Beautifully explained! I'm an American uni student that has probably bit off more than I can chew in terms of my Real Analysis class so I turned to your video for help and I cannot thank you enough for taking the time to explain each and every step from scratch. Wikipedia, Math exchange, lecture notes, etc. have all failed me due to omitted "known" information but you sir, have not. Admittedly, I was a little confused at the part in which you compared the coefficients of both sequences, but I eventually got it. Your video put a smile on my face by making me understand maths again! Thanks you!!
It's a beautiful proof, mostly well-stated. The missing piece is that one has to impose that sin(x)/x =1 at x=0. That is the reason it's divided through by n^2pi^2. For example, a polynomial p(x)=2x^2-8x+6 has zeros 1 and 3. But then it is incorrect to assume p(x)=(x-1)(x-3). Knowing p(0)=6 demands the (x-1)(x-3) be scaled up by a factor of 2.
I do remember both. 🤣🤣 And such facts are true because you are doing these observations for a neighborhood of 0, not for itself hahaha. Btw, good job on the proof ✌🏼
Not rigorous without a proof of the Weierstrass factorization thm. I think this is the reson why Euler's contemporaries did not accept his first proof. Weierstrass came much later. How do we know, for example, that the product you reference does not converge to e^(-x^2) times sin(x)/x ? I do the same thing in a math class I teach, but always am concerned because of this missing piece. That class doesn't have the analytic function theory needed for the complete proof and I thought this video would show me a different way.
That was amazing,sir.Thank you for your time and effort.I don't usually comment under videos,but yours was really useful.You just gained a new subscriber.
fourier series direct method: the fourier series of |x| on [-pi, pi] is pi/2 - (4/pi) sum(n odd >= 1) cos(nx) / n^2, this converges absolutely to |x|, if you let x=0 and do some algebra you get basel sum (its actually not too difficult to show that the fourier series converges uniformly to the original function if the sum of the coefficients converges absolutely, i would argue the analytical machinery behind that is much simpler than whats needed for weierstrass product) residue theorem method: the function f(z)=cot(pi z)/z^2 has a triple pole at 0 and a simple pole at other integers, the residue at 0 is -pi/3, the residue at n not equal to 0 is 1/(n^2 pi). it can be shown with a square contour that if g(z) is O(1/z^2), then the sum of the residues of g(z)cot(pi z) is equal to 0, hence -pi/3 + (2/pi) * sum from n=1 to infty of 1/n^2 = 0. (this method generalizes very neatly for calculating many different infinite series of the form 1/P(n) where deg P >= 2)
@@Packerfan130 I’m sorry that you feel that way, I made that video 2 years ago and I can’t remember all the steps - why don’t you have a go and prove it by yourself and tell me where I made the mistake
Perhaps another commenter noted this already but you should have sin(x)/x = (1-x/π)(1+x/π)(1-x/2π)... Now when you take the limit as x approaches 0, you get 1 on both sides. As you had it, you would get 1 on the left side and an undefined infinite product on the right hand side. In particular, on the right side you'll have the infinite product π(-π)(2π)(-2π)... after you take the limit as x tends to 0.
you have constructed a polynomial whose zeros match those of the function, but multiplying that polynomial by a constant k would also satisfy the condition, but you have choosed k=1. Why?
The Weierstrass expansion needs to be explained better. Otherwise sin(x) = 2sin(x) = sin(x)^30 = exp(x) sin(x) because their roots coincide.
Beautifully explained! I'm an American uni student that has probably bit off more than I can chew in terms of my Real Analysis class so I turned to your video for help and I cannot thank you enough for taking the time to explain each and every step from scratch. Wikipedia, Math exchange, lecture notes, etc. have all failed me due to omitted "known" information but you sir, have not.
Admittedly, I was a little confused at the part in which you compared the coefficients of both sequences, but I eventually got it. Your video put a smile on my face by making me understand maths again! Thanks you!!
It's a beautiful proof, mostly well-stated. The missing piece is that one has to impose that sin(x)/x =1 at x=0. That is the reason it's divided through by n^2pi^2. For example, a polynomial p(x)=2x^2-8x+6 has zeros 1 and 3. But then it is incorrect to assume p(x)=(x-1)(x-3). Knowing p(0)=6 demands the (x-1)(x-3) be scaled up by a factor of 2.
Thank you, yes perhaps I could have explained this more clearly!
Thumbnail said "Proven carefully", the first opperation the dude does is to divide by x, without assuming that x is not 0. (Contains irony xD)
Unless you remember that the limit as x approaches 0 of sin(x)/x = 1!
And you’ll also remember that the Maclaurin expansion is the Taylor Expansion about the point x=0 😬
I do remember both. 🤣🤣 And such facts are true because you are doing these observations for a neighborhood of 0, not for itself hahaha. Btw, good job on the proof ✌🏼
@@IsnardLopes thank you 🙏 this is true, but 0 is contained within that neighbourhood
@@MrH1207 sinx/x is not defined at 0, so more work (which I also don't know) needs to be done, to make the proof more rigorous.
