Wikipedia:Reference desk/Archives/Mathematics/2007 December 19

= December 19 =

3 apples?
If I have 3 apples and I eat 2, how many apples will I have left?!! Please help!!! this question has had me thinking ever since I heard it!!! Superawesomgoat (talk) 01:36, 19 December 2007 (UTC)


 * You have three apples, since you apparently ate the number two, and by doing so screwed up all of mathematics. Great job. Now how am I going to finish my dissertation? mattbuck (talk) 01:45, 19 December 2007 (UTC)
 * 2 is overrated :) -- Meni Rosenfeld (talk) 11:38, 19 December 2007 (UTC)


 * The good thing is, it has you thinking. Keep it up! --Lambiam 15:58, 19 December 2007 (UTC)


 * You will have one apple left. 75.170.41.88 (talk) 16:53, 19 December 2007 (UTC)
 * Probably, but to prove it you need a result by Paul Erdős. Goochelaar (talk) 17:19, 19 December 2007 (UTC)
 * As long as you can demonstrate a non-trivial mapping from the set of apples to the natural numbers, and then define an appropriate operator to represent the action of eating apples (I thought I had it, then I realised I needed to extend the domain). Confusing Manifestation (Say hi!) 01:11, 20 December 2007 (UTC)
 * Ow, Come on nerd's! Remember simple maths. 3-2=1 1 apple left! --ジェイターナー ✉／✐ 21:37, 20 December 2007 (UTC)
 * Simple... maths? What is this "simple maths"? Oh, I know! You mean $$3-2 = \sum_{n=0}^{\infty}\frac{2}{2n+1}\left(\frac{\sum_{k=0}^{\infty}\int_2^3\frac{x^k}{k!}dx}{\sum_{k=0}^{\infty}\frac{3^k+2^k}{k!}}\right)^{2n+1}=1$$, right? -- Meni Rosenfeld (talk) 22:46, 20 December 2007 (UTC)
 * You know, like $$1+1=2$$ and $$ 1+2=3 $$. The stuff they teach you in pre-school.
 * None of this crap....
 * $$3-2 = \sum_{n=0}^{\infty}\frac{2}{2n+1}\left(\frac{\sum_{k=0}^{\infty}\int_2^3\frac{x^k}{k!}dx}{\sum_{k=0}^{\infty}\frac{3^k+2^k}{k!}}\right)^{2n+1}=1$$
 * --ジェイターナー ✉／✐ 12:01, 21 December 2007 (UTC)
 * You will never be a good mathematician if THAT'S your attitude. mattbuck (talk) 12:02, 21 December 2007 (UTC)


 * Hey, it's a simple question. HE's asking $$ 3-2 $$...... --ジェイターナー ✉／✐ 12:04, 21 December 2007 (UTC)
 * ITS 1!!! ジェイターナー ✉／✐ 12:06, 21 December 2007 (UTC)
 * How do you know he's asking $$3-2$$ and not $$3-0$$? Consider: he has 1 apple uneaten, as well as 2 apple eaten, in his digestive system at the time of asking. –Pomte 12:15, 21 December 2007 (UTC)
 * OK, If you thinkj of it like that than, yeah, your right, but, What would a 5 year old say? Lets ask Gordan Brown, he has the mentle age of a 5 year old. --ジェイターナー ✉／✐ 12:23, 21 December 2007 (UTC)


 * OK, we're not here to answer homework questions. Besides, if someone is clever enough to use wikipedia, they're clever enough to know how to do addition and subtraction of numbers below 5. Thus, the question is facetious, which means it's time for us to have fun. mattbuck (talk) 12:38, 21 December 2007 (UTC)
 * I am sure ジェイターナー understood this, and is bringing this to a new level of humour, as his jocular spelling and "Gordan Brown" non sequitur testify, isn't he? Goochelaar (talk) 13:16, 21 December 2007 (UTC)
 * I'm not completely sure. When there's so much sarcasm flying around, some people get confused. Perhaps ジェイターナー acknowledges that we are mocking the OP, but still believes the question was serious. -- Meni Rosenfeld (talk) 13:25, 21 December 2007 (UTC)
 * Hey that was a typo. I was just simplyfying the question! --ジェイターナー ✉／✐ 12:17, 22 December 2007 (UTC)
 * (unindent) looks like we're still not all on the same page. The OP did not post this as a serious question, but rather as some sort of humor \ trolling. As such, it does not deserve a serious answer. You can be quite sure that the OP (as well as anybody else who reads this) knows that 3-2=1 and that, under some assumptions, you are left with 3-2 apples if you start with 3 and eat 2. There is therefore no point whatsoever in replying "1". The only options are either ignoring the question completely, or coming up with humorous responses. Your attempts to "simplify" the question are unnecessary and uninsightful, as it is already as simple as it gets. -- Meni Rosenfeld (talk) 12:32, 22 December 2007 (UTC)
 * You can talk! --ジェイターナー ✉／✐ 19:12, 23 December 2007 (UTC)
 * Yes, I am physically able to pronounce syllables that form meaningful sentences in at least one natural language. What does this have to do with anything? -- Meni Rosenfeld (talk) 19:21, 23 December 2007 (UTC)
 * Well,...


