Wikipedia:Reference desk/Archives/Mathematics/2007 February 5

= February 5 =

What is a tangent plane to z=f(x,y) at (a,b)?
My grand children just finished calculus of functions of one real variable. They can take derivatives. The tangent at a point on a curve is defined intuitively as the limiting position of secant line through the point. I am supposed to introduce functions of two variables without dot-cross products. Tangents to curves in spaces and differentiability are not in the syllabus. Question: With my hands tied, how can I convince my kids, intuitively psychologically and emotionally but need not be logically mathematically, that the plane z=f(a,b)+D_x f(a,b)(x-a)+D_y f(a,b)(y-b) is usually frequently TANGENT to the NICE surface z=f(x,y) at (a,b)? What does it mean by a tangent plane? It appears that I have only one option, that is


 * Shut up and accept it because I am the grand father! The appearance is an obvious extension of y=f(a)+f'(x)(x-a).

What is your alternative solutions to this teaching situation? Thank you in advance. Twma 07:06, 5 February 2007 (UTC)


 * Your option sounds like a great idea. Let me know how it works out! ;-)
 * But seriously, if it is "an obvious extension", what is so obvious about it? I claim that the core intuition is that we are always looking for a local linear approximation. For a curve, we seek a line; for a surface, a plane.
 * How do we find the plane? Run a vertical slice through the surface; we should see a curve (the cut through the surface) and a tangent line (the cut through the plane). Make one slice parallel the x axis, and another parallel the y axis. That should give us two independent lines through a common point, and thus determine a plane, the tangent plane.
 * A nice general description of a linear subspace (sometimes called a flat), such as a line or a plane, is a point plus a vector space. The vector space gives all allowed displacements from the point, and a basis for that vector space allows us to move anywhere we like in the flat.
 * It could help to show
 * radial symmetry, like x2+y2
 * reflection symmetry, like xy
 * asymmetry, like x+y3
 * a crease, like |x|+y2
 * a point, like (x2+y2)1/2
 * a nasty point, like |x+y|+|x−y|
 * I see no need for cross products and "normal vectors"; the latter are not truly vectors, and fail to generalize.
 * Incidentally, we can attempt higher-order fits; that's what an osculating circle is. --KSmrqT 08:32, 5 February 2007 (UTC)

Linear subspaces, flats were not available to me for explanation. Exactly two slices as you described were at my disposal. I did that and followed up with ice cream, candy and cookies. My grandchildren were happy. At the end in a subtle way, I pleased them to be good drivers (get good marks) without knowing anything about petrol chemistry (mathematical theory). Am I an acceptable grand father? Thanks. Twma 00:50, 7 February 2007 (UTC)

This is a retarded question, but...
if $$a^m$$ X $$a^n$$ = $$a^{m+n}$$, then $$a^m$$ X $$b^m$$ = $$ab^m$$? In other words, do I multiply numbers where the exponents(?) are equal? Is that against any mathematical laws? ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:43, 5 February 2007 (UTC)
 * Also, in an unrelated question, what is the square root of this whole thing: $${{16^{4x}}^2}$$? ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 14:02, 5 February 2007 (UTC)

am x bm = (ab)m is true.

and the square root is 162x^2 - you half the exponent - (taking logs, dividing by two - then exponentiating again..)

square root (nx) = nx/2

In fact using the first rule you supplied (which is correct) you can get

(nx/2)2 = nx/2 x nx/2 = nx/2+x/2 = nx so: square root (nx) = square root ((nx/2)2) = nx/2

(I get to say QED now)87.102.8.103 15:45, 5 February 2007 (UTC) Hope that helps.87.102.8.103 15:34, 5 February 2007 (UTC)


 * I'm not sure why you characterize the question as "retarded". In fact, I consider it a good question that brings up some valuable mathematics.
 * We can expand am×an as m copies of a followed by n more, which is m+n copies, so equal to am+n. But am×bm expands to m copies of a followed by m copies of b, which is something completely different from m copies of a×b. We need something extra, that a×b = b×a, to complete a proof. In many important mathematical situtations, with objects more sophisticated than ordinary numbers for a and b, this commutative law of multiplication does not hold, and the proof does not go through.
 * Your second question requires better typesetting to be seen clearly, asking what is the square root of
 * $$ \left( 16^{4x} \right)^2 . \,\!$$
 * This also rewards careful examination. Is it the case that the square root of a2 equals a, for all a? The answer is, no. For example, −2 squares to 4, and we conventionally take the square root of 4 to be +2. We can expect this issue to arise any time more than one quantity can square to the same thing, which is common. In the given example, we should ask if 164x can ever be negative; if not, then the answer will be 164x. So, try it. --KSmrqT 22:05, 5 February 2007 (UTC)

