Re: function of a function
- To: mathgroup at smc.vnet.net
- Subject: [mg62668] Re: function of a function
- From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
- Date: Wed, 30 Nov 2005 05:40:40 -0500 (EST)
- Organization: Uni Leipzig
- References: <dmha20$932$1@smc.vnet.net><dmhfhd$bit$1@smc.vnet.net> <dmjrv9$638$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi, "Narasimham" <mathma18 at hotmail.com> schrieb im Newsbeitrag news:dmjrv9$638$1 at smc.vnet.net... | Jens-Peer Kuska wrote: | | > it can't work because f [0] ==1 given in your differential equation | > f ' [0]==f [1] and NDSolve[] can't find the value for | > f[1] until it has integrated the equation. | | ??? | You don't understand, that in EQ= { f'[x] == f[f[x]], f[0]== 1} ; NDSolve[EQ,f,{x,0,2}]; the right hand side must be evaluated by NDSolve[] at x==0 and this gives f'[x]==f[f[x]] /. x->0 f'[0]==f[f[0]] and with your initial condition f[0]==1 one ends up with f'[0]==f[1] and this can only evaluated when NDSolve[] has escaped from x==0 but it can't, because it is not able to get a numerical value for the derivative at x==0 ??? | > The nested dependence is equivalent to an infinite | > system of ordinary differential equations and it seems to be | > hard to do this by a finte computer. | | I cannot understand this. In the following two examples the first one | works, not the second. | | Clear[x,f,EQ]; | EQ={f'[x] == f[Cos[x]],f[0]== 1}; | NDSolve[EQ,f,{x,0,4}]; | f[x_]=f[x]/.First[%]; | Plot[f[x],{x,0,4}]; | | Clear[x,f,EQ]; | EQ={f'[x] == Cos[f[x]],f[0]== 1}; | NDSolve[EQ,f,{x,0,4}]; | f[x_]=f[x]/.First[%]; | Plot[f[x],{x,0,4}]; the first example can't work (and it does not) because EQ={f'[x] == f[Cos[x]],f[0]== 1}; at x==0 gives f'[0]==f[Cos[0]] and f'[0]==f[1] but nobody can find f[1] That has *nothing* to do with Mathematica because it is a functional equation that can't be solved. Even the simpler case f'[x]==g[f[x-tau]] has an infinite number of degrees of freedom because you have to supply a function for f[] in the interval [x0-tau,x0] to integrate the equation numerical. Regards Jens