Wikipedia:Reference desk/Archives/Mathematics/2008 July 26

= July 26 =

Integrals
If you have$$\int{u}{v}\,dx$$, where u and v are both functions of x, can you split it up into $$\int{u}\,dx * \int{v}\,dx$$? 92.0.139.45 (talk) 10:39, 26 July 2008 (UTC)


 * No, not usually. For example, if u(x)=x and v(x)=x then
 * $$\int{u}{v}\,dx = \int{x^2}\,dx = \frac{x^3}{3}$$
 * $$\int{u}\,dx \int{v}\,dx = \left( \int{x}\,dx \right)^2 = \frac{x^4}{4}$$
 * See integration by parts for the rule that actually applies here. Gandalf61 (talk) 10:51, 26 July 2008 (UTC)


 * It does work for addition, though. $$\int (u+v)\,dx=\int u\,dx+\int v\,dx$$ --Tango (talk) 15:18, 26 July 2008 (UTC)

3d annulus?
What is the name for the 3-d equivalent of an annulus, ie. the region between two spheres, not the same size, with a common centre? 203.221.127.38 (talk) 17:16, 26 July 2008 (UTC)
 * If there is a name for such, it is not in List of geometric shapes but probably should be. GromXXVII (talk) 17:52, 26 July 2008 (UTC)


 * "Thick shell" would be a good name eg see Shell theorem              87.102.86.73 (talk) 17:58, 26 July 2008 (UTC)


 * "solid of revolution of the annulus" would be a fairly unambiguous interpretable description.87.102.86.73 (talk) 18:41, 26 July 2008 (UTC)


 * If we are to be rather pedantic, this is somewhat ambiguous, since if the axis of revolution does not go through the center of the annulus, then you get something else. I would use "symmetric difference of two concentric balls". Oded (talk) 00:59, 27 July 2008 (UTC)


 * "Hollow Sphere" is in fairly common use http://www.google.co.uk/search?ndsp=20&um=1&hl=en&q=hollow%20sphere&ie=UTF-8&sa=N&tab=iw 87.102.86.73 (talk) 18:56, 26 July 2008 (UTC)
 * I would interpret "hollow sphere" to mean a shape with zero thickness, in other words a 2-sphere. I think "thick shell" is the best term. Solid of revolution is ambiguous if you don't specify the axis - unless you say otherwise, everyone is going to assume you mean a diameter, but it's still ambiguous. --Tango (talk) 23:01, 26 July 2008 (UTC)


 * As you might expect, it's called an annulus. --C S (talk) 17:41, 27 July 2008 (UTC)
 * A 3-annulus? I've never heard the phrase before, but I suppose it's pretty clear what it would mean. --Tango (talk) 18:41, 27 July 2008 (UTC)


 * This term is used in all dimensions, not just 2 and 3. For example, see annulus theorem, one of the most famous results in geometric topology.  Sadly, WP seems lacking in this department, but I'll get around to it eventually.  (I don't watch this page regularly.  Leave a note on my talk page if you have further questions)--C S (talk) 19:11, 27 July 2008 (UTC)

Thanks, folks. You always deliver :) 202.89.166.179 (talk) 11:58, 28 July 2008 (UTC)

In a nutshell, its gotta be a shell, surely

Simultaneous equations
I have three simultaneous equations in three variables. I either have to 'solve the simultaneous equations or prove that they have no solution'. After some investigation, it becomes apparent that the first two equations can only have a solution if they are the same equation. This now leaves me with two equations in three unknowns. I can tell from this that there is no unique solution to these equations, however does it count as solving the equations if I put one variable in terms of another, which I can do? Thanks 92.3.123.172 (talk) 18:45, 26 July 2008 (UTC)
 * Is this hypothetical or do you have the equations? Because - After some investigation, it becomes apparent that the first two equations can only have a solution if they are the same equation doesn't make much sense to me, are they not linear equations?
 * If you have two equations with three unknowns then you should do what you suggest, which is as near to a solution as you may be able to get.87.102.86.73 (talk) 19:02, 26 July 2008 (UTC)
 * 'Solving equations' means to find the solutions to the equations. If there is no solution, (which is usually the case when the equations contradicts one another, such as 0x=1), it means stating that fact. If there is only one solution, (which is usually the case when the equations are linear, such as x+y=2; x&minus;y=0), it means to state that solution. If there is a finite number of solutions, (which is usually the case when the equations are of finite degree > 1, such as x2=x+1), it means to list these solutions. If there is a continuum of solutions, (as is usually the case when there are more unknown than equations, such as x=y), it is not possible to list all the solutions. The best one can do is to put one variable in terms of another, so yes, it counts! Bo Jacoby (talk) 19:09, 26 July 2008 (UTC).


