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## 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 previous question which I already proved may be useful

$$

\frac{d^{n}}{dx^{n}} \left[ f(x)g(x) \right] = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x)g^{(k)}(x)

$$

As well as this identity:

$$

\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}

$$

## The Attempt at a Solution

I go out to prove this by induction. I've already shown that the base case is true; that is easy and I do not have problems with that.

So thus I go about assuming this is true for some ##n##, then to show truth for ##n+1## I derive the whole thing and rearrange a bit

$$

f(x)g^{(n+1)}(x) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{d^{n-k+1}}{dx^{n-k+1}} \left[ f^{(k)}(x)g(x) \right] - f'(x)g^{(n)}(x)

$$

Because we assumed that the ##n## case is true, we get:

$$

f(x)g^{(n+1)}(x) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{d^{n-k+1}}{dx^{n-k+1}} \left[ f^{(k)}(x)g(x) \right] - \frac{f'(x)}{f(x)} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}} \left[ f^{(k)}(x)g(x) \right]

$$

Then, because shift ##k## up in the second term (I don't know the proper name for this process). Note that in doing so I can change the sign as ##(-1)^{k-1}=-(-1)^k##

$$

f(x)g^{(n+1)}(x) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{d^{n-k+1}}{dx^{n-k+1}} \left[ f^{(k)}(x)g(x) \right] + \frac{f'(x)}{f(x)} \sum_{k=1}^{n+1} (-1)^k \binom{n}{k-1} \frac{d^{n-k+1}}{dx^{n-k+1}} \left[ f^{(k-1)}(x)g(x) \right]

$$

This is where I have become stuck. I am trying to get to this point, after which I can finish the rest of the problem:

$$

f(x)g^{(n+1)}(x) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{d^{n-k+1}}{dx^{n-k+1}} \left[ f^{(k)}(x)g(x) \right] + \sum_{k=1}^{n+1} (-1)^k \binom{n}{k-1} \frac{d^{n-k+1}}{dx^{n-k+1}} \left[ f^{(k)}(x)g(x) \right]

$$

As far as I know (this may the point where I am completely overseeing something), to get there, this would require that:

$$

\frac{f'(x)}{f(x)} \frac{d^{n-k+1}}{dx^{n-k+1}} \left[ f^{(k-1)}(x)g(x) \right]=\frac{d^{n-k+1}}{dx^{n-k+1}} \left[ f^{(k)}(x)g(x) \right]

$$

,which doesn't seem to make much sense to me. Anyway, I tried breaking down the above equation further anyway, so that I get:

$$

\frac{f'(x)}{f(x)} \sum_{l=0}^{n-k+1}f^{(k+n-l)}(x)g^{(l)}(x) = \sum_{l=0}^{n-k+1}f^{(k+n+1-l)}(x)g^{(l)}(x)

$$

I can't find how this statement can be true, because it seems like to me that the only way to make this true is if ##\frac{f'(x)}{f(x)} f^{(a-1)}=f^{(a)}## for some natural number ##a##, which is not true. (In general, I am baffled on how I'm supposed to get rid of the ##\frac{f'(x)}{f(x)}## term)

I've tried re-doing this and checking my work but I always end up at the same point. I am wondering if there is some glaring mistake or oversight that I missed (perhaps in doing the derivatives in the sums?), or some principle that I overlooked along the way trying to do this problem.

Thanks in advance~