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February 4

matrix/vector algebra problem

I am reading a paper, and there is one part that I cannot follow.

Here we have an equation

${\displaystyle \alpha \log(h)=\alpha \epsilon _{i}-G+\log(p_{i})-(1-\alpha )\sum _{j=1}^{n}w_{ij}\log(p_{j})+(1-\alpha )H_{i}}$

where ${\displaystyle w_{ij}}$ are the entries of the ${\displaystyle n\times n}$ stochastic matrix ${\displaystyle W_{n}}$, ${\displaystyle G}$ is a constant, ${\displaystyle \alpha }$ is a constant between 0 and 1, and ${\displaystyle H_{i}\equiv \sum _{j=1}^{n}w_{ij}\log(w_{ij})}$

The above equation can be written in vector form, which I assume for ${\displaystyle n=2}$ will look like:

${\displaystyle {\begin{bmatrix}\alpha \log(h)\\\alpha \log(h)\end{bmatrix}}={\begin{bmatrix}\alpha \epsilon _{1}\\\alpha \epsilon _{2}\end{bmatrix}}+{\begin{bmatrix}\log(p_{1})\\\log(p_{2})\end{bmatrix}}+{\begin{bmatrix}-G\\-G\end{bmatrix}}+{\begin{bmatrix}-(1-\alpha )(w_{11}\log(p_{1})+w_{12}\log(p_{2}))\\-(1-\alpha )(w_{21}\log(p_{1})+w_{22}\log(p_{2}))\end{bmatrix}}+{\begin{bmatrix}(1-\alpha )H_{1}\\(1-\alpha )H_{2}\end{bmatrix}}}$

Now a row vector ${\displaystyle v'_{n}}$ is introduced where ${\displaystyle v'_{n}={\tfrac {\alpha }{n}}{\begin{bmatrix}1&1\end{bmatrix}}^{-1}}$

The entries of this row vector sum to one.

Both sides of the equation are then premultiplied by this vector which apparently gives ${\displaystyle \log(h)=v'_{n}\epsilon +{\tfrac {1}{n}}\sum _{i=1}^{n}\log(p_{i})-{\frac {G}{\alpha }}+{\frac {1-\alpha }{\alpha }}v'_{n}H}$

where ${\displaystyle H='}$

Now I cannot figure out how they got this equation from multiplying both sides by ${\displaystyle v'_{n}}$. Seemingly both sides have also been divided by ${\displaystyle \alpha }$, so I can understand what happened with the left hand side, and the ${\displaystyle \epsilon ,\;G}$ and ${\displaystyle H}$ terms. But I cannot understand what happened to the ${\displaystyle \log(p_{i})}$ and ${\displaystyle \log(p_{j})}$ terms, so that nothing else was left except ${\displaystyle {\tfrac {1}{n}}\sum _{i=1}^{n}\log(p_{i})}$

The paper does not explain this. Does it follow from the information given? —Preceding unsigned comment added by 130.102.78.164 (talk) 04:44, 4 February 2011 (UTC)

Yes, it's clearly designed to. Let P be the (column) vector of log(p_i)'s. Then in the original equation, the terms ${\displaystyle \textstyle \log(p_{i})-(1-\alpha )\sum _{j=1}^{n}w_{ij}\log(p_{j})}$ give the elements of the vector ${\displaystyle P-(1-\alpha )WP=(I-(1-\alpha )W)P}$. The left factor of this cancels out with ${\displaystyle ^{-1}}$ in the defintion of v', so what is left after you multiply is ${\displaystyle {\frac {\alpha }{n}}P}$. –Henning Makholm (talk) 09:50, 4 February 2011 (UTC)

Thanks for that, but unfortunately no sooner had I continued reading that I got stuck on the next line...

Again I have this vector ${\displaystyle v'_{n}={\tfrac {\alpha }{n}}{\begin{bmatrix}1&1\end{bmatrix}}^{-1}}$

Then I am told that the equation

${\displaystyle s_{i}=h/n+(1-\alpha )\sum _{j=1}^{n}s_{j}w_{ji}}$

implies that

${\displaystyle s'=(h/n){\begin{bmatrix}1&1\end{bmatrix}}^{-1}=(h/\alpha )v'_{n}}$

Now I am trying to get this myself, but not sure how...

I see that the first equation implies something like

${\displaystyle s=h/n{\begin{bmatrix}1\\1\end{bmatrix}}}$

Then do I premultiply both sides giving

${\displaystyle s=^{-1}h/n{\begin{bmatrix}1\\1\end{bmatrix}}}$

What is the next step? I forgot all my rules about matrix algebra... —Preceding unsigned comment added by 130.102.158.15 (talk) 03:03, 5 February 2011 (UTC)

Note the position of the indices in ${\displaystyle \Sigma _{j}s_{j}w_{ji}}$. Since you're now summing over the first index of w, you have a row vector multiplied by W on the right, so your defining equation for s' is

${\displaystyle s'=(h/n)+(1-\alpha )s'W}$

and then when you solve for s' you automatically get the ${\displaystyle ^{-1}}$ on the right side of the . Remember that ${\displaystyle 1-\alpha }$ and ${\displaystyle h/n}$ are scalars and therefore commute with everything. –Henning Makholm (talk) 09:17, 5 February 2011 (UTC)

ah, I was stupid, I didnt notice that s could just be a row vector to start with... thanks again. —Preceding unsigned comment added by 118.208.112.152 (talk) 09:25, 6 February 2011 (UTC)

Love heart

The graph of

${\displaystyle x^{2}+(y-{\sqrt{x^{2}}})^{2}=1}$

looks like a love heart. Is there a more proper name for this curve, or for a general family to which it belongs? Thanks. (For reference: WolframAlpha.) —Anonymous Dissident 12:27, 4 February 2011 (UTC)

If you want to say "heart-shaped" and still sound like an intellectual, you can say "cardioid" :) But cardioid has a specific meaning which is not the same as your curve. Tinfoilcat (talk) 12:53, 4 February 2011 (UTC)

See also the MathWorld page.--RDBury (talk) 14:20, 4 February 2011 (UTC)

Finding range of 7+5cos5x divided by 8-6cos6x

Normally to find critical points we use dy/dx=0 then from critical points and the monotonicity of the function we try to find the range, but here, I am unable to find the critical points.

I got dy/dx =  150(sin5x)(cos6x) - 200sin5x - 252sin6x - 180(sin6x)(cos5x)  whole divided by  (8-6cos6x)^2
From here, I am unable to find the value of x for which it is 0

Can you please help me with this? — Preceding unsigned comment added by Krishnashyam1994 (talk • contribs) 15:11, 4 February 2011 (UTC)

I don't see why this can't be solved by taking the range of the numerator and denominator since both are cyclic and not synchronized with one another. The numerator has a range of . The denominator has a range . So, the function's range is = . Doing a quick plot, it does hit both of those extremes, but does not cross them. -- kainaw™ 15:25, 4 February 2011 (UTC)

y=6 is reached at x=0, but the low limit of 2/14 is not actually reached. For that to happen, cos(5x) and cos(6x) must both be -1, so 5x and 6x would both have to be π modulo 2π, so (6-5)x must be a multiple of 2π, which contradicts 5x being π module 2π. Furthermore, since 2π is obviously a period of the function, its range has to be closed, so its lower limit must be strictly larger than 2/14. –Henning Makholm (talk) 19:39, 4 February 2011 (UTC)

For the lower bound, you can only hope for a numerical solution. The actual value is a root of a polynomial of degree 10 (which cannot be resolved by radicals). Newtons method gives a value of 0.15292 for the lower bound. 166.137.140.202 (talk) 22:50, 4 February 2011 (UTC)

See http://www.wolframalpha.com/input/?i=(7%2B5cos5x)%2F(8-6cos6x) Bo Jacoby (talk) 10:40, 5 February 2011 (UTC).

Coin nominal values

Why is it not good to have both 20 cent and 25 cent coins in a single currency? --84.61.176.167 (talk) 15:33, 4 February 2011 (UTC)

This is a classic example of the greedy algorithm when it comes to making change. If you want to give 40 cents in change and you have standard American currency of 1, 5, 10, and 25 cents, the least number of coins will be 1x25, 2x10. If you add a 20 cent coin, the least number of coins will be 2x20 - which requires use of something other than the greedy algorithm. -- kainaw™ 15:36, 4 February 2011 (UTC)

You have to define what you mean by "good". You could have a coin in every denomination from 1 cent to 99 cents, and then you could always make change with a single coin. But that benefit would be outweighed by the problems of having 99 different types of coins to keep track of. At the other extreme you just have a 1 cent coin and it would take up to 99 to make change. There are other factors such as people find it easier to do arithmetic with multiples of 5 and 10. With various conflicting objectives, some of which are cultural, the question doesn't make sense mathematically. You could ask a more specific question like "Given there are to be 4 denominations, what should they be to minimize the average number of coins needed to make change?" Then you might then be able to show there is no solution that uses both a 20 and 25 cent coin.--RDBury (talk) 17:27, 4 February 2011 (UTC)

We need an article on coin nominal value. The simpler solution to the problem of designing currency value systems is to have only the denominations 1, 10, 100, 1000 and so on. Then the representation of a number in terms of coins and banknotes matches the decimal positional notation of the number. Intermediate denominations such as 2, 5, 20, 25, 50, 200, 250, 500 and so on complicates the system but also has a benefit in terms of carrying fewer coins and notes. I agree that this is not a mathematical question. Bo Jacoby (talk) 11:53, 5 February 2011 (UTC).

