Power Series - Calculus Explorer

Definition & Notation

What is a Power Series?

A power series is a series of the form:

\(\sum_{n=0}^{\infty} c_n(x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \ldots\)

where:

\(a\) is a fixed number (the center of the power series)
\(c_n\) are constants (the coefficients of the power series)
\(x\) is a variable

A power series is a function of \(x\). Unlike the series we've studied previously, the convergence of a power series depends on the value of \(x\).

Example 1: Power Series Notation

The power series \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\) can be written as:

\(\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots\)

This particular power series represents the exponential function \(e^x\).

Special Case: Maclaurin Series

When the center of a power series is \(a = 0\), the power series takes the form:

\(\sum_{n=0}^{\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \ldots\)

This special case is called a Maclaurin series.

Radius of Convergence

For a power series \(\sum_{n=0}^{\infty} c_n(x-a)^n\), there is a number \(R\) (possibly 0, possibly \(\infty\)) such that:

The series converges absolutely for \(|x-a| < R\)
The series diverges for \(|x-a| > R\)

This number \(R\) is called the radius of convergence of the power series.

Finding the Radius of Convergence

The radius of convergence \(R\) of a power series \(\sum_{n=0}^{\infty} c_n(x-a)^n\) can be found using:

The Ratio Test: \(R = \frac{1}{\rho}\) where \(\rho = \lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right|\) (if this limit exists)
The Root Test: \(R = \frac{1}{\rho}\) where \(\rho = \lim_{n \to \infty} \sqrt[n]{|c_n|}\) (if this limit exists)

If \(\rho = 0\), then \(R = \infty\) (the series converges for all \(x\)).

If \(\rho = \infty\), then \(R = 0\) (the series converges only at \(x = a\)).

Example 2: Finding the Radius of Convergence

Find the radius of convergence of the power series \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\).

Solution:

Let \(a_n = \frac{x^n}{n!}\). We apply the ratio test to the series terms:

\begin{align} \rho &= \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \\ &= \lim_{n \to \infty} \left|\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}\right| \\ &= \lim_{n \to \infty} \left|\frac{x^{n+1} \cdot n!}{(n+1)! \cdot x^n}\right| \\ &= \lim_{n \to \infty} \left|\frac{x}{n+1}\right| \\ &= |x| \cdot \lim_{n \to \infty} \frac{1}{n+1} \\ &= |x| \cdot 0 \\ &= 0 \end{align}

For convergence, we need \(\rho < 1\). Since \(\rho = 0 < 1\) for all values of \(x\), the series converges for all \(x\), and the radius of convergence is \(R = \infty\).

Interval of Convergence

The interval of convergence of a power series is the set of all values of \(x\) for which the series converges.

If the radius of convergence is \(R\), then the interval of convergence includes at least the open interval \((a-R, a+R)\).

To determine the full interval of convergence, we need to check the endpoints \(x = a-R\) and \(x = a+R\) separately.

Possible Intervals of Convergence

The interval of convergence can be one of the following:

Just the point \(\{a\}\) (when \(R = 0\))
The open interval \((a-R, a+R)\)
The half-open interval \([a-R, a+R)\) or \((a-R, a+R]\)
The closed interval \([a-R, a+R]\)
The entire real line \(\mathbb{R}\) (when \(R = \infty\))

Example 3: Finding the Interval of Convergence

Find the interval of convergence of the power series \(\sum_{n=1}^{\infty} \frac{(x-2)^n}{n}\).

Solution:

First, let's find the radius of convergence using the ratio test:

\begin{align} \rho &= \lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| \\ &= \lim_{n \to \infty} \left|\frac{\frac{1}{n+1}}{\frac{1}{n}}\right| \\ &= \lim_{n \to \infty} \frac{n}{n+1} \\ &= 1 \end{align}

So the radius of convergence is \(R = \frac{1}{1} = 1\).

This means the series converges absolutely for \(|x-2| < 1\), or \(1 < x < 3\).

Now we need to check the endpoints:

At \(x = 1\) (or \(x-2 = -1\)), the series becomes \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\), which is the alternating harmonic series. This series converges by the alternating series test.

At \(x = 3\) (or \(x-2 = 1\)), the series becomes \(\sum_{n=1}^{\infty} \frac{1}{n}\), which is the harmonic series. This series diverges.

Therefore, the interval of convergence is \([1, 3)\).

