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#51 2006-07-31 13:01:26

Zhylliolom
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Re: Riemann hypothesis

Ricky laid it out on the first page, but I'll write it out for you (I like to use LaTeX):



From this you can say that the non-trivial zeros of ζ(s) lie on the "critical line" σ = ½ + it. ζ(s) has zeros at all negative even integers, so the zeros s = -2, -4, -6 are referred to as the trivial zeros of ζ(s).

I'm not sure what you know about the zeta function so I'll write a few things extra for you.

For integral n,



For real x,



For complex z,



One last note:

Γ(x) is the gamma function, defined by



Γ(n) is related to the factorial for integral n by



and may be considered a continuous extension of the factorial.

Last edited by Zhylliolom (2006-07-31 18:14:52)

#52 2006-07-31 13:14:57

mikau
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Re: Riemann hypothesis

this is a hypthosis, so it hasn't been proven?


A logarithm is just a misspelled algorithm.

#53 2006-07-31 13:15:26

Zhylliolom
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Re: Riemann hypothesis

Correct, it remains to be proved.

#54 2006-07-31 17:54:13

se7en
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Re: Riemann hypothesis

Okay, I have two questions about the Riemann hypothesis.

The Riemann hypothesis states that...

-----

Let a be a complex number such that a is not a negative even number.

If zeta(a) = 0, then Re(a) = 1/2, or rather, the real part of a is 1/2.

-----

But how can a equal a negative even number? A negative even number is an INTEGER, not a complex number.

The only way I can see that a could be a negative even number is if we have the product of two conjugate complex numbers... but if a = 89 say, what is the real part of a? a would equal (5 + j8)(5 - j8)... there are two real parts.

Here's a guess...

(5 + j8)(5 - j8)
= 25 + j40 - j40 - j^2(64)
= 25 + j^2(64)

So does the real part equal 25 and the imaginary part equal j(64)?

But 89 also equals 90 + j^2

so we could also say that the real part of 89 is 90 and the imaginary part is j.

I guess we could also say that the real part of (5 + j8)(5 - j8) is 5, because both real parts are 5.

So we have three totally different answers that all sound right heh.

I suppose for my proof I could sort of just brute force it out... say that if x = a + jb and the real part of x is 1/2 (a = 1/2) then zeta(x) = 0... even if we are talking about two or more totally different numbers that have real part 1/2. Yeah... that could work.

Okay, now my other question...

The generalized Riemann hypothesis conjectures that neither the Riemann zeta function nor any Dirichlet L-series has a zero with real part larger than 1/2.

But does this mean that for every x such that x is the real part of some imaginary number a, if x is less than 1/2 then there is DEFINITELY a zero? In other words, is it possible, for example, for there to be a zero when the real part equals 1/4 but there's no zero when the real part equals 1/8?

"when do you think your algorithm will be sufficiently used by you so you'd feel comfortable publishing some proofs and thus releasing it to the mathematical community?" - Zhylliolom

I'll go public when I've proved Riemann and I've written out neat copies of my proofs. There are also some programs I want to write first. Basically, I want my thesis to be finished and polished before I go public. Thanks for the information btw.

Last edited by se7en (2006-07-31 17:59:26)

#55 2006-07-31 18:01:45

luca-deltodesco
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Re: Riemann hypothesis

se7en wrote:

But how can a equal a negative even number? A negative even number is an INTEGER, not a complex number.

lol. so your saying that integers dont form part of the complex number set?


The Beginning Of All Things To End.
The End Of All Things To Come.

#56 2006-07-31 18:24:50

MathsIsFun
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Re: Riemann hypothesis

The complex numbers are like cartesian coordinates if you imagine (ha!) that y is imaginary.

So the reals are along the "x" axis (y=0) and the complex numbers are the whole plane (including the reals).

