This actually comes out faster than the proof given too. 1-e^i = -2ie^(i/2)*sin(1/2) so -ln(1-e^i) = -ln(-i*e^(i/2)) - ln(2sin(1/2)) = -(-i*pi/2 + i/2) - ln(2sin(1/2)) the real part gives the answer. This also gives sum_n sin(n)/n = (pi-1)/2 by taking the imaginary part.
One can prove it simply (without the issue on the geometric series) by computing the series of z^n/n (which gives you -Log(1-z)) for |z| < 1. Then, since this series converges at z=e^i, this power series is continuous at this point (on a certain angular domain, Abel's theorem). Anyway, the series of (e^i)^n/n is just -Log(1-e^i) . Then, just compute and take the real part.
@@marinchan6334 You are right, I should have mentioned it. Since the limit when z goes to e^i of -Log(1-z) exists, it also follows from the (strong) Tauberian theorem that the series converges at z=e^i, but this is more complicated.
How are you integrating from a complex imaginary number to a real number? Hypothetically wouldn't the result depend on the path you take between the endpoints?
This is one of the really nice things about complex analysis: it turns out to not depend on the path at all, only the endpoints! To be precise, this relies on the fact that e^iθ is an analytic function everywhere, meaning it can be written as a power series at any point θ in the complex plane. If you have this, then any path between 1 and i*t will have the same value when you integrate along that path. Then just take the limit of this value as t goes to ∞
@@ethanwinters1519 But the integrand here is not simply e^i theta, but e^i theta/(1 - e^i theta), i. e. it is a meromorphic function, not an analytic function. So one _does_ have to pay attention to the integration path.
@@bjornfeuerbacher5514 It's technically not relevant in this example because we only take the real part of the integral. And since the real part of a closed contour integral over a meromorphic function with real residues (such as this one) along any path is always zero, the path of integration does not matter. However, it would definitely matter for the imaginary part of the integral because Im{ln(z)} is arg(z), which is of course a multi-valued function.
We are not. The integral is complex and we then took the real part. Integrals of analytic functions are independent of the path... as long as the path encloses no poles.
@@byronwatkins2565 You're talking about closed contour integral, which is not the case in this example. If the function has any poles at all, the path of integration matters even if the path does not enclose any poles. For example, if you integrate 1/z from -i to i clockwise along the unit circle, you get a different result than if you integrated counterclockwise around the unit circle. By necessity, the difference between these integrals is non-zero because that is equivalent to a closed contour integral around the pole at z=0. However, as I pointed out above, any closed contour integral is purely imaginary for functions with only real residues as in the example. Hence the real part of a closed contour integral over such functions will always be 0 and thus the path of integration truly does not matter.
There is a potential serious issue when using geometric series formula. In this case your r has modulus 1! Also, the accompanying sine series sum is well known and is equal to (\pi-1)/2. This is easily found using Fourier expansion. Looking at your method I can't see the same result if you take the imaginary part. Please, look at this and clarify if you can.
I stumbled across the very same thing when I watched this video. The value however is correct. Also if you use this approch for the related sum with sines, you do get the correct value. We are looking for -b in the end in this case. We can divide the equations at 10:50 to get tan(b) = -sin(1)/(1-cos(1)), so -b = - arctan(-sin(1)/(1-cos(1))) which indeed yields (pi-1)/2.
@@christophdietrich4240 Thanks for response. I assumed the result is still correct but the issue should have been addressed in any case, even if just passing
@@MrMctasticsWhen i saw your comments I thought it was a mistake in the proof, where we sum infinity and minus infinity, but looking more closely I saw it, definitley should have been pointed in the video
Hi you probably won’t see this but I’ve been busy the past few months with exploring Latin squares using group theory (permutations of the rows and columns, which form a group, and it turns out that the permutations which leave a square looking the same form a subgroup, meaning you can use langrange’s theorem to get the number of unique Latin squares you can achieve for a certain normalized Latin square, etc.) it would be interesting if you made a video about it! Maybe not this exactly, but it would be really cool if you’d make a video about Latin squares. They are such incredibly interesting mathematical objects.