K Thanks very much. ► Adriaan90 ( Talk ♥ Contribs ) ♪♫ 13:23, 6 February 2007 (UTC)

Curves of quickest descent
What precisely are bernoulli's curves of quickest descent?? —Preceding unsigned comment added by 59.183.58.204 (talk • contribs)


 * See our article on the brachistochrone curve. Gandalf61 15:56, 5 February 2007 (UTC)

mark up question
I have been told that when we markup costs for our work instead of just for example muliplying the cost times 1.20 for a 20% makup we should divide by 0.80. Which is correct?


 * Multiplying by 1.2 gives a 20% markup; dividing by 0.8 is the same as multiplying by 1.25, so gives a 25% markup. All on cost, though - if you define markup as wrt the derived figure, naturally the arithmetic will be different.…86.146.174.174 17:43, 5 February 2007 (UTC)


 * Remember that dividing is the same as multiplying by one over the number. Mathmo Talk 20:09, 5 February 2007 (UTC)

Infinity + 1
This isn't a homework question, but it's based on something our teacher explained in class. I now have some idea why he didn't continue any farther. It's a geometric coloring problem. Take a circle of radius r, a point on that circle, and two colors. Every point on the circle must be a different color from any point on the circle that's unit distance away from it. Is it, or is it not, possible to color this circle? In some cases, where r=1 for instance, if you start coloring clockwise from a point, you return to your starting point within a finite number of steps. If this covers an odd number of points, the coloring is impossible, otherwise it's easy. What about when r is irrational, though? If you have to cover an infinite number of points before you return to the starting point, do you wind up with a contradiction, or not? Or is it like quantum mechanics, where you get both? I guess what I'm asking is, is infinity even or odd? Black Carrot 20:15, 5 February 2007 (UTC)


 * There's a paradox known as Thomson's lamp which addresses this issue; if you can flick a light switch an infinite number of times in a second, is it on (odd) or off (even)? I think that the current opinion is that it is both, and that defining infinity as either odd or even is meaningless (if it's odd, you can just add one to make it even and vice-versa). See also Hilbert's paradox of the Grand Hotel, which considers a similar problem. Laïka  20:58, 5 February 2007 (UTC)


 * You should not confuse philosophical issues with actual mathematics. The question you asked is about a concrete mathematical problem (the circle part, not the "is infinity even" part). Unless I am somehow mistaken, the answer is yes, but this may require using the axiom of choice. And no, it certainly cannot be both (in classical logic, anyway, and unless we are using an inconsistent system), though it can be independent from a given axiom system (for example, it might be independent of ZF sans choice). By the way, you meant irrational circumference, not irrational radius. -- Meni Rosenfeld (talk) 22:27, 5 February 2007 (UTC)

No, I mean an irrational radius, if I worked the formula for chord through correctly. The problem says "unit distance", not "unit distance along the circle". I think this is slightly different from Thomson's lamp and the Grand Hotel, and you're right, the divisibility of infinity may not be related either. I can't imagine, though, what "yes" even means, let alone how you got it as an answer. Could you give more detail? Black Carrot 22:46, 5 February 2007 (UTC)
 * What you really want to know is not whether r is rational or irrational, but whether a chord of unit length subtends a rational or irrational angle (as a fraction of the circle, not in radians). In the case of r=1, the answers are the same, but this will not be true in general.
 * Anyway, suppose the angle is irrational. Then you get a group action from the integers to the circle, generated by the operation "move counterclockwise by an angle with a chord length of 1", and the orbit of the action is countably infinite. From each orbit, pick one point (this is where you use the axiom of choice) and call it red; then the points one step away are black, the points two steps away are red, and so on.
 * Is the axiom of choice necessary to prove this result? My guess is yes, but I don't know for sure. You could prove it, if you could show, say, that the set of all red points is not Lebesgue measurable, or lacks the property of Baire (though these proofs assume the consistency of large cardinals). --Trovatore 23:16, 5 February 2007 (UTC)


