KindiakMath

Who Cares About Binomials?

As part of the school curriculum, students are taught the binomial theorem and use solve various meaningless problems. I even asked the following question out of pure curiosity: what is the expansion of (1+x)^5? But why should we even bother with such a question in the first place?

As a pure math student by training, I find no application of the binomial theorem at the secondary school level, except for the proof of one concept: derivatives. Let me use the examples n = 1,2,3 to illustrate.

Among other crucial ideas, one key component in differentiation is to evaluate the expression

\displaystyle \frac{(x+h)^n - x^n}{h}.

If we consider n = 1, 2, 3, then leave it as an exercise to verify the following simplifications:

\begin{aligned} \frac{(x+h)^1 - x^1}{h} &= \frac{(x+h)-x}{h} = 1, \\ \frac{(x+h)^2 - x^2}{h} &= \frac{(x^2 + 2xh + h^2) - x^2}{h} = 2x + h, \\ \frac{(x+h)^3 - x^3}{h} &= \frac{ x^3 + 3x^2h + 3xh^2 + h^3 }{ h } = 3x^2 + (3x + h) \cdot h. \end{aligned}

In the differentiation process, we would set h = 0 on the right-hand side and obtain the expressions 1, 2x, 3x^2 respectively. Continuing the pattern, we would expect that

\displaystyle \frac{(x+h)^n - x^n}{h} = nx^{n-1} + (\text{finite stuff}) \cdot h,

so that setting h = 0, we obtain the expression nx^{n-1}. This is better known as the derivative of the function x^n, but how can we know this information for sure?

It boils down to expanding the expression (x+h)^n, how do we do so?

Before we answer this question more completely, we remark that the binomial expansion does get extended to the cases when n is not a positive integer in various ways. To achieve that goal, however, would require us to explore some A-Level calculus, and we will delay that task for now.

The content in this post will mirror the discussion in this one, which ties in the binomial expansion with counting problems in probability, but perhaps with less technical terminology.

Firstly, by observation, the highest power of x in the expansion (1+x)^n is n. We see this observation play out in the expansions

(1+x)^2 = 1+ 2x + x^2,\quad (1+x)^3 = 1+ 3x + 3x^2 + x^3.

Therefore, we will suppose that for any k, there will be constants a_{k,0}, a_{k,1},\dots, a_{k,k} such that

(1+x)^k = a_{k,0} + a_{k,1} x + a_{k,2} x^2 + \cdots + a_{k,k} x^k.

How do we expand (1+x)^{k+1}? A power of (k+1) denotes multiplying the expression (1+x) a total of k times, so that expanding the expression,

\begin{aligned} (1+x)^{k+1} &= (1+x)(1+x)^k \\ &= (1+x)(a_{k,0} + a_{k,1} x + a_{k,2} x^2 + \cdots + a_{k,k} x^k) \\ &= (a_{k,0} + a_{k,1} x + a_{k,2} x^2 + \cdots + a_{k,k-1} x^{k-1} + a_{k,k} x^k) \\ &\phantom{====}+ (a_{k,0}x + a_{k,1} x^2 + a_{k,2} x^3 + \cdots + a_{k,k-1}x^k + a_{k,k} x^{k+1}) \\ &= \underbrace{ a_{k,0} }_{ a_{k+1,0} } + \underbrace{ (a_{k,1} + a_{k,0}) }_{ a_{k+1, 1} }x + (\underbrace{ a_{k,2} + a_{k,1} }_{a_{k+1,2}} )x^2 + \cdots + ( \underbrace{ a_{k,k} + a_{k,k-1} }_{ a_{k+1,k} } )x^k + \underbrace{ a_{k,k} }_{ a_{k+1,k+1} } x^{k+1} \\ &= a_{k+1, 0} + a_{k+1, 1} x + a_{k+1, 2} x^2 + \cdots + a_{k+1, k+1} x^{k+1}. \end{aligned}

Therefore, our expansion of (1+x)^{k+1} is given by the coefficients a_{k+1, 0} = a_{k, 0}, and for any 0 \leq i \leq k,

a_{k+1, i+1} = a_{k, i+1} + a_{k, i}.

For the very special case k = 1, (1+x)^1 = 1+ x = a_{1,0} + a_{1,1}x. Comparing coefficients, we set a_{1,0} = a_{1,1} = 1. We will declare a_{0,0} = 1 and a_{k, k+1} = 0 for convenience / convention / simplicity in formulas, so that for k = 0, a_{1,0} = a_{0,0} = 1, and

a_{1, 1} = a_{0, 1} + a_{0, 0} = 0 + 1 = 1,

as expected.

Lemma 1. For any k, a_{k, 0} = a_{k,k} = 1.

Proof. We have a_{0,0} = 1 and a_{k+1, 0} = a_{k,0} = \cdots = a_{0,0} = 1. Furthermore,

a_{k+1,k+1} = a_{k, k+1} + a_{k,k} = 0 + a_{k,k} = a_{k,k},

so that a_{k+1,k+1} = a_{k,k} = \cdots = a_{0,0} = 1.