@byronwatkins2565 True, but the original integral was on the real axis and unrigorously, it had been transformed into an integral from 1 on the real axis to infinity on the imaginary axis. I would have used an integral on a quarter circle, indented at 0, with a radius let tend to infinity. I abhor hand waving and leaps of faith.
Michael, could you do a follow up which looks in more detail at the criteria for convergence of the geometric series when the first term and common ratio are those complex exponentials. Also a future suggestion for a video that examines rigorously when it is allowed to interchange the order of a sum and an integration. Perhaps some examples of when that doesn't work. Love your content. Thanks.
2:27 Why can't the lower bound just be negative infinity? 5:04 Is it even okay to use the geometric series formula? How can we know that e^ i theta will never have a modulus greater than (or equal to) 1 at more than a set of points with negligible measure? What does it even mean to take an integral in that portion of the complex plane. Is it a straight line or region? Does it matter if it's not a straight line? I have so many questions... Edit: Perhaps it's time to finally study complex analysis.
2:27 Because if you insert negative infinity, you'll get e to the power of -i times infinity, which isn't well defined. However, when you insert i times negative infinity, you'll get e to the power of -infinity, which gives zero - as required. 5:04 To do this in a sensible way, you'll have to specify an actual integration path in the complex plane, which Michael does not do here. But I think it's no big problem to find such a path such that the modules is always smaller than 1, except at the endpoint of the path - which is a set with measure zero.
Dear Michael Penn It is not known whether the Flint Hills series converges. Sum (1/(n^3*Sin^2(x)) It seems not difficult, but no one knows about it. Could you break down this example and show why this task is very difficult?
Since we're integrating, take the limit as the value approaches one. (omitting the endpoint of an interval won't change its value since the function is continuous and otherwise "nice".)
I believe in this case we're simply looking at the behaviour of the function as the lower bound approaches the point at infinity, and this behaviour is different depending on which direction you approach the point at infinity. For example approaching it going along the negative imaginary axis will have e^iθ to diverge instead of approach 0. Basically, thinking of e^iθ as a function on the riemann sphere like you're suggesting, the point at infinity is a discontinuity, so there is no single value for it. You must instead specify in what way you're approaching infinity.
In R you can go to infinity in the positive or negative direction, and the limit of a function to infinity in general depends on the direction. In C it's similar, but more complex 😁
-∞ as the bound would be a analogous to a path with a horizontal asymptote as x->-∞. i*∞ is more analogous to a path with a vertical asymptote as x->0.
When integrating with real numbers you integrate over an interval, but with complex numbers you integrate over a path in the complex plane. For example integrating from 2+i to -3+2i you could integrate over all sorts of different paths. There isn't just one interval like with the reals
@@Happy_Abe Because then you have a meaningful limit as the input θ approaches i∞, namely it will be zero since it is approaching e^-∞. But if you have θ approach -∞, then the limit no longer exists as you're essentially moving around the unit circle over and over, and never settling down at any particular value.
Can you help me with this problem ? Let I = [a1,b1]×…×[an,bn] be a n dimensional rectangle Let f : I → IR be continuous and f ≥ 0 on I Suppose ∫f = 0 over I Show that f = 0 on I. I have referenced some posts about this problem and most of them are proving with contradiction : suppose ∃a ∈ I : f(a) > 0, then by continuousness, f > 0 on I∩B where B = B(a,|f(a)|/2). (*) Since f > 0 on I∩B, ∫f over I∩B > 0 But ∫f = 0 over I and ∫f over I ≥ ∫f over I∩B, a contradiction. My question is how to prove (*) ? And besides, this proof requires using the concept of integral over a domain I∩B that is not a rectangle, so my next wonder is there another way to solve this problem which only using the concept of integral over a rectangle ? Thanh you
Hold on a minute. At the end, he sets a+bi = ln(1-e^i) and therefore the value of a is the answer. But in the next line, he is comparing a+bi to just 1-e^i. He still needs to take a log at some point right?
