Taylor Series

How does a chip compute $\sin(37°)$ when it only knows $+$ and $\times$?

The problem. A processor has no "sine" organ inside it. It can add, multiply, and that's essentially all. Yet you ask it for $\sin$, $\cos$, $e^x$, $\ln$ a billion times a second and it answers instantly. How?

The move. It evaluates the Taylor polynomial. To get $\sin x$ for $x$ in radians, sum $x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots$, a handful of multiplies and adds. Because the remainder shrinks like $\frac{x^{n+1}}{(n+1)!}$, just a few terms nail full machine precision for small $x$.

Why it's load-bearing. The real implementations add two tricks the series hands them: range reduction (use $\sin$'s periodicity to fold any angle into a small interval where the series converges fastest — the same "make $x$ small first" move as the binomial root), and a few tuned coefficients (minimax polynomials) that squeeze the error flat across the interval. But the skeleton is pure Taylor. Every transcendental function in every math library, every calculator, every GPU, is a polynomial standing in for the real thing.

Why does almost everything behave like a spring when you nudge it?

The problem. A pendulum, a guitar string, a molecule's bond, an atom in a crystal, a ship rolling at anchor — wildly different objects, yet all of them oscillate, and at small amplitude they all obey the same simple law. Why should the universe be so repetitive?

The move. Take any potential energy $V(x)$ with a stable minimum at $x_0$ and Taylor-expand it there. The constant term is just a reference level; the linear term vanishes because the slope is zero at a minimum (that's what "minimum" means); so the first thing that survives is the quadratic:

The punchline. That is the energy of a spring, $\frac12 k x^2$, with stiffness $k=V''(x_0)$. So every smooth potential, viewed up close near its bottom, is a parabola — and a parabola means simple harmonic motion, a perfect sinusoid. This is why the simple-harmonic-oscillator is the most important toy in physics: it's not that everything is a spring, it's that the second Taylor term of everything is. Quantum field theory itself starts here — particles are the small oscillations of fields about their minimum.

How do you solve a differential equation a computer can't integrate?

The problem. Most differential equations — weather, orbits, chemical kinetics — have no formula for the answer. You can only step forward in time numerically. But how do you take a step into a future you don't have a formula for?

The move. Taylor tells you where you'll be a small step $h$ later, using only what you know now: $y(t+h)\approx y(t)+h\,y'(t)+\frac{h^2}{2}y''(t)+\cdots$. Keep one term and you have Euler's method; keep more, or sample cleverly, and you have Runge–Kutta — the workhorses of every simulation on Earth.

Why the remainder is the whole game. The error you make per step is the first dropped term, of size $\sim h^2$. Halve the step and you quarter the error — that exact scaling, read straight off the Taylor remainder, is what lets engineers trade compute time for accuracy with their eyes open. "Second-order accurate," "fourth-order accurate" — those phrases are just counting how many Taylor terms the method respects.

How does a learning machine know which way is downhill?

The problem. Training a model means minimizing a loss over millions of parameters. You can't see the whole landscape — you can only feel the ground right under your feet. How do you decide where to step?

The move. Taylor-expand the loss around where you stand. The first-order term, the gradient, points uphill — so step the other way: that's gradient descent. Keep the second-order term too — the curvature, the Hessian — and you can jump straight toward the bottom of the local parabola: that's Newton's method.

The punchline. Every optimizer behind modern AI is a strategy for using one or two Taylor terms of a landscape it can never fully see. First-order methods are cheap and everywhere; second-order methods converge in spectacularly few steps when you can afford the curvature. The whole field is an argument about how many Taylor terms are worth their cost.

How precisely does an error in the input become an error in the output?

The problem. You feed a measured quantity $x$, uncertain by $\Delta x$, through some function $f$. How uncertain is the result $f(x)$ — and which inputs is your answer most sensitive to?

The move. The first Taylor term answers it on sight: $\Delta f \approx f'(x)\,\Delta x$. The derivative is the amplification factor. Several inputs? Sum their first-order contributions in quadrature. This is the entire foundation of error propagation across every experimental science.

Why it matters. It tells you where to spend your money: the input with the biggest $f'$ is the one worth measuring more carefully. And when $f'$ happens to vanish — a "sweet spot" where the output is insensitive to wobble — you've found a design that's robust by construction. Engineers hunt for those flat spots precisely because the first Taylor term goes quiet there.

How do you steer a wildly nonlinear system?

The problem. A rocket, a power grid, a chemical reactor — the equations governing them are ferociously nonlinear, and nonlinear systems are hard to analyze and harder to control. Yet we fly rockets and run grids reliably. How?

