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August 13

Convolution

I know that the convolution with the Delta function has the property (I'm working in the frequency domain): $G(f)*\delta (f-a)=G(f-a)$

But what will $G(kf)*\delta (f-a)$ yield? I tried to search through convolution properties but I couldn't find an answer. Thank you. 212.98.136.42 (talk) —Preceding undated comment was added at 12:15, 13 August 2008 (UTC)

If you define, say, g(f) = G(kf), then

$g(f)*\delta (f-a)=g(f-a)=G(kf-ka).\,$

Hope that helps. siℓℓy rabbit (talk) 12:20, 13 August 2008 (UTC)

Yes I imagined it was $G(k(f-a))\,$

Thank you. 212.98.136.42 (talk) 12:34, 13 August 2008 (UTC)

Note that your $G(f)*\delta (f-a)=G(f-a)$ is something of an abuse of notation. Convolution is normally defined as an operator acting on functions, such that it would, for example, be meaningful to write $(G*\delta )(f)=G(f)$ or just $G*\delta =G$. (Indeed, this is basically the definition of the delta function.) Your expression only makes sense if one interprets $G(f-a)$ as a shorthand for the composite function $G\circ \tau _{a}$, where $\tau _{a}(f)=f-a$. Without such shorthand, it could be written as $G*(\delta \circ \tau _{a})=G\circ \tau _{a}$, with $\tau _{a}$ as defined above. It then clearly follows that $(G\circ q)*(\delta \circ \tau _{a})=(G\circ q)\circ \tau _{a}=G\circ q\circ \tau _{a}$, for all functions $q$, including the particular case $q(f)=kf$. —Ilmari Karonen (talk) 19:30, 13 August 2008 (UTC)

Marge Inoverra election

If, in a two-candidate race, a poll says candidate A has 45% of the vote, and candidate B has 55% of the vote, with a 5% margin of error, does that 5% apply to each candidate, so that A has 40-50% of the vote and B has 50-60% of the vote, or should the 5% be split by the two candidates to 2.5% each, for a range of 42.5-47.5% for candidate A and 52.5-57.5% for candidate B ? Also, how does it work with 3 or more candidates ? StuRat (talk) 15:27, 13 August 2008 (UTC)

The first case. The usual meaning of this would be that A has a 95% chance of getting 40-50% of the vote. See Margin of error for more details. Dmcq (talk) 22:26, 13 August 2008 (UTC)

Oh come on someone should have got at me for that :) What it really means is they asked a question on the vote to about a thousand people. Since about half said either way the standard deviation is sqrt(.45 * .55 /1000) or about 1/60. 95% is about two standard deviations or about 1/30 which is a bit over 3% but they just said 5% as it looks good and there's whole lots of problems with the procedure. To get a decent estimate they'd have to do stratified sampling and make sure their question was phrased just right. Even so there'd be lots of people who, for example, see the nice researcher as left or right wing and give the answer they think will make the researcher pleased. An opinion poll gives an answer to a question at the time it's run. As to how the actual vote will go, that may well just depend on something like the weather on the day. The pollsters aren't going to specify any tighter interval because they don't like people making fun of them after misunderstanding what they're doing. They do also sometimes make predictions and I suppose these will get better and better as they start using all sorts of other measures as well, e.g. pro and anti posts on the web, but I don't think you'll find margins of error on those yet. Dmcq (talk) 07:54, 14 August 2008 (UTC)

August 14

Euler's theorem

a and n are positive integers.

Claim: a is coprime to n if and only if $a^{\varphi (n)}\equiv 1{\pmod {n}}$.

Prove or disprove. —Preceding unsigned comment added by 93.172.15.144 (talk) 09:56, 14 August 2008 (UTC)

The article you've linked to in the title has two proofs of the non-trivial implication. Algebraist 12:06, 14 August 2008 (UTC)

Deceptively distributed decimal digits

X and Y are independent continuous random variables both uniformly distributed between 0 and some upper limit a. Z is the leading (non-zero) digit in the decimal expansion of Y/X. So, for example, if X=2 and Y=5 then Z=2; if X=5 and Y=2 then Z=4. What is the probability that Z=1 ?

I expected Benford's law to apply, because I thought this was analogous to measuring an observable of magnitude Y in units of magnitude X. So I expected to find P(Z=1) = log₁₀(2). But I think I have a geometric argument that shows that P(Z=1) is 1/3. Is this correct ? And, if so, why doesn't Benford's law apply ? Gandalf61 (talk) 10:17, 14 August 2008 (UTC)

The probability is independent of a, so set a=1. This gives some simplification. Within the unit square in the X,Y-plane consider the triangles for which 0.1≤X/Y<0.2, 1≤X/Y<2, 10≤X/Y<20, 100≤X/Y<200 etc and sum their areas. Is this what you did? Bo Jacoby (talk) 11:00, 14 August 2008 (UTC).

