Calculus speedrun

Exam-cram notes from February 2023, rebuilt and carefully checked afterwards.

Derivatives

Differentiability and the differential

A function $f$ is differentiable at $x_0$ when some constant $k$ satisfies

$$\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)-k\,\Delta x}{\Delta x}=0.$$

That $k$ is the derivative $f'(x_0)$. The definition says $f$ can be approximated near $x_0$ by an affine function whose error shrinks faster than $\Delta x$ does.

The derivative and the differential are different objects. The differential is the linear part, $\mathrm{d}f=f'(x_0)\,\mathrm{d}x$, a function of $\mathrm{d}x$. In one variable the distinction buys little and the two get used interchangeably. In several variables it stops being optional.

Continuity and differentiability

Differentiable implies continuous, and the converse fails.

Take $f(x)=|x|$ at $x=0$. It is continuous there and not differentiable there: the left derivative is $-1$, the right derivative is $+1$, and a derivative exists exactly when the one-sided derivatives exist and agree.

The Weierstrass function is continuous on all of $\mathbb{R}$ and differentiable nowhere, so continuity constrains differentiability hardly at all.

The mean value theorem

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, some $c\in(a,b)$ has

$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

The average rate of change across the interval is attained as an instantaneous rate somewhere inside it. Rolle’s theorem is the case $f(a)=f(b)$, and each implies the other in a line.

Limits

Equivalent infinitesimals

Two infinitesimals $\alpha(x)$ and $\beta(x)$ as $x\to x_0$ are equivalent, written $\alpha\sim\beta$, when their ratio tends to one:

$$\lim_{x\to x_0}\frac{\alpha(x)}{\beta(x)}=1.$$

Substituting equivalents is legitimate for factors of a product or quotient. Across a sum the leading terms can cancel, leaving the answer in the part that was discarded. With $\tan x\sim x$ and $\sin x\sim x$ as $x\to 0$, replacing both in

$$\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}$$

gives $0$, while the limit is $\tfrac12$. Subtraction annihilates the linear terms and what survives is cubic.

L’Hôpital’s rule

The rule applies to $\frac{0}{0}$ and $\frac{\infty}{\infty}$, given $f$ and $g$ differentiable near the point with $g'\neq 0$ there. Every other indeterminate form has to be pushed algebraically into one of those two shapes first.

It concludes only when $\lim f'/g'$ exists or is infinite. Where that limit does not exist, the rule yields no information about $f/g$, which may still be perfectly well behaved:

$$\lim_{x\to\infty}\frac{x+\sin x}{x}=1,$$

whose ratio of derivatives is $1+\cos x$, oscillating in $[0,2]$ forever.

Each application costs a differentiation, and on $\frac{\sin x}{x}$ the rule argues in a circle, since the derivative of $\sin$ is the thing being established.

Taylor expansion

For $f$ with enough derivatives at $a$,

$$f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+R_n(x).$$

Maclaurin’s expansion is this with $a=0$.

The two standard remainders answer different questions. The Peano remainder is qualitative,

$$R_n(x)=o\big((x-a)^n\big)\quad (x\to a),$$

and needs only that $f$ be $n$ times differentiable at $a$. It says the error is of smaller order than the last term kept, which is enough to compute a limit and gives no numerical bound anywhere.

The Lagrange remainder is quantitative,

$$R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$$

for some $\xi$ between $a$ and $x$, and needs $n+1$ derivatives across the interval. Both the derivative order and the power are $n+1$.

