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- เผยแพร่เมื่อ 11 ก.ย. 2024
- Though I'm sure these results are somewhat well known, (in fact I firmly believe Euler derived these fascinating identities as some side quest) it is, nonetheless, a lovely little exercise showcasing how boredom can lead to productivity.
My complex analysis lectures:
• Complex Analysis Lectures
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Advanced MathWear:
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NOTE: x belongs to (0,2π) to ensure that the complex logarithm is continuous. This also justifies the closed form derived for the cosine series towards the end; x approaching 0 or 2π yeilds the equation for ζ(2).
When a mathematician gets bored, he summons Euler's Identity one too many times
Looks like it😂😂
He summons it one, two, three, four, five, six thousand times and makes it obvious for each one of us.😂
really need to see ln(ln(
Ooh, i. Love e^(i(pi)) +1 =0!
@@Tryh4rd3rr 0! = 1 ))
9:07 I'm so happy for you bro, you got your first sponsor on this channel 🎉
9:05 all of advanced maths is sponsored by euler
Hi,
5:50 : I don't agree, that is an awesome chimical joke! I was just waiting for Na 🤣
6:22 : missing i in front of the sinus,
7:05 : missing d theta,
"ok, cool" : 0:32 , 3:34 , 4:35 , 5:32 , 8:07 , 9:52 ,
"terribly sorry about that" : 1:12 , 3:58 , 9:39 .
The scope of timestamps is evolving 😂
Ou khei khuoul
Time stamp for 9:39 is not correct
5:32 and 5:50 joke is also incorrect
Gotcha!
As someone who just returned to watching this channel after a while, on one hand, i regret having missed out on so many puns and "ok coooool"s but on the other, atleast now i have the knowledge to be able to comprehend these random functions you keep invoking out of nowhere(zeta, reimann,etc)(i passed high school!) Thanks for being an awesome man, math guy
Congrats on graduating
@@maths_505 💜💜
@@maths_505 💜
Wow, first time Ive saw the poly logarithm function used outside of weird wolfram answers
@@Nottherealbegula4 wanna see it again in the next video or so?
Pretty please!
1:20 The integral operator is not countably additive, so you need to prove this equality
More specifically, the series are not absolutely convergent. So, the reordering he performed might be off.
I cant believe Euler came back to sponsor this video, what a legend
Moving integral sign through the summation sign requires the series to be absolute convergent.
i knew this would impress me, but it honestly exceeded my expectations, well done kamal
Thanks mate
*Complicated mathematics*
"Terribly sorry about that..."
*Complicated mathematics*
"Ooooookkkkk cool!"
Love this guy
This so damn accurate I'm tempted to write this as the channel description 😂
"Im just gonna leave it as the zeta function cause it looks really cool" is the most relatable math tick I have ever heard
you lost me at 0:03😂
😂😂😂
The thumbnail was correct, I now love the email person who sent the gaussian integral solution for last video
Funny thing is I almost forgot I had written this up. I posted these results to patreon a while back and only last night decided to finally make a video.
I don't understand most of this, but it is fascinating, I guess.
It's cool after a few maths 505 videos this is gonna seem trivial😂
@@maths_505 facts.
@@maths_505 That's true, but the Clausen function is new for me
@@Mario_Altare it was the functions debut today
@@maths_505 Great! Always glad to know these functions
Sorry to be a Debby downer, but the very first expansion is wrong. The series for ln(1-z) is correct, but only if |z|
Same as at 3:16. The series expansion is written for |z|
Nothing wrong with the series since it can always be fixed by a limiting case: replace e^(iθ) by εe^(iθ) and let ε approach 1 from the left. As far as the polylogarithm is concerned, I explicitly mentioned that for integers s \ge 1, the series converges for abs(z)=1 as well.
The series converges to ln(1-z) for |z|
@@maths_505 As you noticed I didn't say the series is divergent, just that convergence should be shown. Moreover, not only convergence is known, but also the exact value was calculated long time ago. If you split it into \sin and \cos series, then the first one has the value (\pi-\theta)/2 and the second one -ln(\sin(\theta/2)-ln2, that's all.
What about the bounds of integrition? I've derived as an exercise the radius of convergence of taylors expansion of log as 1 unit, but i didn't study complex analysis enough to really play with convergence in the complex plane.
By the hindsight lemma i guess the radius is at least 1 in the complex plane so x can be any real number, but pointing that out would be nice.
The restriction on x is mentioned in the pinned comment
Ok, terribly cool about that 😃
Indeed😂
"am I missing any negative signs?.. no way" - famous last words
havent watched the video but picturing this in my head it will be a circle in the complex plane with center at -1 so re^itheta where r is -2costheta because r=-2costheta looks like the right circle in polar coordinates so its -2costheta*e^itheta
In the description you say that you believe Euler may have already derived these identities.
