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[MUG] more about implicitdiff:complexity and recursiveness

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[MUG] more about implicitdiff:complexity and recursiveness
Author: Charles James Leonardo Quarra Cappiello    Posted: Thu, 21 Nov 2002 12:26:09 -0400
>> From: "Charles James Leonardo Quarra Cappiello" /> Hi, I've noted that the time of calculation of implicitdiff( eq(z,w) , z , w$n ); grows very quick with n, since in most calculations of the n-th implicit derivative of z respect of w, one has usually: n n-1 n-2 d z d z d z dz --- = f ( ----- , ----- , -- , z , w ); n n-1 n-2 dw dw dw dw now, in a particular calculation im in need of calculating all k-th implicit derivatives from 1 to n, but it looks like each timestep, each k=1..n-1 derivatives are recalculated, amounting to an enourmous waste. Any ideas how could implicitdiff reuse the results of the n-1 previous implicit derivatives to calculate the n-th one? Greetings, Charles Quarra _________________________________________________________________ The new MSN 8: smart spam protection and 2 months FREE* http://join.msn.com/?page=features/junkmail

[MUG] Re: more about implicitdiff:complexity and recursiveness
Author: Preben Alsholm    Posted: Tue, 26 Nov 2002 10:38:18 +0100
>> From: Preben Alsholm /> Here are two solutions. As they are written the output is a Taylor polynomial. The last one is by far the fastest. It makes use of implicitdiff just once and then dsolve/series. The only problem with that one is that dsolve/series has a bug that makes it unreliable. The case below is not affected by that bug. restart; > implicittaylor1:=proc(eq::equation,L::list(name=algebraic),n::posint) > local x,y,x0,y0,EQ,ym,k; > x,y:=op(map(lhs,L)); > x0,y0:=op(map(rhs,L)); > EQ:=subs(y=y(x),eq); > for k to n do > EQ:=diff(EQ,x); > ym[k]:=eval(solve(EQ,diff(y(x),x$k)),{x=x0,y(x)=y0,seq(diff(y(x),x$i)=ym[i],i=1..k-1)}) > end do; > y0+add(ym[k]*(x-x0)^k/k!,k=1..n) > end proc: > eq1:=x^2*y+x^3+y^4-3*y+2 = 0: > time(implicittaylor1(eq1,[x=0,y=1],25)); 30.754 > implicittaylor2:=proc(eq::equation,L::list(name=algebraic),n::posint) local x,y,x0,y0,ODE,ym,k; > x,y:=op(map(lhs,L)); > x0,y0:=op(map(rhs,L)); > ODE:=diff(y(x),x)=subs(y=y(x),implicitdiff(eq,y,x)); > Order:=n+1; > convert(rhs(dsolve({ODE,y(x0)=y0},y(x),type=series)),polynom) > end proc: > time(implicittaylor2(eq1,[x=0,y=1],25)); 1.102 The bug in dsolve/series is exhibited here: > restart; > Order:=2: This is fine: > dsolve({diff(y(x),x)=f(x,y(x)),y(x0)=y0},y(x),type=series); 2 y(x) = y0 + f(x0, y0) (x - x0) + O((x - x0) ) this is not > dsolve({diff(y(x),x)=f(x+y(x)),y(x0)=y0},y(x),type=series); 2 y(x) = y0 + f(y0) (x - x0) + O((x - x0) ) nor is this > dsolve({diff(y(x),x)=x*f(x+y(x)),y(x0)=y0},y(x),type=series); 2 y(x) = y0 + x0 f(y0) (x - x0) + O((x - x0) ) Preben Alsholm Department of Mathematics Technical University of Denmark Charles James Leonardo Quarra Cappiello wrote: >>>From: "Charles James Leonardo Quarra Cappiello" /> >> > > Hi, > > I've noted that the time of calculation of implicitdiff( eq(z,w) , z , w$n > ); grows very quick with n, since in most calculations of the n-th implicit > derivative of z respect of w, one has usually: > > n n-1 n-2 > d z d z d z dz > --- = f ( ----- , ----- , -- , z , w ); > n n-1 n-2 > dw dw dw dw > > now, in a particular calculation im in need of calculating all k-th implicit > derivatives from 1 to n, but it looks like each timestep, each k=1..n-1 > derivatives are recalculated, amounting to an enourmous waste. Any ideas how > could implicitdiff reuse the results of the n-1 previous implicit > derivatives to calculate the n-th one? > > Greetings, > > Charles Quarra > > > > > _________________________________________________________________ > The new MSN 8: smart spam protection and 2 months FREE* > http://join.msn.com/?page=features/junkmail > > >

[MUG] Re: more about implicitdiff:complexity and recursiveness
Author: Charles James Leonardo Quarra Cappiello    Posted: Thu, 28 Nov 2002 12:33:48 -0400
>> From: "Charles James Leonardo Quarra Cappiello" /> Thanks Preben, I come out with a solution in the meantime (this mailing list is SLOOOW) EfficientImplicitDiff:=proc( Limplicitderivatives::name , f::name , z::name , w::name , ord::integer ) local j,k,temp1,temp2; Limplicitderivatives:=[ diff(z(w),w)=-1/f(z(w)) ]; for k from 2 to ord do; temp1:=diff( f(z(w))*diff(z(w),w) , w$(k-1) ); temp2:=isolate( temp1 , diff(z(w),w$k ) ); for j from 1 to k-1 do; temp2:=op(1,temp2)=subs( Limplicitderivatives[k-j] , op(2,temp2) ); end do; temp2:=op(1,temp2)=normal(op(2,temp2)); Limplicitderivatives:=[ op(op(Limplicitderivatives)) , temp2 ]; end do; end proc; The idea is that it uses L as a list where it puts the implicit derivatives of z respect to w of increasing order, and f ,z ,w as names for f(z(w)). In this examples, i assume that f(z(w))*diff(z(w),w) + 1 = 0; and i put that assumption in the first entry of the list, but it could be generalized for the case where the dependence on w is a function g(z(w)) Greetings, Charles Quarra _________________________________________________________________ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus

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