The Riemann Hypothesis Explained.....

[andyandy]

I'm currently trying to write something that explains the basics of the Riemann Hypothesis in such a way as to be accessible for high achieving 17-18 students - so I'm working my way through the wiki explanation at the moment. My first question is related to why the riemann zeta function has trivial roots only at negative values of s (ie -2,-4,-6 etc) - surely positive s = 0,2,4 etc also give zeroes of the functional equation, so why are these not also regarded as trivial zeroes?

And anyone got any good web links for a good basic introduction on this (somewhat complicated) topic?

Thanks (there will likely be some more questions later )

 

[andyandy]

Seeing if I can copy the latex from wiki (not sure why that's not showing up in equation form - anyone know? )

[latex]$$\zeta(s) = \sum_{n=1}^\infty n^{-s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \;\;\;\;\;\;\; \sigma = \mathfrak{R}(s) > 1. \! $$[/latex]

For this we have the domain s being Re(s) > 1 so how does this get extended such that the domain is for all the complex plane? And if it is extended to the whole complex plane, why are the solutions s = 0,2,4 etc to:


[latex]$$ \zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s) \!, $$[/latex]

not also trivial solutions?

Additionally, given that the Riemann hypothesis is assumed to be true what impact would an actual proof actually have? And what impact would a proof that it is not correct have?

Thanks

 

[Meridian]

For s=2,4,6,... the value of R(s) is given (as it is for any s with Re(s)>1) by the sum, and in this case (as for any real s>1) is clearly a positive real number.

I think what you are missing looking at the functional equation is that Gamma(z) has a pole at negative integers. For integer s>=2 the pole in Gamma(1-s) `cancels' the zero of the sin bit. More precisely, you can't just plug in values of s where one or more terms in the equation is not defined, but have to take some kind of limit, and as mentioned in the wikipedia article, as s approaches an integer >=2 the product of the Gamma and sin terms approaches something non-zero.

 

[andyandy]

For s=2,4,6,... the value of R(s) is given (as it is for any s with Re(s)>1) by the sum, and in this case (as for any real s>1) is clearly a positive real number.

I think what you are missing looking at the functional equation is that Gamma(z) has a pole at negative integers. For integer s>=2 the pole in Gamma(1-s) `cancels' the zero of the sin bit. More precisely, you can't just plug in values of s where one or more terms in the equation is not defined, but have to take some kind of limit, and as mentioned in the wikipedia article, as s approaches an integer >=2 the product of the Gamma and sin terms approaches something non-zero.

OK that explains that bit - thanks

I don't really get why in the original sum to infinity it requires re(s) > 1 but then this function is transformed into one which has a domain for all complex numbers....

If the sum to infinity definition is:

\zeta(s) = \sum_{n=1}^\infty n^{-s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + ... where Re(s) > 1

then I can see why Re(s) (with no imaginary part say) needs to be greater than 1 - otherwise the sum to infinity will diverge and will have no zeroes. And with a complex number shouldn't the domain also remain bounded (eg. abs(s) > 1 ? )

So what trick then allows for Re(s) to be less than 1? Wiki mentions analytic continuation - any easy-(ish!) explanation of how this works?

 

[ctamblyn]

Roughly speaking: (complex-)analytic functions are remarkably "rigid" in a certain precise sense, in that if you constrain their values at enough points (e.g. in an open set in the complex plane, or along an interval of the real line) it turns out that there is only one way of extending the function just beyond that set which preserves analyticity.

So, it happens that exactly one analytic function defined on the whole complex plane (excluding a pole singularity at z=1) equals the sum defining zeta for Re(s) > 1, and this is how zeta is defined for Re(s) <= 1 (again, excluding z=1).

 

[andyandy]

Roughly speaking: (complex-)analytic functions are remarkably "rigid" in a certain precise sense, in that if you constrain their values at enough points (e.g. in an open set in the complex plane, or along an interval of the real line) it turns out that there is only one way of extending the function just beyond that set which preserves analyticity.

So, it happens that exactly one analytic function defined on the whole complex plane (excluding a pole singularity at z=1) equals the sum defining zeta for Re(s) > 1, and this is how zeta is defined for Re(s) <= 1 (again, excluding z=1).

ok that kind of makes sense - but how can I then explain what the Riemann function is? Is it correct to say that:

\zeta(s) = \sum_{n=1}^\infty n^{-s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + ..... where Re(s) > 1

represents the zeta function?

If so, is there a way of then understanding how complex values for s are substituted into this equation? Do the values for s need to be transformed to result in a Re(s) > 1 ? If not, how can there be any zeroes?

And is there an alternative form of the zeta function that makes more intuitive sense?