Trying to be mathematical about it, to get maximum coverage over a wide range of values, it seems that the ideal split is logarithmic ( X A's make a B, X B's make a C, X C's make a D ...). Your 1, 10, 100 system does that, but the jumps of ten are too big. Pay for a 0.99& candy bar with a 10& bill, and you get ten coins in change, nine of them all of the same denomination. Humans are better dealing with smaller numbers; three to four is about the limit we can handle intuitively. A binary system would be ideal for that (denominations 1, 2, 4, 8, 16, ...), but you also want to mesh with our base ten number system. You can't just alter the multiplier on the logarithmic system (10 is about 2, so we could have 1, 2.15, 4.64, 10, 21.54, 46.41, 100), because now you can't make change easily. A compromise is needed, so if you round the 10 logarithmic scale to even multiples of ten, you get 1, 2, 5, 10, 20, 50, 100 ... . I'll note that this matches the system used in most modern currencies such as Euro coins and Euro banknotes, as well as Coins of the Australian dollar, Banknotes of the Australian dollar, and even Banknotes of the pound sterling and US Federal Reserve Notes (although the $2 note is rarely used). The presence of the 0.25 coin is somewhat of an aberration to this trend in United States coinage (it's also present in UK coins, but unused in favor of the 0.20 coin), but that's historical, as the US Quarter is it is derived from two pieces of eight, which was the commonly used coinage in the US at one time. The 25p piece of the UK has a similar history, being the post-decimalisation equivalent of the pre-decimilisation crown. -- 174.24.195.38 (talk) 21:50, 5 February 2011 (UTC)

We do have a nice article on Preferred_number. Bo Jacoby (talk) 20:26, 6 February 2011 (UTC).

February 5

maximization

How do I do this maximization problem?

${\displaystyle \max _{c_{i}}{\frac {1}{n}}\sum _{i=1}^{n}\log(c_{i})}$

subject to

${\displaystyle \sum _{i=1}^{n}p_{i}c_{i}=h}$

Cannot use first order conditions here. But because it is maximizing over an average of a sum of logarithms, I can see that all the ${\displaystyle c_{i}}$ should be equal. The answer is ${\displaystyle c_{i}=h/(np_{i})}$ but I am unsure how to verify this. —Preceding unsigned comment added by 130.102.158.15 (talk) 01:32, 5 February 2011 (UTC)

Seems like a pretty straightforward application of Lagrange multipliers. I'd write more, but the article explains it better. Invrnc (talk) 04:04, 5 February 2011 (UTC)

The Arithmetic-geometric mean inequality shows that you have achieved the maximum. Sławomir Biały (talk) 13:23, 5 February 2011 (UTC)

gradient of scalar product

hi, any help with proving that grad (v_ . r_) = v_ using spherical polars, where v_ is a uniform vector field would be great it is trivial to prove using summation convention or Cartesian coordinates but having to use spherical polars looks messy... v_.r_ alone gives lots of terms containing sines and cosines of thetas and phis... i have read the wikipage on it, but perhaps i have not fully grasped the way it works or what the question is after... —Preceding unsigned comment added by 131.111.222.12 (talk) 02:17, 5 February 2011 (UTC)

Since it is indeed easy in Cartesian coordinates, what makes you want to do it in spherical? This looks like a prime candidate for the "convert to a frame where things are easy, then convert back again if necessary" strategy.

If this is homework, the point of the exercise may be to allow you to see for yourself how most of the horrible trigonometric terms cancel each other out once you differentiate. So carry on, and if you actually get stuck instead of just losing courage, then show some work here. –Henning Makholm (talk) 09:39, 5 February 2011 (UTC)

thanks, henning. i have proved it. —Preceding unsigned comment added by 131.111.222.12 (talk) 17:30, 5 February 2011 (UTC)

Multiplicative inverse function?

Hi. This is not a homework question. What is the term used to describe the following type of math done to a quadratic function: ${\displaystyle nx^{n-1}-mx^{m-1}}$? I think it may be a "multiplicative inverse", but this is done to find the highest possible value of the area, and what is a simplification thereof? Thanks. ~AH1 02:39, 5 February 2011 (UTC)

• Based on the context of maximizing, you might be thinking of the derivative, i.e., the derivative of ${\displaystyle x^{n}-x^{m}}$ with respect to x is ${\displaystyle nx^{n-1}-mx^{m-1}}$. --Kinu /_c 02:48, 5 February 2011 (UTC)

A calculus problem

Dear Wikipedians:

I am working on the a problem that seems to employ the fundamental theorem of calculus. The problem asks me to show that

${\displaystyle \int _{0}^{x}f(u)(x-u)du=\int _{0}^{x}\left(\int _{0}^{u}f(t)dt\right)du}$

given a continuous function f.

My basic approach is to let ${\displaystyle G(x)=\int _{0}^{x}f(u)(x-u)du}$ and ${\displaystyle F(x)=\int _{0}^{x}\left(\int _{0}^{u}f(t)dt\right)du}$ and to show that ${\displaystyle G''(x)=F''(x)}$.

For ${\displaystyle F(x)}$, we have:

${\displaystyle F'(x)=\int _{0}^{x}f(t)dt}$

${\displaystyle F''(x)=f(x)}$

However, for ${\displaystyle G(x)}$ I am encountering great difficulty in finding its second derivative. How do I show that ${\displaystyle G''(x)=f(x)}$?

Thanks,

70.29.26.56 (talk) 05:30, 5 February 2011 (UTC)

Let's try something:

${\displaystyle \int _{0}^{x}f(u)(x-u)\,du=\int _{0}^{x}f(u)\left(\int _{u}^{x}\,dt\right)\,du=\int _{0}^{x}\left(\int _{u}^{x}f(u)\,dt\right)\,du}$

The last step relies on the fact that ƒ(u) does not change as t goes from u to x, so ƒ(u) is a "constant" as a function of t. Then we can reverse the order of integration if we're careful about the bounds:

${\displaystyle \int _{0}^{x}\int _{0}^{t}f(u)\,du\,dt.}$

And that's the same thing as what you've got. Michael Hardy (talk) 07:11, 5 February 2011 (UTC)

Thanks Michael! Your solution is so beautiful! 174.88.241.210 (talk) 16:13, 5 February 2011 (UTC)

Primitive Roots

Just learning about basic number theory and primitive roots and I wrote a program to find a first few primes that have two/three as a primitive root. But when I started for primes which have 4 as a primitive root, I can't find any solution. I tried up to some large numbers but couldn't find a single solution. So my question is, is it possible for a prime to have 4 as a primitive root? How can we do this analytically? How can we find a prime for which 4 is a primitive root? Or if there doesn't exist one, then how would I go about proving it? Thanks!174.29.78.245 (talk) 06:25, 5 February 2011 (UTC)

Suppose 4 is a primitive root. Then 2 is a primitive root (if ${\displaystyle 4^{m}=x}$, then ${\displaystyle 2^{2m}=x}$). So 2 has order ${\displaystyle p-1}$. But ${\displaystyle p-1}$ is even, so ${\displaystyle 4=2^{2}}$ has order ${\displaystyle (p-1)/2}$, contradicting its being a primitive root. So there are no such primes.--203.97.79.114 (talk) 08:07, 5 February 2011 (UTC)

proper class in ZFC

Let r be some arbitrary real number, which might be undefinable.

Is "all elements of V except for r" a proper class? That is, are there uncountably many proper classes? One could even say "arbitrary ordinal" instead of "arbitrary real", so there would be more classes than sets.

The identification of classes with formulas doesn't seem all that satisfying. 71.141.88.54 (talk) 18:39, 5 February 2011 (UTC)

Yes, V\{r} has to be a proper class. It cannot be a set, because the union of two sets is a set, and V=(V\{r})u{r} is definitely not a set.

Strictly speaking, ZFC does not allow you to even speak about proper classes in formal statements, so "uncountably many proper classes" is not a meaningful concept. NBG set theory is a conservative extension of ZFC in which one can speak about (and quantify over) proper classes, but even there it is unclear which formal sense "uncountably many proper classes" might have.

If we ask for a bijection between proper classes and the integers, it is clear that such a bijection cannot exist inside the theory, because it would need to contain proper classes as elements, and nothing can do that. In NBG one could ask whether there is a logical formula ${\displaystyle \phi (n,X)}$ that bijectively assigns a proper class X to each integer n and vice versa. I don't think such a formula can possibly exist, but I cannot prove it offhand. (It's certainly impossible in Morse-Kelley set theory, but that might conceivably be because MK has more proper classes than NBG).

