Oh, I forgot to specify that ##f^{(0)}(x)=f(x)##. I apologise; I didn't know that this was non-standard notation, I've seen it quite a few times, so it must be a recent development or something.
For the ##n=2## example:
The left side is ##f(x)g''(x)##.
Thus the right side is: ##(-1)^0 (1)...
For the first equation, what happens is that if you evaluate the sum on the right side, the alternating sum causes all of the terms to cancel out except for ##f(x)g^{(n)}(x)##.
*Edit*: In other words, if you write out the left hand side as ##\sum_{k=0}^{n} a_k f^{(n-k)}(x) g^{(k)}(x)##, ##a_k =...
I actually tried this formula raw with a few values of n, and it actually does hold for the various values of n I tested. Obviously that doesn't prove anything rigourously... but it should be a valid problem I think. I did find a pattern arising from doing this manually that I can't seem to put...
Homework Statement
In the problem, I should provide proof for the statement, where ##f^{(n)}(x)## denotes the ##n##th derivative of the function ##f(x)##:
$$
f(x)g^{(n)}(x) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}} \left[ f^{(k)}(x)g(x) \right]
$$
Homework Equations
The...