This one was so beautiful. In just one problem there were the basel problem, the laplace transform, geometric series, feynman integrals, euler's formula and even one of the weirdest versions of zero. Very nice.
Agreed. His visuals really good -- both the graphics and the clarity of characters in the equations. I didn't know the chalk is new, but I don't recall seeing some of the colors before and their brightness is helpful. Following the logic in a math presentation can be tough enough. If I had to struggle to read the images or equations, it would be tougher.
@@MichaelPennMath And why dont you write sine of x as [e^ix minus cos x]/i since if you rearrange and solve for sine x that is whatyou get--sine is not only the imaginary part once you divide by the imaginary number..
I could see the complex exponentials coming to unify the trigonometric and hyperbolic functions. The Laplace Transform and Basel Problem were icing on the cake.
Small mistake at 10.58 U cannot put 1/n outside the summation and multiply it with x=t/n U should put x=t/n inside the summation and multiply it with 1/n Then u will get the same solution in the rest of the video
Introducing s, taking its antiderivative, and then using that to complete a known Laplace transform was an absolutely slick move. I had to pause and go over it to make sure it was valid -- it's almost magic!
Wow this problem had almost everything. Having said that, I think it's high time we saw some differential equations with Bessel function solutions on the channel!?
At 12:20 when you pull the derivative out of the integral, why is that possible? Does it have anything to do with the dominated convergence theorem? Do you have a video where you explain how that actually works/when it can and can't be applied?
I was nervous when the denominator was rolled into Imaginary operator at 3:45. Also, at 7:35: don't we need to confirm that we are inside the radius of convergence when applying the formula for the sum of a geometric series? The role of 'r' in a geometric series seems to be played by "e^(-x) * e^(ix)" -- how do we know this value remains less than one, especially when it is in an integral where the value of 'x' can be anywhere between zero and infinity? I see the formula for a geometric series used similarly on other channels and everyone treats it as legit, but I haven't worked through the details to stop me from wondering when I see it used in integrals over ranges much greater than from -1 to 1.
Okay - love this way of doing math. Sometimes math elsewhere but here has a step by step linearity constrained by the topic or theme. In these videos the mathematician professor shows how every tool in the Math toolbox can be used working laterally (I mean lateral and linear thinking coexisting dynamically!). But the main reason I write is continuum. Can anything really be distinct in a continuum? As soon as there is a partitioning of a continuum doesn't that mean we have two continuums separated at a point of partition? For example each graph to add context and interpretation needs to have a scalar somewhere along each axis or a vertex and line segment or ... . Isn't that a partitioning by labeling? And if so doesn't it imply that meaningful interpretation is only added after a partitioning process? I think I may be overthinking this or alternatively overlooking something trivial with aim being that only the quotients seem to give ordering and clumping of things together as meaningful.
EDIT: The integrand x^k Sin(x)/( Cos(x) + Cosh(x)) can be solved in closed form with some straightforward extensions to this derivation. The key is that the x->t/n substitution changes the Laplace transform to L(t^k Sin(t)) instead of L(t Sin(t)) and the power of (1/n)^2 in the sum goes to (1/n)^(k+2). The Laplace transform yields 2^(-(1/2) - k/2) Gamma[1 + k] Sin[1/4 (1 + k) Pi]. And the sum yields 2^-j (-2 + 2^j) Zeta[j] where j is k+1. The value of the integral of x^k Sin(x)/(Cos(x)+Cosh(x)) is the Laplace transform value times the sum value. x^1 gives Pi^2/12 as here. x^2 -> 3/4 Zeta(3), x^3->0, x^4-> 45/8 Zeta(5), x^5-> 31/1008 Pi^6, x^6-> 2835/32 Zeta(7), and x^7 ->0. Original comment: It's interesting that a similar method works for x^k in the numerator instead of just x^1. The Laplace Transform becomes more complicated but has a closed form 2^(-(1/2) - k/2) Gamma[1 + k] Sin[1/4 (1 + k) Pi]. The sum on n is trivial to change since the x->t/n substitution just changes the power of n in the sum which works out to 2^-j (-2 + 2^j) Zeta[j] where j is the power of (1/n)^j in the sum. The only changes to the video are the Laplace transform being for t^k Sin[t] instead of t Sin[t] and the power of (1/n) in the sum.
One of the things I first thought of doing was to write the denominator at e^x+e^(ix)+e^(-x)+e^(-ix). I'm blanking if there is a cyclic function that works with those, though.
I can appreciate the desire for practical applications, but imo separation of theory and examples is the most efficient method of conducting science and math. It's better for theoreticians to spit out the theory without any filler and practicians to seek out the relevant theory for their specific application. Hopefully the latter group also makes their work public so other people can learn from the examples.
Mathematica was able to show that Integrate[x*Sin[x]/(Cos[x]+Cosh[x]),{x,-Infinity,Infinity}] = Pi^2/6. When I used the residual theorem to evaluate: Integrate[X*Exp[I*x]/(Cos[x]+Cosh[x]),{x,-Infinity,Infinity}] using all the poles in the upper half plane I got: 2*Pi*I *Sum[2*(1+2*n)*Pi/(1+Exp[Pi+2*n*Pi]),{n,0,Infinity}]. Mathematica was not able to evaluate this Sum. However Numercally I was able to confirm this was the right answer. This Sum is very rapidly converging which makes it interesting.
