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/

**References**:**Intra functional relations***From:*mathma18@hotmail.com (Narasimham G.L.)