Power Series: Building Functions from Powers

By Vivek Sharma·MATHEMATICS·Ch. 1 · Infinite Series and Power SeriesSunday, October 18, 2026· 15 min read

Turn the ideas of infinite series into a way to represent and approximate functions.

A polynomial such as $1+x+x^2$ is easy to calculate. A power series extends that idea with infinitely many powers. Its value may depend on where you choose $x$ : the same expression can converge at one value and diverge at another. Read the Infinite Series lesson first; only the optional Taylor section uses derivatives.

Before you begin: Infinite Series: Adding Without an End →

1. A polynomial with infinitely many terms

A power series adds increasing whole-number powers of $(x-c)$ . The number $c$ is the centre, and $a_n$ is the coefficient multiplying the $n$ th power. The coefficients do not depend on $x$ .

The subscript in $a_n$ labels a coefficient, while the superscript in $(x-c)^n$ is an exponent. The first term is the constant $a_0$ . At $x=c$ , every positive-power term vanishes, so the series always converges there to $a_0$ .

\sum_{n=0}^{\infty}a_n(x-c)^n=a_0+a_1(x-c)+a_2(x-c)^2+\cdots

Worked example · Read the ingredients

In $\sum_{n=0}^{\infty}(x-2)^n/3^n$ , the centre is $c=2$ and the coefficients are $a_n=1/3^n$ .
The first terms are $1+(x-2)/3+(x-2)^2/9+\cdots$ . At $x=2$ , the value is $1$ .
At a different value, such as $x=3$ , it becomes the numerical geometric series $1+1/3+1/9+\cdots$ , which sums to $3/2$ .

Remember: Fixing a value of $x$ turns a power series into an ordinary numerical series.

2. One series, different behaviour at different inputs

The simplest power series is $1+x+x^2+\cdots$ . It is geometric with first term $1$ and ratio $x$ . The geometric formula therefore gives a function representation, but only where $|x|<1$ .

At $x=1/2$ , its sum is $2$ . At $x=1$ , it becomes $1+1+1+\cdots$ and diverges. At $x=-1$ , its partial sums oscillate. At $x=2$ , the terms grow rather than approach zero.

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

Remember: Always attach a convergence condition to a power-series representation.

3. Radius and interval of convergence

A power series has a radius of convergence $R$ . It converges absolutely for $|x-c|<R$ and diverges for $|x-c|>R$ . At the boundary $|x-c|=R$ , test each endpoint separately.

Think of $R$ as how far you can move from the centre before reaching a boundary. If $R=0$ , only the centre works. If $R=\infty$ , every real value works. We discuss real inputs in this lesson.

The interval of convergence includes exactly the real values that work. Round brackets exclude an endpoint; square brackets include it. For example, $[-1,1)$ includes $-1$ and excludes $1$ .

Worked example · A shifted centre

For $\sum_{n=0}^{\infty}(x-2)^n/3^n$ , the geometric ratio is $(x-2)/3$ .
Solve $|(x-2)/3|<1$ : this gives $|x-2|<3$ , or $-1<x<5$ . The radius is $R=3$ .
At $x=-1$ , the series is $1-1+1-\cdots$ and diverges. At $x=5$ , it is $1+1+1+\cdots$ and diverges. The interval is therefore $(-1,5)$ .

Remember: The radius gives the distance from the centre; the interval also records which endpoints converge.

4. Find a radius, then check both endpoints

When a power series is not obviously geometric, the ratio test often helps. Treat the whole term, including its power of $x$ , as $u_n$ . Here is an example where only one endpoint belongs to the interval.

\sum_{n=1}^{\infty}\frac{x^n}{n}

\lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to\infty}|x|\frac{n}{n+1}=|x|\qquad(x\ne0)

Worked example · One endpoint included

At $x=0$ , every term is zero, so convergence is immediate. For other values, the ratio test gives convergence when $|x|<1$ and divergence when $|x|>1$ . Thus $c=0$ and $R=1$ .
At $x=1$ , the series becomes $\sum1/n$ , the divergent harmonic series.
At $x=-1$ , the series becomes $\sum(-1)^n/n$ . The magnitudes $1/n$ decrease to zero, so the alternating-series test proves convergence.
Combine the interior with the endpoint results: the interval of convergence is $[-1,1)$ . The ratio test alone could not settle the endpoints.

Remember: Use a three-step routine: find the interior, substitute the left endpoint, then substitute the right endpoint.

5. Approximate a function with a short polynomial

A partial sum of a power series is a polynomial. Truncating means stopping after finitely many terms. This gives an approximation; it is not usually an exact equality.

For $1/(1-x)$ , retaining the terms through $x^N$ leaves the exact remainder shown below. Inside $|x|<1$ , this error tends to zero as $N$ increases. Near the boundary, more terms may be needed.

\frac{1}{1-x}=\underbrace{\sum_{n=0}^{N}x^n}_{\text{approximation}}+\underbrace{\frac{x^{N+1}}{1-x}}_{\text{remainder}}\qquad(|x|<1)

Worked example · An approximation with a known error

Choose $x=0.1$ . The exact function value is $1/0.9=1.111111\ldots$ .
Keep terms through $x^3$ : $1+0.1+0.01+0.001=1.111$ .
The absolute error is $0.1^4/0.9=1/9000\approx0.0001111$ . Using the remainder tells us how accurate the answer is.

Remember: A finite sum becomes useful when you also know how much error remains.

6. Familiar functions as power series

Many familiar functions have power-series representations. The formulas below are centred at zero and converge for every real $x$ . In the sine and cosine formulas, $x$ is measured in radians.

The factorial $n!$ means $1\times2\times\cdots\times n$ , with $0!=1$ . For example, $4!=24$ . Factorials in these denominators make the terms shrink rapidly for a fixed $x$ .

e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots

\sin x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots

\cos x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots

Optional bridge · Where Taylor coefficients come from

If you know derivatives, a Taylor series at $c$ uses the coefficients $a_n=f^{(n)}(c)/n!$ , where $f^{(n)}$ means the $n$ th derivative and $f^{(0)}=f$ .
When $c=0$ , it is called a Maclaurin series. All derivatives of $e^x$ equal $e^x$ , so at zero each derivative is $1$ , giving $a_n=1/n!$ .
A function having derivatives of every order does not alone guarantee it equals its Taylor series. The remainder must tend to zero. The three expansions above do satisfy this condition.

Remember: Power series let us compute complicated functions with polynomial sums; convergence and error determine where the approximation is valid.

Power Series: Building Functions from Powers

1. A polynomial with infinitely many terms

Worked example · Read the ingredients

2. One series, different behaviour at different inputs

3. Radius and interval of convergence

Worked example · A shifted centre

4. Find a radius, then check both endpoints

Worked example · One endpoint included

5. Approximate a function with a short polynomial

Worked example · An approximation with a known error

6. Familiar functions as power series

Optional bridge · Where Taylor coefficients come from

Objective questions