First time I watch a math video on my own free will without having to freak out over a test. I love your take on maths with the memes and genuinely being funny in your videos. I am learning math with a smile on my face. Keep it up.
Booy, am I excited seeing double integral and not only that, but the el classico intro + most importantly, that one constant guy, flammy, who makes me smile at every of his videos :)
Hello! Pappa Flammy, I just wanted to say thanks for your work! I love the effort and passion that you put on every single video. Most importantly the sense of humor, I believe that it is the seasoning of the whole channel. Greetings from Mexico! : )
Oh man.... this did it. Integral eta on the square. This is nuclear fire. A glowing core of mathematics. Math fucking rules. I love you Flammy. I got the Math high from this one.
This is amazing! I made a proof recently proving all negative even zeta values are 0 without using Bernouilli numbers! I couldn't have done it without you
hey flammy, you proved the identity for the positive integers right? and i wonder if you can extend the identity to the complex numbers at least in the right semiplane, obviously for the exponent you can take a suitable branch of log. However, you dont have the identity holding a priori in a set with a condensation point. Is the identy true for other values?
I thought a bit about it, we have f(z)=g(z) holomorphic in the right semiplane, and i am not sure, but i think that atleast the right hand-side doesnt vanish in an interval of the positive real axis so maybe we can set h(z)=f(z)/g(z) there and say that if the interval contains an integer in the interior, then h assumes the value 1 and if we can prove that h
Enzo Giannotta You are overthinking it gravely. The reason his method appears to be restricted to natural numbers specifically is because the derivative operator D appears to only be iterable by a number n of times, where n is a natural number. However, the reality is that the linear operator D^s is well-defined for all complex of values of s, so in fact, for all values of s where the right-hand side of the equation exists, the identity holds. He did not state this explicitly, but it is not technically necessary to do so if you assume it a priori. We can assume it a priori because of the existence of fractional calculus.
hey papa could u solve the gaussian integral by fubini:ing it and then taking the limit of the series, it looks really fucking ridiculous so id love to see it
Archmaster Colesław In reality, he showed it works for any C+ implicitly, since the formula D^s(e^kx) = (k^s)e^kx still holds true for complex s. Remember that fractional derivatives exist. He did not mention it, but the demonstration still holds equally for those values because the derivative behaves identically for those values.
Johannes H That's fine, (-1)^s is well defined for all complex s as well. And that factor would still pop up from fractionally differentiating the right hand side. (-1)^s = e^(πis)
In fact, I can give a more general solution, by substituting πi/2 in the final equation by ln(i), and rather than using the principal value, just choose all the values for the multifunction, and then calculate W(-ln(i))/(-ln(i)) for each value of ln(i). I just gave every single solution for the original equation with a single expression. x = W[-ln(i)]/[-ln(i)].
Why does everybody writes an n-th derivative like this f⁽ⁿ⁾ (x) ? I find this notation awful, and I personnaly think that the notation dⁿ/dxⁿ is better and way more appropriate
micrapop _ I agree. The reason seems to be a historical artifact, and not a rigorous with any justification whatsoever. Modern mathematical papers actually use the notation D^n, since it is much more useful as it can also be utilized in generalizations of calculus - other notations cannot - it does not involve pesky parenthesis or fractional notation, it is more compact, it emphasizes the linear nature of the operator, and it also is very non-ambiguous as there is no other operator with standard notation D or D^n with which to confuse it with.
Golam Martuza Hossain Why not Wikipedia? Wikipedia is actually a really great introductory source for these topics. And then, Wikipedia has references, which are sources in themselves, which you can read for a more detailed treatment of it all. Those articles tend to cite a lot of books too.
You’re doing amazing work man! Never stop.🤩
First time I watch a math video on my own free will without having to freak out over a test. I love your take on maths with the memes and genuinely being funny in your videos. I am learning math with a smile on my face. Keep it up.
Booy, am I excited seeing double integral and not only that, but the el classico intro + most importantly, that one constant guy, flammy, who makes me smile at every of his videos :)
Hello! Pappa Flammy, I just wanted to say thanks for your work! I love the effort and passion that you put on every single video. Most importantly the sense of humor, I believe that it is the seasoning of the whole channel. Greetings from Mexico! : )
You're doing amazing work man, don't let anything/anyone negative bring you down!
Remember: *no i dont remember*
>mfw the video has 3 hours worth of prereqs
Papa going buckwild in this one.
Oh man.... this did it. Integral eta on the square. This is nuclear fire. A glowing core of mathematics. Math fucking rules. I love you Flammy. I got the Math high from this one.
Great video my boi ! You're doing such a great work, this kind of stuff it's really amazing! Keep doing it!
no fucking way, this is just what I needed for a homework of statistical mechanics THANKS!
12:42 Is this an audio glitch or is Flammy turning into an alien?
Ik I was like wtf
ahh...das hawt,daas hawt
I m dead 😂
This is amazing! I made a proof recently proving all negative even zeta values are 0 without using Bernouilli numbers! I couldn't have done it without you
Flammy’s spicy math went over my head 🥵😔
spending some time with the one and only mathy boi for my heart ❤ papa flammy
FAN MEETUP NOW! 👏👏👏
Oh my Papa!
