Newton and Wallis

In January 1663/4 Newton dated his extracts from Wallis’s procedures for squaring the parabola and hyperbola by means of indivisibles; a little later he was annotating Wallis’s treatment of series in Arithmetica infinitorum and applying these procedures also to quadratures.⁴ All the time he was thinking and inventing, for example: “Thus Wallis doth it, but it may bee done thus . . .”⁵ The study of series brought him to the problem of how to interpolate terms. The point of interpolation – that is, finding the value of a new term, inserted between other terms in a series whose values were already established – was that by this method Newton knew he would be able to ‘square’ the general sector of the circle, for example, or the segment under an hyperbola, in terms of the two limiting ordinates. In algebra, Wallis had got so far as discovering the limits of π between two infinitely extendable series.⁶ He had also established generally that the area under any curve y = xⁿ was

$$ \frac{x^{n+1}}{n+1} $$

Now the equation for the circle is y = (1 − x²)^(1/2) which can be put in Wallis’s form as y = z^(1/2). Wallis did not know how to deal with this fractional exponent, but Newton now supplied the answer by interpolation. Following Wallis, Newton wrote down a regular arithmetical progression thus:

1. □ · $\frac{2}{3}$ · \* · $\frac{8}{15}$ · \* · $\frac{48}{105}$ · \* · $\frac{384}{945}$

“’Twixt which termes \[he continued] if the intermediate termes □ \[and] \* can bee found the 2d □ will give the area of the line y = √aa − xx , the circle.”⁷

In effect, working (as Wallis, Pascal and others had done before) with arrays of tabulated values of coefficients, Newton arrived at a version of the bounding series used by Wallis:

$$ \frac12 \times \frac24 \times \frac36 \times \frac58 \times \frac710 \times \frac912 \times \frac{11}{14} \times \ldots $$

Multiplying each successive fraction into the preceding product yields another series of numbers:

1, $\frac12$, $\frac18$, $\frac1{16}$, $\frac5{128}$, $\frac7{256}$, $\frac{21}{1024}$, $\frac{33}{2048}$

Applying these fractions as coefficients to the series of even powers Newton produced the equivalence:

$$ \frac{x}{2} \sqrt{1 - x^2} = \frac{x}{2} - \frac{x^3}{4} - \frac{x^5}{16} - \frac{x^7}{32} - \frac{5x^9}{256} - \ldots $$

From this expression the quadrature of the circle can be derived, the number of terms being extendable to the limits of the calculator’s patience. Newton worked out similar power series for sin⁻¹x and the area under the hyperbola, computing the latter at Boothby Pagnell to 51 decimal places.

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Please explain Newton’s discovery in the text above.

Here’s what’s going on, in plain terms:

• Starting point (Wallis): John Wallis had shown how to “square” (i.e., find areas under) curves of the form $y=x^n$ for whole-number $n$, giving the rule

  $$ \int_0^x t^n\,dt=\frac{x^{n+1}}{n+1}. $$

  He also produced product/series tables that bounded numbers like $\pi$, but he couldn’t yet handle curves with fractional powers like $y=\sqrt{1-x^2}$ (the circle).

• Newton’s idea (interpolation): Newton looked at Wallis’s arrays of coefficients (those strings of fractions) and asked: if we “fill in” the missing intermediate entries so the pattern works not just for integers but also for half-integers, what must the coefficients be? This “filling in” is what he called interpolation. Concretely, he used the regular product pattern

  $$ \frac12\times\frac24\times\frac36\times\frac58\times\frac7{10}\times\cdots $$

  to generate a sequence of coefficients

  $$ 1,\;\frac12,\;\frac18,\;\frac1{16},\;\frac5{128},\;\frac7{256},\ldots $$

  which he then attached to the even powers of $x$.

