Differentiation operators similar to exponents?

[Gib Z]

I was just messing around on another problem I was trying to solve for someone in the homework forums when I stumbled on this:

If we denote the n-th derivative the same way we do for exponents, eg the second derivative of f will be denoted by f^2, and the original function as its zero-th derivative, and define a function f, that is the product of two functions u and v, ie $$f(x) = u(x)\cdot v(x)$$.

Then the following holds:

$$f^n (x) = \sum_{k=0}^n \frac{n!}{k! (n-k)!} u^{n-k} v^k$$

I'm guessing if we chose to use induction, that theorem would be not too difficult to prove since it would be analogous to the proof of the binomial theorem. Also in the deeps on my mind it looks somewhat similar to a repeated differentiation formula, but that may have been more general (any function f(x), not necessarily product of u and v) or even something else.

Could someone either tell me that repeated differentiation formula, post a nice proof, or just comment, because commentaries are welcome =]

EDIT: I was thinking of making my reply to you the 2000th post, but I thought best leave it till tomorrow (its getting quite late here) to use it on a post that my enlighten someone else who needs help =] Thank you very much for the congrats arildno, and everyone else on these forums that makes it the wonderful, pleasant and enlightening place it is! Heres to 2000 more =D

PS. For frequent posters, if you are able to, send it some Contributions to Greg, it feels nice giving a little bit back !

 

[arildno]

Pre-congratulations for post 2000, GibZ!

 

[daveb]

This doesn't hold true for all functions, though. Take u(x) = e^x and v(x) = e^-x so that f(x) = 1, and the summation doesn't hold.

 

[D H]

This doesn't hold true for all functions, though. Take u(x) = e^x and v(x) = e^-x so that f(x) = 1, and the summation doesn't hold.

The summation still holds. With $f(x)= e^xe^{-x}=1$, the n^th derivative of $f(x)$ just generates a lot of different expressions for zero:
$$f^{(n)} (x) = 0 = e^xe^{-x}\sum_{k=0}^n \frac{n!}{k! (n-k)!} (-1)^{(k)}$$

 

[daveb]

DOH! I knew I should have worked that out before I opened my big mouth!

 

[Rainbow Child]

...I'm guessing if we choose to use induction, that theorem would be not too difficult to prove since it would be analogous to the proof of the binomial theorem. ...

You are guessing right, because all we need is Pascal's recurrence relation, i.e.

$$\left( \begin{array}{c}n+1\\k\end{array}\right)=\left( \begin{array}{c}n\\k-1\end{array}\right)+\left( \begin{array}{c}n\\k\end{array}\right)$$

We want to prove

$$f^{(n)} (x) = \sum_{k=0}^n \frac{n!}{k! (n-k)!} u^{(n-k)} v^{(k)}\Rightarrow f^{(n)}(x)=\sum_{k=0}^n \left( \begin{array}{c}n\\k\end{array}\right)\,u^{(n-k)}\,v^{(k)} \quad (*)$$
​

(I prefer parenthesis )

• For $$n=1, \quad f'(x)= \left(\begin{array}{c}1\\0\end{array}\right)\,u^{(1)}\,v^{(0)}+\left(\begin{array}{c}1\\1\end{array}\right)\,u^{(0)}\,v^{(1)}\Rightarrow f'(x)=u'(x)\,v(x)+u(x)\,v'(x) \quad \text{True}$$
  
• We assume that $(*)$ is true for some $n$. We will show that for $n+1$ is also true.
  For $n+1$, $(*)$ reads
  $$\begin{array}{llll}f^{(n+1)}(x)&=&\sum_{k=0}^{n+1} \left( \begin{array}{c}n+1\\k\end{array}\right)\,u^{(n+1-k)}\,v^{(k)} & \\&=& \sum_{k=1}^{n} \left( \begin{array}{c}n+1\\k\end{array}\right)\,u^{(n+1-k)}\,v^{(k)}+u^{(n+1)}\,v+u\,v^{(n+1)} &\\ &=&\sum_{k=1}^{n} \left( \begin{array}{c}n\\k-1\end{array}\right)\,u^{(n+1-k)}\,v^{(k)}+ \sum_{k=1}^{n} \left( \begin{array}{c}n\\k\end{array}\right)\,u^{(n+1-k)}\,v^{(k)}+u^{(n+1)}\,v+u\,v^{(n+1)} &\text{Pascal's rule} \\ &=& \sum_{l=0}^{n-1} \left( \begin{array}{c}n\\l\end{array}\right)\,u^{(n-l)}\,v^{(l+1)}+ \sum_{k=1}^{n} \left( \begin{array}{c}n\\k\end{array}\right)\,u^{(n+1-k)}\,v^{(k)}+u^{(n+1)}\,v+u\,v^{(n+1)} &\text{define l=k-1} \\ &=& \sum_{l=0}^{n} \left( \begin{array}{c}n\\l\end{array}\right)\,u^{(n-l)}\,v^{(l+1)}+ \sum_{k=0}^{n} \left( \begin{array}{c}n\\k\end{array}\right)\,u^{(n+1-k)}\,v^{(k)} &\text{combine 1st sum - 4rd term, 2nd sum - 3rd term} \\ &=& \sum_{k=0}^{n} \left( \begin{array}{c}n\\k\end{array}\right)\,(u^{(n-k)}\,v^{(k+1)}+u^{(n+1-k)}\,v^{(k)}) &\text{remane l to k} \\ &=& \sum_{k=0}^{n} \left( \begin{array}{c}n\\k\end{array}\right)\,\frac{d}{d\,x}u^{(n-k)}\,v^{(k)} & \\ &=& \frac{d}{d\,x}\sum_{k=0}^{n} \left( \begin{array}{c}n\\k\end{array}\right)\,u^{(n-k)}\,v^{(k)} & \\ &=& \frac{d}{d\,x}f^{(n)}(x) &\text{using} \, (*) \, True\end{array}$$

