RE: First collisions at the LHC with unprecedented Energy! (Ask a particle physisicist)
September 2, 2015 at 10:21 am
(This post was last modified: September 2, 2015 at 10:27 am by Alex K.)
@poca
You mean x^n * exp(-x), surely, otherwise it's not finite.
Look up the Gamma function which is directly related to n!, the above is more or less its definition.
But I'll tell you a secret (the most useful trick in the book when working with moments of exponentials): such identities and similar ones can for example be proven by introducing an auxiliary variable in the exponent:
exp(-a x)
and realizing that n times the derivative with respect to a and then setting a=1 gives you (-1)^n x^n exp(-x).
You can therefore simply calculate the integral
int_0^infty exp(-a x)
and *then* perform the derivatives. All you have to take care of is to put the appropriate minus sign to compensate for the (-1)^n.
But it turns out that simply,
int_0^infty exp(-a x)= a^-1
Taking the derivatives of that with respect to "a" the required number of times automatically gives you the combinatorial factor n! from the rule for calculating derivatives of monomials, and the alternating minus sign which cancels the extra (-1)^n above.
You mean x^n * exp(-x), surely, otherwise it's not finite.
Look up the Gamma function which is directly related to n!, the above is more or less its definition.
But I'll tell you a secret (the most useful trick in the book when working with moments of exponentials): such identities and similar ones can for example be proven by introducing an auxiliary variable in the exponent:
exp(-a x)
and realizing that n times the derivative with respect to a and then setting a=1 gives you (-1)^n x^n exp(-x).
You can therefore simply calculate the integral
int_0^infty exp(-a x)
and *then* perform the derivatives. All you have to take care of is to put the appropriate minus sign to compensate for the (-1)^n.
But it turns out that simply,
int_0^infty exp(-a x)= a^-1
Taking the derivatives of that with respect to "a" the required number of times automatically gives you the combinatorial factor n! from the rule for calculating derivatives of monomials, and the alternating minus sign which cancels the extra (-1)^n above.
The fool hath said in his heart, There is a God. They are corrupt, they have done abominable works, there is none that doeth good.
Psalm 14, KJV revised edition