 * Simple... maths? What is this "simple maths"? Oh, I know! You mean $$3-2 = \sum_{n=0}^{\infty}\frac{2}{2n+1}\left(\frac{\sum_{k=0}^{\infty}\int_2^3\frac{x^k}{k!}dx}{\sum_{k=0}^{\infty}\frac{3^k+2^k}{k!}}\right)^{2n+1}=1$$, right? -- Meni Rosenfeld (talk) 22:46, 20 December 2007 (UTC)


 * See...  --ジェイターナー ✉／✐ 16:55, 24 December 2007 (UTC)
 * Whatever... -- Meni Rosenfeld (talk) 16:58, 24 December 2007 (UTC)


 * Sure it's simple. In fact, it's "elementary", since it only uses real numbers. My dear Watson. Black Carrot (talk) 05:37, 25 December 2007 (UTC)

Derivative of projectile function
I have come up with this equation for the range of a projectile for a game/demo (using simple newton mechanics). It is for a projectile fired into the air from high ground to lower ground. $$s = u \cos(a) \left( \frac{u \sin(a)}{g} + \frac{ \sqrt{2g \left( \frac{\left( u \sin(a) \right)^2}{2g} + d\right)}}{g} \right)$$

The varibles are:
 * s=range
 * a=angle of fire
 * g=gravity
 * d=distance above target
 * u velocity of projectile.

I am trying to get a function so that given gravity, velocity and drop only, you get the firing angle that results in maximum possible range.

I think i know what to do. The maximum s, will be when the derivative of the function with respect to angle is zero. So I have to do $$\frac{d}{da}$$ of the function and it will equal zero but I need some help simplifying the function down before I can do the derivative.--Dacium (talk) 03:54, 19 December 2007 (UTC)


 * Alright, Dacium, in this way you are finding a candidate for an interior maximum (well, could be also a minimum!). In fact, I have arrived as something like


 * $$u \cos (2 a) \left(2 u \sin (a)+\sqrt{2} \sqrt{-\cos (2 a) u^2+u^2+4 d

g}\right)=4 d g \sin (a)$$
 * It's not possible, up to my knowledge (and up to my CAS!), to reduce it to a (closed-form-expressed) function a=f(u,g,d), I'm afraid. But you can compute it numerically for values of u, g and d, and even make graphs, if you like. Pallida  Mors  04:19, 19 December 2007 (UTC)
 * It is actually possible: The optimum is $$a = \arccos\left(\sqrt{\frac{2dg+u^2}{2dg+2u^2}}\right)$$. If memory serves, you are using Mathematica, like myself; the interesting point is that it apparently cannot solve the simplified equation you have written here, but it can solve the raw expression you get from differentiating.
 * Is this a game you (Dacium) are developing yourself? If so, you may want to add air drag if you want any kind of realism. Otherwise, which game is this about? -- Meni Rosenfeld (talk) 11:53, 19 December 2007 (UTC)
 * A neater expression for the same is $$\arcsin\left(\frac{u}{\sqrt{2(dg+u^2)}}\right)$$. -- Meni Rosenfeld (talk) 12:02, 19 December 2007 (UTC)
 * A few words about air drag, if it is in any way relevant: If the shape of the projectile isn't too complicated (say, a perfect ball), then up to a very good approximation, the effect of air is a force with direction opposite to that of the velocity, and magnitude which depends on the speed. The dependence is usually said to be quadratic, but according to my own calculations, there is a linear term which is much stronger unless the speed is on the order of 1000 m/s or higher. Also, solving the movement equation is more or less impossible for quadratic resistance, but it is possible for linear resistance. Therefore, if for any reason you will ever be interested in equations describing projectile motion more realistically, treating air drag as directly proportional to the velocity is a good compromise. -- Meni Rosenfeld (talk) 12:20, 19 December 2007 (UTC)


