Multiplicative Versions of Infinitesimal Calculus What happens when you replace the summation of standard integral calculus with multiplication? Compare the abbreviated definition of a standard integral f( x ) dx= lim f ( xi )* D x ò Dx ®0å With ÕÕf( x )­ dx = lim f ( xi ) ­ D x Dx ®0 and (1+f () x dx ) = lim (1 + f ()* xi D x ) ÕÕDx ®0 Call these later two “integrals” multigrals of Type I and II. (Note: unlike “normal” products, these products are not discrete but continuous over an interval). Consider each in turn. Multigrals (Type I) By standard operations Õ f( x )­ dx = e ­ (ò ln( f ( x )) dx ) By not taking limits, a finite product approximation can be obtained. For example, let f(x)= x from 0 to 1. Then the Type I multigral of x from 0 to 1 is: 1 1 Õ x­ dx = e ­(ò ln( x ) dx ) = 1/ e 0 0 This can be approximated by the sequence 1 2 1 [( )( )]­ ( ) = 0.4714... 3 3 2 1 2 3 1 [( )( )( )]­ ( ) = 0.4543... 4 4 4 3 1 2 3 4 1 [( )( )( )( )]­ ( ) = 0.4427... 5 5 5 5 4 .... 1 2 999 1 [( )( )....( )]­ ( ) = 0.369123...etc 1000 1000 1000 999 Which tends to 1/e=0.36788… (use Stirling’s Formula to support this). pi / 2 For f(x)=tan(x) in radians from 0 to pi/2, Õ tan(x )­ dx = e0 = 1 0 And 1 [tan(p / 6)*tan(2 p / 6)] ­ ( ) = 1 2 1 [tan(p /8)*tan(2 p / 8)*tan(3 p / 8)] ­ ( ) = 1...etc . 3 The above approximations of the multigral can be likened to the mid-point-rule when approximating standard integrals. Like standard integrals, multiplicative analogs of the Trapezoidal Rule and Simpson’s Rule can be found, like: “Simpson’s” Product: b Õ f() x­ dx » a ì[f ( a )* f ( b )]* ü ï ï ï ï Dx í{[f ( a+ D x )* f ( a + 3 D x )*...] ­ 4} * ý ­ ( ) ï ï 3 îï{[f ( a+ 2 D x )*...] ­ 2} þï Consider the following approximations: 2 2 Y=Õ x ­ dx = e ­(ò ln( x ) dx = e ­ (2ln(2) - 2 + 1) = 4 / e = 1.471517765... 1 1 Multiplicative Analog of …. Mid-point Rule Trapezoidal Rule Simpson’s Rule ∆x=1 1.5 [(1)(2)]^(1/2) n.a. =1.4142…. ∆x=1/2 [(1.25)(1.75)]^(1/2) [(1)(2)]^(1/4) [(1)(2)]^((1/2)(1/3)) =1.4790199… *[1.5]^(1/2) *[1.5]^((4/3)(1/2)) =1.4564753… =1.47084… ∆x=1/3 [(7/6)(9/6)(11/6)]^(1/3) [(1)(2)]^(1/6) n.a. =1.474890668… *[(4/3)(5/3]^(1/3) =1.46476345…. ∆x=1/4 [(9/8)(11/8)(13/8) [(1)(2)]^(1/8) [(1)(2)]^((1/4)(1/3)) (15/8)]^(1/4) *[(5/4)(6/4)(7/4)]^(1/4) *[(1.25)(1.75)]^((4/3)(1/4)) =1.473423… =1.4677043…. *[1.5]^((2/3)(1/4)) =1.471466559…. Like standard calculus you can define a multiplicative analog of the derivative ( the m-derivative), construct a multiplicative version of the Fundamental Theorem of Calculus, construct a multiplicative analog of Maclaurin’s