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# How Isaac Newton Discovered the Binomial Power Series

Isaac Newton was not known for his generosity of spirit, and his disdain for his rivals was legendary. But in one letter to his competitor Gottfried Leibniz, now known as the *Epistola Posterior*, Newton comes off as nostalgic and almost friendly. In it, he tells a story from his student days, when he was just beginning to learn mathematics. He recounts how he made a major discovery equating areas under curves with infinite sums by a process of guessing and checking. His reasoning in the letter is so charming and accessible, it reminds me of the pattern-guessing games little kids like to play.

It all began when young Newton read John Wallis’ *Arithmetica Infinitorum,* a seminal work of 17th-century math. Wallis included a novel and inductive method of determining the value of pi, and Newton wanted to devise something similar. He started with the problem of finding the area of a “circular segment” of adjustable width *$latex x$*. This is the region under the unit circle, defined by $latex y=\sqrt{1-x^2}$, that lies above the portion of the horizontal axis from 0 to *$latex x$*. Here *$latex x$* could be any number from 0 to 1, and 1 is the radius of the circle. The area of a unit circle is pi, as Newton well knew, so when *$latex x=1$*, the area under the curve is a quarter of the unit circle, $latex\frac{π}{4}$. But for other values of *$latex x$*, nothing was known.

If Newton could find a way to determine the area under the curve for every possible value of *$latex x$**,* it might give him an unprecedented means of approximating pi. That was originally his grand plan. But along the way he found something even better: a method for replacing complicated curves with infinite sums of simpler building blocks made of powers of *$latex x$*.

Newton’s first step was to reason by analogy. Instead of aiming directly for the area of the circular segment, he investigated the areas of analogous segments bounded by the following curves:

$latex y_0=(1-x^2)^\frac{0}{2}$,

$latex y_1=(1-x^2)^\frac{1}{2}$,

$latex y_2=(1-x^2)^\frac{2}{2}$,

$latex y_3=(1-x^2)^\frac{3}{2}$,

$latex y_4=(1-x^2)^\frac{4}{2}$,

$latex y_5=(1-x^2)^\frac{5}{2}$,

$latex y_6=(1-x^2)^\frac{6}{2}$.

Newton knew that the areas under the curves in the list with whole-number powers (like $latex \frac{0}{2}=0$ and $latex \frac{2}{2} = 1$) would be easy to calculate, because they simplify algebraically. For example,

$latex y_0=(1-x^2)^\frac{0}{2}=(1-x^2)^0=1$.

Similarly,

But no such simplification is available for the circle’s equation — $latex y_1 = \sqrt {1-x^2}=(1-x^2)^\frac{1}{2}$— or the other curves with the half powers. At the time, no one knew how to find the area under any of them.

Fortunately, the areas under the curves with whole-number powers were straightforward. Take the curve $latex y_4=1-2x^2+x^4$. A well-known rule at the time for such functions allowed Newton (and anyone else) to find the area quickly: For any whole-number power $latex n\ge 0$, the area under the curve $latex y=x^n$ over the interval from *$latex 0$* to *$latex x$* is given by $latex \frac{x^{n+1}}{n+1}$. (Wallis had guessed this rule with his inductive method, and Pierre de Fermat proved it conclusively.) Armed with this rule, Newton knew that the area under the curve $latex y_4$ was $latex x- \frac{2x^3}{3} + \frac{x^5}{5}$.

The same rule allowed him to find the area under the other curves with whole-number powers in the list above. Let’s write $latex A_n$ for the area under the curve $latex y_n = (1-x^2)^\frac{n}{2}$, where $latex n= 0, 1, 2, …$ . Applying the rule yields

