Great Formulas: the Binomial Series
In learning algebra, we all encounter the Binomial Theorem. As you probably recall, it shows us how to raise a binomial to a power. For example,
\[ (1+x)^6 = 1 + 6x + 15x^2 + 20 x^3 + 16 x^4 + 6 x^5 + x^6\tag{1}\]
We learn how to write out the expanded form without doing all the tedious multiplication and combining of like terms. The number of terms in the expanded form, 7 in this case, is one more than the exponent.
The textbook gives us a formula for directly computing each coefficient. Suppose that our exponent is \(n\), and that we number the terms of the expansion from \(k=0\) through \(k=n\). Then, the coefficient for term \(k\) is:
\[{{n}\choose{k}}=\frac{n!}{k!(n-k)!}\tag{2}\]
These values are called Binomial Coefficients. Although defined as a fraction, the actual value comes out to be an integer. As an example, in equation (1), we have \( n=6 \). The 4th coefficient (\(k=3\)) is 20, and we can compute it as
\[ {{6}\choose{3}}=\frac{6!}{3!(6-3)!} = \frac{720}{6 \cdot 6} = 20\tag{3}\]
That is basically the binomial theorem as taught in algebra classes. For any binomial raised to a power, formula (2) lets us compute the \(n+1\) coefficients of the expansion. Obviously, the exponent must be a positive integer, else (2) does not work.
The Binomial Series, a Brilliant Generalization
The Binomial Theorem had been known for about 600 years when Isaac Newton, age 22, had a look at it in 1664. Newton believed in patterns in mathematics, and he suspected there was a way to make the theorem work for any exponent, not just positive integers. For example, there should be a way to expand \(\sqrt{1+x}=(1+x)^{1/2}\), even though the exponent is a fraction.
Newton’s first problem was that formula (2) cannot be used for the coefficients in this case because it would require us to evaluate \(\left(1 / 2\right)!\), and we don’t know how to do that. However, Newton readily saw that he could rewrite (2) without the troublesome factorials. The term \((n-k)!\) in the denominator can be divided into the \(n!\) in the numerator, leaving \(n (n-1)\cdots (n-k+1)\). That is,
\[{{n}\choose{k}}=\frac{n (n-1)\cdots (n-k+1)}{k!}\tag{4}\]
Computed this way, the coefficient in (3) above comes out the same as before:
\[ {{6} \choose{3}}=\frac{6(6-1)(6-2)}{3!}=20 \]
One elegant thing about formula (4): if we try using it to generate “extra” coefficients, they will be zero. For example,
\[ {{6} \choose{7}}=\frac{6(6-1)(6-2)(6-3)(6-4)(6-5)(6-6)}{7!}=0\]
All such coefficients will have a zero in the numerator, making the result zero. In a sense, we don’t have to define the number of terms, since everything is zero after \(n+1\) terms. Finally, with this new formula, we need to define the first value to be \({{n} \choose{0}}=1\).
Now there was nothing to stop Newton from letting the exponent \(n\) be anything, so he tried,
\[ \begin{align*} (1+x)^{\frac{1}{2}}& =\\&{{\frac{1}{2}} \choose{0}} x^0 + {{\frac{1}{2}} \choose{1}} x^1 + {{\frac{1}{2}} \choose{2}} x^2 + {{\frac{1}{2}} \choose{3}} x^3 + {{\frac{1}{2}} \choose{4}} x^4 + \cdots \\&= 1+\frac{\frac{1}{2}}{1!}x+\frac{\frac{1}{2}\left(\frac{1}{2} - 1\right)}{2!}x^2+\frac{\frac{1}{2}\left(\frac{1}{2} - 1\right) \left(\frac{1}{2}-2\right)}{3!}x^3+\\ &\frac{\frac{1}{2}\left(\frac{1}{2} - 1\right) \left(\frac{1}{2}-2\right) \left(\frac{1}{2}-3\right)}{4!}x^4+\cdots\\&= 1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16} x^3 - \frac{5}{128} x^4 + \cdots\tag{5} \end{align*} \]
No zero ever appears in the numerator for the coefficients, so the number of terms is endless!
