Finding Taylor or Maclaurin series for a function — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Definition of Taylor and Maclaurin series, direct coefficient calculation via the definition, constructing series via substitution, term-by-term differentiation, integration, and scaling by powers of x, aligned to AP CED requirements.

You should already know: How to compute higher-order derivatives of elementary functions, geometric series formulas, and term-by-term operations for convergent power series.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Finding Taylor or Maclaurin series for a function?

Finding Taylor or Maclaurin series means constructing an infinite power series representation of a given infinitely differentiable function, where the series coefficients are defined to match the function’s value and all orders of derivative at a chosen center point (a). A Maclaurin series is simply a Taylor series with center (a = 0), the most common center tested on the AP exam. This topic is a core skill in Unit 10 (Infinite Sequences and Series), which makes up 17-18% of the AP Calculus BC exam overall; specific questions on finding series appear in both multiple-choice (MCQ) and free-response (FRQ) sections, and are often combined with related skills like finding radius of convergence or approximating functions with series. Unlike finite Taylor polynomials, a Taylor series is infinite, and converges to the original function on its interval of convergence. Mastery of this topic is required for almost all other series application questions on the exam.

2. Direct Computation of Taylor Coefficients (The Definition Method)

The formal definition of a Taylor series centered at (a) for a function (f(x)) is: $$f(x)=\sum _{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(x-a{)}^n$$ where (f^{(n)}(a)) is the (n)-th derivative of (f) evaluated at (a), and by convention (f^{(0)}(a) = f(a)) (the 0-th derivative is the function itself). For a Maclaurin series, this simplifies to: $$f(x)=\sum _{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}{x}^n$$ The intuition behind this formula is that we build the series term by term to match all properties of (f(x)) at (a): the constant term matches (f(a)), the linear term matches the slope (first derivative) at (a), the quadratic term matches the concavity (second derivative), and so on for all higher derivatives. The (n!) in the denominator comes from differentiating the power series term: when you differentiate ((x-a)^n) (n) times, you get (n!), so dividing by (n!) cancels this out to leave exactly (f^{(n)}(a)). The direct method requires computing the first few derivatives, identifying a pattern in coefficients, and writing the general term of the series. This method is most often tested when the function does not fit a standard known series, or when the center is non-zero.

Worked Example

Problem: Find the first four non-zero terms of the Maclaurin series for (f(x) = \sin(3x)) using the definition.

1. Compute the first four non-zero derivatives: (f^{(0)}(x) = \sin(3x)), (f^{(1)}(x) = 3\cos(3x)), (f^{(2)}(x) = -9\sin(3x)), (f^{(3)}(x) = -27\cos(3x)), (f^{(4)}(x) = 81\sin(3x)), (f^{(5)}(x) = 243\cos(3x)).
2. Evaluate derivatives at (a=0): (f^{(0)}(0) = 0), (f^{(1)}(0) = 3), (f^{(2)}(0) = 0), (f^{(3)}(0) = -27), (f^{(4)}(0) = 0), (f^{(5)}(0) = 243).
3. Compute coefficients as (\frac{f^{(n)}(0)}{n!}): For (n=1): (\frac{3}{1!} = 3); for (n=3): (\frac{-27}{3!} = -\frac{9}{2}); for (n=5): (\frac{243}{5!} = \frac{81}{40}); next non-zero is (n=7): (\frac{-3^7}{7!} = -\frac{2187}{5040} = -\frac{81}{186}).
4. Write the first four non-zero terms: (\sin(3x) = 3x - \frac{9}{2}x^3 + \frac{81}{40}x^5 - \frac{81}{186}x^7 + \dots)

Exam tip: Always count non-zero terms carefully: even/odd functions like sine and cosine will have half their terms equal to zero, so it is easy to accidentally stop too early or too late.

