Math Newsletter number 16; Wednesday, November 10, 2010.
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In calculus, the first derivation of x**n is n*x**(n-1).
Applying the same rule: Multiply the coefficient by the
exponent, and reduce the exponent by 1,
gives second derivative of
n*(n-1) * x**(n-2).
Third derivative is n*(n-1)*(n-2) * x**(n-3).
How would we write the kth derivative for some positive
kth derivative is:
n*(n-1)*(n-2)*...(n-k+1) * x**(n-k)
We can write this more compactly as:
kth derivative is:
(n! / (n-k)! ) * x**(n-k).
Now that we have this compact form, we can consider
what it would mean if k is not an integer.
Following this form, we can say that the (1/2) derivative
is (n!/(n-1/2)! ) * x**(n-1/2)
and the (1/2) derivative of the (1/2) derivative is:
(n!/(n-1/2)!) * ( (n-1/2)! / (n-1)! ) * x**(n-1/2 - 1/2)
= (n! / (n-1)! ) * x**(n-1)
which is consistent with the first derivative.
Is it possible to assign a meaning to (n-1/2)! ?
Yes. The gamma function extends the factorial
function to non-integers.
For integer values of k, gamma(k+1) = k!.
kth derivative of x**n is:
(n! / (n-k)! ) * x**(n-k)
= (gamma(n+1)/gamma(n-k+1) ) * x**(n-k)
(1/2) derivative of x**n is:
(gamma(n+1)/gamma(n+1/2)) * x**(n-1/2).
It happens that gamma(n+1/2) can be calculated exactly.
Note that gamma(n+1) = n! = n*(n-1)*(n-2)*...2*1
(n+1) gamma(n+1) = (n+1)*n! = (n+1)!
(n+1) gamma(n+1) = gamma(n+2)
In general, for any number z, for which gamma(z) is defined,
z*gamma(z) = gamma(z+1).
Thus gamma(1/2 + 1) = (1/2) gamma(1/2)
gamma(3/2) = (1/2) gamma(1/2)
gamma(5/2) = (3/2)(1/2) gamma(1/2)
gamma(7/2) = (7/2)(3/2)(1/2) gamma (1/2)
Thus we can calculate gamma(n+1/2), for any not too large
integer n, if we can calculate gamma(1/2).
It is, in fact easy to calculate gamma(1/2).
gamma(1/2) = sqrt(pi).