On the other hand, by the Skolem-Löwenheim theorem, if NBG is consistent, then it has a countable model, and in this model there would of course be meta-countably many proper classes -- see Skolem's paradox. –Henning Makholm (talk) 20:42, 5 February 2011 (UTC)

Thanks. I was thinking about the ZFC-centric view that classes are not formalized in ZFC itself, but are defined in terms of formulas in the metalanguage. But I guess that means classes are relative to specific models; for undefinable r, there has to be a constant symbol for r in the metalanguage before we can say "V\{r}". \ is a set operation anyway, so we can only use it in the context where V denotes the elements of a model, if I have it right. The apparent contradiction between "classes are denoted by formulas" and "uncountably many classes" is resolved by the idea that since the metalanguage can have uncountably many constant symbols, it can have uncountably many formulas. 71.141.88.54 (talk) 23:59, 5 February 2011 (UTC)

Normally, when you talk about definable proper classes, you mean definable with parameters. If you like, sure, you can add a constant symbol to represent the parameter, but it's slightly the long way around.

Then there are proper-class many possible values for the parameter. --Trovatore (talk) 17:43, 7 February 2011 (UTC)

Algorithmic complexity

Are weakly polynomial algorithms in P? —Preceding unsigned comment added by 207.210.136.252 (talk) 20:50, 5 February 2011 (UTC)

Yes, at least according to how "weakly polynomial" is defined in our Time complexity article. –Henning Makholm (talk) 21:09, 5 February 2011 (UTC)

Strictly speaking, P is a class of problems, not algorithms. —Bkell (talk) 04:27, 6 February 2011 (UTC)

Seemingly simple problem

A flag is going up a pole at 1 m/s. What is the rate of increase of the distance from the flag to an observer 3m from the base of the pole at time t?

This is a problem from my computer science class, and we're supposed to solve it both numerically, with a simulation, and mathematically, to demonstrate the accumulation of errors over time. The numerical part is easy, but how am I supposed to solve it mathematically? --99.237.234.245 (talk) 23:20, 5 February 2011 (UTC)

Hint: Pythagoras. —Preceding unsigned comment added by 68.9.151.24 (talk) 23:32, 5 February 2011 (UTC)

Hint 2: d/dt (Pythagoras). 71.141.88.54 (talk) 23:43, 5 February 2011 (UTC)

If the OP can do a numerical solution, he probably doesn't need to be reminded about Pythagoras. But he'll want to refresh his memory about derivatives. –Henning Makholm (talk) 23:48, 5 February 2011 (UTC)

The article related rates solves a very similar problem. Sławomir Biały (talk) 23:54, 5 February 2011 (UTC)

OK, so this requires calculus. I was kind of hoping for a geometrical solution. The problem is, I'm a grade 9 student listening in on a university computer science class, so while the professor can probably assume his students know calculus, that doesn't apply to me.

To Slavomir: using the related rates article, I decided to take a stab at the problem. According to the article, x*dx/dt + y*dy/dt = h*dh/dt. I want dh/dt, and I'm guessing dx/dt=0 while dy/dt=1 m/s. Can I plug in h=sqrt(x^2+y^2) to get dh/dt=(y*dy/dt)/sqrt(x^2+y^2)? --99.237.234.245 (talk) 00:53, 6 February 2011 (UTC)

Yes, that's the right formula.

However, let me suggest that you devote some time and energy to learning some calculus before you break your neck getting too far ahead on computer science. I don't know what kind of CS course you're taking, but if it's serious numerical methods stuff, you will need calculus as a background skill. On the other hand, if it's a generic CS 101 introductory programming thing, you can probably learn that just as well or better toying around with a computer and a programming language on your own, especially if you're bright enough to be hanging around universities in grade 9. –Henning Makholm (talk) 01:20, 6 February 2011 (UTC)

Thanks. The course is supposed to be an introduction to algorithms and algorithm analysis, with only a chapter or so on numerical methods. As far as I can tell, the math isn't difficult: "step A takes log(n) computations, but this is repeated n^2 times, so running time is some constant times n^2*log(n)". --99.237.234.245 (talk) 05:51, 6 February 2011 (UTC)

Okay. My advice still stands, speaking as a computer scientist. Algorithmics is an excellent touchstone for mathematical aptitude, so if you can wrap your head around that, learning some basic calculus by self-study would likely take you very little effort compared to the problem-solving power it will bring you. Good luck! –Henning Makholm (talk) 09:20, 6 February 2011 (UTC)

February 6

Un-deducting percentages

I have a net amount of $150.98, and need to find out the gross amount that 4.95% and 1.73% were deducted from to reach the net of $150.98. Can you help — Preceding unsigned comment added by Nannyno1 (talk • contribs) 06:25, 6 February 2011 (UTC)

This looks like homework. We will not do people's homework for them. If you try to solve the problem yourself and get stuck, feel free to show your work up to that point, and we'll try to point you in the the right direction. In this case, your first thing to do would be to figure out whether the two percentages were deducted sequentially (first take away 4.95% of the gross, then 1.73% of what remains) or at once (take away (4.95+1.73)% of the gross in a single operation). –Henning Makholm (talk) 08:51, 6 February 2011 (UTC)

Click here for a basic explanation. The second example is particularly relevant. Have a go at following that, and see if you can work out how to solve your problem. You are very welcome to ask for further help when you've had a go, perhaps explaining where you are stuck or what you've tried so far. 86.164.58.119 (talk) 11:42, 6 February 2011 (UTC)

If you know algebra, these sorts of problems are somewhat easy to work out. If you start with the original value (call it x), subtract off 4.95% and 1.73% (figure out how to write these in terms of x), the result is $150.98. Write that down in equation form and solve for x. -- 174.24.195.38 (talk) 17:58, 6 February 2011 (UTC)

... but if algebra is not to your liking, you can still easily solve the problem by noting what operations you would do on the original "unknown x" to arrive at $150.98, then you can start with $150.98 and just apply the inverse operations in reverse order to find the "unknown x". Division is the inverse of multiplication. Dbfirs 22:51, 6 February 2011 (UTC)

But that is just doing the algebra without calling it that. And being able to see that, say, ${\displaystyle x-7\times (x/100)}$ is the same as ${\displaystyle 0.93\times x}$ is essentially algebra. –Henning Makholm (talk) 09:08, 7 February 2011 (UTC)

Yes, of course, but some people are turned off by algebra, and it it not necessary to set up an equation to solve this type of problem. In the UK, the "School Mathematics Project" used to introduce algebraic solution of equations through inverse operations. From my experiments (OR) I observed that students grasped this method more quickly, but were subsequently slower to learn the "rules of algebra". Dbfirs 10:36, 7 February 2011 (UTC)

Oddly, I've found that many of those who are capable of inverse operations, but panic at algebra, are quite capable of algebra if you use symbols that are not letters. It's a very specific block that I assume must come from our surrounding culture somehow. Little pictograms are easily handled, but throw in an x and they don't understand :/ 212.183.128.67 (talk) 11:59, 7 February 2011 (UTC)

Yes, I've found that, too. An empty square is often used for an "unknown". I don't use algebra myself for reverse percentages, perhaps because I learnt to calculate them before I had ever heard the word "Algebra". Dbfirs 01:17, 8 February 2011 (UTC)

I don't think there's any argument that for a question like the OP's, explicitly writing down an equation with a named variable is unnecessary overhead. -- Meni Rosenfeld (talk) 07:32, 8 February 2011 (UTC)

Integers

Are integers defined by our experience of reality, or are they something more innate? I.e. Is there a definition which appeals to something more fundamental than that they are the numbers which indivisible 'stuff' comes in? If we lived in a universe with a fractional dimensionality, would our experience of which numbers are integers be different? Would indivisible 'stuff' come in different quantities? —Preceding unsigned comment added by 129.67.37.227 (talk) 13:13, 6 February 2011 (UTC)

Our articles on the Natural Numbers and the Integers have some explanations for this. I would say that in short, yes, there is something fundamental about the natural numbers (and the integers). If something is 'indivisible', then it really is counted by the natural numbers (although, I have to admit, I'm not sure about this concept of living in 'fractional dimensions'). Invrnc (talk) 14:05, 6 February 2011 (UTC)

I can't answer the question, but I'll just note that there are many mathematical concepts which on the surface seem only applicable to integers but in fact can be generalized, such as fractional Hausdorff dimension and fractional calculus. -- Meni Rosenfeld (talk) 14:53, 6 February 2011 (UTC)

It's a partial generalization. You can have fractional dimensions in certain senses (self-similarity or the behavior of the measure of balls of a certain radius), but you can't have one-and-a-half mutually orthogonal directions.

Things that you do one at a time are really, inherently different from things that vary continuously. I remember as a kid I didn't like this and it took me a long time to come to terms with, but it's really true.

