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MathGroup Archive 2004

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Re: Intra functional relations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg48008] Re: Intra functional relations
  • From: mathma18 at hotmail.com (Narasimham G.L.)
  • Date: Wed, 5 May 2004 08:11:14 -0400 (EDT)
  • References: <200404141116.HAA27283@smc.vnet.net> <c5lgsf$c4e$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote in message news:<c5lgsf$c4e$1 at smc.vnet.net>...
> On 14 Apr 2004, at 20:16, Narasimham G.L. wrote:
> 
> > This may be almost frivolous: If f[x_]+f[y_]=f[x*y],then,
> > can Mathematica prove or solve f[x_]= C Log[x],( C arbitrary constant)
> > ?
> > Also, if g[x_+y_]=g[x]*Sqrt[1- g[y]^2] + Sqrt[1- g[x]^2]*g[y],then,
> > can it be proved that g[x_]=Sin[x]? Can the function capabilities of
> > Mathematica in some way be put to use here?  Regards.
> 
> I hope you do not expect Mathematica to do any proving by itself? But 
> if you mean whether it can be helpful in proving things of this kind 
> then the answer is certainly. I will just indicate what can be done in 
> your second case (the first is much easier).
> 
> But first of all, both your statements are not true without additional 
> conditions on the function g. For example, if the first rule is to hold 
> for all x than there is no solution at all. And in the second case, 
> g[x]=0 for all x is at least one other solution. Besides, even the 
> solution you are thinking of is g[x]:= Sin[ c x], where c is any 
> constant. So you certainly need more assumptions on g. But anyway, this is 
> essentially how one can try to solve such problems:

I left out what appeared trivial or obvoius at the posting time,
aiming for main definition. I thought it an intersting way to
synthesize new "derived functions"
> 
> The idea is to define a function of two variables:
> 
> In[1]:=
> h[x_, y_] := g[x + y] - (g[y]*Sqrt[1 - g[x]^2] +
>      g[x]*Sqrt[1 - g[y]^2])
> 
> and then use the fact that h[x,y] is identically zero to obtain 
> information about g. The first thing to do is to find the value of 
> g[0]. We can do it as follows:
> 
> Solve[{h[0, 0] == 0}, g[0]]
> 
> {{g[0] -> 0}, {g[0] -> -(Sqrt[3]/2)}, {g[0] -> Sqrt[3]/2}}
> 
> This gives three possible values. With additional assumptions on g one 
> can eliminate the last two but I shall leave it to you ;-) So let's 
> consider only the case
> 
>   g[0]=0;
> 
> We now use differentiation and obtain a differential equation:
> 
> 
> eq = (D[h[x, y], y] /. y -> 0 /. Derivative[1][g][0] -> c) == 0
> 
> (-c)*Sqrt[1 - g[x]^2] + Derivative[1][g][x] == 0
> 
> (I had to substitute some other symbol, I chose c,  for 
> Derivative[1][g][0] since its presence would interfere with the next 
> step).
> 
> We need also to remove the definition g[0]=0 to be able to use it in 
> the differential equation, so:
> 
> g[0] =.
> 
> Now we get our answer:
> 
> DSolve[{eq,g[0]==0},g[x],x]
> 
> Inverse functions are
>        being used by
>        Solve so some solutions may not be found; use Reduce for complete 
>        \ solution information.
> 
> {{g[x]->Sin[c x]}}
> 
> Andrzej Kozlowski
> Chiba, Japan
> http://www.mimuw.edu.pl/~akoz/

It is brilliant,not expected at all... Thanks,  exactly this was what
I wanted to see. Regards (Sorry about the delayed response..)


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