Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

whatz gamma function in easy language

OpenStudy (anonymous):

KYAH PATAH

OpenStudy (experimentx):

a kind of special function ...

OpenStudy (anonymous):

check http://mathworld.wolfram.com/GammaFunction.html

OpenStudy (anonymous):

PLEASE EXPLAIN experimentx

OpenStudy (experimentx):

first of all, you should know what is a function ... then you should know what are elementary functions, and for what values integrand you get output from integration as elementary functions. for example:- you know \( \int e^x dx = e^x +c\) and what is the value of \( \int e^{-x^2} dx =?\)

OpenStudy (experimentx):

the thing is there is no nice form (in elementary functions) for the this integral. but these types of problem occur frequently in many problems of this (or similar) type \[ \int_{-\infty}^\infty e^{-x^2}dx\] but this has nice value define a function like this \[ \int t^{n - 1 } e^{-t}dt\] this function has no elementary form. but this type of function \[ \int_0^{\infty } t^{n - 1 } e^{-t}dt\] has very nice answer for appropriate values of n. since this type of integral frequently occurs in probalems. let's give is a name Gamma and call it Gamma function. and i takes argument 'n'

OpenStudy (anonymous):

WHAT IS RELATION BTWEEN FACTORIAL AND GAMMA FUNCTION???????????PLZ. put it in easy way?????????

OpenStudy (experimentx):

\[ \Gamma (n+1) = n! = \int_0^\infty t^{n + 1 - 1}e^{-t}dt\] using this definition, you can calculate (0.5)!

OpenStudy (anonymous):

any physical significance of (1/2)! ????????????????

OpenStudy (experimentx):

i don't know ...

OpenStudy (amistre64):

if we run a by parts integration using a polynomial as u, we eventually run it down to 0 \[+x^n\\-nx^{n-1}\\+n(n-1)x^{n-2}\\...\\\pm n!x\\\pm n!\\0\] using e^-x as the v parts, they retain their function, but swap out signs to produce an all subtraction result \[~~~~~~~~~~~e^{-x}\\+x^n~~-e^{-x}\\-nx^{n-1}~~e^{-x}\\+n(n-1)x^{n-2}~~-e^{-x}\\...\\\pm n!x~~\mp e^{-x}\\\pm n!~~\mp e^{-x}\\0~~\mp e^{-x}\] taking this from 0 to inf; at e^-inf it all zeros out; this leaves us with the -x=0 parts, which zeroes it all out except for the n! constant next to the last part.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!