- #36
jbriggs444
Science Advisor
Homework Helper
- 12,855
- 7,508
I still do not understand this.Stavros Kiri said:Real continuous functions on the [also continuous] variables and constants, e.g. say (x, A) [continuous quantities (i.e. their domain is a continuous subset of R)], plus f(x;A) [or notation like fA(x)], for one variable and one constant etc.], continuous functions on their domain, a requirement for simplicity to give you easily every choice possible, that would make the puzzle trivial.
A constant function is continuous. But if you allow for arbitrary constant functions to be invoked (specified how?) then the problem becomes trivial. So one must have a limited roster of functions.
If you are allowed to include functions with no arguments (also known as zero-place functions or constants) then the problem becomes trivial. Given one such function (call it k), one can write ##\frac{k+k+k+k}{k+k+k}## or similar to encode any rational number. So we cannot allow zero-place functions . (Not even zero, see below).
If one is given a one-place monotone increasing function f and an unrelated one-place monotone decreasing function g then one can write expressions like f(f(g(g(f(g(x)))))) and use the pattern of f's and g's to encode values. So we must exercise care with the set of one-place functions that are permitted, otherwise the problem becomes trivial. [That was the point of post #26].
For two-place or greater functions, the fact that no constants (not even zero) are allowed is helpful. With only four constant terms (the four fours) to use, a maximum of three two-place functions can possibly be used. e.g. f(4,g(4,h(4,4))). Similarly, at most two three place functions or at most one four-place function. The point is that the user is prevented from chaining together an arbitrarily long string of function compositions.
So. Which functions do you have in mind, consistent with the above guidelines?
Plus, minus, times, divide, exponentiation and unary negation?
Last edited: