Representing Functions as Power Series

AP Course

AP Calculus AB/BC

Study the complete College Board sequence for AP Calculus AB and BC, from limits through infinite series.

Choose an official unit and topic to open its lecture, concept check, or focused practice.

AP Calculus AB/BC · Unit 10 · Topic 10.15 · BC Only

Representing Functions as Power Series

Represent functions by transforming a known power series through algebra, substitution, differentiation, or integration while tracking convergence.

1. Topic Focus

Determine series convergence, estimate error, construct Taylor approximations, and represent functions with power series on valid intervals.

This topic: Represent functions by transforming a known power series through algebra, substitution, differentiation, or integration while tracking convergence.

2. Key Relationship

\(\frac1{1-u}=\sum_{n=0}^\infty u^n\quad(|u|<1)\)

Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.

3. Visual Connection

Power-series representationDerivative patterns or transformations create the coefficients; convergence must travel with the new series.

4. Worked Example

Rewrite the target around a known seed series, transform its general term, and test the transformed endpoints separately.

Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.

5. Concept Development

BC-only topic. Instead of recomputing every derivative, begin with a power series whose sum is known and transform both sides. The goal is to produce a correct general term together with the set of \(x\)-values on which the new series represents the target function.

The geometric seed

The most useful starting identity is

\(\boxed{\frac1{1-u}=\sum_{n=0}^{\infty}u^n=1+u+u^2+u^3+\cdots,\qquad |u|<1.}\)

Here \(u\) may be a constant multiple, a power, or another expression in \(x\). The convergence condition must undergo the same substitution.

Rewrite before expanding

A denominator must match \(1-u\). Factor out its constant first:

\(\frac1{A-Bx}=\frac1A\cdot\frac1{1-(B/A)x}=\sum_{n=0}^{\infty}\frac{B^n}{A^{n+1}}x^n,\qquad \left|\frac{B}{A}x\right|<1.\)

The prefactor \(1/A\) multiplies every coefficient.

Substitution

If \(1/(1-u)=\sum u^n\), replacing \(u\) by \(g(x)\) gives

\(\frac1{1-g(x)}=\sum_{n=0}^{\infty}[g(x)]^n,\qquad |g(x)|<1.\)

For example, \(g(x)=-x^2\) creates alternating even powers. Solve \(|g(x)|<1\) rather than copying \(|x|<1\).

Algebraic operations

Multiplying a known identity by a constant or a power of \(x\) multiplies every term. Sums and differences of two power series are valid wherever both original series converge, so use the intersection of their intervals.

Term-by-term differentiation

Inside its radius of convergence, a power series differentiates like an infinite polynomial:

\(\frac{d}{dx}\sum_{n=0}^{\infty}c_nx^n=\sum_{n=1}^{\infty}nc_nx^{n-1}.\)

Differentiating the geometric identity gives

\(\frac1{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}(n+1)x^n,\qquad |x|<1.\)

Reindex only when it makes the final expression clearer, and verify the first few terms afterward.

Term-by-term integration

Inside the radius, integrate each term and determine the constant from a convenient input:

\(\int_0^x\frac1{1-t}\,dt=\sum_{n=0}^{\infty}\int_0^x t^n\,dt,\)

\(\boxed{-\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n},\qquad -1\le x<1.}\)

The open-radius calculation first gives \(|x|<1\). At \(x=-1\) the series is alternating harmonic and converges; at \(x=1\) it is harmonic and diverges.

Radius versus interval after calculus operations

Term-by-term differentiation and integration preserve the radius of convergence, but they can change endpoint behavior. Always substitute each endpoint into the transformed series and apply an ordinary series test.

Building logarithm and arctangent series

Replacing \(x\) by \(-x\) and integrating produces

\(\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n},\qquad -1<x\le1.\)

Replacing \(x\) by \(-x^2\) and integrating produces

\(\arctan x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1},\qquad -1\le x\le1.\)

In each derivation, using a definite integral from \(0\) to \(x\) automatically handles the constant.

Shifted-center geometric series

A target can be expanded in powers of \(x-a\). Rewrite every occurrence of \(x\) in terms of \(x-a\), then expose \(1-u\). For example,

\(\frac1{4-x}=\frac1{2-(x-2)}=\frac12\sum_{n=0}^{\infty}\left(\frac{x-2}{2}\right)^n,\qquad |x-2|<2.\)

A reliable transformation workflow

Choose a seed series, usually the geometric series.
Rewrite the target so the seed function is visible.
Apply the same substitution, multiplication, differentiation, or integration to both sides.
Simplify the general term and check it against the first three terms.
Transform the interior convergence condition.
Test every finite endpoint in the final series.
State the power-series identity and its complete interval.

Topic 10.14 versus Topic 10.15

Topic 10.14 builds a Taylor series from the derivative values \(f^{(n)}(a)\). This topic usually starts with a known series and transforms it. Both methods can reach the same representation, but the justification and algebraic work are different.

AP-style checklist

Name or display the known seed identity.
Show the algebra that exposes its form.
Transform the function and the general term consistently.
Keep the index and exponent synchronized.
State the condition before endpoint testing.
Check endpoints separately and use brackets only when justified.

6. Detailed Worked Example and Error Check

Example 1: Scale the input

Substitute \(u=3x\) into the geometric identity:

\(\frac1{1-3x}=\sum_{n=0}^{\infty}(3x)^n=\sum_{n=0}^{\infty}3^nx^n,\qquad |x|<\frac13.\)

At \(x=\pm1/3\), the terms do not approach zero, so the interval is \((-1/3,1/3)\).

Example 2: Factor the denominator

Rewrite \(2+x=2[1-(-x/2)]\):

\(\frac1{2+x}=\frac12\sum_{n=0}^{\infty}\left(-\frac x2\right)^n=\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^{n+1}},\qquad |x|<2.\)

Both endpoints fail the nth-term test, so the interval is \((-2,2)\).

Example 3: Create even powers

Use \(u=-x^2\):

\(\frac1{1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n},\qquad |x^2|<1.\)

Thus \(|x|<1\), and both endpoints diverge because the term magnitude remains \(1\).

Example 4: Multiply the whole identity

First substitute \(u=x^2\), then multiply by \(x\):

\(\frac{x}{1-x^2}=x\sum_{n=0}^{\infty}x^{2n}=\sum_{n=0}^{\infty}x^{2n+1},\qquad |x|<1.\)

Multiplication by \(x\) shifts every exponent without changing the interior condition.

Example 5: Differentiate and reindex

Differentiating \(1/(1-x)=\sum x^n\) gives

\(\frac1{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}(n+1)x^n,\qquad |x|<1.\)

The expanded form \(1+2x+3x^2+\cdots\) confirms the reindexing.

Example 6: Differentiate after substitution

Since \(1/(1+x)=\sum_{n=0}^{\infty}(-1)^nx^n\), differentiating and multiplying by \(-1\) yields

\(\frac1{(1+x)^2}=\sum_{n=0}^{\infty}(-1)^n(n+1)x^n,\qquad |x|<1.\)

Example 7: Integrate to obtain a logarithm

Integrate the geometric identity from \(0\) to \(x\):

\(-\ln(1-x)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=\sum_{n=1}^{\infty}\frac{x^n}{n}.\)

The radius is \(1\). Endpoint testing gives the interval \([-1,1)\).

Example 8: Integrate to obtain arctangent

From Example 3, integrate from \(0\) to \(x\):

\(\arctan x=\int_0^x\frac{dt}{1+t^2}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}.\)

Both \(x=1\) and \(x=-1\) produce convergent alternating series, so the interval is \([-1,1]\).

Example 9: Shift the center

To expand around \(a=2\), write

\(\frac1{4-x}=\frac1{2-(x-2)}=\frac12\cdot\frac1{1-(x-2)/2}=\sum_{n=0}^{\infty}\frac{(x-2)^n}{2^{n+1}}.\)

The condition \(|x-2|<2\) gives \((0,4)\); at both endpoints the terms fail to approach zero.

Example 10: Combine two known series

On the intersection \(|x|<1\),

\(\frac1{1-x}+\frac1{1+x}=\sum_{n=0}^{\infty}[1+(-1)^n]x^n=2\sum_{k=0}^{\infty}x^{2k}.\)

Algebra confirms the same result because the left side is \(2/(1-x^2)\).

Common errors

Using \(1/(1+u)=\sum u^n\) instead of inserting \(u\mapsto -u\).
Forgetting a constant factored from the denominator.
Changing the function but not the general term.
Copying the original condition \(|x|<1\) after a nonlinear substitution.
Dropping the chain factor when differentiating a substituted identity.
Forgetting the constant of integration.
Changing an index without changing the exponent or coefficient.
Assuming differentiation or integration preserves endpoint inclusion.
Using the union rather than the intersection when adding two series.
Writing an identity outside the interval where the transformed series converges.
Giving only a few terms when a general term is requested.