Regarding the main identity you have derived around 8:48 of the video: The German wikipedia lists exactly that equation as "Kummer's identity" (after Ernst Kummer), see here:
de.wikipedia.org/wiki/Clausen-Funktion
I stumbled upon it when I wanted to know what the Cl_2 functions is.
Edit: I see, it's also in the English wikipedia: en.wikipedia.org/wiki/Clausen_function#Kummer's_relation
Slick, Sir. Nicely done.
@@mickodillon1480 thanks mate
Doesn't the expansion of ln(1-z) works just when |z|
No problem whatsoever since it can be fixed by a limiting case: replace e^(iθ) by εe^(iθ) and let ε approach 1 from the left.
No that was an incredible chemistry joke 😂
5:42 All the good chemistry jokes Ar.
Cool! I never heard of the Clausen Function.
Middle earth fam we are back
@@surfsidecrayon0702 FOR GONDOR!!!
@@maths_505for souran 😂
If I understood this video correctly, Euler's identity truly is the One ring 😀
Just found this channel, first time I ever heard about polylogs, etc.
I absolutely love this 😀 😂🎉
Your content is good, but please refrain from clickbait thumbnails. I'm sure people who already love math will appreciate this video. However for people who didn't already love math, nothing about the derivations and explanations here will make them love it.
Beginning 4:00, the 1-exp(itheta) could also have been decomposed more easily by factoring out exp(itheta/2)(exp(-itheta/2)-exp(itheta/2) and then using the definition of the sin() in terms of complex exponentials.
"Sponsored by Euler" ❤
Can also pop in a few trivial values for x to pull out some more interesting stuff. (Trivial one being x = pi, letting you start off with ln(2) for a bunch of stuff.)
I have absolutely zero clue what I just witnessed here, but watched until the end in hope to see some kind graphical representation or utility. I wonder if there is some real world phenomenon this function describes?
I'm not sure that switching the order of the sum and integral is fully rigorous at the beginning. Since the sum does not converge absolutely (the absolute values of the terms yield the harmonic series), this is not immediate. (If it did converge absolutely, we could apply Lebesgue's Dominated Convergence Theorem.) Swapping the sum and integral without regard may very well yield an arbitrary incorrect value.
@@uy-ge3dm Nothing wrong with the series since it can always be fixed by a limiting case: replace e^(iθ) by εe^(iθ) and let ε approach 1 from the left.
I know you have told me that you have solved a similar integral in another video, but, as a subscriber's suggestion, I would really recommend to solve the integral from 0 to pi of ((sin(2nx))(cosx))/sin(x). The answer is literally π itself, and all of your fans including me know you love those simple answers with transcendental terms. I know that the integral itself in the thumbnail won't be too attractive, but u can surely manage a way to make some ppl interested on it. Soo, that's pretty much it and I am sorry for all the yappery.
i only see a bunch of random functions, variables and numbers
You wrote abs(z) < 1 not
I also mentioned that for integers s greeter than 1 it converges for abs(z)=1 just like for z=e^(ix)
more interesting than the content is the study of your handwritten symbols! my favourite is the bold integral symbol, followed by the no less artistic THETA.
me when i pull the series through the integral without checking anything
@maths_505 One question to timestamp 6:21: Why do you assume that the logarithm and the exponential function do cancel out exactly? I mean, if we were talking about the real numbers, then yes, of course. But here we're integrating complex-valued functions in the complex plane. And I thought that the complex 'ln' function would be a multi-valued function, so how do we know we get the correct branch of it? So that in fact ln(e^-ix) would give us back the -ix for all x? I don't know that much about complex analysis and feel unsure about that point.
See the pinned comment.
Why do you drop d theta in half of the integrals you write? 0_o
This stuff could actually be really interesting in the context of Fourier series, I think some forms of these answers even play a role in average power calculations for electronic devices (don’t take my word for it 100% tho, I switched from EE to math lol)
3:48 you miss +C 😑 🤣
No
You did not reduce zeta(2) to pi^2/6 at the end of the integral :( I was wondering why not... that's a famous, famous identity! (see Euler and the "Basel problem"!)
When a mathematician gets bored...you are not challenging yourself enough. There are many unsolved problems out there.
Very cool solution. Thank you.
I feel like using Greek letters in maths is ok as long as you are Greek and otherwise is too bombastic :)
Excellent
Ok before watching this, I smell polylogarithms.
Edit: I was right, but this was more of a feast of special functions. Nice work.
6:05 missing an i next to the sin
What is the clausen function?