I came up with a formula very similar to yours (through residues): 2\[Pi]^2 Sum[(2 k + 1)/(1 + Exp[\[Pi] (2 k + 1)]), {k, 0, \[Infinity]}] and Mathematica correctly evaluated it to \[Pi]^2/12. The problem is, I don't know how to evaluate the series by hand.
@@SiphonSoulsX Since Mathematica was able to do your Sum it should have been able to do my Sum since they are so closely related. I already contacted the Mathematica guys about these observations. Unfortunately Mathematica does not reveal the algorithms they are using.
This one was so beautiful. In just one problem there were the basel problem, the laplace transform, geometric series, feynman integrals, euler's formula and even one of the weirdest versions of zero. Very nice.
Hoping no Calc 1 professor puts this on their tests
@@fantiscious laplace transform are not covered not in calc 1 so no they wouldnt imo
and feynmann integrals maybe?
Great integral thanks Michael. BTW I love the new chalk with high contrast bright colors
Agreed. His visuals really good -- both the graphics and the clarity of characters in the equations. I didn't know the chalk is new, but I don't recall seeing some of the colors before and their brightness is helpful. Following the logic in a math presentation can be tough enough. If I had to struggle to read the images or equations, it would be tougher.
Glad you like it! it's just a change in saturation. I did it to make everything "pop" a bit more.
Thanks for noticing.
-Stephanie
MP Editor
@@MichaelPennMath Great job Stephanie. We appreciate your efforts
@@MichaelPennMath Why use the trick at all thoughMichael if no one willthink of that?Thanks for sharing.
@@MichaelPennMath And why dont you write sine of x as [e^ix minus cos x]/i since if you rearrange and solve for sine x that is whatyou get--sine is not only the imaginary part once you divide by the imaginary number..
Amazing integral and cool tricks!
This is something I might imagine would be on a Putnam, especially with that Laplace transform towards the end...
هذا هو عمل الخبراء.
I could see the complex exponentials coming to unify the trigonometric and hyperbolic functions. The Laplace Transform and Basel Problem were icing on the cake.
Small mistake at 10.58
U cannot put 1/n outside the summation and multiply it with x=t/n
U should put x=t/n inside the summation and multiply it with 1/n
Then u will get the same solution in the rest of the video
Introducing s, taking its antiderivative, and then using that to complete a known Laplace transform was an absolutely slick move. I had to pause and go over it to make sure it was valid -- it's almost magic!
Does anyone think the thumbnail looks a little off? The LaTeX integral seems to use expressions like $cos(x)$ instead of $\cos(x)$.
Cool Integral!!! Lots of fun tricks
I remember having to so the Fourier transform of a very similar function. It was a nightmare.
Wow this problem had almost everything. Having said that, I think it's high time we saw some differential equations with Bessel function solutions on the channel!?
Nice cut at 12:30, hard to notice :)
Btw cool integral, hiw did you come up with it?
Thank goodness he closed the parentheses. 😅
Credits to Wolfram Alpha
Gnarly!
Thank you, professor!
Thanks!
Thank YOU for the support!
-Stephanie
MP Editor
15:37
Seeing both cos and cosh led me to think that this won't be easy
Wow, this is a real subtle solution, a very very fascinating solution👌
No need to take the imaginary part before the end. Keeping it as e^(ix) would make the following integral much easier
At 12:20 when you pull the derivative out of the integral, why is that possible? Does it have anything to do with the dominated convergence theorem? Do you have a video where you explain how that actually works/when it can and can't be applied?
Here comes our integral slayer. 🔥🔥🔥
ahahaha love thatt🔥
I was nervous when the denominator was rolled into Imaginary operator at 3:45.
Also, at 7:35: don't we need to confirm that we are inside the radius of convergence when applying the formula for the sum of a geometric series?
The role of 'r' in a geometric series seems to be played by "e^(-x) * e^(ix)" -- how do we know this value remains less than one, especially when it is in an integral where the value of 'x' can be anywhere between zero and infinity?
I see the formula for a geometric series used similarly on other channels and everyone treats it as legit, but I haven't worked through the details to stop me from wondering when I see it used in integrals over ranges much greater than from -1 to 1.
|e^ix| = 1 and |e^-x| < 1 (for x > 0) so we're fine
@@TheEternalVortex42 Thank you. This was clear: |e^-x| < 1 (for x > 1). It was less clear for 0
@@artsmith1347 But the _product_ of |e^ix| and |e^-x| must be strictly less than 1 for all x > 0, so we're good.
@@RexxSchneider Got it. Thank you.
Okay - love this way of doing math. Sometimes math elsewhere but here has a step by step linearity constrained by the topic or theme. In these videos the mathematician professor shows how every tool in the Math toolbox can be used working laterally (I mean lateral and linear thinking coexisting dynamically!).
