blackpenredpen Well, I think for this question, you can multiply by e^(-y^2)dy Let it be a double integral and change it into polar coordinate ,then square root. Is my opinion correct?
@@iamtrash288 well yeah. Gauss was a genius so no surprises there and there can be no overstatement about that either. I was just making a joke about Peyam making it look easy
@@benjaminbrady2385 oh I see. I just thought that I knew it wrong or something and it was an integral that Gauss called impossible. Though I guess it is kinda obvious from my previous coment
I remember when I first learned how to do this. I was still a technician in the Air Force. I would used it to verify bit error rate values. The people around me thought I was a wizard because of this technique!
Another possible way= Integral[e*(-ysquare)]dy gives the same thing. Multiplying them into the double integral(which is double integral of "e to the negative x*2+y*2" ) gives the result's square. Then use polar coordinates and obtain doubleintegral[e*(-r square).r]drdθ. Substitute r*2=u. (Btw θ is from 0 to 2π and r is from 0 to infinity) Solve the integral,get π and take its square root there we have sqrrootπ .
Can you prove that the series goes to sqrt(pi)/2 then x approach infinity? In another words, prove by the series that the integral from 0 to infinity equal to sqrt(pi)/2. Thank you!
So you can prove that as x goes to infinity the power series (-1)^n*x^(2*n+1)/(n!*(2*n+1)) for n from 0 to infinity converges to sqrt(pi)/2, C = 0 and that e^(-x^2) is even function and it will be the 13th way to calculate the Gaussian Integral. Edit: C = 0 is unnecessary and of course it can be any real number
well the integrated series looks awfully a lot like the taylor series of sin(x) at a=0 so that explains where pi comes from this integral lol (but hopefully i will see a proof one day showing the connection xD)
Here's a little something using the 'Taylor Transform' I commented about in previous videos: for ((i)^(z-1) + (-i)^(z-1)) = ((-1)^z)X(z) T[ sin(x) ](z) = ((-1)^z)X(z) / 2*Γ(z+1) and T[ {integral}[e^(-x^2)]dx ](z) = ((-1)^z)X(z) / z*Γ( (z+1)/2 ) so T[ {integral}[e^(-x^2)]dx ](z) / T[ sin(x) ](z) = 2*Γ(z+1) / z*Γ( (z+1)/2 ) = 2^(z) * Γ(z/2) / √π This is not likely to translate into something nice without a better understanding of this 'Taylor Transform' and it's inverse. The interesting part is how much this resembles the functional equation for the Riemann Zeta function.
I've done this with the Euler spiral, using 8-byte floats, and it becomes unusably inaccurate when |x| is bigger than 6. How do you compute this in the tails (the tightly winding spirals of the Euler spiral)?
use the vanilla google android calculator app... Or use GMP. Science and technology GNU Multiple Precision Arithmetic Library Granulocyte-macrophage progenitor Guanosine monophosphate
But can we say exact number of terms to get exact decimal place? Is there a formula? I feel that I had such formula when I learned calculus 9 years ago
Todos sabemos que esta función no tiene una primitiva, pero cómo lo sabemos?, hay una demostración?, si tengo que integrar una función, cómo sé si tiene primitiva? cómo demuestro que no la tiene?
What if we just define a new function Bell(x) or something (cause the curve looks like a bell) to be this anti derivative and then that's the Taylor series for it
Just came from a video about an old Sci-Fi short story in which one of the main characters is an all-powerful entity referred to as "AM" :) Dunno if it could integrate exp(-x^2) though.
Not all taylor series have a radius at infinity. It's dependent on the distance from your point and the nearest essential singularity in the complex plane. so like the geometric series has a radius of convergence = 1 because 1/(1-x) has a singularity at z=1 -- so you can only be sure that it converges for values of |z|
@@ShiaServant I mean come on, there are hundreds of videos about this. This is literally the simplest thing you learn in first year of university, this is your curiosity in high school, and you find the answer in uni. But it's not even that complicated. I just hate this guy makes a pointless video about it when there are tons already.