 * Yes, Meni, you are completely right, that's Mathematica. This astonishes me! How on Earth can't it solve the "simpler" bit? Pallida  Mors  14:35, 19 December 2007 (UTC)
 * It's not uncommon for something apparently more complicated to be actually easier. My favorite example is the trivial integral $$\int x\sin{x^2}\ dx$$ compared to the non-elementary integral $$\int \sin{x^2}\ dx$$ - the extra factor allows a delievering transformation. The same probably happens here - the extra stuff in the original expression (in particular, the denominator which you have cancelled) enables Mathematica to transform it into something it can solve. If you take it away, you get something which can only be solved if you re-intorduce it, but Mathematica isn't smart enough to realize this. -- Meni Rosenfeld (talk) 14:47, 19 December 2007 (UTC)
 * Yeah, Meni, I see your point. That's the explanation to it. Somehow this has chipped my almost blind faith on CAS. I tend to forget they are just a bunch of algorithms. Pallida  Mors  16:51, 19 December 2007 (UTC)

Is there any rule for looking at a fraction and knowing right away if it will terminate or repeat?
Is there any rule for looking at a fraction and knowing right away if it will terminate or repeat? —Preceding unsigned comment added by 74.71.101.164 (talk) 02:42, 19 December 2007 (UTC)


 * I'm not an expert in number theory, but don't worry. Scholars will correct my mistakes.
 * First, let's assume you have an irreducible fraction. That is, you have cancelled every common factor between numerator and denominator.
 * Second, let's fix a base, that is, a numerical system of representation. We normally use the decimal system, but there are many other possibilities (binary, hexadecimal, etc.).
 * So, let's say you have fixed base b, and in that base, your irreducible fraction has the form


 * $$\frac{x}{y}$$


 * If y has no factors which are not divisors of b, then there is a finite representation of the fraction. In any other case, you'll have a(n) (infinite) periodic representation of it.


 * Hence, any "thirds", "sevenths" etc. proper fractions (not 14/7) have infinite representations in our (decimal system). But 2/3 has a finite representation in the base-3 numerical system. Pallida  Mors  03:09, 19 December 2007 (UTC)


 * To clarify it a bit, 3/4 has a finite representation (in decimal)because $$4=1\times 2^2$$, and 2 is a factor of 10. On the other hand, 13/60 has no finite representation in decimal because $$60=1\times 2^2 \times 3 \times 5$$, and 3 does not divide 10. Pallida  Mors  03:16, 19 December 2007 (UTC)
 * I'll translate this into a simple procedure you can follow to know if a fraction will terminate in our decimal system (whether you can do it "right away" is entirely up to you):
 * You are given a fraction $$\frac{m}{n}$$.
 * Reduce it to lower terms, $$\frac{m}{n}=\frac{x}{y}$$, where x and y have no divisor in common.
 * Let z be the number obtained by removing any zeroes at the end of y.
 * If z is a power of 2 or a power of 5, the fraction terminates; otherwise, it repeats.
 * Example: You are given $$\frac{21}{307200}$$. You reduce it to $$\frac{7}{102400}$$. You take 102400 and remove the zeroes at the end: You get 1024 which is equal to $$2^{10}$$, so the fraction terminates. -- Meni Rosenfeld (talk) 12:09, 19 December 2007 (UTC)


 * See also Decimal representation. --Lambiam 16:06, 19 December 2007 (UTC)

Solution to a puzzle...
I was presented with the following puzzle the other day:

Take three nodes (circles, whatever) and then another three below those ones so it looks like this:

. ..

. ..

Now, try and connect each node above to every node below once without drawing a line that crosses through another line.

That's probably a bad explanation but I'm willing to bet someone is aware of the problem I'm describing. Furthermore, I can't seem to find a solution and I was wondering if 1) there is one and 2) can this be solved from a graph-theoretical perspective (edges and vertices)? —Preceding unsigned comment added by Damien Karras (talk • contribs) 18:33, 19 December 2007 (UTC)
 * There isn't.
 * Yes. The requirement that all edges lie in the plane without intersections means this is a Planar graph; Kuratowski's theorem then tells us that it cannot contain $$K_{3,3}$$ (a graph where each of 3 nodes is connected to each of 3 other nodes).
 * -- Meni Rosenfeld (talk) 18:43, 19 December 2007 (UTC)