Yes, in essence that was my geometric argument that led to P(Z=1) = 1/3. Gandalf61 (talk) 11:10, 14 August 2008 (UTC)

Benford's law is just a rule of thumb, and I don't think there's any realistic situation in which it holds exactly. I'd expect it to hold approximately here, and it does: log₁₀(2) ≈ 1/3. A precisely correct statement of the law would be that P(Z=1) is the integrated area from 0 to log₁₀(2) of the distribution of the log₁₀ of your random variable in $\mathbb {R} /\mathbb {Z}$. Often that distribution is roughly flat, so the integral comes out to about 0.301. -- BenRG (talk) 11:31, 14 August 2008 (UTC)

Hmmm. 1/3 doesn't seem very close to log₁₀(2). And the difference gets worse if you increase the base. In hexadecimal, Benford's law says the P(Z=1) should be log₁₆(2) = 1/4, whereas the geometric method gives P(Z=1) = 3/10. In fact, as the base increases, Benford's law says that P(Z=1) should tend towards 0 because

$\lim _{b\rightarrow \infty }\log _{b}(2)=0$

whereas with the geometric method, P(Z=1) tends to a limit of 1/4 (because the area of the largest triangle, between the lines Y=X and Y=2X, is always 1/4 and the areas of all the other triangles become negligible). So I still think that either the geometric method is incorrect or Benford's law doesn't apply - but I don't know which. Gandalf61 (talk) 12:58, 14 August 2008 (UTC)

Your geometric method is fine; it's a proof. Benford's law never applies in the sense that you mean—it's not a theorem (except in base 2) and can't be used to prove anything. It's generally the case that it gets less accurate for higher bases. For this problem it's exactly right (for a leading digit of 1) in bases 2 and 4 and off by less than 1% in base 3. -- BenRG (talk) 14:56, 14 August 2008 (UTC)

But Benford's law isn't just theoretical - Benford supported in with statistical evidence in his original paper, and MathWorld mentions other evidence as well. What I am trying to work out is whether Y/X is a good model for measuring an arbitrary observable in arbitary units. What is the probability that the first significant digit in the base b expansion of an arbitary observable measured in arbitrary units (for example, the speed of light in furlongs per fortnight) is 1 ? Is it log_b(2) as per Benford ? Or is it (b+2)/4(b-1) as per the Y/X model ? Gandalf61 (talk) 15:31, 14 August 2008 (UTC)

Oh, I see what you mean. Benford's law is better. What you're seeing in this problem is edge effects arising from the particular cutoff you've imposed on X and Y. If X is a physical quantity and Y is a unit, it's probably more realistic to assume a uniform distribution on log X and log Y rather than X and Y. The leading-1 case of Benford's law says that the fractional part of log X − log Y will lie in the range [0, 0.301) about 30.1% of the time, which will be true if the fractional part of log X − log Y is uniformly distributed, which will be roughly true as long as X and Y are allowed to range over enough orders of magnitude. -- BenRG (talk) 18:35, 14 August 2008 (UTC)

Gandalf, the answer is P(Z=1) = 1/9. The nine triangles from 0.1 to 1.0 have each the same area. The nine triangles from 0.01 to 0.1 have each the same area. &c. So each of the nine possible leading digits from 1 to 9, have the same probability, = 1/9. You do not even need to sum the geometric series. Bo Jacoby (talk) 14:18, 14 August 2008 (UTC).

Here is how I got P(Z=1) = 1/3. We have one sequence of triangles with bases down the right hand side of the square: 0.1X <=Y<0.2X, 0.01X<=Y<0.02X etc. with bases 1/10ⁿ and height 1 so total area of this sequence is 1/18. We have another sequence of triangles with bases along the top of the square: X <=Y<2X, 10X<=Y<20X etc. with bases 5/10ⁿ and height 1 so total area of this sequence is 5/18. And 1/18 + 5/18 = 1/3. Gandalf61 (talk) 14:35, 14 August 2008 (UTC)

You're correct, the total area is 1/3 (in general, ${\tfrac {b+2}{4(b-1)}}$). The nine triangles corresponding to different leading digits have different areas. -- BenRG (talk) 14:56, 14 August 2008 (UTC)

Sorry, I was too hasty. Bo Jacoby (talk) 17:13, 14 August 2008 (UTC).

When $Y\approx 1$, Benford can't apply at all; the first digit is uniformly distributed because it's uniform on each of $[0.1,1),[0.01,0.1)$, etc. When $Y={\frac {1}{2}}$, on the other hand, you get a full half probability for 1 from $X/Y\in [1,2)$, and then uniform otherwise. The pattern repeats at $Y={\frac {1}{10}},{\frac {1}{20}}$; it only makes sense that the final probabilities resemble Benford's Law (since $Y={\frac {1}{5}}$ benefits 1…4 but not 5…9), but nothing more precise may be assumed because there is a variable amount of uniformity mixed in and because the relative favor given 1 and then 1/2 and then 1/2/3 is not logarithmically distributed. --Tardis (talk) 18:14, 14 August 2008 (UTC)

As BenRG notes above, for Benford's law (in base b) to hold exactly for a random variable R, the variable S = (log R) mod (log b) must be uniformly distributed. If X and Y are both uniformly distributed between 0 and a, then the probability density and cumulative probability functions for X/Y and their logarithmic equivalents are