@@redroach401 en.m.wikipedia.org/wiki/Clausen_function
It's a function that yields delicious pickles from a cucumber
@@maths_505 Thank you
I found this video "ooooookaaaaaay, cool!" 😄 no, honestly: it's a great one🙂. The only thing i would like to criticize: The special formulas could be further explained maybe.
What was the chlorine molecule function? (yes, I am a chemist)
The Clausen function
what software do you use for writing?
If we're middle earth, what does that make you? Gandalf or Saruman?
I think I'd be on the battlefront as one of the men serving under king Aragorn
If really given the choice I'd like to be one of the dunedine.
"Okay.. Cool :)"
i still don‘t love that function - the point where i should start to do that elluded me ;D
this is beautiful
Is there minecraft on in the background?
is it -(pi)ln2 when x=2pi?
you look like , if I tell you to go outside , touch some grass you would probably go there and start making formula about grass distribution in the ground 😂😂
🤣🤣🤣🤣🤣🤣🤣🤣
Top tier math content creator
Thanks mate
what
Dont the equations at 7:17 imply that Cl(x)=0 for all x, since it is real valued?
@@thatdude_93 no ofcourse not. What gives you that idea?
I suppose it is easy to assume that the second representation is all complex-valued, but the Dilogarithm is taking a complex argument, therefore the second equation isn't fully imaginary and Cl_2(x) is allowed to not be zero
@@hyrumhaddox3589 exactly
any good books to learn more about special functions?
@@lambertwfunction I just use the internet tbh
Something is screwy with your final result for the real part. The LHS is bounded as x goes to infinity since the sum of 1/k^2 is finite, but the RHS is not.
Indeed there needs to be further clarification: we're making use of the complex logarithm so x should be contained with the interval (0,2π) so as to ensure the logarithm is indeed continuous.
"Negative chlorine molecule of x", the chemistry between u and math are having negative rizz😂
A cool proof that Li2(1) = dzeta(2) :)
Howd he know that series though
Integrate the geometric series i.e. expand 1/(1-z) as a series and integrate term by term.
Where is the guy in the comments pointing out every "Terribly sorry bout that"
He's way ahead of ya😂
Bro... You're switching integral sign and sum sign on an infinite series without showing the series converges? What kind of mathematician are ya?
One that knows convergence here is pretty trivial. You can see this via a limiting case: replace e^(iθ) by εe^(iθ) with ε \le 1 and let ε approach 1 from the left to get the result.
First view and comment and like
The Holy trinity of TH-cam is achieved
Okay, cool!
Cl, Li, elementary!
A lovely mix indeed
You sure you haven't forgot the constant term?
Wow this one is very cool
Thanks
We're so back
Yeah I feel much better after taking a break and decreasing my upload frequency.
Ok cool. Neat vid
Ok... Cool!
It's actually pronounced th-ay-duh
In my country, everyone pronounce it as "Thii-Tah"
10 minutes of funky number manipulation and that’s your #1 takeaway? (It’s okay I, too, barely understood what i was looking at)
I love cats
🌀 Spoiler Shield 🌀
You skipped lots of convergence issues
Nothing wrong with the first series since it can always be fixed by a limiting case: replace e^(iθ) by εe^(iθ) and let ε approach 1 from the left. As far as the polylogarithm is concerned, I explicitly mentioned that for integers s \ge 1, the series converges for abs(z)=1 as well.
@@maths_505 The integral of 1/(1-x) from 0 to 1 is infinite. The integral of 1/(1-(1-ε)χ) from 0 to 1 converges for any positive small ε, but as a function of ε itself. diverges to infinity as ε tends to zero.
So no, it's not as simple as that.
You need to talk a bit more about what happens when e^(ix)=1 ie x=2πk for integer k, because log is undefined there
Some of the writing is illegible
Βαρετό
when an average fella gets bored:
may I offer these tidbits from the Popcorn Model of Nature's Reality.
This is in the study of the Harmonics and the Harmony of Our Universe in the context of Everything:
so,
lets use a metaphor where 1 musical note, (*) , represents Nature's Reality;
This note, (*) , represents the true existence of Nature's reality.
This is the realm of the lord, the almighty GOOD (not a religion but an attitude). The real note in which everything resides. What follows are just harmonics of the supreme existence of reality.
1st harmonic of reality (hor)* the human mind and the MotherVerse.
2nd harmonic of reality * commonly referred to as our universe and where electromagnetic radiative force is dominant.
3rd hor * dark matter, the strong nuclear force dominates.
4th hor * the weak nuclear force dominates.
5th hor * gravity, where the popcorn really explodes.
6th hor * time, the here and now where the rubber meets the road.
The 3rd, 4th, and 5th combine to create Dark Energy.
This not everything. Undescribed harmonics extend, ad infinitum, above and below the note (*). The harmonics show that space that appears empty is never in fact empty.
Between Nothing and Everything is Something :)