- #1
phoenixthoth
- 1,605
- 2
what f satisfy the following equation: f(f(x)) = 2^x?
g(2)=2^2=4 so 2 is not a fixed point. I don't think that 2^x has a fixed point in the reals.Let g(x) = 2^x
x = 2 is a fixed point of g(x), so it makes sense to define
oops! yeah, the second to the last line is incorrect. as you said, it should be A(16)=A(2^4)=A(4)+1=s+5. then the last line should be, as you said, A(65536)=A(2^16)=A(16)+1=s+7.A(2^(-oo))=A(-oo)+1 yields A(0)=s+1.
A(2^0)=A(0)+1 yields A(1)=s+2.
A(2^1)=A(1)+1 yields A(2)=s+3.
A(2^2)=A(2)+1 yields A(4)=s+4.
A(2^3)=A(3)+1 yields A(8)=s+5.
A(2^4)=A(4)+1 yields A(16)=s+6.
a similar thing was done with abel's equation. now i get how definining f or the solution to abel's/schroeder's equation on some interval will then lock in place all other values but i don't think this can be done in an arbitrary way. how is it ensured, for example, that the arbitrary assignment of values to f over the range [-oo,b] will lead to satisfaction of f(f(x))=2^x?Assign any values you like to f(x) over the range [-oo, b] satisfying f(-oo) = b, f(b) = 0, and f continuous on [-oo, b].
A half iterate is a mathematical function that takes the value of 2^x and halves it, giving the result of f(x) = 2^(x/2). This function can be iterated to find values of f(f(x)), which is equivalent to 2^x.
To find the function f, we can use the property of logarithms that states log(base b)(x^y) = y*log(base b)(x). In this case, since we are looking for f(x) = 2^(x/2), we can rewrite this as f(x) = 2^(log(base 2)(x)/2). This gives us the function f(x) = sqrt(x).
The half iterate of 2^x can be applied in various fields such as computer science, physics, and engineering. It can be used to model exponential decay or growth, calculate the doubling time of a population or bacteria, and in computer algorithms that involve halving numbers.
The relationship between f(f(x)) and 2^x is that they are equivalent functions. This means that for any value of x, f(f(x)) will give the same result as 2^x. This can be seen by substituting the function f(x) = sqrt(x) into f(f(x)) and simplifying to 2^(x/2).
One limitation of using the half iterate of 2^x is that it only works for positive values of x. This is because taking the square root of a negative number would result in imaginary numbers, which are not allowed in this context. Additionally, this function may not be appropriate for all applications and may need to be modified or combined with other functions to accurately model a real-life scenario.