Empty
  • Empty
Quick little thing explaining how to get the gamma function integral @astrophysics
Mathematics
chestercat
  • chestercat
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Empty
  • Empty
Ok solve this integral, a is a constant with respect to x. $$\int_0^\infty e^{-ax}dx$$
Astrophysics
  • Astrophysics
Wait wait do this
Astrophysics
  • Astrophysics
|dw:1437992640744:dw| i saw this and wanted to do it

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Empty
  • Empty
Yeah I know, trust me
Empty
  • Empty
lol
Astrophysics
  • Astrophysics
by parts
Astrophysics
  • Astrophysics
ok
Astrophysics
  • Astrophysics
ima do it then
Empty
  • Empty
By parts is one way to do it, but this way is much cooler
Empty
  • Empty
Hahaha trust me you will be doing the same thing but cooler
Empty
  • Empty
don't you wanna be cool
Empty
  • Empty
all the kids are doing it
Empty
  • Empty
it's just one integral
Empty
  • Empty
one can't hurt
Empty
  • Empty
Woah this is OPEN STUDY not PEER PRESSURE ANSWER
Empty
  • Empty
c'mon what's the integral of e^-ax already hurrrrry
Empty
  • Empty
ty
Astrophysics
  • Astrophysics
|dw:1437993094157:dw| I forgot to take the limit the first time ok, relax it's been a while since I've done an improper integral
Empty
  • Empty
haha yeah no problem this looks perfect!
Astrophysics
  • Astrophysics
Jk I did one earlier...|dw:1437993189968:dw|
Astrophysics
  • Astrophysics
|dw:1437993463484:dw|
Empty
  • Empty
Boom ok perfect now we have this nice looking thing: $$a^{-1} = \int_0^\infty e^{-ax}dx$$ This is where the magic happens, this is the cool way. We differentiate both sides WITH RESPECT TO A. This is weird if you're not used to it, so it'll be good practice. If you don't make a mistake doing this, then you will be officially better than me cause I messed this up the first time I saw it lol.
Astrophysics
  • Astrophysics
I am better
Empty
  • Empty
lolol woahhhh champ
Empty
  • Empty
I was gonna give you a medal + fan but now I'm considering giving it to Jhan
ParthKohli
  • ParthKohli
JUST DO IT.
Jhannybean
  • Jhannybean
|dw:1437993589194:dw|
Astrophysics
  • Astrophysics
I take that back, please I need my medals and fans ok leggo
Empty
  • Empty
lolol
Astrophysics
  • Astrophysics
I tried putting the a over the t JB
Empty
  • Empty
Parth gets the medal
Empty
  • Empty
sry
Empty
  • Empty
I haven't got all night
ParthKohli
  • ParthKohli
It's because I'm better than him.
Empty
  • Empty
I'm gonna pull a Michele_Laino on you if you don't hurry up
Astrophysics
  • Astrophysics
Go do your discriminants and quit discriminating against me so I can do this
Astrophysics
  • Astrophysics
|dw:1437993707615:dw|
Astrophysics
  • Astrophysics
Differentiate both sides you say?
Empty
  • Empty
yeah with respect to a
Empty
  • Empty
and with respect to me, your sensei
Astrophysics
  • Astrophysics
|dw:1437993761110:dw| I'm better
ParthKohli
  • ParthKohli
FLAWLESS VICTORY
Jhannybean
  • Jhannybean
Why take the limit when you can differentiate - his logic.
Empty
  • Empty
|dw:1437994060905:dw| You evaluated the left side, but not the right side and now Parth is enjoying his birthday wish pssssh
ParthKohli
  • ParthKohli
Yes, my birthday wish was a free medal. <3
Astrophysics
  • Astrophysics
That was the right side
Astrophysics
  • Astrophysics
|dw:1437994295728:dw|
Empty
  • Empty
$$\frac{d}{da} \int_0^\infty e^{-ax}dx = \int_0^\infty \frac{\partial}{\partial a} (e^{-ax})dx$$
Astrophysics
  • Astrophysics
Yeah ok I see what you're doing, you want me to do use leibniz rule pretty much and not just plug and chug haha
Jhannybean
  • Jhannybean
defining a is indeed a constant and not just another variable
Empty
  • Empty
a is not a constant, it is just an independent variable from x
Astrophysics
  • Astrophysics
|dw:1437994507765:dw| oOoOoOo
Empty
  • Empty
Constant with respect to x though, you could say that, also https://www.youtube.com/watch?v=Lcw6xBaCaXM
Empty
  • Empty
Boom good job, ok now we have something good
Jhannybean
  • Jhannybean
Ah, I see.
Empty
  • Empty
\[a^{-1} = \int_0^\infty e^{-ax}dx\] \[a^{-2} = \int_0^\infty xe^{-ax}dx\] ...? Keep going do this 2 or 3 more times until you get the picture.
Empty
  • Empty
differentiate with respect to a I mean haha
Astrophysics
  • Astrophysics
Integrate it, or keep taking derivatives?
