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June 19[edit]

Riemann Zeta Function[edit]

I cannot understand why Zeta[0] is -1/2. One would think that it is infinity but obviously this is not the case. 69.120.50.249 (talk) 01:45, 19 June 2009 (UTC)[reply]

One would only think it was infinity if one was using the definition

$\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1^{s}}}+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+\cdots .\!$

But that definition is valid only for s with real part greater than 1. Algebraist 01:51, 19 June 2009 (UTC)[reply]

What is the definition for other values? —Preceding unsigned comment added by 69.120.50.249 (talk) 02:01, 19 June 2009 (UTC)[reply]

As stated in our article, the zeta function is defined elsewhere by analytic continuation. Algebraist 02:03, 19 June 2009 (UTC)[reply]

Furthermore, the article gives several formulas for ζ(s), some of which may be useful when Re s < 1. For example, the formula

\zeta (s)={\frac {1}{1-2^{1-s}}}\sum _{n=0}^{\infty }{\frac {1}{2^{n+1}}}\sum _{k=0}^{n}(-1)^{k}{n \choose k}(k+1)^{-s}.\!

in Riemann zeta function#Globally convergent series readily gives

\zeta (0)={\frac {1}{-1}}\sum _{n=0}^{\infty }{\frac {1}{2^{n+1}}}\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{k}=-\sum _{n=0}^{\infty }{\frac {1}{2^{n+1}}}(1-1)^{n}=-{\frac {1}{2}}.

Having said that, I do not understand how this formula is supposed to work for all s ≠ 1 as claimed there. The fraction

1/(1-2^{1-s})

is infinite whenever s = 1 − 2πik/log(2) for some integer k. — Emil J. 15:56, 22 June 2009 (UTC)[reply]

Integral and series[edit]

While punching buttons on my calculator recently, I noticed that (using Mathematica syntax):

Integrate[x^x,{x,0,1}] ≈ 0.783430510909

and

-Sum[(-x)^(-x),{x,1,Infinity}] ≈ 0.783430510712.

I was wondering: is this near-miss a near miss because they are actually the same and the calculator is making rounding errors, or is it a near-miss because the integral and the sum actually give different results?

If it helps, the calculator is a TI-89 Titanium, OS version 3.0, hardware version 4.0. --72.197.202.36 (talk) 02:42, 19 June 2009 (UTC)[reply]

See Sophomore's dream. J Elliot (talk) 08:43, 19 June 2009 (UTC)[reply]

Congratulations. And another one if you can prove the result, think how to convert it into a series. You might also like Lambert W function. Dmcq (talk) 14:44, 19 June 2009 (UTC)[reply]

Spoiler alert: If you want to try proving the equality (which is correct, if it wasn't clear) yourself, don't follow the link to Sophomore's dream. -- Meni Rosenfeld (talk) 15:42, 19 June 2009 (UTC)[reply]

Two side remarks. The integral

\scriptstyle \int _{0}^{1}(x\log \,x)^{n}\,dx

, that appears in the quoted computation, may also be evaluated by a change of variable

\scriptstyle x=e^{-t}

: then another linear change of variables leads to the Euler integral

\scriptstyle \int _{0}^{\infty }u^{n}e^{-u}du

. Second remark: the same computation gives, more generally, the power series expansion

\scriptstyle z\int _{0}^{1}x^{-zx}dx=\sum _{n=1}^{\infty }{\frac {z^{n}}{n^{n}}}

(the linked article reports

\scriptstyle z=1

and

\scriptstyle z=-1

). --pma (talk) 19:22, 19 June 2009 (UTC)[reply]

Bernoulli numbers[edit]

Does anybody know a nice, closed-form way of generating Bernoulli numbers? I.e., B_n = what? --72.197.202.36 (talk) 03:59, 19 June 2009 (UTC)[reply]

Is this a serious question? I cannot believe it. Please read the article. There are several closed form formulas given there. —Preceding unsigned comment added by 92.227.144.52 (talk • contribs)

If no such method is available, are there any relatively simple algorithms to generate them? --72.197.202.36 (talk) 04:00, 19 June 2009 (UTC)[reply]

Bernoulli_number#Efficient_computation_of_Bernoulli_numbers contains an algoritm. Bo Jacoby (talk) 08:32, 19 June 2009 (UTC).[reply]

But this is not a 'relatively simple algorithm'. The Akiyama-Tanigawa algorithm is one and is given in the article in full length in pseudo code. Obviously people are talking here about articles without reading them. This is annoying and frustrating. —Preceding unsigned comment added by 92.227.144.52 (talk • contribs)

Why the disbelief of the OP being serious? WP:AGF and all that... I skimmed through the article, and didn't really see the nice, closed form formula, that I'd easily be able to implement (preferring simplicity above speed). Would anyone care to spell it out? --NorwegianBlue^talk 21:12, 19 June 2009 (UTC)[reply]

Why the disbelief of the OP being serious? Are you serious? I copy verbatim from the article. If you want I also copy the formula. If you want I also read it loud. Here it is:

These relations lead to a simple algorithm to compute the Bernoulli number. The input is the first row, A_0,m = 1/(m + 1) and the output are the Bernoulli number in the first column A_n,0 = B_n . This transformation is shown in pseudo-code below.

