# A *simple* question on zeta

1. Mar 10, 2005

### franic

Hallo
Can one use ***simple*** words to intuitively explain what does it mean that the (analytic continuation of the) Riemann zeta function yields

zeta(0) = 1+1+1+1+1+.... = -1/2 ?
and
zeta(-1) = 1+2+3+4+5+.... = -1/12 ???

Thank you!
Francesco

2. Mar 10, 2005

### matt grime

Well, firstly the sums you've written down aren't zeta of 0 or -1

The zeta function has a power series that is valid only for certain inputs.

Every power series has a radius of convergence, right?

Well, in some cases one can extend functions onto other patches so that on the overlap of two patches the functions agree.

For instance

x - x/2 + x/3 - x/4 +...

is the expansion of log(1+x) about x=0, and os only valid for |x|<1

Obviosuly one may extend this to a function, log(1+x) for all x not equal to -1.

The extension to other patches will have different power series expansions inside those patches. The expansion you've used for zeta is not valid in the area in which you're evaluating zeta, you'd have to use a different power series there, ie zeta(0) ISN'T 1+1+1......

I can't really think of anything else to say about it

3. Mar 10, 2005

### shmoe

You've presumable seen that $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$, valid when the real part of s is greater than 1. If you try to put in something with real part less than 1, this series diverges, I'm sure you've seen the harmonic series, this is just $$\zeta(1)$$.

Using some analytical trickery (which you may or may not find simple or intuitive), it turns out it's possible to find a function f(s) that is valid for any complex number s (except a "pole" at s=1) and has the property that $$f(s)=\zeta(s)$$ whenever the real part of s is greater than 1. Since this function f agrees with Zeta where Zeta is defined by it's Dirichlet series (that infinite sum above), we're going to use this f to define Zeta for all complex values. This what's meant by analytic continuation, and it turns out there's only one analytic continuation of Zeta to the entire complex plane.

So, using this f(s) it's possible to any value of Zeta we like, such as $$\zeta(0)\equiv f(0)=-1/2$$, $$\zeta(-1)\equiv f(-1)=-1/12$$, $$\zeta(-2)\equiv f(-2)=0$$ (I'm using $$\equiv$$ to mean "defined as"). If you were to blindly substitute these s values into the Dirichlet series and ignore matters of convergence, you get the equations you've written.

Don't read much into this. For a simple analogy, consider the function g(x)=x/x. This is undefined at x=0 and the constant 1 everywhere else. Set f(x)=1. In a similar sense to our analytic continutation of Zeta above, f(x) is a continuation of g(x) to the entire real line, so we can now define some reasonable value of g at x=0, namely $$g(0)\equiv f(0)=1$$. Same as above, if we ignore the initial defects our formula for g(x) had at 1, we get the equation 0/0=g(0)=1. This is a bit of nonsense of course, if you've taken any calculus you should know the dangers of 0/0.

Last edited: Mar 10, 2005
4. Mar 11, 2005

### franic

Thanks a lot and CIAO from Italy!
Francesco

5. Mar 21, 2005

### guitry

riemann h

As far as I can tell Riemann h states that he conjectures that his formula determines the EXACT quantities of primes up to a given whole number .Does this mean that if a proof is found, that a prime generator will be uncovered along with the proof?

6. Mar 21, 2005

### shmoe

I think you've got something confused here. The Riemann hypothesis is equivalent to saying the folowing is true for every $$\epsilon>0$$:

$$\pi(x)=Li(x)+O(x^{1/2+\epsilon})$$

Where Li(x) is the usual logarithmic integral and $$\pi(x)$$ is the usual prime counting function. Notice the big-O error term, that can keep you a long way off from using this to find the exact number of primes up to a given x.

However it's not clear to me that this is what you mean by "his formula". If you mean Riemann's so-called explicit formula for $$\pi(x)$$, then this creature already expresses $$\pi(x)$$ exactly as an infinite sum over the zeros of zeta without any additional assumptions. It's just not an easy thing to work with due to this infinite sum. The more we know about the location of the zeros (such as if the Riemann hypothesis is true) the better capable we are of dealing with this nasty sum that's involved, and we can get a better error term in the prime number theorem (such as the form I've stated above).

It will probably take some very deep and high powered (probably currently unknown) tools to solve, so who knows what other consequences will follow? However I doubt it will produce the kind of prime generator you might be looking for.