Oh man, I had no idea this was going to turn out like that. Fubini's theorem came in way more naturally than I'd anticipated. Well-done, Kamaal! I'd love to see more stuff like this. Historically, I've told my Calc II students that infinite series are a completely different beast from improper integrals, but have never been able to give great examples beyond the Basel problem. On the flip side, since I haven't worked with exact values of non-power series I am not familiar with many techniques beyond Parseval's theorem for computing actual values. Even that method I avoid using because it's still not my field.
First, we convert the series into a function f(x)=Σn!/(2n+1)!x^(2n+1) Our goal is to find the function. We can use a differential equation to do this f'(x)=Σn!/(2n)!x^(2n) f''(x)=Σn!/(2n-1)x^(2n-1) After performing simple calculations, we arrive at the following differential equation 2y''-xy'-y=0 It is a linear ordinary differential equation with non-constant coefficients d/dx(2y'-xy)=0 which can be easily solved 2y'-xy=a I think everyone can solve this differential equation y=e^(x²/4)(a/2*√π*erf(x/2)+b) Well, we can easily determine the unknown coefficients a=2 and b=0 y=f(x)=√π*e^(x²/4)(erf(x/2)) Σn!/(2n+1)!x^(2n+1)=√π*e^(x²/4)(erf(x/2)) as a result Σn!/(2n+1)!=√(π√e)*erf(1/2)
ah yes, the gamma and error functions in their natural habitat.
Man I swear you gotta make a video deriving all those formulas of the Beta function.
This isn’t factorial based, this is just based
it isn't just based, it's based!
But it isn't just based, it's based!
You brought the error down to earth.
Dear Kamal, I deeply appreciate the ERFfort you put into the evaluation of this sum!
tbh the hardest part of this problem was completing the square 😭😭
not that hard, using normal distribution is easy in this case
@@ayushrudra8600 you think that's hard? try completing the circle 😶
@@GeraldPreston1 lol
if you have ax²+bx+c, then it can be rewritten as a(x+(-b/2a))²+smth else
the smth else can be easily figured out but I'm too lazy to do it
Hi,
"Terribly sorry about that" : 3:02 , 3:38 , 6:01 .
Love it😂
There's nothing we can do
Why did the x in xᵏ decompose?
Because it was put "to decay" (to the k).
@@GeoPeron 😭😭😭
Have this like and get out
error function 🤮👎👎👎
why does the thumbnail have index n but video has index k? clickbait?!?!
Yes
Fr!!! Scam 2024
Oh man, I had no idea this was going to turn out like that. Fubini's theorem came in way more naturally than I'd anticipated. Well-done, Kamaal!
I'd love to see more stuff like this. Historically, I've told my Calc II students that infinite series are a completely different beast from improper integrals, but have never been able to give great examples beyond the Basel problem. On the flip side, since I haven't worked with exact values of non-power series I am not familiar with many techniques beyond Parseval's theorem for computing actual values. Even that method I avoid using because it's still not my field.
I'm sorry, but how is x^k(1-x^k) = (x(1-x))^k ?
Notice that also
It's a mistake in the video. The correct equation is x^k*(1-x)^k (see min 2:47) and it is obviously equal to (x(1-x))^k.
He's terribly sorry about that
Indeed I am
I stick to this channel because the math is so amazing.
finally the erf is coming home
If only the English could say something about like that about a mojor trophy 😭
Very interesting infinite series. Thank you for you innovative solution.
sqrt(pisqrt(e)) is quite the mouthful 😂 great video
First, we convert the series into a function
f(x)=Σn!/(2n+1)!x^(2n+1)
Our goal is to find the function. We can use a differential equation to do this
f'(x)=Σn!/(2n)!x^(2n)
f''(x)=Σn!/(2n-1)x^(2n-1)
After performing simple calculations, we arrive at the following differential equation
2y''-xy'-y=0
It is a linear ordinary differential equation with non-constant coefficients
d/dx(2y'-xy)=0
which can be easily solved
2y'-xy=a
I think everyone can solve this differential equation
y=e^(x²/4)(a/2*√π*erf(x/2)+b)
Well, we can easily determine the unknown coefficients
a=2 and b=0
y=f(x)=√π*e^(x²/4)(erf(x/2))
Σn!/(2n+1)!x^(2n+1)=√π*e^(x²/4)(erf(x/2))
as a result
Σn!/(2n+1)!=√(π√e)*erf(1/2)
Love ur vids man ❤
At 3:16 did I miss something? Isn't the integrand supposed to be x^k*(1-x)^k ... not x^k*(1-x^k)?
Whoops ... saw this many times in previous comments 🙂
But at 3:30 a reverse mistake which makes all well 🙂
Nice result❤
sums are just budget integrals
nice vid
3:30 is that incorrect? Should be (x^k)*(1-x^k) = x^k - x^(2k); not (x-x^2)^k
Fr bro im looking at it like am I retarded or whats this
He mistakenly put power on just k while it should have been on whole (1-x)
there is a mistake in the previous step while writing k in the power , remaining all steps are correct
You guys are right, I missed the previous mistake/typo.
Genial
❤
This wasnt the hardest one its all basic properties of integral and gamma functions😊😊
It's 1.18459307294.....