$latex A_0=x$

$latex A_1 = \hspace{.295em}?$

$latex A_2 = x -\frac{1}{3}x^3$

$latex A_3 = \hspace{.295em}?$

$latex A_4 = x -\frac{2}{3}x^3 + \frac{1}{5}x^5$

$latex A_5 =\hspace{.295em}? $

$latex A_6 = x -\frac{3}{3}x^3 + \frac{3}{5}x^5 – \frac{1}{7}x^7$

and so on. Newton’s crafty idea was to fill in the gaps, hoping to guess $latexA_1$ (the series for the unknown area of the circular segment) based on what he could see in the other series. One thing was immediately clear: Each $latexA_n$ began simply with $latex x$ . That suggested amending the formulas like so:

$latex A_0=x$

$latex A_1 = x\hspace{.247em}-\hspace{.247em}?$

$latex A_2 = x -\frac{1}{3}x^3$

$latex A_3 = x\hspace{.247em}-\hspace{.247em}?$

$latex A_4 = x -\frac{2}{3}x^3 + \frac{1}{5}x^5$

$latex A_5 = x\hspace{.247em}-\hspace{.247em}?$

$latex A_6 = x -\frac{3}{3}x^3 + \frac{3}{5}x^5 – \frac{1}{7}x^7$.

Then, to replace the next batch of question marks, Newton looked at the $latex x^3$ terms. With a little license, we can see that even $latexA_0$ had one of these cubic terms, since we can rewrite it as $latex A_0 = x-\frac{0}{3}x^3$. As Newton explained to Leibniz, he observed “that the second terms $latex \frac{0}{3}x^3, \frac{1}{3}x^3, \frac{2}{3}x^3, \frac{3}{3}x^3$ etc., were in arithmetical progression” (he was referring to the 0, 1, 2, 3 in the numerators). Suspecting that this arithmetic progression might extend into the gaps as well, Newton guessed that the entire sequence of numerators, known and unknown, ought to be numbers separated by $latex \frac{1}{2} (0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3 …)$ “and hence that the first two terms of the series” he was interested in — the still unknown $latex A_1$, $latex A_3$ and $latex A_5$ — “ought to be $latex x- \frac{1}{3}(\frac{1}{2}x^3), x-\frac{1}{3} (\frac {3}{2}x^3), x-\frac{1}{3}(\frac{5}{2}x^3)$, etc.”

Thus, at this stage the patterns suggested to Newton that $latex A_1$ should begin as

$latex A_1 = x-\frac{1}{3}(\frac{1}{2}x^3) + …$.

This was a good start, but he needed more. As he hunted for other patterns, Newton noticed that the denominators in the equations always contained odd numbers in increasing order. For instance, look at $latex A_6$, which has 1, 3, 5 and 7 in its denominators. That same pattern worked for $latex A_4$ and $latex A_2$. Simple enough. That pattern apparently persisted in all the denominators of all the equations.

What remained was to find a pattern in the numerators. Newton examined $latex A_2$, $latex A_4$ and $latex A_6$ again and spotted something. In $latex A_2 = x-\frac{1}{3}x^3$ he saw a 1 multiplying the $latex x$ and another 1 in the term $latex\frac {1}{3}x^3$ (he ignored its negative sign for the time being). In $latex A_4 = x-\frac{2}{3}x^3 + \frac{1}{5}x^5$, he saw numerators of 1, 2, 1. And in $latex A_6=x-\frac{3}{3}x^3 + \frac{3}{5}x^5 -\frac{1}{7}x^7$ , he saw numerators 1, 3, 3, 1. These numbers should be familiar to anyone who’s ever studied Pascal’s triangle, a triangular arrangement of numbers that, at its simplest, is created by adding together the numbers above it, starting with 1 at the top.

Instead of invoking Pascal, Newton referred to these numerators as “powers of the number 11.” For example, 11^{2} = 121, which is the second row in the triangle, and 11^{3} = 1331, which is the third. Nowadays these numbers are also called binomial coefficients. They arise when you expand the powers of a binomial like ($latex a +b$), as in $latex (a+b)^3 = 1a^3 + 3a^2b+3ab^2 +1b^3$. With this pattern in hand, Newton now had an easy way of writing out $latex A_2, A_4, A_6$, and all the other even-numbered *A*’s.