 

[ctamblyn]

ok that kind of makes sense - but how can I then explain what the Riemann function is? Is it correct to say that:

\zeta(s) = \sum_{n=1}^\infty n^{-s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + ..... where Re(s) > 1

represents the zeta function?

If so, is there a way of then understanding how complex values for s are substituted into this equation? Do the values for s need to be transformed to result in a Re(s) > 1 ? If not, how can there be any zeroes?

And is there an alternative form of the zeta function that makes more intuitive sense?

You could say that the zeta function is the unique analytic function on C (excluding 1) equal to that sum for Re(s)>1, so it represents it in that sense.

You can't put values of s with Re(s)<=1 into that sum, because it diverges. It's rather like the way you can't put values of z with |z|>=1 in the series expansion

1/(1-z) = 1 + z + z² + ... + zⁿ + ...

even though the left hand side is defined outside the unit disc, and gives the unique analytic continuation of the right hand side.

If you look at the following section of the wiki article, they give some other forms of zeta which are defined for larger sets:

http://en.wikipedia.org/wiki/Riemann_zeta_function#Representations

E.g. the formula

ζ(s) = 2^(s-1)/(s - 1) - 2^s∫sin(s arctan t)/((1 + t²)^s/2(e^πt + 1)) dt

(where the integral is from 0 to infinity) is valid for all s.

 

[Farsight]

Seeing if I can copy the latex from wiki (not sure why that's not showing up in equation form - anyone know?

The latex here no longer works, andy, and the admin don't seem to know how to fix it. However you can use an image workaround which accesses John Forkosh's public latex rendering. Like so:

[img]http://www.forkosh.com/mimetex.cgi?$$\zeta(s) = \sum_{n=1}^\infty n^{-s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \;\;\;\;\;\;\; \sigma = \mathfrak{R}(s) > 1. \! $$[/img]

This gives you your latex:

 

[andyandy]

You could say that the zeta function is the unique analytic function on C (excluding 1) equal to that sum for Re(s)>1, so it represents it in that sense.

You can't put values of s with Re(s)<=1 into that sum, because it diverges. It's rather like the way you can't put values of z with |z|>=1 in the series expansion

1/(1-z) = 1 + z + z² + ... + zⁿ + ...

even though the left hand side is defined outside the unit disc, and gives the unique analytic continuation of the right hand side.

If you look at the following section of the wiki article, they give some other forms of zeta which are defined for larger sets:

http://en.wikipedia.org/wiki/Riemann_zeta_function#Representations

E.g. the formula

ζ(s) = 2^(s-1)/(s - 1) - 2^s∫sin(s arctan t)/((1 + t²)^s/2(e^πt + 1)) dt

(where the integral is from 0 to infinity) is valid for all s.

Thanks for your explanation....

Doing some googling, it appears that it can also be represented as:

((1-2^(1-s))^-1).sum(n= 1 to inf) (-1)^n / n^s for Re(s) > 0

http://people.math.gatech.edu/~ecroot/riemann_notes.pdf (page 4)

which might be slightly less daunting.... with this equation you could therefore sub in the actual complex solutions to the zeta function and get zeroes?

 

[andyandy]

The latex here no longer works, andy, and the admin don't seem to know how to fix it. However you can use an image workaround which accesses John Forkosh's public latex rendering. Like so:

[qimg]http://www.forkosh.com/mimetex.cgi?$$\zeta(s) = \sum_{n=1}^\infty n^{-s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \;\;\;\;\;\;\; \sigma = \mathfrak{R}(s) > 1. \! $$[/qimg]

This gives you your latex:

[qimg]http://www.forkosh.com/mimetex.cgi?$$\zeta(s) = \sum_{n=1}^\infty n^{-s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \;\;\;\;\;\;\; \sigma = \mathfrak{R}(s) > 1. \! $$[/qimg]

we lost Latex?! When did that happen?

Thanks for the link - I'll have a play around on that later....

 

[ctamblyn]

Thanks for your explanation....

Doing some googling, it appears that it can also be represented as:

((1-2^(1-s))^-1).sum(n= 1 to inf) (-1)^n / n^s for Re(s) > 0

http://people.math.gatech.edu/~ecroot/riemann_notes.pdf (page 4)

which might be slightly less daunting.... with this equation you could therefore sub in the actual complex solutions to the zeta function and get zeroes?

Sadly not, because that formula doesn't converge for Re(s) <= 0 (which is where all of the trivial zeroes are). If you look at the end of section 2.3 in that PDF, they give a proof that there are zeroes at -2, -4, ... by making use of the relation

π^-s/2Γ(s/2)ζ(s) = π^{-(1 - s)/2}Γ((1 - s)/2)ζ(1 - s),

together with the fact that the gamma function has singularities at 0, -1, -2, ... (so that Γ(s/2) is singular at 0, -2, -4 etc.).

