1.3 Taylor series and linearisation

The previous two lessons developed derivatives and integrals as the two inverse operations of calculus. This third lesson sits on top of both: a Taylor series uses derivatives at a single point to approximate a function in a neighbourhood. The first-order truncation — keeping only the constant and linear terms — is linearisation, the single most-used technique in physics. Acoustics is the linearised theory of fluid mechanics. Optics is the linearised theory of Maxwell’s equations. The harmonic oscillator is the linearised pendulum. Newton’s law of cooling is the linearised Stefan–Boltzmann law.

This lesson develops Taylor series and the linearisation it makes precise.

The Taylor expansion

For a smooth function $f$ and a small displacement $\varepsilon$ from a base point $t_0$ , the Taylor expansion writes the value at the displaced point $t_0 + \varepsilon$ as a power series in $\varepsilon$ :

\boxed{\;f(t_0 + \varepsilon) \;=\; f(t_0) \;+\; \varepsilon\, f'(t_0) \;+\; \tfrac{1}{2!} \varepsilon^2\, f''(t_0) \;+\; \tfrac{1}{3!} \varepsilon^3\, f'''(t_0) \;+\; \cdots\;}

or, compactly,

f(t_0 + \varepsilon) \;=\; \sum_{n=0}^{\infty} \frac{\varepsilon^n}{n!}\, f^{(n)}(t_0).

Here $t_0$ is the base point, $\varepsilon = t - t_0$ is the small displacement from it, and $f^{(n)}(t_0)$ is the $n$ -th derivative of $f$ evaluated at $t_0$ , with the conventions $f^{(0)} = f$ and $0! = 1$ . The series gives $f$ in a neighbourhood of $t_0$ entirely in terms of $f$ and its derivatives at $t_0$ : all the local information about a smooth $f$ is carried by those derivatives.

Truncating the series at degree $N$ gives the Taylor polynomial $T_N(t; t_0)$ , the polynomial of degree $N$ that matches $f$ and its first $N$ derivatives at $t_0$ . The construction has a clean logic. A degree- $N$ polynomial has $N+1$ free coefficients — exactly enough to fix the value and the first $N$ derivatives at one point. Degree 0 matches only the value, a horizontal line through $(t_0, f(t_0))$ ; degree 1 also matches the slope, the tangent line; degree 2 also matches the curvature, the best-fitting parabola; each higher degree pins down one further derivative. The weighting $\varepsilon^n/n!$ is what keeps the terms from interfering: the $n$ -th term contributes to the $n$ -th derivative at $t_0$ and to no lower one, so adding it corrects $f^{(n)}$ without disturbing what the earlier terms already matched. The error left after degree $N$ is of order $\varepsilon^{N+1}$ — written $\mathcal{O}(\varepsilon^{N+1})$ , meaning it is bounded by a constant times $\varepsilon^{N+1}$ as $\varepsilon \to 0$ — small for small $\varepsilon$ and growing as $t$ moves away from $t_0$ .

function:

expansion point x₀ = 0.00 degree N = 3

The interactive builds the Taylor polynomial $T_N(x; x_0)$ of a chosen function. Slide the expansion point $x_0$ and the degree $N$ and watch the red approximation track the black true curve near $x_0$ and diverge from it far away. Two things to look for:

Near the expansion point, even low-order polynomials are nearly perfect. Linear ( $N=1$ ) suffices for very small $\varepsilon$ ; quadratic ( $N=2$ ) extends the agreement; higher orders are needed only for larger neighbourhoods or for the tails of the function.
Far from the expansion point, the approximation worsens and may diverge entirely. The series converges only within its radius of convergence, defined below.

The Maclaurin series: Taylor at zero

When the base point is $t_0 = 0$ , the Taylor series is called the Maclaurin series,

f(x) \;=\; \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\, x^n.

