Differentiating and Integrating Power Series

You can differentiate or integrate a power series term by term, just like a polynomial. If f(x) = \sum c_n x^n, then f'(x) = \sum n c_n x^{n-1} and \int f(x)\,dx = C + \sum \frac{c_n x^{n+1}}{n + 1}. The radius of convergence stays the same, but the endpoints must be checked again.

Before this lesson: Power Series, Radius and Interval of Convergence, Maclaurin Series and Antiderivatives

Why term by term works

A polynomial is differentiated one term at a time with the power rule. Inside its interval of convergence, a power series behaves the same way: the derivative of the sum is the sum of the derivatives, and the same is true for integrals.

This gives new series without new derivative patterns. Start from a series you know, such as the geometric series, and differentiate or integrate it. It also lets you integrate functions like e^{-x^2} that have no antiderivative made of basic functions.

Differentiating and integrating power series

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

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

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

• Both new series have the same radius R.
• The interval can gain or lose an endpoint, so test the endpoints again.
• After differentiating, the n = 0 term (a constant) becomes 0, so the sum starts at n = 1.
• After integrating, find C by putting in x = a.

Term-by-term examples

Example 1: Differentiate the geometric series

Find a power series for \dfrac{1}{(1 - x)^2}.

\dfrac{1}{1 - x} = \sum_{n=0}^{\infty} x^n, \quad |x| \lt 1

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

\dfrac{d}{dx}\sum_{n=0}^{\infty} x^n = \sum_{n=1}^{\infty} n x^{n-1}

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

Answer: \frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots for |x| < 1.

Example 2: Integrate to get ln(1 + x)

Integrate the series for \dfrac{1}{1 + x} to find a series for ln(1 + x).

\dfrac{1}{1 + x} = \sum_{n=0}^{\infty} (-1)^n x^n, \quad |x| \lt 1

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

\ln(1 + 0) = C + 0 \quad \Rightarrow \quad C = 0

\ln(1 + x) = \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1} x^n}{n}

Answer: \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots. The radius is still 1, and the series now also converges at the endpoint x = 1, so it is valid for −1 < x ≤ 1.

Example 3: Integrate to get arctan x

Find a series for arctan x, and use it to write a series for π.

\dfrac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \quad |x| \lt 1

\arctan x = C + \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{2n + 1}

\arctan 0 = 0 \quad \Rightarrow \quad C = 0

\arctan 1 = \dfrac{\pi}{4} = 1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \cdots

Answer: \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots for −1 ≤ x ≤ 1, and \pi = 4\left(1 - \frac{1}{3} + \frac{1}{5} - \cdots\right).

Example 4: A definite integral with no basic antiderivative

Estimate \int_0^{0.5} e^{-x^2}\,dx with four terms of its series.

e^{-x^2} = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n}}{n!}

\int_0^{0.5} e^{-x^2}\,dx = \sum_{n=0}^{\infty} \dfrac{(-1)^n (0.5)^{2n+1}}{(2n + 1)\,n!}

0.5 - \dfrac{0.125}{3} + \dfrac{0.03125}{10} - \dfrac{0.0078125}{42}

0.5 - 0.0416667 + 0.003125 - 0.0001860 \approx 0.4612723

\dfrac{(0.5)^9}{9 \cdot 4!} \approx 0.0000090

Answer: about 0.46127, with error less than 0.00001. (The true value is 0.461281.)

Example 5: Expected number of coin flips (word problem)

You flip a fair coin until it lands heads. The expected number of flips is \sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^n. Find this sum.

\sum_{n=1}^{\infty} n x^{n-1} = \dfrac{1}{(1 - x)^2}

\sum_{n=1}^{\infty} n x^n = \dfrac{x}{(1 - x)^2}

\sum_{n=1}^{\infty} n\left(\dfrac{1}{2}\right)^n = \dfrac{1/2}{(1/2)^2}

\dfrac{1/2}{1/4} = 2

Answer: on average it takes 2 flips to get heads.

Common mistakes

Related methods

Checking with a known series

Differentiating the series for e^x gives back the same series, and differentiating the sine series gives the cosine series. These checks are a quick way to see that a series is right; see the Maclaurin series table.

Estimating the error

When an integrated series alternates, as in Example 4, the next term bounds the error. When it doesn’t, use the Lagrange error bound or compare the tail with a geometric series.

Practice problems

1. Differentiate \sum_{n=0}^{\infty} \dfrac{x^n}{n!} term by term. What do you notice?
2. Find a power series for \dfrac{1}{(1 + x)^2} by differentiating the series for \dfrac{1}{1 + x}.
3. Find a power series for ln(1 − x) by integrating, and give its interval of convergence.
4. Find a power series for \int_0^x \sin(t^2)\,dt.
5. Use three terms to estimate \int_0^1 \sin(x^2)\,dx, and bound the error.
6. Find the interval of convergence of \sum_{n=0}^{\infty} \dfrac{x^{n+1}}{n + 1}, the integral of the geometric series.
7. Use the series for \frac{1}{(1 - x)^2} to find \sum_{n=0}^{\infty} \dfrac{n + 1}{3^n}.
8. Differentiate f(x) = \sum_{n=1}^{\infty} \dfrac{x^n}{n^2} term by term, and give the radius of convergence of f′.
9. Use three terms of the arctan series to estimate arctan 0.2.
10. A sensor’s signal strength at time t is e^{-t^2}. Use three terms of a series to estimate the total signal \int_0^{0.2} e^{-t^2}\,dt.

