The Princeton Review, Cracking the Math GRE Subject Test
“Calculus: The Greatest Hits”, good breadth.
Shallow treatment of Algebra, Real Analysis, Topology, Number Theory.
Five Official Practice Exams (with Solutions)
- GR 1268
- GR 0568
- GR 9367
- GR 8767
- GR 9768
All old and significantly easier than exams in recent years.
Aim for 90th percentile in $< 2$ hours.
- Computing limits
- Showing continuity
- Computing derivatives
- Rolle’s Theorem
- Mean Value Theorem
- Extreme Value Theorem
- Implicit Differentiation
- Related Rates
- Optimization
- Computing Taylor expansions
- Computing linear approximations
- Riemann sum definition of the integral
- The fundamental theorem of Calculus (both forms)
- Computing antiderivatives
- $u\dash$substitutions
- Partial fraction decomposition
- Trigonometric Substitution
- Integration by parts
- Specific integrands
- Computing definite integrals
- Solids of revolution
- Series (see real analysis section)
- Tools for finding $\lim_{x\to a} f(x)$, in order of difficulty:
- Plug in: equal to $f(a)$ if $f \in C^0(N_\varepsilon(a))$
- Algebraic Manipulation
- L’Hopital’s Rule (only for indeterminate forms $\frac 0 0, \frac \infty \infty$)
- For $\lim f(x)^{g(x)} = 1^\infty, \infty^0, 0^0$, let $L = \lim f^g \implies \ln L = \lim g \ln f$
- Squeeze theorem
- Take Taylor expansion at $a$
- Monotonic + bounded (for sequences)
When possible, of course.
$$\frac{a}{b+\sqrt{c}} = \frac{a}{b+\sqrt{c}} \left( \frac {b-\sqrt c} {b-\sqrt c} \right) = \frac {a(b-\sqrt c)} {b^2 - c} \\ \frac{1}{ax^2 + bx + c} = \frac{1}{(x-r_1)(x-r_2)} = \frac{A}{x-r_1} + \frac{B}{x-r_2}$$
$$\begin{align*} \frac{d}{dx} \int_a^x f(t)~dt &= f(x) \\ \\ \int_a^b \dd{}{x} f(x)~dx &= f(b) - f(a) \\ \end{align*}$$
First form is usually skimmed over, but very important!
$$\frac{\partial}{\partial x} \int_{a(x)}^{b(x)} g(t) dt = g(b(x))b'(x) - g(a(x))a'(x)$$
Commuting $D$ and $I$
Commuting a derivative with an integral
$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(x,t) dt = \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x, t) dt \\ + f(x, b(x))\frac{d}{dx}b(x) - f(x, a(x))\frac{d}{dx}a(x)$$
(Derived from chain rule)
Set
$$
a(x) = a, b(x) = b, f(x,t) = f(t) \implies \dd{}{x} f(t) = 0,
$$
then commute to derive the FTC.
- Solids of Revolution
- Disks: $A = \int \pi r(t)^2 ~dt$
- Cylinders: $A = \int 2\pi r(t)h(t) ~dt$
- Arc Lengths
- $ds = \sqrt{dx^2 + dy^2},\qquad L = \int ~ds$
There are 6 major tests at our disposal:
- Vectors, div, grad, curl
- Equations of lines, planes, parameterized curves
- And finding intersections
- Multivariable Taylor series
- Computing linear approximations
- Multivariable optimization
- Arc lengths of curves
- Line/surface/flux integrals
- Green’s Theorem
- The divergence theorem
- Stoke’s Theorem
Lines
$$Ax + By + C = 0,~ \vector x = \vector p + t\vector v,\\ \vector x \in L \iff \inner{\vector x-\vector p}{\vector n} = 0$$
Planes
$$A x + B y + C z + D = 0,~ \vector x(t,s) = \vector p + t\vector v_1 + s\vector v_2 \\ \vector x \in P \iff \inner{\vector x - \vector p}{\vector n} = 0$$
Distances to lines/planes: project onto orthogonal complement.
Let $S \subseteq \RR^3$ be a surface. Generally need a point $\vector{p} \in S$ and a normal $\vector{n}$.
Key Insight: The gradient of a function is normal to its level sets.
$$\text{Case 1: } S = \theset{[x,y,z] \in \RR^3 \mid f(x,y, z) = 0}$$
i.e. it is the zero set of some function $f:\RR^3 \to \RR$
- $\nabla f$ is a vector that is normal to the zero level set.
- So just write the equation for a tangent plane $\inner{\vector n}{\vector x - \vector p_0}$.
$$\text{Case 2: } S \text{ is given by } z = g(x,y)$$
-
Let $f(x, y, z) = g(x,y) - z$, then
$$\vector p \in S \iff \vector p \in \theset{[x,y,z] \in \RR^3 \mid f(x,y, z) = 0}.\\$$
-
Then $\nabla f$ is normal to level sets, compute $\nabla f = [\dd{}{x}g, \dd{}{y}g, -1]$
-
Proceed as in previous case.
Single variable: solve $\dd{}{x} f(x) = 0$ to find critical points $c_i$ then check min/max by computing $\dd{^2}{x^2}f(c_i)$.