3. Constructing Series by Substitution

Once you memorize the standard Maclaurin series for common functions, you do not need to recompute coefficients from scratch for most exam problems. The most common shortcut is substitution, where you replace (x) in a known series with a function of (x) (like (kx), (x^k), or (-x^2)) to get the series for a new related function. The key rule is that substitution preserves the structure of the series: if (f(u) = \sum_{n=0}^{\infty} c_n u^n), then (f(g(x)) = \sum_{n=0}^{\infty} c_n (g(x))^n) within the new interval of convergence. The radius of convergence will adjust to match the substitution, but the general term pattern just follows directly from the substitution. This method is tested more frequently than the direct definition method on the AP exam, because it tests your ability to manipulate known series rather than just compute derivatives.

Worked Example

Problem: Find the Maclaurin series for (g(x) = \frac{1}{1 + 5x^2}), including the general term.

1. Recall the standard geometric Maclaurin series for (\frac{1}{1 - u} = \sum_{n=0}^{\infty} u^n), valid for (|u| < 1).
2. Rewrite (g(x)) to match the geometric series form: (\frac{1}{1 + 5x^2} = \frac{1}{1 - (-5x^2)}), so we substitute (u = -5x^2) into the geometric series.
3. Substitute (u) into the general term, applying the exponent to the entire expression: (u^n = (-5x^2)^n = (-1)^n 5^n x^{2n}).
4. Write the full series: (\frac{1}{1 + 5x^2} = \sum_{n=0}^{\infty} (-1)^n 5^n x^{2n}), valid for (|x| < \frac{1}{\sqrt{5}}).

Exam tip: Always wrap the entire substituted expression in parentheses before applying the exponent. This avoids forgetting to raise constants and negative signs to the (n)-th power, the most common mistake with substitution.

4. Constructing Series by Differentiation and Integration

Power series can be differentiated and integrated term-by-term within their interval of convergence, so you can use this property to derive a new series from a known one. If (f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n), then: $$f^{\prime }(x)=\sum _{n=1}^{\infty }n{c}_n(x-a{)}^{n-1}$$ $$\int f(x)dx=C+\sum _{n=0}^{\infty }\frac{c_n}{n+1}(x-a{)}^{n+1}$$ Term-by-term differentiation and integration do not change the radius of convergence of the original series; only the convergence at the endpoints may change, which is covered in a later topic. This method is used when your target function is the derivative or integral of a function you already know the series for, which is common for functions like (\arctan(x)), (\ln(1+x)), and rational functions with powers in the denominator.

Worked Example

Problem: Find the Maclaurin series for (h(x) = \arctan(2x)) for (|x| < 1/2), including the general term.

1. Notice that the derivative of (h(x)) is (h'(x) = \frac{2}{1 + (2x)^2} = \frac{2}{1 + 4x^2}), which we can get via substitution.
2. Using the substitution method from the previous section, we know (\frac{1}{1 + 4x^2} = \sum_{n=0}^{\infty} (-1)^n 4^n x^{2n}), so multiply by 2 to get (h'(x) = \sum_{n=0}^{\infty} (-1)^n 2 \cdot 4^n x^{2n} = \sum_{n=0}^{\infty} (-1)^n 2^{2n+1} x^{2n}).
3. Integrate term-by-term from 0 to (x) to recover (h(x) = \arctan(2x) = \int_0^x h'(t) dt).
4. Swap the sum and integral, then integrate term-by-term: (\sum_{n=0}^{\infty} (-1)^n 2^{2n+1} \int_0^x t^{2n} dt = \sum_{n=0}^{\infty} (-1)^n 2^{2n+1} \frac{x^{2n+1}}{2n+1}).
5. Evaluate at (x=0): (\arctan(0) = 0), so the constant of integration is 0. The final series is (\arctan(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n+1}}{2n+1}).

Exam tip: When integrating to find a Maclaurin series, always confirm the constant of integration by plugging in (x=0) to the target function. For most common functions like logarithms and inverse trigonometric functions, (f(0) = 0) so (C=0), but always check for trick questions.

5. Adjusting Series by Multiplying by a Power of (x)

A common AP exam problem requires finding the series for a function that is a known function multiplied by (x^k) (or ((x-a)^k) for a general center). This simple manipulation is often tested because it checks your understanding of how power series terms work. When you multiply a power series (\sum_{n=0}^{\infty} c_n (x-a)^n) by ((x-a)^k), every exponent in the series increases by (k), and the coefficients do not change in value. This manipulation is almost always combined with substitution or integration to get the final series.