![Gaussian Integral [Int{e^-x^2} from -inf to inf]](/img/n.gif)
Dr. P in the house!!!
@blackpenredpen, hi. Im subscribed to you, Black Pen Red Pen. And Im subscribed to Dr Peyam, and to your son Fematika. And to Flammable Maths.
WHOA!!!
He used blue pen and red pen
:(
blackpenredpen Well, I think for this question, you can multiply by e^(-y^2)dy
Let it be a double integral and change it into polar coordinate ,then square root.
Is my opinion correct?
Makes silly joke. Laughs at own joke. Explains own joke. Elaborates explanation of own joke.
I love it...termwise
Gauss: This integral is impossible
Dr. Peyam: Hold my beer
I don't remember well but didn't Gauss use the very neat trick of using multiple integrals and then converting it to polar coordinates?
@@iamtrash288 well yeah. Gauss was a genius so no surprises there and there can be no overstatement about that either. I was just making a joke about Peyam making it look easy
@@benjaminbrady2385 oh I see. I just thought that I knew it wrong or something and it was an integral that Gauss called impossible. Though I guess it is kinda obvious from my previous coment
I remember when I first learned how to do this. I was still a technician in the Air Force. I would used it to verify bit error rate values. The people around me thought I was a wizard because of this technique!
for the power expansion of e^-x^2 why is it -x^6/6! ? shouldnt it be -x^6/3! ?
Pretty sure he just got confused since 3! = 6
I think it's supposed to be 3!
@@mountainc1027 Or he though it was (-1)^n x^2n /2n!
Hi where is that vedio (expansion of e^-x/2)
When I saw the thumbnail I was like
SAY WHAAAAAAAAAAT?
Of course he’s gonna do some πM magic
2:16 This has to be 3!, right?
Other than that, fun video :)
I'm sure Dr Peyam expanded it to 6 and was so surprised that it was so easy that he just put a normal exclamation mark after it.
he meant a shouted 6, not 6 factorial, of course.
@@Tomaplen It is "shouted 6" but pronounced as "6 factorial" :)
I noticed the same thing haha
Hey dr peyam now do basel problem in 12 ways.
Brother what are you currently studying for?
@@Praddyy29 I am going to class XI this year and I am a jee aspirant
@@rishavgupta2117 all the best bro I gave jee this year.
@@ddm1912 Thanks and best of luck for Advance
Another possible way= Integral[e*(-ysquare)]dy gives the same thing. Multiplying them into the double integral(which is double integral of "e to the negative x*2+y*2" ) gives the result's square. Then use polar coordinates and obtain doubleintegral[e*(-r square).r]drdθ. Substitute r*2=u. (Btw θ is from 0 to 2π and r is from 0 to infinity) Solve the integral,get π and take its square root there we have sqrrootπ .
Yup I exactly done with integration of e^x^2 using maclaurin series. SUM x^(2n-1) / (2n-1) n!
There's a minor mistake at 2:16 on the 4th term of the Taylor polynomial for e^-x^2. It should be -(x^6)/(3!) not -(x^6)/(6!).
Dr. P you're an example for me, you seem to be a really good person, love from Bologna!!
Grazie 😄
@@drpeyam figurati, continua così 💪❤
Dr. Peyam the integral is a beautiful calculus, thank you for introducing us 👍.
R.I.P. erf(x)
Can you prove that the series goes to sqrt(pi)/2 then x approach infinity?
In another words, prove by the series that the integral from 0 to infinity equal to sqrt(pi)/2.
Thank you!
If you haven't found it in the past 11 months, he actually already did it here: ua-cam.com/video/kpmRS4s6ZR4/v-deo.html
@@cadekachelmeier7251 That's not what he did
Dr Peyam The Mathemagic Man! LOVE this channel, thank you!
Bro this is amazing, the happiest mathematician :)
If mathematics have iron man it would be black pen red pen... dr. P is doctor Strange
∫e^(x²)dx: I am inevitable
Dr. P: I am Peyam
Hi dr peyam if ln(0) doesn't exist then How we calcul the integral from 0 to π/2 of tan(x) using series ?