Multivariable: solve $\nabla f(\vector x) = 0$ for critical points $\vector c_i$, then check min/max by computing the determinant of the Hessian:
$$H_f({ \mathbf { a } }) = \left[ \begin{array} { c c c } { \frac { \partial ^ { 2 } f } { \partial x _ { 1 } \partial x _ { 1 } } ( \mathbf { a } ) } & { \dots } & { \frac { \partial ^ { 2 } f } { \partial x _ { 1 } \partial x _ { n } } ( \mathbf { a } ) } \\ { \vdots } & { \ddots } & { \vdots } \\ { \frac { \partial ^ { 2 } f } { \partial x _ { n } \partial x _ { 1 } } ( \mathbf { a } ) } & { \cdots } & { \frac { \partial ^ { 2 } f } { \partial x _ { n } \partial x _ { n } } ( \mathbf { a } ) } \end{array} \right].$$
Lagrange Multipliers:
$$\text{Optimize } f(\mathbf x) \text{ subject to } g(\mathbf x) = c \\ \implies \nabla \vector f = \lambda \nabla \vector g$$
- Generally a system of nonlinear equations
- But there are a few common tricks to help solve.
To get any one derivative, sum over all possible paths to it:
$$\begin{align*} \left(\dd{z}{x}\right)_y &= \left(\dd{z}{x}\right)_{u,y,v} \\ & + \left(\dd{z}{v}\right)_{x,y,u} \left(\dd{v}{x}\right)_y \\ & + \left(\dd{z}{u}\right)_{x,y,v} \left(\dd{u}{x}\right)_{v,y} \\ & + \left(\dd{z}{u}\right)_{x,y,v} \left(\dd{u}{v}\right)_{x,y} \left(\dd{v}{x}\right)_y \end{align*}$$
Subscripts denote variables held constant while differentiating.
Just use Taylor expansions.
$$f(x) = f(p) + f'(p)(x-p) \\+ f''(p)(x-a)^2 + O(x^3)$$
$$f(\vector x) = f(\vector p) + \nabla f(\vector p)(\vector x - \vector a) \\+ (\vector x - \vector p )^T H_f(p)(\vector x - \vector p) + O(\norm{\vector x - \vector p}_2^3)$$
-
The fundamental theorem of arithmetic:
$$n\in\mathbb Z \implies n = \prod_{i=1}^n p_i^{k_i}, \quad p_i \text{ prime}$$
-
Divisibility and modular congruence:
$$x\mid y \iff y = 0 \mod x \iff \exists c \suchthat y = xc$$
-
Useful fact:
$$x = 0 \mod n \iff x = 0 \mod p_i^{k_i} ~\forall i$$
(Follows from the Chinese remainder theorem since all of the $p_i^{k_i}$ are coprime)
- GCD, LCM
$$xy = \gcd{(x,y)}~\mathrm{lcm}{(x,y)} \\ d\mid x \and d\mid y \implies d \mid \gcd(x,y) \\ \quad \and \gcd(x,y) = d\gcd(\frac x d, \frac y d)$$
- Also works for $\mathrm{lcm}(x,y)$
- Computing $\gcd(x,y)$:
- Take prime factorization of $x$ and $y$,
- Take only the distinct primes they have in common,
- Take the minimum exponent appearing
Computes GCD, can also be used to find modular inverses:
$$\begin{aligned} a & = q _ { 0 } b + r _ { 0 } \\ b & = q _ { 1 } r _ { 0 } + r _ { 1 } \\ r _ { 0 } & = q _ { 2 } r _ { 1 } + r _ { 2 } \\ r _ { 1 } & = q _ { 3 } r _ { 2 } + r _ { 3 } \\ & \vdots \\r_k &= q_{k+2}r_{k+1} + \mathbf{r_{k+2}} \\ r_{k+1} &= q_{k+3}r_{k+2} + 0 \end{aligned}$$
Back-substitute to write $ax+by = \mathbf{r_{k+2}} = \gcd(a,b)$.
(Also works for polynomials!)
-
Coprime
$$a\text{ is coprime to } b \iff \gcd(a,b) = 1$$
-
Euler’s Totient Funtion
$$\phi(a) = \abs{\theset{x \in \NN \suchthat x \leq a \and \gcd(x,a) = 1}}$$
Know some group and ring theoretic properties of $\ZZ/n\ZZ$
Chinese Remainder Theorem
The system
$$\begin{array} { c } { x \equiv a _ { 1 } \quad \left( \bmod m _ { 1 } \right) } \\ { x \equiv a _ { 2 } \quad \left( \bmod m _ { 2 } \right) } \\ { \vdots } \\ { x \equiv a _ { r } \quad \left( \bmod m _ { r } \right) } \end{array}$$
has a unique solution $x \mod \prod m_i$ iff $\gcd(m_i, m_j) = 1$ for each pair $i,j$.
Chinese Remainder Theorem
The solution is given by
$$x = \sum_{j=1}^r a_j \frac{\prod_i m_i}{m_j} \left( \left[ \frac{\prod_i m_i}{m_j} \right]^{-1}_{\mod m_j}\right)$$
Seems symbolically complex, but actually an easy algorithm to carry out by hand.
Chinese Remainder Theorem
Ring-theoretic interpretation: let $N = \prod n_i$, then
$$\gcd(i,j) = 1 ~~\forall (i,j) \implies \ZZ_N \cong \bigoplus \ZZ_{n_i}$$
-
Fermat’s Little Theorem and Euler’s Theorem
$$a^p = a \mod p \\ p \not\mid a \implies a^{p-1} = 1 \mod p \\ \text{and in general, } \\ a^{\phi(p)} = 1 \mod p$$
-
Wilson’s Theorem
$$n \text{ is prime } \iff (n-1)! = -1 \mod n$$
- Mobius Inversion
- Quadratic residues
- The Legendre/Jacobi Symbols
- Quadratic Reciprocity