# A real number definition involving Bruijn-Newmann constant

1. Nov 8, 2006

A "real" number definition involving Bruijn-Newmann constant..

Ok, ths isn't anything new, but i would like to discuss this possibility, taking into account the function:

$$\xi(1/2+iz)=A(\lambda)^{-1/2}\int_{-\infty}^{\infty}dxe^{(-\lambda)^{-1} B (x-z)^{2}}H(\lambda, x)$$

then with the expression above we could study all the values of "lambda" so the Wiener-Hopf integral above involving a symmetryc tranlational Kernel has only real roots, or use it to prove that for `$$\lambda >0$$ has always real roots so RH would be proved and Bruijn constant would be $$6.10^{-9}<\Lambda <0$$:grumpy:

Last edited: Nov 8, 2006
2. Nov 8, 2006

### arildno

Nice try, eljose, but I'm sure this doesn't work either.

3. Nov 10, 2006

It would be nice if someone could explain me the trick Newmann use to prove that $$\lambda >1/2$$ to take a look at it... i'm not a mathematician but i believe that a real function should always have real roots am i wrong??

4. Nov 10, 2006

### Office_Shredder

Staff Emeritus
x2+1=f(x) is a real function with no real roots

5. Nov 10, 2006

oh..sorry you are rigth and also the functions (non polynomials) have complex roots, such us:

$$exp(x)+x+1=0$$ $$e^{x^2}+1=0$$

amazingly the complex function $$exp(2i \pi x)-1=0$$ has only real roots.

the definition of a function f(x) is more than 300 years old, i don't know why mathematicians don't have some criteria to decide wether a real function has only real roots or complex appart from knowing that if f(x) and f(x*) (complex conjugate) are equal then there are pairs or complex roots changing only their imaginary part b to -b

6. Nov 10, 2006

### matt grime

We do have a simple criterion: f has real zeroes if and only if f(x)=0 implies x is in R. You're once more confusing several different issues, Jose. Testing when something satisfies some property is (always?) a hard problem except in toy examples as anyone can tell just by thinking about it for a few seconds, instead of offering yet another damning indictment of the stupidity of mathematicians.

7. Nov 11, 2006

I didn't want to offend mathematician i only questioned that such a "easy" (in appearance) question was not answered and that you could find some theorems for much more difficult questions,.. that's all, in fact Newmann could prove that for $$\lambda > 1/2$$ H(z,\lambda) had real roots the question is why this can't be applied for the other positive values of Lambda