The value "-2" could be thought of as -2+0i


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#57 2006-07-31 18:30:22

Zhylliolom
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Re: Riemann hypothesis

se7en, (Edit: Ricky didn't leave it out in his post, I just didn't see it. And yes, any real number a is a complex number with Im(a) = 0. By the way, I just want to be friendly and correct another statement of yours, to save you some confusion: you said 89 = 90 + i², which is true, but no, you cannot say Re(89) = 90 and Im(89) = i, because by definition a complex number is expressed in the form z = a + bi, where a and b are real numbers, and Re(z) = a and Im(z) = b. Since a and b are real you cannot let b = i.  This, along with the previous statement about how Re(z) and Im(z) must be real numbers, explains why Re(89) ≠ 25 and Im(89) ≠ 64i in your other problem (even though following your logic Im(89) should have been -64i, I assume it was just a simple mistake on your part with forgetting the subtraction though)) (refer to my post at the top of this page if you wish for another statement of the hypothesis). The thing is, it is conjectured that Re(s) = 1/2 for all non-trivial zeros s. As my post explained, ζ(s) = 0 for every negative even integer, and thus those zeros are considered trival.

In regard to the generalized Riemann hypothesis, no, it does not say that ζ(s) = 0 (or L(χ, s) = 0, since you mentioned the Dirichlet L-series and this series is really what makes this different than the plain old Riemann hypothesis) if Re(s) ≤ 1/2. What it is saying is that if ζ(s) = 0 or L(χ, s) = 0, then Re(s) ≤ 1/2 (basically it's the other way around from what you said). Proof of the Riemann hypothesis would help this hypothesis as the mathematical world knows that the trivial zeros of ζ(s) have the property Re(s) ≤ 1/2 since they are all negative even integers, and if one proves that all other zeros of ζ(s) (the non-trivial zeros) have the property Re(s) = 1/2, then all the bases are covered, we would be certain that Re(s) is either 1/2 or a negative even integer if ζ(s) = 0. Now L-functions are very similar to Riemann's zeta function (for example, L(χ, s) = ∑χ(n)/ns, and for χ(n) = 1, L(χ, s) = ζ(s)), and thus properties may easily be compared between the two, but this is only a fraction of the power of the Riemann hypothesis.

So, using the previous paragraph this question can be answered:

se7en wrote:

...is it possible, for example, for there to be a zero when the real part equals 1/4 but there's no zero when the real part equals 1/8?

No, the conjecture says that it is only possible for a zero s to have the property Re(s) = 1/2 or s = -2, -4, -6... It cannot be 1/4, or 1/8 or anything other than a complex number with real part 1/2 or a negative even integer.

Last edited by Zhylliolom (2006-07-31 20:46:22)

#58 2006-07-31 22:57:39

John
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Re: Riemann hypothesis

Okay, I don't really understand it so I'm probably wrong but what se7en claims is that he can prove any true statement but didn't Gödel prove that there are true statements that cannot be proven hence disproving se7en's claim.

Sorry if what I just said was completely wrong but like I said I don't really understand Gödel's Incompleteness Theorem.

#59 2006-07-31 23:56:19

ben
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Re: Riemann hypothesis

se7en wrote:

I'll go public when I've proved Riemann and I've written out neat copies of my proofs. There are also some programs I want to write first. Basically, I want my thesis to be finished and polished before I go public. Thanks for the information btw.

Ha! You are in a worse hole than I had at first thought. Not only do you not understand the theorem you are trying to prove, you seem to know squiddly-dit about the set it is defined over! Pretty well everything you did invovling complex numbers was wrong, as a glance at any grade school text would show you.

How do you expect to be taken seriously?

#60 2006-08-01 00:06:57

se7en
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Re: Riemann hypothesis

I just read about complex numbers in an engineering text book this morning... so I'm anything but an expert on them...

And complex numbers is university level stuff, at least here in South Africa. I don't think I'd find anything about them in a grade school book.

Last edited by se7en (2006-08-01 00:08:52)

#61 2006-08-01 00:17:46

se7en
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Re: Riemann hypothesis

"Okay, I don't really understand it so I'm probably wrong but what se7en claims is that he can prove any true statement but didn't Gödel prove that there are true statements that cannot be proven hence disproving se7en's claim.

Sorry if what I just said was completely wrong but like I said I don't really understand Gödel's Incompleteness Theorem." - John

That's actually very interesting, I'll have to think about that.