Empty
  • Empty
keep taking derivatives :D
Astrophysics
  • Astrophysics
K good lol
Astrophysics
  • Astrophysics
|dw:1437994844847:dw|
Empty
  • Empty
noice
Astrophysics
  • Astrophysics
I think we can write a series for this...
Empty
  • Empty
no don't!
Empty
  • Empty
just keep taking derivatives
Jhannybean
  • Jhannybean
how do partials work again? \(a^{-2} = \int_0^\infty \color{red}{x}e^{-ax}dx\) work again? The highlighted red portion I mean...
Astrophysics
  • Astrophysics
Ok I'll take one more lol
Empty
  • Empty
x is a constant when we're differentiating with respect to a @Jhannybean since a and x are independent of each other nothing happens
Jhannybean
  • Jhannybean
Oh that's right,
Jhannybean
  • Jhannybean
I kept reading \(\frac{ \partial}{\partial x}\) instead of \(\frac{\partial}{\partial a}\) and I was confusing myself. -_-
Astrophysics
  • Astrophysics
Taking derivatives in integrals is amazing, I saw this technique once before, did not know you would treat it like a constant in the integral, awesome. |dw:1437995191725:dw|
Astrophysics
  • Astrophysics
Oki, what's next
Empty
  • Empty
Do one more!
Astrophysics
  • Astrophysics
|dw:1437995573741:dw|
Astrophysics
  • Astrophysics
|dw:1437995632368:dw|
Empty
  • Empty
\[a^{-1} = \int_0^\infty e^{-ax}dx\] \[a^{-2} = \int_0^\infty xe^{-ax}dx\] \[6a^{-3} = \int_0^\infty x^2e^{-ax}dx\] \[24a^{-4} = \int_0^\infty x^3e^{-ax}dx\] ...?
Empty
  • Empty
What are we doing again
Empty
  • Empty
something something gamma is the continuous version of the factorial something something whatever I already forgot
Astrophysics
  • Astrophysics
Ok I see, there's a factorial relationship, so there's a reoccurrence relationship.
Empty
  • Empty
Reoccurrence relationship?
Empty
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\[a^{-1} = \int_0^\infty e^{-ax}dx\] \[a^{-2} = \int_0^\infty xe^{-ax}dx\] \[6a^{-3} = \int_0^\infty x^2e^{-ax}dx\] \[24a^{-4} = \int_0^\infty x^3e^{-ax}dx\] ...? This is the pattern, what's the general form?
Astrophysics
  • Astrophysics
Haha recurrence*
Empty
  • Empty
No recurrence relation here either
Empty
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Whoops I had been writing it wrong lol oh well don't look at what I write you wrote it yourself correctly anyways \[a^{-1} = \int_0^\infty e^{-ax}dx\] \[a^{-2} = \int_0^\infty xe^{-ax}dx\] \[2a^{-3} = \int_0^\infty x^2e^{-ax}dx\] \[6a^{-4} = \int_0^\infty x^3e^{-ax}dx\] \[24a^{-5} = \int_0^\infty x^4e^{-ax}dx\] ... \[\frac{n!}{a^{-???} }= \int_0^\infty ???dx\] Here's my hint, what the hell is this last thing!
Astrophysics
  • Astrophysics
(n-1)!
Empty
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Stop looking at what it should be and look at what it actually is, forget the gamma function for a minute
Empty
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Just find the truth don't worry about being right right now
Empty
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idk check it: Here's what you KNOW \[2a^{-3} = \int_0^\infty x^2e^{-ax}dx\] try out your false formula to find out how to fix it
Astrophysics
  • Astrophysics
Yeah I tried, that didn't work..one sec let me go over it again..6 am xD
Empty
  • Empty
Pythagoras didn't have anyone to check his work I think on the whole a^2+b^2 thing lol so you're at least fortunate :P
Astrophysics
  • Astrophysics
\[\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^{n+1} e^{-ax} dx\]
Astrophysics
  • Astrophysics
I tried by plugging in integers haha
Empty
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Known: \[\frac{2!}{a^3}= \int_0^\infty x^2e^{-ax}dx\] Yours: \[\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^{n+1} e^{-ax} dx\] Plugged in n=2 \[\frac{ 2! }{ a^{-3} } = \int\limits_{0}^{\infty} x^{3} e^{-ax} dx\] Same or not?
Astrophysics
  • Astrophysics
x^n but the known should be a^-(...) ye?
Empty
  • Empty
???
Empty
  • Empty
Sorry I don't know what you're saying X_X
Astrophysics
  • Astrophysics
Oh no I was just looking at when you said \[\frac{n!}{a^{-???} }= \int\limits_0^\infty ???dx\]
Astrophysics
  • Astrophysics
\[\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^n e^{-ax} dx\]
Empty
  • Empty
better but why are you just going with my mistake?
