within the sum is a finite real number (it doesn't need to be zero), therefore the limit of the partial sums is also a finite real number, which when multiplied by 2i becomes pure imaginary. Since you can move the Re function inside the sum (it is continuous), it makes the first sum over k collapse into a sum of zeroes, which is zero.
It is infact how sin works. He hasn't included a bracket which is a choice of notation. It should be sin{(2k+1)*(π/2)}. You can verify for yourself that sin{(2k+1)*(π/2)} = (-1)^k for integer values of k. This is why everyone should write sin(x) instead of sin x to avoid confusion.
The super non-rigorous part is 1:45 where you replace one term with a non-divergent series. There must be behind the scenes some nontrivial reason why this nonsense worked out in the end. Maybe using divergent sums methods we could save this approach, but that would need some unusual theory. That's why you have doubts about the innocent idea at 5:10.
@@hayksargsyan4416 I have noticed that even if an integral doesn't converge, sometimes using a damping factor like e^(-λx) and taking the limit as λ→0 results in a finite value. Such an approach is often used to define the Fourier transform of functions in L²\L¹.
@@maths_505 What about the cosine integral did you worry about? Since it was multiplied by 2i it was part of the complex part that you safely threw away. The only real part was the integral with sine in it.
The more you think about it the more mathematically sound it becomes... Maybe it was just a "feeling" that went hold up....it really is zero right? Thankfully, wolfram alpha says so
@@maths_505 The cosine integral itself is not zero for all k. But when you take the infinite sum from k=0 to k=infinity, the sum converges to zero. The sum has negative and positive terms which somehow cancel out to zero in the end.
Correction: the sum of cosine integrals at the 5:00 mark does indeed evaluate to zero so yes the approach is 100% rigorous
How do you make the geometric series at 1:20 rigorous?
Komisch, in Theorie und Anwendung der unendlichen Reihen von Konrad Knopp wird dieses G nicht erwähnt. Erst im Transkript konnte ich den Namen CATALAN lesen. Sie sprechen ihn soooo schnell aus, dass ich den guten CATALAN ni ht wahrnehmen konnte. Wieso so rasant?
0:00 Form Zero
0:30 sin(x) = f(e^ix) esoteric identity
0:49 multiply top and bottom by e^-ix
1:15 rewrite division as multiplication
1:20 series expansion of recognised component
Thank you for your continued brilliant challenges and refreshing my memory! Trust me - - you are appreciated for this site!😊
I wonder why you think your approach of evaluating the real parts of the integrals on both sides of the equation should not be not rigorous. The only thing which is not rigorously proven is the change between summation and integration, which most people including myself as a mathematician usually don‘t want to spend a lot of time on… :-)
Wow.
Amaizing integral with interesting solution! 💚
I can not wait for Hankel contour.
Nice. You can modify the integral slightly to eliminate the Catalan constant: Zeta(3)=(2Pi^3/7)Int_0^1/2 x(x-1)csc(Pi x). I derived this by a more circuitous route. You can also show Zeta(3)=(4Pi^3/7)Int_0^1/2 x(1-2x)tan(Pi x)=(4Pi^3/7)Int_0^1/2 x(1-2x)cot(Pi x).
Thanks a lot for this brilliant solution👍
You cant just replace 1/(1-e^-2ix) with a geometric series because that only works for |r|
Where is the alternate solution video?
Didn't need that cuz this approach turns out to he rigorous as the sum of cosine integrals is indeed zero
@@maths_505 the geometric series step doesn’t seem rigorous. The exponential has modulus 1.
One way to analyze it is to multiply the common ratio by \epsilon and let \epsilon approach 1 from the left....
5:00 this step is 100% rigorous, why are you doubting yourself? If z=w then Re(z)=Re(w).
Yes you could show the steps one by one by taking the summation k=0..N, THEN let N-->infinity, which would 100% show that the Re part of the first integral is zero. But other than this minor question of the convergence of the first sum, there's no doubt as to the validity of this operation.