It’s an improper integral
Dr Peyam what that means ? Diverges !
Best duo in UA-cam .
Dr.Peyam is very nice and funny guy..and very good mathematician as well.
This is a pretty better camera angle... : )
Integral of x^x: laughs
So you can prove that as x goes to infinity the power series (-1)^n*x^(2*n+1)/(n!*(2*n+1)) for n from 0 to infinity converges to sqrt(pi)/2, C = 0 and that e^(-x^2) is even function and it will be the 13th way to calculate the Gaussian Integral.
Edit: C = 0 is unnecessary and of course it can be any real number
Three eighths of the comments are people talking about how the denominator should have been 3! instead of 6!.
Yeah, I know 🙄
Isn't is π/8? All the complaints are irrational!
@@dhunt6618 lmfaooo
Didnt u make a 12 part series about this already
He only calculated value of improper integral, he didn't find an actual antiderivative
Anu I did, but this is the director’s cut 😉
well the integrated series looks awfully a lot like the taylor series of sin(x) at a=0 so that explains where pi comes from this integral lol (but hopefully i will see a proof one day showing the connection xD)
Here's a little something using the 'Taylor Transform' I commented about in previous videos:
for ((i)^(z-1) + (-i)^(z-1)) = ((-1)^z)X(z)
T[ sin(x) ](z) = ((-1)^z)X(z) / 2*Γ(z+1)
and
T[ {integral}[e^(-x^2)]dx ](z) = ((-1)^z)X(z) / z*Γ( (z+1)/2 )
so
T[ {integral}[e^(-x^2)]dx ](z) / T[ sin(x) ](z) =
2*Γ(z+1) / z*Γ( (z+1)/2 ) =
2^(z) * Γ(z/2) / √π
This is not likely to translate into something nice without a better understanding of this 'Taylor Transform' and it's inverse. The interesting part is how much this resembles the functional equation for the Riemann Zeta function.
You doing the impossible.
Yes, I am. Cuz I am π am
(This is the best part) 😂😂😂
What we used to do is consider 2 integrals I(X) and I(y) then convert it to cylindrical coordinates using x²+y²
Check out the playlist
Well we have a function defining the last sum made by the power series, it's 2/√π ×erf(x) also known as error function
Yes
I love these integrals 😘😘😘😘
This is awesome!
@Dr.Peyam I have the question of math, but I can't solve it. Can you help me ?
Send me
@@Magic73805 P(x) : x^2+4 remainder (2x+3), P(x) : x^2+6 remainder (6x-1). If P(x) : x^4+10x^2+24 remainder s(x),then s(4)?
@@Magic73805 -109
It seems the series has a slow convergence.
Maybe
Today's lecture is very funny thanks a lot D peyam السلام عليكم
Good Instruction
Sou muito fã do seu canal, não perco um vídeo seu, parabéns
Obrigado!!! 😄
Dr. P, we can only plug in because the radius is infinity, right?
In addition, why do you separate the (-1)^n out before integrating
Yep, and it’s because (-1)^n is a constant, but we don’t really need to separate it out
@@drpeyam so I could integrate without doing that?
But it looks like it converges when we intgrate from 0 to inf
So impressive
I've done this with the Euler spiral, using 8-byte floats, and it becomes unusably inaccurate when |x| is bigger than 6. How do you compute this in the tails (the tightly winding spirals of the Euler spiral)?
use the vanilla google android calculator app... Or use GMP. Science and technology
GNU Multiple Precision Arithmetic Library
Granulocyte-macrophage progenitor
Guanosine monophosphate
This is amazing
But can we say exact number of terms to get exact decimal place? Is there a formula? I feel that I had such formula when I learned calculus 9 years ago
The solution is: sqrt(pi) * erf(x) / 2 where erf is the error function. ... What did I do? I used Mathcad.
Congrats again on 20K
Thank you!!!! 😄
Todos sabemos que esta función no tiene una primitiva, pero cómo lo sabemos?, hay una demostración?, si tengo que integrar una función, cómo sé si tiene primitiva? cómo demuestro que no la tiene?
