Exercises — Linear approximation and differentials

What we need: $L(x)=f(a)+f^{\prime }(a)(x-a)$, so we need two numbers: the height $f(2)$ and the slope $f^{\prime }(2)$.

• Height: $f(2)=2^3=8$.
• Slope: $f^{\prime }(x)=3x^2$, so $f^{\prime }(2)=3(4)=12$.

Assemble: $L(x)=8+12(x-2)$. That is the tangent line to $y=x^3$ at the point $(2,8)$.

$f(1)=\ln 1=0$. Slope $f^{\prime }(x)=\frac{1}{x}$, so $f^{\prime }(1)=1$. $L(x)=0+1(x-1)=x-1.$ Now let $x=1+u$ (so the "step" $x-1=u$): $\ln (1+u)\approx u$. This is the standard near-zero rule.

Why $dy$: $dy=f^{\prime }(x)dx$ is "rise along the tangent = slope × run." $f^{\prime }(x)=-\sin x$, so $dy=-\sin xdx$. At $x=\frac{\pi }{3}$: $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}\approx 0.8660$. $dy=-0.8660\times 0.01=-\mathrm{0.008660.}$ The minus sign says cosine is decreasing there, so a small step right drops $y$.

Choose $f,a$: $f(x)=x^{1/3}$, $a=8$ because $\sqrt[3]{8}=2$ exactly and $8.2$ is close. Height: $f(8)=2$. Slope: $f^{\prime }(x)=\frac{1}{3}{x}^{-2/3}$, so $f^{\prime }(8)=\frac{1}{3}\cdot 8^{-2/3}=\frac{1}{3}\cdot \frac{1}{4}=\frac{1}{12}$. Line & evaluate: step $=8.2-8=0.2$. $\sqrt[3]{8.2}\approx 2+\frac{1}{12}(0.2)=2+0.016\overline{6}=\mathrm{2.01667.}$ (True value $2.01653\dots$ — off by about $0.00014$.)

Choose: $f(x)=x^5$, base $a=2$ (since $2^5=32$ exact), $dx=1.97-2=-0.03$. Slope: $f^{\prime }(x)=5x^4$, $f^{\prime }(2)=5(16)=80$. Differential: $dy=80\times (-0.03)=-2.4$. $(1.97{)}^5\approx f(2)+dy=32-2.4=\mathrm{29.6.}$ (True $\approx 29.584$.) Note $dx<0$ handled cleanly — the step just points left.

Convert: $45^{\circ }=\frac{\pi }{4}$, and the step is $1^{\circ }=\frac{\pi }{180}\approx 0.017453$ rad. Choose: $f(x)=\tan x$, $a=\frac{\pi }{4}$ (there $\tan =1$ exactly). Slope: $f^{\prime }(x)={\sec }^{2}x=\frac{1}{{\cos }^{2}x}$. At $\frac{\pi }{4}$, $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, so ${\cos }^2=\frac{1}{2}$ and $f^{\prime }=2$. Line: $\tan (46^{\circ })\approx 1+2(0.017453)=1+0.034907=1.034907$. (True $\tan 46^{\circ }=1.03553\dots$ — off by $0.0006$; tangent curves upward fast, hence a slightly bigger miss.)

Second derivative test of shape: $f^{\prime }(x)=\frac{1}{2}{x}^{-1/2}$, $f^{\prime \prime }(x)=-\frac{1}{4}{x}^{-3/2}<0$ for $x>0$. Negative second derivative $\Rightarrow$ the curve is concave down — it bends below its tangent lines. See the figure: the teal curve sits under the orange tangent to the right of $a$.

So the tangent line lies above $\sqrt{x}$, and $L(4.1)=2.025$ is an over-estimate of the true $\sqrt{4.1}=2.02485\dots$. Confirmed: $2.025>2.02485$. This is the Concavity and second derivative connection in action.

