3.3.1 Maclaurin Series

Maclaurin Series

Maclaurin hypothesized that every function could be written as an infinite-order polynomial.

He assumed for any $f(x)$ :

f(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots + c_n x^n + \cdots

Assuming this is true, to find the values of the constants $c_n$ , we do the following:

$f(0) = c_0$
$f'(x) = c_1 + 2c_2 x + 3c_3 x^2 + 4c_4 x^3 + 5c_5 x^4 + \cdots \Rightarrow f'(0) = c_1$
$f''(x) = 2c_2 + 6c_3 x + 12c_4 x^2 + 20c_5 x^3 + \cdots \Rightarrow f''(0) = 2c_2 \Rightarrow c_2 = \frac{f''(0)}{2}$
$f'''(x) = 6c_3 + 24c_4 x + 60c_5 x^2 + \cdots \Rightarrow f'''(0) = 6c_3 \Rightarrow c_3 = \frac{f'''(0)}{6}$
$f''''(x) = 24c_4 + 120c_5 x + \cdots \Rightarrow f''''(0) = 24c_4 \Rightarrow c_4 = \frac{f''''(0)}{24}$

\therefore \text {in general }\ c_n = \frac{f^{(n)}(0)}{n!}

Therefore.

f(x) = 1 + f'(0)x + \frac{f''(0)x^2}{2} + \frac{f'''(0)x^3}{6} + \cdots + \frac{f^{(n)}(0)x^n}{n!} + \cdots

Or equivalently:

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n

lightbulbExample

Example: Find the first three terms of the Maclaurin expansion of $f(x) = e^x$ :

f(0) = e^0 = 1

f'(x) = e^x \Rightarrow f'(0) = 1

f''(x) = e^x \Rightarrow f''(0) = 1

Thus:

f(x) \approx 1 + \frac{1 \times x}{1!} + \frac{1 \times x^2}{2!} = 1 + x + \frac{x^2}{2}

lightbulbExample

Example: Use the above example to find the expansion up to and including $x^2$ of $e^{1 - x^2}$ :

Let $x = 1 - x^2$

f(1 - x^2) \approx 1 + (1 - x^2) + \frac{(1 - x^2)^2}{2}

= 1 + 1 - x^2 + \frac{1 - 2x^2 + \cancel{x^4}}{2}

= 1 + 1 - x^2 + \frac{1 - 2x^2}{2}

= \frac{5}{2} - 2x^2

In the formula booklet, we are given that:

\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{(-1)^{r+1} x^r}{r} + \cdots

lightbulbExample

Example: Use this expansion to expand $ln(\color {red}2\color {black}+x)$ up to and including the $x^3$ term.

$\color {red}2$ is not of the form $\ln(1 + \square)$ , so we need to rearrange to get it in this form.

\ln(2 + x) = \ln\left[2\left(1 + \frac{x}{2}\right)\right]

= \ln 2 + \ln\left(1 + \frac{1}{2}x\right) \quad (\text{using } \log(ab) = \log a + \log b)

= \ln 2 + \left(\frac{1}{2}x\right) + \frac{\left(\frac{1}{2}x\right)^2}{2} + \frac{\left(\frac{1}{2}x\right)^3}{3} + \cdots

= \ln 2 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{24}x^3 + \cdots

infoNote

Use the given expansion of $e^x$ to find the expansion of $e^{e^x}$ up to and including the $x^2$ term.

e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots

Write as a product of known expansions:

e^{e^x} = e^{1 + x + \frac{x^2}{2} + \dots} \approx e e^x e^{\frac{x^2}{2}}

= e \left(1 + x + \frac{x^2}{2}\right) \left(1 + \frac{x^2}{2} + \cancel{\frac{\left(\frac{x^2}{2}\right)^2}{2!}} + \dots \right)

ignoring powers greater than 2

= \left(e + ex + \frac{ex^2}{2}\right)\left(1 + \frac{x^2}{2}\right)

= e + ex + \frac{ex^2}{2} + \frac{ex^2}{2} +\cancel {\frac{ex^3}{2}} +\cancel {\frac{ex^4}{4!}}

(ignoring $x^4$ and higher terms)

= e + ex + ex^2