Newton’s method, question 15: UBC MATH 100 practice

Question 15 of 16Capstone6 marks · 10 minTheory UBC Math 100 §10.1Key results

The equation $e^{x} = 3 - x$ has exactly one solution, $r$ . Write it as $f(x) = 0$ and find $x_{1}$ and $x_{2}$ exactly by Newton’s method from $x_{0} = 0$ . Then use the concavity of $f$ to decide, without finding $r$ , whether $x_{2}$ is above or below $r$ .

Explain also why every later estimate lies on the same side of r, and why each is closer than the one before. Compare your working with the model solution.

Your answer

Theory UBC Math 100 §10.1Key results

Worked solution

Strategy

Compute the two steps exactly. Then compare each tangent line with the curve: concavity says which is higher, which fixes the sign of f at the next estimate, and the sign of f fixes the side of r.

Working

$f(x) = e^{x} + x - 3, \qquad f'(x) = e^{x} + 1, \qquad f''(x) = e^{x}$
Every term on one side; r is the zero of f.
$x_{1} = 0 - \frac{f(0)}{f'(0)} = 0 - \frac{-2}{2} = 1$
f(0) = 1 − 3 = −2 and f′(0) = 2.
$x_{2} = 1 - \frac{e - 2}{e + 1} = \frac{(e + 1) - (e - 2)}{e + 1} = \frac{3}{e + 1} \approx 0.807$
f(1) = e − 2 and f′(1) = e + 1; over a common denominator the e terms cancel.
$f(x_{n+1}) > f(x_{n}) + f'(x_{n})\,(x_{n+1} - x_{n}) = 0$
f″ = e^x > 0, so the curve lies above the tangent at x_n, and x_n+1 is where that tangent is zero.
$f'(x) = e^{x} + 1 > 0 \implies \bigl(f(x) > 0 \iff x > r\bigr)$
f is increasing and zero only at r, so it is positive exactly to the right of r.
$x_{1},\ x_{2},\ x_{3},\ \ldots > r$
Every estimate from x₁ on overshoots. In particular x₂ = 3/(e + 1) is above r, which is 0.7921 to four places.
$x_{n} > r \implies x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} < x_{n}$
Above r, f and f′ are both positive, so each step goes left: the estimates decrease while staying above r.

Answer

x_{1} = 1, \quad x_{2} = \frac{3}{e + 1} \approx 0.807 > r; \quad \text{later estimates stay above } r \text{ and decrease}

What earns the marks

Writes f(x) = e^x + x − 3, or an equivalent with every term on one side, and f′(x) = e^x + 1.
Finds x₁ = 0 − (−2)/2 = 1 and x₂ = 1 − (e − 2)/(e + 1) = 3/(e + 1) ≈ 0.807.
Notes that f″(x) = e^x > 0, so f is concave up and its graph lies above every tangent line; at the point where a tangent line crosses the x-axis, f is therefore positive.
Uses f′ > 0 (f is increasing, so it is positive only to the right of r) to conclude that x₁, x₂ and every later estimate are larger than r.
Explains that at an estimate above r, f and f′ are both positive, so the step f/f′ is positive and the next estimate is smaller: the estimates decrease while staying above r.

Check

f(x₂) ≈ 0.048 is positive and f is increasing, so x₂ ≈ 0.807 does lie above r, as concavity predicted.

Where this goes wrong

Concluding from x₀ < r that the estimates approach r from below. The first step jumps past r, because a concave-up curve lies above its tangent lines, and from then on every estimate overshoots.