Wikipedia:Reference desk/Archives/Mathematics/2006 September 7

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September 7

I'm absolutely stumped by this one

$y^{5}+3x^{2}y^{2}+5x^{4}=12^{x}$

Solve for y.

--Ķĩřβȳ♥ŤįɱéØ 05:00, 7 September 2006 (UTC)

is this a serious question? --SeanMcG 06:00, 7 September 2006 (UTC)

Yes, I am wondering how (if it is even possible) to solve this equation in terms of y. --Ķĩřβȳ♥ŤįɱéØ 06:19, 7 September 2006 (UTC)

You're done. The polynomial $y^{5}+ay^{2}+c=0$ is not solvable in radicals. The most compact answer to the question is: "the roots of $y^{5}+3x^{2}y^{2}+5x^{4}=12^{x}$ in the variable y". See the Abel–Ruffini theorem and note that the stated problem is a quntic in y (since the exponential on the right hand side is independent of y). The result can be expressed in terms of Jacobi's elliptic functions and/or hypergeometric functions, but this isn't normally any more informative than the original statement of hte problem. -- Fuzzyeric 06:36, 7 September 2006 (UTC)

What about using implicit differentiation?--Ķĩřβȳ♥ŤįɱéØ 07:20, 7 September 2006 (UTC)

Do you really mean $12^{x}$ on RHS, or should that be $12x$, that term would make it almost impossible to solve analytically. --Salix alba (talk) 07:37, 7 September 2006 (UTC)

If you just need a solution: x = 0 and y = 1 is one. --Lambiam^Talk 08:30, 7 September 2006 (UTC)

The OP's intention was to find y as a function of x, so the 12^x term alters nothing. As for implicit differentiation, it's just that - differentiation. It may tell you something about the derivative of y, but it will not give you an expression for y itself. The latter will require, as was mentioned, the usage of non-elementary functions. -- Meni Rosenfeld (talk) 10:37, 7 September 2006 (UTC)

Globe

How many spindles must a globe have for a point on its surface to describe any path? Rentwa 06:03, 7 September 2006 (UTC)

Could you rephrase the question, or give some context? What is a "spindle"? —Tamfang 07:02, 7 September 2006 (UTC)

If you tried to predict the motion of earth (as a 'stationary' observer) using a single N Pole - S Pole axis you'd find it wasn't sufficient (my house would appear to 'wobble' as it went past).

My Q is: how many axes (poles, spindles, whatever) are required to describe the wobbliest motion? I.e. is there a point at which adding poles becomes redundant and a finite set of poles can describe any motion, or is an infinite number necessary? Rentwa 07:09, 7 September 2006 (UTC)

If my understanding of the term 'path' here is correct, then you would need every possible axis: given any randomly chosen axis of rotation, then for the path, just take a circle around this axis. Doesn't that force you to have that (and hence every) axis? Madmath789 07:33, 7 September 2006 (UTC)

Can you supply an outline of a proof then? Rentwa 07:43, 7 September 2006 (UTC)

The Rotation group of a sphere can be described by three angles, two specifying the axis of rotation (equivilent to defining a point of the sphere) and 1 for the angle of rotation about it, which is SO(3,R). So that gives you a lower bound on number of spindles, the Euler angles may do you. Whether all positions can be arrived at in a continuous manner is a problem, indeed Charts on SO(3) I think describe the know possible solutions. --Salix alba (talk) 07:53, 7 September 2006 (UTC)

A cursory glance seems to me to suggest that the angular momentum velocity function in the Euler angles article can only account for a 'wobble' consisting of a finite number of parts, which may supply a proof that the function required is a polynomial of degree greater than n for some n in N, which is part way to constructing some more elaborate proof, but I'm waffling really until I've read all the articles more fully. Thanks anyway, Salix alba, I'll have a think :) . Rentwa 11:35, 7 September 2006 (UTC)

I might be answering a different question, to what your looking for. If you know that the sphere is constrained into a particular set of configurations, say if the north pole does not rotate past the equator, then you can simplify things. If you are further not interesed in changes is angular velocity around the rotation axis, that is you are basically only interested in the position of the axis, just two parameters will suffice, to descibe all positions, basically two angles, similar to polar coordninates of the sphere. Have you spotted the precession article? This is basically the same as the wobble you describe. --Salix alba (talk) 13:28, 7 September 2006 (UTC)

He he! I think precession is a better word than 'wobble'. :D Rentwa 15:35, 7 September 2006 (UTC)

Swimming Pool

How is the water sloped to make one end shallow and the other end deep? Rentwa 06:03, 7 September 2006 (UTC)

