Product Rule

\[ \begin{aligned} \Delta (ab)_n &= a_n \Delta b_n + b_n \Delta a_n + \Delta a_n \Delta b_n \\ \triangledown (ab)_n &= a_n \triangledown b_n + b_n \triangledown a_n - \triangledown a_n \triangledown b_n \\ \\ \Delta (uv) &= u \Delta v + Ev \Delta u \\ \end{aligned} \]

If \(a\) is a polynomial of degree \(p\), then

\[ \begin{aligned} \Delta^{p+1} a &= (0) \\ \end{aligned} \]

Falling Power Rule

\[ \begin{aligned} n^{\underline k} &= n (n-1) \cdots (n-k+1), \hskip4em n^\underline{0} = 1 \\ \Delta n^{\underline k} &= k \, n ^{\underline{k-1}}\\ n^{- \underline k} &= \frac{1}{(n+1) (n+2) \cdots (n+k)} \\ \end{aligned} \]

Sequence operators

\[ \begin{aligned} (Ia)_n &= a_n \\ (Ea)_n &= a_{n+1} \\ (\Delta a)_n &= a_{n+1} - a_n, \hskip4em \Delta = E - I \\ (\triangledown a)_n &= a_n - a_{n-1}, \hskip4em \triangledown = I - E^{-1} \\ (\triangledown \Delta a)_n &= a_{n+1} - 2a_n + a_{n-1}, \hskip1em \text{second central difference} \\ \triangledown \Delta \, a &= \Delta \triangledown a \\ \end{aligned} \]

Higher Derivatives

\[ \begin{aligned} \Delta^k &= (E - I)^k = \sum_{i=0}^k (-1)^{k-i} {k \choose i} E^i \\ \end{aligned} \]

Anti-difference

\[ \begin{aligned} \Delta^{-1} &= (E - I)^{-1} = -(I - E)^{-1} = -(I + E + E^2 + E^3 + \cdots) + C \\ \end{aligned} \]

so long as the sequence \(a_n\) is eventually 0

Discrete Fundamental Theorem of Integral Calculus

\[ \begin{aligned} \sum_{n=A}^B \Delta u &= u \, \bigg \rvert_{n=A}^{B+1} \\ \sum_{n=A}^B \triangledown u &= u \, \bigg \rvert_{n=A-1}^{B} \\ \end{aligned} \]

Alternatively,

\[ \begin{aligned} \sum_{n=A}^B u &= \Delta^{-1} u \, \bigg \rvert_{n=A}^{B+1} \\ \sum_{n=A}^B u &= \triangledown^{-1} u \, \bigg \rvert_{n=A-1}^{B} \\ \end{aligned} \]

Integration by parts

\[ \begin{aligned} \Delta (uv) &= u \Delta v + Ev \Delta u \\ \sum_{n=A}^B u \Delta v &= uv \, \bigg \rvert_{n=A}^{B+1} - \sum_{n=A}^B Ev \Delta u \\ \end{aligned} \]

Differential equations

Linear first-order recurrence relation:

\[ \begin{aligned} u_{n+1} &= \lambda u_n \\ E \, u &= \lambda u \\ (E - \lambda \, I) \, u &= 0 \\ \implies u_n &= u_0 \, \lambda^n \\ \end{aligned} \]

Linear first-order difference equation:

\[ \begin{aligned} \Delta u &= \lambda u \\ (E - I) \, u &= \lambda u \\ (E - (\lambda + 1) \, I) \, u &= 0 \\ \implies u_n &= u_0 \, (\lambda + 1)^n \\ \end{aligned} \]

Numerical ODE's

Many differential equations of the form \(\displaystyle \frac{dx}{dt} = f(x,t) \) cannot be solved exactly. Hence the use of numerical ODE's to approximate a solution.

Euler's method

Uses a difference equation to approximate the solution of an initial value problem. Essentially it's Taylor series 1st order approximation for the recursion relation. Given \(\displaystyle \frac{dx}{dt} = f(x,t) \) , and initial value \( x_0 = x(t_0) \) , Euler's method approximates \( x(t_*) \) for some \(t_* > t_0 \).

\[ \begin{aligned} x_{n+1} = x_n + h\,f(x_n, t_n) \\ t_{n+1} = t_n + h \\ h = \frac{t_* - t_0}{N} \\ \end{aligned} \]

where N is the number of intervals.

Numerical Integration

How to approximate the definite integral \(\displaystyle \int_a^b f(x)\,dx\) via finite sums which approximates the area under the curve ? Common techniques: Riemann sums, trapezoid rules and Simpson's rule.

Left Riemann Sum

\[ \begin{aligned} \int_a^b f(x)\,dx \approx \sum_{n=0}^{N-1} f_n \cdot (\Delta x)_n = \sum_{n=0}^{N-1} f_n \cdot (x_{n+1} - x_n) \\ \end{aligned} \]

Right Riemann Sum

\[ \begin{aligned} \int_a^b f(x)\,dx \approx \sum_{n=1}^{N} f_n \cdot (\triangledown x)_n = \sum_{n=1}^{N} f_n \cdot (x_n - x_{n-1}) \\ \end{aligned} \]

Trapezoid Rule

An improvement over the left and right Riemann sums by averaging them.

\[ \begin{aligned} \int_a^b f(x)\,dx \approx \sum_{n=0}^{N-1} \frac{1}{2} (f_n + f_{n+1}) \cdot (\Delta x)_n = \sum_{n=0}^{N-1} \frac{1}{2} (f_n + f_{n+1}) \cdot (x_{n+1} - x_n) \\ \end{aligned} \]

If sample points are evenly spaced:

\[ \begin{aligned} \int_a^b f(x)\,dx \approx h \, \bigg(\frac{1}{2}(f_0 + f_N) + \sum_{n=1}^{N-1} f_n \bigg) \\ \end{aligned} \]

where \(\displaystyle h = \frac{b-a}{N} \)