%This macro provides the text about the finite calculus %on the first horiyontal area of the page 8. %The macro has one parameter % 1( The width of the mathematical text \newcommand\TEightFiniteCalculusA[1]{% \deflength{\HSpace}{#1}% \parbox[t]{\HSpace}{% \TEightSeriesFiniteCalculusFontSize \DisplaySpace{\TEightDisplaySpace}{\TEightDisplayShortSpace} \TEightTitle{Difference, shift operators:} \begin{DisplayFormulae}{1}{\SpaceBeforeFormula}{\TEightBaselineSkipFormulaeB}{\SmallChar}{\StyleWithoutNumber} \Fm{\Delta f(x) = f(x+1) - f(x)i} \Fm{\E f(x) = f(x+1)} \end{DisplayFormulae} \TEightTitle{Fundamental Theorem:} \begin{DisplayFormulae}{1}{\SpaceBeforeFormula}{\TEightBaselineSkipFormulaeB}{\SmallChar}{\StyleWithoutNumber} \Fm{f(x) = \Delta F(x) \Leftrightarrow \sum f(x) \delta x = F(x) + C} \Fm{\sum_a^b f(x) \delta x = \sum_{i=a}^{b-1} f(i)} \end{DisplayFormulae} \TEightTitle{Differences:} \begin{DisplayFormulae}{1}{\SpaceBeforeFormula}{\TEightBaselineSkipFormulaeB}{\SmallChar}{\StyleWithoutNumber} \Fm{\Delta(c u) = c \Delta u} \Fm{\Delta(u + v) = \Delta u + \Delta v} \Fm{\Delta(uv) = u\Delta v + \E v \Delta u} \Fm{\Delta(x^{\underline n}) = n x^{\underline n - 1}} \Fm{\Delta(H_x) = x^{\underline{-1}}} \Fm{\Delta(2^x) = 2^x} \Fm{\Delta(c^x) = (c-1)c^x} \Fm{\Delta\binom{x}{m} = \binom{x}{m-1}.} \end{DisplayFormulae} \TEightTitle{Sums:} \begin{DisplayFormulae}{1}{\SpaceBeforeFormula}{\TEightBaselineSkipFormulaeB}{\SmallChar}{\StyleWithoutNumber} \Fm{\sum c u \, \delta x = c \sum u \, \delta x} \Fm{\sum (u + v) \, \delta x = \sum u \, \delta x + \sum v \, \delta x} \Fm{\sum u \Delta v \, \delta x = uv - \sum\E v \Delta u \, \delta x} \Fm{\sum x^{\underline n}\, \delta x = \frac{x^{\underline{n + 1}}}{m + 1}} \Fm{\sum x^{\underline{-1}} \, \delta x = H_x} \Fm{\sum c^x \, \delta x = {c^x \over c-1}} \Fm{\sum \binom{x}{m} \, \delta x = \binom{x}{m+1}} \end{DisplayFormulae} \TEightTitle{Falling Factorial Powers:} \begin{DisplayFormulae}{1}{\SpaceBeforeFormula}{\TEightBaselineSkipFormulaeB}{\SmallChar}{\StyleWithoutNumber} \Fm{x^{\underline n} = x(x-1) \cdots (x-n+1), \quad n > 0} \Fm{x^{\underline 0} = 1} \Fm{x^{\underline n} = \frac{1}{(x+1) \cdots (x+\vert n \vert)}, \quad n < 0} \Fm{x^{\underline{n+m}} = x^{\underline m}(x-m)^{\underline n}} \end{DisplayFormulae} \TEightTitle{Rising Factorial Powers:} \begin{DisplayFormulae}{1}{\SpaceBeforeFormula}{\TEightBaselineSkipFormulaeB}{\SmallChar}{\StyleWithoutNumber} \Fm{x^{\overline n} = x(x+1) \cdots (x+n-1), \quad n > 0} \Fm{x^{\overline 0} = 1} \Fm{x^{\overline n} = \frac{1}{(x-1) \cdots (x-\vert n \vert)}, \quad n < 0} \Fm{x^{\overline{n+m}} = x^{\overline m}(x+m)^{\overline n}} \end{DisplayFormulae} \TEightTitle{Conversion:} \AdjustSpace{-2ex plus .5ex minus .5ex}% \begin{DisplayFormulae}{1}{\SpaceBeforeFormula}{\TEightBaselineSkipFormulaeB}{\SmallChar}{\StyleWithoutNumber} \def\FirstPart{x^{\underline n}=} \FmPartA{\FirstPart (-1)^n(-x)^{\overline n}=(x-n+1)^{\overline n}=} \FmPartB{\FirstPart}{1/(x+1)^{\overline{-n}}} \def\FirstPart{x^{\overline n} =} \FmPartA{\FirstPart (-1)^n(-x)^{\underline n}=(x+n-1)^{\underline n}=} \FmPartB{\FirstPart}{1/(x-1)^{\underline{-n}}} \Fm{x^n = \sum_{k=1}^n \SousEnsemble{n}{k} x^{\underline k}= \sum_{k=1}^n \SousEnsemble{n}{k} (-1)^{n-k} x^{\overline k}} \Fm{x^{\underline n} = \sum_{k=1}^n \Cycle{n}{k} (-1)^{n-k} x^k} \Fm{x^{\overline n} = \sum_{k=1}^n \Cycle{n}{k} x^k} \end{DisplayFormulae} } }