Water flows down-hill, and water seeks its own level (giving a flat surface). Sloped bottom and flat top yields a wedge-shaped chamber of water, with a shallow end and a deep end. There's less depth from the top to a high spot on the bottom than to a low spot. DMacks 06:10, 7 September 2006 (UTC)

|                           |
|                           |
|                           |
|_________                  |
  shallow \                 |
           \                |
            \               |
             \______________|
                deep

When the water is poured in, it first fills the deep, then it fills the shallow. Of course, the slope of the descent from shallow into deep is greatly exaggerrated in my crude drawing due to technical restrictions. --Ķĩřβȳ♥ŤįɱéØ 06:19, 7 September 2006 (UTC)

Reversing skill to predict

Imagine I have a game where people try to guess the outcome of some event - for simplicity's sake, let's say it results in a number up to 100, and this game involves skill and analysis. A tally of results is produced, and statistics are made with it, the most useful of which might be the average error of each person away from the actual result. So now I have a database of people, their average error, and the number of times they've played. They've also entered their guesses for the next round.

How do I convert this data into an overall prediction for the number in the next round? (If there is a name for this process, could someone point me towards it?) Thanks! —AySz88\^-^ 06:46, 7 September 2006 (UTC)

Well, there's linear prediction, which kind of assumes that your "some event" process is predictable using a "simple" model. See also forecasting and predictive analytics for a broader survey of several methods. —The preceding unsigned comment was added by Fuzzyeric (talk • contribs) 07:07, September 7, 2006 (UTC).

You need some model. Here is one possible model: A participant's prediction is the true future outcome + an error term. For each participant, the error term is a random variable with a Gaussian distribution with fixed, participant-dependent mean and variance. In a formula:

$P_{i}=F+\mu _{i}+\sigma _{i}Z_{i},\,$

in which the $Z_{i}$ are independent random variables with mean 0 and variance 1.

From the past performance you can estimate the parameters μ_i and σ_i. Now, given all P_i, the maximum likelihood estimator for F is a weighted sum of the quantities P_i − μ_i, where the weight of the i-th entry is inversely proportional to the variance σ_i².  --Lambiam^Talk 08:47, 7 September 2006 (UTC)

I don't think I'm looking for linear prediction - I'm not saying that each outcome is in any way connected to a previous outcome, just that the participants of the game have enough data to make a prediction for the outcome. In other words, the rounds are all independent of each other.

But I think the second bit was what I was trying to get an idea of... thank you! —AySz88\^-^ 01:38, 8 September 2006 (UTC)

Determinant of an orthogonal matrix

A is orthogonal matrix.then show that det A=+_1 70.179.185.146 11:33, 7 September 2006 (UTC)

Try this. Rentwa 16:49, 7 September 2006 (UTC)

There are several ways to prove this. Intuitively, you can think that an orthogonal nxn matrix means the columns give an orthonomal basis for n dim. space. Since a determinant finds the volume of shape with edges given by the columns of the matrix and parallel lines to those, the determinant of an orthogonal matrix is just the unit hypercube in n dimensions. As such it has volume 1 or negative 1 (depedning on the orientation of the axes).

A slicker proof, which is less intuitive, is to note that det(A) = det(A^T). Thus, since an orthogonal matrix is defined as A such that A*A^T = I, the identity matrix, we have

$\det(AA^{T})=\det(I)$

$\det(A)*det(A^{T})=1$

$\det(A)^{2}=1$

$\det(A)=\pm 1$

Hope this helps. --TeaDrinker 19:01, 7 September 2006 (UTC)

differentiation

2^x=2x,FIND X80.32.12.22 17:38, 7 September 2006 (UTC)DEREK,SPAIN

How would you use differentiation to solve this question? --HappyCamper 17:43, 7 September 2006 (UTC)

If you consider the function y=2^x-2x, the solution to the above equation is the root of this transendental equation. As such, I don't think there is an algebraic answer. You can, however, find the root approximately using Newton's method. You can also show that a solution exists (which it does) by proving the above function takes on both positive and negative y values (since it is continious, this shows a root exists). --TeaDrinker 18:43, 7 September 2006 (UTC)

See Lambert's W function. (By the way, the answer is trivial.) - Fredrik Johansson 18:56, 7 September 2006 (UTC)

Some numbers among the two first integers give trivial answers. -- DLL ^{.. T} 20:02, 7 September 2006 (UTC)

You can solve it generally, but if you need only real answers x = 1 and x=2 by graph intersection.M.manary 23:58, 7 September 2006 (UTC)

Or if you want to be strictier, you can use differentiation to show that they are all the real roots there can be. --Lemontea 15:19, 8 September 2006 (UTC)

That is the sound of me slapping my forehead--I didn't actually work out the problem, just thought of how I would solve it... --TeaDrinker 01:56, 8 September 2006 (UTC)

Over the Complexes, there are an infinite number of solutions. ${\frac {-1}{\ln 2}}W({\frac {-\ln 2}{2}})$ (which has an infinite number of values because W() is infinitely multi-valued. Q.v. Lambert's W function for an introduction to the W() function. Only one of those values is in the Reals (and is the one given above). -- Fuzzyeric 03:20, 8 September 2006 (UTC)

... and for extra credit try $2^{x}=x^{2}$. Gandalf61 15:57, 8 September 2006 (UTC)

Actually how would one solve analytically once the solution ${\frac {-1}{\ln 2}}W{\bigg (}{\frac {-\ln 2}{2}}{\bigg )}$ is obtained ?