${\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} z}}\;\mathrm {P} \left({\frac {X}{Y}}<z\right)&={\begin{cases}1/2&{\mbox{for }}0\leq z\leq 1,\\1/(2z^{2})&{\mbox{for }}z>1,\end{cases}}\\\mathrm {P} \left({\frac {X}{Y}}<z\right)&={\begin{cases}z/2&{\mbox{for }}0\leq z\leq 1,\\1-1/(2z)&{\mbox{for }}z>1,\end{cases}}\\\mathrm {P} \left(\log {\frac {X}{Y}}<\zeta \right)&={\begin{cases}e^{\zeta }/2&{\mbox{for }}\zeta \leq 0,\\1-e^{-\zeta }/2&{\mbox{for }}\zeta >0,\end{cases}}\\{\frac {\mathrm {d} }{\mathrm {d} \zeta }}\;\mathrm {P} \left(\log {\frac {X}{Y}}<\zeta \right)&={\begin{cases}e^{\zeta }/2&{\mbox{for }}\zeta \leq 0,\\e^{-\zeta }/2&{\mbox{for }}\zeta >0.\end{cases}}\end{aligned}}$

Thus, the logarithmic probability density peaks at 0, reflecting the fact that the construction makes it fairly likely that X and Y are of similar magnitude. Writing β = log b, the probability density function for log X/Y mod β is then

${\frac {\mathrm {d} }{\mathrm {d} \eta }}\;\mathrm {P} \left(\log {\frac {X}{Y}}\ {\mbox{mod}}\ \beta <\eta \right)={\frac {1}{2}}\sum _{k\in \mathbb {N} }\left(e^{-k\beta +\eta }+e^{(1-k)\beta -\eta }\right)={\frac {1}{2}}(e^{\eta }+e^{-\eta })\sum _{k\in \mathbb {N} }e^{-k\beta }+{\frac {1}{2}}e^{-\eta }.$

This cannot be constant for any b; therefore Benford's law (at least in its extended, multi-digit form) will not hold exactly in any base for this distribution. The probability density function does become approximately constant (over its domain 0 ≤ η ≤ β) as β approaches 0, so the law will more or less hold for small b. For large b, however, the probability density peaks sharply at the ends of the range; in particular, P(log X/Y mod β < η) tends (pointwise) to 1/2 - e^-η/2 as β tends to infinity. (The second peak at β moves out to infinity, taking half of the probability mass with it.) Thus, as the geometric reasoning above suggests, P(log X/Y mod β < log 2) indeed approaches 1/2 - e^{-log 2}/2 = 1/4 for large β. —Ilmari Karonen (talk) 22:16, 14 August 2008 (UTC)

Thank you for all the responses. Obviously if we start with a uniform distribution for log X and log Y over a wide enough range then a Benford-type distribution for Z follows almost trivially. I was hoping that there might be a simple explanation of why Benford's law is empirically seen to apply in less restricted scenarios. As far as I know, the only explanation of Benford's law that does not start out by assuming an underlying uniform log distribution or something close to that is Hill's 1996 paper on samples from "random" distributions, which is a fairly technical explanation. Gandalf61 (talk) 09:21, 15 August 2008 (UTC)

Well, a slightly better explanation is that, even if the log distribution isn't that close to uniform, "rolling it up" modulo log b will tend to make it more uniform if b is small enough. (Even your sharply peaked example gets pretty close for small b, the relative error P(log X/Y mod log b < log 2) / (log_b 2) - 1 being, coincidentally, about b/100 for b up to 60 or so.) In particular, if the variable is approximately log-normally distributed (which the law of large numbers says is likely for variables that are products of many independent underlying factors) with a standard (log-)deviation of more than a few orders of magnitude, then Benford's law is likely to hold quite well. —Ilmari Karonen (talk) 12:46, 15 August 2008 (UTC)

2012

what day of the week willb 18th August 2012 fall on —Preceding unsigned comment added by 86.135.111.151 (talk) 10:30, 14 August 2008 (UTC)

Saturday. You can doubleclick on Windows Clock on taskbar. See Calculating the day of the week. —Preceding unsigned comment added by 93.172.15.144 (talk) 10:47, 14 August 2008 (UTC)

You can also visit 2012 here, and click on the link to leap year starting on Sunday. -- Coneslayer (talk) 16:04, 14 August 2008 (UTC)

86.135.111.151: What do you expect will happen on August 18, 2012? None of these, I hope? --Bowlhover (talk) 19:09, 15 August 2008 (UTC)

A question about terminology in elementary category theory

How is it named, or how would you name, the following property for a map h:x-->y in some category C:

(P) for any f:x-->x there exists a g:y-->y such that gh=hf.

In other words, if h is put at the two horizontal edges of a square diagram, and f is put at the left edge, we can close a commutative diagram putting g on the right edge.

In fact an analog property may be considered in any algebraic context, that is f,g,h may be elements of a group or so, but has this a name? —Preceding unsigned comment added by 79.38.22.37 (talk) 13:48, 14 August 2008 (UTC)

How fast?

This is an odd question. I purchased the The Flash (TV series) on dvd a while back and one episode had the superhero running 10 miles in 30 seconds to escape a radio signal that would blow up a bomb. I'm sorry to say that I don't remember the formula for calculating speed. How fast is that anyway? --Ghostexorcist (talk) 19:52, 14 August 2008 (UTC)

Distance = (Speed) x (Time) and Speed = (Distance)/(Time). After you do the math, you figure that the superhero was going at 1200 mph.--El aprendelenguas (talk) 20:45, 14 August 2008 (UTC)

Which is still slower than the speed of light, which radio wave (being a very low frequency light) travel. --antilived^{T | C | G} 08:52, 15 August 2008 (UTC)