Five cases recur throughout physics, each obtained by the same routine — compute the derivatives $f^{(n)}(0)$ and read off the coefficients $f^{(n)}(0)/n!$ .

e^x \;=\; 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots

▶ Maclaurin series of eˣ Derivation

Every derivative of $f(x) = e^x$ is again $e^x$ , so $f^{(n)}(x) = e^x$ and $f^{(n)}(0) = e^0 = 1$ for every $n$ . The coefficients are $f^{(n)}(0)/n! = 1/n!$ :

e^x \;=\; \sum_{n=0}^{\infty} \frac{x^n}{n!} \;=\; 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots. \;✓

\sin x \;=\; x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

▶ Maclaurin series of sin x Derivation

Differentiating $\sin$ cycles with period four: $f = \sin x$ , $f' = \cos x$ , $f'' = -\sin x$ , $f''' = -\cos x$ , $f^{(4)} = \sin x$ , and so on. At $x = 0$ , $\sin 0 = 0$ and $\cos 0 = 1$ , so the derivative values cycle $0,\,1,\,0,\,-1,\,0,\,1,\dots$ . The even-order terms vanish; the odd-order terms alternate in sign:

\sin x \;=\; \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}\, x^{2k+1} \;=\; x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots. \;✓

\cos x \;=\; 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

▶ Maclaurin series of cos x Derivation

The same four-cycle, started from $\cos$ : $f = \cos x$ , $f' = -\sin x$ , $f'' = -\cos x$ , $f''' = \sin x$ . At $x = 0$ the values cycle $1,\,0,\,-1,\,0,\dots$ , so only the even-order terms survive:

\cos x \;=\; \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}\, x^{2k} \;=\; 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots. \;✓

Equivalently, differentiate the $\sin x$ series term by term.

(1 + x)^p \;=\; 1 + p x + \frac{p(p-1)}{2!} x^2 + \frac{p(p-1)(p-2)}{3!} x^3 + \cdots \qquad (\text{binomial series})

▶ The binomial series Derivation

For $f(x) = (1+x)^p$ each derivative brings down a factor and lowers the exponent by one:

f^{(n)}(x) \;=\; p(p-1)\cdots(p-n+1)\,(1+x)^{p-n}.

At $x = 0$ the power factor $(1+x)^{p-n}$ is $1$ , so $f^{(n)}(0) = p(p-1)\cdots(p-n+1)$ . Dividing by $n!$ gives the binomial coefficient $\binom{p}{n}$ :

(1 + x)^p \;=\; \sum_{n=0}^{\infty} \binom{p}{n} x^n, \qquad \binom{p}{n} = \frac{p(p-1)\cdots(p-n+1)}{n!}.

When $p$ is a non-negative integer the chain $p(p-1)\cdots$ hits zero and the series terminates, reproducing the ordinary binomial theorem; for any other $p$ it is a genuine infinite series. ✓

\ln(1 + x) \;=\; x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots

▶ Maclaurin series of ln(1+x) Derivation

Here $f(x) = \ln(1+x)$ with $f(0) = \ln 1 = 0$ , and $f'(x) = (1+x)^{-1}$ . Differentiating the power repeatedly,

f^{(n)}(x) \;=\; (-1)^{n-1}\,(n-1)!\,(1+x)^{-n} \quad (n \ge 1), \qquad f^{(n)}(0) = (-1)^{n-1}(n-1)!.

The coefficient is $f^{(n)}(0)/n! = (-1)^{n-1}/n$ :

\ln(1 + x) \;=\; \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\, x^n \;=\; x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots. \;✓

Equivalently, integrate the geometric series $\tfrac{1}{1+x} = 1 - x + x^2 - \cdots$ term by term from $0$ to $x$ .

Substituting $x = i\theta$ into the $e^x$ series — and using that the powers of $i$ cycle $1, i, -1, -i$ with period four — sorts the terms into the real $\cos\theta$ series and the imaginary $i\sin\theta$ series, giving Euler’s formula $e^{i\theta} = \cos\theta + i\sin\theta$ . This is developed in Foundations 3.1.

Radius of convergence

A power series $\sum_{n} a_n (t - t_0)^n$ need not converge for every displacement $\varepsilon = t - t_0$ . The radius of convergence $R$ is the number for which the series converges whenever $|t - t_0| < R$ and diverges whenever $|t - t_0| > R$ (behaviour exactly at $|t - t_0| = R$ has to be checked case by case). For the five series above:

$e^x$ , $\sin x$ , $\cos x$ converge for all $x$ : $R = \infty$ .
$\ln(1+x)$ and $(1+x)^p$ (with non-integer $p$ ) converge only for $|x| < 1$ : $R = 1$ .

