Riemann's Hypothesis

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• Dec 22nd 2009, 03:53 PM
Vitruvian
Riemann's Hypothesis
Today I was speaking to a friend of mine, and he brought up something called Riemann's Hypothesis. He told me that it was a concept relating to the distribution of primes and has yet to be proven, and is evidently somehow related to modern financial security systems.

However, upon researching the topic I've found very little in the area of practical application of the Hypothesis... and add to the fact I find the hypothesis confusing to understand.

Would anyone care to explain the hypothesis to me in layman's terms, and what it means holistically speaking?
• Dec 22nd 2009, 04:25 PM
wonderboy1953
A little help
This involves the zeta function
• Dec 22nd 2009, 04:38 PM
VonNemo19
The simply is no "lay" explaination for this particular hypothesis. The very nature of the conjecture requires that you know someting of several different areas of advanced mathematics. I'm not too bad at mathematics, and I've tried to interpret the meaning with no success.

Good luck.
• Dec 22nd 2009, 04:41 PM
Vitruvian
Quote:

Originally Posted by wonderboy1953
This involves the zeta function

This I have read and am trying to do more research to understand it.

Quote:

Originally Posted by VonNemo19
The simply is no "lay" explaination for this particular hypothesis. The very anture of the conjecture requires that you know someting of several different areas of advanced mathematics. I'm not too bad at mathematics, and I've tried to interpret the meaning with no success.

Good luck.

You have confirmed my suspicions. I suppose I will lay this particular problem to rest.

Thank you both for your replies.
• Dec 22nd 2009, 08:18 PM
chiph588@
The Riemann Zeta function is defined by $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$

Note that $\zeta(s)$ is defined only for $s$ with $\Re(s) > 1$.

One can extend $\zeta(s)$ by noticing $\zeta(s) = \frac{1}{2^{1-s}-1}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}$ (this can be shown fairly easily). This is called an extension because now $\zeta(s)$ is defined for $\Re(s)>0$ (except for $s=1$ where there is a simple pole).

The Riemann Hypothesis states that the only zeros of $\zeta(s)$ in the strip $0<\Re(s)<1$ have real part equal to $\frac{1}{2}$ i.e. $s=\frac{1}{2}+it$.
• Dec 22nd 2009, 08:22 PM
VonNemo19
Quote:

Originally Posted by chiph588@
The Riemann Zeta function is defined by $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$

Note that $\zeta(s)$ is defined only for $s$ with $\Re(s) > 1$. <---What do you mean by this?

One can extend $\zeta(s)$ by noticing $\zeta(s) = \frac{1}{2^{1-s}-1}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}$ (this can be shown fairly easily). This is called an extension because now $\zeta(s)$ is defined for $\Re(s)>0$ (except for $s=1$ where there is a simple pole).

The Riemann Hypothesis states that the only zeros of $\zeta(s)$ in the strip $0<\Re(s)<1$ have real part equal to $\frac{1}{2}$ i.e. $s=\frac{1}{2}+it$.

.
• Dec 22nd 2009, 08:26 PM
chiph588@
Let $s=\sigma+it$, $\Re(s) = \sigma$.

$\Re(s)$ is the real part of $s$.
• Dec 22nd 2009, 08:34 PM
VonNemo19
Quote:

Originally Posted by chiph588@
Let $s=\sigma+it$, $\Re(s) = \sigma$.

$\Re(s)$ is the real part of $s$.

OK, so the sum in quetion parallels the p-series. So, can you provide an argument that

$\sum_{n=1}^{\infty}\frac{1}{n^s}$ converges if the real part of $s>1$?

I'm sorry if I sound silly right now, it's just that I'd really lke to understand the hypothesis.
• Dec 22nd 2009, 09:08 PM
chiph588@
I'd be happy to explain.

Let $s=\sigma+it$.

We want to know when $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$ converges. For this we'll need a couple facts about complex numbers.

1.) $\left|e^{x+iy}\right| = \left|e^x\right|\cdot\left|e^{iy}\right| = e^x\cdot\left|\cos(y)+i\sin(y)\right| = e^x\cdot (\cos(y)^2+\sin(y)^2) = e^x$

2.) $n^s = e^{\log(n^s)} = e^{s\log(n)} = e^{(\sigma+it)\log(n)} = e^{\sigma\log(n)+it\log(n)}$.
Hence $\left|n^s\right| = n^{\sigma}$.

So $\sum_{n=1}^{\infty} \left|\frac{1}{n^s}\right| = \sum_{n=1}^{\infty} \frac{1}{n^\sigma}$. Which is convergent for $\sigma >1$ by the p-test.

