Chapter 11 — Exercises

Practice with linear approximation, differentials and error propagation, Newton's method by hand and in Python, convergence and failure analysis, and applications. Work the problems before opening the solutions. A calculator is fine for checking, but most of these are designed to be done with arithmetic you can carry on paper.

These problems follow the chapter in order: writing the linearization $L(x)$ (§11.2–11.3), using differentials to estimate change (§11.4), propagating measurement error (§11.5), running Newton's method by hand (§11.6–11.8), diagnosing failure (§11.9), and applying all of it across fields (§11.10).

Difficulty tiers: ⭐ routine · ⭐⭐ standard · ⭐⭐⭐ challenging · ⭐⭐⭐⭐ deep / multi-step.

Part A — Linearization by Hand (§11.2–11.3)

A1. ⭐ Find the linearization $L(x)$ of $f(x) = \sqrt{x}$ at $a = 9$, and use it to estimate $\sqrt{9.2}$.

A2. ⭐ Find $L(x)$ for $f(x) = x^3$ at $a = 2$, and estimate $(2.05)^3$.

A3. ⭐ Find $L(x)$ for $f(x) = \ln x$ at $a = 1$ and estimate $\ln(0.97)$.

A4. ⭐ Find $L(x)$ for $f(x) = \sin x$ at $a = 0$ and estimate $\sin(0.2)$ (radians). State the error against the true value $\sin(0.2) = 0.198669\ldots$

A5. ⭐ Use linear approximation at a convenient anchor to estimate $\sqrt[3]{27.5}$.

A6. ⭐ Find $L(x)$ for $f(x) = e^x$ at $a = 0$ and estimate $e^{0.1}$.

A7. ⭐ Find $L(x)$ for $f(x) = 1/x$ at $a = 2$ and estimate $1/2.1$.

A8. ⭐ Use the linearization of $f(x) = (1+x)^k$ at $a = 0$ to write the general "small-$x$" rule $(1+x)^k \approx 1 + kx$. Then estimate $(1.01)^{20}$.

A9. ⭐ For $f(x) = \cos x$ at $a = 0$, show that the linearization is the constant $L(x) = 1$. Explain in one sentence why the linear term vanishes here.

Part B — Differentials and Estimating Change (§11.4)

B1. ⭐⭐ For $y = x^4$, find $dy$ in terms of $x$ and $dx$. Evaluate $dy$ at $x = 3$ with $dx = 0.01$.

B2. ⭐⭐ The edge of a cube is measured as $s = 10$ cm and increases by $ds = 0.1$ cm. Use the differential $dV$ to estimate the change in volume $V = s^3$, then compare to the exact $\Delta V$.

B3. ⭐⭐ A circular oil slick has radius $r = 50$ m, growing by $dr = 0.5$ m. Estimate the change in area $A = \pi r^2$ using $dA$.

B4. ⭐⭐ For $y = \tan x$, find $dy$. If $x = \pi/4$ and $dx = 0.02$, estimate $\Delta y$.

B5. ⭐⭐ Use a differential to estimate how much $\sqrt{x}$ changes as $x$ goes from $100$ to $101$. Compare with the exact change.

B6. ⭐⭐ The period of a pendulum is $T = 2\pi\sqrt{\ell/g}$. With $g = 9.8\ \mathrm{m/s^2}$ fixed, find $dT$ in terms of $\ell$ and $d\ell$. If $\ell = 1$ m grows by $d\ell = 0.01$ m, estimate $\Delta T$.

B7. ⭐⭐⭐ A spherical balloon is inflated so that at the instant $r = 6$ cm the volume is increasing at $dV = 30\ \mathrm{cm^3}$ per second. Using $dV = 4\pi r^2\, dr$, estimate $dr$ (the rate the radius grows). Comment on how this is the differential form of a related-rates problem (Chapter 8).

Part C — Error Propagation (§11.5)

C1. ⭐⭐ A square's side is measured as $s = 20.0$ cm with possible error $\pm 0.1$ cm. Estimate the absolute and relative error in the area $A = s^2$.

C2. ⭐⭐ A sphere's radius is $r = 5$ cm with error $\pm 0.05$ cm. Estimate the absolute and percentage error in the volume $V = \frac{4}{3}\pi r^3$.

C3. ⭐⭐ Using the power-law rule $\dfrac{\Delta y}{y} \approx |n|\dfrac{\Delta x}{x}$ from §11.5, find the percentage error in $y = x^{1/2}$ when $x$ has a $4\%$ error. Then for $y = x^{-2}$ with the same $4\%$ input error.

C4. ⭐⭐ A resistor's power dissipation is $P = V^2/R$. If $V$ is known to $1\%$ and $R$ to $2\%$, estimate the worst-case percentage error in $P$. (Hint: relative errors of independent multiplicative factors add, and the exponent on $V$ is $2$.)