Worked Example

Problem: Write the first three non-zero terms of the Maclaurin series for (f(x) = x^3 e^{-2x}).

1. Recall the standard Maclaurin series for (e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots).
2. Substitute (u = -2x) to get (e^{-2x} = 1 - 2x + \frac{(4x^2)}{2} - \frac{(8x^3)}{6} + \dots = 1 - 2x + 2x^2 - \frac{4}{3}x^3 + \dots).
3. Multiply the entire series by (x^3), which adds 3 to every exponent: (x^3 e^{-2x} = x^3 \left(1 - 2x + 2x^2 - \frac{4}{3}x^3 + \dots\right)).
4. Simplify to get the first three non-zero terms: (x^3 - 2x^4 + 2x^5 + \dots).

Exam tip: When asked for the coefficient of (x^m) in (x^k f(x)), remember the coefficient comes from the (x^{m-k}) term in the original (f(x)) series, not the (x^m) term. Don't shift the index wrong.

6. Common Pitfalls (and how to avoid them)

• Wrong move: When substituting (u = -x^2) into (e^u = \sum \frac{u^n}{n!}), writing the general term as (\frac{-x^{2n}}{n!}) instead of (\frac{(-1)^n x^{2n}}{n!}). Why: Students forget to apply the exponent to the negative sign, only applying it to the (x^2) term. Correct move: Always wrap the entire substituted expression in parentheses before applying the exponent: ((-x^2)^n = (-1)^n (x^2)^n = (-1)^n x^{2n}).
• Wrong move: When computing a Taylor series centered at (a = 2), using memorized Maclaurin coefficients and just replacing (x) with (x-2) without re-evaluating derivatives. Why: Students confuse the general form of a Taylor series with pre-memorized Maclaurin coefficients, assuming shifting (x) is enough for a non-zero center. Correct move: For a non-zero center, either recompute coefficients using the definition at (a), or rewrite the function in terms of (u = x - a) and build the series around (u).
• Wrong move: When integrating term-by-term to find the series for (\ln(1+x)), leaving a non-zero constant of integration as the first term of the series. Why: Students forget that (\ln(1+0) = 0), so the constant of integration evaluated at the center is zero. Correct move: Always evaluate the integrated function at the center (a) to solve for (C); for Maclaurin series, plug in (x=0) to the target function to get (C).
• Wrong move: When computing the n-th derivative of (e^{kx}) for direct coefficient calculation, writing (f^{(n)}(0) = 1) instead of (k^n). Why: Students get used to derivatives of (e^x) and forget the chain rule gives an extra constant factor for the linear inner function. Correct move: For composite functions (f(kx)), always apply chain rule n times, which gives an extra factor of (k^n) for the n-th derivative.
• Wrong move: When asked for the coefficient of (x^5) in (x^2 \sin(x)), taking the coefficient of (x^5) from the (\sin(x)) series as the answer. Why: Students forget that multiplying by (x^2) shifts all exponents up by 2. Correct move: If multiplying by (x^k), the coefficient of (x^m) in the new series is equal to the coefficient of (x^{m-k}) in the original series. For this example, the coefficient of (x^5) in (x^2 \sin(x)) is the coefficient of (x^3) in (\sin(x)).

7. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following is the coefficient of (x^4) in the Maclaurin series for (f(x) = x e^{-2x})? A) (\frac{2}{3}) B) (-\frac{2}{3}) C) (\frac{4}{3}) D) (-\frac{4}{3})

Worked Solution: We start with the known Maclaurin series for (e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}). Substitute (u = -2x) to get (e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-2)^n x^n}{n!}). Multiply by (x) to get (f(x) = x e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2)^n x^{n+1}}{n!}). To find the coefficient of (x^4), set (n+1 = 4), so (n=3). Calculate the coefficient: (\frac{(-2)^3}{3!} = \frac{-8}{6} = -\frac{4}{3}). The correct answer is D.

