Wikipedia:Reference desk/Archives/Mathematics/2008 July 13

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July 13

Powers

Hello. If 60^a = 3 and 60^b = 5, then why does $12^{\frac {1-a-b}{2(1-b)}}=2$? Thanks in advance. --Mayfare (talk) 02:23, 13 July 2008 (UTC)

The reason for that is because

$60^{a}=3\Rightarrow a={\frac {\ln 3}{\ln 60}}$

$60^{b}=5\Rightarrow b={\frac {\ln 5}{\ln 60}}$

by taking the natural log of both sides, using their properties and then dividing. After that you just plug in these values of a and b into $12^{\frac {1-a-b}{2(1-b)}}$ and simplification (making extensive use of properties of logs) will give you 2. Is there a particular step that is giving you trouble?--A Real Kaiser (talk) 02:48, 13 July 2008 (UTC)

That'll get you to the answer, but it looks neater if you work with $a={\frac {\log _{12}3}{\log _{12}60}}$ and $b={\frac {\log _{12}5}{\log _{12}60}}$. Also, simplify the rational expression ${\frac {1-a-b}{2(1-b)}}$ first so you don't have to substitute b in 2 places. --tcsetattr (talk / contribs) 06:05, 13 July 2008 (UTC)

Question about baseballs and wind resistance

If x were to represent velocity, angle, and wind resistance, would it travel in a standard trajectory? My understanding is no.--Baseball and and and Popcorn Fanatic (talk) 06:28, 13 July 2008 (UTC)

Are you asking whether a baseball that travels with a non-zero velocity at a non-zero angle in air would follow a symmetrical path? If so, the answer is indeed "no". Here's an analysis of the ball's trajectory:

(1) At the moment the ball is hit, it is both moving vertically and horizontally. Both the horizontal and vertical speeds decrease due to air resistance.

(2) The vertical speed eventually becomes 0 and the ball begins to accelerate in its fall until its terminal velocity is reached, if it ever is reached. It will stop when it hits the ground. The acceleration while the ball was climbing was -9.8 m/s^2 plus the acceleration applied by air resistance; the acceleration during the fall is 9.8 m/s^2 minus the acceleration applied by air resistance. As you can see, the ball takes longer to reach the ground from its maximum point, and therefore end its movement, then it took to reach the maximum point.

(3) The horizontal speed of the ball continues to decrease until the ball hits the ground. Since both the horizontal speed and the vertical speed decrease, the path might be parabolic under certain circumstances, but I'm unsure about whether this is true. --Bowlhover (talk) 07:25, 13 July 2008 (UTC)

I think Baseball and Popcorn fanatic was not asking about symmetry, but about consistency. That is to say, if you through again with the same velocity and angle, would you get the same results? There are several factors effecting the ball's motion: initial velocity and angle, spin, wind, altitude. All these effect the ball's trajectory. The more of these that are accounted for, the more close would be the theoretical trajectory to the actual trajectory. If we try to be very precise, then wind itself is not constant over time or space, hence rather complicated. So I think that the answer is that the motion is not really standard. Oded (talk) 08:30, 13 July 2008 (UTC)

There are some dynamical systems that exhibit sensitivity to initial data. This is sometimes called the butterfly effect. The meaning of this is that event if you now all the parameters (in this case, the wind, initial velocity, spin, etc) but you knew them only up to some finite accuracy (which is practically always the case), you could not predict the resulting motion. I would guess that the motion of a baseball through the air is to a high degree of accuracy not sensitive to initial data, and does not suffer from this effect. But I might be wrong. Is there an expert out there who might know if that is correct? Oded (talk) 17:39, 13 July 2008 (UTC)

Further problems with exponential growth model

I've tried modelling Wikipedia's growth with two different exponential functions, but both functions are inappropriate for my use. I've tried an exponential of a quadratic function: $e^{ax^{2}+bx+c}=y$ and a quadratic of an exponential function: $ae^{2x}+be^{x}+c=y$ but both functions have the quadratic coefficient as negative, meaning the curve turns downwards, and according to the models, Wikipedia will never reach five million articles.

