A slightly simpler way of getting a geometric series from the start is to multiply both numerator and denominator by e^-x. The denominator becomes 1-(-e^(-x)), which can be expanded into a geometric series. This makes the resulting series+integrals easier to work with as well.
I love the more subtle humor of Michael Penn and the more chaotic humor of our dear editor, its really charming and another reason why i love this channel so much :) (beyond the hard math of course)
at 3:00 it's trivial that e^-2x is between 0 and 1 for x > 0, because since 0 < e^-x < 1 and e^-2x = (e^-x)^2 it follows easily, we all know that if you square a number that is between 0 and 1 you get another number between 0 and 1
Ah yes. An integral related to the Fermi-Dirac statistics. Also, is clever what he is doing here. Doing that difference of squares, avoid a (-1)^n term in the series that makes the work more difficult. Also, we can avoid the D-I method, considering the integral of x e^(-ax) as a Laplace trasform.
Oh this is an example of an absolutely gorgeous integration result that connects the gamma and zeta functions!...I solved that integral on my channel and it was marvelous!!!
ive been messing around with dirichlet series recently, and from this you can see quite quickly that this is 1/1^s - 1/2^s + 1/3^s -.. at s=2 by a well known integral form of these series, then also that the terms with even denominator are 1/2^s times zeta(s), then that this series is the zeta function minus twice these terms, and so is half zeta(2) and this also lets you find this same integral when the x is raised to some power s in terms of zeta(s)
that was nice! i didn't know that integral so no spoilers here. what a great idea to use a geometric series that converges to use dominated convergence theorem, how instructive, and again wow!
It would be interesting to see how a "rough draft solution" might start. (I assume it's just working with exp(-x) and then noticing a difference of squares would be helpful several steps along.)
Funnily enough my normal routine involves various exercises, then doing maths and waiting to lunchtime to eat (12:8 diet). Then I settle down to watch RUclips, starting with a Michael Penn video... with a smoothie. I've already sent my beautiful spouse the link to the store so maybe after my birthday I can have a smoothie covered math hoodie too
It could be, but then the video would need to actually give an answer. Yes, that series gets us to pi/6, but why? You have to go to a 3Blue1Brown video to get the answer. Look for something about the zeta function and pi.
I've got to say that it is very frustrating to see a title like that but then the video doesn't actually answer the question. I was expecting to see an explanation that provides an intuition for why.
I use to solve it in a slightly different (but totally equivalent) manner. Firstly, one recognizes that the 1/(1+exp(x)) is (barring a sign) the derivative of ln(1+exp(-x)). Integrating by parts, the boundary term vanishes and one is left with the integral of ln(1+exp(-x)). Taking some license at x=0 (heck, after all I am a poor Physicist!) rewrite this log as a taylor series in exp(-x). Swap summation with integration (each term converges very quickly) one finds a nice power series which upon simple rearrangements is just the one written by Penn.
It's interesting that If we raise x in the numerator to some power k - 1, we will get integral that equals to (1-2^(1-k))ζ(k) Γ(k) which is nice connection to the Riemann's Zeta and Gamma functions. I even thought to propose you that integral to show that it is connected to Basel problem for k=2, and was very surprised to see that video.
Hello professor, could you make a video about the dominated convergence theorem? We always admit that everything converges nicely but how would someone prove it rigorously? Some examples would be nice
after three semesters of calculus and a separate course in analysis and topology i can tell you that it's just something you feel and wave your hands saying that "it clearly follows from [insert appropriate convergence theorem] that this converges nicely"
As an engineer, I would integrate it numerically from 0 to a very large number assuming the integral converges and then take the first few significant digits as my answer😂
A bit disappointed by the clickbait title. You showed what the answer was, but there was little discussion about why. The title implies some insight that was not there.
I would have written x/(1+e^x)= (x*e^-x)/(e^(-x)+1) =(x*e^(-x))/(1+e^(-x)) and then expanded 1/(1+e^(-x)) using the familiar expansion of 1/(1+u) for |u|
At 11:30, he makes the assumption that you can add and subtract extra copies without changing the sum. I forget the name of the theorem, but I know I've seen a video explaining that there are infinite sums where, by messing with the order of the terms, you can make it equal literally any result. Is just adding new terms meaningfully different from pairing terms in different ways, and what's the criteria for knowing when you can and can't do something like that? Does anyone know?
