Question about functions and defining new functions. im thinking about a kind of problem where you are given some property of a function, and the solution is to find what functions satisfy that property. for example: f[x+n*2pi]=f[x] for all integer n the solution is the trig functions. f[x+c]^n=f[nx]*f[nc] for all c and n the solution is all functions in the form of f[x]=a^x a is any number g[(cx)^n]=n(g[x]+g[c]) for all c and n the solution is all functions in the form of loga [x] a is any number x*f[x]=f[x+1] for all integer n the solution is the gamma function. (interesting. most other solutions are classes of functions, but as far as i know there is only one gamma function) in the simplest case, f[x*n]=n*f[x] and f[x+c]=f[x]+f[c] the solution is functions in the form of f[x]=a*x (commutative property and distributive property) All these problems are easy to solve because we already know the functions and due to us knowing the functions we know the properties of the functions. but how do i define a new function merely by establishing that it has a certain property? for example, there may be a class of functions that has the property f[x^n]=nx^(n-1)*f[x], or class of functions with the property f[x+n]=f[x]^(2^n). infact there are probably infinitely many functions such as these with unique propertys. So; how do i determine whether there is a class of functions that obeys some property? How do i know whether that class is singular, or if there are many functions in that class? and most importantly, how do i determine what those functions are, and how to calculate them? differentiating(even known functions) yeilds equations with terms in both df/dx and df/d(g[x]). i cant figure out how to solve these. The only thing i can think of is setting the taylor series equal, and then solving an nxn system with cramers rule (as n approaches infinity, nonetheless) for the coefficients, and then use those coefficients to formulate the taylor series of a function that satisfies the property that was already established. seems like a lot of work.