Is there any other way I can have a function either in the form $e^{f(x,a,b,c)}=y$ or in the form $f(e^{x},a,b,c)=y$ so I could express x in terms of a, b, c and $\ln y$ and all parameters a, b and c would be independent of each other? JIP | Talk 09:20, 13 July 2008 (UTC)

I would question the assumption that Wikipedia's growth is (or will permanently remain) exponential. I'd expect the rate of growth to slow once these events occur:

1) We already have articles for most topics in which people have an interest.

2) Most people interested in participating in Wikipedia have already done so, and perhaps they've even got bored and moved on.

So, trying to model the growth with the wrong type of equations would then lead to poor results. StuRat (talk) 14:25, 13 July 2008 (UTC)

For e^f(x) you want df(x)/dx to tend to zero as x tends to infinity? there are so many functions I wouldn't know where to start (maybe f(x) = ax/(bx+c) or variations of that)

Once again I suggest looking at the logistic function which is derived from dA/dt = kA(A_max-A) where A=articles..87.102.86.73 (talk) 14:31, 13 July 2008 (UTC)

I tend to agree that Wikipedia growth will probably not be exponential. It was an idea worth considering anyway. I abandoned the idea of exponential growth and tried a simple quadratic model: $ax^{2}+bx+c=y$ but it suffered from the same problem: the curve turns downwards, and the apex of the curve is below five million, so according to the model, Wikipedia will never reach five million articles. I tried formulas ${\frac {ax}{bx+c}}=y$ and ${\frac {ax+b}{cx}}=y$, but they are equivalent to $ax-ybx=cy$ and $ax-cyx=-b$ respectively, and the only possible solution to either of them is a=0, b=0, c=0. I think there has to be at least one term which has at least one of x and y but none of a, b, and c. I shall have to look into logistic functions but I do not understand enough of them yet. JIP | Talk 16:49, 13 July 2008 (UTC)

Why not try y=e^ax/(bx+c)? I think that could give a good fit..

By the way the logistic function is quite simple - simpler than the article makes it seem.. Can you solve equations of the form dy/dx=kx to show that y=e^kx? If so you should have no problem with the logistic function.

The idea behind the logistic function (in this case) is that

a. more articles mean more readers

b. more readers means more editors

c. more editors means more articles

d. But. it's assumed that there is an absolute limit on how many articles there can be - this is the origin of the (A_maxM-A) term - meaning that the chances of an editor being able to write a new article is proportional to (A_maxM-A)/A_maxM .87.102.86.73 (talk) 16:58, 13 July 2008 (UTC)

$y=e^{\frac {ax}{bx+c}}$ won't work. If $y=f(x,a,b,c)$ already gives a degenerate solution, then $y=e^{f(x,a,b,c)}$ isn't going to make it any better, because it's equivalent to $\ln y=f(x,a,b,c)$. But thanks for the information about logistic functions, I'll have to see what I can come up with. JIP | Talk 17:18, 13 July 2008 (UTC)

I'm not sure you're quite right, or maybe I haven't understood properly?

you say "...they are equivalent to ax − ybx = cy and ax − cyx = − b respectively, and the only possible solution to either of them is a=0, b=0, c=0" but if y=1 when x=0 then ln(y)=0 when x=0 so the equation becomes ax - bln(y) = cln(y) or a.0-b.0=c.0 which means that a,b,c can be anything from the initial conditions? Also the limit is y=e^a/b at infinite x (time). Is this not good?. I would ignore the behaviour when x<0 because then wikipedia does not exist??87.102.86.73 (talk) 17:41, 13 July 2008 (UTC)

As the size of wikipedia, y, is nonnegative, choose a growth model that gives nonnegative values only. So look at log y. At time x=0 when wp was created the size was zero, and in the far future the size of wp may be limited by some value. So try the function log(y)=a−b/x. Plot log(y) as a function of 1/x. Fit a and b. Then y=e^a−b/x is your growth model. Bo Jacoby (talk) 10:28, 14 July 2008 (UTC).

