You won't believe all the tricks for this integral!!
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- Опубліковано 18 сер 2024
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Continuity has never been great in these videos but 11:04 proves that Michael Penn's clothing is a discontinuous function.
To one of the best UA-camrs: Keep making videos. Everything about them is amazing.
11:05 when the Cauchy product so strong you gotta take your jacket off
18:53
The left box should read "The integral from 0 to 2*pi of cos(mx)*cos(nx) is equal to pi if m=n."
... with the exception of m=n=0 (e.g. when m,n are from N0). I always find that exception quite inconvenient when looking for an orthonormal basis set of periodic functions and prefer to use exp(i*n*x) instead.
Bro where is dx in original integral ?
Teacher: "Nice 19min video, but no dx. Zero points."
2:55 absolutely?
I don't see either how `\sum_n exp(i2nx)/n` is supposed to converge absolutely ...
Another approach is to consider integrals of (log(sin(x)+log(cos(x))^2 and (log(sin(x)-log(cos(x))^2, we end up with the original integral represented in terms of:
- integral of log(sin(x)), which is easy to calculate
- integral of log(tan(x))^2. Substitute t=tanx, we get integral of log(t)^2/(1+t^2), which can be solved using integral on a semi-circle contour of log(z)^2/(1+z^2).
Hello Riad! consider look to my channel too for similar math olympiad problems. Thanks and regards.
4:50 here it is assumed that the geometric series will converge, and this happens when |u| strictly less than 1, and in the case shown, e^(i something real), the modulus is one. Am I wrong, or is the condition on the modulus circumvented in some way that is not explained? Anyone who can exxplain a bit more this part I would be grateful.
The to be integrated function is not defined for integer multiples of pi and pi/2. Therefore x=/=k*pi/pi/2 can be assumed and for these values e^(i2x)=/=1.
valid point
You're right, it doesn't converge. However the series u^n/n (its integral ) does converge for |u|=1 and u != 1, which can be written in terms of log function.
For converging integrals, generally the values (real values, not infinity) at bounds can be omited. For example, one use geometric series to prove that \int_0^1 log(x)/(1-x)dx=-zeta(2) and 1/(1-x) cannot be expanded in x=1 using usual formula.
@@IsomerSoma Remember that if f is continue on [a,b], then |int_a^b f(x)dx|
To compute this, Euler beta function could be useful. Another way to do is to use ab=1/4((a+b)^2-(a-b)^2)
Exactly! Given the trigonometric representation of the beta function, the above integral can be obtained by differentiating B(x,y) with respect to x and y and then set x=y=1/2. Then using a certain Identity for the mixed derivative of the beta function the result follows readily!
@@damosdamianos8731 Very cool! I see I'm going to have to study up on the properties of the polygamma function.
@@michaelguenther7105 Yes! Many integrals can be computed easily using the properties of Gamma, Beta and also polygamma functions!
Best part of my day is when professor penn uploads video!
11:04 THIS CAUCHY PRODUCT SO FIRE MICHAEL TOOK OFF HIS HOODIE
Beautiful demonstration. Thanks,...
12:40 You shouldn't have halved the range too, otherwise you'd be right
In a previous video, the integral of log(cos(x))^2 from 0 to pi/2 was calculated. This result can be used to our advantage by noticing that it's the same integral as log(sin(x))^2 from 0 to pi/2. Then the integral in this video is the middle part of the expansion of 1/2*(log(cos(x)) + log(sin(x)))^2 from 0 to pi/2. The integral of that binomial simplifies down to 3*pi/4*log(2)^2 + integral of 1/2*log(sin(x))^2 from 0 to pi/2. The rest is easy.
It's not that easy. The problem is to evaluate int_0^1 ln^2 x/(1+x^2)dx This is a non trivial integral.
may I ask, exactly which prev. video ?
@@richardheiville937 I say the rest is easy because Michael Penn calculated it in a previous video, ua-cam.com/video/ikyVHEHmgP8/v-deo.html That particular integral you just described can be done by converting 1/(1+x^2) to a geometric series, though.
@@charlievane ua-cam.com/video/ikyVHEHmgP8/v-deo.html
@@charlievane Here is the link. It's a very old one, so you may not have watched Michael at the time. ua-cam.com/video/7wiybMkEfbc/v-deo.html
2:55 this doesn't seem obvious to me at all. Isn't the absolute equivalent of that series just the harmonic series which diverges?
It seems like you don't need absolute convergence here, though, but it's unclear to me how they converge at all... they do look like they converge, but it doesn't seem that obvious.
I believe you're right. Instead, we can apply Dirichlet's test to see that sigma((z^n)/n) converges everywhere on the unit circle except at z=1. IMO that would make a nice video in itself.
