Taylor Series Practice Problems with Answers and Solutions

These Taylor series practice problems cover building Maclaurin and Taylor series, Taylor polynomials, estimates with error bounds, and using series to find limits, derivatives and sums. Try each one first. The answers come first, then full solutions.

The formula:

For the reasoning and more examples, read the Taylor series guide.

Problems

Building series

1. Find the first four nonzero terms of the Maclaurin series for e^{2x}.
2. Find the first three nonzero terms of the Maclaurin series for \cos(x^2).
3. Find the first four nonzero terms of the Maclaurin series for x\,e^{-x}.
4. Find the Taylor series for \ln x centered at a = 1.
5. Find the degree 3 Taylor polynomial for \sqrt{x} centered at a = 4.
6. Use the series for \frac{1}{1 + x^2} to find the Maclaurin series for \arctan x.

Estimates

7. Use the degree 2 Taylor polynomial from problem 5 to estimate \sqrt{4.1}.
8. Estimate \sin(0.5) with x - \frac{x^3}{6}, and bound the error.

Using series

9. Find f^{(10)}(0) for f(x) = x^2\cos x.
10. Find \displaystyle\lim_{x\to 0}\frac{\sin x - x}{x^3}.
11. Find \displaystyle\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}.
12. Find the sum \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n\pi^{2n}}{(2n)!}.

Answers

Step-by-step solutions

1. Replace x with 2x in e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots:

2. Replace x with x^2 in \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots:

3. e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots. Multiply each term by x.

4. For f(x) = \ln x: f(1) = 0, f'(1) = 1, f''(1) = -1, f'''(1) = 2, and in general f^{(n)}(1) = (-1)^{n+1}(n - 1)!. Divide by n! to get \frac{(-1)^{n+1}}{n}:

5. Take derivatives at x = 4:

6. \frac{1}{1 + x^2} = 1 - x^2 + x^4 - \cdots for |x| < 1. Integrate term by term from 0 to x:

7. With x - 4 = 0.1:

The true value is 2.0248457\ldots, so the estimate is off by about 0.000002.

8. 0.5 - \frac{0.125}{6} \approx 0.479167. The series alternates with decreasing terms, so the error is less than the next term, \frac{0.5^5}{120} \approx 0.00026. The true value is 0.479426, off by 0.000259.

9. x^2\cos x = x^2 - \frac{x^4}{2!} + \frac{x^6}{4!} - \frac{x^8}{6!} + \frac{x^{10}}{8!} - \cdots. The coefficient of x^{10} is \frac{1}{8!}, and it also equals \frac{f^{(10)}(0)}{10!}:

10. \sin x - x = -\frac{x^3}{6} + \frac{x^5}{120} - \cdots. Divide by x^3: -\frac{1}{6} + \frac{x^2}{120} - \cdots \to -\frac{1}{6}.

11. e^x - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \cdots. Divide by x^2: \frac{1}{2} + \frac{x}{6} + \cdots \to \frac{1}{2}.

12. This is the cosine series \sum\frac{(-1)^n x^{2n}}{(2n)!} with x = \pi, so the sum is \cos\pi = -1.

Common mistakes on these problems

• Forgetting the factorial. The coefficient is \frac{f^{(n)}(a)}{n!}, not f^{(n)}(a). In problem 5, f''(4) = -\frac{1}{32} but the coefficient is -\frac{1}{64}.
• Taking derivatives when substitution works. Differentiating \cos(x^2) eight times is slow. Substitute into the known series.
• Using x instead of x - a. A Taylor series at a = 4 uses powers of (x - 4).
• Mixing up which coefficient gives which derivative. In problem 9, the x^{10} coefficient comes from the n = 4 term of cosine, and you multiply by 10!.

Keep practicing

For unlimited problems with instant feedback, try the Taylor polynomials lesson, the Maclaurin series lesson and the estimating with Taylor lesson. Then try the parametric and polar problems, which finish the same Calculus 2 unit.