 * Oh, actually you don't need any large cardinals to show AC is necessary if you can show the set of all red points lacks the property of Baire. If all you can show is it's not Lebesgue measurable, you need an inaccessible cardinal. --Trovatore 00:29, 6 February 2007 (UTC)

You're right, I copied the formula wrong. So, supposing it's irrational. I can pretty much understand the group action page, it seems to mean "shuffle everything in the set around." Naturally, we can say more specifically that the action consists of rotating the circle by a particular angle, and mapping everything from before to after. We chose irrational angles specifically to make the orbit infinite, so naturally the orbit is infinite. No problems so far. That's about where you lose me, though. How is it "countably" infinite? I could have sworn it would hit every point on the circle if we chose the angle/circumference relation properly, which would make it "uncountably" infinite, right? A complete line segment (and therefore the circumference of a circle) contains uncountably many points, corresponding to the real numbers on the interval [0,1). Then, you mention the axiom of choice, defined in the article as


 * "Let X be a set of non-empty sets. Then we can choose a single member from each set in X."

It says nothing about what that choice would entail, and reading related articles, I don't think it's supposed to. This can be applied to the current situation, as far as I can tell, by saying "if I'm trying to color this set of points, then at each and every point, there exists at least one color I can specifically choose." I kind of already knew that. So, anyway, we move on with the proof. We pick an orbit (by my reckoning, the orbit, but whatever) and pick a point in it. We then move along the orbit, coloring each point, alternating (in this case) red and black. ...Then it stops. How has that gotten me farther along? As far as I can tell, you're interpreting the axiom of choice as saying


 * "Let X be a set of non-empty sets. Then we can choose a single member from each set in X to satisfy any random conditions we can think of."

...which is ridiculous. But hopefully I misunderstood. Black Carrot 01:46, 6 February 2007 (UTC)


 * So your basic misunderstanding is that you think you can get to every point on the circle, starting from a particular point. That's not true. You can't get from x to y unless there's a finite sequence of points on the circle, the first being x and the last being y, such that successive points are distance 1 from each other. That's only countably many points, for each starting point.
 * That means there are in fact uncountably many orbits, not just one. To choose one point from each orbit requires AC. --Trovatore 02:11, 6 February 2007 (UTC)


 * The axiom of choice is not used to choose a color for every point, but rather a point for every orbit. To clarify the argument, the main observations are:
 * Two points in the same orbit can be reached (one from the other) by a sequence of chords of length 1.
 * Two points in different orbits cannot be reached (one from the other) by a sequence of chords of length 1.
 * Therefore, the selection of colors for the points of one orbit do not pose any restrictions on the selections of colors for other orbits.
 * Therefore, the problem reduces to finding a coloring for each orbit.
 * Every orbit can be put in a one-to-one correspondence with $$\mathbb{Z}$$ (which, by the way, means it is countable), by fixing some point and mapping it to 0, and proceeding naturally.
 * Since $$\mathbb{Z}$$ can be colored (black for even numbers, red for odd numbers), so can an orbit.
 * To have a coloring which applies to all points, we just need to choose, for every orbit, a correspondence between it and $$\mathbb{Z}$$.
 * This means that we need to consistently choose a point from any orbit.
 * This can be done using the axiom of choice.
 * To find the color of any point: Take its orbit; Take the point in the orbit given by our choice function; Take the induced correspondence with $$\mathbb{Z}$$; Take the integer to which our original point corresponds; Color it accordingly.
 * This is guaranteed to satisfy our requirement.
 * -- Meni Rosenfeld (talk) 09:36, 6 February 2007 (UTC)