 * There is.
 * No. Hint: You have to "think outside the box" of graph theory; the "lines" connecting the nodes are not the same as edges. It is easier to solve this if your nodes are not dots, as in the diagram you drew, but have a shape with a definite extension. Imagine you have the municipal Gas, Water, and Electricity works, and three Homes that need to get connected to all three. First make the gas and water connections. Now how can you next ensure all three will get connected to the electricity network? --Lambiam 19:01, 19 December 2007 (UTC)

+-+    +-+    +-+     |  G  |     |  W  |    |  E  | +-+    +-+    +-+     +-+     +-+    +-+     |  H  |     |  H  |    |  H  | +-+    +-+    +-+
 * Are you perhaps suggesting that the lines will pass through the nodes? I don't think that is allowed by the OP's problem description. -- Meni Rosenfeld (talk) 19:14, 19 December 2007 (UTC)
 * Yes, that is what I am suggesting; it is why the phrasing of the question has "(circles, whatever)". I don't see how the problem description in any way disallows a line passing through such a shape. The version with gas, water and electricity needing to come to three homes without the lines crossing is how I heard this puzzle first, in primary school. (Actually, the solution requires you to think very much inside the box.) --Lambiam 22:02, 19 December 2007 (UTC)


 * Sorry for not making it clear, but the lines cannot pass through the "works"; in which case I assume there is no solution? —Preceding unsigned comment added by Damien Karras (talk • contribs) 08:51, 20 December 2007 (UTC)
 * That is correct. -- Meni Rosenfeld (talk) 09:35, 20 December 2007 (UTC)


 * If I remember my graph theory right, this is also a Bipartite graph. The graph in question is usually notated, $$K_{3,3}$$ A math-wiki (talk) 21:00, 19 December 2007 (UTC)


 * I dimly recall this problem can be solved on the surface of a torus. Unfortunately, I have not got the time right now to verify this, but I will do so later. --Cookatoo.ergo.ZooM (talk) 22:17, 22 December 2007 (UTC)


 * It's easy on a torus. Just draw the graph with a single line crossing over others, then lift it from the plane and embed it in a handle. Black Carrot (talk) 05:41, 25 December 2007 (UTC)

Graph Theory Question
Recently I have tried to come up with a method of describing the structure of graphs. I have been rather unsuccessful and I wonder if such a method already exists. My method (still needs work) is based on listing the degrees of the vertices in decending order and then notating if necessary what degree's of the vertice's that the first few vertices are connected to. So the graph with five vertices and edges (AB,AC,AD,AE,BC,BD,BE) would be notated (4,4,2,2,2) which means it has two vertices of degree 4 and three of degree 2. You can deduce that this graph, one that is described as (4,4,2,2,2) is unique in structure (though it may be named 5! ways it's 'physical structure' is still the same) the structure can be described as follows. The graph (4,4,2,2,2) has a vertex (which shall be called A now) of degree 4 that is connected to all of the other vertices, and one of those other vertices has degree 4 and as thus is connected to A and the remaining 3 others (it shall be called B now). The remaining vertices are of degree 2 and therefore are connected to A and B only (Note that I am not considering the possibility of graphs with several edges between two vertices, as this notation is simply incorrect for a situation that calls for such graphs, a more complex version of this system may work) An example of a non-unique graph would be (4,2,2,2,1,1) which has two forms (if I haven't missed any) I notate these two forms as (4,2,2,2,1,1:2-2-2-1) and (4,2,2,2,1,1:2-2-1-1) the part after the colon tells me the degree's of the vertices connected to the vertex of degree 4. I should also note that I notate a single vertex with no edge as (0) and I describe unconnected graphs by summing there connected subgraphs and singular vertices. An example would be (4,4,2,2,2)+(2,2,2)+3(0) where 3(0)=(0)+(0)+(0). A math-wiki (talk) 22:13, 19 December 2007 (UTC)

I have also discovered a graph which has two forms that are indistinguishable from one another in this notation. Notated (3,3,3,3,2,2:3-3-2). A math-wiki (talk) 22:40, 19 December 2007 (UTC)


 * You're right that the degree sequence does not in general determine the graph. However, if you know in advance that you are looking at a threshold graph, then the degree sequence does completely determine the structure of the graph.  There's even a nice formula for the number of spanning trees of a threshold graph, based on taking the conjugate of the degree sequence.  If you're interested in the question of when a sequence must correspond to an actual graph, there is an explicit condition known (apparently due to Ruch and Gutman 1979, but I'm looking at Merris 1994), which again depends on taking the conjugate of the degree sequence. Michael Slone (talk) 23:20, 19 December 2007 (UTC)

