Newton’s Generalized Binomial Theorem

December 9, 2023 Shreyas Comments 0 Comment

1. Introduction

$\begin{equation*} (x + y)^2 = x^2 + 2xy + y^2 \end{equation*}$

$\begin{equation*} (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \end{equation*}$

$\begin{equation*} (x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 \end{equation*}$

$\begin{equation*} (x + y)^\frac{1}{2} =\ ??? \end{equation*}$

The binomial theorem famously allows one to fully expand out expressions of the form $(x + y)^n$ , where $n$ is a natural number. In these cases, the justification seems intuitive enough: if we write out a product of $(x + y)$ with itself a total of $n$ times, each individual term is produced by choosing either an $x$ or a $y$ from each factor, and multiplying them together. Then, for any $0 \leq i \leq n$ , the term $x^iy^{n - i}$ can be produced in $n\choose i$ ways, so

$\begin{equation*} (x + y)^n = \sum_{i = 0}^n {n\choose i} x^iy^{n - i} \end{equation*}$

But what happens when $n$ isn’t a natural number? Does it make sense to write an expansion for $(x + y)^r$ for any real number $r$ ? I decided to give my own shot at a discovery proved long ago by Newton: it does.

2. Power Rule

Firstly, let us investigate something many might have learned in introductory calculus classes without a formal justification: the power rule for derivatives. To provide that justification, consider the function $f(x) = x^a$ , and begin by restricting $a$ to be a natural number. By the definition of the derivative,

$\begin{equation*} f'(x) = \lim_{h\rightarrow0} \frac{(x + h)^a - x^a}{h} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{\left(\sum_{i = 0}^a {a\choose i} x^ih^{a - i}\right) - x^a}{h} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{\sum_{i = 0}^{a - 1} {a\choose i} x^ih^{a - i}}{h} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\sum_{i = 0}^{a - 1} {a\choose i} x^ih^{a - i - 1} \end{equation*}$

Setting $h = 0$ in the above expression, the only nonzero term left is the case of $i = a - 1$ , which yields

$\begin{equation*} f'(x) = {a\choose a - 1} x^{a - 1} = ax^{a - 1} \end{equation*}$

As expected, we have arrive at the familiar expression for the power rule. Note that in the first step, we used the ordinary binomial theorem itself; this is not circular reasoning, because in this case $a$ was restricted to a natural number, for which the binomial theorem can already be proved in many independent ways.

Moving to our next case, what if $a$ is a negative integer? Then $-a$ is a positive integer, and we have a similar proof:

$\begin{equation*} f'(x) = \lim_{h\rightarrow0} \frac{(x + h)^a - x^a}{h} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{\frac{1}{(x + h)^{-a}} - \frac{1}{x^{-a}}}{h} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{x^{-a} - (x + h)^{-a}}{hx^{-a}(x + h)^{-a}} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{-\sum_{i = 0}^{-a - 1} {-a\choose i} x^ih^{-a - i}}{hx^{-a}(x+h)^{-a}} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{-\sum_{i = 0}^{-a - 1} {-a\choose i} x^ih^{-a - i - 1}}{x^{-a}(x + h)^{-a}} \end{equation*}$

Again, we used the binomial theorem for $-a$ (a natural number) in the third step. Setting $h = 0$ , the only remaining term on top is the case of $i = -a - 1$ , and we have

$\begin{equation*} [f'(x) = \frac{-{-a\choose -a - 1}x^{-a - 1}}{x^{-a}(x+0)^{-a}} = \frac{ax^{-a - 1}}{x^{-2a}} = ax^{a - 1} \end{equation*}$

Once again, this is consistent with the power rule. Finally, we will consider the case when $a$ is a rational number. Let $a = \frac{p}{q}$ , where $p$ and $q$ are positive integers. We have

$\begin{equation*} f'(x) = \lim_{h\rightarrow0} \frac{(x + h)^{\frac{p}{q}} - x^{\frac{p}{q}}}{h} = \lim_{y\rightarrow x} \frac{y^{\frac{p}{q}} - x^{\frac{p}{q}}}{y - x} \end{equation*}$

where we have made the variable substitution $y = x + h$ . We will make another substitution: let $Y = y^{\frac{1}{q}}$ and $X = x^{\frac{1}{q}}$ . Now, we have

$\begin{equation*} f'(X) = \frac{Y^p - X^p}{Y^q - X^q} \end{equation*}$

$\begin{equation*} =\frac{(Y - X)\left(\sum_{i = 0}^{p - 1} Y^iX^{p - 1 - i}\right)}{(Y-X)\left(\sum_{j = 0}^{q - 1} Y^jX^{q - 1 - j}\right)} \end{equation*}$

$\begin{equation*} f'(x) = \lim_{h\rightarrow0} \frac{(x + h)^a - x^a}{h} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{\frac{1}{(x + h)^{-a}} - \frac{1}{x^{-a}}}{h} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{x^{-a}-(x + h)^{-a}}{hx^{-a}(x + h)^{-a}} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{-\sum_{i = 0}^{-a - 1} {-a\choose i} x^ih^{-a - i}}{hx^{-a}(x + h)^{-a}} \end{equation*}$

$\begin{equation*} =\lim_{h\rightarrow0}\frac{-\sum_{i = 0}^{-a - 1} {-a\choose i} x^ih^{-a - i - 1}}{x^{-a}(x + h)^{-a}} \end{equation*}$

Again, we used the binomial theorem for $-a$ (a natural number) in the third step. Setting $h = 0$ , the only remaining term on top is the case of $i = -a - 1$ , and we have