C5. ⭐⭐⭐ A pendulum gives $g = 4\pi^2 \ell / T^2$. The length $\ell$ is measured to $0.5\%$ and the period $T$ to $0.3\%$. Estimate the percentage error in $g$, and explain which measurement deserves more care.

C6. ⭐⭐⭐ The volume of a right circular cylinder is $V = \pi r^2 h$. If $r$ is measured to $1\%$ and $h$ to $1\%$, what is the approximate worst-case percentage error in $V$? Show how the exponents weight each input's relative error.

C7. ⭐⭐⭐⭐ An experimenter measures gravitational acceleration with a pendulum and wants the final result good to $1\%$. She can time a single swing (period $\approx 2$ s) to $\pm 0.1$ s, or time $N$ swings and divide. Using $\Delta g/g \approx \Delta\ell/\ell + 2\,\Delta T/T$ and assuming the length contributes $0.3\%$, find the smallest number of swings $N$ she must time so the total error stays under $1\%$. (Timing $N$ swings divides the per-swing timing error by $N$.)

Part D — Newton's Method by Hand (§11.6–11.8)

D1. ⭐⭐ Write the Newton iteration $x_{n+1} = x_n - f(x_n)/f'(x_n)$ for $f(x) = x^2 - 5$. Simplify it to the "averaging" form, then carry out two steps from $x_0 = 2$.

D2. ⭐⭐ Set up Newton's method to compute $\sqrt[3]{10}$. State $f$, $f'$, the simplified iteration, and one step from $x_0 = 2$.

D3. ⭐⭐ Apply Newton's method to $f(x) = x^2 - x - 1$ (whose positive root is the golden ratio $\varphi = 1.618\ldots$). Starting from $x_0 = 1.5$, compute $x_1$ and $x_2$.

D4. ⭐⭐ Use one Newton step on $f(x) = \cos x - x$ from $x_0 = 1$ to estimate the Dottie number (§11.10). Recall $f'(x) = -\sin x - 1$; use $\cos 1 = 0.5403$, $\sin 1 = 0.8415$.

D5. ⭐⭐⭐ For $f(x) = x^3 - 2x - 5$ (Newton's own historical example), start at $x_0 = 2$ and compute $x_1$ and $x_2$ to four decimals. The true root is $2.0945514\ldots$ — count the correct digits at each step and comment on the doubling.

D6. ⭐⭐⭐ The reciprocal $1/d$ can be found without dividing by applying Newton's method to $f(x) = 1/x - d$ (§11.7). Show that the iteration simplifies to $x_{n+1} = x_n(2 - d\,x_n)$, then compute $1/3$ to four decimals starting from $x_0 = 0.3$.

Part E — Convergence and Failure Analysis (§11.8–11.9)

E1. ⭐⭐⭐ The error of Newton's method near a simple root obeys $e_{n+1} \approx C\,e_n^2$. If a step has error $e_n = 10^{-2}$ and $C \approx 1$, estimate $e_{n+1}$, $e_{n+2}$, and $e_{n+3}$. How many steps to reach machine precision ($\approx 10^{-16}$)?

E2. ⭐⭐⭐ Explain, with reference to the iteration formula, what goes wrong if an iterate lands at a point where $f'(x_n) = 0$. Give a specific $f$ and starting point where this happens on the first step.

E3. ⭐⭐⭐ Apply Newton's method to $f(x) = x^3 - 2x + 2$ from $x_0 = 0$ by hand for three steps. Show that the iterates cycle, and identify the period of the cycle.

E4. ⭐⭐⭐ Apply Newton's method to the double root of $f(x) = (x-3)^2$ from $x_0 = 4$. Compute three iterates, show the error is halving (linear convergence), and explain why the quadratic-convergence guarantee fails here.

E5. ⭐⭐⭐⭐ Consider $f(x) = x^{1/3}$ (cube root), whose only root is $x = 0$. Write the Newton iteration and simplify it. Show that from any nonzero $x_0$ the iterates satisfy $x_{n+1} = -2x_n$, so the method diverges away from the root. Explain geometrically what the tangent line to $x^{1/3}$ does near $0$.

Part F — Applied / Synthesis (⭐⭐⭐⭐)

F1. ⭐⭐⭐⭐ (Astronomy — Kepler's equation, §11.10.) For a satellite with eccentricity $e = 0.2$ and mean anomaly $M = 0.8$ rad, solve Kepler's equation $M = E - e\sin E$ for the eccentric anomaly $E$ using Newton's method on $g(E) = E - e\sin E - M$. (a) Write the iteration $E_{n+1} = E_n - \dfrac{E_n - e\sin E_n - M}{1 - e\cos E_n}$. (b) Starting from $E_0 = M = 0.8$, carry out two iterations to four decimals. (Use $\sin 0.8 = 0.7174$, $\cos 0.8 = 0.6967$.) (c) The true answer is $E = 0.9643\ldots$ — comment on how fast you converged.

