Calculus

Taylor and Maclaurin series

Writing a function as a power series makes approximations and indeterminate limits much easier to handle.

Basic, Standard, Advanced: University Year 1 · 1st semester

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Taylor and Maclaurin series

Formula

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots

The case $a = 0$ is the Maclaurin series. The coefficient of $x^n$ is $\frac{f^{(n)}(0)}{n!}$ , so from an expansion you can read off $f^{(n)}(0) = n! \times (\text{coefficient of } x^n)$ .

Basic expansions

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots,\qquad \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots,\qquad \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots

\log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots,\qquad \frac{1}{1 - x} = 1 + x + x^2 + \cdots,\qquad (1 + x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \cdots

The expansions of $\log(1 + x)$ , $\frac{1}{1 - x}$ and $(1 + x)^{\alpha}$ hold for $|x| < 1$ .

Products and compositions

For $e^{x^2}$ , substitute $t = x^2$ into the series of $e^t$ ; for $e^x\sin x$ , multiply the two series and collect terms up to the required degree. This is much faster than computing each $f^{(n)}(0)$ .

e^x\sin x = \left(1 + x + \frac{x^2}{2} + \cdots\right)\left(x - \frac{x^3}{6} + \cdots\right) = x + x^2 + \frac{x^3}{3} + \cdots

Limits by series

Expanding the numerator and denominator makes $\frac{0}{0}$ limits transparent:

\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{6} - \frac{x^5}{120} + \cdots}{x^3} = \frac{1}{6}

Tip

It is enough to expand accurately up to the degree of the denominator (here $x^3$ ).

Q1BasicUniv Y1 · Sem 1Taylor and Maclaurin series

Use the Maclaurin series up to the $x^{2}$ term to approximate $\sqrt[3]{1.06}$ .

Answer

1.0196

Solution

For $x$ near $0$ ,
$\sqrt[3]{1 + x} \approx 1 + \frac{x}{3} - \frac{x^2}{9}$
Substituting $x = 0.06$ ,
$\sqrt[3]{1.06} \approx 1.0196$

Q2StandardUniv Y1 · Sem 1Taylor and Maclaurin series

Find the Maclaurin series of $e^{x}\log(1 + x)$ up to the $x^{3}$ term.

Answer

e^{x}\log(1 + x) = x + \frac{1}{2}x^{2} + \frac{1}{3}x^{3} + \cdots

Solution

Using the basic expansions,
$e^{x}\log(1 + x) = \left(1 + x + \frac{x^2}{2} + \cdots\right)\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\right)$
Collecting terms up to $x^3$ ,
$e^{x}\log(1 + x) = x + \frac{1}{2}x^{2} + \frac{1}{3}x^{3} + \cdots$

Q3AdvancedUniv Y1 · Sem 1Taylor and Maclaurin series

Find the constants $a$ and $b$ for which $\displaystyle \lim_{x \to 0} \frac{e^{3x} - (1 + ax + bx^2)}{x^3}$ is finite, and find the value of the limit.

(1)the values of $a$ and $b$
(2)the value of the limit

Answer

(1) $a = 3,\ b = \frac{9}{2}$
(2) $\frac{9}{2}$

Solution

By the Maclaurin series,
$e^{3x} = 1 + 3x + \frac{9}{2}x^{2} + \frac{9}{2}x^{3} + \cdots$
The numerator is
$(3 - a)x + \left(\frac{9}{2} - b\right)x^2 + \frac{9}{2}x^3 + \cdots$
The limit is finite exactly when the coefficients of $x$ and $x^2$ vanish:
$a = 3,\ b = \frac{9}{2},\quad \text{limit} = \frac{9}{2}$

Q1BasicUniv Y1 · Sem 1Taylor and Maclaurin series

Find the Maclaurin series of $\cos x$ up to the $x^{4}$ term.

Q2StandardUniv Y1 · Sem 1Taylor and Maclaurin series

In the Taylor series of $\sqrt{x}$ about $x = 9$ , find the coefficient of $(x - 9)^{2}$ .

Q3AdvancedUniv Y1 · Sem 1Taylor and Maclaurin series

Find the constants $a$ and $b$ for which $\displaystyle \lim_{x \to 0} \frac{e^{-2x} - (1 + ax + bx^2)}{x^3}$ is finite, and find the value of the limit.

(1)the values of $a$ and $b$
(2)the value of the limit

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Related topics

Limits of functions

Partial derivatives

Improper integrals