 * Please keep all advice as general as possible, I want to do all the hard work for myself. The equations are:
 * $$x+y+az=2$$


 * $$x+ay+z=2$$


 * $$2x+y+z=2b$$
 * If you make y the subject of the first equation and sub it into the second, then equality can only hold if either a=1 or y=z. Subbing both of those conditions into equations 1 and 2 reduces both of them to $$x+2y=2$$.
 * While I appreciate your advice 87.102.86.73, I find it a bit unhelpful because the questions tells me to either solve the simultaneous equations or prove they have no solution. You seem to be suggesting something that is in between these two things. Does giving the equations make it any clearer? 92.3.123.172 (talk) 19:13, 26 July 2008 (UTC)


 * x+y+az=2=x+ay+z
 * y+az=ay+z
 * (a-1)z=(a-1)y
 * x=y y=z (or a=1 as you say)
 * so the equations become
 * x+(1+a)z=2
 * 2x+2z=2b
 * Which you can solve as you said - BUT you have 5 unknowns and only three equations.87.102.86.73 (talk) 19:26, 26 July 2008 (UTC)
 * You must have made a mistake because from the first two I get x=y. Did that make sense? Can you take it from there.87.102.86.73 (talk) 19:23, 26 July 2008 (UTC)
 * No, you made the mistake; you eliminated x in your first step, leaving everything in terms of y and z, and then somehow added in back in in the last line. Try subbing in x=y and then try y=z. I also forgot to mention that a and b are both constants. I think I should probably just go with putting one variable in terms of another. 92.3.123.172 (talk) 19:28, 26 July 2008 (UTC)
 * Yes, (corrected) - you still have too many unknowns.87.102.86.73 (talk) 19:31, 26 July 2008 (UTC)
 * I wrote in my last post that a and b are both constants, not variables. Unless, by 'too many unknowns', you mean 3 variables for two equations, which I mentioned when I started this thread. 92.3.123.172 (talk) 19:33, 26 July 2008 (UTC)
 * After solving $$x+2y=2$$ and $$2x+y+z=2b$$ simultaneously, the only way they can both have a solution is if they are the same equation. So I take it that I just have to prove the three original equations have no solution, which I think I've basically done by showing that the only solutions make them into the same equation. 92.3.123.172 (talk) 19:40, 26 July 2008 (UTC)
 * After solving $$x+2y=2$$ and $$2x+y+z=2b$$ simultaneously, the only way they can both have a solution is if they are the same equation. So I take it that I just have to prove the three original equations have no solution, which I think I've basically done by showing that the only solutions make them into the same equation. 92.3.123.172 (talk) 19:40, 26 July 2008 (UTC)


 * (that should have been an edit conflict... somehow it just put my comment below the ones that weren't present when I started editing) Here's your mistake: you proved that a=1 OR y=z, which is correct. Then you assumed that a=1 AND y=z, which is incorrect. When y=z, a doesn't have to be 1. When a=1, y doesn't have to equal z. When you have an "OR" like that, you need to divide the problem into 2 subproblems and solve them separately. First substitute a=1 (leaving y and z as independent unknowns) and solve (or prove unsolvable) the resulting three equations. Then go back to the start, substitute y=z (but leave a unknown) and do it again. Your final answer will be something like "If a=1, there are _____ solutions. If a!=1, there are _____ solutions." --tcsetattr (talk / contribs) 19:51, 26 July 2008 (UTC)
 * Ah, I see. It does make sense not to solve with a=1 and y=z at the same time. It'll still leave me with three unknowns for two equations but perhaps the equations won't have to be identical for there to be a solution. Thanks. 92.3.123.172 (talk) 19:58, 26 July 2008 (UTC)


 * You correctly deduced that either a=1 or y=z. What you should do now is split the rest of the solution into two cases, like this: "Case 1: a=1. Then [...]. Case 2: a≠1. Then it follows that y=z and [...]." You shouldn't substitute both a=1 and y=z at the same time, since that will get you only a subset of the solutions. Note that in case 1 you won't be able to solve the equations completely—you'll be able to find y+z but not y or z individually. -- BenRG (talk) 20:03, 26 July 2008 (UTC)


 * You have as one option x+(1+a)z=2 and 2x+2z=2b (or x+z=b)
 * Therefor 2-(1+a)z=x=b-z
 * 2-az=b
 * 2-b=az
 * (2-b)/a=z ?! This is a solution?