And how is it possible to solve Gandalf61's equation ? --Xedi 21:36, 14 September 2006 (UTC)

Calculus help - Limits

I am trying a few practice problems and I was wondering if someone could help me.

The questions state: Evaluate the limit, if it exists.

1.) lim (t^2 - 9)/ (2t^2 + 7t + 3)

x -> -3

I factored the top to (t + 3)(t - 3) and then I tried factoring the bottom, but I couldn't.

2.) lim (x^3 - 1)/(x^2 - 1)

 x -> 1

3.) lim (1/t square root(1+t)) - (1/t)

 t -> 0

Thank you! —The preceding unsigned comment was added by 151.213.201.30 (talk • contribs) .

Generally we're not here to answer your homwork questions, so how about a hint? Your approach is quite feasible in the first two cases. A hint on factoring them is to note that (if the limit exists), there is a common factor in the top and bottom. That is, -3 must be a root of both the top and bottom of number 1, and 1 must be a root of the top and bottom in the second problem.

For the third problem, try writing it with a common denominator. Does the limit still go to an indefinite form?

--TeaDrinker 18:29, 7 September 2006 (UTC)

Also, did they specify whether the limits are approached from numbers greater than, or less than, each limit ? In some cases, this can make a difference. For example 1/t, as t -> 0, would be either +inf or -inf, depending on the direction of the approach to the limit. StuRat 18:34, 7 September 2006 (UTC)

Why don't you use the law of the hospital and differentiate it and then take the limit? You can differentiate using http://www.calc101.com/webMathematica/derivatives.jsp Ohanian 23:54, 7 September 2006 (UTC)

Yes, if you are familiar with derivatives, L'Hôpital's rule will also work. --TeaDrinker 01:55, 8 September 2006 (UTC)

The first example seems inconsistent, you have variable t in the expression and x in a limit. I asume both shoul be t.

If you can verify that the denominator becomes zero for t=-3, then you may try to factorize it by dividing polynomials (2t^2 + 7t + 3) / (t + 3). --CiaPan 08:02, 8 September 2006 (UTC)

Looking For Assistance To A Vexing Problem

I would like to know if a formula exists by which I can calculate the specific diamter that a circular piece of .037 gauge/mill uncoated polystyrne material must be so that the piece of material will "slowly" sink through the liquid. I am attempting to replicate the effect of how "snow" or glitter travels through liquid in a "snowglobe" which is commonly purchased in souvenir and gift shops. Any assistance of guidance would be most appreciative! I thank you in advance for your interest and willingness to reply! Thank you!Topromote 18:55, 7 September 2006 (UTC)Frederic I. Goldsmith, C.A.S.

What liquid is it in? Perhaps you can use some formulas that Einstein derived for viscosity? If we can assume those snowflakes form a dilute solution of hard spheres, there might be something interesting to be said. this might be interesting, but you'll need to sift through some of the formulas to see if anything is useful. Just some ideas off the top of my head. --HappyCamper 21:48, 7 September 2006 (UTC)

recurring

what is the mathmatical sign for recurring eg o.33333333

If the pattern is obvious, I'd just append three dots: "333...". Fredrik Johansson 20:11, 7 September 2006 (UTC)

Traditionally, the notation used is to use one dot over a single recurring digit, and if there is a group of recurring digits, then put a dot over the first and last in the repeating group. Madmath789 20:14, 7 September 2006 (UTC)

The ellipse method is acceptable, but for mathematical papers you should write a three with a bar over it as long as you know what exactly repeats. For instance:

${\frac {1}{3}}=0.33333...\ \ \ \ \ \ {\frac {1}{7}}=.142857142857...$
but it is better to write ${\frac {1}{3}}=0.{\overline {3}}\ \ \ \ \ \ \ {\frac {1}{7}}=0.{\overline {142857}}$

You can see where that notation is useful, especially for fractions etc. that repeat more than one different digit. M.manary 20:20, 7 September 2006 (UTC)

But for a fraction with denominator with a factor or two or five, the entire thing does not recur. For example:

${\frac {3}{14}}=0.2142857142857...$

we really need to write

${\frac {3}{14}}=0.2{\overline {142857}}$

--MathMan64 00:21, 8 September 2006 (UTC)