First, thanks for your previous help. It was greatly needed and appreciated.

So we have proved [A,B^{n}]=nB^{n-1}[A,B]

Next we have: If F(x)=$\displaystyle \sum\limits_n f_{n} x^{n}$ is any Taylor Series expandable function with an appropriate radius of convergence, and A and B satisfy the conditions already stated, show that the operator function F(B) obeys $\displaystyle [A,F(B)]=F^{'}(B)[A,B]=[A,B]F^{'}(B)$, where $\displaystyle F^{'}(x)=dF/dx$.

$\displaystyle F(x)=\sum\limits_n f_{n} x^{n} so F^{'}(x)=n f_{n} x^{n-1}$

Using earlier proof

$\displaystyle [A,F(B)]=[A,\sum\limits_n f_{n} B^{n}]=\sum\limits_n f_{n} B^{n-1}[A,B]=[A,B]\sum\limits_n f_{n} B^{n-1}=F^{'}(B)[A,B]=[A,B]F^{'}(B)$

Now the last part: Use prior result to evaluate $\displaystyle [X,e^{i K_{x} L}] $where L is a constant having units of length.

I got $\displaystyle [X,e^{i K_{x} L}]=[X,K_{x}] \sum\limits_x (iLe^{i K_{x} L})$

Would someone be able to verify the second part and the last part?

Thank you