This representation may give you what you're looking for:

ζ(s) = π^s/2(Π_ρ(1 - s/ρ)) / (2(s - 1)Γ(1 + s/2)),

where the product (Π_ρ...) is over the non-trivial zeroes of ζ (i.e. the ones with Re(ρ) = 1/2 ETA: if, of course, the RH is correct!). Again, the presence of the trivial zeroes follows from the singularities of the gamma function.

ETA2: Sorry, I didn't think properly. Although you couldn't use your formula to prove that zeta vanishes at the negative even integers, you could in principle use it to evaluate zeta at the non-trivial zeroes. I don't know how easy it would be to prove that zeta vanished exactly at those points, though.

 

[ctamblyn]

we lost Latex?! When did that happen?

Thanks for the link - I'll have a play around on that later....

There's a thread in the Help section:

http://www.internationalskeptics.com/forums/showthread.php?t=246584

 

[andyandy]

Sadly not, because that formula doesn't converge for Re(s) <= 0 (which is where all of the trivial zeroes are). If you look at the end of section 2.3 in that PDF, they give a proof that there are zeroes at -2, -4, ... by making use of the relation

π^-s/2Γ(s/2)ζ(s) = π^{-(1 - s)/2}Γ((1 - s)/2)ζ(1 - s),

together with the fact that the gamma function has singularities at 0, -1, -2, ... (so that Γ(s/2) is singular at 0, -2, -4 etc.).

This representation may give you what you're looking for:

ζ(s) = π^s/2(Π_ρ(1 - s/ρ)) / (2(s - 1)Γ(1 + s/2)),

where the product (Π_ρ...) is over the non-trivial zeroes of ζ (i.e. the ones with Re(ρ) = 1/2 ETA: if, of course, the RH is correct!). Again, the presence of the trivial zeroes follows from the singularities of the gamma function.

ETA2: Sorry, I didn't think properly. Although you couldn't use your formula to prove that zeta vanishes at the negative even integers, you could in principle use it to evaluate zeta at the non-trivial zeroes. I don't know how easy it would be to prove that zeta vanished exactly at those points, though.

Many thanks for your contributions to the thread - yes I didn't explain my post properly but you're right with the last paragraph - I want to be able to show a formula which in principle allows people to find the non-trivial roots of the form 1/2 + ai (or others if they do exist...!) I can brush over the trivial roots by reference to the sine bit in the functional equation.....

If I used the original equation for the infinite summation with Re(s) > 1 then that's going to cause too much confusion....

It doesn't matter if in practice that this particular version of the function wouldn't be the easiest to use in trying to actually find the roots - as the maths involved in doing that is way beyond anything the students (or I!) are going to be capable of. It's enough I think to be able to point to a function - which itself isn't too complicated - and say, "the Riemann Hypothesis says that all non-trivial zeroes of this function have real part 1/2...." - ie. an introduction to the problem not an introduction to how to start solving it

Anyone any thoughts on the importance of Riemann? If the Riemann hypothesis is already assumed to be true what impact would an actual proof actually have? And what impact would a proof that it is not correct have?

 

[ctamblyn]

Many thanks for your contributions to the thread - yes I didn't explain my post properly but you're right with the last paragraph - I want to be able to show a formula which in principle allows people to find the non-trivial roots of the form 1/2 + ai (or others if they do exist...!) I can brush over the trivial roots by reference to the sine bit in the functional equation.....

If I used the original equation for the infinite summation with Re(s) > 1 then that's going to cause too much confusion....

It doesn't matter if in practice that this particular version of the function wouldn't be the easiest to use in trying to actually find the roots - as the maths involved in doing that is way beyond anything the students (or I!) are going to be capable of. It's enough I think to be able to point to a function - which itself isn't too complicated - and say, "the Riemann Hypothesis says that all non-trivial zeroes of this function have real part 1/2...." - ie. an introduction to the problem not an introduction to how to start solving it

OK, then your formula from the PDF would be absolutely fine

ζ(s) = (1 - 2^1-s)^-1 Σ (-1)ⁿ/n^s __ (summed over n = 1, 2, 3, ...).​

It might (or might not) be interesting to touch on the notion of analytic continuation, depending on the level of the class, so you could explain why the "original" series can't be used to find the non-trivial zeroes. Just my 2 pennies worth.

Anyone any thoughts on the importance of Riemann? If the Riemann hypothesis is already assumed to be true what impact would an actual proof actually have? And what impact would a proof that it is not correct have?

I'd be interested to hear from some professional mathematicians on this (I'm certainly not one). John Baez wrote about it a few years ago in his "This Week's Finds in Mathematical Physics" series, perhaps you'd find it interesting:

http://math.ucr.edu/home/baez/week216.html

It links to a nice animation showing how zeta is related to the problem of counting the number of primes less than a given integer.