Question

How do you take factorials of a fraction?

Answer 1

\[\left( \frac{ 1 }{ 2 } \right)!, \left( \frac{ 3 }{ 2 } \right)!\] etc @ganeshie8
Trying to understand this for gamma functions haha

Answer 2

\(\color{#000000 }{ \displaystyle \Gamma (x)=\int\limits_{0}^{\infty}z^{x-1}e^{-z}~dz }\)

If I recall correctly, and then you can express the integral as a power series, and use partial sums to evaluate.

Answer 3

Do you know Euler reflection formula ?

Answer 4

\[\Gamma(1-z)\Gamma(z) = \dfrac{\pi}{\sin(\pi z)}\] for \(0\lt z\lt 1\)

Answer 5

plugin \(z=\frac{1}{2}\)

Answer 6

That's right, but I'm not sure how to take factorials of fractions, that's the part I'm not entirely sure about. \[\Gamma (n) = (n-1)!\]

Answer 7

Ooh interesting

Answer 8

correct me if i am wrong, but isn't \(\Gamma(n)=(n-1)!\) ?

Answer 9

right, euler reflection formula and the definition of factorial as recurrence relation allow you to compute the value of factorual of any fraction

Answer 10

I'm still not sure how that helps

Answer 11

and \[\Gamma(\frac{1}{2})\] is something goofy like \(\sqrt{\pi}\)

Answer 12

that is about all i remember

Answer 13

\[\Gamma (3/2) = (3/2-1)! = 1/2! = ?\]

Answer 14

http://www.wolframalpha.com/input/?i=%281%2F2%29%21 how

Answer 15

I mean is there a way to do that without integrating

Answer 16

recurrence relation gives you

\[\Gamma(x+1) = x\Gamma(x) \]

Answer 17

i guess we could try to compute the integral \[\int_0^{\infty}x^{-\frac{1}{2}}e^{-x}dx\]

Answer 18

without doing it my guess is parts, but i could be way off

Answer 19

\[\Gamma(\frac{3}{2}) = \Gamma(\frac{1}{2}+1) =\frac{1}{2}\Gamma(\frac{1}{2}) \]

Answer 20

You can find the value of \(\Gamma(\frac{1}{2})\) using euler reflection formula easily

Answer 21

\[\Gamma(1-z)\Gamma(z) = \dfrac{\pi}{\sin(\pi z)}\]

plugin \(z=\frac{1}{2}\)

Answer 22

ok then what?

Answer 23

oh nvm

Answer 24

has a fit of stupidity there, yes, that works nicely for \(\frac{1}{2}\)

Answer 25

\[(\frac{1}{2})! =(\frac{3}{2}-1)! = \Gamma(\frac{3}{2}) = \Gamma(\frac{1}{2}+1) =\frac{1}{2}\Gamma(\frac{1}{2}) \]

Answer 26

With the integral I guess we can use \[\int\limits_{0}^{\infty} x^n e^{-ax} dx = \frac{ n! }{ a^{n+1} }\]

Answer 27

https://math.stackexchange.com/questions/396889/how-to-find-the-factorial-of-a-fraction

Answer 28

got that link through bibby

Answer 29

yeah but that is just begging the question because your numerator still has a factorial in it

Answer 30

Ooh ok I see what you did ganeshie

Answer 31

Thanks :P

Answer 32

Above works only when you can reduce a fraction down to \(\frac{1}{2}\) using the recurrence relation

Answer 33

Yup I see, \[\Gamma (1/2) = \sqrt{\pi}\] is the only one I see in my formulae sheet, so is that the only one defined?

Answer 34

For example, you can't find \(\Gamma(1/3)\) using just the above two goodies

Answer 35

Yeah you can find the gamma value of all half integers using above recurrence relation

Answer 36

Ah gotcha, good stuff! Thank you

Answer 37

See if you can you find a formula for \(\Gamma(\frac{n}{2})\)

Answer 38

\(n\in \mathbb{Z^+}\)

Answer 39

That seems tricky, I'll try

Answer 40

\[\Gamma(\frac{n}{2}) =(\frac{n}{2}-1)\Gamma(\frac{n}{2}-1) \]

Answer 41

Yeah haha, I was just going to ask you that, if I could start with \[\Gamma(\frac{ n }{ 2 }) =(\frac{ n }{ 2 }-1)!\] sort of deal

Answer 42

We use the symbol \(!\) only for integers

use \(\Gamma\) for reals

Answer 43

Ah ok, I've never used gamma functions before I don't think until QM

Answer 44

I actually needed to know how to use this for here haha

Answer 45

Let \(ax^2 = u \\\implies 2axdx = du \\\implies dx =\dfrac{du}{2a \sqrt{u/a}} \)

Answer 46

\[\color{#000000 }{ \displaystyle \Gamma (x)=\int\limits\limits_{0}^{\infty}z^{x-1}e^{-z}~dz } \]

Answer 47

Nice!

Answer 48

the integral then becomes
\[\int\limits_0^{\infty} (\sqrt{u/a})^me^{-u} \dfrac{du}{2a\sqrt{u/a}}\\~\\=\frac{1}{2a^{(m+1)/2}}\int\limits_0^{\infty} u^{(m-1)/2}e^{-u}du\\~\\
=\frac{1}{2a^{(m+1)/2}}\int\limits_0^{\infty} u^{\color{red}{(m+1)/2}-1}e^{-u}du\\~\\
\]

Compare that integeral with the gamma integral

Answer 49

Haha, now I know how they got that formula, that seems super simple..in hindsight

Answer 50

are you also working with beta function ?

Answer 51

there is a nice relation between gamma and beta functions

Answer 52

\[\beta(a,b) = \dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\]

Answer 53

Nope, we're really not even taught these gamma functions, but rather given a formula sheet with bunch of integrals on them...that's physics for you! Formulas, no need to prove

Answer 54

That is not physics, that is more like engineering.

Physicists do care for proofs. Only dumb engineers look for formulas w/o proofs ;)

Answer 55

Haha, well yeah I guess I should've clarified, it's not that there is no need to prove, I suppose they assume you already know the proof at least in quantum mechanics!

Answer 56

It's weird...like these guassian integrals, it's so different from classical mechanics, all about probability

Answer 57

These special functions are not so hard. They are defined using definite integrals and if you're in QM, the definition is all you care for, I guess..

Answer 58

I still like the proofs though, but these are handy only because otherwise each problem will take like 10 years to do without them

Answer 59

Like check this integral out, what in the world

Answer 60

I don't like to mindlessly use these integrals, but sometimes I think it's pretty important haha.

Answer 61

That integral is easy if you're familiar with the beta function

Answer 62

But I feel that you wont learn anything by evaluating that integral on your own. It's just algebra and you will get tired for sure. Better to use the result directly :)