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June 19

Riemann Zeta Function

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

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

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

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?

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

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]

June 20

Approximation

Hi. Just did a show that involving an approximation and I'm not convinced that what I've done is rigorous. Could someone please check it for me?

"Let ${\frac {1}{y}}={\frac {2+3x^{2}}{3(1+x^{2})^{1.5}}}+{\frac {1}{3}}$ . Show that, for large positive x, $y\approx 3-{\frac {9}{x}}$ ."

My method goes like this. for large positive x $1+x^{2}\approx x^{2}$ and so $(1+x^{2})^{1.5}\approx x^{3}$ . Also for large positive x, $2+3x^{2}\approx 3x^{2}$ so ${\frac {2+3x^{2}}{3(1+x^{2})^{1.5}}}\approx {\frac {3x^{2}}{3x^{3}}}={\frac {1}{x}}$ so ${\frac {1}{y}}\approx {\frac {1}{x}}+{\frac {1}{3}}$ . Rearranging this gives $y\approx {\frac {3x}{3+x}}={\frac {3x+9}{3+x}}-{\frac {9}{3+x}}=3-{\frac {9}{3+x}}$ . Finally, for large positive x, $3+x\approx x$ so $y\approx 3-{\frac {9}{x}}.$

Is this a solid argument? Thanks 92.2.16.39 (talk) 10:09, 20 June 2009 (UTC)[reply]

Yes, I think that's spot on.. Rkr1991 (talk) 14:15, 20 June 2009 (UTC)[reply]

Yes, that looks fine to me. The statement "for large positive x,

y\approx 3-{\frac {9}{x}}

" isn't a rigorous statement, so there is no way your "proof" could be rigorous. What you've got is good enough, in my opinion. If the question had been to show that

y=3-{\frac {9}{x}}+O({\frac {1}{x^{2}}})

or something (or whatever the appropriate notation is, I can never remember the differences between big-O, little-O, etc. without looking them up), then you could get a proper rigorous proof (probably along the same lines as the argument you gave). --Tango (talk) 14:26, 20 June 2009 (UTC)[reply]

Hm, either two curves are asymptotic or they're not, meseems. —Tamfang (talk) 04:54, 25 June 2009 (UTC)[reply]

You may prefer to simplify first and approximate afterwards.

y={\frac {3(1+x^{2})^{1.5}}{2+3x^{2}+{(1+x^{2})^{1.5}}}}=3-9x^{-1}\cdot {\frac {2x^{-2}+3}{6x^{-3}+9x^{-1}+3{(1+x^{-2})^{1.5}}}}\approx 3-9x^{-1}

.

Bo Jacoby (talk) 14:37, 20 June 2009 (UTC).[reply]

Car dynamics

Please tell me about the algorithm of finding International Roughness Index(of a road surface)from a longitudinal profile data. —Preceding unsigned comment added by 113.199.158.140 (talk) 12:36, 20 June 2009 (UTC)[reply]

Googleing 'International Roughness Index' leads to the fortran program at http://www.umtri.umich.edu/content/IRIMain.f . Bo Jacoby (talk) 13:51, 20 June 2009 (UTC).[reply]

I wanted to have the algorithm and not the program. —Preceding unsigned comment added by 113.199.170.182 (talk) 02:47, 21 June 2009 (UTC)[reply]

What's the difference? A programming language is just a language for writing algorithms. --Tango (talk) 03:09, 21 June 2009 (UTC)[reply]

The program contains the following comment: 'For more information about the IRI and how this program works, see Sayers, M. W., "On the Calcluation of International Roughness Index from Longitudinal Road Profile." Transportation Research Record 1501, (1995) p. 1-12. See http://spyder.umtri.umich.edu/erd_soft/erd_file.html on the web for a description of ERD files'. So now it is time for the OP to do some homework. Bo Jacoby (talk) 10:20, 21 June 2009 (UTC).[reply]

Necessary and sufficient conditions

I'm a little hazy on N&S conditions so wanted to check my thoughts with people who can tell me if I'm right or wrong. Let $f(x)=x^{2}-2bx+c$ . I have to find N&S conditions on b and c a) for f(x) to have two distinct real roots and b) for f(x) to have two distinct positive real roots.

For a) I get that $c-b^{2}<0$ just from using the fact that if the turning point is below the x axis, you've got two distinct roots.

For b) I again get $c-b^{2}<0$ , for the same reason as before. I then get c>0, which ensures that the roots are of the same sign and b>0, which says the minimum point occurs in the first quadrant (at least I think that's its name; I mean bottom right). So the y-intercept is positive, both the roots are of the same sign and the turning point is in the first quadrant, so between them, they say that the roots are distinct and positive.

Are they necessary and sufficient or not? Thanks 92.4.255.16 (talk) 17:52, 20 June 2009 (UTC)[reply]

"A is necessary for B" means "B implies A". "A is sufficient for B" means "A implies B". Therefore to prove that something is necessary and sufficient you need to prove it in both directions. Sometimes you can do both directions at once, but if you're not sure it is best to do them separately. You need to come up with a sequence of implications from A to B and then a sequence from B to A. While considering the graphs is a good idea in order to get an intuitive answer, your rigorous proof should probably be purely algebraic. --Tango (talk) 18:01, 20 June 2009 (UTC)[reply]

The quadratic formula gives you the roots of f as

x={\frac {2b\pm {\sqrt {4b^{2}-4c}}}{2}}=b\pm {\sqrt {b^{2}-c}}

. The answer to both questions follow naturally from there. For question a you are correct; the expression in the radical must be positive for the radical to be real. For question b, it is sufficient and necessary for the smaller root to be real and positive. I'll let you work the algebra to see where that gets you. --COVIZAPIBETEFOKY (talk) 18:24, 20 June 2009 (UTC)[reply]

~~Your solution to (b) is almost right. The x-coordinate of the vertex is ${\frac {-b}{2a}}$ (note the minus sign). So to be in the fourth quadrant, you need $b<0$~~ . -- Meni Rosenfeld (talk) 19:37, 20 June 2009 (UTC)[reply]

Where has the 'a' come from? Are you possibly confusing the f(x) I gave above with the general form of a quadratic equation? 92.7.54.6 (talk) 19:41, 20 June 2009 (UTC)[reply]

Yes, COVIZAPIBETEFOKY was correct, and Meni is correct to say that you meant the fourth quadrant, and your conclusions were also correct, but see Tango's suggestion for a rigorous proof. Dbfirs 00:12, 21 June 2009 (UTC)[reply]

Of course, silly me. I was thinking about the general case, and didn't notice the differences in your case, most importantly that your b is already negated. So indeed it should be

b>0

. -- Meni Rosenfeld (talk) 08:50, 21 June 2009 (UTC)[reply]

Maximizing the volume in a certain class of sets

Has anybody already seen an optimization problem like this one:

Find the maximum value of the 3-dimensional volume of the subset of the unit cube:

{\hat {A}}:=\{(x,y,z)\in [0,1]\times [0,1]\times [0,1]\,:(x,y)\in A,(y,z)\notin A\}

among all measurable subsets $\textstyle A$ of the unit square $[0,1]\times [0,1]$ .

