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You are not logged in. #1 20060713 10:06:19
ζ(2n).For a time I wondered how one could possibly obtain exact values for various zeta functions, such as ζ(2) = π^{2}/6 and ζ(4) = π^{4}/90. Then, one fateful day, by luck or extreme insight, I realized that Fourier series could possibly do the job. Upon testing, I was pleasantly led the to correct answer. Due to the nature of the method, only values for ζ(2n) are obtainable. Of course, as far as I know, ζ(2n + 1) (odd values, basically) always appears unable to be expressed in terms of known constants. Anyway, let us move on to the method. For this problem, we will have f(x) = x^{2} and L = 2π. Then and The the Fourier series expansion for f(x) is Now when x = 0, the series becomes Now, for the convergence of Fourier series, there are 3 conditions called the Dirichlet conditions: 1. f(x) is defined, expect for possibly at a finite number of points in (0, 2L). 2. f(x) is 2Lperiodic outside (0, 2L). 3, f(x) and f'(x) are piecewise continuous in (0, 2L). Now, our function is definitely defined all throughout (0, 2L). We defined f(x) as being 2L periodic, so it is a parabola from x = 0 to x = 2π, and this is repeated periodically(so f(x + 2π) = f(x)). On (0, 2L), f(x) is a continuous parabola and f'(x) is a continuous straight line. So the Dirichlet conditions are satisfied. Now that the Dirichlet conditions are known to be satisfied, what does that do for us? It does something very good. It tells us that the Fourier series for f(x) will converge to Now let us return to the problem at hand. We had just let x = 0. This means we can find what the Fourier series will converge to when x = 0, thanks to the Dirichlet conditions: Now we may write the Fourier series at x = 0 as Now we may easily solve the equation for ζ(2): And so it is done. In general, to find ζ(2n), write the Fourier series for a 2πperiodic function f(x) = x^{2n} defined on (0, 2π) and work out the problem as I have above. Does anyone else have methods for finding exact values of infinite series? Last edited by Zhylliolom (20060713 10:10:36) #2 20060713 10:51:33
Re: ζ(2n).Sure, there are many ways. Alternating series can be pretty easy. For example: I already gave you a major hint that it is an alternating series. But in general, what you try to do is take the first few summations, and come up with a sequence. Then you must prove that it actually is the sequence through induction (which normally isn't too bad). Then you just have to show what number that sequence converges to, which can be very painful or even just about impossible. "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..." #3 20060713 15:04:39
Re: ζ(2n).Ricky, your sum is divergent. Is i supposed to start at 1 instead? I'll assume this is the case and try out the problem. Also, I don't know what you mean by an alternating series, don't alternating series have (1)^{i} in them? After these first few partial sums, a pattern is noted: We will now prove this by induction. We start with the base case: So the base case holds. Now assume that Then we now want to show that the equality holds for n + 1: So by mathematical induction, Now the question is, "what do the partial sums converge to?" To find this out we take the limit as n approaches infinity: So we conclude that the series converges to the value of 1. Last edited by Zhylliolom (20060809 19:58:03) #4 20060713 23:35:41
Re: ζ(2n).I didn't really read through your proofs, but you got the answer right. But there is an easier way to do it. As I said, this is an alternating series, though it doesn't look like it. Now when you write the sum, it will look like: But we can't reorder terms until we prove that the series is convergent. So let's prove the series is convergent: And now we're done. We can regroup terms, and so the sum is 1, since all the fractions cancle themselves out. "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..." #6 20060714 00:04:37
Re: ζ(2n).Fourier series are so difficult. X'(yXβ)=0 #7 20060714 08:04:12
Re: ζ(2n).Ricky: I feel a little silly now that I didn't approach it that way. But I arrived at the same answer, so I suppose it isn't too bad. I like your method better though. #8 20060714 09:15:58
Re: ζ(2n).You need a good book on real analysis. In my experience, infinite sums are either really easy, or almost impossibly hard.
Great! Now you know how I felt when I had my advanced calculus final a while back with this exact problem on it, and I was the last one in the room with only this problem to go. "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..." #9 20060725 22:50:01
Re: ζ(2n).The discovery of another method of finding ζ(2) has led me to post on this topic again. I find this one enjoyable, as it doesn't seem like a standard approach. Now divide by x to obtain This function will have zeros at nπ, so factor it according to that fact: This simplifies to Now if some crazy old man were to multiply this all out, the coefficient of x² would be But in our original expression for sin(x)/x, the coefficient for x² was 1/3! = 1/6, so the sum must be equal to 1/6: Now, simplify: #10 20060726 01:56:01
Re: ζ(2n).It most certainly isn't standard. How in the world did you come up with that? "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..." #12 20060809 19:56:41
Re: ζ(2n).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). 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_{2n + 1} = 0, or in other words, odd values of n greater than 1 for B_{n} 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_{n}'s generating function: Now we already established that for all odd positive integers n > 1, B_{n} = 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_{2n + 1} = 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_{2n + 1} = 0 for all integers n > 0. Recall the expression from earlier. We know that B_{1} is an odd value of n in B_{n} 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)^{n}B_{n} = B_{n}, 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_{2n + 1} = 0 for integers n > 0 and ζ(2n) is always zero. Last edited by Zhylliolom (20060813 08:10:31) #13 20060809 20:04:23
Re: ζ(2n).reading that makes me dizzy sorry Presenting the Prinny dance. Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! #14 20060810 00:46:53
Re: ζ(2n).Just need a bit of help with your variables. k, n are natural, x is real, and z is complex, right? To Is the identity supposed to be: "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..." #15 20060810 00:56:11
Re: ζ(2n). IPBLE: Increasing Performance By Lowering Expectations. #16 20060810 08:38:54
Re: ζ(2n).Ok, I admit I rushed the post a little(I was excited to share this with everyone), so a few things might be confusing. In my notes, I used z and s (Riemann style) for the variables. Consider x and z to be both complex. Should I change x to s in my original post?
No, I did a few steps in one there(probably the only part of my proof where I made a small jump, I think). I substituted x = z/π to get this: Then I multiplied through by x = z/π to get Next, z²/π² was moved inside the sum, and inside the sum we clear fractions by multiplying by π²/π²: So I think the part that confused you there was me getting from π cot πx to z cot z, but as you can see, the substitution cleared the π and gave us a z inside the cot, and the multiplication of both sides by z/π put a z outside the cot and cancelled out the π outside, giving us z cot z. Last edited by Zhylliolom (20060810 10:24:59) #17 20060810 18:16:29
Re: ζ(2n).i give up i cant understadnd anything ur saying Presenting the Prinny dance. Take this dood! Huh doood!!! HUH DOOOOD!?!? DOOD HUH!!!!!! DOOOOOOOOOOOOOOOOOOOOOOOOOD!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! #18 20060810 21:47:59
Re: ζ(2n).Zhylliolom, or maybe someone else could explain what happened here, I'm a bit puzzled : I noticed the geometric sum after that, but I've never worked with double sums before, ever. But does it basically mean that you take the sum over the sum, as in (k=1 to ∞) first and then (n=1 to ∞) over that previous sum? If so, I learned something new by reading your post Bang postponed. Not big enough. Reboot. #19 20060811 00:33:43
Re: ζ(2n).numen, he took a negative sign out of the inside of the sum, and put it on the outside. And as for the 2nd part, in short, yes. "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..." #20 20070512 01:30:07
Re: ζ(2n).Zhylliolom wrote: */ I give up, would you give me a hint? #21 20070512 01:31:27
Re: ζ(2n).Zhylliolom wrote: */ I give up, would you give me a hint? 