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MathGroup Archive 2004

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Re: Intra functional relations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47543] Re: [mg47493] Intra functional relations
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 15 Apr 2004 03:40:50 -0400 (EDT)
  • References: <200404141116.HAA27283@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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:

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/


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