Incidentally, I know the answer (it is hidden here in case somebody prefers not to see it), but I'd like to settle this and other similar but more difficult problems into a general theory/method, if there is any available. --pma (talk) 21:17, 20 June 2009 (UTC)[reply]

Cute problem. I don't recall having seen it before. Maybe I'll try to work out where it came from. I wonder if thinking of it in the language of probability could shed any light. Michael Hardy (talk) 17:48, 21 June 2009 (UTC)[reply]

In the most general sense, this seems like calculus of variations. But that answer is probably useless. 67.122.209.126 (talk) 08:53, 22 June 2009 (UTC)[reply]

June 21

Finding the median of a continuous random variable

I am told that I need to solve $\int _{m}^{\infty }f(x)\,dx=0.5$ , and in the textbook example they get a polynomial with only m⁴, m² and m⁰ terms in it, which lets them solve a quadratic in m², take a square root and compare the results to the range which you are given at the start to find the right one, but I keep getting equations like $\int _{m}^{3}{\frac {2}{9}}x\left(3-x\right)\,dx=0.5$ , where the p.d.f f(x) of the continuous random variable X was the above for $0\leq x\leq 3$ , which eventually gets me to ${\frac {2}{27}}m^{3}-{\frac {1}{3}}m^{2}+0.5=0$ , but I don't know how to solve this. In another example, I had the p.d.f. f(y) of the continuous random variable Y being $12y^{2}\left(1-y\right)$ for $0\leq y\leq 1$ , which got me to $\int _{m}^{1}12y^{2}\left(1-y\right)\,dy=0.5\Rightarrow 3m^{4}-4m^{3}+0.5=0$ , but, again, I don't know what to do with this. It Is Me Here ^{t / c} 08:58, 21 June 2009 (UTC)[reply]

Well,

\int _{0}^{3}{\frac {2}{9}}x\left(3-x\right)\,dx=1

, and the symmetry of

x\left(3-x\right)

suggests that

\int _{0}^{3/2}{\frac {2}{9}}x\left(3-x\right)\,dx=\int _{3/2}^{3}{\frac {2}{9}}x\left(3-x\right)\,dx=0.5

so try m = 3/2 in your first problem. Your second problem is more difficult, as

y^{2}\left(1-y\right)

doesn't have an obvious symmetry to exploit. Gandalf61 (talk) 12:21, 21 June 2009 (UTC)[reply]

What's worse, Mathematica seems unable to find a representation of the solution simpler than direct substitution in the quartic formula. Is there any chance they want you to find a numerical solution (which is 0.614272...)?

By the way, the first problem can also be solved by the rational root theorem. -- Meni Rosenfeld (talk) 13:21, 21 June 2009 (UTC)[reply]

See root-finding algorithm for methods to solve equations numerically. Or draw the conclusion that the median is not that interesting anyway and use the mean value instead! Your examples are beta distributions. By the way, 0.614272 is an approximate solution to

3m^{4}-4m^{3}+0.5=0

. The J program p. 1 0 0 _8 6 does the trick. Bo Jacoby (talk) 13:42, 21 June 2009 (UTC).[reply]

OK, the actual question states: "The continuous random variable Y has p.d.f. f(y) defined by f(y) = 12y²(1 - y) for 0 ≤ y ≤ 1; f(y) = 0 otherwise. Show that, to 2 decimal places, the median value of Y is 0.61." It Is Me Here ^{t / c} 15:09, 21 June 2009 (UTC)[reply]

That's easy then. You know that

F(y)

(the cumulative distribution function) is continuous and increasing, so you only need to show that

F(0.605)<0.5

and

F(0.615)>0.5

. The intermediate value theorem will then say that there is a root

0.605<y<0.615

, which is thus equal to 0.61 to two decimal places. -- Meni Rosenfeld (talk) 18:47, 21 June 2009 (UTC)[reply]

Induction

Not a massive fan of induction and so I would like to check my solution to a problem with you guys.

"Let $S_{n}(x)=e^{x^{3}}{\frac {d^{n}}{dx^{n}}}(e^{-x^{3}})$ . Prove by induction on n that $S_{n}(x)$ is a polynomial."

Assume that $S_{n}(x)$ is a polynomial for n=k.

$e^{x^{3}}{\frac {d^{k}}{dx^{k}}}(e^{-x^{3}})=f(x)$

So ${\frac {d^{k}}{dx^{k}}}(e^{-x^{3}})=e^{-x^{3}}f(x)$

Differentiate the above wrt x.

${\frac {d^{k+1}}{dx^{k+1}}}(e^{-x^{3}})=e^{-x^{3}}f'(x)-3x^{2}e^{-x^{3}}f(x)$

Factorise

${\frac {d^{k+1}}{dx^{k+1}}}(e^{-x^{3}})=e^{-x^{3}}(f'(x)-3x^{2}f(x))$

$f'(x)-3x^{2}f(x)$ is still a polynomial for any f(x), so let us call it g(x).

So we have

${\frac {d^{k+1}}{dx^{k+1}}}(e^{-x^{3}})=e^{-x^{3}}g(x)$

So if the result is true for k, it is also true for k+1. Now let n=0 (the question doesn't actually state which set of numbers 'n' belongs to but I assume it's the non-negative integers), which gives us the case where $S_{0}(x)=1$ . So by induction, $S_{n}(x)$ is a polynomial for all integers n, n≥0.

Is that airtight? Thanks. asyndeton talk 14:15, 21 June 2009 (UTC)[reply]

Yes. —JAO • T • C 14:42, 21 June 2009 (UTC)[reply]

It looks fine to me, although I would have started with n=1. While defining the zeroth derivative to be the identity makes sense, I would normally think of n be positive in the context of n-th derivatives. --Tango (talk) 14:56, 21 June 2009 (UTC)[reply]

well, I think yours is mainly a psycological limit. But if you work on it a bit, you too will be able to start with n=0 ;-) --pma (talk) 19:11, 21 June 2009 (UTC)[reply]

It's obvious that is does work with n=0, I just don't normally think of the identify being a derivative. It's a matter of definition, rather than any mathematical, of course. --Tango (talk) 00:05, 22 June 2009 (UTC)[reply]

The definition A⁰ = 1 apply very generally. See exponentiation. And so

\scriptstyle {\frac {d^{0}}{dx^{0}}}=1

is what you should normally think. What happens if you don't differentiate? The same thing as if you multiply by one. Nothing happens! Bo Jacoby (talk) 05:08, 22 June 2009 (UTC).[reply]

When you say

\scriptstyle {\frac {d^{0}}{dx^{0}}}=1

, do you mean this is true in general no matter what it is you're differentiating or is it just true for what I gave above? asyndeton talk 07:44, 22 June 2009 (UTC)[reply]

Yes, in general, for a linear operator A, the useful notation is A⁰=I (the identity operator, that we think to as the multiplication by 1 and write indeed 1) and A^k=A...A k-times for all positive integers k. --pma (talk) 10:06, 22 June 2009 (UTC)[reply]

In case there is any confusion, note that it is

\scriptstyle {\frac {d^{0}}{dx^{0}}}=1

, not

\scriptstyle {\frac {d^{0}}{dx^{0}}}f(x)=1

. It's the identity operator, not a constant operator. --Tango (talk) 15:09, 22 June 2009 (UTC) [reply]

I don't see the difference between the two. asyndeton talk 16:54, 22 June 2009 (UTC)[reply]