August 15

Projective Space

When you apply the construction of a projective space to Rn, by removing the origin and identifying all points that are constant multiples of each other as vectors, the space you get resembles R(n-1) plus a bit. This "bit", in the case n=2, seems to be beyond the other numbers of R1 in a way that resembles infinity, so it's convenient to call it the point "at infinity". This terminology behaves well with respect to the field operations. So, that, for instance, infinity times anything besides zero is itself. That extends into more dimensions, and into other fields, like the complex numbers. The question I have is, how well does it extend to fields with positive characteristic, especially the finite fields? In this case, there is no real sense of "size" to anything, so is there any sense in which the new element or subspace is "at infinity"? Black Carrot (talk) 05:29, 15 August 2008 (UTC)

Your point about the field operations still holds, I believe. Algebraist 09:31, 15 August 2008 (UTC)

Yes, you can do the same construction as well. Although for finite fields the topological aspect vanishes and the algebraic one seems rather poor, still the combinatoric is very rich. Geometrically, you can start from a vector space V over a field K: the corresponding projective space (that you can identify with the set of 1 dimensional subspaces of V) is of great interest also in case of finite fields, e.g. in combinatorics. Notice that all lines in V are similar to each other (in a sense that one could make precise), thus no element of the projective space is "at infinity" more than others. On the other hand, if you also fix a hyperplane W of V (say, an affine hyperplane not containing the origin) then you can see the projective space as a superset of W (identifying any point of W with a line of V in the obvious way), and you have the right to say that what is out of W is "at infinity". But it seems to me that the corresponding enlargement of a finite field, e.g. Z_2, by means of a point at infinity is somehow of less algebraic interest. 79.38.22.37 (talk) 14:47, 15 August 2008 (UTC)

Well, I know they're of some theoretical interest in algebra, since some of the finite simple groups are represented as projective matrices over finite fields. I'm especially curious about infinite fields with finite characteristic, though, which can be given interesting topologies. Does anyone know of an interpretation of one of these topologies, that would be consistent with the interpretation of points in its projective space being at infinity? Perhaps a topology based on a metric, like in the complex numbers, where division by a deleted neighborhood of zero gives a deleted neighborhood of infinity. Black Carrot (talk) 12:08, 16 August 2008 (UTC)

First Principle

How to find the derivative of $x^{3}\cos x$ USING FIRST PRINCIPLE? —Preceding unsigned comment added by Ftbhrygvn (talk • contribs) 09:38, 15 August 2008 (UTC)

Write down the definition of the derivative and work it out. You'll want the addition formula for cos and the fact that sin(x) is approximately x for small x, which can both be proven geometrically. Algebraist 10:18, 15 August 2008 (UTC)

Just to be more clear, in general when you see using first principles, means the problem can't be solved using all of the nice tricks you've learned for doing derivatives (like the chain rule) and are forced to use the original limit formula for a derivative. As the above poster hinted at, in order to do the limit, you'll need some trigonometric identities. Anythingapplied (talk) 13:22, 15 August 2008 (UTC)

...although in this case the problem can be easily solved using the product rule and a knowledge of the derivatives of trigonometric functions, so I imagine the "first principles" instruction is just to give the reader experience of taking derivatives "by hand", so that they will appreciate the usefulness of the "tricks". Gandalf61 (talk) 13:32, 15 August 2008 (UTC)

August 16

Induced coloring method

Where can I find information about the induced coloring method in Ramsey theory. It is used to prove the generalized Ramsey Theorem, Van der Waerdens theorem, Hales Jewett theorem etc). I want to understand the underlying idea of the method. Thanks--Shahab (talk) 06:08, 16 August 2008 (UTC)

Polynomials

How do you find the coefficients of a polynomial (of degree n-1) that passes through n points ((x₁, y₁), (x₂, y₂) . . .)? Thanks *Max* (talk) 06:22, 16 August 2008 (UTC).

Check out Polynomial interpolation. Best, RayAYang (talk) 06:31, 16 August 2008 (UTC)

Let the polynomial be $a_{0}+a_{1}x+a_{2}x^{2}\cdots a_{n-1}x^{n-1}$. Now substitute $(x_{1},y_{1})$ in it. You will get the linear equation $y_{1}=a_{0}+a_{1}x_{1}+a_{2}x_{1}^{2}\cdots a_{n-1}x_{1}^{n-1}$ which has n variables $a_{0},a_{1}\cdots a_{n-1}$. In this way get n linear equations and solve for the coefficients. That's actually what happens in interpolation--Shahab (talk) 06:38, 16 August 2008 (UTC)

And have also a look to the Lagrange polynomial method, that gives you immediately the interpolating polynomial, although not explicitely with its coefficients. If you need the coefficients, you can expand it and write them as linear combinations of the symmetric functions of x₁,.., x_n. But I guess this is not better than the previous method.79.38.22.37 (talk) 07:54, 16 August 2008 (UTC)

Thank you. *Max* (talk) 15:25, 16 August 2008 (UTC).