On the other hand, things that you do one at a time are not limited to things you can index by the integers. Set theorists routinely use transfinite recursion to do uncountably many things, one at a time. --Trovatore (talk) 19:26, 7 February 2011 (UTC)

I don't know if this is "more innate" than the "numbers which indivisible 'stuff' comes in", but to construct the integers, one needs only the notion of the empty set, and the notion of set inclusion. This construction of natural numbers has the same general idea. Presumably the empty set and set inclusion exist in any universe, but whether you agree may depend on your philosophy of mathematics.SemanticMantis (talk) 03:23, 7 February 2011 (UTC)

“God gave us the integers; the rest is the work of Man.” —Leopold Kronecker. Sławomir Biały (talk) 14:02, 7 February 2011 (UTC)

I like that quote too, but once you've accepted the existence of the empty set and set inclusion, the existence of real numbers becomes a necessary truth as well ;-)SemanticMantis (talk) 18:36, 7 February 2011 (UTC)

Kronecker's God was too small. (Or, more precisely, Kronecker had too small a conception of God.) --Trovatore (talk) 18:52, 7 February 2011 (UTC)

Differential equations are probably more innate than integers. Your dog probably can't count past 5 so, but if you throw a stick, s/he can predict where it will land and run towards that spot. 71.141.88.54 (talk) 07:52, 8 February 2011 (UTC)

mathematics

50oranges were in a bag,25 is bad.what is the probability of picking a bad oranges at random.what is also the probability of choosen a good one without replacement. —Preceding unsigned comment added by 41.206.12.1 (talk) 14:48, 6 February 2011 (UTC)

If I take your first question literally, the answer is trivial: 50%. Your second question is nonsense, because you haven't specified how many oranges you remove without replacement. However, suppose you remove n oranges. Then the probability of all of them being bad is:

${\displaystyle {\frac {\binom {25}{n}}{\binom {50}{n}}}}$

That is, the number of ways of choosing from just the 25 bad oranges, over the total number of ways of choosing from 50 oranges. So the probability of any one of them being good is 1 minus that:

${\displaystyle 1-{\frac {\binom {25}{n}}{\binom {50}{n}}}}$

See Binomial Coefficient if the notation is unfamiliar to you. In short, ${\displaystyle {\binom {n}{k}}={\frac {n!}{k!(n-k)!}}}$ is the number of ways of choosing k things out of a collection of n things. HTH. --COVIZAPIBETEFOKY (talk) 16:24, 6 February 2011 (UTC)

Calculus

What's the integral of (f(x))^2? --75.15.161.185 (talk) 16:33, 6 February 2011 (UTC)

That depends on what f(x) is. There isn't a general formula. Perhaps you really want the integral of ${\displaystyle f(x)^{2}f'(x)}$. To do that, use Integration by substitution (u-substitution), which is the analogue of the chain rule for integration. In that case, you should get ${\displaystyle (1/3)(f(x))^{3}}$. Staecker (talk) 18:03, 6 February 2011 (UTC)

Input-to-state stability

What is input-to-state stability? I discovered this article while new-page patrolling; the language is confusing, and it has no references or links that can help me understand the topic. It looks mathematical, and Google results for the phrase are mostly from university math department websites, but anything beyond that is more than I can understand. Nyttend (talk) 20:28, 6 February 2011 (UTC)

Hmm, I just discovered that the article is a copyvio of one of the top Google results. However, since pretty much that entire page was copied to here, it doesn't help me understand at all. Nyttend (talk) 20:30, 6 February 2011 (UTC)

If you still want to know, it looks like a topic in dynamical systems theory. It resembles Lyapunov stability which I see has a redlink to it. There is a book by Hirsch, Smale, and Devaney (ISBN 0123497035) that should be a very good introduction to the general subject. I'm only familiar with the older edition which had a somewhat different approach, but I liked that one a lot, and the Amazon reviews say that the new one is more accessible. 71.141.88.54 (talk) 12:13, 7 February 2011 (UTC)

I don't understand the terminology in that handout, but I think it basically describes a system like a rocking chair, which is at equilibrium at a certain displacement (call this 0), and if you push it 3 inches with your finger and then let go, it will start rocking back and forth, but the amplitude of the rocking will be bounded. That contrasts with an umbrella balanced on its tip, so if you poke it, it falls over (unbounded displacement). 71.141.88.54 (talk) 14:06, 7 February 2011 (UTC)

Compact manifold without boundary not contractible

Could anyone please link me to a proof that a compact manifold without boundary is not contractible? I would prefer if it was not based on things like cohomology but rather closer to first principles, though fairly basic or fundamental theorems in differential geometry such as the inverse function theorem/preimage theorem/Sard's theorem are okay. I am teaching myself material from a set of online notes alongside a number of other courses, but unfortunately the proofs are omitted, and while i would like to derive the proof myself, I'm afraid I don't really have the time currently. Any help greatly appreciated, Estrenostre (talk) 23:12, 6 February 2011 (UTC)

Sorry, I don't know a proof from just calculus, but this follows quite easily from the homotopy-invariance of the n-th singular homology group over Z₂. If I were pressed to not use any topology, then I might try to construct a nonconstant harmonic function on a given contractible manifold. I have some ideas how this can be done, but nothing that seems particularly easy to me. Sławomir Biały (talk) 13:50, 7 February 2011 (UTC)

The way it's stated, the claim is false: A point is a zero-dimensional compact manifold without boundary, and it's contractible by definition. To make a working argument you somehow need to use the positive dimensionality; which leads us to homology (or maybe homotopy) because it's an obvious way of getting at positive dimensional information. I have a feeling that every proof we might come up with in some way comes down to homology. Ozob (talk) 12:05, 8 February 2011 (UTC)

Could you not use the fact that every open cover of a compact space has a finite subcover? Assuming the space to be Hausdorff should give a contradiction some how. If you assume a homotopy does exist then you could prove that the homotopy wasn't continuous, and that's a contradiction. I know it's very hand-wavy. It's just an idea of what I might try to do. — Fly by Night (talk) 15:41, 8 February 2011 (UTC)

February 7

Human Population - Two Questions

Can human population be graphed using an exponential function? Say there were about 4 million people in The United States in 1790, and now there are 310 million. Would that be consistent with the function? Also, using the same function, what year was it when there was only about 1 Human? Thanks. 129.120.195.10 (talk) 19:25, 7 February 2011 (UTC)

As to the second question, exponential models tend to fall down before going back to 10,000 BC or so; they grow too quickly compared to human growth. Thus most models I've seen do not consider the case for world population being one (bearing in mind that would be an odd thing to find, given it's widely believed more than one human evolved, or else two). Grandiose (me, talk, contribs) 20:03, 7 February 2011 (UTC)

Exponential growth means that the logarithm grows linearly. The log of 4 million is 15.2 and the log of 310 million is 19.6. So the log has grown by 4.4 during the 220 years that have passed since 1790. So log of the population grows 4.4/220=0.02 per year. An approximate formula for the population N in year y is N(y)=e. Using this formula you get that N=1 in the year 1030, long before the States of America were United. Bo Jacoby (talk) 21:51, 7 February 2011 (UTC).

I thought the OP meant world population growth, my mistake. Clearly it depends on the population as to what additional problems there might be, as below. Grandiose (me, talk, contribs) 18:07, 8 February 2011 (UTC)

One could view the matter like this: The number of years from 1790 until year x is x − 1790. It has now been 221 years since then, and the number of 221-year periods from then until the year x is (x − 1790)/221. Every time a 221-year period passes, the population multiplies by 310/4. The number of times it multiplies by 310/4 from 1790 until the year x is therefore the number of 221-year periods. So it multiplies by (310/4) between the year 1790 and the year x. It starts at 4 in the year 1790. So the population in the year x is 4 × (310/4).

That's provided growth is exponential. But I think the rate of growth relative to the size of the whole population varies, so that equation is probably not realistic.

So year, the numbers you give are consistent with exponential growth (look at that article if you haven't already), but that doesn't make that equation a realistic model. Michael Hardy (talk) 22:47, 7 February 2011 (UTC)

A logistic function is often a more accurate tool to model population growth. In some regimes, the growth looks exponential, but in either extreme (very far in the past or future), the behavior is more realistic. Sophisticated population models are often multi-parameter models that are optimized to fit the measured data. Nimur (talk) 23:09, 7 February 2011 (UTC)

Quibble: A logistic function behaves just like an exponential in the far past. -- Meni Rosenfeld (talk) 07:26, 8 February 2011 (UTC)

OP here (different IP). So if a logistic function is more appropriate, what would the function be using that type of math? Also, I stopped liking math after I learned that y=mx+b, so if you can keep it simple, I would appreciate it. Thanks. 71.21.143.33 (talk) 02:01, 8 February 2011 (UTC)

It's not realistic to think you can describe human population with a simple curve. There are too many fluctuations due to epidemics, climate change, techological progress like the green revolution or the earlier invention of agriculture, etc. There is genetic evidence that humans had a population bottleneck a few hundred thousand years ago, in which the species almost collapsed. Some related branches like the Neanderthals actually did collapse. But you can look at the graph of logistic function: it shows basically a population stabilizing after an exponential ramp-up, as it hits the limits imposed by some resource. 71.141.88.54 (talk) 03:14, 8 February 2011 (UTC)

Yet it's completely reasonable to model population growth. All models are wrong to some degree, that is why we call them models. The logistic curve is one of the simpler models, but it can be a good fit for certain populations and time periods. Similarly, exponential growth is always a good model on short time scales. However, even stochastic processes like the ones you mention above can be incorporated into more sophisticated models. Complaining that the logistic model is too simple is to completely miss the spirit of using mathematical models to understand population growth. The logistic model is the classic 'next best' choice after exponential growth has been discarded. SemanticMantis (talk) 14:54, 8 February 2011 (UTC)

(ec) The problem is one of interpolation. Having only the two datapoints (1790,4000000) and (2010,310000000) you should not use a logistic function which has three parameters. This means that infinitely many logistic curves fit the data exactly, while only one linear function, and only one exponential function, fits the data. Linear interpolation does not take into account the fact that no population is negative. So it is a good idea to take the logarithm and interpolate between the two datapoints (1790,15.2) and (2010,19.6). The exponential growth is not realistic for sufficiently late times where the predicted population grows unrealisticly large, nor for sufficiently early times where the predicted population grows unrealisticly small. Bo Jacoby (talk) 15:45, 8 February 2011 (UTC).

