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\begin{document}

\begin{center}
  {\Large\bfseries Calculus and Linear Algebra} \quad
  {\normalsize Course 101, Your Name}
\end{center}

\vspace{0.5ex}

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\section*{Limits}

$\lim_{x \to a} f(x) = L$ means $f(x)$ can be held as close to
$L$ as you like by taking $x$ close enough to $a$.

\begin{align*}
  \lim_{x \to 0} \frac{\sin x}{x} &= 1 \\
  \lim_{x \to 0} \frac{1 - \cos x}{x} &= 0 \\
  \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x} &= e
\end{align*}

L'Hopital's rule, for $0/0$ or $\infty/\infty$ only:
\[
  \lim_{x \to a} \frac{f(x)}{g(x)}
    = \lim_{x \to a} \frac{f'(x)}{g'(x)} .
\]

\section*{Derivatives}

\begin{align*}
  (fg)' &= f'g + fg' \\
  \left(\frac{f}{g}\right)' &= \frac{f'g - fg'}{g^2} \\
  \bigl(f(g(x))\bigr)' &= f'(g(x))\, g'(x)
\end{align*}

\begin{align*}
  (x^n)' &= n x^{n-1} \\
  (e^x)' &= e^x \\
  (a^x)' &= a^x \ln a \\
  (\ln x)' &= \frac{1}{x} \\
  (\sin x)' &= \cos x \\
  (\cos x)' &= -\sin x \\
  (\tan x)' &= \sec^2 x \\
  (\arctan x)' &= \frac{1}{1 + x^2}
\end{align*}

\section*{Integrals}

By parts and by substitution:
\begin{align*}
  \int u \, dv &= uv - \int v \, du \\
  \int f(g(x)) g'(x) \, dx &= \int f(u) \, du
\end{align*}

\begin{align*}
  \int x^n \, dx &= \frac{x^{n+1}}{n+1}, \quad n \neq -1 \\
  \int \frac{1}{x} \, dx &= \ln |x| \\
  \int e^{ax} \, dx &= \frac{e^{ax}}{a} \\
  \int \sec^2 x \, dx &= \tan x \\
  \int \frac{1}{1 + x^2} \, dx &= \arctan x \\
  \int \frac{1}{\sqrt{1 - x^2}} \, dx &= \arcsin x
\end{align*}

Every indefinite integral carries $+ C$.

\section*{Series}

\begin{align*}
  e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!} \\
  \sin x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \\
  \cos x &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \\
  \frac{1}{1 - x} &= \sum_{n=0}^{\infty} x^n, \quad |x| < 1 \\
  \ln(1 + x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}
\end{align*}

Taylor series of $f$ about $a$:
\[
  f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n .
\]

\section*{Trigonometry}

\begin{align*}
  \sin^2 x + \cos^2 x &= 1 \\
  1 + \tan^2 x &= \sec^2 x \\
  \sin(x \pm y) &= \sin x \cos y \pm \cos x \sin y \\
  \cos(x \pm y) &= \cos x \cos y \mp \sin x \sin y \\
  \sin 2x &= 2 \sin x \cos x \\
  \cos 2x &= \cos^2 x - \sin^2 x
\end{align*}

\section*{Vectors}

\begin{align*}
  \mathbf{a} \cdot \mathbf{b}
    &= |\mathbf{a}| \, |\mathbf{b}| \cos \theta \\
  |\mathbf{a} \times \mathbf{b}|
    &= |\mathbf{a}| \, |\mathbf{b}| \sin \theta
\end{align*}

\[
  \mathbf{a} \times \mathbf{b} =
  \begin{vmatrix}
    \mathbf{i} & \mathbf{j} & \mathbf{k} \\
    a_1 & a_2 & a_3 \\
    b_1 & b_2 & b_3
  \end{vmatrix}
\]

The vectors are orthogonal when $\mathbf{a} \cdot \mathbf{b} = 0$ and
parallel when $\mathbf{a} \times \mathbf{b} = \mathbf{0}$.

\section*{Matrices}

\begin{align*}
  (AB)^{\mathsf{T}} &= B^{\mathsf{T}} A^{\mathsf{T}} \\
  (AB)^{-1} &= B^{-1} A^{-1} \\
  \det(AB) &= \det A \, \det B \\
  \det(A^{\mathsf{T}}) &= \det A
\end{align*}

For a $2 \times 2$ matrix:
\[
  A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},
  \quad \det A = ad - bc,
\]
\[
  A^{-1} = \frac{1}{ad - bc}
    \begin{pmatrix} d & -b \\ -c & a \end{pmatrix},
    \quad ad - bc \neq 0 .
\]

Expanding a $3 \times 3$ determinant along the first row:
\[
  \begin{vmatrix}
    a & b & c \\ d & e & f \\ g & h & i
  \end{vmatrix}
  = a(ei - fh) - b(di - fg) + c(dh - eg) .
\]

\section*{Eigenvalues}

$\lambda$ is an eigenvalue of $A$ with eigenvector
$\mathbf{v} \neq \mathbf{0}$ when
\[
  A \mathbf{v} = \lambda \mathbf{v},
  \quad \text{that is} \quad
  \det(A - \lambda I) = 0 .
\]

\begin{itemize}
  \item The eigenvalues add up to $\operatorname{tr} A$.
  \item The eigenvalues multiply to $\det A$.
  \item $A$ is invertible exactly when no eigenvalue is $0$.
  \item $\operatorname{rank} A + \dim \ker A = n$ for an
        $n \times n$ matrix $A$.
\end{itemize}

\end{multicols}

\end{document}