The geometric meaning is exact: $R$ is the distance from the base point $t_0$ to the nearest singularity of $f$ in the complex plane. The functions $e^x$ , $\sin x$ , $\cos x$ have no singularities anywhere — they are entire — so $R = \infty$ . Both $\ln(1+x)$ and $(1+x)^p$ misbehave at $x = -1$ (a logarithmic divergence in the first, a branch point in the second), a distance $1$ from the origin, so $R = 1$ . The same rule explains a function that is perfectly finite on the real line: $1/(1+x^2)$ expanded about $0$ has $R = 1$ , because it has singularities at $x = \pm i$ , a distance $1$ away in the complex plane even though nothing goes wrong for real $x$ . Physically, “Taylor expand about equilibrium” is trustworthy only within this radius; a perturbation large enough to reach the nearest singularity requires a different base point or a different method entirely.

Linearisation: first-order Taylor

The first-order truncation of the Taylor expansion is

f(t_0 + \varepsilon) \;\approx\; f(t_0) + \varepsilon\, f'(t_0) + \mathcal{O}(\varepsilon^2).

This is the linearisation of $f$ around $t_0$ : a good approximation for small $\varepsilon$ , useless for large. It is the move that turns a nonlinear physics problem into a linear one. The full pendulum equation,

\ddot \theta \;+\; \frac{g}{L} \sin\theta \;=\; 0,

with $\theta$ the angular displacement from vertical, $g$ the gravitational acceleration, and $L$ the pendulum length, is nonlinear because $\sin\theta$ depends nonlinearly on $\theta$ . For small angles,

\sin\theta \;=\; \theta - \frac{\theta^3}{6} + \cdots \;\approx\; \theta + \mathcal{O}(\theta^3),

so the linearised pendulum is

\ddot \theta \;+\; \frac{g}{L}\, \theta \;=\; 0,

the simple-harmonic-motion equation of Foundations 5.3, with the closed-form solution $\theta(t) = A \cos(\omega_0 t + \varphi)$ — amplitude $A$ , phase $\varphi$ , and natural frequency $\omega_0 = \sqrt{g/L}$ . Below about $30°$ this is essentially exact; above that, the cubic correction begins to matter.

▶ Linearising sin θ: the full procedure with error bounds Worked Example

The linearisation $\sin\theta \approx \theta$ is one line of algebra in the main text, but the procedure that produces it generalises to every linearisation in physics.

Step 1 — Identify the function and the base point. We want to approximate $f(\theta) = \sin\theta$ near $\theta_0 = 0$ (the pendulum’s equilibrium). The small displacement is $\varepsilon = \theta - \theta_0 = \theta$ .

Step 2 — Compute the function value at the base point.

f(\theta_0) = \sin(0) = 0.

Step 3 — Compute the first derivative at the base point.

f'(\theta) = \cos\theta, \qquad f'(\theta_0) = \cos(0) = 1.

Step 4 — Assemble the first-order Taylor expansion.

\sin\theta \;\approx\; \underbrace{f(\theta_0)}_{0} + \varepsilon\,\underbrace{f'(\theta_0)}_{1} \;=\; \theta.

This is the small-angle approximation in the main text.

Step 5 — Estimate the error using the next term. The Taylor remainder after truncating at order $N$ is $\mathcal{O}(\varepsilon^{N+1})$ . For a generic function the leading error of a linearisation is the $\varepsilon^2$ term, $\tfrac{1}{2}\varepsilon^2 f''(\theta_0)$ . For $\sin\theta$ that term vanishes — $\sin$ is odd, so all its even-order derivatives at $0$ are zero ( $f''(0) = -\sin 0 = 0$ ) — and the leading error is the $\varepsilon^3$ term:

\sin\theta \;=\; \theta \;-\; \frac{\theta^3}{6} \;+\; \mathcal{O}(\theta^5).