Interesting, I'm still curious if there is a general way to notate the possible structure that graphs can have. A math-wiki (talk) 20:21, 20 December 2007 (UTC)
 * Graph isomorphism is a hard problem, so it is unlikely there is a pleasing notation that unambiguously decides isomorphism (whether two graphs have the same structure) and yet can be written down easily from the graph. JackSchmidt (talk) 21:45, 20 December 2007 (UTC)

Approximate angular size/distance revisited
Hi. I've thought about this for a while, and revised my hypopthetical formulas. So, here goes. Remember this formula is only in its early stages, and it is meant to be a way to figure out the approximate distance of an object by knowing its apparent angular size and its dimentions, and vise versa, without having to use the sine, cosine, and tangent functions. To find the distance of an object if you know its size, divide 52 by the apparent angular size in degrees of the object. Now multiply the approximate known dimention (eg. height, diameter, etc) by this number, and you get its distance. To find out its size by knowing its distance, again divide 52 by its angular size, and this time divide the known distance by this number, to get the size for the scelected dimention. Make sure the dimention is linear, not square or cubic. 52 is the approximate angle in degrees rounded, I assume by making more assumptions than calculations, that the only non-isoceleusic angle of the isoceles triangle would have if its base length were equal to its height. Also, this formula works by assuming that doubling the base length of an isoceles triangle also doubles its non-isoceleusic angle, which I'm not sure if it works or not in practise, and maybe it only works if you cut an isoceles triangle in half, turning it into a right-angled triangle, and the angle mentioned would be cut in half, or if this only works this way and not by expanding the base length and if it exponentially decreases in doublingness as the angle increases, etc. If this is the case, please adjust my formula, but keep its simplicity. Thanks. ~ A H  1 (TCU) 23:53, 19 December 2007 (UTC)


 * $$d = \dfrac{x}{2\tan\theta}$$ where d is distance from you to object, x is the object width and 2θ is the angular width, assuming you are standing so that the angle between the line normal to the object's face passing through you passes through the centre of the width. mattbuck (talk) 01:03, 20 December 2007 (UTC)


 * Hi. I've tried doing it with trigonometry, and it requires a scientific calculator. I'm asking if there is a way to do it without using a scientific calculator (estimation). So, theta is half the angular width? In what unit? Does my formula work (approximately), and if there is a problem with it, where's the problem? So, do you multiply the distance by twice the tangent of half the anglular size? In what unit? Does this work when you're trying to find the distance by knowing the size? Would whether the line normal to the object's face or not matter if the object you were viewing were a sphere? Does my half/double the icosoles angle thingy work, or is it fundamentally wrong and the source of the error? If my formula is innacurate, at what angular size would the estimation be the most accurate (linear and exponential accuracy)? Thanks. ~ A H  1 (TCU) 01:50, 20 December 2007 (UTC)
 * see rational trigonometry--Dacium (talk) 02:12, 20 December 2007 (UTC)


 * Let's put everything here in order. We will denote the distance by d, the diameter by x, the angle by $$2\theta$$, and the angle in degrees by &phi;. It follows that $$2\theta=\varphi^{\circ}$$, and since $$\pi = 180^{\circ}$$ you have $$\theta=\frac{\varphi \pi}{360}$$. The correct formula is now $$d=\frac{x}{2\tan \theta}$$. We know that if &theta; is very small, then $$\tan \theta \approx \theta$$ (in particular, this means that if &theta; is very small, then doubling the angle roughly doubles the tangent. This should answer your question about the isoceles triangle). Therefore you can also write $$d = \frac{x}{2\tan \theta} \approx \frac{x}{2\theta}$$. The error you get from this approximation is $$\left|\frac{x}{2\tan \theta} - \frac{x}{2\theta}\right| \approx \left| \frac{x}{2(\theta+\theta^3/3)} - \frac{x}{2\theta}\right| = \frac{x}{2\theta} \left|\frac{1}{1+\theta^2/3}-1\right| \approx \frac{x \theta}{6}$$. So the error with this calculation is small when &theta; is small. Now, if you're more comfortable using degrees, you have $$d \approx \frac{x}{2\theta} = \frac{180x}{\pi \varphi} \approx \frac{57.3x}{\varphi}$$ (I'm not sure where you got the number 52 from). This last approximation gives a relative error of $$\frac{57.3\pi}{180}-1 \approx 0.007%$$, which can be a large absolute error if the distance is large. -- Meni Rosenfeld (talk) 10:09, 20 December 2007 (UTC)


 * See also Angular diameter, which gives formulas in the other direction, which however are easily inverted. --Lambiam 12:41, 20 December 2007 (UTC)