#62 2006-08-01 03:49:23

ben
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Re: Riemann hypothesis

se7en wrote:

I just read about complex numbers in an engineering text book this morning... so I'm anything but an expert on them...

Sorry, I don't mean to be rude, but you do need to be expert on the complexes to even get a hint at what the Riemann hypothesis is all about.

I'm not going to pursue this, because I detest number theory, but maybe this will help you.

Years and years ago, mathematicians stumbled upon a neat trick. It was well known in the 17th centuary that you cannot define the square root of a negative number. But these guys found that if you "pretended" such a thing exists, held you breath and plodded on with your calculation regardless, everything came out as expected in the end.

This led to the introduction of the imaginary unit i, that
being defined as follows: it is always the case that i² = -1.
(Therefore, equivalently i can be defined as √-1, but it's best not to start there)

Now a complex number is simply a number which has two parts, real and imaginary (I assume we all know what a real number is). It is written thus: z = x + iy, where x and y are real numbers.

Perhaps confusingly, x is called the real part Re(z) and y the imaginary part Im(z). For the most part, compex numbers can be manipulated just like real numbers, as long as we remember to replace with -1 wherever it appears. However, division requires some work, but it is well worth it, as it throws us a very useful gadget. Let me explain.

First, take it from me that multiplication works as follows:


as is easily checked.
It is less easy to derive the following:
.
This suggests that
.
Now c² + d² = (c + id)(c - id), so that, for
any z = x + iy, one says there is a complex conjugate z* = x
- iy
such that zz* = x² + y²
So, now we have the complex conjugate in our possession, we can do
this. Note that z = x + iy can be written z = Re(z) + Im(z),
which suggests we can think of z as a point on a plane with coordinates
Re(z) and Im(z). Moreover, the existence of the complex
conjugate allows us to think of this plane as a vector space with a
notion of distance between points - a so-called metric space. Why is
this?

First, note that in general, for two complex numbers z ≠ w we
will have zw ≠ wz, as is easily shown. But it is the case that
(zw) = (wz)*. Translated into the language of vector spaces we
will say that the inner product (w·z) = (z·w)*
Now recall that in a real metric space, Pythagoras says that the
distance between two points a and b is given by √(a² +
b²). This implies that the "length" of a one-dimensional vector x is found by
√|x²|. But suppose we had no complex conjugate, then on the
complex plane we would have √(ix²) = √-x², an imaginary
length! But if we may have a x* for each x, we may have
√(x·x*) as the "length" of this vector in the complex plane.

I don't know why I made that detour, just general education, I guess. But as has been pointed out to you, unless z is an even,  negative real, the zeros of Riemann's zeta function occur (by hypothesis) when Re(z) = ½. That's all I know, or care about the subject

I've run out of puff for now, let me know if you need more

Last edited by ben (2006-08-01 06:14:24)

#63 2006-08-01 04:03:02

se7en
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Re: Riemann hypothesis

To the administrators of this forum...

I would appreciate it if you would delete this whole topic. I have already deleted most of the content of my posts (I didn't select the 'Delete' option because I think it's nifty that there are so many posts in a topic I started heh, even so I want this topic to be deleted... so I just edited my posts, as you can see, and I think this way it will also prevent any confusion)... I'm just not comfortable with too many people even knowing what I'm working on. I'll come back here when my thesis is finished (I'll open up a site for it when it's finished with a link to a download and maybe also a forum) but until then I would prefer that this topic is gone.

Thanks to everyone who has posted information about the Riemann hypothesis and complex numbers and what all.

Btw, I didn't come here to try and impress anyone or convince anyone that my thesis works. All I wanted was information on the Riemann hypothesis heh.

Oh, and Ricky, no... I don't even really need to understand a problem in order to solve it. So I could probably prove the Poincaré conjecture even so I don't really understand it at all. You'll understand why when you read my thesis. (I've written a whole section on this today because of you! Heh.)