August 17

Connecting points without intersecting lines

Is there an algorithm out there, or some sort of theory for me to read up on, which gives a method of connecting n points using n-1 lines without any of these lines intersecting? Thanks in advance. Not homework, just a curiosity. x42bn6 Talk Mess 04:13, 17 August 2008 (UTC)

The first thing that comes to mind is convex hulls. That is, you could take the convex hull of the set, remove any points connected by that, and interate inwards. Once you have a sequence of nonintersecting loops that cover all the points, it's just a matter of removing a link from each loop and connecting them up going outwards to get a nice spiral. Black Carrot (talk) 04:33, 17 August 2008 (UTC)

Here's another approach (I'm assuming you're in two dimensions). First suppose that no two points have the same y-coordinate. Order the points by increasing y-coordinate, and then connect consecutive points in this list. If some two points have the same y-coordinate value, then you can (1) rotate all points (around the origin, say) by some angle such that no two points have the same y-coordinate (such an angle must exist), then (2) connect the points as above, then (3) unrotate the points by the same angle. A similar approach should work in any number of dimensions of Euclidean space.

If you specifically want to find the connections that minimize the total length of the n-1 lines, then look at minimum spanning trees. Eric. 80.101.159.204 (talk) 12:49, 17 August 2008 (UTC)

A minimum spanning tree of a planar graph weighted by Euclidean distance will not self-intersect. Proof: suppose it does, and let AC and BD be intersecting line segments in the MST. Remove them from the graph. Since the original tree was connected there must still be a path connecting (A,B) or (C,D) or (B,C) or (D,A). By symmetry we may as well suppose it's (A,B). Case 1: A, B, C, D are in general position. Then adding the edges (A,D) and (B,C) to the reduced graph yields another spanning tree, and by the triangle inequality it's shorter than the one we started with, which is a contradiction. Case 2 (three collinear and one not) and case 3 (all collinear) yield similar contradictions, and we're done. This proof assumes all the points are distinct. If they're not, there might be no non-self-intersecting spanning tree depending on how you define self-intersection. -- BenRG (talk) 14:59, 17 August 2008 (UTC)

I don't think that's quite enough for a proof: the new edges might create new intersections, which would then have to be resolved. To complete the proof, you'd actually have to show that the process eventually terminates (or, for infinite graphs, that any finite subset of the graph will stabilize with no intersections after finitely many steps).

It's fairly simple to prove (by building the tree one node at a time) that any set of points on the plane has a non-intersecting spanning tree. It's slightly harder to show that the minimum-distance tree must be non-intersecting, though it seems intuitively obvious. —Ilmari Karonen (talk) 18:30, 17 August 2008 (UTC)

No, that proof works fine. The point is that any spanning tree with self-intersections is not of minimal weight, so any minimal spanning tree must be non-self-intersecting. There's no step-by-step process involved. It only works for finite graphs, but that's all the OP asked about. Of course, finding an MST is going to be far slower than Eric's trivial algorithm. Algebraist 18:39, 17 August 2008 (UTC)

Square root of i

I have read the article on i and have seen what the square root of i is but the article only discusses how to show that such and such is the square root of i; it doesn't actually show how you can determine the square root by yourself. Can anyone shed some light on the issue? 92.4.196.6 (talk) 18:55, 17 August 2008 (UTC)

There are a number of ways. For example, you could write the square root of i as a+bi, deduce that a²=b² and 2ab=1, and solve for a and b. Or you could write i as $e^{\frac {i\pi }{2}}$, and deduce that $e^{\frac {i\pi }{4}}$ is a square root of i, and hence that $-e^{\frac {i\pi }{4}}$ is the other one. Algebraist 18:59, 17 August 2008 (UTC)

When dealing with complex numbers it's often useful to think geometrically (see argand diagram). In that context, multiplying by i means "rotate 90 degrees anticlockwise". Then it's clear that multiplying by the square root of i must be a 45 degree rotation (so that when you do it twice (ie. square it) you get a 90 degree rotation). Then you can use basic trig to work out what point you get to by rotating 1 by 45 degrees anticlockwise, and you find that it's $\cos(45)+i\sin(45)$ (actually, you don't need trig, Pythagoras' theorem combined with some observations about symmetries is enough - try drawing a diagram). --Tango (talk) 21:12, 17 August 2008 (UTC)

To be precise, I ought to say either 45 degrees, or 225 degrees - there are, of course, two square roots. --Tango (talk) 21:14, 17 August 2008 (UTC)

There is an easy way to figure out the square root of i.

• In the mathematics of complex numbers, cis θ is an abbreviation for cos θ + i·sin θ —Preceding unsigned comment added by 211.28.51.233 (talk) 10:39, 18 August 2008 (UTC)

First, imagine that multiplying by a complex number z = x + y i = r cis(theta) can be represented geometrically by two geometric operations.

Operation one is ROTATION

Operation two is SCALING

Now first z = i means that x=0 and y=1 also be represented as r = 1 and theta = 90deg

Now find the solution zans = rans cis( thetaans ) such that

2 * thetaans = 90deg + k * 360deg where k is an integer

and

rans^2 = 1

Basically what I am saying is "find a rotation angle (thetaans) such that when it is rotated twice, it had rotated 90 degrees".

Next I say "find a non-negative scaling factor (rans) such that when it scale up twice in a row, it scales up by 1".

