A conjecture about the roots of real functions

[eljose]

All the roots of a real function f(x) are real unless.
1.K(x) is a Polynomial of degree k
2.f(x)=exp(g(x)) where g(x) is different from ln of something
3.f(z) with z=u+iv is invariant under the transformation of v=-v with f(u+iv)=F(u-iv)..
4.the function f includes some of the functions above named

In this cases there can be complex roots,excluding this all the roots are real...

 

[fourier jr]

All the roots of a real function f(x) are real unless.
1.K(x) is a Polynomial of degree k
2.f(x)=exp(g(x)) where g(x) is different from ln of something
3.f(z) with z=u+iv is invariant under the transformation of v=-v with f(u+iv)=F(u-iv)..
4.the function f includes some of the functions above named
In this cases there can be complex roots,excluding this all the roots are real...

1. polynomials with even degree might not have any real roots. what about x^2 + 1 ? if the degree of a polynomial is odd then it will always have a real root. if i the degree is even it might have real roots, it might not.

 

[fourier jr]

2. set g(x) = ln(sin(x)) + 1, which doesn't equal ln of something but exp(g(x)) has infinitely many real roots, at multiples of pi

edit: i guess i didn't consider domain/range with this counterexample but i don't think it's really important.

 

[Hurkyl]

All the roots of a real function f(x) are real unless.

There is no unless. The roots of a real function are always real.

The real function defined by g(x) = x^2 + 1 has no roots.

The complex function defined by h(x) = x^2 + 1, of course, has two complex roots.

 

[HallsofIvy]

In light of Hurkyl's point, eljose, exactly what do you mean by a "real function"? Hurkyl is clearly taking it to mean "a real valued function of a single real variable" which, by definition, cannot have any non-real roots- so your statement is wrong.
I might be inclined to take it to mean "a real valued function of a single complex variable" (the "real function" simply meaning real valued) so that it is possible to have non-real roots.
However, in that case, your statement is still wrong.
Counterexample: take f(x+ iy)= y- 1.
1) f is not a polynomial.
2) f is not exp(f(z)) (in what I understand to be your sense).
3) f(x-iy)= -y-1 so it is not invariant under complex conjugate.
4) f is not a combination of the above.
But the roots of f are all numbers of the form x+ i.

 

[eljose]

the case f(x+iy)=y-1 makes no sense..i am supposing HallsoftIvy that f is real so f(x-y) for real x and y will be real,however according to your definition:
f(x-y)=iy-1 is complex but if f,x and y are real also f(x-y) should be real.

The case exp(f(x))-1=0, i am supposign that f(x) is not ln(g(x)+1)

the counterxamples i have proposed can have or can not have real roots

this conjecture is realted to Riemann hypothesis in the sense that in both cases for $$\zeta(1/2+is)$$ if s is a root also s* is another root,and the Riemann function has no symmetry under the change of s=-s.

 

[HallsofIvy]

the case f(x+iy)=y-1 makes no sense..i am supposing HallsoftIvy that f is real so f(x-y) for real x and y will be real,however according to your definition:
f(x-y)=iy-1 is complex but if f,x and y are real also f(x-y) should be real.

Apparently you don't understand the function I used. The function I defined: f(x+iy)= y-1 means exactly what it says: if z= x+ iy then f(z)= y-1.
f(z)= the imaginary part of z, minus 1. If x and y are real then x-y has 0 imaginary part so f(x-y)= 0-1= -1, not a complex number.

The case exp(f(x))-1=0, i am supposign that f(x) is not ln(g(x)+1)
the counterxamples i have proposed can have or can not have real roots
this conjecture is realted to Riemann hypothesis in the sense that in both cases for $$\zeta(1/2+is)$$ if s is a root also s* is another root,and the Riemann function has no symmetry under the change of s=-s.