The sums must absolutely converge, which is to say, that when you take the absolute value of each term, their sum also converges. In this case it's ok! Look up absolute convergence of series to find out more information.
If a series is absolutely convergent, the order of the terms does not matter and any rearrangement will converge to the same sum. Since these sums are all of positive terms, they are most definitely absolutely convergent. See en.wikipedia.org/wiki/Riemann_series_theorem
You know, I feel like I have seen something like this many, many times before. To anyone more knowledgeable than I: Is there a theory behind these integrals?
I know them because they appear in statistical mechanics when you work in the Fermi-Dirac statistics (when You work with electrons, for example) or the Bose-Einstein statistics (in this case is with photons, for example), and want to find expected values of physical observables.
Maybe a bit out of place to comment here when my problem is deeply rooted in other videos, but I still wanted to ask. Are there any solutions on the internet or from you on the problems you assign in the number theory playlist at the end of every video? I have an exam this year and would like to know if what I'm doing is right. Your problems are way harder than the ones in class, but also more interesting.
This is cool but the title implies that you'd be giving some sort of understanding as to why there's a pi appearing; i.e. exposing the "hidden circle", so it comes off as misleading.
It’s the same reason the Basel problem has pi appearing, which as he mentioned, he’s covered before. So you can check out those videos if you want an explanation
@@General12th You can't have it both ways. If this problem is congruent to the Basel problem, then "Why do we find pi in the solution to the Basel problem?" is exactly the same question as "Why do we find pi in the solution to this problem."
Please stop flashing up those “subscribe!!” banners. They are off-putting, and also insulting to viewers who are already subscribed. I am perfectly capable of subscribing to things I like without being cajoled.
I really dislike the inline advertising in your video. I can not pay to get rid of it. I am paying google to watch you without ads. Doesn't google pay you any of that to make a profit without selling out? I know others do that, but I appreciate when you don't. I think it is weird to assume that your audience would be ok to have ads forced upon them. I'm not. I don't think I am alone. To be clear: I will not watch TV. I plug my ears when advertising is blasted when out. I have decided not to be lazy and let others fill my free time with their personal interests. So I will NOT be watching this channel if inline advertising continues. I am just one, but I hope there are others who will not stand for "free" content at the expense of time with random comments from random people who I am uninterested in listening to. I would like you to take a stand. Stop doing it and say so. This would make your audience very loyal. Like I want to me. PS: I am a big fan. I love your show. Please don't take it away.
At the end of the day people like Micheal have to get money from making good quality shows like this or they wouldn’t do it. If the ads bother you then skip through them. If the thought of a 1 minute ad read is too much for you to mentally handle than maybe the internet is not the best place to spend your time 😅
![Blox Fruits Dragon Rework Update [Full Stream]](http://i.ytimg.com/vi/EqDAp8udhm0/mqdefault.jpg)
i would love for the smoothie hoodie to be a recurring hoodie in the hoodieverse of michael penn
A slightly simpler way of getting a geometric series from the start is to multiply both numerator and denominator by e^-x. The denominator becomes 1-(-e^(-x)), which can be expanded into a geometric series. This makes the resulting series+integrals easier to work with as well.
That was my first idea upon seeing the integral, too.
@@cheedozer7391 And mine. 😀
Me too.
I agree!
Yep, I really did not get why it was better to multiply by e^x - 1 at the beginning...
13:28 Indeed, 1 minus 1/2 is 1 makes all of maths substantially easier.
I replayed that moment multiple times to make sure
I heard him correctly.
After feeling foolish for replaying it, I found it interesting.
Cheers
HAHAHAHAHA, I laughed really hard at the merch monologue at the beginning XXXXDDD that presentation was just superb!!
I love the more subtle humor of Michael Penn and the more chaotic humor of our dear editor, its really charming and another reason why i love this channel so much :) (beyond the hard math of course)
"π is that one guy who is never invited to parties,yet still shows up."
- a brilliant commenter who's not me
I love how you levitate through the algebraic transformations.
@ 8:50, Blackpen/Redpen... yaaaaaayyyy!!!
