Proving the connection between the Binomial Theorem and the product rule for derivatives

Question

Let $a(x)$ and $b(x)$ be smooth functions, i.e they are infinitely times differentiable.

I have made the assumption that the derivative for the function

$$f(x)= (a\cdot b)(x)$$

can be given by

$$f^{(n)}(x)=\sum_{k=0}^{n}\binom{n}{k}a^{(k)}b^{(n-k)}(x)$$

Where I have defined $g^{(n)}$ as the $n-$ th derivative of some function.

I have made the observation when I was trying to calculate the derivative of such a function and I noticed that the derivatives are in the form:

$$f^{(1)}(x)=a'b+b'a$$

$$f^{(2)}(x)=ab''+2a'b'+a''b$$

$$f^{(3)}(x)=ab'''+3a'b''+3a''b'+a'''b$$

What I have tried so far is induction but I don't know how to manipulate the formula to get the result I want

$${f^{(n+1)}=f^{(n)}}^{'}=(\sum_{k=0}^{n}\binom{n}{k}a^{(k)}b^{(n-k)})^{'}=(\sum_{k=0}^{n}\binom{n}{k}[a^{(k+1)}b^{(n-k)}+a^{(k)}b^{(n-k+1)}])$$

Now I cannot say something like :

$$[\uparrow]=(\sum_{k=0}^{n}\binom{n}{k}[a^{(k)}b^{(n-k)}(a+b)]) $$

Because they are derivatives and they do not necessarily behave like exponentials. But I don't know if it even would help me

Because in the end I want to have a term that Looks like

$$\sum_{k=0}^{n+1}\binom{n+1}{k}a^kb^{n-k}$$

But I don't know how to do it.

Answer 1

The proof is identical to the proof of the Binomial Theorem:

$$\left(\sum_{k=0}^{n}\binom{n}{k}[a^{(k)}b^{(n-k)}(a+b)]\right) = \sum_{k=0}^{n}\binom{n}{k}a^{(k+1)}b^{(n-k)}+\sum_{k=0}^{n}\binom{n}{k}a^kb^{(n+1-k)} =$$

$$a^{(n+1)}b^{(0)}+\sum_{k=0}^{n-1}\binom{n}{k}a^{(k+1)}b^{(n-k)}+\sum_{k=1}^{n}\binom{n}{k}a^kb^{(n+1-k)}+b^{(n+1)}a^{(0)}=$$

Careful!: $a^{(0)},b^{(0)}$ does not vanish!

$$a^{(n+1)}b^{(0)}+\sum_{k=1}^{n}\binom{n}{k-1}a^{(k)}b^{(n+1-k)} +\sum_{k=1}^{n}\binom{n}{k}a^kb^{(n+1-k)}+b^{n+1}a^{(0)}= $$

$$\binom{n+1}{0}a^{(n+1)}b^{(n+1-(n+1))}+\sum_{k=1}^{n}\left(\binom{n}{k}+\binom{n}{k-1}\right)a^{(k)}b^{(n+1-k)}+\binom{n+1}{n+1}a^{(0)}b^{(n+1)}=$$

$$\sum_{k=0}^{n+1}\binom{n+1}{k}a^kb^{n-k}$$

Since

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

Proof:

For the Edge cases $k=0$ and $k=n$

we get $$\binom{n+1}{1}=\binom{n}{0}+\binom{n}{1}=n+1$$ and $$\binom{n}{n}+\binom{n}{n+1}=\frac{n(n-1)...(n-n+1)}{n!}+\frac{n(n-1)...(n-n)}{(n+1)!} =\frac{n!}{n!}+\frac{0}{(n+1)!}=1+0=1=\binom{n+1}{n+1}$$

now for $k>0$ and $<n$ actually also $=n$:

$$\binom{n}{k}+\binom{n}{k+1}= \frac{n(n-1)…(n-l+1)}{k!}+\frac{n(n-1)…(n-k)}{k!(k+1)}\overset{\text{arithmetic rules}}{=} \frac{n(n-1)…(n-k+1)(k+1+n-k)}{(k+1)!}=\binom{n+1}{k+1}. \square$$

To understand the arithmetic manipulation it helps to substitute a specific part with some variable and think about it as just a number and then read the proof on arithmetic properties of General numbers in $\mathbb{Z}$.