\dfrac{d}{dx}\sum_{n=0}^{\infty} \dfrac{x^n}{n!} = \sum_{n=1}^{\infty} \dfrac{n x^{n-1}}{n!}

\sum_{n=1}^{\infty} \dfrac{n x^{n-1}}{n!} = \sum_{n=1}^{\infty} \dfrac{x^{n-1}}{(n - 1)!} = \sum_{n=0}^{\infty} \dfrac{x^n}{n!}

\dfrac{d}{dx}\sum_{n=0}^{\infty} (-1)^n x^n = \sum_{n=1}^{\infty} (-1)^n n x^{n-1}

\dfrac{d}{dx}\left[\dfrac{1}{1 + x}\right] = -\dfrac{1}{(1 + x)^2}

\dfrac{1}{(1 + x)^2} = \sum_{n=1}^{\infty} (-1)^{n+1} n x^{n-1} = \sum_{n=0}^{\infty} (-1)^n (n + 1)x^n

\ln(1 - x) = -\int \dfrac{1}{1 - x}\,dx = C - \sum_{n=0}^{\infty} \dfrac{x^{n+1}}{n + 1}

\ln 1 = C - 0 \quad \Rightarrow \quad C = 0

\text{At } x = -1: \; -\sum \dfrac{(-1)^{n+1}}{n + 1} \text{ converges}; \quad \text{at } x = 1: \; -\sum \dfrac{1}{n + 1} \text{ diverges}

\sin(t^2) = \sum_{n=0}^{\infty} \dfrac{(-1)^n t^{4n+2}}{(2n + 1)!}

\int_0^x \sin(t^2)\,dt = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{4n+3}}{(4n + 3)(2n + 1)!}

\dfrac{1}{3} - \dfrac{1}{42} + \dfrac{1}{1320}

0.3333333 - 0.0238095 + 0.0007576 \approx 0.3102814

\dfrac{1}{15 \cdot 7!} = \dfrac{1}{75{,}600} \approx 0.0000132

L = \lim_{n \to \infty} |x| \cdot \dfrac{n + 1}{n + 2} = |x|

\text{At } x = 1: \; \sum \dfrac{1}{n + 1} \text{ diverges}; \quad \text{at } x = -1: \; \sum \dfrac{(-1)^{n+1}}{n + 1} \text{ converges}

\sum_{n=0}^{\infty} (n + 1)x^n = \dfrac{1}{(1 - x)^2}

\sum_{n=0}^{\infty} \dfrac{n + 1}{3^n} = \dfrac{1}{(1 - \frac{1}{3})^2} = \dfrac{1}{4/9}

\dfrac{1}{4/9} = \dfrac{9}{4}

f'(x) = \sum_{n=1}^{\infty} \dfrac{n x^{n-1}}{n^2} = \sum_{n=1}^{\infty} \dfrac{x^{n-1}}{n}

L = \lim_{n \to \infty} |x| \cdot \dfrac{n}{n + 1} = |x|

\arctan 0.2 \approx 0.2 - \dfrac{0.008}{3} + \dfrac{0.00032}{5}

0.2 - 0.0026667 + 0.000064 \approx 0.1973973

\int_0^{0.2} e^{-t^2}\,dt = 0.2 - \dfrac{(0.2)^3}{3} + \dfrac{(0.2)^5}{10} - \cdots

0.2 - 0.0026667 + 0.000032 \approx 0.1973653

Differentiating and integrating power series FAQ

Can you differentiate a power series term by term?

Yes. Inside its interval of convergence, a power series can be differentiated one term at a time, and the result has the same radius of convergence.

Does the radius of convergence change when you integrate a power series?

No, the radius stays the same. The endpoints can change, though: the integral of the geometric series converges at x = −1, but the geometric series does not.

How do you find the constant of integration?

Put the center (often x = 0) into both sides. Every term with a power of (x − a) becomes 0, so you can solve for C.

How do you integrate e^(−x²)?

There is no basic antiderivative, but you can integrate its series term by term: \int e^{-x^2}\,dx = C + \sum \frac{(-1)^n x^{2n+1}}{(2n + 1)\,n!}.

Summary

• Differentiate or integrate a power series one term at a time, like a polynomial.
• The radius of convergence stays the same; retest the endpoints.
• After differentiating, start at n = 1. After integrating, find C at the center.
• Integrating geometric series gives ln(1 + x) and arctan x.
• Series let you estimate integrals like \int e^{-x^2}\,dx and find sums like \sum n/2^n.

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