I'll second SemanticMantis' contributions regarding the purpose of mathematical modeling. We have an article on this subject: mathematical model. The original questioner should be aware that no model is intended to be a 100.0% exact predictor or representation of the time-history of the population: the point of the model is to elucidate the way the population trends are behaving, and to quantify some parameters that might be affecting it. A very simple model, like an exponential growth or a logistic curve-fit, will have very limited accuracy and almost no useful capability for extrapolation far from the measured data. Nimur (talk) 19:51, 8 February 2011 (UTC)

Modelling is fine over small intervals. The problem is the OP's wish to extrapolate all the way back to when there was just one person. A more extreme example would be something like weather prediction. If it's sunny now, it will probably still be sunny 2 minutes from now. With enough measurements and fancy models you can predict fairly accurately about tomorrow or even next week. But beyond that, say if you want to know if it will rain on the 4th of July, you can only make a general guess based on climate. 71.141.88.54 (talk) 21:44, 8 February 2011 (UTC)

February 8

Difficulty in finding Indefinite integral

I am a graduate student of computer science and I pursue integration as a hobby /passion.I find it impossible to find Integral f(x)dx where f(x)=(x^4-1)^(-1/4) by a way other than the following f(x)=(x^-1)(1-x^(-4)) or f(x)=x^4/(x^5)(1-x^(-4)) or Integral f(x)dx=4*Integral f(x)/4 dx =4* Integral ^-1/2Is there any elegant and logical way of solving the problem by method of substitution or integration by parts.More specifically is there any good heuristic way of determining what substitution function to be used in the method of substitution.Ichgab (talk) 15:04, 8 February 2011 (UTC)

Let's try and put this together. I think the lack of replies is because it's quite hard to understand mathematics when it's written using a keyboard. Am I right in assuming that you want to calculate the following indefinite integral:

${\displaystyle \int \!{\frac {\operatorname {d} x}{\sqrt{x^{4}-1}}}\,.}$

As far as I remember, this doesn't have an explicit anti-derivative in terms of elementary functions. That means that there is no expression using elementary functions whose derivative gives you (x – 1). In fact, sadly, the same is true for

${\displaystyle \int {\frac {\operatorname {d} x}{\sqrt {\sin x}}}\,.}$

The anti-derivatives of these two integrands are expressible in terms of hypergeometric functions and elliptic integrals, but not as elementary functions. There may be techniques to evaluate definite integrals with these integrands, but there are no elementary anti-derivatives. — Fly by Night (talk) 21:41, 8 February 2011 (UTC)

The Risch algorithm is a decision procedure (incorporated into some symbolic algebra software) that tells you exactly whether an integral can done as elementary functions. The wider topic area for differential equations is differential Galois theory, which I just came across a few days ago and don't know anything about. 71.141.88.54 (talk) 22:26, 8 February 2011 (UTC)

Thank you for the answer,my doubt has been cleared.Ichgab (talk) 05:26, 9 February 2011 (UTC)

The integral is elementary, albeit complicated. Try http://www.wolframalpha.com integrate . Bo Jacoby (talk) 09:43, 9 February 2011 (UTC).

That's a hand link that you supply; I'll have to use that in the future. It says that

${\displaystyle \int \!{\frac {\operatorname {d} x}{\sqrt{x^{4}-1}}}={\frac {x{\sqrt{1-x^{4}}}\,{}_{2}F_{1}\!\left({\frac {1}{4}},{\frac {1}{4}};{\frac {5}{4}};x^{4}\right)}{\sqrt{1-x^{4}}}}\,.}$

The ₂F₁ is a hypergeometric function. I didn't think that they were elementary functions. It seems that some are and some aren't; it depends on the arguments. So the question is: Is ₂F₁(1/4,1/4;5/4;x) one of the special hypergeometric functions that reduces to an elementary function? I would guess not; for if it did then that link would hopefully present the elementary function and not ₂F₁(1/4,1/4;5/4;x). — Fly by Night (talk) 14:28, 9 February 2011 (UTC)

Having said that, we have ₂F₁(1/4,1/4;5/4;x). — Fly by Night (talk) 14:48, 9 February 2011 (UTC)

The integral is expressed in terms of the elementary functions logarithm and arctangent like this

integral 1/(x^4-1)^(1/4) dx = ((1-x)^(1/4) x (x+1)^(1/4) (x^2+1)^(1/4) (-(tan^(-1)(1-x/(sqrt(2) (1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4)), -x/(sqrt(2) (1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))))/(2 sqrt(2) x)-(tan^(-1)(x/(sqrt(2) (1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))+1, -x/(sqrt(2) (1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))))/(2 sqrt(2) x)-(log(x^2/(sqrt(1-x) sqrt(x+1) sqrt(x^2+1))-(sqrt(2) x)/((1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))+1))/(4 sqrt(2) x)+(log(x^2/(sqrt(1-x) sqrt(x+1) sqrt(x^2+1))+(sqrt(2) x)/((1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))+1))/(4 sqrt(2) x)))/(x^4-1)^(1/4)+constant

Bo Jacoby (talk) 01:27, 10 February 2011 (UTC).

Yes, we know. I already linked to that 11 hours ago. Here it is again, in case you missed it: ₂F₁(1/4,1/4;5/4;x). This is somewhat unusual, however, since not all hypergeometric functions reduce to elementary functions. I mentioned that too. — Fly by Night (talk) 01:56, 10 February 2011 (UTC)

Always assume that people will not follow links. If you said you didn't think the function was elementary and then wanted to retract it, a better form would be "Turns out that it is elementary - see ₂F₁(1/4,1/4;5/4;x)". Since you did not explicitly say it was elementary, Bo was in the right to do so. -- Meni Rosenfeld (talk) 09:48, 10 February 2011 (UTC)

Can game theory help with reverse auctions?

Hello. Does game theory provide any "best strategy" in a reverse auction? Let's imagine the following case:

• A public administration puts to tender the construction of a stadium for an approximate value of $25 million. Its main criterion for assessing bids is cost (it will choose the lowest bid);
• All bidders provide the same service with the same level of quality, so that variable is irrelevant here;
• Each bidder wants to win the auction, but they also want to maximise the price they will be paid for the works if they win;
• Bidders do not know what price the other bidders have submitted.

If I were a bidder, what would be the sweet spot which would give me the most chances of winning the auction, while at the same time getting a high enough price to make a decent profit? Is there such a sweet spot at all? Thanks in advance. Leptictidium (mt) 16:26, 8 February 2011 (UTC)

I think there is something optimal about the different auction types. Indeed if you look at Auction theory there is the Revenue equivalence theorem which states all auction types will give the same result for the seller. I think the upshot is to bid what you think it is worth, bid less and another bidder will win.--Salix (talk): 17:47, 8 February 2011 (UTC)

I think that as far as the theory is concerned, reverse auctions are completely equivalent to regular auctions.

The key issue here is what each contractor knows about the project cost for himself and for others. If all costs are known the result is simple - the contractor with the lowest cost wins, with a bid slightly less than the second best's cost. Otherwise each contractor needs to estimate (with probability distributions, preferably) his own cost and what the others will bid, and solve the resulting optimization problem. -- Meni Rosenfeld (talk) 20:17, 8 February 2011 (UTC)

You also need to know the cost of performing the service to a bidder (you said they're equivalent); no one will bid below that because they would do better not to bid at all. You could also incorporate the cost of not getting the bid: salaries may need to be paid anyway, and another bid will have to be prepared. In that case you can actually get bids slightly below cost because the alternative is (say) paying your staff to do nothing. --Tardis (talk) 21:39, 8 February 2011 (UTC)

Conversely, there may be an opportunity cost of taking on the job at a low price, if it ties up resources that could otherwise be used on more profitable work. AndrewWTaylor (talk) 09:46, 9 February 2011 (UTC)

I would be surprised if people behave the same way in reverse auctions as in regular auctions. Many auctions that are theoretically equivalant have known behavioral tendencies in participants, while prospect theory demonstrates that people's economic decisions under uncertainty are affected by context (here, whether the bids increase or decrease over time). Vernon Smith has written much about this; if this is more than an academic question, I would check out his writings. —Preceding unsigned comment added by 12.186.80.1 (talk) 15:34, 9 February 2011 (UTC)

