Question

Show that d^l-m/dx^l-m (x²-1)^l=(l-m)!/(l+m)! (x²-1)^m d^l+m/dx^l+m (x²-1)^l.

Hint: Write(x²-1)^l=(x-1)^l(x+1)^land find the derivatives by Leibniz' rule.

Short answer

The derivative d^l-m/dx^l-m (x²-1)^l=(l-m)!/(l+m)! (x²-1)^m d^l+m/dx^l+m (x²-1)^l. is proved.

Step-by-step solution

Concept of Leibnitz rule:

Consider the Leibnitz equation:

d^l+m/dx^l+m (x²-1)^l = d^l+m/dx^l+m ((x-1)^l (x+1)^l)

d^l+m/dx^l+m (x²-1)^l = ∑_r=0^l+m (l+m)!/ r! (l+m-r)! d^r/dx^r [(x-1)^l d^l+m-r/dx^l+m-r (x+1)^l]

d^l+m/dx^l+m (x²-1)^l = ∑_r=m^l (l+m)!/ r! (l+m-r)! d^r/dx^r l!/(l-r)! (x-1)^l-r l!/(l-r)!(x+1)^r-m]

Assume, s=r-m and r=s+m.

If r=m then s=0 and

if r=l then s=l-m

Thus,

Again, d^l-m/dx^l-m [(x²-1)^l]=d^l-m/dx^l-m ((x-1)^l (x+1)^l)solve as:

Replace r by s in equation (1):

On replacing r by s in equation (1):

Now, is solved as:

Therefore,

d^l-m/dx^l-m (x²-1)^l=(l-m)!/(l+m)! (x²-1)^m d^l+m/dx^l+m (x²-1)^l