Thus the square root of i is zans

I find (Please note that there are TWO solutions)

thetaans = 45deg (the obvious answer) or thetaans = -135deg (the not so obvious answer)

rans = 1

So zans = 1 cis( 45deg) or zans = 1 cis ( -135deg )

211.28.51.233 (talk) 10:37, 18 August 2008 (UTC)

Infinity

I just been thinking about infinity. And I wondered what is the infintith root (know what i mean) of infinity?--79.76.130.174 (talk) 19:09, 17 August 2008 (UTC)

$\infty ^{0}$ is an indeterminate form, q.v. That's not exactly an answer to your question, but it's probably the closest thing to an answer that you're really going to get. --Trovatore (talk) 19:24, 17 August 2008 (UTC)

A related question might be whether $\lim _{x\to \infty }x^{1/x}=1$. It seems to approach 1, but I am not certain that it monotonically decreases as x increases. - Rainwarrior (talk) 19:50, 17 August 2008 (UTC)

Correct me if I make any mistakes...

$\lim _{x\to \infty }x^{1/x}=\exp \left[\lim \ln \left(x^{1/x}\right)\right]$

$=\exp \left[\lim {\frac {\ln x}{x}}\right]$

$=\exp \left[\lim {\frac {(1/x)}{1}}\right]$

$=\exp \left[0\right]$

$=1$

I used L'hopitals rule in there. The limits are all taken as x goes to infinity. Eric. 80.101.159.204 (talk) 20:18, 17 August 2008 (UTC)

That's fine if you assume that base and exponent approach infinity and zero in such a way that they have a constant product. But what if the base increases as a^x while the exponent decreases as 1/x ? In that case the limit is a. So you can make the limit equal whatever you like - as -Trovatore said, it is indeterminate. Gandalf61 (talk) 20:38, 17 August 2008 (UTC)

But the base is x, which increases as x, not as anything else... I don't understand you... --Tango (talk) 21:06, 17 August 2008 (UTC)

The assumption was that the "infinitieth root of zero" can be interpreted as

$\infty ^{0}=\lim _{x\to \infty }x^{1/x}=1$

but an equaly valid interpretation of the "infinitieth root of zero" is

$\infty ^{0}=\lim _{x\to \infty }(a^{x})^{1/x}=a$

Gandalf61 (talk) 21:14, 17 August 2008 (UTC)

Rainwarrior only claimed it was a related question, not the same question. --Tango (talk) 21:15, 17 August 2008 (UTC)

Fine. Ignore what I said. Gandalf61 (talk) 21:20, 17 August 2008 (UTC)

I'm not certain I know what you mean. Viewing the problem in terms of limits, you can ask what happens as each argument grows without bound, in which case the answers are all over the map. On the other hand, you can also view 'infinity' as a number in its own right, or at least a formal symbol, in which case the infinity-ith root of infinity would be exactly that. Black Carrot (talk) 07:55, 18 August 2008 (UTC)

August 18

Sum of two disks

Does anyone know a proper reference to the following result:

Let $i\in \{1,2\}$ be an integer and $a_{i}$, $b_{i}$, $r_{i}$, $x_{i}$, $y_{i}$ be real numbers with $r_{i}>0$. Denote $B_{i}=B(r_{i},x_{i},y_{i})$ closed, convex disk in 2-dimensional euclidean space, with and radius $r_{i}$ and center $(x_{i},y_{i})$ and denote $(a_{i},b_{i})$ some point in $B_{i}$. Define $S$ to be the sum of two disks $B_{i}$ ie. $S=\{(a_{1}+a_{2},b_{1}+b_{2})|(a_{i},b_{i})\in B_{i},i=1,2\}$. Then the following holds:

Theorem: $S=B(r_{1}+r_{2},x_{1}+x_{2},y_{1}+y_{2})$.

88.72.195.199 (talk) 05:42, 18 August 2008 (UTC)

Does it really need a reference? It's pretty easy to prove. Let a = x+p, and b = y+q. Then a+b = x+p+y+q = (x+y)+(p+q). So, the x+y part commutes out in all cases. That means that, centered at the point x+y, the question is whether B(r1,0,0)+B(r2,0,0) = B(r1+r2,0,0). Every sum must land within that radius, by the triangle inequality, so all that's left is to show they cover the entire disc. Split it up into straight lines radiating out from the center. Along each line, each point is within r1+r2 from the center, and is therefore the sum of two vectors in the same direction, one of which is within r1 of the center, the other of which is within r2 of the center. So, every point is covered. Black Carrot (talk) 08:08, 18 August 2008 (UTC)

220 pages probability

Suppose that I have a book of 220 pages. I keep opening the book at random. How many times do I have to open the book until the probability that I'm looking at the only page I haven't seen yet becomes greater than 99%? How many times do I have to open the book until the probability that there are only ten pages I haven't seen yet becomes greater than 90%?

• Random here means random. Please ignore the physics of opening a book.
• This isn't the way opening a book usually works, but assume that when you open a book, you only see one page.

Thanks. —Preceding unsigned comment added by 24.7.54.224 (talk) 06:53, 18 August 2008 (UTC)

I don't think the first one will ever happen. For any number of openings, the odds that there is exactly 1 page left are pretty low, and the odds that you happened to land on it are even lower. The chances will be positive from 220 openings onwards, but I don't think they'll ever get that high. Black Carrot (talk) 08:10, 18 August 2008 (UTC)

Given that there is only one page that you haven't seen, the conditional probability that you see that page when you open the book at a random page is 1/220, and the conditional probability that you do not see that page is 219/220. The probability that you will see that one page at least once in the next n attempts is 1-(219/220)ⁿ, but probability of seeing that page on any specific attempt (given that you are only looking for one page) is always 1/220.

