Representing Functions as Power Series (College Board AP® Calculus BC): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 13 February 2025

Power series of composite functions & products

How can I find the power series for a composite function?

A composite function is a ‘function of a function’ or a ‘function within a function’
- For example $\sin (2 x)$ is a composite function,
  - with $2 x$ as the ‘inside function’
  - and $\sin x$ as the ‘outside function’
- Similarly $e^{x^{2}}$ is a composite function
  - with $x^{2}$ as the ‘inside function’
  - and $e^{x}$ as the ‘outside function’
To find the power series for a composite function
STEP 1
Start with the power series for the ‘outside function’
- Usually this will be one of the ‘standard functions’ whose Maclaurin series you should know
  - I.e. $\sin x$ , $\cos x$ , $e^{x}$ and $\frac{1}{1 - x}$
  - See the 'Taylor or Maclaurin series for a function' study guide
STEP 2
Substitute the ‘inside function’ every place that $x$ appears in the series for the ‘outside function’
- For example, for $\sin (2 x)$ you would substitute $2 x$ everywhere that $x$ appears in the series for $\sin x$
  - Put brackets around the 'inside function' when you do this, $(2 x)$
STEP 3
Expand the brackets and simplify the coefficients for the powers of $x$ in the resultant power series
- It may be necessary to use substitution to find the new interval of convergence as well
- See the Worked Example

How can I find the power series for a product of two functions?

To find the power series for a product of two functions
STEP 1
Start with the series for the individual functions
- For each of these series you should only use terms up to an appropriately chosen power of $x$
- See the Worked Example
STEP 2
Put each of the series into brackets and multiply them together
- Only keep terms in powers of $x$ up to the power you are interested in
STEP 3
Collect terms and simplify coefficients

Examiner Tips and Tricks

For composite functions and products you could work out the power series 'from scratch' using the Taylor series formula $\sum_{n = 0}^{\infty} \frac{f^{(n)} (a)}{n!} {(x - a)}^{n}$ . However using standard results and the methods here is usually much quicker.

Worked Example

(a) Find the Maclaurin series for the function $f (x) = \frac{1}{1 + 4 x^{2}}$ , up to and including the term in $x^{8}$ . Be sure to include the interval of convergence.

Start with the standard result for the 'outside function', $\frac{1}{1 - x}$

$\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + x^{4} + . . .$

$\frac{1}{1 + 4 x^{2}} = \frac{1}{1 - (- 4 x^{2})}$ , so substitute the 'inside function' in brackets, $(- 4 x^{2})$ everywhere $x$ appears in the original series and simplify

$\begin{array}{rcl} \frac{1}{1 + 4 x^{2}} & = & 1 + (- 4 x^{2}) + {(- 4 x^{2})}^{2} + {(- 4 x^{2})}^{3} + {(- 4 x^{2})}^{4} + . . . \\ = & 1 - 4 x^{2} + 16 x^{4} - 64 x^{6} + 256 x^{8} - . . . \end{array}$

Now consider the interval of convergence

The series for $\frac{1}{1 - x}$ converges for $|x| < 1$ , so substitute $- 4 x^{2}$ as $x$ into that

$|- 4 x^{2}| < 1 \Leftrightarrow 4 x^{2} < 1 \Leftrightarrow x^{2} < \frac{1}{4} \Leftrightarrow - \frac{1}{2} < x < \frac{1}{2}$

$\begin{array}{rcl} 1 - 4 x^{2} + 16 x^{4} - 64 x^{6} + 256 x^{8} \end{array}$ , with interval of convergence $- \frac{1}{2} < x < \frac{1}{2}$

(b) Find the Maclaurin series for the function $g (x) = e^{x} \sin (2 x)$ , up to and including the term in $x^{4}$ .

First use substitution to find the Maclaurin series for $\sin (2 x)$

The 'outside function' is $\sin x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . . .$ and the 'inside function' is $2 x$ so substitute $(2 x)$ everywhere $x$ appears and simplify

$\begin{array}{rcl} \sin (2 x) & = & (2 x) - \frac{{(2 x)}^{3}}{3!} + \frac{{(2 x)}^{5}}{5!} - . . . \\ = & 2 x - \frac{4 x^{3}}{3} + \frac{4 x^{5}}{15} - . . . \end{array}$

The Maclaurin series for $e^{x}$ is also a standard result

$\begin{array}{rcl} e^{x} & = & 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + . . . \\ = & 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \frac{x^{4}}{24} + . . . \end{array}$

