[Maxima-lang-fr] Encore un probleme (fonction reciproque) !

 [Maxima-lang-fr] Encore un probleme (fonction reciproque) ! From: - 2007-12-20 16:12:22 ```Bonjour, C'est toujours moi ! Je cherche a calculer la fonction reciproque d'une fonction bijective. Quelqu'un pourrait-il m'expliquer pourquoi le programme suivant donne correctement la solution de g(x)=0.2 ou g(x) est l'integrale de f de 0 a x. kill(all)\$ f(x):=x\$ g(A):=romberg(f(q),q,0,A)\$ phi(u):=g(u)-0.2\$ find_root(phi,u,0,4); tandis que celui-ci qui cherche a faire la meme chose ne marche pas. kill(all)\$ f(x):=x\$ g(A):=romberg(f(q),q,0,A)\$ find_root((g(u)-0.2),u,0,4); Comment ecrire un programme donnant la fonction qui a y associe la solution de l'equation g(x)=y ou g(x) est l'integrale de f de 0 a x ? Je ne sais pas comment faire. Le programme suivant ne marche pas... kill(all)\$ f(x):=x\$ g(A):=romberg(f(q),q,0,A)\$ phi(u,v):=g(u)-v\$ psi(v):=find_root(phi(u,v),u,0,4); psi(0.2); Cordialement, Cyril This communication is confidential, may be privileged and is meant only for the intended recipient. If you are not the intended recipient, please notify the sender by reply and delete this message from your system. Any unauthorised dissemination, distribution or copying hereof is prohibited. BNP Paribas Fund Services UK Limited, BNP Paribas Trust Corporation UK Limited, BNP Paribas UK Limited, BNP Paribas Commodity Futures Ltd and Investment Fund Services Limited are authorised and regulated by the Financial Services Authority. BNP Paribas, BNP Paribas Securities Services and BNP Paribas Private Bank are authorised by the CECEI and AMF. BNP Paribas London Branch, BNP Paribas Securities Services London Branch and BNP Paribas Private Bank London Branch are regulated by the Financial Services Authority for the conduct of their UK business. BNP Paribas Securities Services London Branch is also a member of the London Stock Exchange. ```

 [Maxima-lang-fr] Encore un probleme (fonction reciproque) ! From: - 2007-12-20 16:12:22 ```Bonjour, C'est toujours moi ! Je cherche a calculer la fonction reciproque d'une fonction bijective. Quelqu'un pourrait-il m'expliquer pourquoi le programme suivant donne correctement la solution de g(x)=0.2 ou g(x) est l'integrale de f de 0 a x. kill(all)\$ f(x):=x\$ g(A):=romberg(f(q),q,0,A)\$ phi(u):=g(u)-0.2\$ find_root(phi,u,0,4); tandis que celui-ci qui cherche a faire la meme chose ne marche pas. kill(all)\$ f(x):=x\$ g(A):=romberg(f(q),q,0,A)\$ find_root((g(u)-0.2),u,0,4); Comment ecrire un programme donnant la fonction qui a y associe la solution de l'equation g(x)=y ou g(x) est l'integrale de f de 0 a x ? Je ne sais pas comment faire. Le programme suivant ne marche pas... kill(all)\$ f(x):=x\$ g(A):=romberg(f(q),q,0,A)\$ phi(u,v):=g(u)-v\$ psi(v):=find_root(phi(u,v),u,0,4); psi(0.2); Cordialement, Cyril This communication is confidential, may be privileged and is meant only for the intended recipient. If you are not the intended recipient, please notify the sender by reply and delete this message from your system. Any unauthorised dissemination, distribution or copying hereof is prohibited. BNP Paribas Fund Services UK Limited, BNP Paribas Trust Corporation UK Limited, BNP Paribas UK Limited, BNP Paribas Commodity Futures Ltd and Investment Fund Services Limited are authorised and regulated by the Financial Services Authority. BNP Paribas, BNP Paribas Securities Services and BNP Paribas Private Bank are authorised by the CECEI and AMF. BNP Paribas London Branch, BNP Paribas Securities Services London Branch and BNP Paribas Private Bank London Branch are regulated by the Financial Services Authority for the conduct of their UK business. BNP Paribas Securities Services London Branch is also a member of the London Stock Exchange. ```
 Re: [Maxima-lang-fr] Encore un probleme (fonction reciproque) ! From: Robert Dodier - 2007-12-22 00:12:53 ```On Dec 20, 2007 9:07 AM, wrote: > f(x):=x\$ > g(A):=romberg(f(q),q,0,A)\$ > phi(u):=g(u)-0.2\$ > find_root(phi,u,0,4); > f(x):=x\$ > g(A):=romberg(f(q),q,0,A)\$ > find_root((g(u)-0.2),u,0,4); > f(x):=x\$ > g(A):=romberg(f(q),q,0,A)\$ > phi(u,v):=g(u)-v\$ > psi(v):=find_root(phi(u,v),u,0,4); > psi(0.2); Cyril, I apologize for writing English here. I believe these examples all yield the expected result with the most recent version of Maxima (namely, the Maxima 5.14.0 release candidate, which you can obtain from Sourceforge). The evaluation of the romberg function was modified after Maxima 5.13.0. Hope this helps, Robert Dodier PS. Here is what I see: (%i1) build_info (); Maxima version: 5.13.99rc1 Maxima build date: 21:32 12/4/2007 host type: i686-pc-mingw32 lisp-implementation-type: GNU Common Lisp (GCL) lisp-implementation-version: GCL 2.6.8 (%o1) (%i2) f(x):=x\$ (%i3) g(A):=romberg(f(q),q,0,A)\$ (%i4) phi(u):=g(u)-0.2\$ (%i5) find_root(phi,u,0,4); (%o5) 0.63245553203368 (%i6) find_root((g(u)-0.2),u,0,4); (%o6) 0.63245553203368 (%i7) phi(u,v):=g(u)-v\$ (%i8) psi(v):=find_root(phi(u,v),u,0,4); (%o8) psi(v) := find_root(phi(u, v), u, 0, 4) (%i9) psi(0.2); (%o9) 0.63245553203368 ```