21. Differentials · Newton's Method

Adapted from: Classical single-variable calculus (linear approximation and Newton root-finding). Rewritten for teaching — not a verbatim textbook dump.

If $f$ is differentiable at $x$, then

$$f'(x)=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x},\qquad \Delta y=f(x+\Delta x)-f(x).$$

So for small $\Delta x$,

$$\frac{\Delta y}{\Delta x}\sim f'(x) \quad\Rightarrow\quad \Delta y\sim f'(x)\,\Delta x. \tag{21.1}$$

Rearranged:

$$f(x+\Delta x)\sim f(x)+f'(x)\,\Delta x. \tag{21.2}$$

This is the linear (or tangent-line) approximation to $f$ near $x$.

No free error bound

Formula (21.2) does not by itself say how small $\Delta x$ must be for a chosen number of correct decimals. That needs remainder estimates (later: Taylor with remainder).

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Example 21.1 — Estimate $\sqrt{16.2}$

Take $f(x)=\sqrt{x}$, $x=16$, $\Delta x=0.2$. Then $f(16)=4$ and

$$f'(x)=\frac{1}{2\sqrt{x}}\implies f'(16)=\frac18.$$

So

$$\sqrt{16.2}\sim 4+\frac18(0.2)=4.025.$$

(True value $\approx 4.0249$ — three decimals agree.)

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Example 21.2 — Estimate $\sin(0.1)$

Take $f(x)=\sin x$, $x=0$, $\Delta x=0.1$. Then $f(0)=0$, $f'(0)=\cos 0=1$, so

$$\sin(0.1)\sim 0.1.$$

True value $\approx 0.0998$. Same idea: $\sin u\sim u$ for small $u$.

Check: linear approx

Quiz

Linear approximation

1 / 2

Formula (21.2) says f(x+Δx) is approximately…

Pick an answer to see if you are right.

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The differential

The product on the right of (21.1) is the differential of $f$:

$$df = f'(x)\,\Delta x.$$

It depends on both $x$ and $\Delta x$. For small $\Delta x$,

$$f(x+\Delta x)-f(x)\sim df. \tag{21.3}$$

Fig. 21-1 — RT = df approximates RQ = Δy

In the figure, the tangent $\mathscr{L}$ at $P$ has slope $f'(x)$, so the vertical rise over $\Delta x$ is $df$. The true change along the curve is $\Delta y$; they nearly match when $\Delta x$ is small.

Writing $dx$

If $f(x)=x$, then $dx=\Delta x$. Hence we may rewrite

$$df = f'(x)\,dx,$$

and when $\Delta x\neq 0$,

$$\frac{df}{dx}=f'(x).$$

With $y=f(x)$ one writes $dy$ for $df$ — the familiar $dy/dx$.

Algebra of differentials

For functions $u,v$ and constant $c$:

$$\begin{aligned} d(c)&=0,& d(cu)&=c\,du,& d(u+v)&=du+dv,\\ d(uv)&=u\,dv+v\,du,& d\!\left(\frac{u}{v}\right)&=\frac{v\,du-u\,dv}{v^{2}}. \end{aligned}$$

Examples: $d(\tan x)=\sec^{2}x\,\Delta x$, and $d(x^{3}-2x)=(3x^{2}-2)\,\Delta x$.

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Newton's method

Suppose $x_0$ is near a root of $f(x)=0$ and $f$ is differentiable. The tangent at $x_0$ usually hits the $x$-axis at a better guess $x_1$.

Fig. 21-2 — tangent at x₀ meets the axis at x₁

Point-slope for the tangent:

$$y-f(x_0)=f'(x_0)(x-x_0).$$

At the intercept $(x_1,0)$, if $f'(x_0)\neq 0$,

$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}.$$

Repeating from $x_1,x_2,\ldots$ gives

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. \tag{21.5}$$

When Newton fails

If $f'(x_n)\approx 0$, or the seed is far from a simple root, the sequence can diverge, cycle, or jump to another root. Always plot or sanity-check the start.

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Example 21.3 — Approximate $\sqrt{3}$

Solve $f(x)=x^{2}-3=0$. Then $f'(x)=2x$ and (21.5) becomes

$$x_{n+1}=\frac{x_n^{2}+3}{2x_n}. \tag{21.6}$$

With $x_0=1$:

$n$	$x_n$ (approx.)
$0$	$1$
$1$	$2$
$2$	$1.75$
$3$	$1.732142857$
$4$	$1.73205081$
$5$–$6$	$1.732050808$ (stable)

So $\sqrt{3}\approx 1.732050808$ to the digits shown.

Check: Newton

Quiz

Newton's method

1 / 3

The Newton update is…

Pick an answer to see if you are right.

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Takeaways

Remember

• $\Delta y\sim f'(x)\,\Delta x$ and $f(x+\Delta x)\sim f(x)+f'(x)\,\Delta x$
• $df=f'(x)\,dx$; algebra mirrors derivative rules
• Newton: $x_{n+1}=x_n-f(x_n)/f'(x_n)$ — fast near a simple root, not magic from a bad start

Further Reading

• 20. Related Rates — rates linked by an equation
• 20b. Related Rates — Practice — prior drills
• 19. Rectilinear & Circular Motion — $ds/dt$ language
• Computational physics: Newton–Raphson — numeric root finding
• 21b. Differentials · Newton's Method — Practice — approx, $dy$, Newton roots