I appreciate the numbering, that's a lot clearer. And I'm sorry, that makes sense as an application of the axiom of choice. I don't see that fixes the proof, but at least that wasn't the problem. Now, though I still don't quite follow how there are uncountably many countably-sized orbits, let's take that as assumed.
 * 1) From each orbit, you choose a point. Call it Bob. (This is allowed by the axiom of choice.)
 * 2) Proceeding clockwise at intervals of length 1, you color in the points of each orbit with alternating colors.
 * 3) Since the points on an orbit can be put (we're assuming) into one-to-one correspondence with the natural numbers, by virtue of their countability, this is the same as coloring all the even numbers Red and all the odd numbers Black.
 * 4) What, exactly, is the color of the point one unit counterclockwise of Bob, and what's its name in terms of the natural numbers? Or in other (and I'm hoping more precise) words, in $$\mathbb{N}$$, what has 1 as its successor?
 * 5) You use, however, $$\mathbb{Z}$$, which I believe means the integers. This changes things a bit, in that it makes it harder to point out the problem. What, exactly, is the color one step counterclockwise of Bob, in terms of the integers, if Bob = 0? Well, -1, so that would be Black and no problem, an argument which could continue forever. However, in both cases, we're assuming that a set of numbers that's not well-known for its circularity can loop back around on itself without difficulty. I've never heard of anything that fits with that, and without some clear rule about how it loops back on itself, it can't solve the problem. It's obviously not hard to color a line, the question is whether I can color a circle. Black Carrot 20:15, 6 February 2007 (UTC)


 * I do indeed use the integers, $$\mathbb{Z}$$. It makes no sense to use the natural numbers here, since as you pointed out, you can continue either clockwise or counterclockwise. As a side note, I'll mention that $$\mathbb{N}$$ is sometimes taken to include 0, sometimes not. Whatever the case, the first element of the set (be that 1 or 0) clearly is not the successor of anything.
 * Suppose r is such that the angle between successive points is 1 (radian). So you begin with some point and mark it 0. You go counterclockwise an angle of 1, and mark that point as 1. You continue another 1, so that you are now an angular distance of 2 from the original point. Mark that point 2. Continue in this way, marking points 3,4,5, and 6. Now, if you continue another 1, you will go past the 0 point, and arrive at angle 7-2&pi;, or roughly 0.71681. Mark that point as 7. Geometrically, this point is quite close to point 0, but this has absolutely nothing to do with our problem, since we are not imposing any restrictions on "close" points or whatever, only at point at angular distance of exactly 1. So the only way to get from one to the other is by making the complete loop of 7 steps. You can continue, marking as 8 the point at angle 1.71681, 9 at 2.71681,..., 12 at 5.71681, and then 13 at 13-4&pi;, roughly 0.43363. Again, this point is geometrically close to the 0 point but is quite distant in terms of steps of angle 1. You can continue this forever; Since &pi; is irrational, you will never return to the 0 point. You can get arbitrarily close to it, that's for sure - for example, the point marked 710 will be at angular distance roughly 0.00006. But you will never reach it exactly, so you will never be in a situation where you want to color it one way but have already colored it another way.
 * Of course, you can also go clockwise and mark as -1 the point at 5.28318, and so on. Again, you will never reach the starting point.
 * As already mentioned, the fact that geometrically you get arbitrarily close to 0 is irrelevant. We're only interested in exact distances, so as far as we are concerned, the set of points encountered in the process is the same as the set of integers. Of course, my thought experiment above is unnecessary - you can redefine the problem in terms of modular arithmetic, consider the equivalence classes of the relation $$a \equiv b \iff \exists m,n\ \mathrm{s.t.}\ a-b+m=nL$$, and so on. But that's just technical details. The essential part is that since you never reach any conflict, it is possible to have a legal coloring. -- Meni Rosenfeld (talk) 21:53, 6 February 2007 (UTC)

How do you never reach a conflict? Sure, it would take infinitely many steps to get back to your exact starting point, that's why I chose that angle, but what's wrong with that? Black Carrot 05:30, 7 February 2007 (UTC)

And in what may be a related question, does induction extend to uncountably long sequences? The article used to say it doesn't, but now it doesn't mention it, and I never found out why or even whether that's true. Black Carrot 05:30, 7 February 2007 (UTC)


 * You're over-complicating that aspect of it, Black Carrot. The only way there can be a conflict is if two points just one step apart are of the same color. All you need to check is that the assignment prevents that from happening. You don't need to check what happens when they're infinitely many steps apart (or even two steps, for that matter, as long as you prove you're OK for one step). --Trovatore 06:21, 7 February 2007 (UTC)