Thank you, this is definitely a model worth considering. However, I would want to have three independent other parameters instead of just two. But no matter how I try to disguise a quadratic equation ($e^{ax+b+{\frac {c}{x}}}=y$ or $e^{a+{\frac {b}{x}}+{\frac {c}{x^{2}}}}=y$), the discriminant always becomes negative before five million. I don't think functions involving a quadratic function of x are going to work. But I don't know how else to write a function of x with three independent other parameters. My first thought was $e^{a+{\frac {b}{x+c}}}=y$ but then I realised that the equivalent non-exponential formula, $a+{\frac {b}{x+c}}=\ln y$, isn't linear in terms of a, b, and c any more, and thus the linear equation system solver can't solve it any more, and it would be far too much trouble to solve by hand. The linear system was enough trouble as it was. JIP | Talk 18:40, 14 July 2008 (UTC)

In your formula, $a+{\frac {b}{x+c}}=\ln y$, the parameter c has the meaning that x₀=−c is the point of time when y=0. Note that b<0 in order to obtain a steady growth of y from y₀=0 to y_max=e^a when x goes from x₀ to infinity. Setting k=−b the formula is $y={y_{\max }}\cdot e^{-{\frac {k}{x-x_{0}}}}.$ Bo Jacoby (talk) 07:29, 15 July 2008 (UTC).

How to solve the nonlinear equations? Look. Suppose you have three observations, (x₁,y₁), (x₂,y₂), and (x₃,y₃). Then the three unknown parameters (x₀,y_max,k) satisfy the three equations

$y_{i}={y_{\max }}\cdot e^{-{\frac {k}{x_{i}-x_{0}}}}$ for i=1,2,3.

Conventionally, unknowns are called x and y and z and constants are called a and b. So rename the variables: x_i→a_i, ln y_i→b_i, x₀→x, ln(y_max)→y, k→z. The equations look like this:

$b_{i}=y-{\frac {z}{a_{i}-x}}.$

Getting rid of the fraction produces three polynomial equations

$a_{1}b_{1}=-xy+b_{1}x+a_{1}y-z$

$a_{2}b_{2}=-xy+b_{2}x+a_{2}y-z$

$a_{3}b_{3}=-xy+b_{3}x+a_{3}y-z$

Subtracting the first equation from the two other equations gives two linear equations

$(b_{2}-b_{1})x+(a_{2}-a_{1})y=a_{2}b_{2}-a_{1}b_{1}$

$(b_{3}-b_{1})x+(a_{3}-a_{1})y=a_{3}b_{3}-a_{1}b_{1}$

which are solved by the standard method. Substituting the solutions x and y into one of the nonlinear equations gives the last unknown z:

$z=-xy+b_{1}x+a_{1}y-a_{1}b_{1}$

Bo Jacoby (talk) 11:42, 18 July 2008 (UTC).

optimum algorithm for choosing from sequential lineup

I recall once hearing a math puzzle. I think it was framed in terms of the prince of a kingdom needing to choose a bride for his princess. He lines up all the eligible women in his kingdom, and starts going down the line, one at a time. Each woman he looks at, he may choose her for his bride. But if he does not choose her, he must move down the line, and may not come back to her, even if she turns out to be the best choice. So when discarding a choice, he must not only weigh how suitable she is, but also how likely he thinks it is that a better choice will come along. Assume the suitability of the women form a total order.

I don't recall where I heard this puzzle, nor whether I've got the framing correct; neither whether I've described the prince's quandary correctly, nor whether the puzzle actually uses a line of brides for illustration. I also don't recall what the solution was. I seem to think there is some optimum algorithm for the prince. Sounds something like this: he should look at 1/e percent of the choices, then choose the first one that's better than the 1/e'th one. Or something like that.

I can't come up with a google query that finds it for me, but that may be because I'm misremembering the specifics. So does it sound familiar to anyone here? Can you set me straight? -lethe ^{talk +} 16:12, 13 July 2008 (UTC)

Just a guess but if the attractiveness forms a linear order eg one princess has attractiveness 1, another 2 etc, and the prince knows how many princesses there are it should be easy to pick one in the top 3...87.102.86.73 (talk) 16:50, 13 July 2008 (UTC)

This is the secretary problem. Michael Slone (talk) 17:52, 13 July 2008 (UTC)

Yep, that's what I was looking for. Thanks, Michael! 97.113.58.212 (talk) 19:36, 13 July 2008 (UTC)

I've encountered this problem is real life when getting gasoline while driving an unfamiliar route. I have to look at several gas stations to get an idea for what the range of prices are, but must choose one before I run out of gas. I could technically go back to a previously passed station to get gas, but that would waste both time and gas, so usually isn't an option. StuRat (talk) 18:02, 13 July 2008 (UTC)