Es muy dificil demonstrarlo, se necesita la teoria de Galois, no la entiendo
Heard the mathematicians drove crazy but i thought that it was a myth, he's smart af but dude it's scary
I don't know y but Dr Sheldon Cooper came to my mind on watching this vedio....
Legendary.
What if we just define a new function Bell(x) or something (cause the curve looks like a bell) to be this anti derivative and then that's the Taylor series for it
That feels like cheating to me.
He is always happy
Just came from a video about an old Sci-Fi short story in which one of the main characters is an all-powerful entity referred to as "AM" :) Dunno if it could integrate exp(-x^2) though.
AM as in PeyAM? 😂
@@drpeyam Well since the story is set in the future, could be. Maybe this is how it starts :-S
Haha good method I like it 😂
Does tylor series of every function has radius of convergence as infinity ?
No
@@lorenzolevy4708 give some example
@@chandankar5032 the Taylor series of ln(1+X) around 0 has radius of convergence 1
Not all taylor series have a radius at infinity. It's dependent on the distance from your point and the nearest essential singularity in the complex plane. so like the geometric series has a radius of convergence = 1 because 1/(1-x) has a singularity at z=1 -- so you can only be sure that it converges for values of |z|
No for example arctan(x) has a radius of convergence equal to 1
Evaluate: integration of log sinx dx
(Indefinite integration)
erf , Gamma, Can be expressed as series
Coooool! Really interesting and helpful video! By the way. I like your coat! I hope I can get it :D
Amazing!
why does (-x^2)^n = (-1)^n*x^(2n)?
nvm it's because (-x^2)^n = (-1*x^2)^n = (-1)^n*(x^2)^n = (-1)^n*x^(2n)
Great... 👍👍👍👍
Super ....
Its quite the same as beta gamma function way
that is with (changement de variable)
I see this video again. Really cooooool again！
Thanks....
but why is C=0?
Using the Gamma and Beta function to find the result.
Yeah, check out my playlist
Nothing is impossible. We can rather say everything is Peyam Possible (small reference to Kim Possible, the series).
pi M complete
So you can use power series even for integrals? Wow.
But that a serie of taylor this is easy?
We can't permute sum and integral like this we should verify some conditions
Well, things are smooth and decay very quickly to 0, so not a problem at all
Technically use dominated convergence on (1,infty) because exp(-x^2)
We should use theorem integral term to term to verify and I think that verify it
But that’s what I did
So this is like solution XIII
I want that jacket
So if you come in this clothes like this to exam.. what will happen? Naked? Gone? Selfie with lecturer?
Tell me I didn't just see this
Newton or Leibniz invented calculus? Most of the calculus we use today originates from Leibniz.
Yeah, there’s this debate on who invented calculus first
....or you can use (e(u))' = u' e(u)
nice
0:43 NO LEIBNIZ DID TOO AND BETTER THAN NEWTON! Lol
1/sqrt(pi) easy
Final inesperado, UwU
Hubiera sido más inesperado hacerle algo a la integral para que diera un resultado complejo.
makasihh bangett, tapi masih belm pahammmm:(
I AM = PEYAM!
3! Not 6!
wow
it's 2019, we all kind of know how to solve this..
@@ShiaServant I mean come on, there are hundreds of videos about this. This is literally the simplest thing you learn in first year of university, this is your curiosity in high school, and you find the answer in uni. But it's not even that complicated. I just hate this guy makes a pointless video about it when there are tons already.
Awesome, 187 likes, 0 dislikes
Not anymore 😢
Yes, 😢😢
چ بەشەری ئەتوو
Good forgot password
Are you indian
No
Who disliked the video? Undislike!!!!
√π
This is just a Gaussian. All you have to do is the polar sub trick.
And 12 other tricks, check out my playlist
Oh no 10 dislikes
😂😂 Yhep
Bonjing
100th dislike!!! yaay
lol, I ain’t even mad