Operations with Power Series

Operations on Power Series

If \(f(x) = \sum_{n=0}^{\infty} a_n(x-a)^n\) and \(g(x) = \sum_{n=0}^{\infty} b_n(x-a)^n\) are power series with radii of convergence \(R_f\) and \(R_g\) respectively, then:

Addition: \(f(x) + g(x) = \sum_{n=0}^{\infty} (a_n + b_n)(x-a)^n\) with radius of convergence at least \(\min(R_f, R_g)\)
Scalar Multiplication: \(c \cdot f(x) = \sum_{n=0}^{\infty} (c \cdot a_n)(x-a)^n\) with the same radius of convergence as \(f(x)\)
Multiplication: \(f(x) \cdot g(x) = \sum_{n=0}^{\infty} c_n(x-a)^n\) where \(c_n = \sum_{k=0}^{n} a_k b_{n-k}\) with radius of convergence at least \(\min(R_f, R_g)\)

Example 4: Adding Power Series

Find the sum of the power series \(f(x) = \sum_{n=0}^{\infty} x^n\) and \(g(x) = \sum_{n=0}^{\infty} nx^n\).

Solution:

First, let's find the radius of convergence for each series:

For \(f(x) = \sum_{n=0}^{\infty} x^n\):

\begin{align} \rho_f &= \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \\ &= \lim_{n \to \infty} \left|\frac{1}{1}\right| \\ &= 1 \end{align}

So \(R_f = 1\).

For \(g(x) = \sum_{n=0}^{\infty} nx^n\):

\begin{align} \rho_g &= \lim_{n \to \infty} \left|\frac{b_{n+1}}{b_n}\right| \\ &= \lim_{n \to \infty} \left|\frac{(n+1)}{n}\right| \\ &= 1 \end{align}

So \(R_g = 1\).

Now, the sum of the two power series is:

\begin{align} f(x) + g(x) &= \sum_{n=0}^{\infty} x^n + \sum_{n=0}^{\infty} nx^n \\ &= \sum_{n=0}^{\infty} (1 + n)x^n \\ &= 1 + \sum_{n=1}^{\infty} (1 + n)x^n \\ &= 1 + \sum_{n=1}^{\infty} (n+1)x^n \end{align}

The radius of convergence of the sum is \(\min(R_f, R_g) = 1\).

Differentiation & Integration

Differentiation of Power Series

If \(f(x) = \sum_{n=0}^{\infty} c_n(x-a)^n\) has radius of convergence \(R > 0\), then \(f(x)\) is differentiable on \((a-R, a+R)\) and:

\(f'(x) = \sum_{n=1}^{\infty} n \cdot c_n(x-a)^{n-1} = \sum_{n=0}^{\infty} (n+1) \cdot c_{n+1}(x-a)^n\)

The radius of convergence of \(f'(x)\) is also \(R\).

Integration of Power Series

If \(f(x) = \sum_{n=0}^{\infty} c_n(x-a)^n\) has radius of convergence \(R > 0\), then:

\(\int f(x) \, dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1} = C + \sum_{n=1}^{\infty} \frac{c_{n-1}}{n}(x-a)^n\)

where \(C\) is a constant of integration. The radius of convergence of the integrated series is also \(R\).

Example 5: Differentiation of a Power Series

Find the derivative of the power series \(f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}\).

Solution:

\begin{align} f'(x) &= \sum_{n=1}^{\infty} n \cdot \frac{x^{n-1}}{n!} \\ &= \sum_{n=1}^{\infty} \frac{n}{n!} \cdot x^{n-1} \\ &= \sum_{n=1}^{\infty} \frac{1}{(n-1)!} \cdot x^{n-1} \\ &= \sum_{k=0}^{\infty} \frac{x^k}{k!} \quad \text{(substituting } k = n-1 \text{)} \\ &= f(x) \end{align}

Therefore, \(f'(x) = f(x)\), which means \(f(x) = e^x\) (this makes sense since \(f(x) = e^x\) is equal to its own derivative).

Example 6: Integration of a Power Series

Find the indefinite integral of the power series \(f(x) = \sum_{n=0}^{\infty} x^n\) for \(|x| < 1\).

Solution:

First, note that \(f(x) = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}\) for \(|x| < 1\) (this is the sum of a geometric series).

\begin{align} \int f(x) \, dx &= \int \sum_{n=0}^{\infty} x^n \, dx \\ &= \sum_{n=0}^{\infty} \int x^n \, dx \\ &= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} + C \\ &= \sum_{n=1}^{\infty} \frac{x^n}{n} + C \end{align}

We can verify this by differentiating:

\begin{align} \frac{d}{dx}\left(\sum_{n=1}^{\infty} \frac{x^n}{n}\right) &= \sum_{n=1}^{\infty} \frac{d}{dx}\left(\frac{x^n}{n}\right) \\ &= \sum_{n=1}^{\infty} x^{n-1} \\ &= \sum_{n=0}^{\infty} x^n \\ &= f(x) \end{align}

Therefore, \(\int f(x) \, dx = \sum_{n=1}^{\infty} \frac{x^n}{n} + C = -\ln(1-x) + C\) for \(|x| < 1\).

Examples

Example 7: Representing Functions as Power Series

Find a power series representation for \(f(x) = \frac{1}{1-x}\) centered at \(a = 0\).

