# Function multiplied by nth derivative of another function

1. May 25, 2016

### Tcw7468

1. The problem statement, all variables and given/known data

In the problem, I should provide proof for the statement, where $f^{(n)}(x)$ denotes the $n$th derivative of the function $f(x)$:

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

2. Relevant 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:

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

3. 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.

2. May 25, 2016

### Staff: Mentor

I don't think the above is true at all. After all, it's just f(x) times the n-th derivative of g, so how would it be a sum involving derivatives of both f and g?

3. May 25, 2016

### Tcw7468

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 in mathematical terms (the terms resulting from the product rule start to cancel out because of the alternating sign); I'm trying to figure out if I can find out where I went wrong from looking at this pattern...

4. May 25, 2016

### Staff: Mentor

This is what you wrote in post #1.
If we're talking about the same thing, the formula above doesn't make any sense to me. On the left side is, as I said before, nothing more than f(x) times g(n)(x). There's no addition, and definitely no alternating signs.

The formula you said you proved earlier, $\frac{d^n}{dx^n} [f(x)g(x)]$ is a sum; namely $\binom{n}{0} f(x)g^{(n)}(x) + \binom{n}{1} f^{(1)}(x)g^{(n - 1)}(x) + \binom{n}{2} f^{(2)}(x)g^{(n - 2)}(x) + \dots + \binom{n}{n} f^{(n)}(x)g(x)$. Are you trying to solve for (isolate) the first term in the sum I wrote? That's the only thing I can think of that makes any sense.

5. May 25, 2016

### Tcw7468

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 $\sum_{k=0}^{n} a_k f^{(n-k)}(x) g^{(k)}(x)$, $a_k = 0$ for all $k$ except $k=n$, in which $a_n=1$. This is what I have to show, basically.

Last edited: May 26, 2016
6. May 26, 2016

### Staff: Mentor

Let n = 2
Then the left side is $f(x)g^{(2)}(x)$, or f(x)g''(x).
The right side is $(-1)^0 1 \frac{d^2}{dx^2}[f^{(0)}(x)g(x) + (-1)^1 2 \frac{d}{dx}[f^{(1)}(x)g(x)] + (-1)^2 1 \frac{d^0}{dx^0}[f^{(2)}(x)g(x)]$
$= g''(x) - 2f''(x)g(x) - 2f'(x)g'(x) + f''(x)g(x) = g''(x) - f''(x)g(x) + f''(x)g(x)$
I'm interpreting $f^{(0)}(x)$ to mean 1, and I'm interpreting $\frac{d^0}{dx^0}$ to be the identity operator, although I've never run across this notation before.

So I still don't see how you can write $f(x)g^{(n)}(x)$ as a sum.

7. May 26, 2016

### Tcw7468

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: $(-1)^0 (1) \frac{d^2}{dx^2} [f(x)g(x)] + (-1)^1(2)\frac{d}{dx} [f'(x)g(x)] + (-1)^2(1)[f''(x)g(x)] =[f(x)g''(x)+2f'(x)g'(x)+f''(x)g(x)]-2[f'(x)g'(x)+f''(x)g(x)]+[f''(x)g(x)] =(1+0+0)f(x)g''(x) + (2-2+0)f'(x)g'(x) + (1-2+1) f(x)g''(x) =f(x)g''(x)$.

Meanwhile, I think I'm getting closer, if I can find out how to manipulate the binomial coefficients in order for it to turn out to 0 except for one case, I think I can solve the problem without induction... but I would still be curious anyway on what I did wrong on the induction attempt.

8. May 26, 2016

### Staff: Mentor

Sure, that makes sense. I think I mixed this one up in my thinking. I'll take another look later today.