Prelude to Taylor’s theorem

April 1, 2019April 3, 2019 / Michael Xue

As an application of derivative, we may prove the Binomial theorem that concerns the expansion of $(1+x)^n$ as a polynomial. Namely,

$(1+x)^n = 1+ a_1 x + a_2 x^2 + ... + a_n x^n\quad\quad\quad(1)$

where $a_k = \frac{n(n-1)(n-2)...(n-k+1)}{k!}, 1 \leq k \leq n, n \in N$ .

There are two steps:

Step 1) Prove $(1+x)^n$ can be expressed as a polynomial $1+a_1 x + a_2 x^2 + ... + a_n x^n$ , i.e.,

$(1+x)^n = 1 + \sum\limits_{i=1}^{n}a_i x^i$

where $a_k$ s are constants.

Step 2) Show that

$a_k = \frac{n(n-1)(n-2)...(n-k+1)}{k!}$

We use mathematical induction first.

When $n = 1$ ,

$(1+x)^1 = 1+x = 1 +a_1 x$

where $a_1 = 1$ .

Assume when $n=k, (1+x)^k$ is a polynomial:

$(1+x)^k = 1+b_1 x + b_2 x + ... +b_k x^k\quad\quad\quad(2)$

When $n = k+1$ ,

$(1+x)^{k+1} = (1+x)^k (1+x)$ .

By (2), it is

$(1+b_1 x+b_2 x^2+ ...+ b_k x^k) (1+x)$

$= (1+\sum\limits_{i=1}^{k} b_i x^i)(1+x)$

$= 1+ x + \sum\limits_{i=1}^{k}b_i x^i (1+x)$

$= 1+x + \sum\limits_{i=1}^{k}(b_i x^i +b_i x^{i+1})$

$= 1+x +\sum\limits_{i=1}^{k}b_i x^i + \sum\limits_{i=1}^{k} b_i x^{i+1}$

$= 1+x + (b_1 x + b_2 x^2 +... +b_k x^k) + (b_1 x^2 + ... + b_{k-1} x^k + b_k x^{k+1})$

$= 1+ (b_1+1)x + (b_2 + b_1) x^2 + ... + (b_k + b_{k-1}) x^k + b_k x^{k+1}$

$= 1 +a_1 x + a_2 x^2 + ... + a_k x^k + a_{k+1} x^{k+1}$

where $a_1 = b_1+1, a_2=b_2+b_1, ..., a_k = b_k + b_{k-1}, a_{k+1} = b_k$ .

Once (1) is established, we proceed to step 2) to construct $a_k$ :

From (1),

$\frac{d^k}{dx^k}(1+x)^n = \frac{d^k}{dx^k}(1+a_1 x + a_2 x^2 + ... + a_k x^k + ... +a_n x^n)$ .

That is

$n(n-1)(n-2)...(n-k+1)(1+x)^{n-k} = k(k-1)(k-2)...1 \cdot {a_k} + \sum\limits_{i=1}^{n-k} c_i x^i\quad\quad\quad(3)$

where $c_i$ s are constants.

Let $x = 0$ , (3) becomes

$n(n-1)(n-2)...(n-k+1) \cdot 1= k(k-1)(k-2)...1 \cdot {a_k} + 0$

i.e.,

$n(n-1)(n-2)...(n-k+1) = k!\;a_k\quad\quad\quad(4)$

Solving (4) for $a_k$ gives

$a_k = \frac{n(n-1)(n-2)...(n-k+1)}{k!}$ .

Since

$\frac{n(n-1)(n-2)...(n-k+1)}{k!} = \frac{n(n-1)(n-k+1)\boxed{(n-k)(n-k-1)...1}}{\boxed{(n-k)(n-k-1)...1}\;k!}=\frac{n!}{(n-k)!\;k!}$ ,

$a_k$ is often expressed as

$a_k = \frac{n!}{(n-k)!\;k!}$

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Connecting to %s

%d bloggers like this: