Math Newsletter number 16; Wednesday, November 10, 2010.

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Fractional Derivatives

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

integer, k?

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.

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

http://www.sosmath.com/calculus/improper/gamma/gamma.html

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).