SMaC: Statistics, Math, and Computing

APSTA-GE 2006: Applied Statistics for Social Science Research

Eric.Novik@nyu.edu | Summer 2026 | Session 2

Mission: recover lunar gravity



100 m tower · 20 flashes · 1 stopwatch

1 kg ball · unknown \(g\)

\[ (t_k, x_k) \longrightarrow \boxed{g} \longrightarrow \text{home} \]

functions \(\rightarrow\) rates \(\rightarrow\) limits \(\rightarrow\) derivatives

A line is a constant rate


\[y = a + bx\]

\[a = y(0)\]

\[b = \frac{\Delta y}{\Delta x}\]

Your Turn: plot_line()

Write plot_line(slope, intercept).

plot_line <- function(slope, intercept) {
  x <- seq(0, 5, length.out = 100)
  # construct y
  # return a ggplot
}

Test three cases:

\[b > 0 \qquad b = 0 \qquad b < 0\]

Use values that remain visible for \(0 \le x \le 5\).

Exponential growth


\[ \frac{y(t + 1)}{y(t)} = b \]

\[ y(t) = A b^t = A e^{kt}, \qquad k = \log b \]

\[ y'(t) = k y(t) \]

Compound interest

\[ A \longrightarrow A(1+r) \longrightarrow A(1+r)^2 \]

\[ P(t) = A(1+r)^t \]

\[ P_n(t) = A\left(1 + \frac{r}{n}\right)^{nt} \]

\[ P_\infty(t) = A e^{rt} \]

Your Turn: bacterial growth

\(N(0) = 1000\) · doubles every 4 hours


\[ N(10) = \; ? \qquad\qquad N(24) = \; ? \]

\[ N(t) = 1000 \cdot 2^{t/4} \]

\[ N(10) \approx 5657 \qquad\qquad N(24) = 64{,}000 \]

The limit that defines \(e\)


\[ e = \lim_{m \to \infty} \left(1 + \frac{1}{m}\right)^m \]

\[ \lim_{n \to \infty} A\left(1 + \frac{r}{n}\right)^{nt} = A e^{rt} \]

Exponent and log toolkit

\[a^{x+y} = a^x a^y\]

\[a^{x-y} = \frac{a^x}{a^y}\]

\[(a^x)^y = a^{xy}\]

\[(ab)^x = a^x b^x\]

\[\log(uv) = \log u + \log v\]

\[\log(u/v) = \log u - \log v\]

\[\log(u^p) = p\log u\]

\[\log y = x \iff y = e^x\]

\(a>0\) · logarithm arguments \(>0\) · for base-\(a\) logs, \(a \ne 1\)

Your Turn: continuous interest

\(A(0) = \$750\) · \(r = 5.5\%\) · continuous compounding


Find \(A(t)\), \(A(5)\), \(A(10)\), and \(A(50)\).

\[ A(t) = 750e^{0.055t} \]

\[ A(5) = \$987.40 \qquad A(10) = \$1{,}299.94 \qquad A(50) = \$11{,}731.97 \]

Logs turn growth into a line

\[ \beta \text{ on a log scale} \quad\Longrightarrow\quad 100\left(e^\beta - 1\right)\% \text{ change} \]

Softmax can overflow

\[ \operatorname{softmax}(x)_i = \frac{e^{x_i}}{\sum_{j=1}^N e^{x_j}} \]

x <- 1000:1002
exp(x)
[1] Inf Inf Inf
exp(x) / sum(exp(x))
[1] NaN NaN NaN

\[ m = \max_j x_j \]

\[ \operatorname{LSE}(x) = m + \log\sum_{j=1}^N e^{x_j-m} \]

\[ \log\operatorname{softmax}(x)_i = x_i - \operatorname{LSE}(x) \]

Your Turn: stable softmax

  1. Implement LSE(x).
  2. Implement log_softmax(x).
  3. Exponentiate; verify the result sums to 1.

Moon model

downward distance = \(+x\)

\[ m\ddot{x} = mg \quad\Longrightarrow\quad \ddot{x} = g \]

