Taylor polynomials, question 10

Topic: Taylor polynomials
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §9 CLP-1 §3.4Key results

Find the third-order Taylor polynomial $T_{3}(x)$ of $f(x) = \dfrac{1}{x}$ about $x = 2$ .

Your answer

A polynomial in x. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.

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Theory UBC Math 100 §9 CLP-1 §3.4Key results

Worked solution

Strategy

Differentiate the power x⁻¹ three times, evaluate at the centre 2, and divide each derivative by its factorial. The polynomial comes out in powers of x − 2, the step from the centre.

Working

$f'(x) = -x^{-2}, \qquad f''(x) = 2x^{-3}, \qquad f'''(x) = -6x^{-4}$
The power rule three times; each step brings down a negative exponent, so the signs alternate.
$f(2) = \tfrac{1}{2}, \quad f'(2) = -\tfrac{1}{4}, \quad f''(2) = \tfrac{1}{4}, \quad f'''(2) = -\tfrac{3}{8}$
Evaluate at the centre: 2³ = 8 and 2⁴ = 16.
$\begin{aligned} T_{3}(x) &= \tfrac{1}{2} - \tfrac{1}{4}(x - 2) + \frac{1/4}{2!}(x - 2)^{2} \\ &\quad + \frac{-3/8}{3!}(x - 2)^{3} \end{aligned}$
Divide each derivative by its factorial.
$= \tfrac{1}{2} - \tfrac{1}{4}(x - 2) + \tfrac{1}{8}(x - 2)^{2} - \tfrac{1}{16}(x - 2)^{3}$
Expanded, this is 2 − 3x/2 + x²/2 − x³/16: the same polynomial in another form.

Answer

\frac{1}{2} - \frac{1}{4}(x - 2) + \frac{1}{8}(x - 2)^{2} - \frac{1}{16}(x - 2)^{3}

Check

At x = 2.2, T₃ gives 0.5 − 0.05 + 0.005 − 0.0005 = 0.4545, and 1/2.2 is 0.4545 to four places.

Where this goes wrong

Losing the alternating signs. Each derivative of x⁻¹ brings down a negative exponent, so the derivatives at 2 are negative, positive and negative, and the terms of T₃ alternate in sign.

Leans on

The power rule for negative powers, from Differentiation rules

Try a similar question

The third-order Taylor polynomial of a function

f

about

x = -1

is given below. Find

f''(-1)

T_{3}(x) = 6 + 3(x + 1) - \tfrac{9}{2}(x + 1)^{2} + \tfrac{1}{3}(x + 1)^{3}

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Taylor polynomials practice question 10