it's okay except for at the point u=0. when you take the integral of that, that exception has measure zero so the Lebesgue integral is still well-defined. the interior is fine because its absolute value is convergent. my own question is about analytic continuation for the -1/12 result: Mathologer disproved in detail that little myth quite some time ago, but there's an easier way to get there: 1+2+3+4+...= S 0+1+2+3+...= S (by stability under translation) 1+1+1+1+...= 0 (by linearity) 0+1+1+1+...= 0 (by stability under translation) 1+0+0+0+...= 0 (by linearity) 1 = 0 (by regularity) this was pointed out within 72 hours of the original mistake, but it's rarely addressed in popular math circles
@@KD-onegaishimasuyeah there are many different ways in which the sum 1+2+3+4+… and -1/12 are correlated throughout all of math. Look up Ramanujan summation for more
@@NoahPrentice No, I was trying to say why we got away with swapping the order of summation and integration this time. You made a great point, limits don't generally exchange like this.
The dilogarithm is also useful in evaluating integrals with exponentials in the denominator _e.g._ int(x/(exp(x) -1) dx) = x ln(1 - exp(x)) + Li2(exp(x)) - x^2/2
One of my favourite dilog papers is Maximon, Proc R Soc Lond A, 2003. It has five very useful dilog equations, including the one used towards the end of this video. Alas, his paper is not freely accessible. For a readily available journal article which contains these equations (plus further fascinating dilog relationships) there is a more recent paper (Stewart, Irish Math. Soc. Bulletin, Number 89, Summer 2022, 43-49): www.maths.tcd.ie/pub/ims/bull89/wef/Articles/Stewart/Stewart-wef.pdf
@@megauser8512 What I mean is that i^n/n^2 can be separated into terms where n is odd and when n is even. When n is even the term is real, when n is odd the term is imaginary.
Also for who want to know how it get first time he just explore 1. Go from expansion to intégrale by using the integral of x^nlog(n) from 0 to 1 = x^n+1/n+1; 2. Return back to expansion from integral(after making a change variable like x->1-y;
Just a question: is the integral definition well defined? From what I remember of my complex analysis course different path produce different result in the integral (unless the integrand function is analityc?)
Yes. It is well defined if you work inside the disk |z|≤1. In this region exists the series expansion around z=0 of ln(1-z) and remembering that the first term is not constant, you don't have to worry with singularities. Or better, z =0 is a removable singularity. You can say that ln(1-z)/z is analytic because it have a Taylor series expansion around z=0 inside the unitary disk.
This video highlights why I always complain about all of the meaningless video titles. @5:50 he mentions working out Li_2(-1) has been done elsewhere on the channel. Searching his channel yields 2 other dilog videos, neither of which is the one he mentions.
It's pretty clear he was talking about the series, not necessarily in the context of the dilogarithm. (-1)^n/n^2 isn't searchable on YT even if it was in the title, so it doesn't really matter if he put it in there or not.
It's a routine summation from an introductory complex analysis class. This link explains the background well enough: www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf
Thanks for a great video. A quick question: do we want to restrict the modulus of z to be strictly less than 1? I am not sure how the analytic continuation would be possible at z=1, and in fact, Wikipedia also states that [1,\infty) is a branch cut. Thank you.
It's nice having a name for a function when it comes up often in calculations, and I assume this is why the dilogarithm was defined in the first place. According to its wikipedia page (link below), the dilogarithm can be used in particle physics and also to calculate the volume of certain geometric shapes (which is to be expected from a function defined using an integral). en.m.wikipedia.org/wiki/Spence%27s_function
Yeah i suggested to every one goto my favorites functions book : Milton Abramowitz and Irene A.Stegun - Handbook of Mathematical Functions-. Concise Encyclopedia of Mathematics. Jordan C - Calculus of finit Differences -. The best one : E. T. Whittaker and G. N. Watson - A C O U R S E O F M O D E R N A N A L Y S I S.
@@Pablo360able An integral does not change when finitely many points are excluded. Can't we just say it is integrated to 1 without the 1, hence only needing the integrand being defined in interval (0, 1)?