You might have read it in the excellent "Fifty Challenging Problems in Probability With Solutions" by Frederick Mosteller, dirt cheap as it's in Dover. There the problem is stated pretty much as you described it, and the description of the solution is very good (better than the one in the wiki article for the secretary problem). 86.15.141.111 (talk) 10:49, 14 July 2008 (UTC)

Growing Numbers

Someone posed this question to me and I have solved it by writing a program. The problem is this. To "grow" a number is to take a two digit number and add its digits to itself until a three digit number is obtained. For example, 90 and 75 both take two steps to grow because
90+9+0=99 -> 99+9+9=117
75+7+5=87 -> 87+8+7=102.
The question is what is the difference between the largest and the smallest numbers which take exactly three steps to grow? I am just looking for hints as to how to do it analytically. I already have the answer. Thanks people!--A Real Kaiser...NOT! (talk) 04:55, 14 July 2008 (UTC)

I got 65 and 81 for my answers by writing an integer program (where each variable is a digit of some growth):

max: 10*x1 + x2
s.t. 11*x1 + 2*x2 - 10*x3 - x4 = 0
     11*x3 + 2*x4 - 10*x5 - x6 = 0
     10*x5 + x6 <= 99
     11*x5 + 2*x6 >= 100
     0 <= xi <= 9
     xi integer

Maple and many spreadsheet programs will solve such programs.

maple code
maple code
with(Optimization); LPSolve(10*x1 + x2, {11*x1 + 2*x2 - 10*x3 - x4 = 0, 11*x3 + 2*x4 - 10*x5 - x6 = 0, 10*x5 + x6 <= 99, 11*x5 + 2*x6 >= 100, x1 <= 9, x2 <= 9, x3 <= 9, x4 <= 9, x5 <= 9, x6 <= 9}, assume={nonnegative, integer}); [65, [x1 = 6, x2 = 5, x3 = 7, x4 = 6, x5 = 8, x6 = 9]] LPSolve(-10*x1 - x2, {11*x1 + 2*x2 - 10*x3 - x4 = 0, 11*x3 + 2*x4 - 10*x5 - x6 = 0, 10*x5 + x6 <= 99, 11*x5 + 2*x6 >= 100, x1 <= 9, x2 <= 9, x3 <= 9, x4 <= 9, x5 <= 9, x6 <= 9}, assume={nonnegative,integer}); [-81, [x1 = 8, x2 = 1, x3 = 9, x4 = 0, x5 = 9, x6 = 9]]

Often solving analytically just means reducing the problem to a form with some known method of solution, such as an integer program. Compare to using the quadratic formula or writing x = A⁻¹b. -- KathrynLybarger (talk) 05:50, 14 July 2008 (UTC)

Well, this is how I solved the problem except that I used MATLAB. But I was just wondering if I was sitting on my desk without a calculator or a computer with just a pencil and a sheet of paper, how would I figure it out in a few minutes. I think I got it now. I just needed to spend some more time thinking on this problem.--A Real Kaiser...NOT! (talk) 22:04, 14 July 2008 (UTC)

Wright Omega function

Hello,

I have recently created the article about the Wright Omega function, which appears for example in the resolution of x = ln(x).

A definition of the function is :

$\omega (z)=W_{{\big \lceil }{\frac {\mathrm {Im} (z)-\pi }{2\pi }}{\big \rceil }}(e^{z})$

where W_k is the k-th branch of the Lambert W function.

If you look at a few graphs that are on that article, you may notice it looks like there is only one discontinuity, at x < 0, y = ±π, which corresponds to changing the branch. But then most other branch-changes are seamless, as illustrated in this image (computed with the Lambert W definition) : http://s4.tinypic.com/r0bdzo.jpg

(Red is W₀, Blue is W₁, etc...) (Also see http://i34.tinypic.com/2lu8lcw.png which shows how the functions only fit precisely in those intervals of width 2π, by showing them functions in slightly expanded intervals)

I would like to know how come all these branch-changes are seamless, how did the person who first discovered this function realise it was possible to stitch together the different branches of the Lambert W function to give this function ?

Also, does anyone have any references concerning this function (preferably freely available on internet), as I was not able to find much except from a few books.

Thanks. --Xedi^Talk 22:30, 13 July 2008 (UTC)