One can simply this further... 2 - 2 cos(1) = 4 sin²(1/2); so the result will be -ln(2 sin(1/2)).
or as ln(csc(1/2)/2) for extra crispiness
This actually comes out faster than the proof given too.
1-e^i = -2ie^(i/2)*sin(1/2)
so
-ln(1-e^i) = -ln(-i*e^(i/2)) - ln(2sin(1/2))
= -(-i*pi/2 + i/2) - ln(2sin(1/2))
the real part gives the answer.
This also gives sum_n sin(n)/n = (pi-1)/2 by taking the imaginary part.
I don't understand why we aren't taking out the ln2
In my personal opinion, it looks prettier with the 2 inside
@@the_lava_wielder6996 What do you mean with "take out the ln 2"?
One can prove it simply (without the issue on the geometric series) by computing the series of z^n/n (which gives you -Log(1-z)) for |z| < 1. Then, since this series converges at z=e^i, this power series is continuous at this point (on a certain angular domain, Abel's theorem). Anyway, the series of (e^i)^n/n is just -Log(1-e^i) . Then, just compute and take the real part.
Just to say,you can prove the convergence of this series at z=e^i using dirichlets test.
@@marinchan6334 You are right, I should have mentioned it.
Since the limit when z goes to e^i of -Log(1-z) exists, it also follows from the (strong) Tauberian theorem that the series converges at z=e^i, but this is more complicated.
How are you integrating from a complex imaginary number to a real number? Hypothetically wouldn't the result depend on the path you take between the endpoints?
This is one of the really nice things about complex analysis: it turns out to not depend on the path at all, only the endpoints! To be precise, this relies on the fact that e^iθ is an analytic function everywhere, meaning it can be written as a power series at any point θ in the complex plane. If you have this, then any path between 1 and i*t will have the same value when you integrate along that path. Then just take the limit of this value as t goes to ∞
@@ethanwinters1519 But the integrand here is not simply e^i theta, but e^i theta/(1 - e^i theta), i. e. it is a meromorphic function, not an analytic function. So one _does_ have to pay attention to the integration path.
@@bjornfeuerbacher5514 It's technically not relevant in this example because we only take the real part of the integral. And since the real part of a closed contour integral over a meromorphic function with real residues (such as this one) along any path is always zero, the path of integration does not matter. However, it would definitely matter for the imaginary part of the integral because Im{ln(z)} is arg(z), which is of course a multi-valued function.
We are not. The integral is complex and we then took the real part. Integrals of analytic functions are independent of the path... as long as the path encloses no poles.
@@byronwatkins2565 You're talking about closed contour integral, which is not the case in this example. If the function has any poles at all, the path of integration matters even if the path does not enclose any poles. For example, if you integrate 1/z from -i to i clockwise along the unit circle, you get a different result than if you integrated counterclockwise around the unit circle. By necessity, the difference between these integrals is non-zero because that is equivalent to a closed contour integral around the pole at z=0.
However, as I pointed out above, any closed contour integral is purely imaginary for functions with only real residues as in the example. Hence the real part of a closed contour integral over such functions will always be 0 and thus the path of integration truly does not matter.
There is a potential serious issue when using geometric series formula. In this case your r has modulus 1! Also, the accompanying sine series sum is well known and is equal to (\pi-1)/2. This is easily found using Fourier expansion. Looking at your method I can't see the same result if you take the imaginary part. Please, look at this and clarify if you can.
I stumbled across the very same thing when I watched this video. The value however is correct. Also if you use this approch for the related sum with sines, you do get the correct value. We are looking for -b in the end in this case. We can divide the equations at 10:50 to get tan(b) = -sin(1)/(1-cos(1)), so -b = - arctan(-sin(1)/(1-cos(1))) which indeed yields (pi-1)/2.
