Residue theorem coming in to flex its elegance! Combining π from the denominator and the series for csc(x) is such a beautiful move, it’s like watching math poetry in action. Props to our integral suggester for sparking conversations like these. If integrals make you feel like hunting for tricks, tools like SolutionInn can help streamline the process and make every solution feel as satisfying as this one!
Props for aiding search for the referenced video. Regarding the pedogogy, I would love an explanation of *how* the ideas could fall out. For this problem, not so much the "why are we using this series", as why I(a) gets defined the way it does. (granted a lot of Leibniz integral rule applications seems to appear from the aether, but they can't all be "I tried a billion other things, but the billion and first one worked.") Regarding the math, Heaviside is knocking wondering why you've forgotten him, especially considering you made a video with a title where you are chastised by Heaviside for forgetting him.
I may be mistaken so apologies to all parties if this is so but ... Motivation on something like this seems to be: is one sufficiently motivated to do the implied humph work and is one sufficiently motivated to find time to do the humph work. The practical hands-on humphwork induces in one a deeper understanding rather than following working methods like a cooking recipe. In this recipe we use arctan and fraction decomposition and bundling up awkward terms into something easier to handle like, for example, alpha. If I did not have a deficiency in time I would be tempted to delve into complex integration AND do the humphwork to see what dribbled out. Disclaimer: I see pedagogical considerations as an nth divisor of humphwork
Residue theorem with semicircle works without having to hunt for a clever trick. Residue from denominator will contribute π. Residue from numerator at z = (2k+1)i/4π is (-1)^k 2πi/(4π^2 - (2k+1)^2), and you will find Σ_k (-1)^k (4π^2)/(4π^2 - (2k+1)^2) using series for csc(x). Combine both for final answer. Unfortunately, no way to get around needing a helper identity originating from sin(x) infinite product. I'd like to be disproven.
Naively as in very naively ... complex integration looks it would help on this but I wish I could do complex integration to find out! Reziduals at +i and -i ?
I think the original series should have been for sec(pi x) rather than just sec(x) so that the asymptotes line up
Residue theorem coming in to flex its elegance! Combining π from the denominator and the series for csc(x) is such a beautiful move, it’s like watching math poetry in action. Props to our integral suggester for sparking conversations like these. If integrals make you feel like hunting for tricks, tools like SolutionInn can help streamline the process and make every solution feel as satisfying as this one!
3:07 is (x/alpha)^2 correct here? The i^2 changed the negative to positive but what happened to the pi^2...?
I guess what happened is that the series is indeed 4*pi and not 4/pi originally. When you divide 4pi/pi^2 you do get 4/pi
I think the issue here is that the original expression for sec(x) should have been for sec(pi*x)
@@ruilopes6638No, it is 4/pi times the series.
That would also make sense
Can also be evaluated by contour integral with pole at z = i and z = (2n+1)i/4, n = 0, 1, 2 ...
Props for aiding search for the referenced video.
Regarding the pedogogy, I would love an explanation of *how* the ideas could fall out. For this problem, not so much the "why are we using this series", as why I(a) gets defined the way it does. (granted a lot of Leibniz integral rule applications seems to appear from the aether, but they can't all be "I tried a billion other things, but the billion and first one worked.")
Regarding the math, Heaviside is knocking wondering why you've forgotten him, especially considering you made a video with a title where you are chastised by Heaviside for forgetting him.
I may be mistaken so apologies to all parties if this is so but ...
Motivation on something like this seems to be: is one sufficiently motivated to do the implied humph work and is one sufficiently motivated to find time to do the humph work.
The practical hands-on humphwork induces in one a deeper understanding rather than following working methods like a cooking recipe.
In this recipe we use arctan and fraction decomposition and bundling up awkward terms into something easier to handle like, for example, alpha.
If I did not have a deficiency in time I would be tempted to delve into complex integration AND do the humphwork to see what dribbled out.
Disclaimer: I see pedagogical considerations as an nth divisor of humphwork
I posted an alternate solution which requires less motivation
13:57
@ 5:24 b should be equal to (2n+1)*alpha/pi
Thank you - Integral Suggester
4:11 I don‘t understand the numerator of the integrand! The alpha^2 comes outfront, ok. But why this extra (x^2+1)?
He is substituting in his definition of sech() into the original integral in the top left which has an (x^2 + 1) in the denominator.
I think in your tool it should say sec(Pi*x)=4/Pi*sum(...).
I have not understood well some things such as at 3:07
Residue theorem with semicircle works without having to hunt for a clever trick. Residue from denominator will contribute π. Residue from numerator at z = (2k+1)i/4π is (-1)^k 2πi/(4π^2 - (2k+1)^2), and you will find Σ_k (-1)^k (4π^2)/(4π^2 - (2k+1)^2) using series for csc(x). Combine both for final answer.
Unfortunately, no way to get around needing a helper identity originating from sin(x) infinite product. I'd like to be disproven.
Is it possible to integrate x^2 * sech(x)^n with +- infinity boundaries and n being a variable from 0 to infinity??
This looks like it can be done with Abel-Plana formula
Naively as in very naively ... complex integration looks it would help on this but I wish I could do complex integration to find out! Reziduals at +i and -i ?
what happened to the video contrast?
Hi,
Ok, cool!
Why has the image been so pale lately?
infinity， 0