The second one is

\scriptstyle {\frac {d^{0}}{dx^{0}}}f(x)=1\cdot f(x)=f(x)

. Bo Jacoby (talk) 17:05, 22 June 2009 (UTC). [reply]

The first is an operator, the second is an operator acting on a function. When we say an operator is "1" we mean it is the identity - it maps everything to itself. That's very different to saying the result of the operator acting on a function is always equal to the number "1". --Tango (talk) 17:08, 22 June 2009 (UTC)[reply]

N-deck hearts

Do the rules of double-deck cancellation hearts generalize well to hearts games with N decks and 4N-2 to 4N+2 players? Neon Merlin 23:46, 21 June 2009 (UTC)[reply]

Well my first thought is that there are two possible ways in which the game could be generalised, either we stick with the rule that if a second identical card is played then the two cards cancel, which leads to a game where if odd numbers of an identical card are played the last player to play that card (assuming it is the strongest one, otherwise it makes no difference) takes the trick whereas if even numbers are played the cards are ignored, this seems slightly artificial to me and as if it would lead to a confusing game.

Alternatively the other way in which the game could be generalized is to only apply the cancellation rule when all N of an identical card are played in one trick, otherwise if say K (<N) identical cards were played in one trick then the last player to play the identical card would count as having played it and the other players who played it would be ignored. This to some extent removes the excitement that double-deck cancelation hearts has, since especially when N is large the chance of a cancellation occuring are very slim.

Both possibilities have their pros and cons, when N is small (say 3 or 4) I think both variations would make interesting playing. —Preceding unsigned comment added by 86.129.82.211 (talk) 00:52, 22 June 2009 (UTC)[reply]

June 22

Linear topological space

What is the exact definition of a linear topological space (the formal definition). I am unable to find it on the internet. Thanks. —Preceding unsigned comment added by 58.161.138.117 (talk) 04:42, 22 June 2009 (UTC)[reply]

The same thing as a topological vector space. Bo Jacoby (talk) 04:53, 22 June 2009 (UTC).[reply]

The gambler's fallacy

I am an EMT and far too often, I hear my co-workers say, "They didn't get a single [emergency] call during the last shift, we're going to get slammed now". Clearly, each 9-1-1 call we receive is completely independent of the next, making this statement an example of the Gambler's fallacy; however, at what point does the law of large numbers take over? We typically get 15 calls a day. Say, for example, suddenly we get 0 calls over 5 days. Those five days have no bearing on what the 6th day will bring (or the 7th, 8th, etc.). However, is it fair to say that it is probable that we will get a number of calls that "make up" for that five day stretch over the course of the rest of the year? Even this seems to imply a causation effect, but really it's just looking at statistical data and assigning probabilities to it, right? -- MacAddct1984 ^{(talk  contribs)} 05:09, 22 June 2009 (UTC)[reply]

It's important to note that the law of large numbers only provides stability of the mean, not the total. Assume you normally get 6 calls a day. If you get zero calls the first five days, your per-day average is 0. But you should still expect to get on average 5 calls a day for the rest of the year, bringing you to an average of (360*6)/365, which is 5.92. Over an even longer period, the expected average would be even closer to the normal 6, but the expected total will still be 30 calls short—it's just divided by a larger number of days. (This is all assuming that the probabilities really are independent.) —JAO • T • C 05:46, 22 June 2009 (UTC)[reply]

Ah, that makes more sense to me, thanks. -- MacAddct1984 ^{(talk  contribs)} 13:53, 22 June 2009 (UTC)[reply]

Why do you think the 911 calls are independent of each other? Say it's snowing heavily. Everyone stays indoors keeping warm, and you don't get any calls. Then the snow stops, those same people go outside to shovel their driveways or drive someplace, a lot of them have heart attacks or road accidents, and the 911 board lights up. That's hypothetical of course, but I don't have too much trouble with the idea that a flurry of calls is correlated with an earlier lull in them. 67.122.209.126 (talk) 06:55, 22 June 2009 (UTC)[reply]

Assuming independence, the number of 911 calls on a shift is likely to have a poisson distribution. (This is called the law of small numbers, by the way). Your description, that you 'typically get 15 calls a day. Say, for example, suddenly we get 0 calls over 5 days' is very unlikely for a poisson distribution, but unlikely events are remembered very well. You need better historical data. Don't you have a record of the total number of calls over a large number of shifts? Bo Jacoby (talk) 07:21, 22 June 2009 (UTC).[reply]

67.122, that seems true for the most part^{[original research]}, but even in that case the calls can still be considered independent of each other, no? The fact that Joe calls 911 on one side of town after a snowfall still has no bearing on whether Jane, another resident, will.

Bo, we do keep track of call statistics. We've typically seen a 5-10% increase in call volume each year. Assuming the sample size is large enough, is it possible/logical to make predictions based off of that data? -- MacAddct1984 ^{(talk  contribs)} 13:53, 22 June 2009 (UTC)[reply]

No, in the case described different people calling are not independent. The fact that Joe has just called makes it more likely that now is a time of higher-than-usual call density, which makes it more likely that Jane will call. Algebraist 13:57, 22 June 2009 (UTC)[reply]

Perhaps you are getting a tad confused about the meaning of 'independent'. It does not necessarily describe a cause-effect relationship. Obviously, Joe's calling 911 does not cause Jane to call 911 around the same time (at least, not usually), but as Algebraist explained, the knowledge that Joe called does affect the probability that Jane will call about the same time. And that is what it means for the two events to be 'dependent', as opposed to independent. --COVIZAPIBETEFOKY (talk) 14:25, 22 June 2009 (UTC)[reply]

Where the probability is a gauged value from the frame of reference of an observer (us), rather than a descriptor of Jane's actual propensity to make the call. —Anonymous Dissident^Talk 14:54, 22 June 2009 (UTC)[reply]

Have you checked against the TV listings? I wouldn't put it past people to wait till a soap is over to ring up about their heart attack. Dmcq (talk) 12:43, 22 June 2009 (UTC)[reply]

Let the number of calls on day number i be X_i. Compute the number of days S₀ = ΣX_i⁰, the number of calls S₁ = ΣX_i¹, and the square sum S₂ = ΣX_i². Compute the mean value μ = S₁/S₀ and the variance σ² = (S₂/S₀)−(S₁/S₀)². Now, if the number of calls are really independent of one another and of time, the distribution is poisson and μ = σ². If this is not the case, but for example μ < σ², then the distribution may be the negative binomial distribution. (See the article on cumulant for explanation). If the historical data fit reasonably well into a mathematical model, then it is reasonable to make predictions based on that model. Bo Jacoby (talk) 14:30, 22 June 2009 (UTC).[reply]

I think the key thing to remember here is that how many calls you got over the last 5 days isn't a random variable, it's a known value. It's meaningless to talk about the probability distribution for it because it doesn't have one - there is a 100% chance that you saw what you saw and a 0% chance that you didn't. If you expect 15 calls a day then at the start of the year you would expect 365*15 calls that year, but once you've experienced a few days you can improve that estimate by replacing the expected number of calls on days past with the actual number. --Tango (talk) 14:49, 22 June 2009 (UTC)[reply]

Not all calls will be independent. A house starts on fire, and 3 different neighbors could very well call 911 over the same event. Or are such incidents not counted? 65.121.141.34 (talk) 18:49, 22 June 2009 (UTC)[reply]

We're after that step, the calls have already been filtered out by the time we get them. Very rarely will we get a call for, say, a stabbing, and then 30 minutes later we'll get another call because they found the other party involved who's also injured. -- MacAddct1984 ^{(talk  contribs)} 13:28, 23 June 2009 (UTC)[reply]

question on stats

Assume I have a giant jar of jelly beans (say 500 individual beans) that is half red and half green. Without being able to see what I am doing, I randomly select one. I replace it and select another. I note if I have drawn 2 red, a red and a green, or two greens and replace the bean I just drew. Now if I get 50 pairs, Logically I should have 25 red/green, 12.5 red/red and 12.5 green/green (ok so 12 of one and 13 of the other). But I know that this exact outcome by itself is somewhat unlikely. How do I determine the probability of the various combinations, (22rg, 10rr, 18gg), (28rg, 11 rr, 11gg) etc?