Integration

The techniques

1. Basic formulas. $\int\cos x\,\mathrm{d}x=\sin x+C$, $\int\frac{\mathrm{d}x}{x}=\ln|x|+C$, $\int\frac{\mathrm{d}x}{x^2+a^2}=\frac1a\arctan\frac{x}{a}+C$, $\int\frac{\mathrm{d}x}{\sqrt{x^2+a^2}}=\ln\left|x+\sqrt{x^2+a^2}\right|+C$.
2. Substitution, running in either direction. You recognize $\mathrm{d}u$ inside the integrand and collapse it, or you introduce a new variable to clear a radical. Trigonometric substitution is the second direction: $\sqrt{a^2-x^2}$ wants $x=a\sin t$, $\sqrt{a^2+x^2}$ wants $x=a\tan t$, and $\sqrt{x^2-a^2}$ wants $x=a\sec t$.
3. Integration by parts, $\int u\,\mathrm{d}v=uv-\int v\,\mathrm{d}u$. So $\int xe^x\mathrm{d}x=xe^x-e^x+C$ and $\int x\cos x\,\mathrm{d}x=x\sin x+\cos x+C$.
4. Partial fractions for rational integrands. For $\frac{x+2}{x^2+x-6}$ the denominator factors as $(x+3)(x-2)$, and solving for the coefficients gives $\frac{1}{5(x+3)}+\frac{4}{5(x-2)}$.
5. Trigonometric identities, to drop a power before integrating: $\int\sin^2 x\,\mathrm{d}x=\int\frac{1-\cos 2x}{2}\mathrm{d}x$.
6. Reduction formulas, derived by parts. On $[0,\frac{\pi}{2}]$, $\int_0^{\pi/2}\cos^n x\,\mathrm{d}x =\frac{n-1}{n}\int_0^{\pi/2}\cos^{n-2}x\,\mathrm{d}x$.
7. Symmetry: an odd integrand over $[-a,a]$ integrates to zero, an even one to twice the half-interval.

Worked examples

The denominator of $\int\frac{\sin x}{\sin x+\cos x}\,\mathrm{d}x$ has derivative $\cos x-\sin x$, which surfaces once the integral is paired with its companion. Let

$$I=\int\frac{\sin x}{\sin x+\cos x}\mathrm{d}x,\qquad J=\int\frac{\cos x}{\sin x+\cos x}\mathrm{d}x.$$

Then $I+J=\int\mathrm{d}x=x$, while $J-I$ has numerator $\cos x-\sin x$, exactly the derivative of the denominator, so $J-I=\ln|\sin x+\cos x|$. Subtracting,

$$I=\frac{x}{2}-\frac{1}{2}\ln|\sin x+\cos x|+C.$$

For $\int\frac{x^3}{(x^2+1)^2}\mathrm{d}x$, put $u=x^2+1$, so $x\,\mathrm{d}x=\tfrac12\mathrm{d}u$ and $x^2=u-1$. The integral becomes $\frac12\int\frac{u-1}{u^2}\mathrm{d}u =\frac12\int\left(\frac1u-\frac1{u^2}\right)\mathrm{d}u$, giving

$$\frac{1}{2}\ln(x^2+1)+\frac{1}{2(x^2+1)}+C.$$

For $\int\frac{\mathrm{d}x}{\sqrt{4-x^2}}$, put $x=2\sin t$ with $t\in[-\frac{\pi}{2},\frac{\pi}{2}]$, so $\mathrm{d}x=2\cos t\,\mathrm{d}t$ and $\sqrt{4-x^2}=2\cos t$, leaving $\int\mathrm{d}t=\arcsin\frac{x}{2}+C$.

For $\int\sqrt{x^2-1}\,\mathrm{d}x$, put $x=\sec t$, so the differential is $\sec t\tan t\,\mathrm{d}t$ and the radical is $|\tan t|$, making the integrand $\sec t\tan^2 t=\sec^3 t-\sec t$. Integrating $\sec^3 t$ by parts and substituting back,

$$\int\sqrt{x^2-1}\,\mathrm{d}x=\frac{x\sqrt{x^2-1}}{2}-\frac{1}{2}\ln\left|x+\sqrt{x^2-1}\right|+C.$$

Wallis integrals

The Wallis integrals are

$$W_n=\int_0^{\pi/2}\sin^n x\,\mathrm{d}x=\int_0^{\pi/2}\cos^n x\,\mathrm{d}x,$$

the two forms agreeing under $x\mapsto\frac{\pi}{2}-x$. Integration by parts gives $W_n=\frac{n-1}{n}W_{n-2}$ for $n\geq 2$, with $W_0=\frac{\pi}{2}$ and $W_1=1$, so

$$W_2=\frac{\pi}{4},\qquad W_3=\frac{2}{3},\qquad W_4=\frac{3\pi}{16}.$$

The sequence satisfies $nW_nW_{n-1}=\frac{\pi}{2}$ for every $n$, and decays as $W_n\sim\sqrt{\frac{\pi}{2n}}$. Even terms carry a $\pi$ and odd terms are rational, which is the asymmetry the product below is built out of.