If you want to know how many times you expect to have to open the book until there is only 1 or 10 or whatever pages that you haven't seen, see our articles on the coupon collector's problem and coupon collector's problem (generating function approach). Gandalf61 (talk) 09:07, 18 August 2008 (UTC)

I agree. Two different openings of a book are mutually exclusive events and therefore, it does not matter if you have opened the book 1000 times prior to your next opening; the probability will be the same. The problem is analagous to the problem of tossing a coin: how many times would you need to toss a coin (call this number n) to obtain between n/2 - 10 and n/2 + 10 heads (not inclusive). There is no answer to this problem; however you would expect that the number of heads you would obtain is approx. half as many times as you have tossed the coin. In fact, there is a law in probability known as the Law of large numbers. It says (for this particular case), that the larger the number of coin tosses you make, the more likely you are to get half as many heads as the number of times that you have tossed the coin. You should see the link, Law of large numbers if you want a more detailed discussion on this.

I hope this helps.

Topology Expert (talk) 07:03, 19 August 2008 (UTC)

I disagree with what you write on a few technical points. In particular, you write that as the number n of coin tosses increase, the probability to get n/2 heads also increases (let's suppose n is even, of course). This is not true: the probability of exactly n/2 heads is

${\frac {1}{2^{n}}}{\binom {n}{n/2}}\approx {\frac {\sqrt {2}}{\sqrt {\pi n}}}$

which decreases with increasing n (I used Stirling's approximation). This is elaborated upon a bit more in the fourth paragraph of the article on law of large numbers. Eric. 90.184.71.189 (talk) 11:17, 19 August 2008 (UTC)

Clifford algebra

Is it possible to describe universal property of Clifford algebras using adjoint functors? 83.23.214.17 (talk) 14:33, 18 August 2008 (UTC)

a,b,c are three rational numbers..

a,b,c are three rational numbers and a2^1/2+b3^1/2+c5^1/2=0 , then prove that a=b=c=0 —Preceding unsigned comment added by 117.99.17.13 (talk) 15:35, 18 August 2008 (UTC)

As it says at the top of the page, do your own homework. The general case of this (with arbitrarily many primes and 1 if you want) can be solved quite neatly with a little Galois theory, but this instance can be done bare-handed with a bit of effort. Algebraist 15:39, 18 August 2008 (UTC)

I would start by re-arranging to get a2^1/2+b3^1/2 = -c5^1/2 and then squaring both sides. I'll let you take it from there. Gandalf61 (talk) 15:53, 18 August 2008 (UTC)

As written, this problem is quite simple, just substitute in 0 for a, b, and c and solve. However, if the problem is to prove that there are no other solutions, this gets more complicated. StuRat (talk) 05:47, 19 August 2008 (UTC)

Proving that there are no other solutions is indeed required: the task as given is of the form "prove that if A, then B", where A in this case is "a2^1/2+b3^1/2+c5^1/2=0" and B is "a=b=c=0". This requires showing that there are no possible cases where A would be true but B false. I agree that the task would be absolutely trivial if it were the other way around. —Ilmari Karonen (talk) 06:47, 19 August 2008 (UTC)

Tensor products

Hello,

I'm trying to understand tensor products (of vector spaces) but I'm not sure how it works. On the construction given on that page, what additional equivalence relations should be made to recover $V\times W$ as the quotient ?

I'm especially trying to understand vector valued differential forms, as the construction $\Omega ^{p}(M,E)=\Gamma (E\otimes \Lambda ^{p}T^{*}M)$. I'm perfectly fine with ordinary real valued differential forms being $\Omega ^{p}(M)=\Gamma (\Lambda ^{p}T^{*}M)$, which implies $\mathbb {R} \otimes \Lambda ^{p}T^{*}M=\Lambda ^{p}T^{*}M$ which I'm not sure about (it makes some sense from the definition of tensor product with the quotient, but it doesn't fit in with my intuitive understanding of tensor products with identification of $L^{2}(U\times V,X)$ with $L(U\otimes V,X)$.)

So how does it work ? What is, for example, $\mathbb {Q} \otimes V$ compared to $V$ ? And really, why does $\Gamma (E\otimes \Lambda ^{p}T^{*}M)$ mean $E$-valued differential forms ?

Thanks. --Xedi^Talk 15:46, 18 August 2008 (UTC)

Well, to be terribly informal, if V is an m dimensional space and W is an n dimensional space then V⊗W is the space of m×n matrices. V×W would be an m+n dimensional space. You can't recover it as a quotient of V⊗W; it can even have a higher dimension than V⊗W. If V is a vector space over the reals then $\mathbb {R} \otimes V$ is naturally isomorphic to V in the same way that 1×n matrices are naturally isomorphic to row/column vectors. If V is a space of real-valued linear functionals then $\mathbb {Q} \otimes V$ could be seen as the corresponding space of quaternionic-valued linear functionals and $E\otimes V$ could be seen as the space of E-valued linear functionals. Does that answer your questions? -- BenRG (talk) 21:10, 18 August 2008 (UTC)

Where did the quaternions come from? $\mathbb {Q}$ is the rationals... --Tango (talk) 21:58, 18 August 2008 (UTC)

Guh, sorry, I had quaternions on the brain after participating in this marathon thread. $\mathbb {Q} \otimes V\cong V$ if V is over the rationals. -- BenRG (talk) 23:12, 18 August 2008 (UTC)

Fuel economy

From an old British Math book, is there adequate info to answer this?: Automobile A does 30 miles to a gallon of petrol and 500 miles to a gallon of oil. Automobile B does 40 miles to a gallon of petrol and 400 miles to a gallon of oil. If oil costs as much as petrol, which will be the cheaper automobile to run? Jonpol (talk) 15:55, 18 August 2008 (UTC)

Of course there is insufficient data to determine which is cheaper to run (what if one of them has an engine that requires constant expensive maintenance, and the other does not?), but there's certainly enough to determine which will leave you spending less on petrol and oil. Algebraist 15:59, 18 August 2008 (UTC)

If you let p be the price of a gallon of oil or petrol.