Have you worked with Catalan numbers? I’d love to see a related lecture to counting/probability:)
The best all over the world!!!😊👏🏾👏🏾
LOL at first I thought you were saying devilishly eta function ==> devilled egg function
hey flammy, you proved the identity for the positive integers right? and i wonder if you can extend the identity to the complex numbers at least in the right semiplane, obviously for the exponent you can take a suitable branch of log. However, you dont have the identity holding a priori in a set with a condensation point. Is the identy true for other values?
I thought a bit about it, we have f(z)=g(z) holomorphic in the right semiplane, and i am not sure, but i think that atleast the right hand-side doesnt vanish in an interval of the positive real axis so maybe we can set h(z)=f(z)/g(z) there and say that if the interval contains an integer in the interior, then h assumes the value 1 and if we can prove that h
Enzo Giannotta You are overthinking it gravely. The reason his method appears to be restricted to natural numbers specifically is because the derivative operator D appears to only be iterable by a number n of times, where n is a natural number. However, the reality is that the linear operator D^s is well-defined for all complex of values of s, so in fact, for all values of s where the right-hand side of the equation exists, the identity holds. He did not state this explicitly, but it is not technically necessary to do so if you assume it a priori. We can assume it a priori because of the existence of fractional calculus.
@@angelmendez-rivera351 well, i saw it coming. thanks for the clarification
At the start, yousaid 1-e^x for the zeta function, but it's the other way around, e^x -1
Nice to Clarky back in the thumbnail
That is so cool
hey papa could u solve the gaussian integral by fubini:ing it and then taking the limit of the series, it looks really fucking ridiculous so id love to see it
Nice! Keep up the good work.
Papa didn't you derive this before?
Edit: I just realized that it was the Riemann Zeta function you derived, not the Dirichlet Eta function :s
The sign in the integral is also different.
When will Flammoids come? Or Jensoids; can't decide which sounds better :D maybe I just missed them
@@PapaFlammy69 crap
Must've been too focused on the math
“Integerraaalll”
Schönes Video
Good shit
You've shown that it works for s in N-{0}. Does it work for R_+ or C_+? If so, why?
Archmaster Colesław In reality, he showed it works for any C+ implicitly, since the formula D^s(e^kx) = (k^s)e^kx still holds true for complex s. Remember that fractional derivatives exist. He did not mention it, but the demonstration still holds equally for those values because the derivative behaves identically for those values.
what do you get for s=2
alright, seems to hold for integharal s, but what about complex s ^_^
everyone Also holds, because D^s is defined for complex s, not just integral s.
I need a whole lot more fundamentals for this 😅
And btw, Papa>>>>>>>>>imagine who(not humorous people)
Am I the only one who read "Dirac Delta function" for a moment?
But if s is not a natural number?
Johannes H It need not be, since fractional derivatives exist.
@@angelmendez-rivera351the problem is the (-1)^s
Johannes H That's fine, (-1)^s is well defined for all complex s as well. And that factor would still pop up from fractionally differentiating the right hand side. (-1)^s = e^(πis)
@@angelmendez-rivera351 ok, that makes sense. I think he could have mentioned that. Because everything else is real valued
@@angelmendez-rivera351 ok, that makes sense. I think he could have mentioned that. Because everything else is real valued
Integh-aral
Came here from Andrew Dotson (meme war). Maybe I should switch teams from Physics to math???!!!
My nama Fapmid :v
Papa flammy did you just talk to yourself?
Thaa fuccc? That's your channel? What's trivial for then?
So fapmid is your 3rd channel? FML! How could you not tell me before papa?
Are you the Tf2 medic
yeye
I N T E G H A R A L
Find x if i^x=x, it's harder than all that.
Ha- Bouz Not at all. i^x = e^(πix/2) = x => 1 = xe^(-πix/2) => -πi/2 = (-πix/2)e^(-πix/2) => W(-πi/2) = -πix/2 => x = W(-πi/2)/(-πi/2).
In fact, I can give a more general solution, by substituting πi/2 in the final equation by ln(i), and rather than using the principal value, just choose all the values for the multifunction, and then calculate W(-ln(i))/(-ln(i)) for each value of ln(i). I just gave every single solution for the original equation with a single expression. x = W[-ln(i)]/[-ln(i)].
@@angelmendez-rivera351 great solution
Oof
Why does everybody writes an n-th derivative like this f⁽ⁿ⁾ (x) ? I find this notation awful, and I personnaly think that the notation dⁿ/dxⁿ is better and way more appropriate
micrapop _ I agree. The reason seems to be a historical artifact, and not a rigorous with any justification whatsoever. Modern mathematical papers actually use the notation D^n, since it is much more useful as it can also be utilized in generalizations of calculus - other notations cannot - it does not involve pesky parenthesis or fractional notation, it is more compact, it emphasizes the linear nature of the operator, and it also is very non-ambiguous as there is no other operator with standard notation D or D^n with which to confuse it with.
@@angelmendez-rivera351 I can only agree with all what you said ; nothing to add : you're totally right
*NeVeRMiND*
oi flammy,tell me some sources for self-studying these special functions.
For the love of God, don't tell me to read wikipedia pages.
@@PapaFlammy69 Then from where do you study these? :3
Golam Martuza Hossain Why not Wikipedia? Wikipedia is actually a really great introductory source for these topics. And then, Wikipedia has references, which are sources in themselves, which you can read for a more detailed treatment of it all. Those articles tend to cite a lot of books too.
Tf is this man