$\begin{equation*} f'(x) = \frac{-{-a\choose -a - 1}x^{-a - 1}}{x^{-a}(x + 0)^{-a}} = \frac{ax^{-a - 1}}{x^{-2a}} = ax^{a - 1} \end{equation*}$

Once again, this is consistent with the power rule. Finally, we will consider the case when $a$ is a rational number. Let $a = \frac{p}{q}$ , where $p$ and $q$ are integers. We have

$\begin{equation*} f'(x) = \lim_{h\rightarrow0} \frac{(x + h)^{\frac{p}{q}} - x^{\frac{p}{q}}}{h} = \lim_{y\rightarrow x} \frac{y^{\frac{p}{q}} - x^{\frac{p}{q}}}{y - x} \end{equation*}$

where we have made the variable substitution $y = x + h$ . We will make another substitution: let $Y = y^{\frac{1}{q}}$ and $X = x^{\frac{1}{q}}$ . Now, we have

$\begin{equation*} f'(x) = \lim_{Y\rightarrow X}\frac{Y^p - X^p}{Y^q - X^q} \end{equation*}$

$\begin{equation*} =\lim_{Y\rightarrow X}\frac{(Y - X)\left(\sum_{i = 0}^{p - 1} Y^iX^{p - 1 - i}\right)}{(Y - X)\left(\sum_{j = 0}^{q - 1} Y^jX^{q - 1 - j}\right)} \end{equation*}$

$\begin{equation*} =\lim_{Y\rightarrow X}\frac{\sum_{i = 0}^{p - 1} Y^iX^{p - 1 - i}}{\sum_{j = 0}^{q - 1} Y^jX^{q - 1 - j}} \end{equation*}$

Setting $Y = X$ , we have

$\begin{equation*} f'(x) = \frac{\sum_{i = 0}^{p - 1} X^iX^{p - 1 - i}}{\sum_{j = 0}^{q - 1} X^jX^{q - 1 - j}} = \frac{\sum_{i = 0}^{p - 1} X^{p - 1}}{\sum_{j = 0}^{q - 1} X^{q - 1}} = \frac{pX^{p - 1}}{qX^{q - 1}} = aX^{p - q} = ax^{\frac{p}{q} - 1} = ax^{a - 1} \end{equation*}$

This confirms the power rule for all positive rational numbers; the case of negative rational numbers can be proved using essentially the same steps as above, as well as letting $p$ be a negative integer and working with the positive value $-p$ , and will be omitted for brevity.

Although I haven’t yet formalized a proof of the case where $a$ is any real number, I have imagined that it can be done by constructing a sequence of cases where the exponents are rational numbers that converge to $a$ . This post will soon be updated once I have a more rigorous version of this approach. For now, I’ll assume its truth for the next section.

3. The General Binomial Theorem

Now, we will put our generalized power rule to good use. Consider the general expression $(x + y)^n$ , where $n$ is any real number. Letting $r = \frac{y}{x}$ , we can rewrite this as $x^n(1 + r)^n$ . Now, let us focus on the factor $f(r) = (1 + r)^n$ ; specifically, we can expand it as a Taylor series about $r = 0$ . Using our generalized power rule, the $i$ th derivative of $(1 + r)^n$ is

$\begin{equation*} \frac{d^if}{dr^i} = n(n - 1)\dots (n - i + 1)(1 + r)^{n - i} \end{equation*}$

Thus, we have

$\begin{equation*} f(r) = \sum_{i = 0}^\infty \frac{f^{(i)}(0)r^i}{i!} = \sum_{i = 0}^\infty \frac{n(n - 1)\dots (n - i + 1)}{i!} r^i = \sum_{i = 0}^\infty {n\choose i} r^i \end{equation*}$

When $n$ is a integer, this is exactly same as the regular binomial theorem we’re all familiar with. However, in any other case, the expression $n \choose i$ no longer has a combinatorial meaning: it is simply a shorthand for the expression $\frac{n(n - 1)\dots (n - i + 1)}{i!}$ . In these cases, our binomial expansion actually becomes infinite! For positive integer $n$ , all terms in the series after $i = n$ become 0; the coefficient $n\choose i$ will vanish due to the factor of $n-n$ in the numerator. However, for any other $n$ , this will never occur, and the series will continue forever. Among many other things, this sort of expansion was used to derive a series converging to $\pi$ . The area under the semicircle $y = \sqrt{1 - x^2}$ is $\frac{\pi}{2}$ , thus, we have

$\begin{equation*} \pi = 2\int_{-1}^1 (1 - x^2)^{\frac{1}{2}}dx = 2\int_{-1}^1 \sum_{i = 0}^\infty {\frac{1}{2}\choose i} (-x^2)^i dx = \sum_{i = 0}^\infty \frac{4(-1)^n}{2i + 1}{\frac{1}{2}\choose i} \end{equation*}$

We conclude today’s by returning to our general binomial expansion. For any real number $n$ ,

$\begin{equation*} (x + y)^n = x^n(1 + r)^n = x^n \sum_{i = 0}^\infty {n\choose i} r^i = \sum_{i = 0}^\infty {n\choose i} x^n\left(\frac{y}{x}\right)^i = \sum_{i = 0}^\infty {n\choose i} x^{n - i}y^i \end{equation*}$

Quite a remarkable result indeed!

Shreyas' blog – Ignited Minds

I love math and science. This is a blog about things I think of, or work on, in my spare time.

Newton’s Generalized Binomial Theorem

December 9, 2023 Shreyas Comments 0 Comment

1. Introduction

2. Power Rule

3. The General Binomial Theorem

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