 * After solving $$x+2y=2$$ and $$2x+y+z=2b$$ simultaneously I get:
 * x+2y=2 and 2x+y+z=2b
 * 2-2y=b-y/2-z/2 or 1-x/2=2b-z-2x
 * 4-4y=2b-y-z or 2-x=4b-2z-4x
 * z=2b+3y-4 or 2z=4b-3x-2
 * 87.102.86.73 (talk) 19:50, 26 July 2008 (UTC)


 * I think the confusion is because you actually have a family of sets of simultaneous equations. Your final answer is going to something along the lines of "For these values of a and b, there are infinitely many solutions, of the form (whatever), for these other values of a and b, there is a unique solution, (whatever) and for these values of a and b, there are no solutions." It looks like you've made a good start, just keep going like you are but keep in mind that the number of solutions will depend on a and b. --Tango (talk) 23:10, 26 July 2008 (UTC)

Solving n equations in n unknowns is conveniently split into two steps. The first step is to combine the equations, trying to end up with n equations each in one unknown. The second step is to solve these independent equations one by one. When the equations are linear, as in your case, the second step is reduced to making a division. The equations are x+y+az&minus;2 = x+ay+z&minus;2 = 2x+y+z&minus;2b = 0. Eliminate the variable x by subtracting the first expression from the second. As both expressions are 0 the difference is also 0. (x+ay+z&minus;2)&minus;(x+y+az&minus;2) = 0 or (a&minus;1)y+(1&minus;a)z = 0. Consider separately case one: a=1, and case two: a&ne;1. Case one gives x+y+z&minus;2 = x+y+z&minus;2 = 2x+y+z&minus;2b = 0. The two first expressions are identical, so effectively there are only two equations: x+y+z&minus;2 = 2x+y+z&minus;2b = 0. You eliminate the variable y by subtracting the first expression from the second. You get (2x+y+z&minus;2b)&minus;(x+y+z&minus;2) = 0. You eliminate x by subtracting twice the first expression from the second: (2x+y+z&minus;2b)&minus;2(x+y+z&minus;2) = 0. Having done the hard work you end up with one equation in x and another equation in y and z. In the second equation you may consider z to be an independent variable, and y to be defined as a function of z. Then proceed to case two. Bo Jacoby (talk) 05:04, 27 July 2008 (UTC).


 * I know it's been a while, but I would like to suggest some answers. Again, if I'm wrong, please don't tell me the answers.
 * There are only two ways that the first two equations can have a solution; if either a=1 or y=z. In the case a=1, the solution of the whole system is x=2(b-1) and y+z=2(2-b). In the case y=z the solution of the whole system is $$ x=\frac{ab+b-2}{a} $$ and $$ y=z=\frac{2-b}{a} $$. In both cases there are infinitely many solutions but in the case y=z a=0 is not a possible solution. How's that? 92.0.60.66 (talk) 17:30, 2 August 2008 (UTC)

Well done. The solution (x,y,z) = ((ab+b&minus;2)/a, (2&minus;b)/a, (2&minus;b)/a) counts as one solution because a and b are constants. What happens if a=0? The answer depends on whether b=1 or not. Bo Jacoby (talk) 21:35, 2 August 2008 (UTC).


 * Thanks for checking it for me, really appreciate it. I don't see though why b equalling 1 will have any bearing on the solution if a=0; isn't division by zero undefined? 92.0.60.66 (talk) 21:39, 2 August 2008 (UTC)

Yes, division by zero is undefined, so you must insert a=0 in the original equations to solve that case. Bo Jacoby (talk) 08:30, 3 August 2008 (UTC).

Probablity equation?
Is there any probablity equation using which we can assess the probablity of frequent natural events like rainfall on a particular day? If so, what is it?117.201.96.242 (talk) 19:14, 26 July 2008 (UTC)


 * The short answer is not really.
 * But if it rains x days a year you can say the probability of rain is x/365 everyday.
 * If you have many years of data you can assess each day separately for the probability of rain, and get a better guess. But you will need data from previous years etc..87.102.86.73 (talk) 19:29, 26 July 2008 (UTC)


 * And, of course, the probability of rain in the next few days can better be estimated based on current weather patterns (is a warm front moving in ?) than historical data. StuRat (talk) 05:16, 27 July 2008 (UTC)


 * Once again you have done an immense favour to be. I'm absolutely surprised by being familier to your depth of knowledge. Thank you for answering. The common people should learn from you.117.201.96.242 (talk) 20:04, 26 July 2008 (UTC)


 * We ARE the common people. But some of us are more common than the others!

Trigonemtry problem
I've been itching away at this for the post couple hours, but still can't get this figured out. Please help me, and if you can, please tell me how you got to the answer.

tan alpha + cot alpha = 4/2

alpha 's period is (4/7, pi/1)

cos^2(x) - 2sin(x)cos(x) - sin^2(x)

so

given f(x) = cos^2(x) - 2sin(x)cos(x) - sin^2(x)

Find the min positive period of f(x)

Find the max and min values of f(x)  —Preceding unsigned comment added by Benadrill (talk • contribs) 21:35, 26 July 2008 (UTC)


 * You seem to be sometimes saying alpha is a variable, at other times it's a constant. I'll assume it's a constant, in which case the first equation becomes a simple quadratic (two solutions), and the answers follow easily. The last one, try applying some trig identities. -mattbuck (Talk) 21:58, 26 July 2008 (UTC)

The function you've called &fnof;(x) is just cos(2x) &minus; sin(2x), by a couple of standard trigonometric identities. That has period &pi;. Michael Hardy (talk) 18:09, 28 July 2008 (UTC)