I usually skip forward through ads, but managed to hear the fallout about LaTeX support in Squarespace. This is the right channel for me.
at 3:00 it's trivial that e^-2x is between 0 and 1 for x > 0, because since 0 < e^-x < 1 and e^-2x = (e^-x)^2 it follows easily, we all know that if you square a number that is between 0 and 1 you get another number between 0 and 1
Hey Michael, I managed to get a smoothie spill on my math major hoodie that i bought last week.
Ah yes. An integral related to the Fermi-Dirac statistics.
Also, is clever what he is doing here. Doing that difference of squares, avoid a (-1)^n term in the series that makes the work more difficult. Also, we can avoid the D-I method, considering the integral of x e^(-ax) as a Laplace trasform.
Oh this is an example of an absolutely gorgeous integration result that connects the gamma and zeta functions!...I solved that integral on my channel and it was marvelous!!!
thats why it seemed familiar
I've encountered this kind of integrals while dealing with fermi dirac distributions in statistical physics. Really nice stuff!
I found a pie on the kitchen counter and I know where that came from. Your pi came out of nowhere.
OMG!!!!. Que fácil parece todo cuando se ha hecho un montón de trabajo duro. Felicidades
Michael, you make everything look as easy as the square root of six times the solution to the Basel problem.
Love your videos Professor Penn! I'm sure I'll love this one, but I'll watch to make sure.
That's s good place to start. 😄
Great subtle solution- thanks a lot Michael👍
I loved this one! Your channel is amazing. Keep rocking!
This is what I call top quality content
ive been messing around with dirichlet series recently, and from this you can see quite quickly that this is 1/1^s - 1/2^s + 1/3^s -.. at s=2 by a well known integral form of these series, then also that the terms with even denominator are 1/2^s times zeta(s), then that this series is the zeta function minus twice these terms, and so is half zeta(2)
and this also lets you find this same integral when the x is raised to some power s in terms of zeta(s)
I didn’t catch the answer to the clickbait question! Why IS pi here?
Speaking of Merch: I desperately want a shirt that says "Play the same game" with some appropriate image
I appreciate the DIY clothes tip at the beginning.
that was nice! i didn't know that integral so no spoilers here.
what a great idea to use a geometric series that converges to use dominated convergence theorem, how instructive, and again wow!
It would be interesting to see how a "rough draft solution" might start. (I assume it's just working with exp(-x) and then noticing a difference of squares would be helpful several steps along.)
Funnily enough my normal routine involves various exercises, then doing maths and waiting to lunchtime to eat (12:8 diet). Then I settle down to watch RUclips, starting with a Michael Penn video... with a smoothie.
I've already sent my beautiful spouse the link to the store so maybe after my birthday I can have a smoothie covered math hoodie too
That is an integral which occurs in the study of a degenerate gas of fermions!
To answer the question in the title, my local Bob Evans restaurant has a sign that says pi fixes everything 😉
"Why are we finding pi here?" could be a weekly series. Maybe even daily.
There is a fun math game called "find the circle". Whenever there is a Pi in a result, find the circle that generates the Pi
It could be, but then the video would need to actually give an answer. Yes, that series gets us to pi/6, but why?
You have to go to a 3Blue1Brown video to get the answer. Look for something about the zeta function and pi.
I've got to say that it is very frustrating to see a title like that but then the video doesn't actually answer the question. I was expecting to see an explanation that provides an intuition for why.
I use to solve it in a slightly different (but totally equivalent) manner. Firstly, one recognizes that the 1/(1+exp(x)) is (barring a sign) the derivative of ln(1+exp(-x)). Integrating by parts, the boundary term vanishes and one is left with the integral of ln(1+exp(-x)). Taking some license at x=0 (heck, after all I am a poor Physicist!) rewrite this log as a taylor series in exp(-x). Swap summation with integration (each term converges very quickly) one finds a nice power series which upon simple rearrangements is just the one written by Penn.
I haven't seen the smoothie spill, my laptop screen has too many smoothie spills.
Man, I am still waiting to hear why pi is here???
It's interesting that If we raise x in the numerator to some power k - 1, we will get integral that equals to (1-2^(1-k))ζ(k) Γ(k) which is nice connection to the Riemann's Zeta and Gamma functions. I even thought to propose you that integral to show that it is connected to Basel problem for k=2, and was very surprised to see that video.