@@BridgeBumhere's a link between the zeta function and polylogarithms in general: Li_s(z) = sum as k ranges from one to infinity of z^k/k^s Where s and z are complex numbers (only defined if the sum converges) Set z = 1 and allow s to vary and you have the zeta function on (1, inf) I'm not aware of any way of defining the zeta function only using Li_2 though
Error at 9:45. Equality doesn't hold as sums start from different indexes. Can be fixed if we expressed the 2nd sum as sum from n from 0 to infinity of reciprocal of (4n-2)^2
He glosses over the details but if you actually see what he collapses them to by plugging in the respective starting terms the end result is the correct.
I like that you call it "My special function" in the beginning.
👉😌👈
They love each other soso much...
The dilog is saying uwu.
you should definitely do a video on swapping the order of summation and integration. seems to come up a lot!
it's okay except for at the point u=0. when you take the integral of that, that exception has measure zero so the Lebesgue integral is still well-defined. the interior is fine because its absolute value is convergent.
my own question is about analytic continuation for the -1/12 result: Mathologer disproved in detail that little myth quite some time ago, but there's an easier way to get there:
1+2+3+4+...= S
0+1+2+3+...= S (by stability under translation)
1+1+1+1+...= 0 (by linearity)
0+1+1+1+...= 0 (by stability under translation)
1+0+0+0+...= 0 (by linearity)
1 = 0 (by regularity)
this was pointed out within 72 hours of the original mistake, but it's rarely addressed in popular math circles
@@KD-onegaishimasuyeah there are many different ways in which the sum 1+2+3+4+… and -1/12 are correlated throughout all of math. Look up Ramanujan summation for more
@@KD-onegaishimasu I think you maybe replied to the wrong comment, just btw
@@NoahPrentice No, I was trying to say why we got away with swapping the order of summation and integration this time. You made a great point, limits don't generally exchange like this.
@@KD-onegaishimasu oh, gotcha. thanks!
That's crazy how evaluating a generalization of a logarithm at the imaginary unit gives back pi squared and Catalan's constant.
That's 4 constants involved directly or indirectly 😂
The dilogarithm is also useful in evaluating integrals with exponentials in the denominator _e.g._ int(x/(exp(x) -1) dx) = x ln(1 - exp(x)) + Li2(exp(x)) - x^2/2
I stumbled upon this function by accident and easily became one of my favorite ones. Great to see you cover it!
Thanks!
FWIW, the "Li_1(z)" series on the unit disk: sum_n=1^infinity z^n/n = -ln(1-z).
مايكل يتعامل مع الأعداد العقدية بكل أريحية.
كنا نخاف من هذه العمليات أيام الدراسات الجامعية
One of my favourite dilog papers is Maximon, Proc R Soc Lond A, 2003. It has five very useful dilog equations, including the one used towards the end of this video. Alas, his paper is not freely accessible. For a readily available journal article which contains these equations (plus further fascinating dilog relationships) there is a more recent paper (Stewart, Irish Math. Soc. Bulletin, Number 89, Summer 2022, 43-49): www.maths.tcd.ie/pub/ims/bull89/wef/Articles/Stewart/Stewart-wef.pdf
Looked a little harder and found the Maximon paper online at: tsapps.nist.gov/publication/get_pdf.cfm?pub_id=150847
One can also evade paywalls using sci-hub
@ 10:38 Personally I would have separated the terms so into odd terms and even terms, rather than factor by their remainder when dividing by 4.
But then you would have to separate those 2 sums further into real and imaginary parts.
@@megauser8512 What I mean is that
i^n/n^2 can be separated into terms where n is odd and when n is even. When n is even the term is real, when n is odd the term is imaginary.
13:49 is also a kind of Euler's reflection identity.
Also for who want to know how it get first time he just explore
1. Go from expansion to intégrale by using the integral of x^nlog(n) from 0 to 1 = x^n+1/n+1;
2. Return back to expansion from integral(after making a change variable like x->1-y;
Just a question: is the integral definition well defined? From what I remember of my complex analysis course different path produce different result in the integral (unless the integrand function is analityc?)
Integrand is entirely analytic inside the unit disk. There is no issue with that definition.
Yes. It is well defined if you work inside the disk |z|≤1. In this region exists the series expansion around z=0 of ln(1-z) and remembering that the first term is not constant, you don't have to worry with singularities. Or better, z =0 is a removable singularity. You can say that ln(1-z)/z is analytic because it have a Taylor series expansion around z=0 inside the unitary disk.