The relative error from using $\sin\theta \approx \theta$ is therefore

\frac{\theta - \sin\theta}{\sin\theta} \;\approx\; \frac{\theta^3/6}{\theta} \;=\; \frac{\theta^2}{6}.

Step 6 — Plug in numbers and see when the approximation breaks.

| $\theta$ | $\theta$ (rad) | $\sin\theta$ | absolute error | relative error | |---|---:|---:|---:|---:| | $1°$ | 0.01745 | 0.01745 | $9 \times 10^{-7}$ | 0.005% | | $5°$ | 0.08727 | 0.08716 | $1.1 \times 10^{-4}$ | 0.13% | | $10°$ | 0.17453 | 0.17365 | $8.9 \times 10^{-4}$ | 0.51% | | $30°$ | 0.52360 | 0.50000 | $2.4 \times 10^{-2}$ | 4.7% | | $45°$ | 0.78540 | 0.70711 | $7.8 \times 10^{-2}$ | 11% | | $90°$ | 1.57080 | 1.00000 | $5.7 \times 10^{-1}$ | 57% |

The relative error grows as $\theta^2/6$ , exactly as predicted. At $30°$ it has reached $\approx 5\%$ — the threshold where the linear pendulum stops being an honest model. By $90°$ the linearisation is useless.

The procedure generalises to any linearisation: identify $f$ , pick $t_0$ , compute $f(t_0)$ and $f'(t_0)$ , assemble, then look at the next non-zero term to estimate the error. Every “linearised theory” in the bookshelf — acoustics, optics, elasticity, weak-field gravity — is these six steps applied to a different $f$ .

Linearisation is the master move of acoustics

Every time the Sound book writes “small perturbation” or " $|p'| \ll p_0$ " or “linearised theory,” it is invoking a first-order Taylor expansion. The wave equation itself is what comes out of linearising the full nonlinear fluid-mechanics equations around a still-air equilibrium. Write each field as a constant equilibrium value plus a small perturbation,

\rho = \rho_0 + \rho', \qquad \mathbf{v} = \mathbf{v}', \qquad p = p_0 + p',

with $\rho_0$ and $p_0$ the equilibrium density and pressure (constants), the primed quantities the small perturbations, and the equilibrium velocity zero — still air. Substituting these into each governing equation and keeping only first-order terms linearises it, as the three worked examples show.

▶ Linearising the continuity equation Worked Example

The continuity equation expresses conservation of mass:

\partial_t \rho + \nabla \cdot (\rho \mathbf{v}) = 0,

with $\rho$ the density and $\mathbf{v}$ the velocity. It is exact but nonlinear, through the product $\rho\mathbf{v}$ .

Substitute $\rho = \rho_0 + \rho'$ and $\mathbf{v} = \mathbf{v}'$ . Since $\rho_0$ is constant, $\partial_t\rho = \partial_t\rho'$ , and the flux term expands as

\nabla \cdot (\rho \mathbf{v}) = \nabla \cdot \big[(\rho_0 + \rho')\,\mathbf{v}'\big] = \rho_0\,\nabla\cdot\mathbf{v}' + \underbrace{\nabla\cdot(\rho'\mathbf{v}')}_{\text{second order}}.

The term $\rho'\mathbf{v}'$ is a product of two small quantities — second order in the perturbation — so it is dropped. What remains is linear:

\partial_t \rho' + \rho_0\,\nabla\cdot\mathbf{v}' = 0. \;✓

▶ Linearising Euler's equation Worked Example

Euler’s equation is Newton’s second law for a fluid element:

\rho\,(\partial_t \mathbf{v} + \mathbf{v}\cdot\nabla\mathbf{v}) = -\nabla p,

with $p$ the pressure. The convective term $\mathbf{v}\cdot\nabla\mathbf{v}$ is nonlinear.