7. AP Reasoning Routine

Check the nth-term condition first, match the series structure to a justified test, state convergence type, and test power-series endpoints separately.

Identify the representation and requested quantity.
State the rule or theorem and verify its conditions.
Keep exact values until the final requested approximation.
Interpret sign, units, interval, and context.

AP Checkpoint

Represent each function as a power series and state its interval of convergence unless another instruction is given.
(a) \(\dfrac1{1-4x}\).
(b) \(\dfrac1{5+x}\).
(c) \(\dfrac{x^2}{1-x^3}\).
(d) \(\dfrac1{1+x^2}\).
(e) Use differentiation to represent \(\dfrac1{(1-x)^3}\).
(f) Use integration to represent \(\ln(1+x)\).
(g) Use integration to represent \(\arctan x\).
(h) Represent \(\dfrac1{6-x}\) in powers of \(x-2\).
(i) Combine geometric series to represent \(\dfrac1{1-x}+\dfrac1{1+x}\).
(j) Explain why term-by-term integration preserves a radius of convergence but may change the endpoint brackets.
(k) Use a power series to represent \(\displaystyle\int_0^{1/2}\frac{dt}{1+t^2}\).
(l) Represent \(\dfrac{x}{2-x}\) as a power series and give a complete endpoint analysis.

Check the solution

(a) Substitute \(u=4x\): \(\dfrac1{1-4x}=\sum_{n=0}^{\infty}4^nx^n\). The condition is \(|x|<1/4\), and both endpoint terms fail to approach zero, so the interval is \((-1/4,1/4)\).
(b) \(\dfrac1{5+x}=\dfrac15[1-(-x/5)]^{-1}=\sum_{n=0}^{\infty}\dfrac{(-1)^nx^n}{5^{n+1}}\), valid on \((-5,5)\).
(c) \(\dfrac{x^2}{1-x^3}=x^2\sum_{n=0}^{\infty}x^{3n}=\sum_{n=0}^{\infty}x^{3n+2}\), valid on \((-1,1)\). At \(x=\pm1\), the terms do not approach zero.
(d) \(\dfrac1{1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\), valid on \((-1,1)\).
(e) Differentiate the geometric series twice: \(\dfrac{2}{(1-x)^3}=\sum_{n=2}^{\infty}n(n-1)x^{n-2}\). Thus \(\dfrac1{(1-x)^3}=\sum_{n=0}^{\infty}\dfrac{(n+2)(n+1)}2x^n\), valid on \((-1,1)\).
(f) Integrating \(1/(1+x)=\sum_{n=0}^{\infty}(-1)^nx^n\) from \(0\) to \(x\) gives \(\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}x^n/n\). The interval is \((-1,1]\).
(g) Integrating \(1/(1+x^2)=\sum(-1)^nx^{2n}\) from \(0\) to \(x\) gives \(\arctan x=\sum_{n=0}^{\infty}(-1)^nx^{2n+1}/(2n+1)\), valid on \([-1,1]\).
(h) Since \(6-x=4-(x-2)\), \(\dfrac1{6-x}=\dfrac14\sum_{n=0}^{\infty}[(x-2)/4]^n=\sum_{n=0}^{\infty}\dfrac{(x-2)^n}{4^{n+1}}\). The condition \(|x-2|<4\) gives \((-2,6)\); both endpoints diverge by the nth-term test.
(i) \(\dfrac1{1-x}+\dfrac1{1+x}=\sum[1+(-1)^n]x^n=2\sum_{n=0}^{\infty}x^{2n}\), valid on \((-1,1)\).
(j) Integration divides coefficients by increasing powers but leaves the interior radius unchanged. At a boundary, this extra factor can turn a divergent numerical series into a convergent one, so each transformed endpoint needs a new test.
(k) \(\displaystyle\int_0^{1/2}\frac{dt}{1+t^2}=\sum_{n=0}^{\infty}\frac{(-1)^n(1/2)^{2n+1}}{2n+1}=\arctan(1/2)\).
(l) \(\dfrac{x}{2-x}=\dfrac{x/2}{1-x/2}=\sum_{n=1}^{\infty}\dfrac{x^n}{2^n}\) for \(|x|<2\). At \(x=2\), every term is \(1\); at \(x=-2\), the terms alternate between \(\pm1\). Neither term sequence approaches zero, so the interval is \((-2,2)\).