Hello professor, could you make a video about the dominated convergence theorem? We always admit that everything converges nicely but how would someone prove it rigorously? Some examples would be nice
after three semesters of calculus and a separate course in analysis and topology i can tell you that it's just something you feel and wave your hands saying that "it clearly follows from [insert appropriate convergence theorem] that this converges nicely"
That would be nice
Dr. Peyam has a video on it if you’re interested
@@kkanden Nah, you can justify it and you should. In a lot of cases Dominated Convergence Theorem works easily
As an engineer, I would integrate it numerically from 0 to a very large number assuming the integral converges and then take the first few significant digits as my answer😂
That's just fascinating. Thanks man!
ngl, i would love to have that exclusive "smoothie stain" merch xD
thank you very much sir, but I want to know why the number Pi appear where there is a log or exponential function ?
The joke at the beginning got me dead
nonono, we need *WHY* there's a pi, not *HOW* we got here.
Gives my thinking meat a pleasant feeling.
A bit disappointed by the clickbait title. You showed what the answer was, but there was little discussion about why. The title implies some insight that was not there.
He mentions Basel at 13:39 & bing, bong, boom Pi appears
Womp womp
Plus the e^x-e^2x is basically like inviting it straight in, pi loves when you compare two exponential functions, especially in an integral
At 11:25, that's just eta(2), which is just 1/2 zeta(2) and thus (pi^2)/12. Maybe a video on the relationship between eta and zeta functions?
Hi Dr. Penn!
Very cool!
Michael enjoys doing math like a kid playing with a dear toy! 😂
I misread that as "deer toy" and was wondering why you were being so specific. 🤣
@@beeble2003 I said DEAR toy meaning a beloved toy.
@@behnamashjari3003 Yes, I know. And I said I misread your comment.
Dirichlet Eta Function for the win
You are an international teacher!
I would have written
x/(1+e^x)= (x*e^-x)/(e^(-x)+1)
=(x*e^(-x))/(1+e^(-x)) and then expanded 1/(1+e^(-x)) using the familiar expansion of 1/(1+u) for |u|
Yup, this path is shorter and more obvious. I did the same
Well the (-1)^n is a pain
Just the final left in calc 2 for me, so i would’ve divided the top and bottom by e^(-x/2) and turned the integrand into (xe^(-x/2)sech(x/2))/2
“But then what?”, you ask?
*good question*
this is how i got it before watching, i was surprised when i saw the approach he took
Which Calculus level course would this be from? I remember a lot of crazy stuff from back in the day, but this one has me wondering. 😁😁😵💫😵💫
"Is this a particularly hard integral? --- No."
He's right, but he decided to solve it the hard way anyway xD
At 11:30, he makes the assumption that you can add and subtract extra copies without changing the sum. I forget the name of the theorem, but I know I've seen a video explaining that there are infinite sums where, by messing with the order of the terms, you can make it equal literally any result. Is just adding new terms meaningfully different from pairing terms in different ways, and what's the criteria for knowing when you can and can't do something like that? Does anyone know?
The sums must absolutely converge, which is to say, that when you take the absolute value of each term, their sum also converges.
In this case it's ok!
Look up absolute convergence of series to find out more information.
If a series is absolutely convergent, the order of the terms does not matter and any rearrangement will converge to the same sum. Since these sums are all of positive terms, they are most definitely absolutely convergent. See en.wikipedia.org/wiki/Riemann_series_theorem
always thankful and helpful
That was fun to watch!
I need to buy some merch soon.
but WHY are we finding pi HERE?
"bing bong boom, I'm following arbitrary directions for collective, parasocial fun", oh wait a second....
I just had this same question in difeq
Hi,Michael we can also use 1/1+e^-x
Clear and clean
I am everywhere.
bing bong boom, I'm following arbitrary directions for collective, parasocial fun
You know, I feel like I have seen something like this many, many times before. To anyone more knowledgeable than I: Is there a theory behind these integrals?
I know them because they appear in statistical mechanics when you work in the Fermi-Dirac statistics (when You work with electrons, for example) or the Bose-Einstein statistics (in this case is with photons, for example), and want to find expected values of physical observables.
If you muliply the zeta function and the gamma function you will get this integral
رائع جدا كالعادة.
طريقة رائعة ستحل العديد من التكاملات
I was thinking it would be a hyperbolic function
Dear Micheal, nice presentation but i can not find the answer of the question "WHY are we finding pi HERE?".
