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Calculus II · Power & Taylor Series

Taylor Series — Practice Problems and Worked Solutions

Derivatives at one point build a polynomial model.

The formulas you need here

Taylor series

f (x) = n = 0 \sum \infty \frac{f ^{(n)} ( c )}{n !} (x - c)^{n}

What it says: Rebuild a function from its derivatives at a single point.
When to reach for it: Approximating a function near a point, or turning an unintegrable expression into a polynomial.
Watch out: The n! is not optional and it is a factorial, not a product of the derivatives. c = 0 gives the Maclaurin series.

e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots

Worth memorising alongside sin x, cos x and 1/(1−x).

16 worked problems

Every step is generated and checked by a computer algebra system.

1.
Find the degree-2 Maclaurin polynomial.
$f (x) = e^{x}$
Answer: $\frac{x ^{2}}{2} + x + 1$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{2} (x) = \frac{x ^{2}}{2} + x + 1$
2.
Find the degree-3 Maclaurin polynomial.
$f (x) = e^{x}$
Answer: $\frac{x ^{3}}{6} + \frac{x ^{2}}{2} + x + 1$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{3} (x) = \frac{x ^{3}}{6} + \frac{x ^{2}}{2} + x + 1$
3.
Find the degree-3 Maclaurin polynomial.
$f (x) = sin (x)$
Answer: $- \frac{x ^{3}}{6} + x$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{3} (x) = - \frac{x ^{3}}{6} + x$
4.
Find the degree-5 Maclaurin polynomial.
$f (x) = sin (x)$
Answer: $\frac{x ^{5}}{120} - \frac{x ^{3}}{6} + x$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{5} (x) = \frac{x ^{5}}{120} - \frac{x ^{3}}{6} + x$
5.
Find the degree-4 Maclaurin polynomial.
$f (x) = cos (x)$
Answer: $\frac{x ^{4}}{24} - \frac{x ^{2}}{2} + 1$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{4} (x) = \frac{x ^{4}}{24} - \frac{x ^{2}}{2} + 1$
6.
Find the degree-3 Maclaurin polynomial.
$f (x) = lo g (x + 1)$
Answer: $x (\frac{x ^{2}}{3} - \frac{x}{2} + 1)$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{3} (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x$
7.
Find the degree-4 Maclaurin polynomial.
$f (x) = e^{x}$
Answer: $\frac{x ^{4}}{24} + \frac{x ^{3}}{6} + \frac{x ^{2}}{2} + x + 1$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{4} (x) = \frac{x ^{4}}{24} + \frac{x ^{3}}{6} + \frac{x ^{2}}{2} + x + 1$
8.
Find the degree-2 Maclaurin polynomial.
$f (x) = cos (x)$
Answer: $1 - \frac{x ^{2}}{2}$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{2} (x) = 1 - \frac{x ^{2}}{2}$
9.
Find the degree-2 Maclaurin polynomial.
$f (x) = sin (x)$
Answer: $x$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{2} (x) = x$
10.
Find the degree-2 Maclaurin polynomial.
$f (x) = lo g (x + 1)$
Answer: $\frac{x ( 2 - x )}{2}$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{2} (x) = - \frac{x ^{2}}{2} + x$
11.
Find the degree-4 Maclaurin polynomial.
$f (x) = lo g (x + 1)$
Answer: $x (- \frac{x ^{3}}{4} + \frac{x ^{2}}{3} - \frac{x}{2} + 1)$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{4} (x) = - \frac{x ^{4}}{4} + \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x$
12.
Find the degree-6 Maclaurin polynomial.
$f (x) = cos (x)$
Answer: $- \frac{x ^{6}}{720} + \frac{x ^{4}}{24} - \frac{x ^{2}}{2} + 1$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{6} (x) = - \frac{x ^{6}}{720} + \frac{x ^{4}}{24} - \frac{x ^{2}}{2} + 1$
13.
Find the degree-3 Maclaurin polynomial.
$f (x) = e^{2 x}$
Answer: $\frac{4 x ^{3}}{3} + 2 x^{2} + 2 x + 1$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{3} (x) = \frac{4 x ^{3}}{3} + 2 x^{2} + 2 x + 1$
14.
Find the degree-3 Maclaurin polynomial.
$f (x) = sin (2 x)$
Answer: $- \frac{4 x ^{3}}{3} + 2 x$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{3} (x) = - \frac{4 x ^{3}}{3} + 2 x$
15.
Find the degree-5 Maclaurin polynomial.
$f (x) = e^{x}$
Answer: $\frac{x ^{5}}{120} + \frac{x ^{4}}{24} + \frac{x ^{3}}{6} + \frac{x ^{2}}{2} + x + 1$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{5} (x) = \frac{x ^{5}}{120} + \frac{x ^{4}}{24} + \frac{x ^{3}}{6} + \frac{x ^{2}}{2} + x + 1$
16.
Find the degree-4 Maclaurin polynomial.
$f (x) = cos (2 x)$
Answer: $\frac{2 x ^{4}}{3} - 2 x^{2} + 1$
Step-by-step solution
1. $P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( 0 )}{k !} x^{k}$
2. $P_{4} (x) = \frac{2 x ^{4}}{3} - 2 x^{2} + 1$

The formulas you need here

Taylor series

16 worked problems

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