Taylor polynomials, question 14

Topic: Taylor polynomials
Difficulty: Challenging
Suggested time: 6 min
Marks: 4
Theory: UBC Math 100 §9 CLP-1 §3.4Key results

Let $f(x) = x^{2}\cos\bigl(x^{2}\bigr)$ . Find $f^{(10)}(0)$ , the tenth derivative of $f$ at $0$ .

Your answer

A whole number.

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

Worked solution

Strategy

A polynomial of degree at most 10 that differs from f by at most a constant times x¹¹ near 0 is f’s Maclaurin polynomial of degree 10, however it was found. Build one from the polynomial of cos u by substitution, keeping a bound on the error, then read the tenth derivative off the coefficient of x¹⁰.

Working

$\cos u = 1 - \frac{u^{2}}{2} + \frac{u^{4}}{24} + R(u), \qquad |R(u)| \le \frac{u^{6}}{720}$
The Maclaurin polynomial of cos of degree 5, whose u⁵ term is 0, with the Lagrange remainder: |cos⁽⁶⁾| ≤ 1.
$x^{2}\cos\bigl(x^{2}\bigr) = x^{2} - \frac{x^{6}}{2} + \frac{x^{10}}{24} + E(x), \qquad |E(x)| \le \frac{x^{14}}{720}$
Put u = x², so u⁴ = x⁸ and |R(x²)| ≤ x¹²/720, then multiply by x².
$\frac{f^{(10)}(0)}{10!} = \frac{1}{24} \quad\Longrightarrow\quad f^{(10)}(0) = \frac{10!}{24} = \frac{3628800}{24} = 151200$
The polynomial has degree 10 and differs from f by at most x¹⁴/720, so it is f’s Maclaurin polynomial of degree 10, whose x¹⁰ coefficient is f⁽¹⁰⁾(0)/10!.

Answer

f^{(10)}(0) = \frac{10!}{24} = 151200

Check

The same reading gives f″(0) = 2!·1 = 2, which two rounds of the product rule confirm, and f⁽⁸⁾(0) = 0, since the polynomial has no x⁸ term.

Where this goes wrong

Stopping at the coefficient 1/24. The coefficient of xⁿ is f⁽ⁿ⁾(0)/n!, so the tenth derivative is 10! times it.

Leans on

Higher-order derivatives, from Differentiation rules

Try a similar question

The third-order Maclaurin polynomial of a function

f

is given below. Find

f'''(0)

T_{3}(x) = 2 - x + 5x^{2} - \tfrac{4}{3}x^{3}

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Taylor polynomials practice question 14