Thus $\zeta(s)$ is absolutely convergent for $\Re(s) > 1$.
• Dec 23rd 2009, 08:03 AM
VonNemo19
Quote:

Originally Posted by chiph588@
I'd be happy to explain.

Let $s=\sigma+it$.

We want to know when $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$ converges. For this we'll need a couple facts about complex numbers.

1.) $\left|e^{x+iy}\right| = \left|e^x\right|\cdot\left|e^{iy}\right| = e^x\cdot\left|\cos(y)+i\sin(y)\right| = e^x\cdot (\cos(y)^2+\sin(y)^2) = e^x$

2.) $n^s = e^{\log(n^s)} = e^{s\log(n)} = e^{(\sigma+it)\log(n)} = e^{\sigma\log(n)+it\log(n)}$.
Hence $\left|n^s\right| = n^{\sigma}$.

So $\sum_{n=1}^{\infty} \left|\frac{1}{n^s}\right| = \sum_{n=1}^{\infty} \frac{1}{n^\sigma}$. Which is convergent for $\sigma >1$ by the p-test.

Thus $\zeta(s)$ is absolutely convergent for $\Re(s) > 1$.

Please don't laugh, but... How is it the case that

$|e^{iy}|=|\sin{y}+i\cos{y}|$ ?

And moreover,

$|\sin{y}+i\cos{y}|=\sin^2{y}+\cos^2{y}$ ?

I'm good with the rest of the argument.
• Dec 23rd 2009, 08:21 AM
shawsend
Hi. I think there is a layman's way of explaining it. Start with:

When is $x+3=0$

Alright, you're insulted right now. How then about: $z+i=0$. Same dif I know. Well, we can also consider integral functions like:

$f(x)=\int_0^x 6t+2 dt$

Then when is that one zero? Well, piece of cake. Integrate it, it's a quadratic in x, then find the roots via the quadratic formula. Well, then, that's what the Riemann Hypothesis is about except the integral function is more complicated and it deals with complex numbers:

$\zeta(s)=\Gamma(1-s)\frac{1}{2\pi i} \mathop\int\limits_{H^-}\frac{z^{s-1}}{e^{-z}-1}dz$

Now, for which values of $s$ is that one zero? The Riemann Hypothesis claims that in the half-plane, $\text{Re}(s)>0$, all of them have the form $s=1/2+it$
• Dec 23rd 2009, 08:33 AM
tonio
Quote:

Originally Posted by VonNemo19
Please don't laugh, but... How is it the case that

$|e^{iy}|=|\sin{y}+i\cos{y}|$ ?

And moreover,

$|\sin{y}+i\cos{y}|=\sin^2{y}+\cos^2{y}$ ?

I'm good with the rest of the argument.

It's the definition of modulus of complex number: if $z=x+iy\in\mathbb{C}$, then $|z|=|x+iy|=\sqrt{x^2+y^2}$

Since, by definition, $e^{iy}=\cos y+i\sin y\,,\,\,y\in\mathbb{R}$ , we get the above.

Tonio

Addition: of course, it should be $|\cos y+i\sin y|=\sqrt{\cos^2y+\sin^2y}=\sqrt{1}=1$
• Dec 23rd 2009, 08:55 AM
wonderboy1953
Further comment
"I've found very little in the area of practical application of the Hypothesis..."

A simple check by Googling would confirm that (my attitude is that if it isn't on the internet, it doesn't exist). When you say "practical application", do you mean within math or outside of it?

"...and is evidently somehow related to modern financial security systems."

That would be because of the primes.

• Dec 23rd 2009, 09:01 AM
VonNemo19
Ok. I feel good about that, but

http://www.mathhelpforum.com/math-he...82031c40-1.gif (which can be shown fairly easily)

Can you run me through this? Starting with

$\sum_{n=1}^{\infty}\frac{1}{n^s}$
• Dec 23rd 2009, 09:27 AM
chiph588@
$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s} = \sum_{n=1}^{\infty} \frac{1}{(2n)^s}-\sum_{n=0}^{\infty} \frac{1}{(2n+1)^s}$

$=\sum_{n=1}^{\infty} \frac{1}{(2n)^s}+\sum_{n=1}^{\infty} \frac{1}{(2n)^s}-\sum_{n=1}^{\infty} \frac{1}{(2n)^s}-\sum_{n=0}^{\infty} \frac{1}{(2n+1)^s}$

$= 2\sum_{n=1}^{\infty} \frac{1}{(2n)^s} - \zeta(s)$

$= \frac{2}{2^s}\sum_{n=1}^{\infty} \frac{1}{n^s}-\zeta(s)$

$= (2^{1-s}-1)\zeta(s)$

Therefore $\zeta(s) = \frac{1}{2^{1-s}-1}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}$.
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