Empty
  • Empty
like negative in the denominator is like "double dividing"
Astrophysics
  • Astrophysics
Haha, I get scared so I thought it was right
Astrophysics
  • Astrophysics
That works out
Astrophysics
  • Astrophysics
Tried n = 2
anonymous
  • anonymous
My gawd this is a long question
Empty
  • Empty
Hahaha ok that's what I like to hear, everything leading to this point doesn't really matter now, it's like the ladder we can kick out from under us and if you like you can pretend we were never wrong on the path to this. @isaac4321 yeah we should probably have flushed halfway through for courtesy :(
anonymous
  • anonymous
It's all good at least your gettin edumacated
Empty
  • Empty
tru
Astrophysics
  • Astrophysics
|dw:1437998982910:dw|
anonymous
  • anonymous
Um ok
Astrophysics
  • Astrophysics
Wow, empty, this happens every time spend so much time on one thing..then...yeah it doesn't matter xD
Astrophysics
  • Astrophysics
You're like a real prof
anonymous
  • anonymous
Who is that
Astrophysics
  • Astrophysics
google your name sir
anonymous
  • anonymous
Ok I thought it was him
Empty
  • Empty
ok so now you can derive the gamma function, you're on your own, you can think more about why this works
anonymous
  • anonymous
👍
Empty
  • Empty
It should be really easy
Empty
  • Empty
and if it isn't, wellllll think about it until you do
Astrophysics
  • Astrophysics
Ok so I don't really know what exactly the gamma function is, I mean is it actually the thing I posted above, I thought that was just a problem that was a gamma function haha..well that's cool, learned so much
anonymous
  • anonymous
Obviously it is very easy 😉
Astrophysics
  • Astrophysics
|dw:1437999236907:dw|
Empty
  • Empty
Yeah, the gamma function is basically defined as \[\Gamma(n)=(n-1)!\] for positive integers but extends further.
Empty
  • Empty
do you see where the n-1 is coming into play now?
Astrophysics
  • Astrophysics
Yeah I think so
Empty
  • Empty
ok
Empty
  • Empty
|dw:1437999792138:dw|
Astrophysics
  • Astrophysics
lmao
Astrophysics
  • Astrophysics
Howd you get so many medals
Empty
  • Empty
it's a thumbs up not a pile of poop for the record
Astrophysics
  • Astrophysics
Why do people always observe you teaching me
Empty
  • Empty
they're creepers
Empty
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creepin'
Empty
  • Empty
idk because this site is boring normally lo
Astrophysics
  • Astrophysics
So I don't really see where exactly to use this function, like would you use it in QM, looks like it may have to do with statistics/ probability.
Empty
  • Empty
Yeah I have used it in QM which is a statistical theory, so yes to both. You can use it anywhere you would use a factorial.
Astrophysics
  • Astrophysics
There's also the beta functions...
Empty
  • Empty
Yeah beta functions are just like the continuous binomial coefficients
Astrophysics
  • Astrophysics
Well, did a lot of math today, I didn't think this would take this long, was about to go to bed before the post came up xD
Empty
  • Empty
Yeah if you want some exercises go find the pdf online of Mary Boas book Mathematical Methods in the Sciences, there's a whole section on gamma functions and beta functions, fun exercises really nothing too hard that you could work through to get some basic exposure to finding stuff that fits the form for integrating. Consider them new closed form solutions to some integrals that you can get to from doing u-substitution and stuff
Astrophysics
  • Astrophysics
Sounds like fun, thanks, I think we should also start working on some relativity/ QM that would be fun, I'm not so good at finding general forms and such, unless I plug integers in xD, and understanding the theory without it being of use really...
Astrophysics
  • Astrophysics
as far as this math goes
Astrophysics
  • Astrophysics
I should say I'm not used to it*
Empty
  • Empty
yeah actually the QM book we can start going through, the guy reccomends Boas' book in the beginning. It's called Griffith's Intro to QM.
Astrophysics
  • Astrophysics
The one with the cat
Astrophysics
  • Astrophysics
Schrodingers cat :(
Empty
  • Empty
Yeah that one
Empty
  • Empty
lol
Astrophysics
  • Astrophysics
I like how it starts with schro eq
Empty
  • Empty
Yeah it's basically like starting a classical physics class with F=ma
Astrophysics
  • Astrophysics
Ah so this means avg < > you know what I mean xD
Empty
  • Empty
Yeah, expected value means average, $$\langle x \rangle$$
Empty
  • Empty
Standard deviation of some value is \(\sigma_x\) and satisfies this nice equation (which is very nicely shown and explained I thought in his book) \[\sigma_x^2 = \langle x^2 \rangle - \langle x \rangle^2\]
Astrophysics
  • Astrophysics
Ok I'll go through this book, it seems pretty straight forward...but lets see how long that lasts haha.
Empty
  • Empty
well when it stops that's why we're reading together lol
Astrophysics
  • Astrophysics
Sounds good xD
Astrophysics
  • Astrophysics
Ok I'm going to sleep, later and thanks!!
Empty
  • Empty
laters

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