So it is half the Basel summation. But why does this one and Basel have a pi?
Maybe a bit out of place to comment here when my problem is deeply rooted in other videos, but I still wanted to ask.
Are there any solutions on the internet or from you on the problems you assign in the number theory playlist at the end of every video?
I have an exam this year and would like to know if what I'm doing is right. Your problems are way harder than the ones in class, but also more interesting.
This is great although i don't understand why you bothered with DI method rather than simple integration by parts
6:50 I stopped the video ... Now what do I do 😨
Ouch, my head hurts
This is cool but the title implies that you'd be giving some sort of understanding as to why there's a pi appearing; i.e. exposing the "hidden circle", so it comes off as misleading.
It’s the same reason the Basel problem has pi appearing, which as he mentioned, he’s covered before. So you can check out those videos if you want an explanation
I love all the gibberish you speak and how convinced you are that it actually means something! 😊😂
do you have any hoodies that say "math minor"??
Nice , i solved by calling zeta(2)
You literally solved the integral
Who is writing the descriptions?
but why are we finding pi here?????
Because this problem is congruent to the Basel problem.
Why do we find pi in the solution to the Basel problem? That's a different question.
@@General12th You can't have it both ways. If this problem is congruent to the Basel problem, then "Why do we find pi in the solution to the Basel problem?" is exactly the same question as "Why do we find pi in the solution to this problem."
3blue1brown has a nice video on why pi is in the basel problem
@@frankhenigman5117 yeah just rewatched it. I just wondered whether there is any geometric motivation to this integral.
How about x=lnu and papa Faynman handle the rest.
bing bong boom, I'm following arbitrary directions for collective, parasocial fun
i didn't like the video though, see how much of a rebel am i?
You never answered the question in the title of the video.
it's left as an exercise to the viewer
Yep, that’s a good place to stop
Their repeated appearance in unexpected places indicates that the universe is spherical and will one day return to the state from which it began
Meh. Honestly, feeling clickbaited by the title.
[I've deleted the accusation that's rebutted in the reply from the channel editor.]
Michael does all the ad reads. I think he was sick when he recorded this one, that's why it doesn't sound like him.
-Stephanie
MP Editor
@@MichaelPennMath Thanks for the clarification. That being the case, I'll delete that part of my original comment.
Who is making these descriptions 😂😂😂
I am
-Stephanie
MP Editor
At this point just substitute the value for x and draw graph for the equation and find area by some other method
What other method?
Lieber Michael! Immer wieder ein Vergnügen, Deine Videos anzusehen! Vielen Dank!
but WHY is pi there?
Please stop flashing up those “subscribe!!” banners. They are off-putting, and also insulting to viewers who are already subscribed. I am perfectly capable of subscribing to things I like without being cajoled.
I’m sure you are capable of that, he’s just playing the social media game! Don’t take it personally man
bad title if you aren't actually going to explain the pi at all >:(
cool
obligatory description comment
الله يحفظكم
π is everywhere in Mathematics because circles and periods are everywhere.
Just as there are 66 books in the Bible pi is in the Bible and 9900 times out of pi it's because GOD has a plan for it.
I really dislike the inline advertising in your video. I can not pay to get rid of it.
I am paying google to watch you without ads. Doesn't google pay you any of that to make a profit without selling out?
I know others do that, but I appreciate when you don't.
I think it is weird to assume that your audience would be ok to have ads forced upon them. I'm not. I don't think I am alone.
To be clear: I will not watch TV. I plug my ears when advertising is blasted when out. I have decided not to be lazy and let others fill my free time with their personal interests.
So I will NOT be watching this channel if inline advertising continues. I am just one, but I hope there are others who will not stand for "free" content at the expense of time with random comments from random people who I am uninterested in listening to.
I would like you to take a stand. Stop doing it and say so. This would make your audience very loyal. Like I want to me.
PS: I am a big fan. I love your show. Please don't take it away.
At the end of the day people like Micheal have to get money from making good quality shows like this or they wouldn’t do it. If the ads bother you then skip through them. If the thought of a 1 minute ad read is too much for you to mentally handle than maybe the internet is not the best place to spend your time 😅
because pi is everywhere...
pi is god
@blackpenredpen wow 08:50