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Function multiplied by nth derivative of another function

  • Thread starter Tcw7468
  • Start date
  • #1
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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~
 

Answers and Replies

  • #2
33,168
4,852

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)##:

$$
f(x)g^{(n)}(x) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}} \left[ f^{(k)}(x)g(x) \right]
$$
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?
Tcw7468 said:

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~
 
  • #3
4
0
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?
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
33,168
4,852
This is what you wrote in post #1.
Tcw7468 said:
In the problem, I should provide proof for the statement, where ##f^{(n)}(x)## denotes the ##n##th derivative of the function ##f(x)##:
$$
f(x)g^{(n)}(x) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}} \left[ f^{(k)}(x)g(x) \right]
$$
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...
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
4
0
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.

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.
 
Last edited:
  • #6
33,168
4,852
Tcw7468 said:
$$f(x)g^{(n)}(x) = \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}} \left[ f^{(k)}(x)g(x) \right]$$
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
4
0
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
33,168
4,852
Tcw7468 said:
Oh, I forgot to specify that ##f^{(0)}(x)=f(x)##.
Sure, that makes sense. I think I mixed this one up in my thinking. I'll take another look later today.
 

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