A Gem from Isaac Newton |

The Binomial Theorem (see “Double Feature on Christmas Day“, “Prelude to Taylor’s theorem“) states:

$(x+y)^n = \sum\limits_{i=0}^{n} \binom{n}{i}x^{n-i}y^i, \quad n \in \mathbb{N}.$

Provide $x$ and $y$ are suitably restricted, there is an Extended Binomial Theorem. Namely,

$(x+y)^r = \sum\limits_{i=0}^{\infty} \binom{r}{i}x^{r-i}y^i, \underline{|\frac{x}{y}|<1}, r \in \mathbb{R}$

where $\binom{r}{i} \overset{\Delta}{=} \frac{r(r-1)(r-2)...(r-i+1)}{i!}.$

Although Issac Newton is generally credited with the Extended Binomial Theorem, he only derived the germane formula for any rational exponent (i.e., $r \in \mathbb{Q}$ ).

We offer a complete proof as follows:

Let

$f(t) = (1+t)^r,\;|t|<1, r \in \mathbb{R}.\quad\quad\quad(1)$

Differentiate (1) with respect to $t$ yields

$f'(t) = r(1+t)^{r-1}.$

We have

$(1+t)f'(t) = (1+t)\cdot r(1+t)^{r-1} = r(1+t)^r.$

That is,

$(1+t)f'(t) = rf(t).$

From

$f(0) = (1+0)^r =1,$

we see that $f(t) = (1+t)^r$ is a solution of initial-value problem

$\begin{cases} (1+t)z'(t)=rz(t) \\ z(0) = 1 \end{cases}\quad\quad\quad(2)$

By $(A-1),$ we also have

$g(t) = \sum\limits_{i=0}^{\infty}\binom{r}{i}\cdot t^{i}, |t|<1, r \in \mathbb{R}.\quad\quad\quad(3)$

Express (3) as

$g(t) = \binom{r}{0}t^0 + \sum\limits_{i=1}^{\infty}\binom{r}{i}t^i=1+ \sum\limits_{i=1}^{\infty}\binom{r}{i}t^i$

and then differentiate it with respect to $t:$

$g'(t) = \sum\limits_{i=0}^{\infty}\binom{r}{i}\cdot i\cdot t^{i-1}\overset{(A-2)}{=} r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^{i-1}$

gives us

$(1+t)g'(t) = (t+1)\cdot r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^{i-1}$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^{i-1}$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\sum\limits_{k=0}^{\infty}\binom{r-1}{k}t^{k}$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\left(\binom{r-1}{0}t^0+\sum\limits_{k=1}^{\infty}\binom{r-1}{k}t^k\right)$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\left(1+\sum\limits_{i=1}^{\infty}\binom{r-1}{i}t^i\right)$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r + r\sum\limits_{i=1}^{\infty}\binom{r-1}{i}t^i$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\sum\limits_{i=1}^{\infty}\binom{r-1}{i}t^i+r$

$= r\sum\limits_{i=1}^{\infty}\left(\binom{r-1}{i-1} + \binom{r-1}{i}\right)t^i+r$

$\overset{(A-3)}{=} r\sum\limits_{i=1}^{\infty}\binom{r}{i}t^i+r$

$= r\sum\limits_{i=1}^{\infty}\binom{r}{i}t^i+r\binom{r}{0}t^0$

$= r\left(\sum\limits_{i=1}^{\infty}\binom{r}{i}t^i+\binom{r}{0}t^0\right)$

$= r\sum\limits_{i=0}^{\infty}\binom{r}{i}t^i$

$\overset{(3)}{=} r\cdot g(t).$

i.e.,

$(1+t)g'(t) = rg(t).$

Since

$g(0) = 1+\sum\limits_{i=1}^{\infty}\binom{r}{i}0^i = 1,$

we see that $g(t)$ is also a solution of initial-value problem (2).

Hence, by the Uniqueness theorem (see Coddington: An Introduction to Ordinary Differential Equations, p. 105),

$f(t) = g(t).$

And so,

$(1+t)^r = \sum\limits_{i=0}^{\infty}\binom{r}{i}t^i.$

Consequently, for $|\frac{x}{y}| <1,$

$(1+\frac{x}{y})^r = \sum\limits_{i=0}^{\infty}\binom{r}{i}(\frac{x}{y})^i.$

Multiply $y^r$ throughout, we obtain

$(x+y)^r = y^r\sum\limits_{i=0}^{\infty}\binom{r}{i}x^iy^{-i}.$

i.e.,

$(x + y)^r = \sum\limits_{i=0}^{\infty}\binom{r}{i}x^iy^{r-i}.$

Prove $(A-1)$ is convergent:

$\sum\limits_{i=1}^{\infty}\binom{r}{i}\cdot t^{i}, |t|<1\quad\quad\quad(A-1)$

Proof

$\frac{\binom{r}{i+1}t^{i+1}}{\binom{r}{i}t^i}$

$= \frac{\frac{r(r-1)(r-2)...(r-i+1)(r-(i+1)+1)}{(i+1)!}}{\frac{r(r-1)(r-2)...(r-i+1)}{i!}}\cdot t$

$=\frac{r(r-1)(r-2)...(r-i+1)(r-(i+1)+1)}{(i+1)!}\cdot \frac{i!}{r(r-1)(r-2)...(r-i+1)}\cdot t$

$= \frac{r-(i+1)+1}{(i+1)i!}i!\cdot t$

$=\frac{r-(i+1)+1}{i+1}\cdot t$

$\lim\limits_{i\rightarrow \infty}\bigg|\frac{\binom{r}{i+1}t^{i+1}}{\binom{r}{i}t^i}\bigg|=\lim\limits_{i\rightarrow \infty}|\frac{r-(i+1)+1}{i+1}\cdot t|=\lim\limits_{i\rightarrow \infty}|\frac{i-r}{i+1}|\cdot |t|=|t|\overset{}{<}1$

Prove

$\sum\limits_{i=1}^{\infty}\binom{r}{i}\cdot i\cdot t^{i-1} = r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}\cdot t^{i-1}\quad\quad\quad(A-2)$

Proof

$\binom{r}{i} = \frac{r(r-1)(r-2)...(r-i+1)}{i!}$

$= \frac{r}{i}\cdot\frac{(r-1)(r-1-1)...(r-1+1-i+1)}{(i-1)!}$

$= \frac{r}{i}\cdot\frac{(r-1)(r-1-1)...(r-1-i+1+1)}{(i-1)!}$

$= \frac{r}{i}\cdot\frac{(r-1)(r-1-1)(r-1-2)...(r-1-(i-1)+1)}{(i-1)!}$

$= \frac{r}{i}\binom{r-1}{i-1}$

that is,

$\binom{r}{i} = \frac{r}{i}\binom{r-1}{i-1}.$

Therefore,

$\sum\limits_{i=1}^{\infty}\boxed{\binom{r}{i}}\cdot i\cdot t^{i-1} =\sum\limits_{i=1}^{\infty}\boxed{\frac{r}{i}\binom{r-1}{i-1}}\cdot i\cdot t^{i-1}=r\sum\limits_{k=1}^{\infty}\binom{i-1}{i-1}t^{i-1}.$