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How do you manipulate power series to represent new functions and evaluate hard integrals?

Topic 10.15 Representing Functions as Power Series: manipulate known power series by substitution, multiplication, differentiation and integration to represent new functions and evaluate otherwise intractable integrals (BC).

A focused answer to AP Calculus BC Topic 10.15, manipulating known power series by substitution, term-by-term differentiation and integration to represent new functions and evaluate integrals with no elementary antiderivative, with worked examples.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. The manipulation toolkit
  3. Why integrating a series beats hunting for an antiderivative
  4. A worked integral via series
  5. A worked function representation
  6. Combining series and the interval of validity

What this topic is asking

The College Board (Topic 10.15, BC only) is the capstone of Unit 10: using the standard power series as building blocks and manipulating them to represent new functions and to evaluate integrals that have no elementary antiderivative. This is where series stop being abstract and become a practical computational tool.

The manipulation toolkit

Why integrating a series beats hunting for an antiderivative

The deepest use of this topic is integrating functions with no elementary antiderivative. Functions like ex2e^{-x^2}, sinxx\frac{\sin x}{x}, and 11+x4\frac{1}{1 + x^4} cannot be antidifferentiated in closed form, so the Fundamental Theorem is no direct help. But each has a simple power series, and a power series can always be integrated term by term, producing a series for the integral. For 01ex2dx\int_0^1 e^{-x^2}\,dx, you write ex2=1x2+x42!e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \cdots, integrate each term to get xx33+x552!x - \frac{x^3}{3} + \frac{x^5}{5\cdot 2!} - \cdots, and evaluate. The result is a convergent series whose partial sums approximate the integral to any desired accuracy, often an alternating series, so the error is bounded by the first omitted term (Topic 10.10). This is the practical reason BC develops series at all.

A worked integral via series

A worked function representation

Combining series and the interval of validity

You can also add, subtract, and multiply series to represent combinations of functions, lining up like powers of xx. When you manipulate a series, the radius of convergence is preserved by substitution (adjusted for the substituted variable), differentiation, and integration, though the endpoints can gain or lose convergence. For example, integrating 11+x=(1)nxn\frac{1}{1+x} = \sum(-1)^n x^n (valid on x<1|x| < 1) gives the series for ln(1+x)\ln(1 + x), which now also converges at the endpoint x=1x = 1 (giving the alternating harmonic series for ln2\ln 2). Always state the interval on which your representation is valid, and remember that the operations of this topic are only legitimate inside the interval of convergence.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (BC, style)1 marksSection I (multiple choice, no calculator). Using 11x=xn\frac{1}{1 - x} = \sum x^n, the power series for 11x3\frac{1}{1 - x^3} is (A) x3n\sum x^{3n} (B) xn/3\sum x^{n/3} (C) 3xn\sum 3x^n (D) xnx3\sum x^n - x^3
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The correct answer is (A), x3n\sum x^{3n}.

Substitute x3x^3 for xx in the geometric series: 11x3=n=0(x3)n=n=0x3n=1+x3+x6+\frac{1}{1 - x^3} = \sum_{n=0}^{\infty}(x^3)^n = \sum_{n=0}^{\infty} x^{3n} = 1 + x^3 + x^6 + \cdots.

AP 2024 (BC, style)4 marksSection II (free response, no calculator). (a) Write the Maclaurin series for sinx\sin x. (b) Use it to write the series for sinxx\frac{\sin x}{x} and hence give a series for 01sinxxdx\int_0^1 \frac{\sin x}{x}\,dx (first two nonzero terms).
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A 4-point integrate-the-series problem.

(a) (1 point) sinx=xx33!+x55!\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots.
(b) (3 points) sinxx=1x23!+x45!\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots. Integrating term by term: 01sinxxdx=[xx333!+]01=1118+\int_0^1\frac{\sin x}{x}\,dx = \left[x - \frac{x^3}{3\cdot 3!} + \cdots\right]_0^1 = 1 - \frac{1}{18} + \cdots.

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