Register to reply 
Complex Partway Functions 
Share this thread: 
#1
Nov1212, 06:16 PM

P: 17

What I mean by a partway function is this:
ff(x)=6x now as you probably know that f(x)=√6(x) or you could argue f(x)=√6(x), with that function that you have just found being the partway function between x and f(x)=6xDo you understand? But what about more complex partway functions like ff(x)=sin(x) so what is f(x)= to?, which is the same as saying what is the partway function between x and sin(x). 


#2
Nov1212, 06:52 PM

Mentor
P: 21,214




#3
Nov1212, 07:01 PM

HW Helper
P: 1,391

$$f(f(x)) = \sin x,$$ then what is ##f(x)##? This particular equation has been studied, but I'm afraid I can't remember the name of the function or the wikipedia article. Edit: Ah! Here's the wikipedia article I was thinking of: Schroder's equation In particular, it discusses the "functional square root", a function such that ##h_{1/2}(h_{1/2}(x)) = h(x)##, which is relevant to the question of ##f(f(x)) = \sin x##. 


#4
Nov1212, 07:01 PM

P: 136

Complex Partway Functions
I would avoid using ff(x) to represent (f(x))^2 since it looks like the composite f(f(x)).
Also, I'm a bit confused about what you mean by complex (complex numbers or complicated). That is, for y^2=6x, x>=0 if x is real or all x if we're using complex numbers. In the case of y^2=sin(x), assuming we're working with real numbers, sin(x)>=0 so x€U(n)[2nπ,(2n+1)π] for n€Z. So, y=+√sin(x) for the same x in that set. (These are two distinct functions.) If we work with complex functions, then y=+√sin(x) for all complex numbers x. Is this what you're looking for? 


#5
Nov1312, 10:52 AM

P: 17




#6
Nov1312, 11:17 AM

HW Helper
P: 1,391




#7
Nov1312, 11:49 AM

P: 136

If you click on the link, "functional square root", you would find that rin(rin(x))=sin(x) where rin(x) is the function you supposedly wanted.
@dalek, in you first post where you said ff(x)=6x, and you also said f(x)=√6(x) which could mean either √(6x) or (√6)x. This was not clear as (f(x))^2 = 6x, and f(f(x))= 6x for the corresponding choice of f(x), respectively. 


#8
Nov1312, 11:56 AM

P: 17




#9
Nov1312, 01:55 PM

P: 17




Register to reply 
Related Discussions  
Complex Analysis  Solving Complex Trig functions  Calculus & Beyond Homework  5  
[complex analysis] domain coloring applet? (visualizing complex functions)  General Math  0  
Help with My complex functions  Precalculus Mathematics Homework  6  
[complex functions] finding complex roots in 1+z+az^n  Precalculus Mathematics Homework  2  
Complex Functions of a complex variable  Calculus & Beyond Homework  3 