Where the error law comes from (Taylor's theorem with Lagrange remainder): if $f$ has two derivatives, then for some point $c$ strictly between $a$ and $x$, $f(x)=\underset{L(x)}{\underbrace{f(a)+f^{\prime }(a)(x-a)}}+\underset{\text{remainder}=\text{error}}{\underbrace{\frac{1}{2}{f}^{\prime \prime }(c)(x-a{)}^2}}.$ This is the standard Taylor series remainder: the leftover after the linear part is exactly one term of the form $\frac{1}{2}{f}^{\prime \prime }(c)(x-a{)}^2$. So $\text{error}=f(x)-L(x)=\frac{1}{2}{f}^{\prime \prime }(c)(x-a{)}^2$, which is $\propto (x-a{)}^2$. Reasoning: since error $\propto (x-a{)}^2$, halving $(x-a)$ multiplies the square by ${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$. So error drops by a factor of 4. Build $L$ (recall where the $\frac{1}{4}$ comes from): $f(x)=\sqrt{x}$, $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$, so $f^{\prime }(4)=\frac{1}{2\cdot 2}=\frac{1}{4}$; with $f(4)=2$ this gives $L(x)=2+\frac{1}{4}(x-4)$. Numeric check:

• Step $0.1$: $L(4.1)=2.025$, true $=2.0248457\dots$, error $=0.0001543$.
• Step $0.05$: $L(4.05)=2.0125$, true $=2.0124612\dots$, error $=0.0000388$. Ratio $=0.0001543/0.0000388\approx 3.98\approx 4$. Matches the quadratic-error prediction — this is exactly why Taylor series adds a $\frac{1}{2}{f}^{\prime \prime }(a)(x-a{)}^2$ term to kill this leading error.

Why differentials: we don't know the true error, only that $ds$ is small, so $\Delta V\approx dV=V^{\prime }(s)ds$. $V^{\prime }(s)=3s^2$, so $dV=3s^{2}ds=3(25)(0.02)=1.5{\text{cm}}^3$ (absolute). Relative error: $\frac{dV}{V}=\frac{3s^{2}ds}{s^3}=3\frac{ds}{s}=3\cdot \frac{0.02}{5}=3(0.004)=0.012=1.2\%.$ The neat pattern: for a power $s^n$, relative error multiplies by $n$. (Sphere $r^3$ gave $3\frac{dr}{r}$ in the parent — same rule.) See Error propagation.

First, the tool we lean on — the $(1+u{)}^n$ rule, derived here: let $h(u)=(1+u{)}^n$ and linearize at $u=0$. Then $h(0)=1$, and $h^{\prime }(u)=n(1+u{)}^{n-1}$ so $h^{\prime }(0)=n$. Hence $L(u)=1+nu$, i.e. $(1+u{)}^n\approx 1+nu(u\text{ small}).$ Piece 1: $\sqrt{4.1}\approx 2+\frac{1}{4}(0.1)=2.025$ (using $f(x)=\sqrt{x}$, $f^{\prime }(4)=\frac{1}{4}$ as in 3.2). Piece 2: apply the boxed rule with $u=0.02,n=3$: $(1.02{)}^3=(1+0.02{)}^3\approx 1+3(0.02)=1.06$. Combine: product $\approx 2.025\times 1.06=2.1465$. (True: $\sqrt{4.1}=2.024846$, $1.02^3=1.061208$, product $=2.148741$.) Error $\approx 0.0022$. Why this is easier than one big linearization: each factor is a clean, memorized near-zero form; multiplying two simple estimates avoids differentiating an ugly product $\sqrt{x}(x{)}^3$ and picking a joint base point. The small cost is that errors from both factors add up.

Linearize $f$ at $x_0$: $L(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)$. $f(x)=x^2-5$, $f(2)=4-5=-1$; $f^{\prime }(x)=2x$, $f^{\prime }(2)=4$. Solve $L(x)=0$: $-1+4(x-2)=0\Rightarrow x-2=\frac{1}{4}\Rightarrow x=2.25$. So $x_1=2.25$. That is exactly the Newton's method update $x_1=x_0-\frac{f(x_0)}{f^{\prime }(x_0)}=2-\frac{-1}{4}=2.25$. (True $\sqrt{5}=2.23607$; one step already good to $0.014$.) Key insight: Newton = repeatedly riding the tangent to the axis, i.e. linear approximation used backwards — solve the line instead of evaluating it.

