In my opinion, I prefer to just use the actual factorial notation. Some people say "that's only defined for nonnegative integers" but I disagree. The Riemann zeta function, through analytic continuation, can be defined almost everywhere. Not all of it will satisfy the initial definition of ζ(x)=sum n≥1 of 1/n^x, but we still call it ζ. ζ(-1) is well defined and equal to -1/12, even if it's not the actual sum n≥1 of n. It's the same here. The factorial function has been analytically continued to all but countably many numbers. And whilst most don't satisfy the original definition of k!, the product n=1 to k of n, it's still the factorial function. (1/2)! is well defined and equal to √π/2, even if it's not the product from n=1 to 1/2 of n.
The factorial function n! is only defined on the non-negative integers n (0,1,2,...), and technically speaking the gamma/pi function is not an analytic continuation of the factorial, but rather an interpolation of the factorial function, as such an extension to the complex plane C is not unique (analytic continuation (AC) applies only to connected domains D, which is not the case here with the non-negative integers - as such, any AC must be unique). The gamma function is what we use by convention in place of this non-unique extension via the integral definition (satisfies gamma(z+1) = zgamma(z), gamma(1) = 1) which indeed only converges for Re(z) > 0. But it is not unique: for example, the so-called pseudogamma function also successfully interpolates the factorial. Hence, denoting something such as the equivalent of gamma(3/2) as (1/2)! can be ambiguous. The AC only applies to the gamma function itself, taking the domain to Re(z) < 0 (on which it is meromorphic, poles at non-positive integers); this is unique by the identity theorem, and this time our domains Di (from Re(z)>0, continued to slices -1 < Re(z) < 0, -2 < Re(z) < 0 etc... until ultimately onto Re(z)
Hi, would't the Pi-function have a z-1 in the exponent of t, because you substituted z -> z-1 with respect of the Gamma-funciotn? Greetings from Germany :)
He did something strange like evaluating π(Z) = √π/2 results instead. I was just as confused but just accepting that the "Indefinite integral" from 0 to infinity of √te^(-t)dt evaluates to √π/2 after substitution of those limits. Hopefully the ending part is correct! 😂
Your handwriting is seriously one of the prettiest I have ever seen. just one minor detail, in the end result, the 27 in the denominator should be +/- 27. Thanks a lot for these videos, amazing quality. As a mechanical engineer, I miss sometimes these math lessons.
Interesting how pi always shows up in integrals 🤔
I think you make a mistake when you put Gamma(-1/2+1) at the end of video. You might say Gamma(1/2+1)
Oops 🙊 you're right.
Please include your logarithms tutorial, for example log(log(logx))
In my opinion, I prefer to just use the actual factorial notation. Some people say "that's only defined for nonnegative integers" but I disagree.
The Riemann zeta function, through analytic continuation, can be defined almost everywhere. Not all of it will satisfy the initial definition of ζ(x)=sum n≥1 of 1/n^x, but we still call it ζ. ζ(-1) is well defined and equal to -1/12, even if it's not the actual sum n≥1 of n.
It's the same here. The factorial function has been analytically continued to all but countably many numbers. And whilst most don't satisfy the original definition of k!, the product n=1 to k of n, it's still the factorial function. (1/2)! is well defined and equal to √π/2, even if it's not the product from n=1 to 1/2 of n.
The factorial function n! is only defined on the non-negative integers n (0,1,2,...), and technically speaking the gamma/pi function is not an analytic continuation of the factorial, but rather an interpolation of the factorial function, as such an extension to the complex plane C is not unique (analytic continuation (AC) applies only to connected domains D, which is not the case here with the non-negative integers - as such, any AC must be unique).
The gamma function is what we use by convention in place of this non-unique extension via the integral definition (satisfies gamma(z+1) = zgamma(z), gamma(1) = 1) which indeed only converges for Re(z) > 0. But it is not unique: for example, the so-called pseudogamma function also successfully interpolates the factorial. Hence, denoting something such as the equivalent of gamma(3/2) as (1/2)! can be ambiguous.
The AC only applies to the gamma function itself, taking the domain to Re(z) < 0 (on which it is meromorphic, poles at non-positive integers); this is unique by the identity theorem, and this time our domains Di (from Re(z)>0, continued to slices -1 < Re(z) < 0, -2 < Re(z) < 0 etc... until ultimately onto Re(z)
Errata - it is gamma(3/2) = 1/2 * gamma(1/2)= root pi/2 .... Not gamma (-half+1)
Hi,
would't the Pi-function have a z-1 in the exponent of t, because you substituted z -> z-1 with respect of the Gamma-funciotn?
Greetings from Germany :)
He did something strange like evaluating π(Z) = √π/2 results instead. I was just as confused but just accepting that the "Indefinite integral" from 0 to infinity of √te^(-t)dt evaluates to √π/2 after substitution of those limits. Hopefully the ending part is correct! 😂
Even from the answer we can tell that the error function is involved in the indefinite integral answer to this.
Please Sir, you forgot to factor out the negative in your multiplcation
I used it to flip the boundaries
Nice teacher
ln(1/x) = -ln(x) so it could be simplier ...
Plus -exp(-u)*du = dx 'cause exp(-u) = x so we can avoid it at all.
Hi, I think I have a fun task: Find all solutions for sin(1/x)=0 in the interval ]0;1]
😊
Excelente!
Γ(z)= ∫₀ ᪲ xᶻ⁻¹ e⁻ˣ dx
Excellent explanation Sir. Thanks 🙏🙏🙏🙏🙏
Can we use Gaussian Integral
Smart change in variable
How groovy is that?!
WOW! Great STUFF! 😊
This was a fun one!
Thanks!
Thanks a lot
I like so much.
Simply Great!!!!!!!!!!!!!!!!
Very interesting
1000th view!
Your handwriting is seriously one of the prettiest I have ever seen.
just one minor detail, in the end result, the 27 in the denominator should be +/- 27.
Thanks a lot for these videos, amazing quality. As a mechanical engineer, I miss sometimes these math lessons.
Should it? He wasnt solving for x and the square root always gives out the absolute value
So nice to see an intelligent black man talking high-level math as opposed to hip-hop. My hat is off to you. sir..
Dafuq?
Slightly racist remark.