I was about to say the same thing. They definitely don't converge absolutely. The whole argument fails. Without absolute convergence, we need something else to show the equality between the sum of sums and the original sum.
@@Alex_Deam How do you show that the sum of z^n/n is bounded?
Totally right. There is no absolute convergence there. In fact the series of the modulus of the terms actually diverges (it's the harmonic series).
@@EebstertheGreat So we use Dirichlet's test, with a_n=1/n and b_n=z^n. There are three criteria for the test to work:
1) a_n must be monotonically non-increasing. We have this, as 1/n > 1/(n+1) for all positive n
2) lim(a_n)=0 as n tends to infinity. Again, we have this, as it's the limit of 1/n.
3) The modulus of the nth partial sum must be less than or equal to some constant M for every n. Our sum of b_n's is just a geometric series, so we have (excuse the messy youtube comment notation lol) mod(sum(b_n)) = mod(z-z^(n+1))/mod(1-z), by the sum of a finite geometric series.
But we can use the triangle inequality to bound the numerator:
mod(z-z^(n+1))/mod(1-z)
This is one of the best Integrations I have ever had.
Your cosine orthogonality is wrong. It should be: int[0, 2pi] cos(mx)cos(nx) = {0, |m| != |n|; pi, |m| = |n|.
Where did the "dx" under the integral go to???
I dont see how this converges absolutely 3:15
When the integral is so complicated you have to change your clothes to finish evaluating it.
At 11:08 You've got the same going from l=1 to l=k and in the denominator you have k-l, what??
Hum. Good catch. I think some messiness occurs because we sum from 1 rather than 0, and at a glance I feel adding 1 to the offending denominator (and anywhere else containing (k-l)) would fix things.
It was clear from the start that we would get a whole lot of zero terms, though, so it's ultimately not a huge deal.
EDIT: we should be summing from k = 2, and lose the first and the last l-term. l should run from 1 to k-1.
I want a T-shirt that says
“… and that’s a good place to stop 🛑 “
👍 😄
Thank you, professor
Hi,
0:18 : dx
About the fact that 1/(1+u)=1-u+u²-u^3+.. works only for |u|
Do you know if these expansions of log sin and log cos were used to construct Napier's original tables of logarithms of sine and cosine? IIRC he constructed some reference values from scratch and then interpolated between them, but I don't know how he constructed the reference values.
great video!
Something went wrong in the Cauchy Product at 10:50. When applying Cuachy product from 1 you have to offset one of summation indices in vaguely unintuitive and annoying way. I think it doesn't affect the subsequent derivation because the orthogonality normalises everything. A great derivation though! Much appreciated!
Thanks
Fun. Thanks.
good old times, integrating using spherical harmonics. a very beautiful method
Is this Spherical harmonics? Don’t spherical harmonics involve Legendre polynomials (I didn’t see them in the video)?
@@edmundwoolliams1240 it's a very special case and here more of a fourier expansion.
@@demenion3521 okay, thanks for the insight
Isn’t it possible to use the coefficient of a Fourier series for the 1sr formula ?
2:56 These sums do not converge absolutely, since the Harmonic series diverges. Maybe it would be better to start with the complex sum, and take the real part at the end?
There's a video making the rounds on tiktok about a math problem involving three circles, with their centres all on the same line, and the outer two circles touch each other at the centre of the inner circle, and the goal is to find the area of the inner circle that isn't covered by either of the outer circles.
Might be up your alley, and a little clickbait to infuse the clicks to this criminally underrated channel. Just a thought.
I beleive the orthogonality of cosines value should be half the interval length if its of integral multie of period(as average value of cos^2x is is half)
Yes... Otherwise the integral around 12:00 would be pi/2 if you substitute 2x with u for example
What is this integral good for? What does it tell us about life and everything?
it tells us that life is difficult.
Ars gratia artis.
Hi Dr. Penn
I sort of understand your vids, but my level of math just isn't at that level. What level of math are these problems??
Anyone who's done a special polynomials class for a physics degree will recall the orthogonality property of cos(n x) cos(m x).
12:21 Why is it pi/4? you are integrating from 0 to pi/2, which is half of the integral in the left box, so why is it not half the answer (pi) since you took half the bounds and be pi/2 instead of pi/4?
Integrating cos(mx)*cos(nx) from 0 to pi gives pi/2 whenever m=n. Thus, letting u=2x means we get 1/2 the integral from 0 to pi of cos(Lu)*cos((L-K)u). Since L=L-K, we obtain 1/2*pi/2, which is pi/4.
@@krisbrandenberger544 But what I am seeing is when he does his setting, this setting puts m=2l=2(k-l)=n, then (int from 0 to pi/2) of cos(2lx)cos(2(k-l))dx is pi/2?
