finite calculus

$\fp{x}{m}$ (“ $x$ to the $m$ falling”) is the falling (factorial) power : $\fp{x}{m}=\overbrace{x(x-1)\dots (x-m+1)}^{m \text{ factors}}$
e.g. $\fp{10}{5}$ = $10\times 9\times 8\times 7\times 6 \text{ , }\qquad$ and $\fp{n}{n}=n!$

$\begin{align*} &\text{(INFINITESIMAL) CALCULUS} && \text{ vs FINITE CALCULUS} \\ & && \\ &\text{derivative operator D} & &\text{difference operator } \Delta \\ &Df(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} & &\Delta f(x)=f(x+1)-f(x) \\ &De^x=e^x && \Delta2^x=2^x \\ &Dx^n=nx^{n-1} & &\Delta \fp{x}{n}=n\fp{x}{n-1} \\ &D(a+bx)^n=bn(a+bx)^{n-1} &&\Delta\fp{(a+bx)}{n}=bn\fp{(a+bx)}{n-1} \\ & Dcf(x)=cDf(x)&& \Delta cf(x)=c\Delta f(x)\\ &D\Big(f(x)+g(x)\Big)=Df(x)+Dg(x) && \Delta \Big(f(x)+g(x)\Big)=\Delta f(x)+\Delta g(x)\\ &\text{product rule:}&& \\ & D\Big(f(x)g(x)\Big) && \Delta\Big(f(x)g(x)\Big)\\ & \text{ }\quad =g(x)Df(x)+f(x)Dg(x) && \text{ }\quad =g(x)\Delta f(x)+f(x+1)\Delta g(x) \\ &\text{quotient rule:}&& \\ &D\frac{f(x)}{g(x)}=\frac{g(x)Df(x)-f(x)Dg(x)}{g^2(x)} &&\Delta \frac{f(x)}{g(x)}=\frac{g(x)\Delta f(x)-f(x)\Delta g(x)}{g(x)g(x+1)}\\ &\qquad(\text{if } g(x)\neq 0) && \qquad(\text{if } g(x)g(x+1)\neq 0) \end{align*}$

$\begin{align*} &\text{integration operator }\int & &\text{summation operator }\sum &\\ \intertext{Fundamental Theorem : } &g(x)=Df(x) && g(x)=\Delta f(x) &\\ &\text{ iff } \int g(x) \;\text{d}x=f(x)+C &&\text{ iff } \sum g(x)\;\delta x=f(x)+C &\\ & && &\\ &\text{Definite integral:} && \text{Definite sum:} &\\ &\text{If }g(x)=Df(x)\text{, then} && \text{If }g(x)=\Delta f(x)\text{, then} &\\ &\int_a^b g(x) \;\text{d}x=f(x)\big|_a^b=f(b)-f(a)&& \sum_a^b g(x) \;\delta x = f(x) \big|_a^b=f(b)-f(a) &\\ &&&&\\ &\int cf(x)\;\text{d}x=c \int f(x)\;\text{d}x && \sum cf(x)=c\sum f(x)&\\ &\int \Big( f(x)+g(x)\Big)\;\text{d}x &&\sum \Big( f(x)+g(x)\Big)& \\ & \text{ }\quadd=\int f(x)\;\text{d}x+\int g(x)\;\text{d}x&&\text{ }\quadd=\sum f(x)+\sum g(x) &\\ &\int f(x)Dg(x) && \sum f(x) \Delta g(x)&\\ &\text{ } =f(x)g(x)-\int g(x)Df(x)&&\text{ }=f(x)g(x)-\sum g(x+1) \Delta f(x)& \end{align*}$

The inverse of $D$ is the anti-derivative (integration) operator $\int$
$\int g(x)\;\text{d}x$ , the indefinite integral of $g(x)$ , is the class of functions whose derivative is $g(x)$ .
The “ $C$ ” for indefinite integrals is an arbitrary constant.

The inverse of $\Delta$ is the anti-difference (summation) operator $\sum$
$\sum g(x)\;\delta x$ , the indefinite sum of $g(x)$ , is the class of functions whose difference is $g(x)$ .
The “ $C$ ” for indefinite sums is any function $p(x)$ such that $p(x+1)=p(x)$ .

