Let me see if I've got this clear.

You want a function such that f(a+b)=f(a)+f(b) and f^2(a)=f(a) for all a,b in the function's domain?

Lets look at it in the operator way. Say a,b are in V and V is a vector space. If P is a linear operator mapping V to V that is idempotent, i.e. P^2=P, i.e. a linear projection operator, then we have of course P(a+b)=P(a)+P(b) and P^2(a) = P(a). If s is a scalar, then additionally P(s a) = s P(a). As regards multiplication of the arguments, this may not be what you want.

Fabricio,

Yes for f(a + b) = f(a) + f(b) and f^2(a) = f(a) is not required, but f(a^2) = f(a)*f(a) is. Further, the same relationship must hold for any function g(x), where f(g(x)). In other words, for any mathematical operator op that takes two inputs (i.e. + , - , etc.), f(a op b) = f(a) op f(b) must hold. Also, df/dx cannot be a constant, meaning f(x) cannot be linear.

Herbert,

Unfortunately, for the first scenario at least, P(ab) = P(a)*P(b) is required. Also, the derivative of f(x) cannot be a constant, thus f(x) cannot be linear. Though it might be possible to work this into the f(x) in the second scenario because it is slightly less restrictive.

The condition f(x^2) = f(x)*f(x) is verified only for linear funcitons. In fact, d/dx( f(x^2) ) = df/dx * 2x and d/dx( f(x)*f(x) ) = 2(df/dx) * f(x).

Then: df/dx * 2x = 2(df/dx) * f(x)

f(x) = x+a

I don't know wether this will help but :

- if you look at the properties your function f must have, for example f(a+b)=f(a)+f(b) and f(a*b)=f(a)*f(b), then, for a such that f(a) is not 0, you must have f(a)=f(a*1)=f(a)*f(1), hence f(1)=1. In the same way, f(2)*f(a)=f(2*a)=f(a+a)=f(a)+f(a)=2*f(a), hence f(2)=2. To generalize, for any integer k, f(k)*f(a)=f(k*a)=f(a+...+a)=f(a)+...+f(a)=k*f(a), hence, for all integer k, f(k)=k. Of course, this works for positive or negative integers (and for 0 as well).

- Now, if you look at numbers in Q*, f(p/q)=f(p)/f(q)=p/q (as p and q are integers)...

- As all real number is the limit of a sequence of decimal numbers, we can extrapolate that for any real number r, we must have f(r)=r. (I guess it can further be extrapolated to complex numbers...)

Then, except for the identity function, I really don't see any function that can fulfill these properties. But of course, this is not what you are looking for... I'm probably missing something here ! (Or maybe I have misunderstood your search...)

By the way, why are you searching such a function ?

Thanks everybody for your responses, but I actually no longer need such a function. I solved my problem a different way. To answer your question Catherine, I am developing a new computer science method that is able to extract an equation from any number of given n-dimensional points. For part of this algorithm, I use a network of operations with each connection weight representing a coefficient. At first the weights (coefficients) are randomized, but they need to be optimized to produce the least amount of error. The way to do this is a process called backpropagation, which distributes the error backwards through the network proportionately based on the derivatives of the activation functions and the weights. Unfortunately, backpropagation doesn't work with a linear activation function. I changed my perspective though. Instead of pushing error backward, I change a weight and observe the change in error and that gives me my change in error with respect to change in weight. This means that I can use a linear activation function, but still have the benefit of systematic training of the weights.

Dear Herman

Your search is not an easy task , such function may be not exist to satisfy

all your requirements to solve and fit your Mathematical Model , so I advise

you to read some thing about Functional Equations and this Text will help

you very much :

Lectures On :

Functional Equations And Their Applications

By J.ACZEL

University Of WATERLLO

WATERLOO,ONTARIO ,CANADA

1966, Academic Press ,NEW YORK And London.

Hope it is useful for you

Wish you good Luck

Bryan: Your algorithm is heuristic!