5.23 General of Binomial Theorem HL ONLY

Consider the infinite geometric series

$1 + x + x^2 + x^3 + \dots$

We know that the series converges for $-1 < x < 1$ and the result is $S_{\infty} = \dfrac{1}{1-x}$.

In this paragraph we have the opposite task: We are given a function, say $f(x) = \dfrac{1}{1-x}$, and we wish to express it as an infinite series (which is called power series) of the form

$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots$

(the power series looks like a polynomial of "infinite degree").

1. The Maclaurin Series

Suppose that a function $f(x)$ has derivatives of every order near $0$. Then $f(x)$ can be expressed as a power series as follows

$f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 + \dots$

This is known as Maclaurin series of the function.

For example, if $f(x) = \dfrac{1}{1-x}$ then

  • $f(x) = (1-x)^{-1} \implies f(0) = 1$
  • $f'(x) = (1-x)^{-2} \implies f'(0) = 1$
  • $f''(x) = 2(1-x)^{-3} \implies f''(0) = 2$
  • $f'''(x) = 6(1-x)^{-4} \implies f'''(0) = 6$
  • and in general $f^{(n)}(x) = n!(1-x)^{-n-1} \implies f^{(n)}(0) = n!$

Thus,

$$ \begin{aligned} f(x) &= 1 + 1x + \dfrac{2!}{2!}x^2 + \dfrac{3!}{3!}x^3 + \dots \\ &= 1 + x + x^2 + x^3 + \dots \end{aligned} $$

as we already know.

Explanation of the Maclaurin series:

We wish to express $f(x)$ as
$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \dots$

Then
$f'(x) = a_1 + 2a_2 x + 3a_3 x^2 + 4a_4 x^3 + \dots$
$f''(x) = 2a_2 + (2)(3)a_3 x + (3)(4)a_4 x^2 + \dots$
$f'''(x) = 3! a_3 + 4! a_4 x + \dots$
etc

Therefore,
$f(0) = a_0 \implies a_0 = f(0)$
$f'(0) = a_1 \implies a_1 = f'(0)$
$f''(0) = 2a_2 \implies a_2 = \dfrac{f''(0)}{2} = \dfrac{f''(0)}{2!}$
$f'''(0) = 3! a_3 \implies a_3 = \dfrac{f'''(0)}{3!}$
etc

In general $\mathbf{a_n = \dfrac{f^{(n)}(0)}{n!}}$

In fact, the partial sums of the Maclaurin series give good approximations of $f(x)$ near $x=0$:

  • $a_0 + a_1 x$ is the tangent line of $f(x)$ at $x=0$
  • $a_0 + a_1 x + a_2 x^2$ is the "best" quadratic that approximates $f(x)$
  • $a_0 + a_1 x + a_2 x^2 + a_3 x^3$ is the "best" cubic that approximates $f(x)$

For $f(x) = \dfrac{1}{1-x}$ (black curve) look at the approximations below:

x y 0 -1 1 1 x=1

$y = 1+x$   |   $y = 1+x+x^2$   |   $y = 1+x+x^2+x^3$

2. Generating New Series

An alternative way to obtain a Maclaurin series is to modify or combine appropriately the already known series.
We can add, multiply, differentiate or integrate series and generate new series.

EXAMPLE 2

Find the Maclaurin series of the function $f(x) = e^{x^2}$

Solution:
Since $$ e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dots $$ we can substitute $x$ by $x^2$ and obtain $$ \mathbf{e^{x^2} = 1 + x^2 + \dfrac{x^4}{2!} + \dfrac{x^6}{3!} + \dots} $$

EXAMPLE 3

Find the Maclaurin series of the function $f(x) = e^x \sin x$

Solution:
We can multiply the series for $e^x$ and $\sin x$. $$ e^x \sin x = \left(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dots\right) \left(x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dots\right) $$ We can find gradually the constant term, the terms in $x$, the terms in $x^2$ etc: $$ \begin{aligned} e^x \sin x &= x + x^2 + \left(-\dfrac{1}{3!} + \dfrac{1}{2!}\right)x^3 + \left(-\dfrac{1}{3!} + \dfrac{1}{3!}\right)x^4 + \dots \\ &= \mathbf{x + x^2 + \dfrac{x^3}{3} + \dots} \quad [\text{there is no } x^4] \end{aligned} $$

Notice also:

  • if we differentiate the series for $\sin x$ (term by term) we obtain the series of $\cos x$.
  • If we differentiate the series for $e^x$ we obtain $e^x$ itself.

In general, we are able to differentiate a series term by term.

EXAMPLE 4

Find the Maclaurin series of the function $f(x) = \dfrac{1}{1+x^2}$

Solution:
Since $$ \arctan x = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \dots $$ Since $\left(\arctan x\right)' = \dfrac{1}{1+x^2}$ $$ \dfrac{1}{1+x^2} = \left(x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \dots\right)' $$ Thus, $$ \mathbf{\dfrac{1}{1+x^2} = 1 - x^2 + x^4 - \dots} $$

3. Differential Equations and Maclaurin Series

Consider a differential equation of the form

$\dfrac{dy}{dx} = F(x,y)$

with boundary condition $y(0)=y_0$

The analytical solution is not always easy or sometimes not possible at all. However, we can easily find the Maclaurin series of the solution $y=f(x)$ (and thus a good approximation of the solution).

$f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 + \dots$

But,

  • $f(0) = y_0$ [by the boundary condition]
  • $f'(0) = \left. \dfrac{dy}{dx} \right|_{x=0, y=y_0} = F(0, y_0)$ [by the D.E. itself]
  • Implicit differentiation on $\dfrac{dy}{dx}$ gives $\dfrac{d^2y}{dx^2}$ and thus $f''(0)$, and so on.

EXAMPLE 6

Find the Maclaurin series up to $x^2$ for the solution of the D.E. $\dfrac{dy}{dx} = x^2 + y^2$ with $y=3$ when $x=0$

Solution:
Clearly $f(0) = 3$
$$ f'(0) = \left. \dfrac{dy}{dx} \right|_{x=0, y=3} = 0^2 + 3^2 = 9 $$ We differentiate implicitly to find the second derivative: $$ \dfrac{d^2y}{dx^2} = 2x + 2y \dfrac{dy}{dx} $$ thus $$ f''(0) = \left. \dfrac{d^2y}{dx^2} \right|_{x=0, y=3} = 2(0) + 2(3)(9) = 54 $$ Therefore $$ \begin{aligned} y &\approx 3 + 9x + \dfrac{54}{2!}x^2 \\ &= \mathbf{3 + 9x + 27x^2} \end{aligned} $$