The move. Linearize. Pick the operating point you care about — cruising altitude, nominal load — and Taylor-expand the dynamics there, keeping only the first-order term. The fearsome nonlinear system becomes a linear one, valid in a neighborhood, and linear systems we understand completely: stability, response, optimal control, all solvable.

Why it's load-bearing. Nearly all of classical control engineering operates on the first Taylor term of reality. The autopilot holding your plane level, the cruise control in your car, the regulator keeping the grid at 50 Hz — each governs a linearized model that's only honest near the operating point. The art is knowing how big a neighborhood the remainder lets you trust before you must re-linearize somewhere new.

You must evaluate $\cos(0.2)$ by hand to six decimal places.

(a) How many terms of the Maclaurin series do you need? (b) Give the answer.

A pendulum is released from $\theta_0 = 20°$. Its true period is longer than the textbook $T_0=2\pi\sqrt{L/g}$ by the correction $T\approx T_0\left(1+\tfrac{\theta_0^2}{16}\right)$, with $\theta_0$ in radians.

(a) By what percentage is the real period longer? (b) A grandfather clock calibrated for tiny swings runs at $20°$ — does it gain or lose time, and how much per day?

You solve $y'=y$, $y(0)=1$, with Euler's method to estimate $y(1)=e\approx2.71828$, using step $h=0.25$.

(a) What value does Euler give? (b) Roughly how does the error change if you halve $h$ to $0.125$?

Evaluate $\displaystyle\lim_{x\to0}\frac{\sin x - x}{x^3}$ without L'Hôpital.

A signal's power is $P = V^2/R$. The voltage $V=5.00$ V is measured with an uncertainty of $\pm0.05$ V; $R$ is exact.

(a) Use the first Taylor term to find the uncertainty in $P$ as a percentage. (b) Why is the percentage uncertainty in power double that in voltage?

You minimize $f(x)=x^4-3x^2+x$ and want Newton's method to take its first step from $x_0=1$.

(a) Write the second-order Taylor model of $f$ about $x_0$. (b) Where does Newton step to (the minimum of that parabola)?

Problem 1 — cos(0.2). Terms: $1$, $-\tfrac{0.2^2}{2}=-0.02$, $+\tfrac{0.2^4}{24}=+6.67\times10^{-5}$, $-\tfrac{0.2^6}{720}=-8.9\times10^{-8}$. (a) The fourth term is already below $10^{-6}$, so three terms suffice. (b) $1-0.02+0.0000667=\mathbf{0.980067}$ (true value $0.9800666$).

What breaks this? The "drop one term" error bound is a gift of the alternating series. For $e^x$, whose terms are all positive, you must bound the tail differently — the remainder doesn't politely cap itself at the next term.

Problem 2 — Pendulum. $\theta_0=20°=0.349$ rad, so $\theta_0^2/16=0.00761$. (a) The period is about 0.76% longer. (b) Longer period → slower → the clock loses time: $0.76\%$ of $86{,}400$ s $\approx \mathbf{657}$ s, nearly 11 minutes a day. (This is exactly why real pendulum clocks keep their swing tiny.)

What breaks this? At large amplitude the single $\theta_0^2/16$ term isn't enough — you need the next correction $\tfrac{11}{3072}\theta_0^4$, and near $180°$ the whole series surrenders and the period heads for infinity. (See Probe 4.)

Problem 3 — Euler. With $h=0.25$, $y(1)=(1+0.25)^4=1.25^4=\mathbf{2.4414}$ — about 10% below $e$. (b) Euler is first-order: error $\propto h$. Halving $h$ to $0.125$ roughly halves the error (to $(1.125)^8=2.566$, now ~5.6% low). To gain a decimal you pay with ten times the steps.

What breaks this? Euler keeps only one Taylor term, so it's crude and, for stiff equations, can go unstable as $h$ grows. The cure is to respect more Taylor terms — that's precisely what Runge–Kutta does.

Problem 4 — Limit. $\sin x = x-\tfrac{x^3}{6}+\tfrac{x^5}{120}-\cdots$, so $\sin x - x = -\tfrac{x^3}{6}+\tfrac{x^5}{120}-\cdots$. Divide by $x^3$: $-\tfrac{1}{6}+\tfrac{x^2}{120}-\cdots$, and as $x\to0$ the limit is $\mathbf{-\tfrac{1}{6}}$. No L'Hôpital, no fuss — the series just hands it over.