Our article on the urn problem is a good starting point. You may also find the article on the binomial distribution helpful. Ray^Talk 21:54, 22 June 2009 (UTC)[reply]

The probability of the combination (22rg, 10rr, 18gg), is $\scriptstyle {\frac {50!\,0.5^{22}\,0.25^{10}\,0.25^{18}}{22!\,10!\,18!}}=0.00385358$ . Bo Jacoby (talk) 08:33, 23 June 2009 (UTC).[reply]

Door codes

In order to get into the building I live in, I need to type in a four-digit code, and today, I noticed something that leads to a problem that I don't know how to solve.

I had assumed that if I typed in an incorrect digit that I would have to wait five seconds or thereabouts so that the pad-thing would reset itself, and then try again. (Trust me, this leads to the problem). So, in this situation, in order to type in every possible combination, I would need 40,000 keystrokes, since there are 10,000 codes, from 0000 to 9999, and each code has four digits. Correct me if I'm wrong, but I believe that this is the minimum number of keystokes to type in every combination.

What I noticed today is that I don't need to wait for the pad-thing to reset itself; I can simply start over without waiting. That is, if the code is 1234, and I type 51234, the door will unlock. As far as I can tell, this means that, as far as the pad-thing is concerned, typing 51234 is essentially the same as entering two codes, to wit, 5123 and 1234. The problem that I cannot solve is: What is the minimum number of keystrokes needed to enter all 10,000 codes if they can be chained up like this. And, if you would continue to humor my curiosity, could you describe what (one of) the minimal patterns would be?

Thank you for your attention to my question. 84.97.254.29 (talk) 23:48, 22 June 2009 (UTC)[reply]

You obviously need at least 10,003 keystrokes, and in fact this is sufficient. See De Bruijn sequence. Algebraist 23:57, 22 June 2009 (UTC)[reply]

That was fast. Thanks. 84.97.254.29 (talk) 00:05, 23 June 2009 (UTC)[reply]

I disagree -- but I may not have understood the question correctly.

My understanding of these machines is that they only store (or pay attention to) the last four digits entered. When you were successful by entering 51234, you did NOT test 5123; rather, the 5 was simply discarded.

(Further, if you know that you've made a keying error before you push enter, you need not wait any time at all. Just continue entering digits; as long as the last four entered are correct, you're in!)

The devices are far less complex than you may be giving them credit for. --DaHorsesMouth (talk) 22:39, 23 June 2009 (UTC)[reply]

There isn't always an "enter" key - when I set or unset the alarm system in my house I just have to press the correct 4 digits. AndrewWTaylor (talk) 22:55, 23 June 2009 (UTC)[reply]

You're wrong, DaHorsesMouth. OP talks about system with no Enter key, so if he enters 51234, he tests 5123 (and the test fails), then enters 4 which causes dropping the first digit, appending 4 at the end and testing a new 1234 code. Consider it as a 4-position shift-register, filled with keystrokes, which tests itself against a pre-set value each time a new digit is entered. CiaPan (talk) 05:36, 24 June 2009 (UTC)[reply]

June 23

Limits

Hi. I haven't done much work on limits and so need a bit of help with this problem.

Given that $\lim _{t\to \infty }e^{-mt}t^{n}=0$ where m is a positive integer and n is a non-negative integer, show that $\lim _{x\to 0}x^{m}(lnx)^{n}=0$ .

Do you just make the substitution $t=-lnx$ ? Thanks 92.7.54.6 (talk) 16:04, 23 June 2009 (UTC)[reply]

Yes, that's one way to look at it, and the method the problem seems to be setting you up for.Ray^Talk 17:25, 23 June 2009 (UTC)[reply]

So do you just make the substitution and then divide by (-1)^n?. On a related note I had to determine

\lim _{x\to 0}f(x)

where

f(x)={\frac {x}{e^{x}-1}}

. Am I correct to say this is 1? After that I had to determine

\lim _{x\to 0}f'(x)

, which I got as -1 (using the approximation

e^{x}\approx x+1

) but then when I graphed it, this was wrong. Any ideas?

You can do that one with L'Hôpital's rule (I'm not a fan of the rule, but I think it is the best method for that limit) and the answer is, indeed, 1. --Tango (talk) 17:50, 23 June 2009 (UTC)[reply]

I've never learnt any rules for limits. I'm working solely with algebraic skills and approximations for small x. 92.7.54.6 (talk) 17:54, 23 June 2009 (UTC)[reply]

For the first question - yes, you should divide by

(-1)^{n}

, which is just a nonzero constant and thus does not change the zero limit.

For the second - the approximation

e^{x}\approx 1+x

is just not accurate enough for calculating

\lim _{x\to 0}f'(x)

here. Using

e^{x}\approx 1+x+{\frac {x^{2}}{2}}

will give the correct result,

-1/2

.

As you may have noticed, knowing when is an approximation good enough is not always trivial. That's why rules like L'Hôpital's are useful. I think proving this limit without it is quite a challenge. -- Meni Rosenfeld (talk) 18:04, 23 June 2009 (UTC)[reply]

Do you only know it's not good enough because you used L'Hôpital's rule first? So was it only luck that I got the limit of f(x) right then? If you used the terms in x^3 or higher powers in the approximation of e^x, would you not get a different answer because you've become more accurate? 92.7.54.6 (talk) 18:08, 23 June 2009 (UTC)[reply]

A simpler questions perhaps; I hadn't heard of L'Hôpital's rule before this thread and looking through the page isn't helping me see how to apply it here. Could someone show me how it is used for the limits of f(x) and f'(x) please? Thanks. 92.7.54.6 (talk) 18:51, 23 June 2009 (UTC)[reply]

I know it's not good enough because it's "just enough" for

\lim _{x\to 0}f(x)

(if you try using

e^{x}\approx 1

you'll obviously fail), and thus you'll need more for

\lim _{x\to 0}f'(x)

(since taking a derivative amplifies errors). Another way is to add a hypothetical term and see if it changes the result (e.g., if you take

e^{x}\approx 1+x+\alpha x^{2}

you'll see that the result depends on

\alpha

). In a more formal proof you would probably use a remainder term with a Big O notation. But if you don't want this kind of sophisticated thinking then yes, you can only know an approximation is not good enough if you see it gives a different result from an exact method (be that using L'Hôpital's rule or otherwise).