Akiyama-Tanigawa algorithm for B_n

Enter integer n.

For m from 0 by 1 to n do

A[m] ← 1/(m + 1)

For j from m by −1 to 1 do

A[j − 1] ← j × (A[j − 1] − A[j])

Return A[0] (which is B_n).

—Preceding unsigned comment added by 92.230.245.194 (talk • contribs)

I wouldn't call that "closed form". Michael Hardy (talk) 22:05, 19 June 2009 (UTC)[reply]

It answers a question of a person who did not read the article: I quote: "If no such method is available, are there any relatively simple algorithms to generate them?" Would you also call it not a simple algorithm, Mr. Hardy? —Preceding unsigned comment added by 92.230.245.194 (talk) 23:45, 19 June 2009 (UTC)[reply]

Certainly simple, but not exactly what we mean by closed form. As far as I know, there are no simple closed forms for Bernoulli numbers. "Closed form" is a relative concept, of course, as relative concept as the one of elementary function &c. However, I met difficulty in following this thread, for one or more posters are not signing their contributions, and it's not clear where a post starts and ends, and who is replying to whom. --pma (talk) 08:51, 20 June 2009 (UTC)[reply]

I have added signatures to the unsigned entries, and also (per WP:IAR) taken the liberty of modifying the indentation, in order to improve the readability of this thread. --NorwegianBlue^talk 11:08, 20 June 2009 (UTC)[reply]

The question may have been inspired by the news story Iraq-born teen cracks maths puzzle which begins

"STOCKHOLM (AFP) – A 16-year-old Iraqi immigrant living in Sweden has cracked a maths puzzle that has stumped experts for more than 300 years, Swedish media reported on Thursday. In just four months, Mohamed Altoumaimi has found a formula to explain and simplify the so-called Bernoulli numbers ..." 208.70.31.186 (talk) 11:04, 20 June 2009 (UTC)[reply]

Some more info. 208.70.31.186 (talk) 11:06, 20 June 2009 (UTC)[reply]

From the generating function formula ${\frac {x}{e^{x}-1}}=\sum _{n=0}^{\infty }B_{n}{\frac {x^{n}}{n!}}$ it follows that $B_{n}=\left({\frac {d^{n}{\frac {x}{e^{x}-1}}}{dx^{n}}}\right)_{x=0}$ which has the closed form the OP was asking for: B_n = what?. But perhaps the OP is not happy because the intermediate results are functions rather than numbers. Bo Jacoby (talk) 15:57, 20 June 2009 (UTC).[reply]

Closed, indeed, as it was Pandora's box... ;-) --pma (talk) 22:14, 20 June 2009 (UTC)[reply]

friction[edit]

knowing that the coefficient of friction between the 25kg block and the incline is 0.25 determine the value of p for which motion of block up the inclined plane is impending —Preceding unsigned comment added by 81.199.50.67 (talk) 12:32, 19 June 2009 (UTC)[reply]

We are not here to do your homework. 65.121.141.34 (talk) 13:31, 19 June 2009 (UTC)[reply]

Especially when you don't post enough of the homework question to work out what you're talking about. Algebraist 13:37, 19 June 2009 (UTC)[reply]

What is p? What is the angle of incline? Describe the problem in full, make an attempt at it (a good start would be to describe the forces on the block), then ask for help.86.132.164.238 (talk) 13:41, 19 June 2009 (UTC)[reply]

For the motion of the block up the plane to be impending, the following equation must be satisfied (the force on the block up the plane must equate to the force on the block down the plane):

m*g*sin(Ə) = f*m*g*cos(Ə)

where m is the mass of the block, g is the acceleration due to gravity, f is the co-efficient of (static) friction and Ə is the angle of inclination. Since all variables except Ə are known, I am assuming Ə is p according to your question. Therefore, if we solve for Ə, we obtain:

tan(Ə) = f, since the m*g factor may be cancelled from both sides of the equation.

