Taylor polynomials, question 15

Topic: Taylor polynomials
Difficulty: Capstone
Suggested time: 10 min
Marks: 5
Theory: UBC Math 100 §9 CLP-1 §3.4Key results

Use the second-order Taylor polynomial of $f(x) = \sqrt{x}$ about $x = 16$ to estimate $\sqrt{17}$ , and use the Taylor remainder to give a bound on the error of your estimate.

Your answer

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

Worked solution

Strategy

Build the quadratic approximation from exact derivatives at 16, evaluate it one unit away, and bound the third-derivative remainder using the worst case on the interval.

Working

$f(16) = 4, \quad f'(16) = \frac{1}{2\cdot 4} = \frac{1}{8}, \quad f''(16) = -\frac{1}{4\cdot 64} = -\frac{1}{256}$
Derivatives at the base point.
$T_{2}(x) = 4 + \frac{1}{8}(x - 16) - \frac{1}{512}(x - 16)^{2}$
f″(16)/2 = −1/512.
$\sqrt{17} \approx T_{2}(17) = 4 + \frac{1}{8} - \frac{1}{512} = 4.123046875$
The true value is 4.1231056…
$|R_{2}(17)| = \frac{|f'''(c)|}{3!}\cdot 1^{3} \leq \frac{1}{6}\cdot\frac{3}{8\cdot 16^{5/2}} = \frac{1}{6}\cdot\frac{3}{8192} = \frac{1}{16384} \approx 6.1\times 10^{-5}$
f‴ is positive and decreasing, so its maximum on (16, 17) is at 16, where 16^5/2 = 1024. The actual error is about 5.9 × 10⁻⁵, inside the bound.

Answer

\sqrt{17} \approx 4.1230, \qquad |\text{error}| \leq \frac{1}{16384}

What earns the marks

Computes f(16) = 4, f′(16) = 1/8 and f″(16) = −1/256, and writes T₂(x) = 4 + (x − 16)/8 − (x − 16)²/512.
Evaluates T₂(17) = 4 + 1/8 − 1/512 ≈ 4.1230.
Bounds the remainder: |R₂(17)| = |f‴(c)|/3! · 1³ with f‴(x) = 3/(8x^5/2) and c in (16, 17), so |R₂(17)| ≤ 3/(8 · 16^5/2 · 6) = 1/16384 ≈ 6.1 × 10⁻⁵.

Check

Squaring back, 4.123046875² ≈ 16.9995, and the actual error, about 0.000059, is inside the bound 1/16384 ≈ 0.000061.

Where this goes wrong

Bounding the remainder with the second derivative instead of the third, or evaluating f‴ at 17 instead of finding its maximum on the interval. The remainder after T₂ uses f‴, and the bound must cover every c in (16, 17).