So in a sense, yes, using it for the first limit without giving a justification was "lucky".

Using more terms than necessary will not give a different result, because we are taking the limit at

x\to 0

. These higher order terms would only make a contribution which is proportional to x and thus vanishes. If you try it for simpler cases, like

\lim _{x\to 0}{\frac {e^{x}-1-x}{x^{2}}}

you'll see what I mean.

As for using L'Hôpital's rule - for the first case, express f as a ratio

f(x)={\frac {g(x)}{h(x)}}

where

g(x)=x

and

h(x)=e^{x}-1

. Then verify that these functions vanish at the limit:

\lim _{x\to 0}g(x)=\lim _{x\to 0}x=0

,

\lim _{x\to 0}h(x)=\lim _{x\to 0}e^{x}-1=0

. Then calculate

g'(x)=1

and

h'(x)=e^{x}

, and finally

\lim _{x\to 0}f(x)=\lim _{x\to 0}{\frac {g(x)}{h(x)}}=\lim _{x\to 0}{\frac {g'(x)}{h'(x)}}=\lim _{x\to 0}{\frac {1}{e^{x}}}=1

.

For the second case you will first need to find an expression for

f'(x)

. -- Meni Rosenfeld (talk) 20:51, 23 June 2009 (UTC)[reply]

Enumerating Rationals

Hi, I am working on qualifying exam problems, so I may start asking a lot of analysis type questions, mostly real analysis. For now, I need to construct a bijection from the natural numbers onto the rationals... in other words, I need to construct a sequence of all rationals without repeating. That's easy. But, I need more. Say g(k) is this function. I need an enumeration such that

$\sum _{k=1}^{\infty }g(k)^{k}$

diverges. And, I also need to know if there exists any such enumeration such that it converges.

My first thought was to just try the obvious enumeration and see if I get anywhere: 0, 1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5... . But, then I could not show that series converged or diverged. I tried root test but you get a limsup of 1 which is inconclusive. Ratio test, limit must exist... or you can use the one in Baby Rudin but it does not work either. I thought about switching it up a bit to make it larger... just for each denominator, write the fractions in decreasing order instead of increasing. But, same thing. Any ideas? Thanks. StatisticsMan (talk) 19:50, 23 June 2009 (UTC)[reply]

Why does the sequence need to converge? DJ Clayworth (talk) 19:52, 23 June 2009 (UTC)[reply]

If you assume that the sequence is a bijection, there's no way it converges. The terms do not go to 0, as after any finite number of steps, there are always infinitely many that are greater than any arbitrary constant. Ray^Talk 20:04, 23 June 2009 (UTC)[reply]

Do you want all rationals or just those in [0,1]? If the former, then what Ray said. But if the latter, you should be able to construct

g(k)

so that

g(k)^{k}<=1/k^{2}\,\!

(since

\lim _{k\to \infty }(1/k^{2})^{1/k}=1

). -- Meni Rosenfeld (talk) 21:01, 23 June 2009 (UTC)[reply]

That is, let

h(k)

be any enumeration of these rationals. Define

g(1)=1

g(k)=

The first (according to h) number which is at most

(1/k^{2})^{1/k}

and has not yet been picked.

Since you have arbitrarily small numbers, this is well-defined. Furthermore, for every

x\in \mathbb {Q} \cap [0,1)

, there is some minimal

K

such that

(1/K^{2})^{1/K}\geq x

. If

x=h(n)

, then

x=g(k)

for some

K\leq k\leq K+n

. -- Meni Rosenfeld (talk) 21:12, 23 June 2009 (UTC)[reply]

For one that diverges, use

g(2m-1)=

The first number which was not yet picked

g(2m)=

The first number which is greater than

(1/2)^{1/(2m)}\,\!

and was not yet picked

-- Meni Rosenfeld (talk) 21:25, 23 June 2009 (UTC)[reply]

Given your study objective, I think it is best to spend quite a long time thinking about this sort of problem before seeking help. There doesn't seem to be any technical machinery involved; it's just a matter of sharpening your ability of attacking such problems with elementary methods. 208.70.31.186 (talk) 23:13, 23 June 2009 (UTC)[reply]

Yea, sorry, the point is the rationals in [0, 1]. I think I get what you're saying Meni. Thanks for the help. As far as the last comment, I disagree. I have only limited time. I do not think I would have ever figured this out. But, now that I have seen a probable solution, and now that I will go through and make sure it makes sense to me and write it up in detail, then I will understand it and have a bit of intuition I didn't have before. If I can do this with lots of problems, then I built lots of little bits of intuition, as well as much technique and theory. As far as spending hours on a problem, I just spent the last two semesters doing that on most homework assignments for real analysis. Once we, often as a group, figured out a problem, in the writing up process is when I actually learned things. Thinking about a problem for hours and not getting anywhere is not helpful to me. I am trying to know how to do as many problems as possible before the test. That will be very helpful. StatisticsMan (talk) 02:26, 24 June 2009 (UTC)[reply]

Your dilemma is understandable. It is often the case that exams, although they test one's ability, are often not very useful. In particular, I think that one should not sacrifice a good understanding for good grades (if I have a perfect understanding and a D grade, I would be happier than if I had a poor understanding and an A grade), but I am not the one in your position! I think that you should attempt a problem for around 30 minutes, and if you cannot solve it, seek help. Often, reading something (such as a solution) does not aid in one's understanding. For instance, if I were memorizing some words from the dictionary, unless I write them down, I would probably forget the words in due time; even if I had looked at the word for at least 30 minutes. Therefore, if you at least attempt a problem (and then ask for the solution), not only will it help your memory (you will not go blank in the exam), but it will also, if your attempt is successful, provide you with another approach to a problem that we may not be able to give you. I also think that successfully solving many problems before an exam boosts your confidence as well as helps your intuition more than a mere solution we may be able to provide you. However, I am probably not the one to tell you what to do, so you should probably continue what you feel as best preparation for the exam. --PS T 10:33, 24 June 2009 (UTC)[reply]

Yea, I agree with you actually. I think it is good for me to try problems for a while. I think it bad for me to get stuck on one problem for hours. So there has to be a balance. We have a study group set up at our school and we basically break up the exams and do our part and then present solutions to each other. I learn a bit but not a huge amount from someone else presenting, unless it's a subject I understand pretty well already. But, in taking their description and rewriting up the whole solution, making sure I understand every detail, then I think I learn a lot. Plus, I am doing some problems on my own. And, I will probably reread the solutions at some point as well to refresh myself on various details. Perhaps, later, I will try to redo some of the problems without looking at the solutions again. StatisticsMan (talk) 00:02, 25 June 2009 (UTC)[reply]

Numerically, the sum for the standard enumeration (the one you gave in the original post) seems to converge to 1.83019344534825... . To prove that it converges I would try to find an upper bound on

d(k)

, the denominator of

g(k)

. Since

g(k)<1-1/d(k)

(for

k>2

), this should allow you to bound the sum by a convergent sum. -- Meni Rosenfeld (talk) 11:50, 24 June 2009 (UTC)[reply]

Inequality

How would you go about showing that $e^{t}(1-t)$ ≤1? I don't know where to start. Thanks Scanning time (talk) 20:24, 23 June 2009 (UTC)[reply]

