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Empty
 one year ago
Quick little thing explaining how to get the gamma function integral @astrophysics
Empty
 one year ago
Quick little thing explaining how to get the gamma function integral @astrophysics

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Empty
 one year ago
Best ResponseYou've already chosen the best response.8Ok solve this integral, a is a constant with respect to x. $$\int_0^\infty e^{ax}dx$$

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Wait wait do this

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437992640744:dw i saw this and wanted to do it

Empty
 one year ago
Best ResponseYou've already chosen the best response.8By parts is one way to do it, but this way is much cooler

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Hahaha trust me you will be doing the same thing but cooler

Empty
 one year ago
Best ResponseYou've already chosen the best response.8all the kids are doing it

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Woah this is OPEN STUDY not PEER PRESSURE ANSWER

Empty
 one year ago
Best ResponseYou've already chosen the best response.8c'mon what's the integral of e^ax already hurrrrry

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw: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
 one year ago
Best ResponseYou've already chosen the best response.8haha yeah no problem this looks perfect!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Jk I did one earlier...dw:1437993189968:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437993463484:dw

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Boom 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8I was gonna give you a medal + fan but now I'm considering giving it to Jhan

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437993589194:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I take that back, please I need my medals and fans ok leggo

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I tried putting the a over the t JB

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1It's because I'm better than him.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8I'm gonna pull a Michele_Laino on you if you don't hurry up

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Go do your discriminants and quit discriminating against me so I can do this

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437993707615:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Differentiate both sides you say?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8and with respect to me, your sensei

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437993761110:dw I'm better

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Why take the limit when you can differentiate  his logic.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8dw:1437994060905:dw You evaluated the left side, but not the right side and now Parth is enjoying his birthday wish pssssh

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yes, my birthday wish was a free medal. <3

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0That was the right side

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437994295728:dw

Empty
 one year ago
Best ResponseYou've already chosen the best response.8$$\frac{d}{da} \int_0^\infty e^{ax}dx = \int_0^\infty \frac{\partial}{\partial a} (e^{ax})dx$$

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.0defining a is indeed a constant and not just another variable

Empty
 one year ago
Best ResponseYou've already chosen the best response.8a is not a constant, it is just an independent variable from x

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437994507765:dw oOoOoOo

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Constant with respect to x though, you could say that, also https://www.youtube.com/watch?v=Lcw6xBaCaXM

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Boom good job, ok now we have something good

Empty
 one year ago
Best ResponseYou've already chosen the best response.8\[a^{1} = \int_0^\infty e^{ax}dx\] \[a^{2} = \int_0^\infty xe^{ax}dx\] ...? Keep going do this 2 or 3 more times until you get the picture.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8differentiate with respect to a I mean haha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Integrate it, or keep taking derivatives?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8keep taking derivatives :D

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437994844847:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I think we can write a series for this...

Empty
 one year ago
Best ResponseYou've already chosen the best response.8just keep taking derivatives

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0how do partials work again? \(a^{2} = \int_0^\infty \color{red}{x}e^{ax}dx\) work again? The highlighted red portion I mean...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ok I'll take one more lol

Empty
 one year ago
Best ResponseYou've already chosen the best response.8x is a constant when we're differentiating with respect to a @Jhannybean since a and x are independent of each other nothing happens

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0I kept reading \(\frac{ \partial}{\partial x}\) instead of \(\frac{\partial}{\partial a}\) and I was confusing myself. _

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Taking 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437995573741:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437995632368:dw

Empty
 one year ago
Best ResponseYou've already chosen the best response.8\[a^{1} = \int_0^\infty e^{ax}dx\] \[a^{2} = \int_0^\infty xe^{ax}dx\] \[6a^{3} = \int_0^\infty x^2e^{ax}dx\] \[24a^{4} = \int_0^\infty x^3e^{ax}dx\] ...?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8something something gamma is the continuous version of the factorial something something whatever I already forgot

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ok I see, there's a factorial relationship, so there's a reoccurrence relationship.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Reoccurrence relationship?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8\[a^{1} = \int_0^\infty e^{ax}dx\] \[a^{2} = \int_0^\infty xe^{ax}dx\] \[6a^{3} = \int_0^\infty x^2e^{ax}dx\] \[24a^{4} = \int_0^\infty x^3e^{ax}dx\] ...? This is the pattern, what's the general form?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8No recurrence relation here either

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Whoops 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!

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Stop looking at what it should be and look at what it actually is, forget the gamma function for a minute

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Just find the truth don't worry about being right right now

Empty
 one year ago
Best ResponseYou've already chosen the best response.8idk 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I tried, that didn't work..one sec let me go over it again..6 am xD

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Pythagoras 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ n! }{ a^{(n+1)} } = \int\limits_{0}^{\infty} x^{n+1} e^{ax} dx\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I tried by plugging in integers haha

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Known: \[\frac{2!}{a^3}= \int_0^\infty x^2e^{ax}dx\] 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
 one year ago
Best ResponseYou've already chosen the best response.0x^n but the known should be a^(...) ye?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Sorry I don't know what you're saying X_X

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Oh no I was just looking at when you said \[\frac{n!}{a^{???} }= \int\limits_0^\infty ???dx\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ n! }{ a^{(n+1)} } = \int\limits_{0}^{\infty} x^n e^{ax} dx\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.8better but why are you just going with my mistake?

Empty
 one year ago
Best ResponseYou've already chosen the best response.8like negative in the denominator is like "double dividing"

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Haha, I get scared so I thought it was right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My gawd this is a long question

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Hahaha 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
 one year ago
Best ResponseYou've already chosen the best response.0It's all good at least your gettin edumacated

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437998982910:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Wow, empty, this happens every time spend so much time on one thing..then...yeah it doesn't matter xD

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0You're like a real prof

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0google your name sir

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok I thought it was him

Empty
 one year ago
Best ResponseYou've already chosen the best response.8ok so now you can derive the gamma function, you're on your own, you can think more about why this works

Empty
 one year ago
Best ResponseYou've already chosen the best response.8It should be really easy

Empty
 one year ago
Best ResponseYou've already chosen the best response.8and if it isn't, wellllll think about it until you do

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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
 one year ago
Best ResponseYou've already chosen the best response.0Obviously it is very easy 😉

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437999236907:dw

Empty
 one year ago
Best ResponseYou've already chosen the best response.8Yeah, the gamma function is basically defined as \[\Gamma(n)=(n1)!\] for positive integers but extends further.

Empty
 one year ago
Best ResponseYou've already chosen the best response.8do you see where the n1 is coming into play now?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Howd you get so many medals

Empty
 one year ago
Best ResponseYou've already chosen the best response.8it's a thumbs up not a pile of poop for the record

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Why do people always observe you teaching me