Power series let us write many functions as infinite polynomials, which makes calculus easier to do and easier to approximate. Term-by-term differentiation and integration mean taking the derivative or antiderivative of each separate power term in the series. This matters because it turns difficult functions into manageable algebraic patterns.
It is a core tool for Taylor series, approximations, differential equations, and physics models near an expansion point.
If a power series converges for |x - a| < R, then its derivative series and integral series have the same radius of convergence R. Inside that interval, the operations are safe because power series converge uniformly on smaller closed intervals. Endpoints must be checked separately because differentiating or integrating can change convergence there.
For example, from 1/(1 - x) = sum from n = 0 to infinity of x^n, we can integrate term by term to get -ln(1 - x) = sum from n = 0 to infinity of x^(n + 1)/(n + 1) for |x| < 1.
Understanding Calculus: Term-by-Term Differentiation and Integration
Each power term carries two kinds of information. Its coefficient sets the size of that contribution. Its exponent controls how the contribution changes as the input moves away from the center of the series.
Differentiation lowers every exponent by one and multiplies the coefficient by the old exponent. This makes higher powers matter more in the derivative. The constant term disappears because it does not change with the input.
Integration does the reverse pattern. It raises each exponent by one and divides by the new exponent. A separate constant must then be included, since differentiation would erase it.
The common radius of convergence is useful, but it does not settle every detail. It tells students where the series behaves reliably in the open interval around its center. At the boundary points, the balance can be delicate.
Consider the geometric series evaluated at negative one. Its terms switch between one and negative one, so the original series does not settle to a single value. After integration, the boundary terms become alternating fractions.
Those fractions shrink, and the resulting series can converge. After differentiation, the terms contain growing whole numbers, so they cannot even approach zero. This is why endpoint testing is a separate step rather than a routine detail.
Series calculus is especially valuable when an exact function is hard to calculate directly. A calculator may evaluate sine, cosine, exponential functions, or logarithms instantly, but a computer chip does this using finite procedures. Near a chosen center, it can use a limited number of power terms.
Keeping more terms usually improves the estimate when the input stays inside the useful range. The first omitted terms give clues about the likely error, though a careful error bound needs more work.
In physics, small displacement models often use only the first few terms because the higher powers are tiny. This is why a complicated curve can behave almost like a line over a short distance.
Power series are closely tied to Taylor series because derivatives determine the coefficients. The value of a function at the center gives the constant coefficient. The first derivative sets the linear coefficient.
Higher derivatives control the higher powers after the appropriate factorial division. Term-by-term work therefore lets students move in both directions. They can start with a known series and produce new functions, or use derivative data to build a series approximation.
In differential equations, substituting a power series often turns one difficult equation into a repeating rule for the coefficients. When learning this topic, keep the center visible, track index changes carefully, and write the first few terms before using a general pattern. Most mistakes come from a shifted exponent, a missing constant, or an untested endpoint.
Key Facts
- If f(x) = sum from n = 0 to infinity of c_n(x - a)^n, then f'(x) = sum from n = 1 to infinity of n c_n(x - a)^(n - 1).
- An antiderivative is integral f(x) dx = C + sum from n = 0 to infinity of c_n(x - a)^(n + 1)/(n + 1).
- The original series, derivative series, and integral series all have the same radius of convergence R.
- The interval endpoints x = a - R and x = a + R must be tested separately after differentiating or integrating.
- Geometric series formula: 1/(1 - x) = sum from n = 0 to infinity of x^n for |x| < 1.
- Worked example: If f(x) = sum from n = 0 to infinity of x^n, then f'(x) = sum from n = 1 to infinity of n x^(n - 1) = 1/(1 - x)^2 for |x| < 1.
Vocabulary
- Power series
- A series of the form sum c_n(x - a)^n, where a is the center and c_n are constants.
- Radius of convergence
- The nonnegative number R such that a power series converges for |x - a| < R and diverges for |x - a| > R.
- Interval of convergence
- The set of x-values where a power series converges, including any endpoints that pass separate tests.
- Term-by-term differentiation
- The process of differentiating each term of a convergent power series to form a new series for the derivative.
- Term-by-term integration
- The process of integrating each term of a convergent power series to form a new series for an antiderivative.
Common Mistakes to Avoid
- Forgetting to start the derivative sum at n = 1 is wrong because the n = 0 constant term differentiates to 0.
- Assuming endpoint behavior is unchanged is wrong because differentiating or integrating can make an endpoint series converge or diverge differently.
- Changing the radius of convergence after term-by-term calculus is wrong for power series because the derivative and integral series keep the same radius R.
- Dropping the constant of integration is wrong because an indefinite integral represents a whole family of functions, so + C is required.
Practice Questions
- 1 Given f(x) = sum from n = 0 to infinity of 3x^n with |x| < 1, find a power series for f'(x) and simplify it as a rational function.
- 2 Find a power series for integral 1/(1 + x^2) dx centered at 0 by using 1/(1 + x^2) = sum from n = 0 to infinity of (-1)^n x^(2n), and state the radius of convergence.
- 3 A power series has radius of convergence R = 4 centered at x = 2. Explain why its derivative and integral series also have radius 4, and describe what still must be checked.