February 9

how long to cross the 0 on a random walk

Whenever you see graphs of random walks, sometimes you'll see them not crossing the zero for a really, really, really long time! How is that possible? Doesn't the law of averages state that it has to hover around zero? If it goes way into negative or way into positive, it needs to go in the other direction for it to be a "random" (instead of a weighted) walk, doesn't it? 217.136.92.148 (talk) 00:04, 9 February 2011 (UTC)

Be careful—that's the gambler's fallacy. If the first step of the random walk takes it up to 1, then the next step is equally likely to take it up to 2 or down to 0. Essentially, after that first step you are starting a brand new random walk, except that you're starting from 1 now instead of 0. So, from that perspective, once the random walk has reached 1, your new expectation should be that it should average out to about 1 from that point on: there is no "force" that should make it more likely to come down to 0 than to go up to 2. —Bkell (talk) 00:48, 9 February 2011 (UTC)

This has something to do with the Levy arcsine law. Apparently the proportion of the time that a random walk with p=1/2 will be in positive territory will have an arcsine distribution, meaning it is much more likely to be close to 0 or 1 than to the intuitive value of 1/2. Although I'm not an expert so I find this quite mysterious myself. For one thing, does the arcsine distribution even integrate to one over the unit interval? For another thing, I find it hard to imagine how this totally independent of the length of your sample path. On a long enough sample I would expect to see 'cycles' between very long periods in positive territory and very long periods in negative territory, which would bring the total back closer to 1/2. I hope someone can give a better answer. —Preceding unsigned comment added by 130.102.78.164 (talk) 01:00, 9 February 2011 (UTC)

${\displaystyle {\frac {2}{\pi }}\arcsin({\sqrt {x}})}$ is the cumulative distribution. The integral of the arcsine distribution over the unit interval is ${\displaystyle {\frac {2}{\pi }}\arcsin({\sqrt {1}})}$, which you can check is 1.

Concerning the law of averages, you can argue that it has to stay relatively close to zero. But here "relatively close" just means less than ${\displaystyle {\sqrt {2n\log \log n}}}$ (and even that not quite).--203.97.79.114 (talk) 02:43, 9 February 2011 (UTC)

Funny enough, while the reasons he gives are iffy (for reasons related to the gambler's fallacy), he's not far from the truth. In fact, a simple random walk on ${\displaystyle \mathbb {Z} }$ is expected to visit every point an infinite number of times (in particular, it returns to 0 in finite time, with probability 1). Our article on random walks addresses this, roughly, and the referenced Mathworld article (in the Higher dimensions section) furthers the discussion. Invrnc (talk) 02:49, 9 February 2011 (UTC)

Also, I'll point out that, though returning to the origin in finite time has probability 1, there is no upper bound on this finite time. Simple random walks can stay away from the origin for arbitrarily long times, but this occurs with arbitrarily small probabilities. SemanticMantis (talk) 15:46, 9 February 2011 (UTC)

Moreover, the expected time between visits to the origin is infinite. Algebraist 15:56, 9 February 2011 (UTC)

WHAT???

How can that possibly be true!! Please elaborate... 217.136.92.148 (talk) 18:04, 9 February 2011 (UTC)

I am not familiar with this result. But it would mean that the distribution of lengths between visits to the origin would have a Long-tail distribution. I am also a bit quesy about the word infinte, probably a better wording is that the expected time is not finitly bounded. Think of it this way, since the walk is recurrent we know that it will cross 0 infinitly often. But we also know that it will cross 10^10 infinitly often. The result just states that large permutations will occur often enough that we can't put a finite expected value for the time until the next crossing of 0. Taemyr (talk) 18:40, 9 February 2011 (UTC)

The word "expected" is being used in the sense of expected value. Just to clarify, this does not mean that the walk will return to the origin "after infinite time", it just means that probabalistically, the expected length of time between visits is really big. How big? Bigger than any fixed real number. Staecker (talk) 19:00, 9 February 2011 (UTC)

The way you stylized "how big" - is that someone in the audience whispering it off-stage? 109.128.101.244 (talk) 20:02, 9 February 2011 (UTC)

There's no reason to be shy about calling the expected time infinite. It is infinite (assuming Algebraist is right, which he usually is). Yes, you have to be careful not to conclude that there might be two individual visits that are separated by infinite time — as the problem is constructed, that can't happen (although you could probably extend the setup to a nonstandard model of Peano arithmetic or something, and in that context it could). --Trovatore (talk) 19:07, 9 February 2011 (UTC)

"The expected time is infinite" is legitimate. If ${\displaystyle p_{i}}$ is the probability that there will be no visit before time i, the expected time of the next visit is ${\displaystyle \sum _{i=1}^{\infty }p_{i}}$, which is ${\displaystyle \infty }$.

To the OP - this does not contradict the fact that ${\displaystyle \lim _{i\to \infty }p_{i}=0}$ and hence there will be a visit with probability 1. -- Meni Rosenfeld (talk) 19:13, 9 February 2011 (UTC)

In some cases, expected value matches with an informal notion of 'average behavior', but it doesn't have to. For instance, if heads is worth 1 and tails is worth 0, then the expected value of a fair coin toss is 1/2, even though that doesn't correspond with any possible event. Likewise, the expected value of a six-sided die is 3.5, and that cannot happen on given throw either. The best intuitive way of understanding the results mentioned above on recurrence of integer simple random walk: we can say with certainty that the walk will return to the origin, and that it will occur in finite time. But there is no finite time after which we can be sure the walker has returned. Does that help? SemanticMantis (talk) 20:03, 9 February 2011 (UTC)

IT helps if you meant to write: "The best intuitive way of understanding the results mentioned above on recurrence of integer simple random walk: we can say with certainty that the walk will return to the origin, and that it will occur in finite time. But there is no finite time after which we can expect that the walker has returned -- i.e. a finite time at which, "on average" he would have returned." —Preceding unsigned comment added by 109.128.101.244 (talk) 20:07, 9 February 2011 (UTC)

The reason that would have helped is, that's what I understand "expected" to mean. (Rather than your introduction of the word "sure"). 109.128.101.244 (talk) 20:09, 9 February 2011 (UTC)

I think your new wording brings up a new issue. I'm not sure what "on the average he would have returned" means exactly, but it sounds like you might be thinking of the median time. The median time is in fact finite (there's a finite time t such that the return is as likely to happen before t as after t). It's the mean time that's infinite (add up all the possible lengths of time, multiplied by their probabilities). --Trovatore (talk) 20:25, 9 February 2011 (UTC)

I meant 'sure' in the sense of Almost_surely, which means probability of the event is 1. I think you're a little hung up on the infinite expectation. We can still answer questions like "after what time can we say the chance of return is at least 60%?" or "after 100 steps, what is the chance that the walker returned?" Anyone have an example of something more concrete with an infinite expectation? The expected degree in some scale-free network perhaps? SemanticMantis (talk) 20:40, 9 February 2011 (UTC)

Yes, you meant that, but weren't you SUPPOSED to mean 0.5 instead of 1? If I walk into a room at a random time, and wait for the minutes on the digit clock in it to advance by one, I have an expected wiat time of 30 seconds. (There is a 50% chance time that I walked in when the seconds were reading between 0-29, giving me a 30+ second wait and a 50% chance that I walked in when they were reading between 30-59, giving me a 30- second wait. Having to waiting 31 seconds is as likely as 29 seconds, and so on all the way down. The expected wait is 30 seconds). But the way that YOU construe the problem, you are saying "I have to wait 60 seconds", since oyu are using "probability one". (and perhaps it just turned the next minute the microsecond I entered the room). So you are in conflict with Trovatore. He is saying, "Median wait" (0.5 probability if you like), and you are saying "Probability 1 time wait". Which of you is correct? 109.128.101.244 (talk) 23:14, 9 February 2011 (UTC)

Sorry, but I just don't understand your first sentence in particular, and where we are miscommunicating in general. Also, I've re-read the whole thread twice, and I see no inconsistencies between what I've said and what Trovatore has. Does Trovatore? ? Perhaps it's best to start with the simpler example of infinite expected value that Algebraist links below. Does that make sense? You could also start a new question about your specific concerns. This is a good question and interesting stuff, but this thread has got a lot of baggage by now. SemanticMantis (talk) 01:04, 10 February 2011 (UTC)

The problem with what you said, "there is no finite time by which we are (almost) sure to return", is that it has nothing to do with the infinite expectation. As long as the time of returning is unbounded, this will be true even if the expectation is finite. -- Meni Rosenfeld (talk) 09:39, 10 February 2011 (UTC)

The classic example is the St. Petersburg game. Algebraist 20:46, 9 February 2011 (UTC)

mathematicians: useful or useless?

I have two kinds of experiences with mathematicians. One, is that they are like magicians, who have a deep understanding of systems that no one in the world has. As long as they can assure that what they have an understanding of indeed corresponds to the conditions of the systems they're applying them to, a mathematician can do magic. But, my other experience is that a mathematician is useuless, and has a body of theoretical knowledge that they cannot apply to anything in any way. What gives? How can I have these two contradictory experiences? 217.136.92.148 (talk) 09:45, 9 February 2011 (UTC)

Mathematical fun and fascination is not necessarily immediately useful. It may find useful or even commercial applications after some time. Bo Jacoby (talk) 09:55, 9 February 2011 (UTC).