But is the infinite series on the right really the same as \(\sqrt{1+x}\)? It was easy for Newton to find out – if the series is equivalent to \(\sqrt{1+x}\), then its square should be \(1+x\). So, he evaluated
\[ \left( 1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16} - \frac{5}{128} x^4 \cdots \right) \left( 1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16} - \frac{5}{128} x^4 \cdots \right) \tag{6}\]
He found that almost every positive term is canceled by an equal negative term. The only terms left uncanceled were \(1\) and \(x\), as required.
Thus encouraged, he tried a negative exponent by expanding
\[\frac{1}{(1+x)^3}=\left(1+x\right)^{-3} \]
The process is the same, and again, the number of terms is infinite:
\[ \begin{align*} \left(1+x\right)^{-3} = &1+\frac{-3}{1!}x+\frac{(-3)(-4)}{2!}x^2+\frac{(-3)(-4)(-5)}{3!} x^3+\cdots \\=& 1-3x+6x^2-10x^3+15x^4 - \cdots \end{align*} \]
If this expansion is valid, we are saying that
\[\frac{1}{(1+x)^3} = \frac{1}{1+3x+3x^2+x^3} = 1-3x+6x^2-10x^3+15x^4 - \cdots \]
To test this result, Newton just cross-multiplied:
\[ (1+3x+3x^2+x^3)(1-3x+6x^2-10x^3+15x^4 - \cdots ) \tag{7} \]
Again, there is wholesale cancellation. The only thing left is 1, as required.
Newton went on to prove the general result, though there is one very important condition: for the infinite series to converge, the absolute value of \(x\) must be less than 1. This is reasonable, because if x is greater than 1, each succeeding term is larger, not smaller, and the sum would be infinite.
In summary, Newton’s procedure for expanding a binomial is this:
\[ (1+x)^y = 1+ \frac{y}{1!} x + \frac{y(y-1)}{2!} x^2 + \frac{y(y-1)(y-3)}{3!} x^3 + \cdots \tag{8}\]
If y is a positive integer, the series terminates. This is usually called the Binomial Series formula, and it’s discovery was a landmark in mathematics. Among other things, it showed that a function can have an alternate existence as an infinite series. The infinite series is often easier to work with than is the original function.
Basic algebra classes don’t teach the Binomial Series, even though it is as easy to describe as the less general Binomial Theorem. I assume that the textbook writers think that delicate young minds are not ready for equations with an infinite number of terms, and perhaps they are right.
In the next post, I will go through three examples of what can be done with this fabulous formula.
Note: All my binomials have been of the form \((1 + x)\). That’s not really restrictive; we can convert any binomial to that form. Suppose we have \( (a + b)^n\). We can factor out the \(a\), and write
\[ (a+b)^n = \left[ a \left(1+\frac{b}{a} \right) \right]^n =a^n \left( 1 + \frac{b}{a} \right)^n \]
If we let \(b / a = x\), then we just have
\[ (a+b)^n = a^n (1 + x)^n \]
The factor \(a^n \) presents no problem, so we can just concentrate on the \( (1+x) \) factor.
References:
There is a good review of the Binomial Theorem at http://www.themathpage.com/aprecalc/binomial-theorem.htm
For an explanation of the Binomial Theorem in video form, watch Sal Khan go through it: http://www.khanacademy.org/math/precalculus/v/binomial-theorem–part-1
Journey through Genius: The Great Theorems of Mathematics by William Dunham. Dunham includes the Binomial Series as part of Chapter 7.
Tags: Math
I'm Larry Phillips, a former engineer, programmer, math teacher, math /physics tutor, and currently owner of a tutoring company. I'm on a mission to show that math is more interesting than the schools made you think it was.