Then the cost per mile for A is p/30 + p/500, same for the other car, and you can then directly compare. As said above, this doesn't account for anything else, as the price of the car for example.--Xedi^Talk 16:04, 18 August 2008 (UTC)

For simplicity, I'd set p to 1 and use my calculator to determine the values of 1/30 + 1/500 and 1/400 + 1/40. --Bowlhover (talk) 19:45, 18 August 2008 (UTC)

The trick is to convert from mpg to gpm. Once you do that, it's quite easy to calculate the cost per mile. 202.147.44.80 (talk) 01:12, 19 August 2008 (UTC)

Hi Thanks, it was not so much the arithmetic but the wording of the question which I find ambiguous. When I first read it, it seemed that it could be interpreted to mean either you run your car on petrol or on oil (e.g. diesel). (There is an answer given at the back, which compares the cost for a 6000 mile journey (presumably using the LCM)). Jonpol (talk) 02:17, 19 August 2008 (UTC)

I don't think it means diesel, I think it means the oil you put in to serve as a lubricant (motor oil). I've not heard of diesel being referred to as "oil". --Tango (talk) 02:44, 19 August 2008 (UTC)

August 19

Mathematicians do not use sliderule??

Is it true that Mathematicians do not use sliderule? I'm talking about back in the days where there is no such thing as pocket (electronic) calculators, and the only device for calculations are sliderule. Heard that Mathematicians do not use them and that most purchasers of sliderule are engineers. 211.28.51.233 (talk) 09:02, 19 August 2008 (UTC)

That is right. See Mathematics, it tends to be about general rules rather than particular numbers and you don't need a slide-rule for that. Engineers are more concerned about the concrete ;-) Dmcq (talk) 09:31, 19 August 2008 (UTC)

By the way you might like Human computer, they'd do the actual calculation very often when someone wanted to work some formulas or statistics out. Even engineers didn't use slide-rules all that much, many used mechanical calculators or could do just as good or better an approximation in their heads. They looked good though. Dmcq (talk) 09:49, 19 August 2008 (UTC)

Pronumerals that satisfy an equation

I need to find all the combinations that satisfy this equation x² + y² + z² = 390. Is there a fast way to do it? 220.244.75.174 (talk) 10:14, 19 August 2008 (UTC)

You will probably notice that all such combinations are points that lie on the surface of a sphere with radius √390. I am not sure what kind of answer you are after but one nice way of characterizing the solutions to this is by using spherical co-ordinates:

x = √390*(cos(θ))*(sin(Φ))

y = √390*(sin(θ))*(sin(Φ))

z = √390*(cos(Φ))

for Φ belonging to [0,π] and θ belonging to [0,2π). Therefore, just choose any Φ and θ in the given range, and you can generate a triple that satisfy your solution. Of course, there may be other ways of characterizing the solutions to your equation but this is one way. For more information, you might like to have a look at Spherical co-ordinates.

I hope this helps.

Topology Expert (talk) 11:20, 19 August 2008 (UTC)

Firstly decide if the order of the numbers and signs matter. Then first simple method is to write a small program to find all the solutions and count them - or just list them all irrespective of order and then decide if the order matters. Second less simple method is to go through by hand taking the squares of 1 to 19 from 390 and check by hand if any of these is the sum of two squares. Third and most complex - learn up some maths Fermat's theorem on sums of two squares. The article doesn't say it but for two squares one can work out from the factorization how many ways of making up the number there are. There's also a theorem by Lagrange which says anything not of the form $4^{a}(8b+7)$ is okay. So with this one can be sure there is some such x,y,z and you just take the squares of 1 to 19 off and use Fermat's theorem. You don't have to work out the numbers but the problem is this last method only works if the order of the numbers matters. Dmcq (talk) 11:07, 19 August 2008 (UTC)

It is not specified whether the solutions to the equation have to be integers or not. However, now that I look at this, it seems more likely that only integer solutions are required. Otherwise, there would be uncountably many solutions.

My method still works though; just work when 390 multiplied with the square of cos (Φ) is a perfect square; this is an easy thing to do since the square of cos (Φ) has an absolute value less than 1. Then for each such angle Φ (there are only 19 such values), work out whether there exists an angle θ in the given range such that (sin (θ))*(sin(Φ)) squared times 390 is a perfect square; i.e work out all such numbers with absolute value less than 1 with the property of sin (θ); from there you can work out θ by using the inverse sine function. From here you reduce the possibilites of θ to ensure that x is also an integer.

I hope this helps.

Topology Expert (talk) 11:38, 19 August 2008 (UTC)

(There is a similar question at Wikipedia:Reference_desk/Computing#Excel_Question) Gandalf61 (talk) 12:59, 19 August 2008 (UTC)