Representing Functions as Power Series

AP Calculus BC · Infinite Sequences and Series

Representing Functions as Power Series

The geometric series is a factory: force the shape, substitute, then differentiate or integrate.

key idea$\frac{1}{1-u}=\sum_{k=0}^{\infty}u^{k}\quad\text{for }\lvert u\rvert\lt 1$

The procedure

The geometric series is a factory. Force a function into the shape $\dfrac{1}{1-u}$, substitute, and you have a power series — then differentiate, integrate or multiply to reach a whole family of related ones, all without computing a single derivative of the original function.

Force the denominator into $1-u$

Factor out whatever is needed so the constant term is exactly $1$. $\dfrac{1}{3+x}=\dfrac{1}{3}\cdot\dfrac{1}{1-\left(-x/3\right)}$.

in a rational function out the substitution $u$

Substitute into the geometric series

$\dfrac{1}{1-u}=\displaystyle\sum u^{k}$, so replace $u$ throughout and simplify the powers.

out a power series

Carry the outside factor in

Any constant or power of $x$ factored out earlier multiplies every term. Distributing it is one line and is easy to forget.

out the corrected series

Differentiate or integrate if the target requires it

Both operations act term by term and preserve the radius. An integration constant is fixed by evaluating at the centre.

out the target function's series

State the interval

Convergence needs $\lvert u\rvert\lt1$. Solve that inequality for $x$ — the radius scales with whatever was substituted.

out series $+$ interval

Target	Rewrite as	$u$	Radius
$\dfrac{1}{1+x}$	$\dfrac{1}{1-(-x)}$	$-x$	$1$
$\dfrac{1}{1-2x}$	already there	$2x$	$\tfrac12$
$\dfrac{1}{3+x}$	$\dfrac13\cdot\dfrac{1}{1-(-x/3)}$	$-\dfrac{x}{3}$	$3$
$\dfrac{x}{1-x^{2}}$	$x\cdot\dfrac{1}{1-x^{2}}$	$x^{2}$	$1$
$\dfrac{1}{(1-x)^{2}}$	differentiate $\dfrac{1}{1-x}$	—	$1$

Row 3 shows the factoring step that makes the constant term $1$, and row 5 shows the derivative trick. Whatever is substituted for $u$ determines the radius: $u=2x$ halves it, $u=-x/3$ triples it.

Substituting $u$changes the radius

Differentiatingkeeps the radius, may lose an endpoint

Integratingkeeps the radius, may gain one

four checks

The denominator must read exactly $1-u$. Factor first so the constant term is $1$.
A factor pulled outside multiplies every term of the resulting series.
Term-by-term differentiation and integration preserve the radius but can change the endpoints.
Solve $\lvert u\rvert\lt1$ for $x$ to get the interval — do not assume it is still $1$.

Worked examples

EX 1

Force the standard form

$\dfrac{1}{3+x}$

Represent $\dfrac{1}{3+x}$ as a power series about $x=0$ and give its radius.

Factor to make the constant $1$

$\dfrac{1}{3+x}=\dfrac{1}{3\left(1+\tfrac{x}{3}\right)}=\dfrac13\cdot\dfrac{1}{1-\left(-\tfrac{x}{3}\right)}$.

Identify the substitution

$u=-\dfrac{x}{3}$.

Expand

$\dfrac13\displaystyle\sum_{k=0}^{\infty}\left(-\dfrac{x}{3}\right)^{k}=\sum_{k=0}^{\infty}\dfrac{(-1)^{k}x^{k}}{3^{k+1}}$.

Find the interval

$\left\lvert-\dfrac{x}{3}\right\rvert\lt1$ gives $\lvert x\rvert\lt3$.

State the result

Radius $R=3$, and the outside $\dfrac13$ merged into $3^{k+1}$ — the step most often dropped.

EX 2

Multiply by a power of $x$

$\dfrac{x^{2}}{1-x^{3}}$

Represent $\dfrac{x^{2}}{1-x^{3}}$ as a power series.

Set aside the factor

$x^{2}\cdot\dfrac{1}{1-x^{3}}$.

Expand the fraction

With $u=x^{3}$: $\displaystyle\sum_{k=0}^{\infty}x^{3k}$.

Multiply back in

$x^{2}\displaystyle\sum_{k=0}^{\infty}x^{3k}=\sum_{k=0}^{\infty}x^{3k+2}$.

Write out a few terms

$x^{2}+x^{5}+x^{8}+\cdots$

Find the interval

$\left\lvert x^{3}\right\rvert\lt1$ gives $\lvert x\rvert\lt1$, so $R=1$.

EX 3

Differentiate to reach the target

$\dfrac{1}{(1-x)^{2}}$

Represent $\dfrac{1}{(1-x)^{2}}$ as a power series.

Spot the relationship

$\dfrac{d}{dx}\left(\dfrac{1}{1-x}\right)=\dfrac{1}{(1-x)^{2}}$.

Start from the geometric series

$\dfrac{1}{1-x}=\displaystyle\sum_{k=0}^{\infty}x^{k}$ for $\lvert x\rvert\lt1$.

Differentiate term by term

$\displaystyle\sum_{k=1}^{\infty}kx^{k-1}$ — the $k=0$ term differentiates to zero and drops out.

Re-index

$=\displaystyle\sum_{k=0}^{\infty}(k+1)x^{k}=1+2x+3x^{2}+\cdots$

State the radius

Still $1$. Differentiation preserves the radius, though the endpoint behaviour can change.

EX 4

Integrate to reach the target

$\arctan x$

Represent $\arctan x$ as a power series.

Differentiate the target

$\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^{2}}$.

Expand that

With $u=-x^{2}$: $\displaystyle\sum_{k=0}^{\infty}(-1)^{k}x^{2k}$.

Integrate term by term

$\arctan x=C+\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^{k}x^{2k+1}}{2k+1}$.

Fix the constant

At $x=0$: $\arctan0=0$ and the series gives $C$, so $C=0$.

State the result

$\arctan x=x-\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}-\cdots$ for $\lvert x\rvert\lt1$. Setting $x=1$ here is the classical series for $\dfrac{\pi}{4}$.

Where it goes wrong

The denominator must be exactly $1-u$

$\dfrac{1}{3+x}$ is not ready as written. Factor the $3$ out first, and remember it multiplies the whole series.

Substitution rescales the radius

$u=2x$ gives $\lvert x\rvert\lt\tfrac12$, and $u=-x/3$ gives $\lvert x\rvert\lt3$. Solve $\lvert u\rvert\lt1$ for $x$ rather than assuming $R=1$.

Carry the outside factor into every term

A $\dfrac13$ pulled out at the start belongs inside the sum at the end. Dropping it scales the entire answer wrongly.

Re-index after differentiating

The constant term vanishes, so the sum starts one index later. Shifting the index makes the pattern readable and prevents an off-by-one.

Course map. Topic 10.15 · LO LIM-8.G — represent a function as a power series using algebraic manipulation, differentiation, or integration. BC-only.