At school we learn to expand $(1+x)^2=1+2x+x^2$ and $(1+x)^3=1+3x+3x^2+x^3$. The coefficients come from the arithmetic triangle, which Pascal studied and which was known long before him. For positive whole-number exponents, the expansion always stops.

In the winter of 1664–1665, still a student, Newton asked a question nobody had asked so boldly: what if the exponent were $\frac12$? Or $-1$? What would $(1+x)^{1/2}$, that is $\sqrt{1+x}$, look like as a sum of powers of $x$?

The rule for the coefficients

Newton was reading John Wallis’s Arithmetic of Infinites, which computed areas by interpolating tables of numbers. Following the same idea, he noticed that the binomial coefficients follow a pattern that does not depend on the exponent being a whole number:

$$(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\cdots$$

Each coefficient comes from the previous one by multiplying by $\dfrac{\alpha-k}{k+1}$. If $\alpha$ is a positive whole number, at some point the factor $(\alpha-k)$ becomes zero, and from then on every term vanishes: that is the school binomial. If $\alpha$ is not a positive whole number, no factor vanishes, and the sum goes on forever.

The series works when $|x|<1$. Outside that range the terms grow instead of shrinking and the sum goes nowhere.

Two examples

With $\alpha=-1$ the coefficients are $1,-1,1,-1,\dots$ and we get a series that was already known, the geometric progression:

$$\frac{1}{1+x}=1-x+x^2-x^3+\cdots$$

With $\alpha=\frac12$, something new appears:

$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\cdots$$

Newton had no proof in the modern sense. He checked: he multiplied the series by itself and saw the terms cancel until only $1+x$ was left. Try it with the first three terms:

$$\left(1+\frac{x}{2}-\frac{x^2}{8}\right)^2=1+x+\left(\frac{x^2}{4}-\frac{x^2}{4}\right)+\cdots=1+x+0\cdot x^2+\cdots$$

The $x^2$ term disappears, as it should. So do the next ones.

Computing a root with sums

To compute $\sqrt{1.21}$ we use $x=0.21$:

$$1+0.105-0.0055125+0.0005788125=1.1000663125.$$

The exact value is $1.1$. With four terms the error is already in the fourth decimal place, and the fifth term, $-5x^4/128\approx-0.000076$, improves it further. A square root, which needs a special algorithm, has become a sequence of multiplications and additions.

Why it mattered

Until then, infinite series had been curiosities. Newton made them a tool: any expression with roots or fractions could be swapped for a sum of powers, and powers are easy to differentiate and integrate. That is how he computed areas under curves that nobody knew how to measure, as we will see in the post on areas.

Newton only explained the method publicly in 1676, in two letters sent to Leibniz through Henry Oldenburg, Secretary of the Royal Society. Today the same formula appears as a special case of the Taylor series.

Exercise: approximate √1.04 with three terms

With $x=0.04$: $1+0.02-\dfrac{0.0016}{8}=1+0.02-0.0002=1.0198$. The next term, $x^3/16$, is only $0.000004$; the true value is $1.019804\ldots$

Exercise: write the first three terms of 1/√(1+x)

Here $\alpha=-\frac12$. The coefficients are $1$, $-\frac12$ and $\dfrac{(-\frac12)(-\frac32)}{2}=\frac38$. So $\dfrac{1}{\sqrt{1+x}}\approx1-\dfrac{x}{2}+\dfrac{3x^2}{8}$.

Sources: mathematical manuscripts at the Newton Project · MacTutor biography.

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