Question 2 (Free Response)

Let (f(x) = \frac{1}{(1 - 3x)^2}). (a) Use the fact that (\frac{d}{dx}\left(\frac{1}{1 - 3x}\right) = \frac{3}{(1 - 3x)^2}) to find the Maclaurin series for (f(x)), including the general term. (b) Write the first four non-zero terms of the series for (f(x)). (c) What is the coefficient of (x^5) in the Maclaurin series for (x^2 f(x))?

Worked Solution: (a) Start with the geometric series: (\frac{1}{1 - 3x} = \sum_{n=0}^{\infty} (3x)^n = \sum_{n=0}^{\infty} 3^n x^n). Differentiate both sides term-by-term: (\frac{d}{dx}\left(\frac{1}{1 - 3x}\right) = \sum_{n=1}^{\infty} n \cdot 3^n x^{n-1} = \frac{3}{(1 - 3x)^2}). Divide by 3 to solve for (f(x)): (f(x) = \frac{1}{(1 - 3x)^2} = \sum_{n=1}^{\infty} n 3^{n-1} x^{n-1}). Reindexing with (k = n-1) gives the general term: (f(x) = \sum_{k=0}^{\infty} (k+1) 3^k x^k). (b) Substitute (k = 0, 1, 2, 3): (k=0: (0+1)3^0 x^0 = 1), (k=1: (1+1)3^1 x^1 = 6x), (k=2: (2+1)3^2 x^2 = 27x^2), (k=3: (3+1)3^3 x^3 = 108x^3). First four non-zero terms: (1 + 6x + 27x^2 + 108x^3). (c) Multiply by (x^2): (x^2 f(x) = \sum_{k=0}^{\infty} (k+1)3^k x^{k+2}). Set (k+2 = 5), so (k=3). Coefficient: ((3+1)3^3 = 4 \cdot 27 = 108).

Question 3 (Application / Real-World Style)

The concentration of a drug in a patient's bloodstream (t) hours after injection is approximated by (C(t) = \frac{1 - e^{-t} \cos t}{t}) for (t > 0), where (C(t)) is measured in milligrams per liter. Find the first three non-zero terms of the Maclaurin series for (C(t)), and use it to approximate (C(0.1)) to 4 decimal places.

Worked Solution: Use the standard Maclaurin series for (e^{-t}) and (\cos t), multiply them: (e^{-t} = 1 - t + \frac{t^2}{2!} - \frac{t^3}{3!} + \frac{t^4}{4!} - \dots), (\cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \dots). Multiplying up to (t^3) gives (e^{-t} \cos t = 1 - t + \frac{1}{3}t^3 - \dots). Subtract from 1: (1 - e^{-t} \cos t = t - \frac{1}{3}t^3 + \frac{1}{6}t^4 - \dots). Divide by (t) to get (C(t) = 1 - \frac{1}{3}t^2 + \frac{1}{6}t^3 - \dots). Substitute (t=0.1): (C(0.1) = 1 - \frac{1}{3}(0.01) + \frac{1}{6}(0.001) \approx 1 - 0.003333 + 0.000167 = 0.996834). Rounded to 4 decimal places, (C(0.1) \approx 0.9968). This means the approximate drug concentration 0.1 hours after injection is 0.9968 milligrams per liter.

8. Quick Reference Cheatsheet

9. What's Next

Now that you can construct Taylor and Maclaurin series for a wide range of functions, the next step in the AP Calculus BC syllabus is to determine the interval and radius of convergence for these series, then use them to approximate functions and bound approximation error. This chapter is an absolute prerequisite for all those topics: you cannot find the radius of convergence of a series you cannot correctly construct, and any approximation question relies on having the correct general term to work with. Beyond series, Taylor series are also used to evaluate limits of indeterminate forms that are difficult to compute with repeated applications of L'Hospital's Rule, so they connect back to the limits and derivatives you learned earlier in the course. The key follow-on topics you will study next are: Radius and interval of convergence for power series, Lagrange error bounds for Taylor polynomials, Approximating functions with Taylor polynomials