Never mind - just found the single turning point at t=0 and then found the value of the second derivative at this point, which is negative so it must be a maximum. Scanning time (talk) 20:28, 23 June 2009 (UTC)[reply]

Right. It may be of interest for you to recall that for all real x the sequence (1+x/n)ⁿ converges to e^x and it is increasing as soon 1+x/n ≥ 0, that is, as soon as n ≥ -x. This gives you a lot of inequalities, for instance, yours 1-t ≤ e^-t (that of course also holds for all t>1) --pma (talk) 09:58, 24 June 2009 (UTC)[reply]

You can just start the inequality

e^{x}\geq 1+x

, substitute

x=-t

to get

e^{-t}\geq 1-t

, then divide by

e^{-t}

(which is always positive) to get

1\geq e^{t}(1-t)

. --76.91.63.71 (talk) 18:14, 24 June 2009 (UTC)[reply]

Empty mean

I'm reading a programming book in which the author gave an example method which takes an array of doubles and returns their mean. If the array is empty, the function returns 0, which the author explains with the comment "average of 0 is 0". This annoyed me quite a bit, since the author is obviously confusing the mean of no terms with the mean of a single zero term. IMO the function should return NaN (or throw an exception), since an empty mean is 0/0.

Does anyone have any further input on the matter? More importantly, is anyone aware of an online resource discussing this (my search in Wikipedia and Google didn't come up with anything)? Thanks. -- Meni Rosenfeld (talk) 22:59, 23 June 2009 (UTC)[reply]

The standard definition certainly breaks down and gives 0/0. There might be a standard convention for how to define it (similar to the conventions for various other empty things), but I don't know what it is and can't think of anything obvious. I think it should probably be left undefined. Consider the average of n 5's. That's obviously 5 for all n>0, so should be 5 for n=0. Similarly the average of n 3's should be 3 for n=0, but no threes and no fives are the same thing. --Tango (talk) 23:07, 23 June 2009 (UTC)[reply]

Yes, saying that the mean of an empty list is 0 certainly sounds wrong. That the sum of an empty list is 0 is standard, but the length is also 0, and 0/0 = NaN. 208.70.31.186 (talk) 23:21, 23 June 2009 (UTC)[reply]

The mean is not just a linear combination of the values; it's an affine combination, i.e. a linear combination in which the sum of the coefficients is 1. As John Baez likes to say, an affine space is a space that has forgotten its origin. It has no "zero". The point is that if you decide, e.g., that the zero point is located HERE, and measure the locations of the points you're averaging relative to that, and I decide the zero point is located somewhere else, and likewise compute the average, then we get the same result! That's not true of sums, and it's not true of linear combinations in which the sum of the coefficients is anything other than 1. If the "empty mean" were "0", then there would have to be some special point called "0". Michael Hardy (talk) 23:30, 23 June 2009 (UTC)[reply]
...Just for concreteness: Say we're measuring heights above sea level, and you get 0, 1, and 5 (after measuring the heights at high tide). Your average is (0 + 1 + 5)/3 = 2 feet above sea level. I measure the heights at low tide when the water is 4 feet lower, so I get 4, 5, and 9. My average is (4 + 5 + 9)/3 = 6 feet above sea level. You got 2 feet. I got 6 feet, when the water is 4 feet lower. So we BOTH got the same average height. But if it were correct to say the average of 0 numbers is 0, then should that 0 be your sea level, or mine? There's no non-arbitrary answer. So it doesn't make sense to say the "empty average" is 0. Michael Hardy (talk) 23:36, 23 June 2009 (UTC)[reply]

I completely agree that the mean value of zero observations is generally undefined. Consider however that you want to estimate a probability by a relative frequency. The probability of success is (the limiting value of) the relative frequency of success for a large number of trials. P = lim i/n. Even if the number of trials is not large it is tempting (and very common) to estimate the unknown probability by the relative frequency: P ~ i/n. This is the mean value of a multiset of i ones (for success) and n−i zeroes (for failure). The case n=i=0 gives P ~ 0/0 which is undefined. However, we know better than that: we know that 0 ≤ P ≤ 1, and that is all we know. So the likelihood function f is f(P)=1 for 0 ≤ P ≤ 1 and f(P)=0 elsewhere. This function has mean value 1/2 and standard deviation about 0.29. So the mean value of zero (0,1)-observations is 1/2. Generally the mean value of the likelihood function, (which is a beta distribution), is (i+1)/(n+2), rather than i/n. The estimate P ~ i/n is a tempting (and very common) error. Note that for large number of observations it doesn't matter: lim (i+1)/(n+2) = lim i/n. Bo Jacoby (talk) 07:01, 24 June 2009 (UTC).[reply]

I think this is equivalent to saying that if we want to use a sample to estimate the population mean, then with no data this will be just the mean of our prior. While certainly worth noting, it is inconsequential with regards to calculating the sample mean. -- Meni Rosenfeld (talk) 08:55, 24 June 2009 (UTC)[reply]

Yes Meni. It often goes without saying that the purpose of computing a sample mean is to estimate the population mean. But the mean of the posterior is not always equal to the mean of the sample. And the mean of the prior probability can never be computed by the undefined mean of the empty sample. Bo Jacoby (talk) 09:22, 24 June 2009 (UTC).[reply]

Ok, thanks for the answers. Looks like we all agree, with different ways to look at it. Perhaps someone will be interested in adding a note about this to one of our articles (personally I've stopped editing articles)? -- Meni Rosenfeld (talk) 08:55, 24 June 2009 (UTC)[reply]

I got the impression that wikipedians have a peculiar taste for trivial objects and trivial cases in mathematics. I remember interesting threads about

\scriptstyle \inf \emptyset

and

\scriptstyle \sup \emptyset

; empty sums and products; empty limits and co-limits in category theory as well as terminal and initial objects, and their ubiquity; the empty function and empty permutation; 0⁰, 0!,

\scriptstyle {0 \choose 0}

; 0-th order iterations of several constructions; derivatives and antiderivatives of order 0; orientations of a zero-dimensional manifold; 0 as a limit ordinal,... and what I'm forgetting? Certainly the empty average deserves a place among them. Moreover, these topics seem to be recurrent here, which is maybe a good enough reason for writing a short article, e.g. Mathematical trivialogy. --pma (talk) 11:12, 24 June 2009 (UTC)[reply]

Are you sure it's only Wikipedians? I am of the opinion that

Properly acknowledging the trivial\empty\limit cases can go a long way towards simplifying notation, definitions and proofs, and
If someone doesn't understand these cases, they don't truly understand the general case either.