Now working on the case k = 1,

\begin{aligned} a_{2,0} = 1, \quad a_{2,1} = a_{1,1} + a_{1,0} = 1+1 = 2, \quad a_{2,2} = 1. \end{aligned}

Hence,

\begin{aligned} (1+x)^2 &= a_{2,0} + a_{2,1} x + a_{2,2} x^2 = 1 + 2 x + x^2, \end{aligned}

as expected.

Remark 1. Since the numbers a_{k,i} are coefficients of the binomial (1+x)^k, they are unsurprisingly called the binomial coefficients, and often denoted \displaystyle a_{k, i} = {k \choose i}. It turns out that the formulas that define these numbers generate Pascal’s triangle.

Theorem 1 (Binomial Theorem). For real numbers a, b and any positive integer n,

\displaystyle (a+b)^n = a^n + {n \choose 1} a^{n-1} b + {n \choose 2} a^{n-2} b^2 + \cdots + b^n.

Furthermore, using ideas in counting problems (i.e. Theorem 4 of this post), for any 0 \leq k \leq n,

\displaystyle {n \choose k} = \frac{ n! }{ k! (n-k)! } = \frac{n \cdot (n-1) \cdot \cdots \cdot (n-k+2) \cdot (n-k+1)}{k \cdot (k-1) \cdot \cdots \cdot 2 \cdot 1},

where 0! := 1 for convenience / convention / simplicity.

Proof. The result is trivial if a = 0. Suppose a \neq 0. Setting a = 1, b = x, we have

\displaystyle (1+x)^n = 1 + {n \choose 1} x + {n \choose 2} x^2 + \cdots + {n \choose n-1} x^{n-1} + x^n.

For the general case, we first factorise to reduce to the special case:

\begin{aligned} (a+b)^n &= \left(a \left(1 + \frac ba \right) \right)^n = a^n \left(1 + \frac ba \right)^n . \end{aligned}

We then use the special case to obtain

\begin{aligned} a^n \left(1 + \frac ba \right)^n &= a^n \left( 1 + {n \choose 1} \cdot \frac ba + {n \choose 2} \frac {b^2}{a^2} + \cdots + {n \choose n-1} \frac { b^{n-1} }{ a^{n-1} } + \frac { b^n }{ a^n } \right) \\ &= a^n + {n \choose 1} \cdot a^n \cdot \frac ba + {n \choose 2} \cdot a^n \cdot \frac {b^2}{a^2} + \cdots + {n \choose n-1} \cdot a^n \cdot \frac { b^{n-1} }{ a^{n-1} } + a^n \cdot \frac { b^n }{ a^n } \\ &= a^n + {n \choose 1} a^{n-1} b + {n \choose 2} a^{n-2} b^2 + \cdots + {n \choose n-1} a b^{n-1} + b^n, \end{aligned}

as required.

Theorem 2. Given n \geq 3, setting h = 0 in the right-hand side of

\displaystyle \frac{(x+h)^n - x^n}{h}

yields nx^{n-1}.

Proof. Using the binomial theorem,

\displaystyle \frac{(x+h)^n - x^n}{h} = {n \choose 1} x^{n-1} + \left( {n \choose 2} x^{n-2} + \cdots + {n \choose n-1} x h^{n-2} + h^{n-1} \right) h.

Setting h = 0 in the right-hand side yields

\begin{aligned} & {n \choose 1} x^{n-1} + \left( {n \choose 2} x^{n-2} + \cdots + {n \choose n-1} x h^{n-2} + h^{n-1} \right) h \\ &\phantom{==} = nx^{n-1} + \left( {n \choose 2} x^{n-2} + \cdots + 0 + 0\right) \cdot 0 = nx^{n-1}.\end{aligned}

A slightly more interesting example arises in the expansion of (1+1/n)^n:

\displaystyle \begin{aligned} \left(1+ \frac 1n \right)^n &= 1 + {n \choose 1} \frac 1n + {n \choose 2} \frac 1{n^2} + \cdots + {n \choose n-1} \frac{1}{n^{n-1}} + \frac 1{n^n} \\ &= 1 + \frac n1 \cdot \frac 1n + \frac{n \cdot (n-1)}{2 \cdot 1} \cdot \frac 1{n^2} + \cdots + \frac{n \cdot (n-1) \cdot \cdots \cdot 2}{(n-1) \cdot \cdots \cdot 2 \cdot 1} \cdot\frac{1}{n^{n-1}} + \frac 1{n^n} \\ &\leq 1 + 1 + \frac 12 + \cdots + \frac{1}{(n-1) !} + \frac{1}{n!}. \end{aligned}

Coupling this calculation with several other high-level convergence ideas, we can derive the definition of e \approx 2.718. This number is crucial in helping us understand exponential functions when coupled with ideas in calculus.

We touch base with the idea of exponential functions next time. For a fully rigorous construction of these objects, see this post involving real analysis.

—Joel Kindiak, 27 Oct 25, 1827H

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