Not rigorous without a proof of the Weierstrass factorization thm. I think this is the reson why Euler's contemporaries did not accept his first proof. Weierstrass came much later. How do we know, for example, that the product you reference does not converge to e^(-x^2) times sin(x)/x ? I do the same thing in a math class I teach, but always am concerned because of this missing piece. That class doesn't have the analytic function theory needed for the complete proof and I thought this video would show me a different way.
That was amazing,sir.Thank you for your time and effort.I don't usually comment under videos,but yours was really useful.You just gained a new subscriber.
Thank you very much! I’m really glad you found the video useful! ☺️
fourier series direct method: the fourier series of |x| on [-pi, pi] is pi/2 - (4/pi) sum(n odd >= 1) cos(nx) / n^2, this converges absolutely to |x|, if you let x=0 and do some algebra you get basel sum (its actually not too difficult to show that the fourier series converges uniformly to the original function if the sum of the coefficients converges absolutely, i would argue the analytical machinery behind that is much simpler than whats needed for weierstrass product)
residue theorem method: the function f(z)=cot(pi z)/z^2 has a triple pole at 0 and a simple pole at other integers, the residue at 0 is -pi/3, the residue at n not equal to 0 is 1/(n^2 pi). it can be shown with a square contour that if g(z) is O(1/z^2), then the sum of the residues of g(z)cot(pi z) is equal to 0, hence -pi/3 + (2/pi) * sum from n=1 to infty of 1/n^2 = 0. (this method generalizes very neatly for calculating many different infinite series of the form 1/P(n) where deg P >= 2)
رائع جدا.
أول فيديو في القناة اشاهده.أعجبني كثيرا.
متتبعك من مدينة تنغير بالمملكة المغربية
6:41 you have to explain in detail，how can we eliminate -pi^2 directly on the right and the left side have no change
Great JOB! Well explained.
How can you set (x + pi)(x - pi)... = 0, then set it equal to sin(x)/x?
At which point?
He really can't. Not like this, without invoking any justification.
Thank you so much !! that was very clear !!
Great video, keep up the good job
how do you go from
sinx / x = (x^2 - pi^2)(x^2 - 4pi^2)...
to
sinx / x = (1 - x^2/pi^2)(1 - x^2/4pi^2)...
where did the -pi^2 in each factor here go?
That is left to you as an exercise! 🤣
@@MrH1207 cool, another math youtube that just writes steps down and doesn't explain them and uses laughing crying emoji randomly, got it.
@@Packerfan130 I’m sorry that you feel that way, I made that video 2 years ago and I can’t remember all the steps - why don’t you have a go and prove it by yourself and tell me where I made the mistake
@@MrH1207 no one here said you made a mistake.
Perhaps another commenter noted this already but you should have
sin(x)/x = (1-x/π)(1+x/π)(1-x/2π)...
Now when you take the limit as x approaches 0, you get 1 on both sides. As you had it, you would get 1 on the left side and an undefined infinite product on the right hand side. In particular, on the right side you'll have the infinite product
π(-π)(2π)(-2π)...
after you take the limit as x tends to 0.
Thank you for commenting on this!
So you should initially have
sin(x)=
x(1-x/π)(1+x/π)(1-x/(2π))(1+x/(2π))...
@@user-lb8qx8yl8k I believe that the proof still works, regardless of this small detail. But thank you for mentioning it anyways
@@MrH1207 -- You're welcome but I botched it a bit myself. I just now edited it.
@@MrH1207 -- A proof should not have any false statements.
sin(x) ≠ x(x-π)(x+π)(
So, someone has proved that the series will converge after expansion? Or do we need to assume x tends to zero?
The Maclaurin expansion is the expansion around the point x = 0. :)
Thank you for this proof, really useful. May i ask where you got it from - which book?
The town is called Buhsel, not Beisel.
Most English speakers are not able to say "Basel"! It's an uncommon sound for them.
why do we need to make the factors into the form of (1-x^2/(npi)^2) but not keeping (x^2-(npi)^2)?
It’s so that when we come to expand it out, we have everything finite! If we didn’t, then our power of x^2 would go racing off to infinity
Brilliant!
Did you publish this proof?
Hahahahahaha. As if this proof is something original ...
Factorization theoreme implies a not justified multiplicative constant ?
Could you explain a little bit more?
you have constructed a polynomial whose zeros match those of the function, but multiplying that polynomial by a constant k would also satisfy the condition, but you have choosed k=1. Why?
@@josepcolominas5616 Consider both sides when x=0 (or more formally limx->0) to get k = lim x->0 of sinx/x = 1
Nice
Well Done
X=0（Yes)
to prove carefully, u need to even prove the intermediary theorems, such as the Weierstrass product formula; this wasn’t done
It's not prononuciated Zahlen but "Sahlen"
Thank you.....😂
Please move the microphone away from the stylus. The clicking is too loud.
X = 0
🤬🤬🤬🤬🤬🤬🤬🤬🤬🤬🤬🤬🤬
Idk x= 0