Substitute the perturbations. The convective term $\mathbf{v}'\cdot\nabla\mathbf{v}'$ is quadratic in the small velocity — drop it. On the left,

\rho\,\partial_t\mathbf{v} = (\rho_0 + \rho')\,\partial_t\mathbf{v}' = \rho_0\,\partial_t\mathbf{v}' + \underbrace{\rho'\,\partial_t\mathbf{v}'}_{\text{second order}},

and the second-order piece is dropped. On the right, $-\nabla p = -\nabla(p_0 + p') = -\nabla p'$ , since $p_0$ is constant. The result is linear:

\rho_0\,\partial_t\mathbf{v}' = -\nabla p'. \;✓

▶ Linearising the equation of state Worked Example

For adiabatic (constant-entropy) sound, pressure is a function of density alone, $p = p(\rho)$ ; for an ideal gas $p \propto \rho^\gamma$ , with $\gamma$ the ratio of specific heats. This is a single-variable function, so linearising it is a one-variable Taylor expansion about the equilibrium density $\rho_0$ :

p = p(\rho_0) + \left.\frac{dp}{d\rho}\right|_{\rho_0}(\rho - \rho_0) + \cdots = p_0 + c^2\rho' + \cdots,

where the displacement is $\rho' = \rho - \rho_0$ and the first-derivative coefficient is named

c^2 \;\equiv\; \left.\frac{dp}{d\rho}\right|_{\rho_0} \;=\; \frac{\gamma p_0}{\rho_0}.

Keeping first order and subtracting the constant $p_0$ leaves a linear relation between the pressure and density perturbations,

p' = c^2 \rho',

and $c$ is the speed of sound. ✓

Combine the three linearised equations and the perturbations collapse into the wave equation $\partial_t^2 p' = c^2 \nabla^2 p'$ . Everything in the Sound book is downstream of this single linearisation. See Sound 4.5 — Linearisation and the wave equation for the full development.

The same move runs through:

Optics. Maxwell’s equations are linear in the EM fields but nonlinear in the medium-dependent permittivity; nonlinear optics treats the higher-order terms.
Elasticity. Hooke’s law $\sigma = E \epsilon$ is the linearisation of stress–strain around the unstressed state ( $\sigma$ stress, $\epsilon$ strain, $E$ the elastic modulus).
Cosmology. Linearised general relativity around a Friedmann–Robertson–Walker background gives the equations for cosmological perturbations.
Chemistry. Linearisation of chemical kinetics around a reaction equilibrium gives the linear-response theory of fluctuations.

In every case the underlying physics is nonlinear; linearisation is what makes it tractable.

Higher-order corrections

When linearisation is insufficient — when the perturbation is large enough that the $\varepsilon^2$ term cannot be ignored — keep more terms. The second-order Taylor expansion is

f(t_0 + \varepsilon) \;\approx\; f(t_0) + \varepsilon\, f'(t_0) + \tfrac{1}{2} \varepsilon^2\, f''(t_0).

This is the quadratic approximation, used whenever the linear term vanishes (e.g. at an extremum) or the first-order correction is too coarse. In acoustics, quadratic terms appear in energy and intensity — kinetic energy is $\tfrac12 \rho v^2$ , quadratic in velocity — and they are exactly what links the linear acoustic field to the nonlinear quantities (energy, momentum, intensity) one ultimately measures.

The third-order corrections appear in nonlinear acoustics — when amplitudes are large enough that even the quadratic terms aren’t sufficient. See Sound 10.4 — Wave steepening for what happens when those corrections compound.

Check yourself

Write the Taylor expansion of a smooth function $f$ around $t_0$ , in compact summation form. What does the $n!$ in the denominator do?

Reveal answer

f(t_0 + \varepsilon) \;=\; \sum_{n=0}^{\infty} \frac{\varepsilon^n}{n!}\, f^{(n)}(t_0).

The $n!$ compensates for the fact that differentiating $\varepsilon^n$ exactly $n$ times gives $n!$ — so the $n$ -th term, evaluated at $\varepsilon = 0$ , must be divided by $n!$ to match $f^{(n)}(t_0)$ on the nose. Without the factorial, only the $n=0,1$ terms would land correctly.

Check yourself

Write the Maclaurin series of $e^x$ , $\sin x$ , and $\cos x$ . Then state the relationship between them that emerges when you substitute $x = i\theta$ into the first series.

