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## #1 Re: Help Me ! » Riemann hypothesis » 2006-08-01 08:23:02

My real name is Oliver Elkington.

## #2 Re: Help Me ! » Riemann hypothesis » 2006-08-01 01:46:45

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.

## #3 Re: Help Me ! » Riemann hypothesis » 2006-08-01 01:40:42

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%. (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.

## #4 Re: Help Me ! » Riemann hypothesis » 2006-07-31 06:03:02

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.

## #5 Re: Help Me ! » Riemann hypothesis » 2006-07-31 02:17:46

"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.

## #6 Re: Help Me ! » Riemann hypothesis » 2006-07-31 02:06:57

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.

## #7 Re: Help Me ! » Riemann hypothesis » 2006-07-30 19:54:13

Okay, I have two questions about the Riemann hypothesis.

The Riemann hypothesis states that...

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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.

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## #12 Re: Help Me ! » Riemann hypothesis » 2006-07-30 04:33:46

Heh, I could have proved that.

Really, I know what I'm talking about, but I can see why people would doubt me. Anyway I didn't come here to try and make people believe me, I just wanted information about the Riemann hypothesis.

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## #23 Help Me ! » Riemann hypothesis » 2006-07-28 05:40:19

se7en
Replies: 88

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basically I just asked for an explanation of the Riemann hypothesis

## #24 Jokes » I heard this joke earlier and had to share it » 2006-02-22 07:45:12

se7en
Replies: 38

An engineer, physicist and topologist are sent to three identical rooms
and presented with the same problem - they have to open a can without a
can opener.

The scientist in charge of the experiment gives them half an hour, and
then goes to check on them.

He visits the engineer first - when he walks in, he see the engineer
lying, exasted on the floor, the walls dented in several places and the
can split and scattered on the floor. He's opened the can by repeatedly
throwing it against the wall. Happy enough with the result, the scientist
moves on to the physicist.

When the scientist entered the physicist's room, he saw that he had made a
number of calculations on the wall with a marker, and that there was a
single, carefuly calculated dent in the wall, with the can open on the floor
beneath it. The scientist ticked off the room and moved onto the topologist.

However, when he entered the third room, the topoligist was nowhere to be
seen - the walls were covered with calculations, but the can was sitting
unopened in the middle of the room. The scientist looked around for a
while, and was about to leave until he heard the can shake about - so he
quickly opened it with his can opener, and the topologist climbed out.

"Thank goodness", the topologist exclaimed, as he gasped for air. "I got a
sign wrong!"

## #25 Puzzles and Games » a cool maths problem I saw in the latest issue of Hugi » 2006-02-21 10:12:42

se7en
Replies: 3

I have a really cool maths problem for you! Pick two random points on the arc of a circle. What's the probability that the straight line that connects these two points is longer than the radius of the circle?

I'm still working on this problem myself. I've been toying around with a lot of different ideas and I've learnt quite a bit about the circle in the process. One thing I tried:

upper semi-circle - lower semi-circle = radius (r)
sqrt(r^2 - x^2) + sqrt(r^2 - x^2) = r
solving for r we get: r = +- 2x/sqrt(3)

if we solve for x we can obtain the values of x for which the upper semi-circle - lower semi-circle = r

therefore, solving for x we get: x = +- r * sqrt(3) / 2

therefore, - r * sqrt(3) / 2 < x < r * sqrt(3) / 2 then sqrt(r^2 - x^2) - -sqrt(r^2 - x^2) > r

now, the number of chords in a circle > radius of the circle (provided the chords are all parallel to the y-axis) = r * sqrt(3) / 2 - -r * sqrt(3) / 2

let the number of chords = n

2 * r * sqrt(3) / 2 = n

r * sqrt(3) = n

that is to say, the number of chords in a circle > radius of the circle = r * sqrt(3) provided the chords are all parallel to the y-axis.

I thought that was really interesting.

Now, the probability of an event is defined as the number of outcomes in the event n(E) over the number of outcomes in the sample space n(S). I've managed to determine the number of outcomes in the sample space for this event... it's simply the circumference of the circle * the circumference of the circle, or 4 * pi^2 * r^2. The remaining piece of the puzzle is what n(E) is. I think the answer lies somewhere in the research I've done... I just have to determine the TOTAL number of chords in a circle > the radius of the circle (i.e. the chords are not limited to being parallel to the y-axis). Then you divide that by n(S) = 4 * pi^2 * r^2 and voila! I'm guessing that n(E) contains the expression r^2 somewhere, so the radius cancels out. (Otherwise P(E) - the probability of the event - will contain the variable r, which is obviously undesirable... we want just a number without any variables.)

I'm going to take an educated guess and say that the TOTAL number of chords in a circle > radius of the circle = r * sqrt(3) * 1/2 * circumference of the circle (the reason that we multiply by only HALF the circumference of the circle is that the second half of the circle contains exactly the same chords as the first half... in actual fact, we should multiply by 1/2 * (circumference of the circle - 1/infinity) because the chords at the highest point of the circle are the same as the chords on the lowest part of the circle, but 1/infinity tends to zero so leaving it out of our calculations is still accurate enough for all practical purposes) = r * sqrt(3) * pi * r = sqrt(3) * pi * r^2

If that is true, then

P(E) = n(E) / n(S)
= (sqrt(3) * pi * r^2) / (4 * pi^2 * r^2)
= sqrt(3) / (4 * pi)
= 1,732 / 12.566
= 0.138

But I don't know if this is right or not. I guess I could write a program to test this out.

All in all, quite an interesting (and frustrating) puzzle! Good luck with it! (This somewhat long rant will probably help get you started out, although I wouldn't be surprised if your method of approach differs entirely from mine. Let me know what you come up with!)