Wallis’s product

$$\frac{\pi}{2}=\prod_{n=1}^{\infty}\frac{2n\cdot 2n}{(2n-1)(2n+1)}=\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdots$$

John Wallis found this in 1655 and published it in Arithmetica Infinitorum the following year, the second infinite product for $\pi$ after Viète’s of 1593.

The proof runs entirely through the integrals above. The sequence $W_n$ is positive and strictly decreasing, so $W_{2n+1}\leq W_{2n}\leq W_{2n-1}$; dividing through by $W_{2n+1}$ and applying the recurrence squeezes $W_{2n}/W_{2n+1}$ to $1$, and writing that ratio out in closed form is the product.

Convergence is slow enough to rule the product out as a way of computing $\pi$. A thousand factors give $3.14081$ against $\pi=3.14159$, with the error falling like $1/n$.

A related identity,

$$\frac{\sin x}{x}=\cos\frac{x}{2}\cos\frac{x}{4}\cos\frac{x}{8}\cdots,$$

yields Viète’s product: at $x=\frac{\pi}{2}$ the left side is $\frac{2}{\pi}$.

Improper integrals

Two kinds, split by where the infinity sits.

The first kind has an unbounded interval. Define $\int_a^{\infty}f=\lim_{b\to\infty}\int_a^b f$ and call it convergent when that limit is finite.

The second kind has an unbounded integrand at a finite point. Approach the bad point with a parameter and take the limit the same way. Where $f$ stays bounded and the discontinuity is removable, the integral is proper and needs no limit.

Either kind is tested by comparison against a known integrand, by limit comparison on the ratio, or by absolute convergence, which implies convergence. The benchmarks are the $p$-integrals, and they point in opposite directions: $\int_1^{\infty}\frac{\mathrm{d}x}{x^p}$ converges exactly when $p>1$, while $\int_0^{1}\frac{\mathrm{d}x}{x^p}$ converges exactly when $p<1$.

Differential equations

Separable equations

For $y'=f(x)g(y)$,

$$\int\frac{\mathrm{d}y}{g(y)}=\int f(x)\,\mathrm{d}x+C.$$

Dividing by $g(y)$ assumes it is nonzero. Each root of $g$ gives a constant solution, and those are read off separately.

Homogeneous equations

For $y'=f\!\left(\frac{y}{x}\right)$, substitute $y=vx$, so $y'=v+xv'$ and

$$x\frac{\mathrm{d}v}{\mathrm{d}x}=f(v)-v,$$

which separates, the substitution having promoted a ratio to a variable.

First-order linear equations

For $y'+p(x)y=q(x)$, the integrating factor is $\mu(x)=e^{\int p(x)\mathrm{d}x}$. It comes from asking for a $\mu$ that makes the left side a single derivative. Expanding,

$$(\mu y)'=\mu y'+\mu' y,$$

which should equal $\mu y'+\mu p y$. Matching forces $\mu'=\mu p$, itself separable, and $\mu=e^{\int p}$ follows.

Then $(\mu y)'=\mu q$, so

$$y=\frac{1}{\mu(x)}\left(\int\mu(x)q(x)\,\mathrm{d}x+C\right)=e^{-\int p}\left(\int qe^{\int p}\mathrm{d}x+C\right),$$

with the outer $e^{-\int p}$ multiplying both terms inside. For constant $p\neq 0$ and constant $q$ this collapses to $y=Ae^{-px}+\frac{q}{p}$.

Constant coefficients

For $y''+ay'+by=0$ with $a,b$ constant, substituting $y=e^{rx}$ gives the characteristic equation $r^2+ar+b=0$. Three cases:

1. Distinct real roots $r_1\neq r_2$: $y=c_1e^{r_1x}+c_2e^{r_2x}$.
2. A repeated root $r$: $y=(c_1+c_2x)e^{rx}$. The factor of $x$ supplies the second solution that a second-order equation needs.
3. Complex roots $\alpha\pm\beta i$: $y=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)$.