@@christophdietrich4240 Thanks for response. I assumed the result is still correct but the issue should have been addressed in any case, even if just passing
You should keep note that theta is not purely real, the imaginery part is greater than 0 (excluding 1) which means e^i*theta has modolus less than one
@@alomirk2812o shiii I think u right
@@MrMctasticsWhen i saw your comments I thought it was a mistake in the proof, where we sum infinity and minus infinity, but looking more closely I saw it, definitley should have been pointed in the video
Could you do a video on why the geometric series converges to 1/(1-r) even for complex r? It's obviously correct for real r, 0
Hi you probably won’t see this but I’ve been busy the past few months with exploring Latin squares using group theory (permutations of the rows and columns, which form a group, and it turns out that the permutations which leave a square looking the same form a subgroup, meaning you can use langrange’s theorem to get the number of unique Latin squares you can achieve for a certain normalized Latin square, etc.) it would be interesting if you made a video about it! Maybe not this exactly, but it would be really cool if you’d make a video about Latin squares. They are such incredibly interesting mathematical objects.
No need to integrate under the sum: ln(1-x) = -\sum_n>=1 (x^n)/n, given -1
What does the integral mean, as it appears to be a path integral in the complex plane?
For perfect differentials, all paths between these points yield the same result.
@byronwatkins2565 True, but the original integral was on the real axis and unrigorously, it had been transformed into an integral from 1 on the real axis to infinity on the imaginary axis. I would have used an integral on a quarter circle, indented at 0, with a radius let tend to infinity. I abhor hand waving and leaps of faith.
I think it means going "straight up" but should definitlry been a partial integral goimg to the limit
@@alomirk2812 Straight up would require 1 + i infinity
Yes, I think it is the path integral on the straight line from i theta to 1. Only then it makes sense.
Michael, could you do a follow up which looks in more detail at the criteria for convergence of the geometric series when the first term and common ratio are those complex exponentials.
Also a future suggestion for a video that examines rigorously when it is allowed to interchange the order of a sum and an integration. Perhaps some examples of when that doesn't work. Love your content. Thanks.
ln(z) = ln|z| + i arg(z)
2:27 Why can't the lower bound just be negative infinity?
5:04 Is it even okay to use the geometric series formula? How can we know that e^ i theta will never have a modulus greater than (or equal to) 1 at more than a set of points with negligible measure? What does it even mean to take an integral in that portion of the complex plane. Is it a straight line or region? Does it matter if it's not a straight line? I have so many questions...
Edit: Perhaps it's time to finally study complex analysis.
2:27 Because if you insert negative infinity, you'll get e to the power of -i times infinity, which isn't well defined. However, when you insert i times negative infinity, you'll get e to the power of -infinity, which gives zero - as required.
5:04 To do this in a sensible way, you'll have to specify an actual integration path in the complex plane, which Michael does not do here. But I think it's no big problem to find such a path such that the modules is always smaller than 1, except at the endpoint of the path - which is a set with measure zero.
@@bjornfeuerbacher5514 Thanks, I was wondering the same as @thesecondderivative8967. Indeed, sum(z^k,0..infinity)=1/(1-z) only if mod(z)
Parabéns, big teacher! As suas aulas são espetaculares e completas!
Interestingly, the sum of sin(n)/n results in an angle, namely arctan(sin(1)/(1-cos(1)), which is 61.35°.
Dear Michael Penn
It is not known whether the Flint Hills series converges.
Sum (1/(n^3*Sin^2(x))
It seems not difficult, but no one knows about it.
Could you break down this example and show why this task is very difficult?
How to check the convergence of a series cos(n)/n , i only need this.
This is a nice introductory problem for use in, say, a course on elementary complex analysis
A beautiful hand weaving symphony
Why can we swap the sum and integral?
Is there no other way than the geometric series from which we can return?
the thumbnail is missing a factor of -1/2
Are you allowed to use an infinite geometric series here since |e^iθ| = 1?
Since we're integrating, take the limit as the value approaches one. (omitting the endpoint of an interval won't change its value since the function is continuous and otherwise "nice".)
|e^iθ| < 1 except at the endpoint so it's fine
@@TheEternalVortex42 |e^iθ| is always 1 though
Edit: I realized that's only for real θ.
Beautiful!
