On a connection between Newton's binomial theorem and general Leibniz rule using a new method.

Question

In calculus the general Leibniz rule asserts that

Let $n$ be a natural numbers, if $f$ and $g$ are $n$-times differentiable functions at a point $x$, then the function $fg$ is also $n$-times differentiable and it's $n$-th derivative at this point is given by $$ (fg)^{(n)}(x)=\sum_{k=1}^n \binom nkf^{(k)}(x)g^{(n-k)}(x) $$

Now a similar theorem in algebra named after Newton asserts that

Let $n$ be a natural number, if $a$ and $b$ are two real numbers, then we have $$ (a+b)^n=\sum_{k=1}^n \binom nka^kb^{n-k} $$

I'm going to show that the Newton's binomial theorem can be deduced from Leibniz general rule.
Let $n$ be a natural number and $a,b$ are two real numbers. And let $f(t)=e^{at}$ and $g(t)=e^{bt}$ and use general Leibniz rule to find $n$-th derivative of the function $f(t)g(t)=e^{(a+b)t}$ at the point $x=0$ to find $$ \begin{align} (fg)^{(n)}(0)&=(e^{(a+b)t})^{(n)}|_{t=0}\\ &=\sum_{k=1}^n \binom nkf^{(k)}(0)g^{(n-k)}(0)\\ &=\sum_{k=1}^n \binom nk(e^{at})^{(k)}|_{t=0}(e^{bt})^{(n-k)}|_{t=0} \end{align} \tag{I}\label{I} $$ On the other hand for every real number $c$ and every natural number $k$ we have $$ (e^{ct})^{(k)}|_{t=0}=c^ke^{ct}|_{t=0}=c^k\tag{II}\label{II} $$ Now apply $\eqref{II}$ in $\eqref{I}$ to find

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

$\square$

The question is to find a proof of Leibniz general rule directly from Newton's binomial theorem.
Thanks in advance...

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Note. You can use this method with other functions to find other interesting formulas.

Answer 1

It is not necessary to reprove the binomial formula if we are willing to wade through some abstract nonsense.

Denote by $X$ the space of sufficiently differentiable functions $x\mapsto f(x)$ defined in some neighborhood $U$ of $a\in{\Bbb R}$. The maps $$p(f,g):=f\cdot g,\quad D_l(f,g):=(f',g),\quad D_r(f,g):=(f,g')$$ are bilinear on $X\times X$ and can therefore be lifted to linear maps on $Y:=X\otimes X$ such that $$p(f\otimes g)=f\cdot g,\quad D_l(f\otimes g)=f'\otimes g,\quad D_r(f\otimes g)=f\otimes g'\ .$$ This means, e.g., that $$p\left(\sum_k \lambda_k (f_k\otimes g_k)\right)=\sum_k \lambda_k\> f_k\cdot g_k\ ,$$ and that $D_l$ and $D_r$ are now maps $Y\to Y$.

The product rule $(f\cdot g)'=f'\cdot g+f\cdot g'$ can be written as $${d\over dx}\bigl( p(f,g)\bigr)=p\bigl(D_l(f,g)\bigr)+p\bigl(D_r(f,g)\bigr)\ ,$$ which lifts to $${d\over dx}\circ p\>(f\otimes g)=p\circ(D_l+D_r)(f\otimes g)\ .$$ As $D_l\circ D_r=D_r\circ D_l$ it then follows by induction that $$\left({d\over dx}\right)^n\circ p=p\circ(D_l+D_r)^n=p\circ\sum_{k=0}^n{n\choose k} D_l^{n-k}\>D_r^k\ .$$ When applied to a single $f\otimes g$ this is Leibniz' formula.

Answer 2

Liebnitz' rule does not follow from the binomial theorem until you add the additional fact that (#) $(fg)' = f'g + g'f = f^{(1)} g^{(0)} + f^{(0)} g^{(1)}$. Once you add that fact, then Leibnitz' rule follows by induction, with the $n=1$ case being $(fg)' = f'g + g'f$, and using the relation $$\binom{n}{m-1}+\binom{n}{m} = \binom{n+1}{m}$$

Proof that ($*_n$) $(fg)^{(n)} = \sum_k \binom{n}{k} f^{(k)} g^{(n-k)}$:

For $n=1$ ($*_1$) is established, being the assumption (#).

Assume ($*_n$). Then $$ (fg)^{(n)} = \sum_k \binom{n}{k} f^{(k)} g^{(n-k)} \\ (fg)^{(n+1)} \sum_k \binom{n}{k} \left[ (f^{(k)})' g^{(n-k)} + f^{(k)} (g^{(n-k)})' \right] \\= \sum_k \binom{n}{k} f^{(k+1)}) g^{(n-k)} + \sum_k \binom{n}{k} f^{(k)} g^{(n-k+1)} \\ = \sum_{j=1}^{n} \binom{n}{j-1} f^{(j)} g^{(n-j+1)} + \sum_k \binom{n}{k} f^{(k)} g^{(n-k+1)} \\ =\sum_{k=0}^{n} \binom{n}{k-1} f^{(k)} g^{(n-k+1)} - \binom{n}{-1} f^{(0)} g^{(n+1)} + \sum_k \binom{n}{k} f^{(k)} g^{(n-k+1)} \\ = \sum_{k=0}^{n} \left[ \binom{n}{k-1} + \binom{n}{k} \right] f^{(k)} g^{(n-k+1)} - \binom{n}{-1} f^{(0)}) g^{(n+1)} \\ = \sum_{k=0}^{n} \binom{n+1}{k} f^{(k)} g^{(n-k+1)} - 0 $$ So $$ (fg)^{(n+1)} = \sum_{k=0}^{n} \binom{n+1}{k} f^{(k)} g^{((n+1)-k)} $$ which is ($*_{n+1}$). So $$ (*_{n}) \implies (*_{n+1}) $$ and induction is established.

Small subtlety: This proof does not require that $g$ be $n+1$ differentiable, since $\binom{n}{-1}$ is inherently zero.