Series, etc. The m-derivative for Type I multigrals is: * f¢() x f ()()x= e ­ I f() x The Fundamental Theorem is: b* b f¢()() x f b (x )­ dx = e ­ (( ) dx ) = ÕÕf I a a f()() x f a Compare with the Fundamental Theorem of Standard Calculus: b ò f¢()()() x dx= f b - f a a Small programs can be written to approximate the above results by finite products for those who doubt. Type I multigrals find application in the area of population dynamics. With stochastic birth- and death- rates, the conventional approach is to use means (ie: expectations). Without migration, mean populations E(P) remain constant iff mean birth-rates E(b)= mean death-rates E(d) under the stochastic recursive equation Pn+1 =(1 + b - d )* P n . But, while mathematically correct, this result is misleading. In certain circumstances, simulations show that mean birth-rates can significantly exceed mean death-rates yet MOST population trials decline, even though the mean population of many trials stays constant. True. Let G(x)= Õ x­ (()) p x dx where X= the random variable of (1+b-d) and p(x) is its probability density function. It can be shown that the MODE of populations (Pn) tends to {G(x)↑n}*P0 as n→∞. In general G(x) is < E(x)=E(1+b-d). Thus when E(b)=E(d), the mode of Pn →0 as n→∞ even though E(Pn) = P0. Thus the stochastic recursive equations (where ran# is a random number between 0 and 1) Pn+1 = (2.718281828....* ran #)* P n Pn+1 =(2* ran # + 0.17696)* P n Pn+1 =( ran # + 0.54421)* P n are all constant (in the long-term mode) unlike Pn+1 = (2* ran #)* P n Pn+1 =( ran # + 0.5)* P n which are constant in the long-term mean but tend to zero in the mode. (Try simulating using Excel if you don’t believe). Now consider… Multigrals (TypeII) 1 Consider Õ(1+ x * dx ) which is the limit of the sequence: 0 æ1 ö ç1+ *1 ÷ = 1.5 è2 ø æ1 1 öæ 3 1 ö ç1+ * ÷ç 1 + * ÷ = 1.546875 è4 2 øè 4 2 ø æ1 1 öæ 3 1 öæ 5 1 ö ç1+ * ÷ç 1 + * ÷ç 1 + * ÷ = 1.573559671 è6 3 øè 6 3 øè 6 3 ø æ1 1 öæ 3 1 öæ 5 1 öæ 7 1 ö ç1+ * ÷ç 1 + * ÷ç 1 + * ÷ç 1 + * ÷ = 1.589455605...etc . è8 4 øè 8 4 øè 8 4 øè 8 4 ø which tends to sqr(e)=1.648721271… 1 This is due to the non-standard integral ò ln(1+x * dx ) = 0.5 0 (which is not of the form ò f() x dx ) and thus 1 1 1 Õ(1+x * dx ) = e ­ (ò ln(1 + x * dx )) = e ­ ( ò x * dx ) = e 0 0 0 In general, Õ(1+f ( x ) dx ) = e ­ (ò f ( x ) dx ) provided ò (f ( x ) dx )­ n = 0 for n Î N ³ 2 . Functions f(x) which fail the later condition appear to be few. 1 dx For instance, f(x)=1/x fails this test from 0 to 1 as ò ( )­ 2 = p 2 / 6 (look at the 0 x limit definition of the integral under