What breaks this? Nothing here — this is Taylor at its cleanest. The only trap is keeping too few terms: you must carry the series at least to the order that survives the division, here $x^3$.

Problem 5 — Power. $\tfrac{dP}{dV}=\tfrac{2V}{R}$, so $\Delta P\approx\tfrac{2V}{R}\Delta V$, giving $\tfrac{\Delta P}{P}=2\tfrac{\Delta V}{V}$. (a) $\tfrac{\Delta V}{V}=\tfrac{0.05}{5}=1\%$, so $\tfrac{\Delta P}{P}=\mathbf{2\%}$. (b) Because $P\propto V^2$ — the exponent 2 multiplies the relative error, the same "errors scale with the power" rule from the binomial section.

What breaks this? First-order propagation assumes the uncertainty is small enough that curvature ($P''$) doesn't matter. For a 1% wobble it's perfect; for a 50% wobble you'd need the second Taylor term, and the symmetric $\pm$ error would skew.

Problem 6 — Newton step. $f'(x)=4x^3-6x+1$, $f''(x)=12x^2-6$. At $x_0=1$: $f'=-1$, $f''=6$. (a) The quadratic model is $f(1)+f'(1)(x-1)+\tfrac{f''(1)}{2}(x-1)^2=-1-(x-1)+3(x-1)^2$. (b) Newton steps to $x_0-\tfrac{f'}{f''}=1-\tfrac{-1}{6}=\mathbf{1.1\overline{6}}$.

What breaks this? Newton trusts the local parabola completely. If $f''<0$ (you're near a maximum, not a minimum) it'll happily step uphill, and far from a root it can overshoot wildly. The second Taylor term is powerful but only locally honest.

A student computes $\ln(3)$ from the series $\ln(1+x)=x-\tfrac{x^2}{2}+\cdots$ by setting $x=2$. (a) What happens? (b) The fix?

(a) With $x=2$ the terms are $2, -2, \tfrac{8}{3}, -4,\ldots$ — they grow, and the series diverges. The Maclaurin series of $\ln(1+x)$ only converges for $-1<x\le1$; at $x=2$ you're far outside its radius. Assumption 1 failed, loudly.

(b) Re-center, or transform. Use $\ln 3 = \ln\frac{1+t}{1-t}$ with $t=\tfrac12$, a fast-converging form, or expand $\ln x$ around $a=e$ instead of $0$. The series isn't wrong — you just asked it about a place it can't see from where it's standing. (Section 5 builds exactly this faster log.)

Consider $f(x)=e^{-1/x^2}$ (with $f(0)=0$). It's infinitely differentiable. (a) What is its Maclaurin series? (b) Why is that a catastrophe?

(a) Every single derivative at $0$ is exactly zero. So the Taylor series is $0+0x+0x^2+\cdots = 0$, identically. (b) But the function is not zero anywhere except at the origin! The series converges beautifully — to the wrong function. This is the nightmare case: smooth but not analytic. The derivatives at one point simply don't carry information about the function elsewhere.

The lesson is humbling: "infinitely differentiable" is not enough for Taylor to work. You need analyticity, a strictly stronger promise. Most functions you meet in physics are analytic, which is why we get away with it — but this little monster is the reason the fine print exists.

To approximate $\frac{1}{1+25x^2}$ on $[-1,1]$, a student fits a high-degree polynomial through evenly spaced points, expecting more degree to mean less error. (a) What goes wrong? (b) What's really happening?

(a) The opposite of what you'd hope: as the degree climbs, the polynomial develops violent oscillations near the ends of the interval, and the error blows up. This is the famous Runge phenomenon.

(b) That innocent function has trouble lurking in the complex plane (poles at $x=\pm i/5$) that pinches its radius of convergence, so a single global high-degree polynomial can't tame it. More terms is not always more accuracy. The real fixes — Chebyshev points, splines, piecewise fits — all amount to not trusting one Taylor expansion across too wide a span.

A student uses $T\approx T_0(1+\theta_0^2/16)$ for a pendulum released from nearly $180°$. (a) What does the formula predict? (b) What actually happens?

(a) The two-term formula gives a finite, modest correction even at $179°$ — utterly wrong. (b) The true period diverges to infinity as $\theta_0\to180°$: balanced exactly at the top, the pendulum takes forever to fall. A couple of Taylor terms around the bottom of the well know nothing about the unstable equilibrium at the top.

This is assumption 2 failing — too few terms — bleeding into assumption 1, since the expansion is only local. A local model is silent about distant features of the landscape, and the top of the swing is as distant as it gets.