@@praharmitra Thank you!
@@MrFtriana Perfect, thank you!
@@davidcroft95 also, there are a theorem in complex analysis that relates analytic functions and power series.
yes, f'(z)=0 => f(z) for some z is equal to some constant C, but this does not mean that this constant is always equal to pi²/6
21:10
Haha you posted this 19min ago, the video was uploaded 19 min ago and you posted a timestamp >19min. You knew what you came here for;)
Is there a Milogarithm(Mi for Mike)?
This video highlights why I always complain about all of the meaningless video titles. @5:50 he mentions working out Li_2(-1) has been done elsewhere on the channel. Searching his channel yields 2 other dilog videos, neither of which is the one he mentions.
It's pretty clear he was talking about the series, not necessarily in the context of the dilogarithm. (-1)^n/n^2 isn't searchable on YT even if it was in the title, so it doesn't really matter if he put it in there or not.
It's a routine summation from an introductory complex analysis class. This link explains the background well enough: www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf
ua-cam.com/video/9-fZzwvYjdc/v-deo.htmlsi=La6qnM7YO_oL21JD
Bprp got video about that identity!
Thanks for a great video. A quick question: do we want to restrict the modulus of z to be strictly less than 1? I am not sure how the analytic continuation would be possible at z=1, and in fact, Wikipedia also states that [1,\infty) is a branch cut. Thank you.
What is the dilogarithm used for?
It's nice having a name for a function when it comes up often in calculations, and I assume this is why the dilogarithm was defined in the first place. According to its wikipedia page (link below), the dilogarithm can be used in particle physics and also to calculate the volume of certain geometric shapes (which is to be expected from a function defined using an integral).
en.m.wikipedia.org/wiki/Spence%27s_function
I feel like there needs some editing?
no no no, editing here was perfect, we love it, warts and all.
What is the inverse of the polylogs
Yeah i suggested to every one goto my favorites functions book :
Milton Abramowitz and Irene A.Stegun - Handbook of Mathematical Functions-.
Concise Encyclopedia of Mathematics.
Jordan C - Calculus of finit Differences -.
The best one :
E. T. Whittaker and G. N. Watson - A C O U R S E O F M O D E R N A N A L Y S I S.
Hi Dr. Penn!
I love swimming in math.
Nice handwriting 👌
5:15 me when zeta
Thought he was talking about “playing possum” or something
If dilog(1) equals the integral from 0 to 1, then how do we justify using a series expansion for -ln(1-z) which is invalid for z = 1?
Something something analytic continuation.
@@Pablo360able An integral does not change when finitely many points are excluded. Can't we just say it is integrated to 1 without the 1, hence only needing the integrand being defined in interval (0, 1)?
Why is your editor putting in all your verbal mistakes?
Based on how it is edited, it seems like his usual editor is on vacation. None of the fancy stuff she(?) normally does is present.
A few verbal mistakes are funny and attention-grabbing.
Too many are annoying, but where you draw the line between them is subjective.
Here is a related video from Michael Penn from the past: ua-cam.com/video/ZizzWhJbkno/v-deo.html
OK... But I can't find any related Michael Penn's video about this identity
ua-cam.com/video/9-fZzwvYjdc/v-deo.htmlsi=elM1wToWKoa6dzx2
I must be 🍃 af but this video was beautiful!!
zeta(2)
I'm not aware of it myself but it feels like there has to be some sort of relationship between Li2 and zeta in general.
@@BridgeBumhere's a link between the zeta function and polylogarithms in general:
Li_s(z) = sum as k ranges from one to infinity of z^k/k^s
Where s and z are complex numbers (only defined if the sum converges)
Set z = 1 and allow s to vary and you have the zeta function on (1, inf)
I'm not aware of any way of defining the zeta function only using Li_2 though
hi...
Dilogarithms are pretty cool. 😎
cool
Error at 9:45. Equality doesn't hold as sums start from different indexes. Can be fixed if we expressed the 2nd sum as sum from n from 0 to infinity of reciprocal of (4n-2)^2
He glosses over the details but if you actually see what he collapses them to by plugging in the respective starting terms the end result is the correct.