 * This again raises the question of whether you want to discuss this as a philosopher or as a mathematician. Philiosophy-wise, you can use words like "never" and "reach". Mathematically, these may only be used for infomral communication, and for actual arguments you should stick to words like "exists" and "for all". Consider the set of angles $$A = \{x|\exists m,n\in \mathbb{Z}\ \mathrm{s.t.}\ x=m+2\pi n\}$$. This is the set of all angles you can reach by steps of angle 1 (which is equivalent to a chord of length 1 for some specific radius). From the irrationality of &pi; it trivially follows that for any $$x \in A$$ there exist unique $$m,n\in \mathbb{Z}$$ such that $$x=m+2\pi n$$. So we may define the color of x to be black if the unique m in its representation is even, and red if it is odd. It is then trivial to show that for all $$x,y \in A$$ with x and y an angle of 1 apart, they have a different color. That's the only restriction we wanted, so this is a legal coloring for A. Trying to find shortcomings in a clearly correct solution is not very helpful.
 * Of course, this only deals with a single orbit. We want to do the same for every orbit. This is easy, but we need the axiom of choice to ensure that every orbit has a consistent origin. This gives a coloring for the entire circle which satisfies what we were after - that any 2 points an angle of 1 apart have a different color. And this can be trivially generalized to different radii, where a different angular distance is sought.
 * About induction - standard induction only deals with proving properties that hold for every natural number. I'm not sure what you mean by "uncountable sequence" - if it is uncountable, it cannot be a sequence in the ordinary sense. There is, however, something called transfinite induction, which can be used to prove properties for every ordinal number, but that is more advanced. -- Meni Rosenfeld (talk) 15:34, 7 February 2007 (UTC)


 * Am I missing something about the problem? It seems to me that there is a fairly simple and elementary solution. As I understand the problem, an equivalent problem is to come up with a colouring of the real numbers modulo 1, such that numbers a distance δ apart are assigned different colours, where 0 < δ ≤ ½. Identifying colours with natural numbers, we are seeking a function g : [0, 1) → N such that g({x}) ≠ g({x+δ}) for all real x, where {x} = x − floor(x) denotes the fractional part of x. A solution is given by g(z) = floor(nz), where n = ceiling(1/δ). --Lambiam Talk  05:56, 13 February 2007 (UTC)


 * How can you color a point, considering that geometric points have no extent? Mr.K. (talk) 18:55, 17 February 2007 (UTC)
 * With my solution you actually colour the half-open intervals [k/n, (k+1)/n) for k = 0, ..., n−1, which do have extent, but only one-dimensionally. For mathematicians, to "colour" a set S simply means to define a function from S to some finite set, subject to certain restrictions. If you are an intuitionist, you believe that all functions are continuous, and so if you colour a connected subset of the real line or the plane, you cannot use more than one colour. --Lambiam Talk  23:11, 19 February 2007 (UTC)


 * Here I'll try to make this clearer to people who may not be familiar with different orders of infinity. The number of points on a circle is the same as the number of real numbers between 0 and 1, which is infinitely larger than the number of integers, as proven by Cantor's diagonal argument.  So if you start at one point on the circle and keep taking steps of 1 unit, colouring each point, after an infinite number of steps it may look as if the whole circle is coloured, but actually you've only coloured a relatively very small subset of all the points on the circle.  If you choose a point at random you will almost certainly find that you haven't coloured it, though if you choose any other nearby point, (even very close nearby), there will be an infinite number of points you've coloured between the two.  In other words, points you have coloured can be found arbitrarily close to the point you chose at random.  --Coppertwig 14:20, 22 February 2007 (UTC)


 * Here's an example of a point which was not coloured: the point half a unit from the first point you coloured.  Proof:  Suppose that point was included in the set of all points you can reach in 1-unit steps from the first point.  Let k be the number of 1-unit steps needed to reach the point, and let n be the number of times you have to go around the circle to get to that point.  Then $$k = 2 n \pi + 0.5$$, therefore $$\pi = (2k - 1)/4n$$, therefore $$\pi$$ is a rational number;  but we know $$\pi$$ isn't a rational number, so that's a contradiction, so the supposition must be false.  A similar proof can be made using any rational number instead of one-half;  but rational-number steps still cover only a relatively small subset of all the real numbers that need to be coloured. --Coppertwig 14:29, 22 February 2007 (UTC)