A real number definition involving Bruijn-Newmann constant

[Karlisbad]

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:

 

[arildno]

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

 

[Karlisbad]

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

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??

 

[Office_Shredder]

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

 

[Karlisbad]

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

 

[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.

 

[Karlisbad]

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