I need a function f(x) such that when f(x) is applied to any function's sub-functions, the result is the same as if f(x) is applied over the entire function. For example: f((a^2)b + ln(c)/d - 2^e) = f( f( f( a^2 )b ) + f( f( f( ln(c) )/d ) - f( 2^e ) ) ). Another requirement is that f(x) cannot be linear. Does such a function exist?
EDIT: I've come up with another acceptable scenario: f((a^2)b + ln(c)/d - 2^e) = g( h( i(a^2)b ) + j( k( l( ln(c) )/d ) - m( 2^e ) ) ), where g(x) is always applied to addition operations, h(x) to multiplication, i(x) to powers, j(x) to subtraction, k(x) to division, and l(x) to logarithms, and m(x) to exponentiation, etc.. With this scenario, f(x) can be linear, but the g(x), h(x), etc. cannot be linear.
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
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