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

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.

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

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.

• You proof for Leibniz general rule doesn't use Newton's binomial coefficient, but your first observation in the answer above is somehow logical. – k1.M Apr 30 '15 at 19:01

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.