F2. ⭐⭐⭐⭐ (Finance — internal rate of return, §11.10.) You pay $\$1000$ now and receive $\$600$ at the end of year 1 and $\$600$ at the end of year 2. The internal rate of return $r$ solves $$-1000 + \frac{600}{1+r} + \frac{600}{(1+r)^2} = 0.$$ (a) Let $u = 1/(1+r)$ and rewrite as a quadratic in $u$. Solve it exactly with the quadratic formula and recover $r$. (b) Now apply Newton's method directly to $P(r) = -1000 + 600(1+r)^{-1} + 600(1+r)^{-2}$ with $P'(r) = -600(1+r)^{-2} - 1200(1+r)^{-3}$, starting from $r_0 = 0.10$. Carry out two iterations and compare with part (a). (c) Explain in two sentences why real bond-pricing software uses Newton's method rather than the quadratic formula. (Hint: what happens with 30 yearly coupons instead of 2?)

Hints and Selected Numerical Answers

A1.

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$f(9)=3$, $f'(9)=\frac{1}{2\cdot 3}=\frac16$, so $L(x)=3+\frac16(x-9)$ and $L(9.2)=3+\frac16(0.2)=3.0333\ldots$ (true $\sqrt{9.2}=3.03315$).

A4.

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$L(x)=x$, so $L(0.2)=0.2$; error $=0.2-0.198669=0.00133$.

A8.

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$(1+x)^k\approx 1+kx$. With $k=20$, $x=0.01$: $(1.01)^{20}\approx 1+0.20=1.20$ (true $1.2202$).

B2.

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$dV=3s^2\,ds=3(100)(0.1)=30\ \mathrm{cm^3}$; exact $\Delta V=10.1^3-10^3=1030.301-1000=30.301\ \mathrm{cm^3}$.

B7.

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$dr=\dfrac{dV}{4\pi r^2}=\dfrac{30}{4\pi(36)}=\dfrac{30}{452.39}\approx 0.0663\ \mathrm{cm/s}$.

C2.

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$dV=4\pi r^2\,dr=4\pi(25)(0.05)=5\pi\approx 15.7\ \mathrm{cm^3}$; $V=\frac43\pi(125)\approx523.6$; relative error $=15.7/523.6=3\%$ (three times the $1\%$ in $r$, the cube law).

C3.

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$y=x^{1/2}$: error $\approx\frac12(4\%)=2\%$. $y=x^{-2}$: error $\approx 2(4\%)=8\%$.

C7.

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Per swing, $\Delta T/T = 0.1/2 = 5\%$; timing $N$ swings gives $\Delta T/T = 5\%/N$. Need $0.3\% + 2(5\%/N) < 1\%$, i.e. $10\%/N < 0.7\%$, so $N > 14.3$ — time at least $N = 15$ swings.

D1.

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$x_{n+1}=\frac12(x_n+5/x_n)$. From $x_0=2$: $x_1=\frac12(2+2.5)=2.25$; $x_2=\frac12(2.25+2.2222)=2.2361$ (true $\sqrt5=2.23607$).

D5.

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$x_1=2-\frac{8-4-5}{12-2}=2-\frac{-1}{10}=2.1$; $x_2=2.1-\frac{9.261-4.2-5}{13.23-2}=2.1-\frac{0.061}{11.23}=2.09457$. Digits: $x_0$ has 1, $x_1$ has 2, $x_2$ has 5 — roughly doubling.

D6.

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$f(x)=1/x-d$, $f'(x)=-1/x^2$; step $=x-\frac{1/x-d}{-1/x^2}=x+x^2(1/x-d)=x+x-dx^2=x(2-dx)$. For $1/3$ ($d=3$) from $0.3$: $x_1=0.3(2-0.9)=0.33$; $x_2=0.33(2-0.99)=0.3333$.

E1.

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$10^{-2}\to10^{-4}\to10^{-8}\to10^{-16}$: three steps reach machine precision.

E5.

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$f'(x)=\frac13 x^{-2/3}$; step $=x-\frac{x^{1/3}}{\frac13 x^{-2/3}}=x-3x=-2x$. So $|x_{n+1}|=2|x_n|$ — the iterates blow up. The tangent line to $x^{1/3}$ gets nearly vertical near $0$, so its $x$-intercept overshoots the root.

F1.

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(b) $E_1=0.8-\dfrac{0.8-0.2(0.7174)-0.8}{1-0.2(0.6967)}=0.8-\dfrac{-0.14348}{0.86066}=0.9667$; one more step gives $E_2\approx0.9643$. Converged to 4 digits in two steps.

F2.

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(a) $-1000+600u+600u^2=0 \Rightarrow 3u^2+3u-5=0$; $u=\frac{-3+\sqrt{9+60}}{6}=\frac{-3+8.3066}{6}=0.8844$, so $1+r=1/0.8844=1.1307$, $r\approx 0.1307=13.07\%$. (b) Newton from $r_0=0.10$ converges to the same $\approx 0.1307$ in two steps.