But the main reason I write is continuum. Can anything really be distinct in a continuum? As soon as there is a partitioning of a continuum doesn't that mean we have two continuums separated at a point of partition?
For example each graph to add context and interpretation needs to have a scalar somewhere along each axis or a vertex and line segment or ... . Isn't that a partitioning by labeling?
And if so doesn't it imply that meaningful interpretation is only added after a partitioning process?
I think I may be overthinking this or alternatively overlooking something trivial with aim being that only the quotients seem to give ordering and clumping of things together as meaningful.
Wonderful job!
Beautiful!
I want the official names of cosh and sinh to be coshine and shine
EDIT: The integrand x^k Sin(x)/( Cos(x) + Cosh(x)) can be solved in closed form with some straightforward extensions to this derivation. The key is that the x->t/n substitution changes the Laplace transform to L(t^k Sin(t)) instead of L(t Sin(t)) and the power of (1/n)^2 in the sum goes to (1/n)^(k+2). The Laplace transform yields 2^(-(1/2) - k/2) Gamma[1 + k] Sin[1/4 (1 + k) Pi]. And the sum yields 2^-j (-2 + 2^j) Zeta[j] where j is k+1. The value of the integral of x^k Sin(x)/(Cos(x)+Cosh(x)) is the Laplace transform value times the sum value. x^1 gives Pi^2/12 as here. x^2 -> 3/4 Zeta(3), x^3->0, x^4-> 45/8 Zeta(5), x^5-> 31/1008 Pi^6, x^6-> 2835/32 Zeta(7), and x^7 ->0.
Original comment: It's interesting that a similar method works for x^k in the numerator instead of just x^1. The Laplace Transform becomes more complicated but has a closed form 2^(-(1/2) - k/2) Gamma[1 + k] Sin[1/4 (1 + k) Pi]. The sum on n is trivial to change since the x->t/n substitution just changes the power of n in the sum which works out to 2^-j (-2 + 2^j) Zeta[j] where j is the power of (1/n)^j in the sum. The only changes to the video are the Laplace transform being for t^k Sin[t] instead of t Sin[t] and the power of (1/n) in the sum.
Can you say that in english?
@@sleepykitten2168 Edited and cleaned up a bit. :)
@@GandalfTheWise0002 Oh its fine lol, I was just making a joke about how little math I knew.
@@sleepykitten2168 :). Actually, it did really need to be cleaned up. Good to see people of all levels of math skills following a channel like this.
Very beautiful integral
Great content!
Thanks
One of the things I first thought of doing was to write the denominator at e^x+e^(ix)+e^(-x)+e^(-ix). I'm blanking if there is a cyclic function that works with those, though.
Same here. I think the beginning would have been a bit less complicated with that - but I did not try doing the computations to see if that works out.
I’d love to see a practical use of this level of math. Just a thought.
I can appreciate the desire for practical applications, but imo separation of theory and examples is the most efficient method of conducting science and math. It's better for theoreticians to spit out the theory without any filler and practicians to seek out the relevant theory for their specific application. Hopefully the latter group also makes their work public so other people can learn from the examples.
Mathematica was able to show that Integrate[x*Sin[x]/(Cos[x]+Cosh[x]),{x,-Infinity,Infinity}] = Pi^2/6. When I used the residual theorem to evaluate:
Integrate[X*Exp[I*x]/(Cos[x]+Cosh[x]),{x,-Infinity,Infinity}] using all the poles in the upper half plane I got:
2*Pi*I *Sum[2*(1+2*n)*Pi/(1+Exp[Pi+2*n*Pi]),{n,0,Infinity}]. Mathematica was not able to evaluate this Sum. However Numercally I was able to confirm this was the right answer. This Sum is very rapidly converging which makes it interesting.
I came up with a formula very similar to yours (through residues): 2\[Pi]^2 Sum[(2 k + 1)/(1 + Exp[\[Pi] (2 k + 1)]), {k, 0, \[Infinity]}] and Mathematica correctly evaluated it to \[Pi]^2/12.
The problem is, I don't know how to evaluate the series by hand.
@@SiphonSoulsX Since Mathematica was able to do your Sum it should have been able to do my Sum since they are so closely related. I already contacted the Mathematica guys about these observations. Unfortunately Mathematica does not reveal the algorithms they are using.
The laplace transform part looks an awful lot like differentiating under the integral...is that how RF figured out the technique?
The technique was known to Leibnitz; RF popularised it after reading it in a book.
@@jip789 well shit. That's nuts lol.
amazing video
When is it safe to swap integration and differentiation signs?
At first look, I guess using Euler relationships between trigonometric and hyperbolic functions ?
Impressive!
What an unusual integral.
Awesome!
complexo-Laplace which is hidden inside an integral.
Very interesting how he used the Im() function
do you not do discrete math on math majors?
You can also solve this using contour integration.
To whoever edited the thumbnail, don’t forget to add your backward slashes before your trig functions in LaTeX! $\sin x$, not $sin x$, and so on
It's actually because of your comment that I know about this now and haven't repeated the mistake. So thank you!
-Stephanie
Editor