Hope it isn' t too mesy !

 

[Gib Z]

Quite the contrary, an excellently set out post, Rainbow Child.

 

[Gib Z]

So anybody seen that expression before (or a generalization)? If not, I might claim it as the Gib Z-Rainbow Child theorem =)

 

[CompuChip]

Hey, I never thought of that
Just one little comment Rainbow Child: the first line of the large formula had me confused at first. I thought you assumed $f^{(n+1)}$ to be what was on the right hand side, though that is actually what you want to prove. Probably you should just start with the sum and in the end find it to be $\frac{d}{dx} f^{(n)} = f^{(n+1)}$, or start from $f^{(n+1)} = \frac{d}{dx} f^{(n)} = \sum \cdots$ and write the whole thing backwards.

But no, it's not messy (rather clear, actually) and yes, it is nice.

 

[d_leet]

So anybody seen that expression before (or a generalization)? If not, I might claim it as the Gib Z-Rainbow Child theorem =)

http://mathworld.wolfram.com/LeibnizIdentity.html

 

[rudinreader]

... the Gib Z-Rainbow Child theorem =)

Hilarious!

 

[Gib Z]

http://mathworld.wolfram.com/LeibnizIdentity.html

DAMN IT!

 

[Rainbow Child]

http://mathworld.wolfram.com/LeibnizIdentity.html

I was sure that I couldn't be the first!

DAMN IT!

Don't worry Gib, we will prove something ...harder!

 

[CompuChip]

Isn't the GibZ-Rainbow Child identity that complicated but intuitive one, from which the Leibniz identity follows?

 

[Gib Z]

Isn't the GibZ-Rainbow Child identity that complicated but intuitive one, from which the Leibniz identity follows?

Are you asking that my theorem is a generalization of the Leibniz Identity? Because if you follow d_leet's link, its exactly the same :( Damn It >.<

 

[Hurkyl]

It's possible to compute this identity by actually using the binomial identity. (Rather than just using the ideas behind the binomial identity)

Let $\mu$ denote the multiplication operator on tensor products of functions, so$\mu(f \otimes g) = f g$. Let $D$ be the differentiation operator on differentiable functions.

Lemma:
$$D \mu = \mu (D \otimes 1 + 1 \otimes D)$$

Because $D \otimes 1$ and $1 \otimes D$ commute, we have:

$$\begin{equation*} \begin{split} D^n \mu &= \mu (D \otimes 1 + 1 \otimes D)^n \\&= \mu \sum_i \binom{n}{i} (D \otimes 1)^i (1 \otimes D)^{n-i} \\&= \mu \sum_i \binom{n}{i} D^i \otimes D^{n-i} \\&= \sum_i \binom{n}{i} \mu (D^i \otimes D^{n-i}) \end{split} \end{equation*}$$

And since the original question is to compute $(D^n \mu) (u \otimes v)$...

 

[CompuChip]

Are you asking that my theorem is a generalization of the Leibniz Identity? Because if you follow d_leet's link, its exactly the same :( Damn It >.<

No, I was saying that the new theorem from the following quote:

Don't worry Gib, we will prove something ...harder!

will be a generalization of it.

 

[Gib Z]

Hurkyl, I am not familiar with a lot of that notation you used, but I think I get the general idea (since the order of operations when multiplying by one and differentiating give the same result, and the binomial theorem applies to things that commute, we use the theorem directly).

CompuChip, perhaps I already know a small and simple generalization, I didn't think it was worth posting though :( It was the n-th derivative of a product of k functions, not just 2. That just goes into the multinomial theorem now, and proof again analogous with the proof of the multinomial theorem.