Next, to extrapolate his results to half-powers and odd-numbered subscripts (and finally get to the series he wanted, $latex A_1$), Newton needed to extend Pascal’s triangle to a fantastic new regime: halfway in between the rows. To perform the extrapolation, he derived a general formula for the binomial coefficients in any given row of Pascal’s triangle — row $latex m$ — and then audaciously plugged in $latex m= \frac{1}{2}$. And amazingly, it worked. That gave him the numerators in the series he was seeking for a unit circle, $latexA_1$.

Here, in Newton’s own words, is his summary to Leibniz of the patterns he noticed inductively up to this stage in the argument:

I began to reflect that the denominators 1, 3, 5, 7, etc. were in arithmetical progression, so that the numerical coefficients of the numerators only were still in need of investigation. But in the alternately given areas, these were the figures of powers of the number 11 … that is, first ‘1’; then ‘1, 1’; thirdly, ‘1, 2, 1’; fourthly ‘1, 3, 3, 1’; fifthly ‘1, 4, 6, 4, 1’ etc. and so I began to inquire how the remaining figures in the series could be derived from the first two given figures, and I found that on putting $latex m$ for the second figure, the rest would be produced by continual multiplication of the terms of this series,

$latex \frac{m-0}{1} \times \frac{m-1}{2} \times \frac {m-2}{3} \times \frac{m-3}{4} \times \frac {m-4}{5}$, etc.

… Accordingly I applied this rule for interposing series among series, and since, for the circle, the second term was $latex \frac{1}{3}(\frac{1}{2}x^3)$, I put $latex m=\frac{1}{2}$, and the terms arising were

$latex \frac {1}{2} \times \frac{\frac{1}{2}-1}{2}$ or $latex -\frac{1}{8}$,

$latex -\frac{1}{8} \times \frac{\frac{1}{2}-2}{3}$ or $latex + \frac{1}{16}$,

$latex \frac{1}{16} \times \frac{\frac{1}{2}-3}{4}$ or $latex – \frac {5}{128}$,so to infinity. Whence I came to understand that the area of the circular segment which I wanted was

$latex x-\frac{\frac{1}{2}x^3}{3}-\frac{\frac{1}{8}x^5}{5}-\frac{\frac{1}{16}x^7}{7}-\frac{\frac{5}{128}x^9}{9}$ etc.

Finally, by plugging in $latex x=1$, Newton could obtain an infinite sum for $latex\frac{π}{4}$. It was an important finding, but it turns out there are better ways to approximate pi by means of an infinite sum, as Newton himself soon discovered after this initial foray into these kinds of infinite sums, now called power series. Eventually he calculated the first 15 digits of pi.

Returning to the problem of the circular segment, Newton realized that the equation for the circle itself (not merely the area underneath it) could also be represented by a power series. All he needed to do was omit the denominators and reduce the powers of $latex x$ by 1 in the power series displayed above. Thus he was led to guess that

To test whether this result made sense, Newton multiplied it by itself: “It became $latex 1-x^2$, the remaining terms vanishing by the continuation of the series to infinity.”

Stepping back a bit from the details, we see several lessons here about problem-solving. If a problem is too hard, change it. If it seems too specific, generalize it. Newton did both and got results more important and more powerful than what he originally sought.

Newton didn’t stubbornly fixate on a quarter of a circle. He looked at a much more general shape, any circular segment of width $latex x$. Rather than sticking to $latex x=1$, he allowed $latex x$ to run freely from 0 to 1. That revealed the binomial character of the coefficients in his series — the unexpected appearance of numbers in Pascal’s triangle and their generalizations — which let Newton see patterns that Wallis and others had missed. Seeing those patterns then gave Newton the insights he needed to develop the theory of power series much more widely and generally.

In his later work, Newton’s power series gave him a Swiss Army knife for calculus. With them, he could do integrals, find roots of algebraic equations, and calculate the values of sines, cosines and logarithms. As he put it, “By their help, analysis reaches, I might almost say, to all problems.”

The moral: Changing a problem is not cheating. It’s creative. And it may be the key to something greater.