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..)