@@ingiford175 Yes.
8:48 an absolute value would be nice: log( |cosx| )
cos x is always nonnegative on the domain of x in this problem, so |cos x| = cos x. Technically we do get the problem of log 0, but that happens almost nowhere (only on the endpoint), so it's not an issue.
Cool 😎 great stuff more real analysis content ❤️⭐
6:20 Are these 'logarithm rules' still true for complex numbers, i.e. is ln(ab) = ln(a) + ln(b) true for complex numbers a, b ?
It kind of does, but because the functions are multivalued, you get a bunch of solutions out of it, so they're only equal up to an integer multiple of 2i*pi.
Sick
I am in std 12 but in India i am an aspirant of jee advanced i do this question in std 11 ... 😂
رائع كالعادة
Here are the videos u may want to know
Part 1:
ua-cam.com/video/f_Sge0HIY8s/v-deo.html
Part 2:
ua-cam.com/video/9-fZzwvYjdc/v-deo.html
My approach
Int(ln^2(tan(x)),x=0..Pi/2)=Int(ln^2(t)/(1+t^2),t=0..infinity)
=Int(ln^2(t)/(1+t^2),t=0..1)+Int(ln^2(t)/(1+t^2),t=1..infinity)
=Int(ln^2(t)/(1+t^2),t=0..1)+Int(ln^2(1/u)/(1+1/u^2)(-1/u^2),u=1..0)
=Int(ln^2(t)/(1+t^2),t=0..1)+Int(ln^2(u)/(u^2+1),u=0..1)
=Int(ln^2(u)/(1+u^2),u=0..1)+Int(ln^2(u)/(u^2+1),u=0..1)
=2Int(ln^2(u)/(1+u^2),u=0..1)
=2Int(sum((-1)^ku^{2k}ln^2(u),k=0..infinity),u=0..1)
=2sum((-1)^k Int(u^{2k}ln^2(u),u=0..1),k=0..infinity)
=2sum((-1)^k (2/(2k+1)u^(2k+1)ln^2(u)|_{0}^{1}-2/(2k+1)Int(u^(2k)ln(u),u=0..1)),k=0..infinity)
=2sum((-1)^k (-2/(2k+1))Int(u^(2k)ln(u),u=0..1),k=0..infinity)
=2sum((-1)^k (-2/(2k+1))(1/(2k+1)u^(2k+1)ln(u)|_{0}^{1}-1/(2k+1)Int(u^{2k},u=0..1)),k=0..infinity)
=2sum((-1)^k (-2/(2k+1))(-1/(2k+1)Int(u^{2k},u=0..1)),k=0..infinity)
=4sum((-1)^k/(2k+1)^3,k=0..infinity)
Int(ln^2(tan(x)),x=0..Pi/2)=Int(ln(sin(x)/cos(x))^2,x=0..Pi/2)
=Int((ln(sin(x))-ln(cos(x))^2,x=0..Pi/2)
=Int(ln^2(sin(x)),x=0..Pi/2)-2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2)+Int(ln^2(cos(x))),x=0..Pi/2)
Int(ln^2(tan(x)),x=0..Pi/2)=Int(ln^2(sin(x)),x=0..Pi/2) - 2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) + Int(ln^2(cos(x))),x=0..Pi/2)
2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(sin(x)),x=0..Pi/2) + Int(ln^2(cos(x))),x=0..Pi/2) - Int(ln^2(tan(x)),x=0..Pi/2)
2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(sin(Pi/2-t)),(-1)t=Pi/2..0) + Int(ln^2(cos(x))),x=0..Pi/2) - Int(ln^2(tan(x)),x=0..Pi/2)
2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(cos(t)),t=0..Pi/2) + Int(ln^2(cos(x))),x=0..Pi/2) - Int(ln^2(tan(x)),x=0..Pi/2)
2Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = 2Int(ln^2(cos(t)),t=0..Pi/2) - 2sum((-1)^k Int(u^{2k}ln^2(u),u=0..1),k=0..infinity)
Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(cos(t)),t=0..Pi/2) - sum((-1)^k Int(u^{2k}ln^2(u),u=0..1),k=0..infinity)
Int(ln(sin(x))ln(cos(x)),x=0..Pi/2) = Int(ln^2(cos(t)),t=0..Pi/2) -2sum((-1)^k/(2k+1)^3,k=0..infinity)
And Michael recorded video for both integral Int(ln^2(cos(t)),t=0..Pi/2) and sum, sum((-1)^k/(2k+1)^3,k=0..infinity)
Great
Whew, played pretty loose with equality until the very end of the vid. 😀
Al min 11 ti è venuto caldo....
kĩ năng tích phân suy rộng.
I jumped at 11:05
Just searching for the dx, someone else too?😑