Am I missing something more subtle? Your certainty is making me doubt...
It's the structure of the cosine integral....it doesn't look like it goes to zero....maybe I'm just paranoid but I'm bothered by it....nd it's all your fault 😂😂😂...you've sparked my interest in mathematical rigor beyond just plain problem solving
@@zunaidparker Because it's not true in general that an integral evaluates precisely to a real result. i.e: it may actually result in a complex number, and you'd be left with an incomplete solution.
Some may argue that it's a real integrand over a defined interval - however, Mathematical rigor simply demands one thing: consistency.
It seems like a hand-wavy argument to me, something which a physicist may do to solve an integral (knowing that it actually converges to the desired value).
This approach assumes that the integral is purely real, which is inconsistent in general.
Atleast that's my two cents.
@@daddy_myers An integral of a real function over a real interval gives a real answer. That is 100% clear, you could go back to the definition of a Riemann integral to see that it is only real multiplication, summation and limits involved, but that feels like an overkill level of fundamentals for where this problem is pitched at.
@@maths_505 each integral within the sum is a finite real number (it doesn't need to be zero), therefore the limit of the partial sums is also a finite real number, which when multiplied by 2i becomes pure imaginary. Since you can move the Re function inside the sum (it is continuous), it makes the first sum over k collapse into a sum of zeroes, which is zero.
9:50 why is sin(2k+1)*pi/2 = (-1)^k? That’s not how the sine works…
It is infact how sin works. He hasn't included a bracket which is a choice of notation. It should be sin{(2k+1)*(π/2)}. You can verify for yourself that sin{(2k+1)*(π/2)} = (-1)^k for integer values of k. This is why everyone should write sin(x) instead of sin x to avoid confusion.
I have a query: the summation that you have is non convergent
I'm not too sure what the requirements for convergence of the geometric series in the complex plane, if it is either abs(Re(z))
@@mimimom0 sounds plausible
why is taking the real part not rigorous?
I like it!
The super non-rigorous part is 1:45 where you replace one term with a non-divergent series. There must be behind the scenes some nontrivial reason why this nonsense worked out in the end. Maybe using divergent sums methods we could save this approach, but that would need some unusual theory.
That's why you have doubts about the innocent idea at 5:10.
Probably the divergent series can be (at least partly) justified by replacing e^(-2ix) with e^(-λ-2ix), where λ>0, and at the end take limits as λ→0.
@@md2perpewow interesting approach
@@hayksargsyan4416 I have noticed that even if an integral doesn't converge, sometimes using a damping factor like e^(-λx) and taking the limit as λ→0 results in a finite value.
Such an approach is often used to define the Fourier transform of functions in L²\L¹.
@md2perpe that's exactly what was in my mind while evaluating the integral using the series
Are you kidding me? Taking real parts of both sides of the equation is totallly fine. In fact almost every complex equation is solved like that.
Oh that was fine....it was the cosine integral I was worried about.....turns out the sum does indeed evaluate to zero
@@maths_505 What about the cosine integral did you worry about? Since it was multiplied by 2i it was part of the complex part that you safely threw away. The only real part was the integral with sine in it.
The more you think about it the more mathematically sound it becomes...
Maybe it was just a "feeling" that went hold up....it really is zero right?
Thankfully, wolfram alpha says so
@@maths_505 The cosine integral itself is not zero for all k. But when you take the infinite sum from k=0 to k=infinity, the sum converges to zero. The sum has negative and positive terms which somehow cancel out to zero in the end.
Here I spent the last 5 days thinking how in the world to solve this integral, and shazam, sin in terms of e comes into the chat...
Complex analysis to the rescue 😂
OK, good solution, but G is not "katlen's" constant. Catalan is pronounced [Kah-teh-'lan].
There they are, officer - the English teacher. 😐👆
@@RM-gc8lx Bro, I don't get what's wrong with y'all 🤣🤣🤣
This is Math for fun, not language class. Stop being massive nitpicks.