Before multiplying those together, think about which terms you actually need

The question only asks for terms up to $x^{4}$

That means you don't need the $x^{5}$ term (or any higher-order terms) from the series for $\sin (2 x)$
Also the smallest power in the $\sin (2 x)$ series is the $2 x$ term
- So you can stop the $e^{x}$ series with the $\frac{x^{3}}{6}$ term
- Because any higher-order term will multiply with $2 x$ to give a term in $x^{5}$ or higher

$\begin{array}{rcl} e^{x} \sin (2 x) & \approx & (1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6}) (2 x - \frac{4 x^{3}}{3}) \\ = & 1 (2 x - \frac{4 x^{3}}{3}) + x (2 x - \frac{4 x^{3}}{3}) + \frac{x^{2}}{2} (2 x - \frac{4 x^{3}}{3}) + \frac{x^{3}}{6} (2 x - \frac{4 x^{3}}{3}) \\ = & 2 x - \frac{4 x^{3}}{3} + 2 x^{2} - \frac{4 x^{4}}{3} + x^{3} - \frac{2 x^{5}}{3} + \frac{x^{4}}{3} - \frac{2 x^{6}}{9} \\ = & 2 x + 2 x^{2} - \frac{x^{3}}{3} - x^{4} - \frac{2 x^{5}}{3} - \frac{2 x^{6}}{9} \end{array}$

For the answer, we only want terms up to $x^{4}$

Note that the terms in $x^{5}$ and $x^{6}$ there are not actually the correct terms for the power series of $e^{x} \sin (2 x)$

To find those terms properly, we would need to use more terms from the two original series in the product

$\begin{array}{rcl} 2 x + 2 x^{2} - \frac{x^{3}}{3} - x^{4} \end{array}$

Examiner Tips and Tricks

If the 'basic' series you are using converge for all values of $x$ , then series found by using substitution or taking products will also converge for all values of $x$ .

Differentiating & integrating power series

How can I use differentiation to find the power series of a function?

If you differentiate the power series for a function $f (x)$ term by term,
- you get the power series for the function’s derivative $f^{'} (x)$
You can use this to find new power series from existing ones
- For example, the derivative of $\sin x$ is $\cos x$
- So if you differentiate the Maclaurin series for $\sin x$ term by term you will get the Maclaurin series for $\cos x$

How can I use integration to find the power series of a function?

If you integrate the power series for a derivative $f^{'} (x)$ term by term
- you get the power series for the function $f (x)$
Be careful however, as you will have a constant of integration to deal with
- The value of the constant of integration will have to be chosen so that the series produces the correct value of
  - $f (a)$ , for a Taylor series about $x = a$
  - $f (0)$ , for a Maclaurin series
- See the Worked Example
You can use this to find new power series from existing ones
- For example, the derivative of $\sin x$ is $\cos x$
- So if you integrate the Maclaurin series for $\cos x$ (and correctly deal with the constant of integration) you will get the Maclaurin series for $\sin x$

Examiner Tips and Tricks

The radius of convergence of a power series found using term-by-term differentiation or term-by-term integration is the same as the radius of convergence of the original power series.

However convergence or divergence at the endpoints is not always the same, and needs to be checked separately.

Worked Example

(a) Confirm that differentiating the Maclaurin series for $\cos x$ term by term yields the Maclaurin series for $- \sin x$ .

The Maclaurin series for $\cos x$ is a standard result, $\cos x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n}}{(2 n)!} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - . . .$

$\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - . . .$

Differentiate that term by term

$\begin{array}{rcl} \frac{d}{d x} (\cos x) & = & \frac{d}{d x} (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - . . .) \\ = & 0 - \frac{2 x}{2!} + \frac{4 x^{3}}{4!} - \frac{6 x^{5}}{6!} + \frac{8 x^{7}}{8!} - . . . \\ = & - x + \frac{x^{3}}{3!} - \frac{x^{5}}{5} + \frac{x^{7}}{7} - . . . \end{array}$

The series for $\sin x$ is also a standard result, $\sin x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . .$

$\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . .$

$\begin{array}{rcl} \Rightarrow & - \sin x = - (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . .) \\ = & - x + \frac{x^{3}}{3!} - \frac{x^{5}}{5!} + \frac{x^{7}}{7!} - . . . \end{array}$

Indeed those two series match

Differentiating the Maclaurin series for $\cos x$ term by term gives the Maclaurin series for $- \sin x$

(b) Find the Maclaurin series for $\frac{1}{1 + x^{2}}$ , and use that result to determine the Maclaurin series for $\arctan x$ up to and including the term in $x^{7}$ . Be sure to include the radius of convergence.