A professor of mine put it this way (quoting Faraday), "Of what use is a baby?" meaning you can't judge the value of something only by its immediate usefulness. The ordinary person doesn't use much math on a daily basis, but without it science and engineering are impossible and without them modern civilization would be impossible.--RDBury (talk) 10:28, 9 February 2011 (UTC)

• The word "mathematicians" generalizes a large group of people with widely varying interests and abilities. Some will proudly say "I have never done anything 'useful'", while others will focus their studies on topics known to be applicable and the the skills of applying them.
• Mathematical discoveries can take time to be assimilated into the theory of other professions, and some more time into their practice.
• Necessarily, some mathematical concepts will be inherently more applicable than others, though it is hard to tell in advance which is which - the cryptographic applications of number theory is a famous example.

-- Meni Rosenfeld (talk) 10:39, 9 February 2011 (UTC)

Calling somebody either "useless" or "useful" are both denigrating (and so not really contradictory except on a purely local scale). Mathematicians are people. –Henning Makholm (talk) 10:42, 9 February 2011 (UTC)

Mathematics is part of what makes life worth living for a mathematician, and they get paid for it, that's no different from than a musician getting paid for their art. Mathematics has helped make the modern world where you can phone up your friends and meet up to have a good time listening to that music if that's what you like. Would you say that the music you don't like is useless? Dmcq (talk) 11:07, 9 February 2011 (UTC)

George Boole's work was "relatively obscure, except among logicians. At the time, it appeared to have no practical uses." In 1900, he probably would have been a good example of a person with a body of theoretical knowledge that couldn't be applied to anything useful. Flash forward 100 years, and Boole's work is fundamental to the use and operation of computers. That is, if computer programmers didn't have Boole's work, someone would have had to invent it first. - That's a repeating theme in mathematics: some esoteric, purely theoretical problem suddenly becomes the focus of practical inquiry, and the results of obscure mathematicians, who during their lifetime never cared about practical applications, suddenly become critical to major industries. In some respects, current theoretical mathematicians you deem as useless are also doing magic, but they're doing it on systems that haven't been invented yet. -- 140.142.20.229 (talk) 01:36, 10 February 2011 (UTC)

I would like to just say if you consider mathematicians useless then perhaps you'd like to first consider what use are philosophers. The view you have towards mathematicians I have towards philosophers (not philsophers of mathematics though). Money is tight (talk) 06:29, 10 February 2011 (UTC)

3,5,7 = triple adjacent primes, are there any others?

Actually, in this case, it's 2,3,5,7 as adjacent primes, with 2 and 3 being the only ones that abutt. I notice adjacent primes like 17,19 and 29,31, but I can't think of series of three such adjacent primes apart from the ones I mentioned. Are there any others? If not, is there are proof that there cannot be? And is there any proof that there is an infinite number of such adjacent primes?

And, if you had a graph of all natural numbers, like the x axis, and you plotted a curve above it which identified the primes, obviously this curve gets steeper quickly as the prime numbers thin out. What can we say about such a curve? Obviously, it cannot be identified by an algorithm, as then we could predict primes in advance. But is there a name for such a curve? (Oh, I'm not a math person really, so go easy on me, please. Myles325a (talk) 11:17, 9 February 2011 (UTC)

There are no more triplets, and the proof is simple - if a is any integer, then exactly one of ${\displaystyle a,\ a+2,\ a+4}$ is divisible by 3. So if ${\displaystyle a>3}$ they cannot all be prime.

Whether there are an infinite number of twin primes is an open problem.

For the second question, you should start with the prime number theorem, which can be used to approximate this curve. Note that the exact curve can most certainly be identified with an algorithm, just not a very quick one. -- Meni Rosenfeld (talk) 11:30, 9 February 2011 (UTC)

The closest admissible constellation of three primes is called a prime triplet. It is of form (p, p + 2, p + 6) or (p, p + 4, p + 6). Four primes is a prime quadruplet and so one. They are all conjectured to have infinitely many occurrences but it hasn't been proved for any of them. The largest known cases are at http://anthony.d.forbes.googlepages.com/ktuplets.htm. Apart from trivial occurrences with primes below 1000, the largest k with a known prime k-tuplet is a 27-digit prime 19-tuplet discovered earlier today after a long search: http://tech.groups.yahoo.com/group/primenumbers/message/22585. PrimeHunter (talk) 14:01, 9 February 2011 (UTC)

Maybe of interest, the Green-Tao theorem shows (nonconstructively) that there are arithmetic sequences of arbitrary length in the primes. I thought some had been discovered of length 23 or so. 71.141.88.54 (talk) 20:44, 9 February 2011 (UTC)

According to The Prime Glossary, the longest known arithmetic sequence of primes is currently of length 25, starting with the prime 6171054912832631 and continuing with common difference 366384*23#*n, found by Chermoni Raanan and Jaroslaw Wroblewski in May 2008. (Here the # symbol denotes the primorial.) —Bkell (talk) 21:50, 9 February 2011 (UTC)

It has since been improved to length 26. See Primes in arithmetic progression#Largest known primes in AP and my website which maintains the current records. PrimeHunter (talk) 00:06, 10 February 2011 (UTC)

Percent of negligible error

I know that there are many ways to estimate error. I would like to know the proper method for this situation. I have a population, say 100 people. Let's assume I'm measuring obesity and 60% are obese. Then, I have a new data point at a later date. The population has grown to 150 people. What percent of difference from 60% should be acceptable or negligible? I need the formula based on original population, original percentage, and new population. I've found dozens of variations on Google, so I don't know which is proper. -- kainaw™ 14:19, 9 February 2011 (UTC)

See Misplaced Pages:Reference_desk/Archives/Mathematics/2011_January_18#Calculating_the_probability_that_two_binomial_distributions_are_in_order. Bo Jacoby (talk) 16:55, 9 February 2011 (UTC).

It appears that it cannot be done using those methods without also knowing the standard deviation. I do not have standard deviation. I only have population and percentage at two different points in time. I would like to know if the second percentage is significantly different than the first based on the change in population size. -- kainaw™ 17:16, 9 February 2011 (UTC)

Consider taking the situation as a series of Bernoulli trials, that is each person is obese independently with some uniform probability p. A simple estimator for p is simply to use the proportion you observe (0.6 in the first population). Then the number of obese people you expect is np, with variance np(1-p). Once you have modeled the initial distribution, you can then calculate the probability of your second sample in that model.

Even if the model of Bernoulli trials is not right, you should attempt to apply some sort of model to your data (put some probability distribution on your population and estimate the parameters). I don't think you can say much about the significance without some knowledge of the variance (standard deviation). Invrnc (talk) 20:10, 9 February 2011 (UTC)

The mean value of the parameter P is ${\displaystyle \mu =E(P)={\frac {k+1}{n+2}}={\frac {61}{102}}=0.598}$ where n=100 people is the size of the sample and k=60 is the number of them being obese. The standard deviation is ${\displaystyle \sigma ={\sqrt {V(P)}}={\sqrt {\frac {\mu (1-\mu )}{n+3}}}=0.0483}$. Bo Jacoby (talk) 01:45, 10 February 2011 (UTC).

Fisher's Least Significant Difference test vs. Tukey's Least Significant Difference test (statistics)

Hello,

Is there a difference between Fisher's Least Significant Difference test and Tukey's Least Significant Difference test, or are they merely two names for the same test? To be clear, by Tukey's LSD test, I'm not referring to Tukey's Honestly Significant Difference test or the Tukey-Kramer test (which are quite different from Fisher's LSD test in that they correct for multiple comparisons).

I ask because the documentation for MATLAB's multcompare function indicates that it can use Tukey's HSD/Tukey-Kramer or Tukey's LSD (among other options; see the "Values of ctype" section about halfway down the page). I've searched for Tukey's Least Significant Difference test on the internet and in the literature but only found references to it in papers that mention its use. In contrast, searches for Fisher's Least Significant Difference test return results that provide information about the actual method, which leads me to suspect that perhaps Tukey's LSD is simply another name for Fisher's LSD or for the process of performing multiple pairwise t-tests.

Any clarification regarding this nomenclature would be appreciated.

Thanks!

142.20.133.215 (talk) 15:40, 9 February 2011 (UTC)

Does such a value exist?

For

x equals x to the cube plus one

Which is

x = x+1

Does such a value exist?

