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Mathematics desk

[edit] April 1

[edit] the number of cows in great britain²

plz

Europe has 130 million cows from the cattle article. Coolotter88 02:06, 1 April 2007 (UTC)

We can't do your homework for you, but the answer can be computed by taking the area integral of the cow density function over the territory of Great Britain. Note that the report was issued April 1st. If you find a different answer than 9,300,454 head of cattle (for 2005), show us your computation and we may be able to spot the error. --Lambiam ^Talk 13:09, 1 April 2007 (UTC)

[edit] Fraction exponents

How do you calculate numbers to fraction exponents, such as 6^1/2? —The preceding unsigned comment was added by 65.30.153.24 (talk) 01:06, 1 April 2007 (UTC).

Well, the simple way is to use your calculator. Keep in mind that raising a base to a fractional exponent, n, is the same as taking the n^th root of that number. So 9^1/2 = $\sqrt{9}$ = 3. or 8^1/3 = $\sqrt[3]{8}$ = 2. See Exponentiation#Rational_exponents for more. (the answer to your question is $\sqrt{6}$ or 2.44948974.)--Ybbor^TSurvey! 01:17, 1 April 2007 (UTC)

Ah, I didn't know that. Thanks much!—The preceding unsigned comment was added by 65.30.153.24 (talk • contribs).

Also, note that raising to a negative exponent (2 raised to the -2) is the same as flipping the base (2/1^-2 = 1/2^2). ST47Talk 01:59, 1 April 2007 (UTC)

Because of awkwardness in typesetting, instead of √6 we sometimes write 6^1/2, which means the same thing. Similarly, we may write (x²+y²)^3/2, which means take the Euclidean length, (x²+y²)^1/2, and cube it (raise it to the power of three). Otherwise, we'd end up with something bloated, like

$\left( \sqrt{x^2 + y^2} \right)^3 . \,\!$

To calculate a numeric value in more general cases, we (and our calculators) typically use logarithms. That is, when presented with b^r, we first compute the (natural) logarithm of b, multiply that by r, and feed the result into the exponential function:

$b^r = \exp(r \log b) . \,\!$

The logarithm and exponential functions are heavily used elementary functions, and we can expect efficient implementations yielding good accuracy to be available in our hand calculator or standard library. For a provocative historical examination of the early computation of tables of logarithms by Henry Briggs, try this narrative by Erik Vestergaard. --KSmrq^T 02:33, 1 April 2007 (UTC)

[edit] toyota symbol

i just noticed that the toyota symbol is three elipses put together, what would be the equation to graph this? —The preceding unsigned comment was added by 72.146.114.13 (talk) 02:23, 1 April 2007 (UTC).

Hi. I got the implicit equation $\frac{(x^2 + 9y^2 - 18y)(4x^2 + y^2 - 4)(x^2 + 4y^2 - 16)}{576} = 0$ by trying. Of course it isn't perfect.

It comes from the product of three implicit equations each corresponding to an ellipse :

$\left(\frac{x}{4}\right)^2 + \left(\frac{y}{2}\right)^2 -1 = 0$

$\left(\frac{x}{3}\right)^2 + \left(y-1\right)^2 -1 = 0$

$\left(x\right)^2 + \left(\frac{y}{2}\right)^2 -1 = 0$

I'm sure someone can achieve better with exact references.

-Xedi 12:53, 1 April 2007 (UTC)

The denominator 576 in the equation is superflous; E/576 = 0 is equivalent to E = 0. The ellipses forming the picture should actually have a varying thickness, and can probably be approximated fairly well as the differences between the areas of an outer and an inner ellipse. The necessary data to get a professional quality result must be present one way or another in this encapsulated PostScript file. --Lambiam ^Talk 13:21, 1 April 2007 (UTC)

[edit] Stuck on proving an inequality

I have a homework problem for my analysis class that I have all worked out, except for one inequality that has resisted all my attempts to prove it. I would like to show that, for all $x\in[0,1]$ and all $n=1,2,3,\ldots$ , the following inequality holds:

$\left(1+{x\over n}\right)^n\le\left(1+{x\over n+1}\right)^{n+1}.$

I have tried to prove this inequality in several ways, but I can't seem to get anywhere useful; I just come back to where I started, or I meet some dead end that I've seen several times before. If anyone can help to point me in the right direction, I'd be very grateful. —Bkell (talk) 07:35, 1 April 2007 (UTC)