Linearize $e^x$ at $a=0$: $f(0)=e^0=1$, $f^{\prime }(x)=e^x$ so $f^{\prime }(0)=1$. Then $L(x)=1+1\cdot (x-0)=1+x$. ✔ Estimates: $e^{0.05}\approx 1.05$; $\ln (1.05)=\ln (1+0.05)\approx 0.05$. Inverse check: $\ln (e^{0.05})$ should be $0.05$. Using estimates: $\ln (1.05)\approx 0.05$. ✔ They undo each other to first order — both are the degree-1 Taylor series of inverse functions, so their leading behaviour matches.

Treat $T$ as a function of $g$: $T(g)=2\pi \sqrt{L}{g}^{-1/2}$. Differentiate: $\frac{dT}{dg}=2\pi \sqrt{L}\cdot \left(-\frac{1}{2}\right)g^{-3/2}=-\pi \sqrt{L}{g}^{-3/2}$. Relative form (cleaner): since $T\propto g^{-1/2}$, $\frac{dT}{T}=-\frac{1}{2}\frac{dg}{g}$. Numerically: $\frac{dT}{T}=-\frac{1}{2}\cdot \frac{0.05}{9.81}=-0.002548=-0.2548\%.$ Absolute: first $T=2\pi \sqrt{1/9.81}=2\pi (0.31944)=2.00709\text{ s}$, so $dT=T\cdot (-0.002548)=-0.005114\text{ s}$. Interpret: the minus sign says a larger $g$ shortens the period; a $0.5\%$ error in $g$ produces only a $0.25\%$ error in $T$ because the square root halves the relative error.

Second derivative: $f^{\prime \prime }(x)=-\frac{1}{4}{x}^{-3/2}$. So $|f^{\prime \prime }(x)|=\frac{1}{4}{x}^{-3/2}$, which is a decreasing function of $x$ (larger $x$ ⇒ smaller $x^{-3/2}$). Therefore on the interval $[4,4.1]$ its maximum sits at the left endpoint $x=4$: $|f^{\prime \prime }(4)|=\frac{1}{4}\cdot 4^{-3/2}$. Now $4^{-3/2}=\frac{1}{4^{3/2}}=\frac{1}{8}$, so $|f^{\prime \prime }(4)|=\frac{1}{4}\cdot \frac{1}{8}=\frac{1}{32}=0.03125$. Bound at step $0.1$: $\frac{1}{2}(0.03125)(0.1{)}^2=\frac{1}{2}(0.03125)(0.01)=0.00015625$. That is $1.56\times 10^{-4}$ — larger than $10^{-5}$, so a guarantee fails (even though the actual error $1.5\times 10^{-4}$ happened to be close to the bound). Required step: solve $\frac{1}{2}(0.03125)h^2\le 10^{-5}$: $h^2\le \frac{2\times 10^{-5}}{0.03125}=6.4\times 10^{-4}$, so $h\le 0.0253$. Conclusion: you'd need $x$ within about $0.0253$ of $4$ (i.e. $\sqrt{4.025}$ or nearer) to guarantee $10^{-5}$ from the linear model — or, keep the step and add the quadratic Taylor series term.

Build $L$ at $a=100$: $f(100)=10$; $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$, $f^{\prime }(100)=\frac{1}{20}=0.05$. $\boxed{\sqrt{x}\approx 10+0.05(x-100)}$ Estimates:

• $\sqrt{102}\approx 10+0.05(2)=10.1$. (True $10.09950$; over-estimate.)
• $\sqrt{99}\approx 10+0.05(-1)=9.95$. (True $9.94987$; over-estimate.) Sign of error: $f^{\prime \prime }(x)=-\frac{1}{4}{x}^{-3/2}<0$ ⇒ concave down ⇒ tangent lies above the curve on both sides ⇒ both estimates are over-estimates. Matches the tiny positive gaps above.