The characteristic equation exists because the coefficients are constants. An equation with variable coefficients admits no such polynomial in $r$, since $r$ would have to depend on $x$ and $e^{rx}$ would lose the derivatives the substitution assumed. Variable-coefficient second-order equations are genuinely harder and have no comparable general method.

Undetermined coefficients

For $y''+ay'+by=f(x)$ the general solution is $y=y_h+y_p$, the homogeneous general solution plus any one particular solution.

Where $f$ has the form $e^{\alpha x}(P(x)\cos\beta x+Q(x)\sin\beta x)$ with $P,Q$ polynomial, guess a $y_p$ of that same shape with undetermined coefficients, substitute, and match. A polynomial $f$ of degree $m$ calls for a general polynomial of degree $m$, lower-order terms included.

Resonance decides whether the guess can work at all. A trial form that already solves the homogeneous equation cannot also produce a nonzero right-hand side, and substituting it returns $0=f(x)$. Multiply the trial form by $x^s$, where $s$ is the multiplicity of $\alpha+\beta i$ as a root of the characteristic equation.

Variation of parameters

Let the constants in the homogeneous solution vary. For $y'+p(x)y=q(x)$, write $y_p=u(x)e^{-\int p}$; substituting collapses the $p$ terms and leaves $u'=qe^{\int p}$, so $y_p=e^{-\int p}\int qe^{\int p}\mathrm{d}x$, agreeing with the integrating factor.

For $y''+ay'+by=f(x)$ with homogeneous solutions $y_1,y_2$, the same move writes $y_p=u_1y_1+u_2y_2$, imposes $u_1'y_1+u_2'y_2=0$, and solves the resulting pair for $u_1'$ and $u_2'$:

$$y_p=-y_1\int\frac{y_2f}{W}\,\mathrm{d}x+y_2\int\frac{y_1f}{W}\,\mathrm{d}x,\qquad W=y_1y_2'-y_1'y_2.$$

Variation of parameters asks only that $f$ be continuous, so it handles right-hand sides like $\tan x$ or $\frac1x$ that no trial form covers. Undetermined coefficients stays confined to exponentials, polynomials, sines and cosines.

Linear systems

A first-order system

$$\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=A\mathbf{x}+\mathbf{f}(t)$$

with $A$ a constant matrix has homogeneous solution $\mathbf{x}_h(t)=e^{At}\mathbf{c}$, and variation of parameters gives

$$\mathbf{x}(t)=e^{At}\left(\mathbf{c}+\int_0^t e^{-As}\mathbf{f}(s)\,\mathrm{d}s\right).$$

Diagonalizing $A$ turns $e^{At}$ into an exponential of each eigenvalue on its own and uncouples the system into independent scalar equations. A defective $A$ needs the Jordan form and produces polynomial factors in $t$, the same phenomenon as the repeated root above.

Curvature

For a curve $\mathbf{r}(t)$,

$$\kappa=\frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}.$$

Parametrized by arc length instead, $\kappa=\left|\frac{\mathrm{d}\mathbf{T}}{\mathrm{d}s}\right|$ with $\mathbf{T}$ the unit tangent, and no denominator is needed, since $\left|\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}s}\right|=1$ is what arc-length parametrization means.

For an explicit curve $y=f(x)$ at the point $(x_0,f(x_0))$,

$$\kappa=\frac{|f''(x_0)|}{\left(1+f'(x_0)^2\right)^{3/2}}.$$

At a critical point, where $f'(x_0)=0$, the denominator is $1$ and $\kappa=|f''(x_0)|$. Large $|f'|$ drives $\kappa$ down, and for twice-differentiable $f$ the curvature is finite everywhere.

A straight line has $\kappa=0$; a circle of radius $r$ has $\kappa=\frac1r$ at every point. The units follow: curvature is an inverse length, and $\frac1\kappa$ is the radius of the circle that best fits the curve there.