equal ∆x subintervals to see this). But for most other functions ò (f ( x ) dx )­ n = 0 for n Î N ³ 2 f¢() x For Type II multigrals, the m-derivative is: f* () x = II f() x b b * f¢()() x f b And the Fundamental Theorem is: ÕÕ(1+fII ( x ) dx ) = (1 + dx ) = a a f()() x f a Higher order m-derivatives can be also used, like: ìf¢() b ü í ý b b f¢¢()()()*() x f ¢ xf* () b f() b f a f ¢ b (1+f** ( x ) dx ) = (1 + ( - ) dx ) =II =î þ = ÕÕII * a a f¢()()()()*() x f x f aìf¢() a ü f ¢ a f b II í ý îf() a þ And so on. In general, the Fundamental Theorem becomes more complicated for higher order m-derivatives, unlike (say) polynomials with standard calculus. For instance, é2 2 ù êf¢¢¢()()()() x f ¢¢ xæ f ¢¢ x ö æ f ¢ x ö ú ê- -ç ÷ + ç ÷ ú êf¢()()()() x f xç f ¢ x ÷ ç f x ÷ ú êè ø è ø ú f*** () x = ë û II éf()() x f x ù ꢢ- ¢ ú ê ú ëf¢()() x f x û And thus é2 2 ù ìêf¢¢¢()()()() x f ¢¢ xæ f ¢¢ x ö æ f ¢ x ö ú ü éf¢¢()() b f ¢ b ù ïê- -ç ÷ + ç ÷ ú ï êf¢()()()() x f xç f ¢ x ÷ ç f x ÷ ú - ** b*** b è ø è ø ê ú ïëê ûú ï ëf¢()() b f b û fII () b (1+ (x ) dx ) =í 1 + dx ý = = ÕÕf II éf¢¢()() x f ¢ x ù ** a a ïê- ú ï éf¢¢()() a f ¢ a ù fII () a êf¢()() x f x ú - ïë û ï êf¢()() a f a ú î þ ë û For example, let f(x)=ln(x+2) then f'(x)=1/(x+2), f''(x)=-1/(x+2)^2, f'''(x)=2/(x+2)^3 and f(0)=ln(2), f(1)=ln(3), etc. Then 1 Õ(1+ f*** ( x ) dx ) 0 ìé2/(x+ 2) ­ 3 1/( x + 2) ­ 2æ - 1/( x + 2) ­ 2 ö ææ 1/( x + 2) ö ö ù ü ïê( )+ ( ) -ç ( ) ­ 2 ÷ +çç ÷ ­ 2 ÷ ú ï 1 ïëê1/(x+ 2) ln( x + 2)è 1/( x + 2) ø èè ln( x + 2) ø ø ûú ï =Õí1 + * dx ý 0 ïé-1/(x + 2) ­ 2æ 1/( x + 2) ö ù ï ê() - ú ï1/(x+ 2)ç ln( x + 2) ÷ ï îëè ø û þ é1 ù ln(x + 2) + 1 1 êln(x + 2) ú =Õ{1 - ê ú * dx} 0 (x + 2) êln(x + 2) + 1 ú ëê ûú ** = f (1) f ** (0) éf¢¢(1) f ¢ (1) ù - êf¢(1) f (1) ú == ê ú f¢¢(0) f ¢ (0) ê- ú ëêf¢(0) f (0) ûú =(2/3)((1 + 1/ ln(3))/(1 + 1/ ln(2))) = 0.5213474447... Approximating using N x subintervals gives: N 10 100 1000 approximation 0.5096103 0.520198 0.5212327 Whacko! Like standard calculus you can change variables in the standard way: ì b f() b du (1+f ( x ) dx ) ® (1 + u ) ïÕÕ ¢ -1 ï a f() a f( f ( u )) ïfrom ï ílet u=f(x) then du=f¢ (x)dx ïx=aÞ u=f(a) ï ïx=bÞ u=f(b) ï -1 îdx=du/f¢ (x)=du/(f ¢(f (u ))) And thus, for example: b 1 ÕÕ(1+xdx ) = (1(( + b - a ) x + a )*( b - a )) dx a 0 Product and Quotient Rules for Type I and II multigrals are: Type I Type II Derivative f¢() x f¢() x f* = e ­ () f * = f() x f() x Product Rule ()fg***= f g ()fg***= f + g * Quotient Rule f f f *** ()* = () =f - g g g* g Surprisingly Type II multigrals have the same sort of “Maclaurin’s” Product as Type I.