Math Is Fun Forum

Wow, the Euclidean algorithm sped this up exponentially. It works a lot better now, it instantly gives the answer.

More clever than Krassi? This may be dangerous.

Edit: Congratulations on post 1600, Krassi!

I can't get it any better on this monitor . I wonder if there is a way to make LaTeX work for everyone's point of view. Equations will wrap, why doesn't \mbox{}?

Ricky, your LaTeX goes off the page, you might want to make it easier to read. Otherwise, everything looks good.

Welcome! Is there a certain type of mathematics that you like to do best?

Yeah, it's Poincaré. I like your last question too. And your new avatar, have you been to that place in person to take the photo or did you get it off of the internet?

You know guys, it's not a lie if he thinks it's true. Of course, this depends on your definition of the term. In my view, a lie is telling false information in order to deceive someone. What if the guy is blind and has never heard of the sky being blue? Ok, I'm just making life hard, of course obvious questions(such as the Poincaré conjecture) should satisfy this riddle.

I think I'm too tired to understand anything right now(it's 6:30 AM now, I've been kept up by an intense theorem). Sorry. There are a few members on this forum who are much more accomplished programmers than I'll ever be, and I think that they will be better suited to help you out. Maybe after I have rested my mind will be working better and I'll be a better help. Until then, good luck.

Oh, the problem is that we are considering two entirely different things. The Hermite polynomials I was talking about are the kind from differential equations.

I've never heard of this "spline technology", but maybe I can try to help still. I have questions though. What do the start/end points and the start/end tangents do for us? Is this something that creates a curve defined by one polynomial between two points, then another curve defined by another polynomial between the next points, and so on?

So you're good? No further assistance is required as of now?

What are you trying to make this for? What is the end product supposed to be?

Great news, folks. After a haunting dream and a good portion of the evening spent sitting furiously with my notebook and pencil, I have found a simple equation which can be used to calculate ζ(2n). No need for Fourier series or wacky factoring approaches, this formula works for all positive integers n. After finding this equation, an immediate consequence was noted: it could be used to find ζ(-n) as well, and also I will show how this miraculous formula can be used to prove a fundamental property of the zeros of the Riemann zeta function(no skipping ahead to see this part, let the suspense build).

Let's start by defining Bernoulli numbers. The Bernoulli numbers are the coefficients B[sub]n[/sub] of this series expansion:

There isn't really a simpler way to define them, aside from a definition using a contour integral. The first few Bernoulli numbers are

From those few values you can probably notice one thing: for integers n > 0, B[sub]2n + 1[/sub] = 0, or in other words, odd values of n greater than 1 for B[sub]n[/sub] give the value zero. This concept will be important much later in this exposition, so try to remember it. Of course, you will be reminded again anyway.

Now

isn't really in a form we can play with much for our present situation. Let's convert it into something we know more about. We can add x/2 to it and get something decent:

So to summarize,

Now we try to work something out for coth using our knowledge of B[sub]n[/sub]'s generating function:

Now we already established that for all odd positive integers n > 1, B[sub]n[/sub] = 0. Thus we can make the substitution n = 2k in the above sum given that we make up for the loss of the term corresponding to n = 1. A quick calculation shows that the term we need is equal to -x/2, so we must add this in for our substitution n = 2k to work:

Alright, great. Now we would prefer something more familiar than the hyperbolic cotangent to work with. Let's use the substitution x = 2iz to get our standard circular cotangent(note that coth ix = -i cot x):

Beautiful series there. But this is no time for sightseeing, we have an equation to derive. Let us consider the known identity(whose proof is left to the reader as an exercise )

Let's clean up the sum a little to make it more workable. Start by making the substitution x = z/π and multiplying through by it:

Now let's push our manipulation of this even farther:

An observant reader may notice that our summand is actually the value of the geometric series

Because of this fact, we will substitute this sum in for the summand and upon manipulation make a marvelous discovery:

So, after all that work, we now have two sums for z cot z, and we may write

Let's kick z cot z out of the picture now, as we have used him as much as we need to:

Now we want to get rid of the summations. It turns out that such a task is easy. We want the limits of summation to be equal, so let's get the lower limit on the left sum to be k = 1. We can do this by simply writing the lower limit as k = 1 and adding the value corresponding to k = 0 to the left side. The corresponding value is 1, so we now write

Now just subtract 1 from each side:

The summands must be equal now:

Solve for ζ(2k) to get

I prefer the expression with the imaginary unit, but some may be more comfortable rewriting it as

So there it is, a nice little formula for ζ(2n). But that was only part of what will be discussed in this post. Let's take things a step farther and find a formula for ζ(-n) as well! To do this we'll use our newly discovered formula for ζ(2n) and the functional equation for the zeta function,

Make the substitution s = 2n:

Use our formula for ζ(2n) and simplify:

Letting k = 2n we get the form

Now substituting n = k - 1, we get

Now we have the power to evaluate the zeta function for negative integers! Here's a few values:

We can also plug in n = 0 to get

Now, as promised, I will prove a fundamental property of the zeros of the Riemann zeta function. The Riemann hypothesis states that for s ≠ -2, -4, -6... such that ζ(s) = 0, Re(s) = 1/2. Let's use our nice equations to figure out why the hypothesis is stated the way it is, and to prove something about ζ(s) = 0.

The statement of the Riemann hypothesis disallows values of s that are negative even numbers. Why is this? Using the derived equation for ζ(-n), we can see that for negative even integers, the Riemann zeta function always seems to have a zero. These zeros must be kept out of the hypothesis because they obviously have real part not equal to 1/2. But one may ask this question about ζ(-2n): "Is it always zero?" The answer is yes, and I shall prove it for you.

From our formula for ζ(-n), we can write ζ(-2n) as

A long while ago, I told you that for all integers n > 0, B[sub]2n + 1[/sub] = 0. So if that is true, ζ(-2n)'s denominator will always be zero and thus ζ(-2n) will always be zero. So our task now is to prove that B[sub]2n + 1[/sub] = 0 for all integers n > 0.

Recall the expression

from earlier. We know that B[sub]1[/sub] is an odd value of n in B[sub]n[/sub] which does not equal zero, so we take its term out of the series and write a little note by our sigma:

Now give x a negative value in the equation and note that coth(-x) = -coth x:

It turns out that x coth x is an even function, and since it is equivalent to the sum, the sum must also be an even function. Thus we require that (-1)[sup]n[/sup]B[sub]n[/sub] = B[sub]n[/sub], which only holds for even n, or in another way of stating the situation, all odd powers of x in the sum must have a coefficient of 0. Thus all B[sub]2n + 1[/sub] = 0 for integers n > 0 and ζ(-2n) is always zero.

or

would look pretty intimidating and exotic, provided you could get "א" to show in the forum directory.

Otherwise something like "The Imaginary Complex" or "The Abelian Group".

You don't have Final Fantasy "3"!? Do yourself a favor and get it! It is a spectacular game.

Also, I'm pretty sure Chrono Trigger (for the Super Nintendo) was a Square game. It is one of my favorites.