# Simplification of Spherical Bessel functions

• Nov 26th 2009, 07:46 AM
MrWaeseL
Simplification of Spherical Bessel functions
From here:

http://i48.tinypic.com/2n7304h.png

Why isn't the $\displaystyle (-x)^n (1/x)^n$ product simplified to $\displaystyle (-1)^n$?

Thanks in advance for any help.
• Nov 26th 2009, 10:32 AM
Jester
Quote:

Originally Posted by MrWaeseL
From here:

http://i48.tinypic.com/2n7304h.png

Why isn't the $\displaystyle (-x)^n (1/x)^n$ product simplified to $\displaystyle (-1)^n$?

Thanks in advance for any help.

Because it's an operator not just an algebraic expression. To demonstrate let $\displaystyle n = 2$ (and $\displaystyle f(x) = \frac{\sin x}{x}$ ) so

$\displaystyle (-x)^2 \left(\frac{1}{x} \frac{d}{dx} \right)^2 f(x) = x^2 \frac{1}{x} \frac{d}{dx} \left(\frac{1}{x} \frac{df}{dx} \right) = \frac{d^2f}{d x^2} - \frac{1}{x} \frac{df}{dx}.$
• Nov 26th 2009, 11:00 AM
MrWaeseL
I see...so it would be wrong to state that $\displaystyle \left(\frac{1}{x} \frac{d}{dx} \right)^n = (\frac{1}{x})^n (\frac{d}{dx})^n$?
• Nov 26th 2009, 12:22 PM
Jester
Quote:

Originally Posted by MrWaeseL
I see...so it would be wrong to state that $\displaystyle \left(\frac{1}{x} \frac{d}{dx} \right)^n = (\frac{1}{x})^n (\frac{d}{dx})^n$?

Absolutely. It only applies when $\displaystyle n = 1$.