1 kg \(\rightarrow\) mass cancels

\[ x(0) = 0 \qquad v(0) = 0 \]

\[ v(t) = gt \]

\[ x(t) = \tfrac{1}{2}gt^2\]

The drop, in data

reference curve: \(g_{\mathrm{ref}} = 1.625\ \mathrm{m/s^2}\) · not an estimate from these observations

Average velocity is a secant slope


\[ \bar v = \frac{\Delta x}{\Delta t} = \frac{x(t_2)-x(t_1)}{t_2-t_1} \]

\[ \bar v = \frac{\tfrac12 g(t_2^2-t_1^2)}{t_2-t_1} = \tfrac12 g(t_1+t_2) \]

From secant to tangent


\[ \text{secant slope} = \frac{x(a+h)-x(a)}{h} \]

\[ v(a) = \lim_{h\to 0}\frac{x(a+h)-x(a)}{h} = x'(a) \]

Animation source: Wikipedia, “Derivative”.

Position \(\rightarrow\) velocity \(\rightarrow\) acceleration

\[ x(t) = \tfrac12 gt^2 \]

\[ \begin{aligned} v(t) &= \lim_{h\to 0} \frac{\tfrac12 g(t+h)^2-\tfrac12 gt^2}{h} \\ &= \lim_{h\to 0} \left(gt+\tfrac12 gh\right) \\ &= gt \end{aligned} \]

\[ a(t) = v'(t) = x''(t) = g \]

\[ v(4) \approx 1.62 \times 4 = 6.5\ \mathrm{m/s}\ \text{downward} \]

Derivative toolkit

\[ \begin{aligned} (c)' &= 0 \\ (x^n)' &= nx^{n-1} \\ (cf)' &= cf' \\ (f+g)' &= f'+g' \end{aligned} \]

\[ \begin{aligned} (e^x)' &= e^x \\ (\log x)' &= \frac{1}{x} \\ (fg)' &= f'g + fg' \\ (f\circ g)' &= (f'\circ g)g' \end{aligned} \]

Your Turn: another moon

\[ x(t) = 4 + 3t + 5t^2 \]

Find \(x(0)\), \(x(1)\), \(v(0)\), \(v(5)\), \(a(5)\), and \(g\).

\[x(0)=4 \qquad x(1)=12\]

\[v(t)=3+10t\]

\[v(0)=3 \qquad v(5)=53\]

\[a(t)=g=10\]

Exact and numerical derivatives

\[ \begin{aligned} \frac{d}{dt}[\sin t\cos t] &= \cos t\cos t + \sin t(-\sin t) \\ &= \cos^2t-\sin^2t \end{aligned} \]

diff(x) / diff(t)

\[ \frac{\Delta x}{\Delta t} \approx \frac{dx}{dt} \]

Your Turn: chain rule

Warm-up:

\[ \frac{d}{dx}\sin(x^2) = 2x\cos(x^2) \]

Now find

\[ \frac{d}{dx}\left(e^{x^2\cos x}\right). \]

\[ e^{x^2\cos x} \left(2x\cos x-x^2\sin x\right) \]

Mission checkpoint


1 drop \(\longrightarrow\) 19 timed marks


\(g\) to one decimal place?

\(g\) to two decimal places?

oxygen \(\leftrightarrow\) precision

Sessions 6–7

Homework: circular motion

\[ x(t)=R\cos(\omega t), \qquad y(t)=R\sin(\omega t), \qquad 0\le t\le \frac{2\pi}{\omega} \]

\(R>0\) · \(\omega>0\) · \(T=2\pi/\omega\)

  • Write position, velocity, and acceleration functions.
  • Plot \(x(t)\) and \(y(t)\) against \(t\); plot \(y\) against \(x\).
  • Plot each velocity and acceleration component with its position component.
  • Verify

\[ \mathbf a(t) = -\omega^2\mathbf r(t). \]

  • Explain the geometry of \(\mathbf r(t)\), \(\mathbf v(t)\), and \(\mathbf a(t)\).