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A conjecture about the roots of real functions

  1. Nov 4, 2005 #1
    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...
     
  2. jcsd
  3. Nov 4, 2005 #2
    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.
     
    Last edited: Nov 4, 2005
  4. Nov 4, 2005 #3
    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.
     
  5. Nov 4, 2005 #4

    Hurkyl

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    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.
     
  6. Nov 5, 2005 #5

    HallsofIvy

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    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.
     
  7. Nov 5, 2005 #6
    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 [tex]\zeta(1/2+is)[/tex] if s is a root also s* is another root,and the Riemann function has no symmetry under the change of s=-s.
     
  8. Nov 5, 2005 #7

    HallsofIvy

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    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.

     
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