This being the case, it is healthy for any mathematician to ponder these matters, and I bet most do. -- Meni Rosenfeld (talk) 11:41, 24 June 2009 (UTC)[reply]

Indeed. I have even seen respect for trivial cases described (in the context of history of mathematics) as one of the characteristic features of the modern trend towards generalization and abstraction. Algebraist 11:49, 24 June 2009 (UTC)[reply]

Hey... I have the feeling that you may have taken my preceding post as hyronic. It was definitely not. I am really thinking to this article as a possibly nice and useful reference, for reasons that you have already expressed (foundational, hystorical, didactic). What maybe causes the misunderstanding is the use of the word "trivial", that sounds reductive, and does not make justice to these fundamental concepts in mathematics (as usual, the most basic and simple ideas came the last). There is one further reason I think. The aesthetic taste, humor included, for them, is one of those things that make a difference between mathematicians and the rest of the world, and one of the most difficult to communicate --I guess, because the elegance of ideas in mathematics relies mostly on how they work, which is of course something that a working mathematician feels better (how to explain to a non-mathematician the comfort of constructing n-spheres starting form the 0-sphere, or the ordinals from the empty set? &c)--pma (talk) 14:23, 24 June 2009 (UTC)[reply]

I understood that you were being serious - the thing is that your usage of words like "Wikipedians", "peculiar", "recurrent here" seemed to suggest that you think dealing with trivial cases is neither common nor particularly desirable. -- Meni Rosenfeld (talk) 19:21, 24 June 2009 (UTC)[reply]

Maybe I should have said we wikipedians --well, I did not for shyness. Yes, I do not see it as a common habit, deserving particular attention to trivial cases in the discussion of definitions and theorems. Though it is desirable for the sake of complete understanding, as we said, and I myself partecipate with pleasure to these debeats. That said, there is also a kind of peculiar humoristic aspect in looking at zero-cases, that I suppose you feel too. --pma (talk) 08:16, 25 June 2009 (UTC) [reply]

Taking the mean of an empty set could be an arbitrary convention, and it is possible that there are circumstances where this would be useful. On the other hand I think it is more likely to arise as a result of a fundamental difficulty which many (perhaps most) people have in connection many ideas related to zero and empty sets. I have had considerable difficulty in trying to convey to non-mathematical people that a value not existing is not the same as the value existing and having the value zero (for example, how many daughters has the king of Germany?). Likewise conveying the distinction between an empty set and no set. A final example is people who are convinced that 12÷0 must be 0. It seems that to many people it is very difficult to conceive of anything involving 0 which produces anything other than 0. On the other hand it once took me a considerable time to persuade someone (of fairly average intelligence in most respects) that 6×0 was 0. He said "if you start with 6 and then do nothing you must still have 6". I eventually persuaded him that it was possible to meaningfully have an operation involving 0 which did not mean "do nothing", but it was hard work. Another example is people who feel 0!=1 is completely unnatural. Zero is a surprisingly confusing and difficult concept for many people. JamesBWatson (talk) 11:47, 24 June 2009 (UTC)[reply]

(indentation confusion: this is to JamesBWatson) In some senses, your question about Germany has to have the answer 0, because the set {girls having a parent that is king of Germany} has no elements. I'm not sure why (aside from having an easier/less-ambiguous formulation in terms of set theory) that question is answerable when "How old is the king of Germany?" isn't. --Tardis (talk) 14:40, 24 June 2009 (UTC)[reply]

Because you've twisted the question. The question was not 'how many girls are there whose parent is a king of Germany' but rather 'how many daughters does the king of Germany have'. The former makes sense if there is no king of Germany (or indeed if there are many), but the latter (by its use of a definite article) presupposes that there is exactly one king of Germany. Algebraist 14:49, 24 June 2009 (UTC)[reply]

Sure, the mean is not really 0 if you have no values, but you have to start from some number that's not NaN or else the recursion algorithm to compute the mean won't work. Zero will do just like any other number. – b_jonas 05:08, 25 June 2009 (UTC)[reply]

A difficulty leaning mathematics is that the meaning of the word "many" changes from signifying three or more to signifying zero or more. Asked "how many?" a mathematician may answer "zero" while a nonmathematician answers "no, not many, not even one". To the nonmathematician "empty set" and "number zero" is nonsense because empty is not a set and zero is not a number, as you don't count to zero. Redefining common words must be carefully explained. Bo Jacoby (talk) 06:48, 25 June 2009 (UTC).[reply]

June 24

Normal distribution curves from two values.

Is it possible to create a normal distribution curve if you only know, say, two of the standard deviation values are?

For example, let's say I know two values, X and Y, and I also know that between these two values 95% of the other values exist; is it possible to figure out what the rest of the standard deviations would be for the graph, and other information? I would thing it would be, but perhaps I'm wrong.

On a side note, does anyone know if there's a way of getting Excel to generate Normal distribution curves? It's been like 5 years since I've had to even think about drawing them and the wikipedia article is confusing me. (I know that's more of a computer sciences question, but I figure Mathematicians probably know the answer to that question as Excel seems very math/statistics oriented)

Thanks in advanced for your help! --Honeymane_{Heghlu meH QaQ jajvam} 03:27, 24 June 2009 (UTC)[reply]

Can't help you with the Excel, but the answer for your question is no. The problem is that a normal distribution is characterized by two parameters (the mean and the variance), and your information only supplies one equation characterizing the relationship between the two (specifically, that 95% of the distribution lies between X and Y). Thus, there are infinitely many possible normal distributions meeting your criteria. You need an additional piece of information. Ray^Talk 05:08, 24 June 2009 (UTC)[reply]

So it's not possible to derive the mean from the part of the standard deviation? --Honeymane_{Heghlu meH QaQ jajvam} 05:35, 24 June 2009 (UTC)[reply]

Well, in your example, the mean will be around midway between X and Y, you just won't be able to figure out exactly where it is between those values. That assumes your "95%" means "exactly 95%" and not "at least 95%". 208.70.31.186 (talk) 06:39, 24 June 2009 (UTC)[reply]

No, that's wrong. There's no reason to think the mean would be half-way between the two values. For example, one of them could be the 99th percentile of the distribution, and the other the 4th percentile. Or one could be the 99.99th percentile (a substantially larger number) and the other the 4.99th percentile---that still adds up to 95% between the two. That's why the answer is not unique. Michael Hardy (talk) 20:49, 24 June 2009 (UTC)[reply]

The anon said around midway but you can't be exact, which is perfectly accurate. The mean is going to be reasonably close to the midpoint. --Tango (talk) 21:04, 24 June 2009 (UTC)[reply]

No, it's not accurate at all. If the upper percentile is close to the 100th percentile (e.g., what if it's the 99.999999999th percentile) then the it will be a very very large number, whereas the lower one will be close to the 5th percentile, not so large at all. The mean will be far closer to the lower endpoint than to the upper one. Michael Hardy (talk) 05:57, 25 June 2009 (UTC)[reply]

There's no reason to believe that unless we know that the distribution has reasonably high variance. If all you know about a normal distribution is that exactly 95% of its mass is between a and b, then the mean could be anywhere in (a,b). Algebraist 21:07, 24 June 2009 (UTC)[reply]

To the OP: Your way of using words is so non-standard as to render your first sentence incomprehensible. Standard deviation is a precisely defined term, and what it means is not at all what you seem to mean. A normal distribution cannot have two different standard deviations; it has only one. Michael Hardy (talk) 20:51, 24 June 2009 (UTC)[reply]

Just for concreteness, for the standard normal distribution, 95% of the probability is between 2.3263 and −1.7507, but also 95% is between 3.719 and −1.6458. In the second case, you certainly don't have the mean halfway between the two, since the mean is 0. Michael Hardy (talk) 20:54, 24 June 2009 (UTC)[reply]

Perhaps I can rephrase then. Is it possible to figure out what the other 'values' in a normal distribution curve are, if you only have access to a limited amount of information? Take this diagram for example, if I know that -2σ is equal to 5, and 2σ is equal to 5, can I figure out what the mean (μ) and Standard Deviation would be?