$\begin{align*} \theta(x)&=\sum f(x)\\ \Delta \theta(x)&=f(x)\\ \Delta \theta(x)&=\theta(x+1)-\theta(x) \end{align*}$

Add together $f(a)$ , $f(a+1)\dots f(a+n-1)$ to get:
Fundamental theorem of the sum calculus:

$\boxed{\sum_{x=a}^{a+n-1} f(x)=\theta(a+n)-\theta(a)=\sum f(x)\Big|_a^{a+n}=\Delta^{-1} f(x) \Big|_a^{a+n}}$

$\boxed{\sum \fp{(a+bx)}{n}=\frac{\fp{(a+bx)}{n+1}}{b(n+1)},\quadd n\neq -1}$

Leibniz’s rule for the $n$ th derivative of the product of two functions $f(x)$ and $g(x)$ :

$\begin{align*} D^n(fg)&=(f)(D^ng)+\binom{n}{1}(Df)(D^{n-1}g) \\ &+\binom{n}{2}(D^2f)(D^{n-2}g)+\dots +\binom{n}{n}(D^nf)(g) \end{align*}$

Leibniz’s rule for differences:

$\begin{align*} \Delta^n(fg)&=(f)(\Delta^n g)+\binom{n}{1}(\Delta f)(\Delta^{n-1}Eg) \\ &+\binom{n}{2}(\Delta^2 f)(\Delta^{n-2}E^2g)+\dots+\binom{n}{n}(\Delta^n f)(E^n g) \end{align*}$

falling powers

$\sum_{0\les k<n} \fp{k}{m} = \frac{\fp{k}{m+1}}{m+1} \bigg|_0^n = \frac{\fp{n}{m+1}}{m+1} \text{for integers }m,n\ges 0$

When $m=1$ , since $\fp{k}{1}=k$ , we get:

$\sum_{0\les k<n} k = \frac{\fp{n}{2}}{2}=\frac{n(n-1)}{2}$

Since $k^2=\fp{k}{2}+\fp{k}{1}$ :

$\sum_{0 \les k<n} k^2=\frac{\fp{n}{3}}{3}+\frac{\fp{n}{2}}{2}=\frac{1}{3}n(n-1)(n-2+\frac{3}{2}) = \frac{1}{3}n(n-\frac{1}{2})(n-1)$

“factorial binomial theorem”

Like $(x+y)^2=x^2+2xy+y^2, \quad \fp{(x+y)}{2} = \fp{x}{2}+2xy+\fp{y}{2}$
And similarly for each $(x+y)^n$ and $\fp{(x+y)}{n}$ .

While $x^{m+n}=x^m x^n, \quad \fp{x}{m+n} = \fp{x}{m}\fp{(x-m)}{n}$

$\begin{align*} \fp{x}{1}&=x \\ \fp{x}{2}&= x^2-x \\ \fp{x}{3}&= x^3-3x^2+2x \\ \fp{x}{4}&= x^4-6x^3+11x^2-6x \\ \fp{x}{5}&= x^5-10x^4+35x^3-50x^2+24x \end{align*}$

The coefficients are the Stirling numbers of the first kind. $s_k^{n+1}=s_{k-1}^n-ns_k^n$

$\begin{align*} x&=\fp{x}{1} \\ x^2&=\fp{x}{2}+\fp{x}{1}\\ x^3&=\fp{x}{3}+3\fp{x}{2}+\fp{x}{1}\\ x^4&=\fp{x}{4}+7\fp{x}{3}+6\fp{x}{2}+\fp{x}{1}\\ x^5&=\fp{x}{5}+15\fp{x}{4}+25\fp{x}{3}+10\fp{x}{2}+\fp{x}{1} \end{align*}$

The coefficients are the Stirling numbers of the second kind. $S_k^{n+1}=S_{k-1}^n+kS_k^n$

Knuth et al – Concrete Mathematics

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finite calculus

falling powers

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