Solution:

We can use the formula for the sum of a geometric series:

\(\sum_{n=0}^{\infty} r^n = \frac{1}{1-r} \quad \text{for } |r| < 1\)

Setting \(r = x\), we get:

\(\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \ldots \quad \text{for } |x| < 1\)

The radius of convergence is \(R = 1\).

Example 8: Using Differentiation to Find Power Series

Find a power series representation for \(f(x) = \frac{x}{(1-x)^2}\) centered at \(a = 0\).

Solution:

We know that \(\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\) for \(|x| < 1\).

Taking the derivative of both sides:

\begin{align} \frac{d}{dx}\left(\frac{1}{1-x}\right) &= \frac{d}{dx}\left(\sum_{n=0}^{\infty} x^n\right) \\ \frac{1}{(1-x)^2} &= \sum_{n=1}^{\infty} nx^{n-1} \\ \frac{1}{(1-x)^2} &= \sum_{n=0}^{\infty} (n+1)x^n \end{align}

Multiplying both sides by \(x\):

\begin{align} \frac{x}{(1-x)^2} &= x \cdot \sum_{n=0}^{\infty} (n+1)x^n \\ &= \sum_{n=0}^{\infty} (n+1)x^{n+1} \\ &= \sum_{n=1}^{\infty} nx^n \end{align}

Therefore, \(f(x) = \frac{x}{(1-x)^2} = \sum_{n=1}^{\infty} nx^n = x + 2x^2 + 3x^3 + 4x^4 + \ldots\) for \(|x| < 1\).

Example 9: Using Integration to Find Power Series

Find a power series representation for \(f(x) = \ln(1+x)\) centered at \(a = 0\).

Solution:

We know that \(\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n\) for \(|x| < 1\) (by substituting \(-x\) for \(x\) in the geometric series).

Integrating both sides:

\begin{align} \int \frac{1}{1+x} \, dx &= \int \sum_{n=0}^{\infty} (-1)^n x^n \, dx \\ \ln(1+x) &= \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} + C \\ \ln(1+x) &= \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} + C \\ &= \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} + C \end{align}

We can determine the constant \(C\) by evaluating at \(x = 0\): \(\ln(1+0) = 0 = 0 + C\), so \(C = 0\).

Therefore, \(f(x) = \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots\) for \(|x| < 1\).

Interactive Visualizations

Explore Power Series Approximations

Use the controls below to explore how power series can approximate functions with increasing accuracy as more terms are included.

Select a function:

Number of terms in approximation: 3

Power Series Information

Function: f(x) = e^x

Power Series: 1 + x + x²/2! + x³/3! + ...

Current Approximation: 1 + x + x²/2

Radius of Convergence: ∞

Practice Exercises

Exercise 1

Find the radius and interval of convergence for the power series \(\sum_{n=1}^{\infty} \frac{(x+2)^n}{n \cdot 3^n}\).

Let's use the ratio test to find the radius of convergence:

\begin{align} \rho &= \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \\ &= \lim_{n \to \infty} \left|\frac{\frac{(x+2)^{n+1}}{(n+1) \cdot 3^{n+1}}}{\frac{(x+2)^n}{n \cdot 3^n}}\right| \\ &= \lim_{n \to \infty} \left|\frac{(x+2)^{n+1} \cdot n \cdot 3^n}{(n+1) \cdot 3^{n+1} \cdot (x+2)^n}\right| \\ &= \lim_{n \to \infty} \left|\frac{(x+2) \cdot n}{(n+1) \cdot 3}\right| \\ &= \lim_{n \to \infty} \frac{|x+2| \cdot n}{(n+1) \cdot 3} \\ &= \frac{|x+2|}{3} \cdot \lim_{n \to \infty} \frac{n}{n+1} \\ &= \frac{|x+2|}{3} \cdot 1 \\ &= \frac{|x+2|}{3} \end{align}

For convergence, we need \(\rho < 1\), which gives us:

\begin{align} \frac{|x+2|}{3} &< 1 \\ |x+2| &< 3 \\ -3 &< x+2 < 3 \\ -5 &< x < 1 \end{align}

So the radius of convergence is \(R = 3\), and the series converges absolutely for \(-5 < x < 1\).

Now we need to check the endpoints:

At \(x = -5\), the series becomes \(\sum_{n=1}^{\infty} \frac{(-3)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\), which is the alternating harmonic series. This series converges by the alternating series test.

At \(x = 1\), the series becomes \(\sum_{n=1}^{\infty} \frac{3^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{1}{n}\), which is the harmonic series. This series diverges.

Therefore, the interval of convergence is \([-5, 1)\).

Exercise 2

Find a power series representation for \(f(x) = \frac{1}{(1-x)^3}\) centered at \(a = 0\).

Exercise 3

Find a power series representation for \(f(x) = \frac{x^2}{1+x^3}\) centered at \(a = 0\).