Start with the standard Maclaurin series result for $\frac{1}{1 - x}$ , and substitute $- x^{2}$ everywhere

$\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + x^{4} + . . .$

$\begin{array}{rcl} \Rightarrow & \frac{1}{1 + x^{2}} = 1 + (- x^{2}) + {(- x^{2})}^{2} + {(- x^{2})}^{3} + {(- x^{2})}^{4} + . . . \\ = & 1 - x^{2} + x^{4} - x^{6} + x^{8} - . . . \end{array}$

Now use the standard integral result, $\int \frac{1}{1 + x^{2}} d x = \arctan x + C$

That means you can integrate the series for $\frac{1}{1 + x^{2}}$ term by term to get the series for $\arctan x$

Just remember to include a constant of integration, which it is easiest to write at the front of the series

$\begin{array}{rcl} \arctan x & = & \int (1 - x^{2} + x^{4} - x^{6} + x^{8} - . . .) d x \\ = & C + x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + \frac{x^{9}}{9} - . . . \end{array}$

To find the value of $C$ , substitute in $x = 0$ ; remember that $\arctan (0) = 0$

$\begin{array}{rcl} \arctan (0) & = & C + (0) - \frac{{(0)}^{3}}{3} + \frac{{(0)}^{5}}{5} - \frac{{(0)}^{7}}{7} + \frac{{(0)}^{9}}{9} - \\ 0 & = & C \end{array}$

Now consider the radius of convergence

The series for $\frac{1}{1 - x}$ converges for $|x| < 1$ , so substitute $- x^{2}$ into that

$|- x^{2}| < 1 \Leftrightarrow x^{2} < 1 \Leftrightarrow - 1 < x < 1 \Leftrightarrow |x| < 1$

That is the radius of convergence for the $\frac{1}{1 + x^{2}}$ series (i.e. the radius of convergence is 1, centered on $x = 0$ ); the series we found by integrating that will have the same radius of convergence

And the question only wants terms up through $x^{7}$ , so

$\begin{array}{rcl} x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} \end{array}$ , with radius of convergence $|x| < 1$

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Representing Functions as Power Series (College Board AP® Calculus BC): Study Guide

Power series of composite functions & products

How can I find the power series for a composite function?

How can I find the power series for a product of two functions?

Differentiating & integrating power series

How can I use differentiation to find the power series of a function?

How can I use integration to find the power series of a function?

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

Unit 1: Limits & Continuity

Limits

Definition of a Limit

Infinite Limits & Limits at Infinity

Properties of Limits

Evaluating Limits Analytically

Evaluating Limits Numerically & Graphically

Squeeze Theorem & Trigonometric Limits

Summary of Limits

Selecting Procedures for Determining Limits

Continuity

Continuity

Removable Discontinuities

Non-removable Discontinuities

Intermediate Value Theorem

Unit 2: Differentiation Definition & Fundamental Properties

Definition of Differentiation

Average Rate of Change

Instantaneous Rate of Change

Derivatives & Tangents

Estimating the Derivative at a Point

Differentiability & Continuity

Fundamental Properties of Differentiation

Derivative Rules

Derivatives of Exponentials and Logarithms

Derivatives of Sine and Cosine Functions

The Product Rule

The Quotient Rule

Derivatives of Tangent and Reciprocal Trig Functions

Unit 3: Differentiation of Composite, Implicit & Inverse Functions

Differentiation of Composite & Inverse Functions

The Chain Rule

The Inverse Function Theorem

Derivatives of Inverse Trigonometric Functions

Higher-Order Derivatives

Summary of Differentiation

Selecting Procedures for Calculating Derivatives

Unit 4: Contextual Applications of Differentiation

Rates of Change & Related Rates

Meaning of a Derivative in Context

Motion in a Straight Line

Related Rates

Linearization

Approximating Values of a Function

L'Hospital's Rule

L'Hospital's Rule

Unit 5: Analytical Applications of Differentiation

Graphs of Functions & Their Derivatives

Mean Value Theorem

Extreme Value Theorem

Critical Points

Increasing & Decreasing Functions

First Derivative Test for Local Extrema

Candidates Test for Global Extrema

Concavity of Functions

Second Derivative Test for Local Extrema

Graphs of f, f' & f''

Optimization Problems

Behaviors of Implicit Relations

Second Derivatives of Implicit Functions

Critical Points of Implicit Relations

Unit 6: Integration & Accumulation of Change

Integration & Antiderivatives

Indefinite Integrals

Derivatives & Antiderivatives

Indefinite Integral Rules

Constant of Integration

Riemann Sums & Definite Integrals

Accumulation of Change

Riemann Sums

Trapezoidal sums