Reveal answer

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots

\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots

Substituting $x = i\theta$ : the even-power terms in $e^{i\theta}$ are real with alternating signs (matching $\cos\theta$ ); the odd-power terms are imaginary with alternating signs (matching $i\sin\theta$ ). Hence $e^{i\theta} = \cos\theta + i\sin\theta$ — Euler’s formula, derived from three Taylor series.

Check yourself

List the six steps of linearising a function $f$ around a base point $t_0$ — the procedure you’d apply to any nonlinear physics problem to get a tractable linear one.

Reveal answer

Identify the function $f$ and the base point $t_0$ . The small displacement is $\varepsilon = t - t_0$ .
Compute $f(t_0)$ . The value at the base point.
Compute $f'(t_0)$ . The slope at the base point.
Assemble the linearisation: $f(t_0 + \varepsilon) \approx f(t_0) + \varepsilon f'(t_0)$ .
Estimate the error from the next non-zero term — typically $\tfrac{1}{2}\varepsilon^2 f''(t_0)$ , unless that term vanishes by symmetry.
Check the regime — for what range of $\varepsilon$ is the error tolerable?

Every linearised theory on the bookshelf — the wave equation, Hooke’s law, the small-angle pendulum, weak-field gravity — is these six steps applied to a different $f$ .

Check yourself

When does the quadratic approximation matter, and the linear not suffice? Give a concrete acoustic example.

Reveal answer

The quadratic correction $\tfrac{1}{2}\varepsilon^2 f''(t_0)$ matters when (a) the linear term vanishes — e.g. at a maximum or minimum — or (b) the linear approximation is too coarse for the problem at hand.

Acoustic example: kinetic energy density is $\tfrac{1}{2}\rho v^2$ , quadratic in velocity. The linear acoustic field gives $v \propto p'$ (small first-order perturbation); plug into the kinetic-energy expression and you get a quantity scaling as $(p')^2$ — i.e. quadratic in the small parameter. Every energy, intensity, or radiation-pressure calculation in acoustics is fundamentally quadratic, even though the underlying wave equation is linear. This is why intensity has units of pressure-squared.

Check yourself

Which of the common Maclaurin series ( $e^x$ , $\sin x$ , $\cos x$ , $\ln(1+x)$ , $(1+x)^p$ ) converge for all $x$ , and which have a finite radius of convergence? Why does the difference matter physically?

Reveal answer

$e^x$ , $\sin x$ , $\cos x$ converge for all $x$ . $\ln(1+x)$ and $(1+x)^p$ converge only for $|x| < 1$ .

Physically: the first three are entire — analytic everywhere on the complex plane. The latter two have a singularity at $x = -1$ (for $\ln$ , a logarithmic divergence; for $(1+x)^p$ with non-integer $p$ , a branch point). The radius of convergence is the distance from the expansion point to the nearest singularity. This is why a Taylor expansion of $1/(1-x)$ around $x=0$ blows up at $x=1$ — and why “Taylor expand around equilibrium” only works near equilibrium; far from it, you need a different expansion point or a different technique entirely.

What we use this for

Calculus operations across the bookshelf:

The wave equation comes from linearising fluid mechanics around a still-air equilibrium (Sound 4.5).
The speed of sound $c = \sqrt{(\partial p / \partial \rho)_s}$ is a derivative evaluated at equilibrium.
Energy, intensity, and power are integrals of pressure-times-velocity over time.
The Fourier coefficients are integrals; the Fourier transform itself is an improper integral (Foundations 7).
Euler’s formula $e^{i\theta} = \cos\theta + i\sin\theta$ is derived from the Taylor series of all three functions (Foundations 3.1).
Every “small-angle approximation,” “weak-field limit,” or “leading-order analysis” in physics is a first-order Taylor expansion.

If any of the above looks unfamiliar, work through a textbook chapter on single-variable calculus before continuing. Spivak’s Calculus is the unsentimental favourite; Strang’s Calculus is the most physics-friendly.

Drill

Rote recall of the standard Maclaurin series, as a spaced-repetition deck. Reveal each card, then grade yourself — Again / Hard / Good / Easy — and SM-2 schedules when it returns. Progress is shared with the Foundations study deck.