• What this unlocked: By doing that, Newton effectively discovered the binomial series for fractional exponents:

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

  (valid for $|u|<1$). Taking $\alpha=\tfrac12$ and $u=-x^2$ gives a power series for the circle’s ordinate:

  $$ \sqrt{1-x^2}=1-\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^6}{16}-\frac{5x^8}{128}-\cdots $$

  Multiplying by $x/2$ (a convenient step for segment/sector areas) yields the series you quoted:

  $$ \frac{x}{2}\sqrt{1-x^2} =\frac{x}{2}-\frac{x^3}{4}-\frac{x^5}{16}-\frac{x^7}{32}-\frac{5x^9}{256}-\cdots $$

• Why that matters (quadrature): Once you have a power series for the curve, you can integrate term-by-term to get areas (quadratures). For the circle, integrating from $0$ to a given $x$ gives the area of a circular segment/sector; integrating all the way to $x=1$ gives the area of a quarter circle (hence a route to $\pi$). Newton applied the same trick to other curves—e.g., the hyperbola—producing the familiar logarithm series, and to inverse trig functions like $\sin^{-1}x$. He pushed these computations astonishingly far (the text notes 51 decimal places for a hyperbolic area).

In one sentence: Newton took Wallis’s patterns and, by interpolating their missing terms, generalized them to fractional exponents—essentially discovering the binomial series—so he could expand curves like $\sqrt{1-x^2}$ into power series and then integrate those series to compute areas with arbitrary precision. This is one of the key early steps toward his calculus of power series.

Why does this work?

Great question. The short answer is: Newton noticed the rule that generates binomial coefficients still makes sense if you let the exponent be a fraction, and the resulting power series converges to the right function on $|x|<1$. That’s why his “interpolated” coefficients actually work.

Here’s the clean modern view:

1. Extend the binomial coefficients. For integers $m$,

$$ (1+u)^m=\sum_{k=0}^m \binom{m}{k}u^k,\qquad \frac{\binom{m}{k+1}}{\binom{m}{k}}=\frac{m-k}{k+1}. $$

Newton keeps the same recurrence law for any exponent $\alpha$ (this is his “interpolation”):

$$ c_0=1,\qquad c_{k+1}=c_k\,\frac{\alpha-k}{k+1}. $$

That uniquely gives

$$ c_k=\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}. $$

For $\alpha=\tfrac12$ these $c_k$ are exactly the numbers he tabulates.

2. Show the series equals $(1+u)^\alpha$. Define $F(u)=\sum_{k\ge0}\binom{\alpha}{k}u^k$. Differentiate term-by-term (justified below):

$$ F'(u)=\sum_{k\ge1}k\binom{\alpha}{k}u^{k-1} =\sum_{k\ge0}(\alpha-k)\binom{\alpha}{k}u^{k}/(1+u). $$

Using the recurrence, this simplifies to the differential equation

$$ (1+u)F'(u)=\alpha F(u),\quad F(0)=1, $$

whose unique solution is $F(u)=(1+u)^\alpha$. So the “interpolated” series is exactly the function with a fractional exponent.

3. Convergence (why you can integrate term-by-term). For fixed $\alpha$,

$$ \left|\frac{\binom{\alpha}{k+1}u^{k+1}}{\binom{\alpha}{k}u^k}\right| =\left|\frac{\alpha-k}{k+1}\right||u|\;\longrightarrow\;|u| \quad(k\to\infty), $$

so the radius of convergence is $1$. On any $|u|\le r<1$ the series converges uniformly, hence you may integrate term-by-term. With $u=-x^2$ and $\alpha=\tfrac12$,

$$ \sqrt{1-x^2}=\sum_{k=0}^\infty \binom{1/2}{k}(-x^2)^k =1-\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^6}{16}-\frac{5x^8}{128}-\cdots, $$

and multiplying by $x/2$ or integrating gives the circular segment/sector areas (the “quadrature”).

4. Why Wallis’s products show up. Newton’s fractions come from the identity

$$ \Bigg|\binom{1/2}{k}\Bigg| =\frac{1\cdot3\cdot5\cdots(2k-1)}{2\cdot4\cdot6\cdots(2k)}\;\frac{1}{2k-1} =\frac{\binom{2k}{k}}{4^k(2k-1)}, $$

i.e., the odd-over-even product (Wallis pattern) determines the next coefficient. That’s exactly the pattern he was “interpolating.”

So it works because Newton preserved the multiplicative law for binomial coefficients, which defines a legitimate power series; that series provably equals the target function on $|x|<1$ and can be integrated term-by-term to get the desired areas.