Thanks for the helpful explanation of complex numbers, ben. I find it interesting that they use j instead of i in engineering text books. smile

#64 2006-08-01 07:46:52

MathsIsFun
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Re: Riemann hypothesis

se7en wrote:

To the administrators of this forum...

I would appreciate it if you would delete this whole topic.

Are you sure? That would be a shame - it was an interesting discussion - I was reading it every day.

But other people have posted here, and it would be unfair to just have their thoughts deleted unless they agree ...


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#65 2006-08-01 07:53:01

ben
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Re: Riemann hypothesis

MathsIsFun wrote:

But other people have posted here, and it would be unfair to just have their thoughts deleted unless they agree ...

I don't agree, MIF

-b-

#66 2006-08-01 07:55:03

MathsIsFun
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Re: Riemann hypothesis

John wrote:

...but didn't Gödel prove that there are true statements that cannot be proven...

For "formal theories" or some such I believe.

My memories of it are a bit hazy, but I read once about computer logic vs human logic ... something about formal systems (which can be computer driven) going into infinite loops whereas a human wouldn't (after a few rounds of the loop a human goes "aha - a loop"). But after reading it I remember thinking that the computer just needed to monitor its own progress to undo the trap. But that goes beyond Gödel's definitions I think.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#67 2006-08-01 10:44:02

Ricky
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Re: Riemann hypothesis

I'm with Ben.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

#68 2006-08-01 12:01:04

George,Y
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Re: Riemann hypothesis

Your request is beyond your rights.


X'(y-Xβ)=0

#69 2006-08-01 14:50:13

Zhylliolom
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Re: Riemann hypothesis

Sad that this thread seems destined to die now due to se7en's choice. I would prefer that this thread is not deleted, I(as well as some others, I'm sure) put some work into my posts and I'd hate to see them thrown away. se7en has already deleted the majority of his "revealing material" anyway. I don't think people who stop by to read now will understand exactly what happened here, and it's not like you can go into our brains and erase our memory of this thread, so I think what's left of it should stay.

Last edited by Zhylliolom (2006-08-01 14:53:53)

#70 2006-08-01 23:40:42

se7en
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Re: Riemann hypothesis

Yeah, I'm okay with this thread staying, especially because of the helpful information that some people have posted about Riemann and complex numbers and stuff, but I feel better now that the revealing material has been deleted. I'm sort of paranoid about someone else stumbling upon my idea and publishing a paper on it before I do.

Anyway, I got my results back for a maths test today. 92%. smile (I actually thought I'd get 100% but I guess 92% is okay.) My teacher actually made a mistake when marking my test... when you solve a quadratic inequality by means of a parabola you only need the x-intercepts and the sign of a to dermine whether the graph is concave up or down, but she went and wrote in the y-intercept and vertex, as if it was needed heh.

#71 2006-08-01 23:46:45

se7en
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Re: Riemann hypothesis

In an effort to keep this monster thread alive a while longer, and because I find the subject interesting, I have a question about Gödel's Incompleteness Theorem.

Can someone give me an example of a true statement that cannot be proven? I want to test this with my algorithm.

#72 2006-08-01 23:59:41

George,Y
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Re: Riemann hypothesis

calculate 1/2+1/4+1/8+... till your program gets 1

I bet it will fail to do so unless it stops with a round-off error.

Or to put it another way, you cannot get it done befor the world ends. big_smilebig_smile


X'(y-Xβ)=0

#73 2006-08-02 00:15:15

Patrick
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Re: Riemann hypothesis

You couldn't even do it if you continued beyond the end of the world :] What's your name btw se7en? And no I'm not going to travel to south africa and steal your documents, I just want to be able to recognize your name when it appears in all the papers.


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#74 2006-08-02 00:26:55

Ricky
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Re: Riemann hypothesis

Patrick, if anything like se7en is proposing comes out, then we can be certain that it was him.  Trust me when I say that no one else would come up with ideas like his that could solve something such as the Goldbach conjecture.

calculate 1/2+1/4+1/8+... till your program gets 1

You never resist from posting that, do you?  tongue


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

#75 2006-08-02 06:23:02

se7en
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Re: Riemann hypothesis

My real name is Oliver Elkington.

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