In view of stereographic projection, there is only one point at infinity in the complex plane. So how is "i * infinity" different from "infinity"?
I believe in this case we're simply looking at the behaviour of the function as the lower bound approaches the point at infinity, and this behaviour is different depending on which direction you approach the point at infinity. For example approaching it going along the negative imaginary axis will have e^iθ to diverge instead of approach 0. Basically, thinking of e^iθ as a function on the riemann sphere like you're suggesting, the point at infinity is a discontinuity, so there is no single value for it. You must instead specify in what way you're approaching infinity.
In R you can go to infinity in the positive or negative direction, and the limit of a function to infinity in general depends on the direction. In C it's similar, but more complex 😁
Why do we write theta approaching i*infinity and not just negative infinity?
What does an angle of i even mean?
-∞ as the bound would be a analogous to a path with a horizontal asymptote as x->-∞. i*∞ is more analogous to a path with a vertical asymptote as x->0.
When integrating with real numbers you integrate over an interval, but with complex numbers you integrate over a path in the complex plane. For example integrating from 2+i to -3+2i you could integrate over all sorts of different paths. There isn't just one interval like with the reals
@@MrMctasticsI understand that but why couldn’t the path go from theta=1 to negative infinity
Why did it need to go i*infinity
@@Happy_Abe Because then you have a meaningful limit as the input θ approaches i∞, namely it will be zero since it is approaching e^-∞. But if you have θ approach -∞, then the limit no longer exists as you're essentially moving around the unit circle over and over, and never settling down at any particular value.
@@ethanwinters1519 why is i*infinity meaningful and -infinity isn’t?
I’m not fully understanding what an argument of i*infinity really is
What does it mean to have complex numbers as bounds of integration? Also, what is i times infinity?🤯
Does the geometric series converge for complex r?
Yup, it is covergent on the closed disk of radius 1 except the point 1
Any physics grad student who used Jackson for E&M is intimately familiar with this method and these results.
I DONT WANT TO DISCOURAGE PEOPLE..
😮
how does an angle approaching i*inf makes sense in any way?
Most simply: Re(z)=(z+z*)/2
Can you help me with this problem ?
Let I = [a1,b1]×…×[an,bn] be a n dimensional rectangle
Let f : I → IR be continuous
and f ≥ 0 on I
Suppose ∫f = 0 over I
Show that f = 0 on I.
I have referenced some posts about this problem and most of them are proving with contradiction :
suppose ∃a ∈ I : f(a) > 0, then by continuousness, f > 0 on I∩B
where B = B(a,|f(a)|/2).
(*) Since f > 0 on I∩B, ∫f over I∩B > 0
But ∫f = 0 over I and ∫f over I ≥ ∫f over I∩B, a contradiction.
My question is how to prove (*) ?
And besides, this proof requires using the concept of integral over a domain I∩B that is not a rectangle, so my next wonder is there another way to solve this problem which only using the concept of integral over a rectangle ?
Thanh you
Oh no, I didn't gently press the like button, I smashed it. Please don't judge me.
How would you prove that this series converges❔ Surely it's not absolutely convergent.
Nice!
Hold on a minute. At the end, he sets a+bi = ln(1-e^i) and therefore the value of a is the answer. But in the next line, he is comparing a+bi to just 1-e^i. He still needs to take a log at some point right?
No. In the next line, he is comparing e to the power of a+bi to 1-e^i. He is _not_ comparing a+bi to 1-e^i.
@@bjornfeuerbacher5514 oh yes, you’re right. That makes more sense.
great
Um...|r|=|exp(i theta)|=1...
No
I destroyed the like button. forgive me please
How do I solve this series?
∑[n=1,∞]n/n^n
hi
I M KING OF ALL THIS....DIMAAG CHHOTAA HAIN....SORRY...TRY
typical american thing, to do a u-sub when its not needed.
Our ability to handle underwater subs is what leads our our dominance as a naval power.
Especially given that no such thing as "u-substitution" exists. You can't find that term in any legitimate Calculus book
Nice!