Thank you very much. --192.197.51.41 (talk) 16:00, 9 February 2011 (UTC)

Yes. Your equation is equivalent to finding the root of f(x)=x-x+1, which is a polynomial and hence a continuous function, which is therefore subject to the intermediate value theorem. Since f(-2)=-5 (i.e. less than ZERO) and f(-1)=1 (greater than ZERO), the IVT guarantees that there exists an x between -1 and -2 such that f(x)=0. For questions like these a quick answer can be gotten from WolframAlpha like so, which shows the answer to be x=-1.32472.... Zunaid 16:15, 9 February 2011 (UTC)

And in general, every polynomial (with real coefficients) of odd degree has a real root. A polynomial of even degree needn't have a real solution, but it must have a complex solution. -- Meni Rosenfeld (talk) 16:39, 9 February 2011 (UTC)

Alright, thank you very much. --192.197.51.41 (talk) 19:49, 9 February 2011 (UTC)

To expand on Meni's answer, see fundamental theorem of algebra, which states that every n-th degree polynomial must have n roots. In the case of your example it has 1 real root and 2 complex roots (as shown in the WolframAlpha link above). Zunaid 21:42, 9 February 2011 (UTC)

Technically, it must have n roots counted with multiplicity. x-2x+1=0 is true only for x=1 but, in a sense, it is true twice for x=1 (1 is also a root of the derivative, if you wish to be precise and what sense we mean). --Tango (talk) 23:46, 9 February 2011 (UTC)

Good ol algebra word problems

Time for some quadratics! I've got a typical plane flies word problem. I know how to solve a quadratic equation, but I'm trying to figure out how to get the quadratic equation.

• A pilot flies from Ottawa to London (5400 km). On the return trip, he travels 50 km/h faster and reduces the travel time by 60 minutes (1 hour). Find his average speeds in each direction.

What I've gotten so far is that the first trip, speed (x1)=5400/y, where y is the time in hours. Therefore, since x2 = x1 + 50:

(5400/y)+50 = 5400/(y-1)

or

(5400/y) - (5400/(y-1)) - 50 = 0

What is the next step? - ʄɭoʏɗiaɲ  _¢ 21:34, 9 February 2011 (UTC)

What I would do next is to clear the fractions: multiply the equation by the common denominator of the two fractions. (Equations are usually easier to work with when they don't have fractions in them.) —Bkell (talk) 21:43, 9 February 2011 (UTC)

discrete fourier transform in mathematica

I am trying to get used to taking discrete fourier transforms with mathematica. The first step I take is to generate some time data from a function whose fourier transform I know, namely I know that the transform of

${\displaystyle f(t)=e^{-{\frac {t^{2}}{2}}-i\omega _{1}t}}$  

is

${\displaystyle {\hat {f}}(\omega )={\sqrt {2\pi }}e^{-{\frac {1}{2}}(\omega -\omega _{1})^{2}}}$  

which is a real-valued function. However when I create a time series of ${\displaystyle f(t)}$, using the following code:

omega1 = 2 Pi;

E3 := Exp Exp;

timelist = Table, {t, -12, 12, .1}];

and then take its DFT by

freqlist = Fourier;

then the result is complex-valued, and I only get the Gaussian function I was expecting if I plot the absolute value. Is this to be expected for the DFT and FT to give different results? Thanks 128.200.11.124 (talk) 22:34, 9 February 2011 (UTC)

Need help with puzzle. No solution please

The following puzzle has given me absolute hell:

An island is inhabited by five men and a pet monkey. One afternoon the men gathered a large pile of coconuts, which they proposed to divide equally among themselves the next morning. During the night one of the men awoke and decided to help himself to his share of the nuts. In dividing them into five equal parts he found that there was one nut left over, which he promptly gave to the monkey. He then hid his one-fifth share, leaving the rest in a single pile. Later during the night, another man awoke with the same idea in mind. He went to the pile, divided it into five equal parts, and found that there was one coconut left over. This he gave to the monkey, and then hid his one-fifth share, restoring the rest to one pile. During the same night, each of the other three men arose, one at a time, and in ignorance of what had happened previously, went to the pie, and followed the same procedure. Each time, one coconut was left over, and it was given to the monkey. The next morning all five men went to the diminished coconut pile and divided it into five equal parts, finding one coconut left over. What is the least number of coconuts the original pile could have contained?

---

Isn't the premise flawed? It seems like if the original pile had n coconuts, where n is a number divisible by 5 with a remainder 1, then each time a man takes n/5 + 1 share away, he makes the pile a multiple of 5, and therefore there would be no coconut left for the monkey. i.e. If there were 51 coconuts originally, the guy divides them into 5 equal piles with 1 left over, takes his 1/5 (10 coconuts) and hides it, then gives 1 to the monkey, you have 40 coconuts, and that is clearly evenly divisible by 5.

If anyone could point me in the right direction, I'd be much obliged. Please no solution though!

Thanks,

170.140.169.129 (talk) 05:10, 10 February 2011 (UTC)

It's no surprise that this puzzle is giving you hell, because in fact this puzzle is specifically designed to give people hell. :-) It's a very famous puzzle; it was first introduced to a wide audience in a short story called "Coconuts" by Ben Ames Williams, published in 1926 in the Saturday Evening Post. In the story, a man named Wadlin is an accountant who works for a building contractor named Dean Story, whose competitor is Marr. Both Story and Marr intend to place bids on a building contract. The night before the bids are due, Wadlin, knowing Marr’s love of puzzles, gives this problem to him. Marr stays up till dawn trying to solve it and thus misses the deadline, allowing Story to win the bid at a comfortable profit. (Mr. Williams didn't include the solution to the puzzle in his story. In the first week after it was published the Post received over 2,000 letters from readers asking for the answer or offering solutions. The editor-in-chief, George Horace Lorimer, sent a desperate telegram to Williams that read, "FOR THE LOVE OF MIKE, HOW MANY COCONUTS? HELL POPPING AROUND HERE.")

Here's an example to see why the premise isn't flawed, at least: Start with 21 coconuts. Then you can give one to the monkey, split the remainder into five piles of 4 coconuts each, take one pile, and put the remaining 16 coconuts back into a pile. Now 16 gives a remainder of 1 when divided by 5, so at least the second man can do his thing. Of course, with this example you run into a problem with the third man, but at least it shows that you don't necessarily end up with a multiple of 5 after the first man is done.

Here is a hint about the solution: It's not unique. If you find one solution to the problem, you can always add a certain number of coconuts to get another valid solution. So there are in fact infinitely many solutions. This is why, if you try to write out equations to solve, you will find that you have one fewer equation than the number of variables, so you can't solve the system explicitly. —Bkell (talk) 05:22, 10 February 2011 (UTC)

By the way, there is an excellent presentation of this problem in the chapter called "The Monkey and the Coconuts" in The Colossal Book of Mathematics by Martin Gardner, including a very clever solution that uses ARTNGVIR PBPBAHGF (possible spoiler there, encrypted with ROT13). —Bkell (talk) 05:27, 10 February 2011 (UTC)

Well, we know that the pile in the morning is one more than a multiple of 5. So it's ${\displaystyle 5n+1}$ for some ${\displaystyle n}$. We also know that it should be a multiple of 4, since that's what the fifth man left us with. So ${\displaystyle 5n+1\equiv 0}$ (mod 4). What form does that tell us ${\displaystyle n}$ needs to be in? Once you have that, what's the expression for the size of the pile before the fifth man got to it (expressed without fractions)? Can you find a pattern?--203.97.79.114 (talk) 08:24, 10 February 2011 (UTC)

What I'd to is to work with fractional coconuts. Start out with X coconuts. Remove one for the monkey, and subtract a fifth of the remainder. Do this five times, then remove one coconut for the monkey and divide by five. The result is how many coconuts each man gets in the final division, and has the form ${\displaystyle {\tfrac {4^{5}X-M}{5^{6}}}}$ for some M that's easy to compute. This fraction, at least, must be an integer, so ${\displaystyle 4^{5}X\equiv M{\pmod {5^{6}}}}$. Because ${\displaystyle 4^{5}}$ and ${\displaystyle 5^{6}}$ are relatively prime, Euclid's algorithm can be used to find a unique solution to that, and thus directly the least positive X that matches. It then remains only to verify that the intermediate divisions also come out even for this X. –Henning Makholm (talk) 09:51, 10 February 2011 (UTC)

(But the strategy sketched by 203.97 is probably simpler in practice, particularly with pen and paper -- it avoids most or all of the long divisions I'd need. –Henning Makholm (talk) 09:59, 10 February 2011 (UTC))

Please calculate the total taxi fare and other fees from Itoman, Okinawa all the way to Cape Soya, Hokkaido.

Taxis are notoriously overpriced in Japan. Therefore, I have a years-old curiosity: How much would it cost to ride a taxi from Itoman City, Okinawa Prefecture, all the way up to Cape Soya? Please calculate the base fare, then add in the other ancillary costs like ferry tickets, highway tolls, feeding and lodging the driver, possibly fuel, and all the other costs to be taxied from the southernmost roadable point in Japan to the northernmost. I would most appreciate any mathematician being able to finally put my curiosity to rest. --70.179.173.95 (talk) 07:22, 10 February 2011 (UTC)

The link http://maps.google.com gives you the distance (3.619 km) and mark which roads are priced. Bo Jacoby (talk) 10:11, 10 February 2011 (UTC).

H2: equidistant curve tangent to a given line

In the hyperboloid model of the hyperbolic plane (where t is imaginary), I have two lines L and M, defined by their unit normals u and v. If |u·v| > 1 they are ultraparallel, and some curve equidistant to L is tangent to M, at point P. Here's my question: Can I find u·P (a function of the signed distance between the nearest points of L and M) without generating P? —Tamfang (talk) 08:00, 10 February 2011 (UTC)

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