To use the example from the SD article, if I told you that 95% of adult men are between 64 and 76 inches in height, is it possible to figure out that the mean value would be 70 inches with a mean of 3 inches? Given Ray's answer, I'm guessing the answer is no.

I realize I may not be using standard math language, it's never been my strong suit.--Honeymane_{Heghlu meH QaQ jajvam} 02:00, 25 June 2009 (UTC)[reply]

A normal distribution is determined by two values, the mean and the standard deviation. That means you cannot uniquely determine a normal distribution with less than two pieces of information. That 95% of the population are in a certain range is one piece of information, so it is not enough. The other example you give is different information, and is contradictory - -2σ and 2σ cannot both equal 5. That diagram is rather poor, anyway, the labels on the x-axis should be "μ+2σ", etc., not just "2σ". If you know that, say, μ+2σ=5 and μ-2σ=-5 then you can just solve the simultaneous equations to get the mean (μ) and standard deviation (σ). Note that there are two equations there, that is two pieces of information so it enough to uniquely determine the normal distribution. (The two pieces of information need to be independent of each other, if one implies the other then you only really have one piece of information.) --Tango (talk) 02:28, 25 June 2009 (UTC)[reply]

To graph a normal distribution in Excel:

Put -4 in A2, -3.9 in A3 and autofill the column to 4 in A82.
Put =EXP(-A2*A2/2)/SQRT(2*PI()) in B2 and autofill the column to B82
Highlight region from the A2 to B82 and click on the chart wizard

--RDBury (talk) 04:46, 25 June 2009 (UTC)[reply]

Bezier curve with width

I want to draw a bezier curve with an arbitrary thickness. How is this done?

The original curve has one control point (making it quadratic, I guess, although I barely know what that means). My idea is to draw two curves either side of it and fill between them. It's easy enough (if I'm doing it right) to find the offset start and end points for the two side-curves, but how can I find suitable control points for them? Offsetting the original control point by the desired width produces curves which are pinched in the middle.

Here's screenshot of this happening. (You have to ignore that the surface is tilted, but I think you can see what's going on.)

I thought of measuring the distance between the start and end points of each of the side-curves, comparing that with the same distance on the original curve, and using the result to scale the offset of the control points. That's just a guess, though, and I think it wouldn't work. 213.122.47.84 (talk) 09:56, 24 June 2009 (UTC)[reply]

The offset curve of a (nontrivial) Bézier curve is not a Bézier curve itself, so you cannot do it in such a way at all. If by "drawing" you mean that you are trying to rasterize the curve, then a simple solution is to use whatever method you used to plot the original curve, except that you draw a (filled) circle instead of each pixel. — Emil J. 12:12, 24 June 2009 (UTC)[reply]

I was afraid so. Yes, I mean rasterize (I nearly posted this to the computing desk), and yes that's simple, but highly inefficient. Thanks for the link. I wonder what to do. Use cubic beziers, maybe. 81.131.10.72 (talk) 12:20, 24 June 2009 (UTC)[reply]

Try doing a google with 'bezier width algorithm' and you'll get a number of ways. Most systems will allow you to just specify the width and they'll do something. If you're doing this yourself you need to decide what you mean by a bezier curve with width and whether you want an ideal or real solution. For instance one definition might be what you get if you move a disc along the line - but then you get round ends. Another might be that you move a segment along the line always at right angle to the line. Dmcq (talk) 12:18, 24 June 2009 (UTC)[reply]

I need a solution similar to this: http://alienryderflex.com/polyspline/ ...which does something magical-seeming (to me) with quadratics to provide the answer for where a spline intersects a scanline. By this means I can fill between curves efficiently (not considering any pixel more than once) and perfectly (not approximating any part of the curve with a line). I would settle for looking right over being exactly correct, though. 81.131.10.72 (talk) 12:34, 24 June 2009 (UTC)[reply]

I think the usual procedure with drawing a Bezier curve is to keep subdividing it until each segment is effectively a line. In other words, if the control points are within 1 pixel of the line between the endpoints then you can assume that segment is a line and apply a line generating algorithm. A line with thickness is just a rectangle so if you want your algorithm to draw a thick Bezier curve then draw rectangles instead of lines in the last step.--RDBury (talk) 04:23, 25 June 2009 (UTC)[reply]

Find an array of numbers, max_i and max_j being given

Hi. I need to find a nXm array (in fact n≤m≤8) of positive numbers x_ij, with max_jx_ij = a_i and max_ix_ij = b_j for all i and j. The positive numbers a_i and b_j are given and max_ia_i = max_jb_j=m. Apparently there are always such arrays, yes? Have you a recepit for it? Thanxx --84.221.68.243 (talk) 16:06, 24 June 2009 (UTC)[reply]

How about

x_{ij}:=a_{i}+b_{j}-m

--84.221.209.203 (talk) 17:54, 24 June 2009 (UTC)[reply]

Isn't

x_{ij}=\min(a_{i},b_{j})

the correct answer (or at least one of correct answers)? The answer may be not unique. Let's take e.g. m=n=2 and all a=b=2. Then

{\begin{bmatrix}2&p\\q&2\end{bmatrix}}

will be a solution for any positive p and q not exceeding 2, as well as

{\begin{bmatrix}p&2\\2&q\end{bmatrix}}

. --CiaPan (talk) 20:16, 24 June 2009 (UTC)[reply]

poker odds

What are the odds that in Texas Holdem poker with 10 players that 2 or more players effectivly have the same initial hand?

S=spade C=club H=heart D=Diamond.

Example,

10D 10H is effectivly the same as 10S 10C

8S 9S is effectivly 8H 9H, but not 8C 9 D 65.121.141.34 (talk) 16:49, 24 June 2009 (UTC)[reply]

A rough approximation is easy: compute the approximate odds that two hands are effectively the same, then use a birthday paradox calculation to find the likelihood of that occuring in the 45 possible pairings of 10 players. Getting everything exactly right sounds messy and if you're mostly interested in the actual number that comes out the problem, it may be easiest to run a computer simulation with a few million random deals and count how many times those ties occur. 67.122.209.126 (talk) 19:39, 24 June 2009 (UTC)[reply]

[ec] I think the probability for two particular players is

{\frac {97}{20825}}\approx 0.466\%

. The calculation is complicated by the fact that the events are not independent for multiple players, but the final result may be close to 20%. -- Meni Rosenfeld (talk) 19:43, 24 June 2009 (UTC)[reply]

June 25

two questions

I have two questions 1:-The internal dimention of an open concrete tank are 1.89 m long,0.77 m wide and 1.23 m high. If the copncrete side walls are 0.1 m thick find in cubic metres ,the volume of concrete used.Give answer to 3 sigficant figures . answer on book is .704 cubic m . but i have solved it and find the answer 0.79989 by multiplying surface area with thickness.

2:-Two trains A and B , are scheduled to arrive at a station at certain time .The time is in  seconds after the scheduled time for each of the 40 days was recorded , mean and standard deviation  are as
A: mean 0.528  S.D  0.155
B: mean 0.498  S.D 0.167 , My question is which train is more consistent in arriving late .why ?

and which train is more punctual on the whole ?why?.