Therefore, tan(Ə) = 0.25 which, with the use of a calculator, gives Ə = 14 degrees approximately. Notice that since the mass of the block (m) cancelled on both sides of the equation, we did not need to know its value for the purposes of this computation. Hope this helps. --PS T 22:17, 19 June 2009 (UTC)[reply]

For all these kinds of questions the method is the same: Resolve the forces parallel and perpendicular to the plane and equate them to the appropriate values using F=ma. Then solve for the value to be found. --Tango (talk) 22:34, 19 June 2009 (UTC)[reply]

Statisics (PCA): Can we have a PCA-based reconstruction that includes negative weighting of some sensors that express only a postive physical quantity?[edit]

Is it automatically unphysical to have a PCA reconstruction that has some stations negatively weighted? Note, that this is more than just a PCA, but PCA followed by some regressions. Would think that it could occur for both degeneracy and anticorrelation with the average (actual physical effects). Of course the summation must be positive, but is it automatically wrong if some of the stations have negative weights?

This is being debated on these blog threads. Unfortunatley, the debate has muddled particular examination of the Stieg Antarctic PCA-based recon with general absolute claims that negative weightings are bad, bad, bad.

Could you please adjuticate?

See here:

http://noconsensus.wordpress.com/2009/06/07/antarctic-warming-the-final-straw/#comment-6727

http://noconsensus.wordpress.com/2009/06/09/tired-and-wrong-again/#comment-6726

http://wattsupwiththat.com/2009/06/10/quote-of-the-week-9-negative-thermometers/#more-8362

http://www.climate-skeptic.com/2009/06/forgetting-about-physical-reality.html

Would appreciate some reference to a citation or an academician and/or a more in-depth answer than just "he or she is right" so that I can either get my opponent to understand where he is wrong, or understand it myself. Muchos gracias. 69.250.46.136 (talk) 18:00, 19 June 2009 (UTC)[reply]

A trig equation[edit]

Hi. I'm halfway through working out the answer to a question and am now stuck. Here's what I have

"Let $f(x)=\sin(2nx)+\sin(4nx)-\sin(6nx)$ where n is a positive integer and $0<x<{\frac {\pi }{2}}$ . Find an expression for the largest root of the equation f(x)=0, distinguishing between the cases where n is even and n is odd. You may assume that $\sin(3\theta )=\sin \theta (4\cos ^{2}\theta -1)$ ."

So after working through this I get solutions as follows:

$x={\frac {p\pi }{2n}}$ provided $0<p<n$

$x={\frac {p\pi }{n}}$ provided $0<2p<n$

$x={\frac {\pi }{n}}({\frac {1}{3}}+p)$ provided $0<{\frac {2}{3}}+2p<n$

$x={\frac {\pi }{n}}({\frac {2}{3}}+p)$ provided $0<{\frac {2}{3}}+p<{\frac {n}{2}}$

where p is some integer.

First of all, is what I have correct? Secondly, how do you distinguish between odd and even? I can't see what difference it will make to my answer. Thanks asyndeton talk 20:51, 19 June 2009 (UTC)[reply]

Have you tried graphing it for a few values of n? -GTBacchus^(talk) 20:48, 19 June 2009 (UTC)[reply]

You aren't allowed access to any graphing applications (or even a calculator) while sitting the paper this question comes from. Slightly hypocritical I know, since you don't have access to Wikipedia mathematicians either, but this way I can see how they intended you to approach it, which is more important to me than just reaching the right answer. asyndeton talk 20:51, 19 June 2009 (UTC)[reply]

If only there was away to draw a graph without a computer or a calculator ... Gandalf61 (talk) 20:56, 19 June 2009 (UTC)[reply]

I have graphed f(x) on my computer in the cases n=2,3. It hasn't helped. asyndeton talk 21:39, 19 June 2009 (UTC)[reply]

Let u = 2nx. Then the function is

{\begin{aligned}&{}\quad \sin u+\sin(2u)-\sin(3u)\\&=(\sin u)(1+2\cos u-(4\cos ^{2}u-1))\\&=(2\sin u)(1+\cos u-2\cos ^{2}u)\end{aligned}}

and this is 0 iff either sin u = 0 (so u is an integral multiple of π) or cos u = either 1 or −1/2 (so that u is an integral multiple of 2π/3).

The condition that x is between 0 and π/2 says that u is between 0 and nπ.

If n is even then the largest multiple of 2π/3 that is less than nπ is π(n − 2/3). If n is odd, then put "1/3" there instead of "2/3". (Just draw the picture on the x-axis with n going from 0 to about 5 or so, and you'll see it. Never mind what the curve looks like; just graph the zeros as I described them above.) Michael Hardy (talk) 21:44, 19 June 2009 (UTC)[reply]