- #1
friend
- 1,452
- 9
Suppose I have a function of two real variables f(x,y). How could I prove that f(x,y) is complex? Thanks.
HallsofIvy said:First, what do you mean by a "complex function"? My first thought was "a function that returns complex numbers", but the set of complex numbers includes the real numbers. Do you mean "a function that returns, for some (x, y), non-real values"?
friend said:Yes, thank you. I don't know if f(x,y) returns a real or complex value. So I think that means I cannot evaluate its complex conjugate. I think I'm looking for some algebraic properties unique to complex numbers (that include the reals) that I can use to prove my function is complex(including reals). Maybe something like f(x,y)≠f(y,x), but f2(x,y)=f2(y,x), or something like that? Or maybe I can somehow show that there is no sense in saying f(x,y)>0?
DonAntonio said:...what are your function's range and definition domains? Why do you suspect the function could possible attain some non-complex values? Without this it's hard to know how to help you.
friend said:Thanks guys for your replies. Sorry it took so long to reply; I've been working.
Suppose I have a set of functions labeled, [itex]\delta (x - {x_0})[/itex], for every value of x0, which have the following property:
[tex]\delta (x - {x_0}) = \int_{ - \infty }^{ + \infty } {\delta (x - {x_1})\delta ({x_1} - {x_0})} \,d{x_1}[/tex]
I wonder if I can prove that this set of functions can be represented by functions that can take on complex values (which includes the real numbers). For example, if I can prove that these functions are multiplicatively commutative and associative, but lack the property of ordering with respect to ≥, then does that prove that they are complex, or at least can be represented by complex numbers since they share the same algebra? This is opposed to say if it were shown that these functions also lack commutativity, they would be represented by quaternions. Or if they also lacked associativity, they'd be represented by octonions. Your comments are appreciated.
DonAntonio said:Once again you don't take the time to tell us what kind of functions those [itex]\,\delta\,[/itex] are: real functions, complex functions...?
You can't take beings out of the blue and decide their functions without first checking some basic stuff: domain, range,...
DonAntonio
friend said:Actually, I'm thinking they are tempered distributions that satisfy, among other things, the Chapman-Kolmogorov equation. But please don't ignor the question: if the algebra is the same as the complex numbers: multiplicably commutative and associative but not ordered wrt ≥, then does that prove it IS complex valued? How could it not be? Aren't complex numbers defined by their algebraic properties? Thanks.
$${}$$friend said:Obviously from the integral, the domain is from plus or minus infinity of the variable x.
$${}$$The range will depend on how you normalize it. But if the output of the function is a complex number with real and imaginary parts, then how can you say which output is more than the other?
The real part of one output may be less than the real part of another output, but the absolute value of one may be more than another because the imaginary part is so much greater. So I'm not sure you can specify a range of the output if you can't say which output is larger than another and you cannot even say yet that its output is complex valued.
To determine if a function is complex, you can look for the presence of the imaginary unit i in the function. If there is an i term in the function, then it is a complex function.
No, the value of a function alone cannot determine if it is a complex function. You must analyze the function itself to determine if it contains complex components.
A real function only takes real numbers as inputs and outputs, while a complex function can take both real and imaginary inputs and outputs. A complex function typically involves the imaginary unit i and may have both real and imaginary parts.
Yes, there are several methods to prove that a function is complex. One common method is to analyze the function algebraically and look for the presence of i terms. Another method is to graph the function and look for any complex or non-real points.
Yes, a function can have both real and complex components. This means that the function has both real and imaginary inputs and outputs. In this case, the function can be classified as a complex function.