Alternatively, I could show that for all $x\in[0,1]$ and all $n=1,2,3,\ldots$ we have

$e^x\ge\left(1+{x\over n}\right)^n$ ,

but I don't see a way to prove this without proving the inequality above. —Bkell (talk) 07:39, 1 April 2007 (UTC)

Taking n-th roots at both sides, the latter inequality becomes

$e^{x/n}\ge 1+{x\over n}\,,$

or, substituting x := nx,

$e^x\ge 1+x\,.$

Using the property that the exponential function is its own derivative, this should not be hard to prove. --Lambiam ^Talk 07:52, 1 April 2007 (UTC)

Aha! Thank you. —Bkell (talk) 08:18, 1 April 2007 (UTC)

Or you could tackle the original inequality using the binomial expansion:

$\left(1+{x\over n}\right)^n=1+x+\frac{n-1}{2!}\frac{x^2}{n}+\frac{(n-1)(n-2)}{3!}\frac{x^3}{n^2} + \ldots$

$\left(1+{x\over n+1}\right)^{n+1}=1+x+\frac{n}{2!}\frac{x^2}{(n+1)}+\frac{n(n-1)}{3!}\frac{x^3}{(n+1)^2} + \ldots$

then show that after the first two terms, each term in the first expansion is smaller than the corresponding term in the second expansion. And in addition there is an extra (positive) term at the end of the second expansion. I think the inequality actually holds for all positive x. It does not necessarily hold for negative x, as the counter example x= -3, n=2 demonstrates. Gandalf61 09:21, 1 April 2007 (UTC)

[edit] Counting, continued

Continued from above, here's my counting problem. I'm trying to count the number of solutions of $a + 2 b + 3 c = n$ where $a, b$ and $c$ are non-negative integers. I can do it, I have this:

$\sum_{c=0}^{\frac{n}{3}-\left(\frac{1}{3}\right)\mathrm{mod}(n,3)}\frac{n-3c}{2}-\left(\frac{1}{2}\right)\mathrm{mod}(n-3c,2)+1$

The rationale behind this foul expression is as follows. Suppose we were looking at counting solutions to $a + 2 b = k$ . Then this is simpler, just count the number of possible values for $b$ . These range from 0 to $\frac{j}{2}$ where $j$ is the greatest $j\leqslant k$ divisible by 2, so the total number is

$\frac{k}{2}-\left(\frac{1}{2}\right)\mathrm{mod}(k,2)+1$

Keeping this in mind, and going back to the original problem, for given $n$ the possible values of $c$ range from 0 to $\frac{m}{3}$ where $m$ is the greatest $m\leqslant n$ divisible by 3. For each of these possible values then, get the remainder $n - 3 c$ and apply the above argument to count how many ways to carve it up. Add up all these bits and we have the answer.

Unless I've made typos, this works. I have my computer churning it out and it agrees with this guy. But there's supposed to be a nicer way - like a simple closed form function of $n$ . How should I proceed? Thanks. --87.194.21.177 10:32, 1 April 2007 (UTC)

Unless I don't understand the problem, the sequence is 1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, 16, 19, 21, 24, 27, 30, 33, 37, 40, 44, 48, 52, 56, ... (sequence A001399 in OEIS). --Lambiam ^Talk 14:11, 1 April 2007 (UTC)

Thanks, that's it. For some reason my computer messed up (okay, I messed up) for n=13 and n=26 so I missed it when I searched OEIS. That's it for sure though, thanks. --87.194.21.177 16:50, 1 April 2007 (UTC)

[edit] Approximating curves

If I pick up a pen and scribble any sort of curve onto a piece of paper, can it definitely be described mathematically, by a function? I would think this would be true, but I can't convince myself with any sort of rigour. If it is the case, how would one even go about approximating the equation of a curve? I've had a go with some online function graphers, but all the curves are too nice and smooth - to really approximate a nasty curve, would you need some horribly complicated piecewise function? Just curious, 81.102.34.92 16:44, 1 April 2007 (UTC)

If you scribble something on paper and I give you a mathematical description, claiming it is a description of your scribbles, how would you determine whether that claim is correct? A mathematical description is or can be expressed, presumably, as a finite sequence of symbols drawn from a finite repertoire of symbols, subject to certain rules to make it meaningful. That means the number of possible descriptions is countable, and since the total number is infinite, countably infinite, the same cardinality as the natural numbers, and even enumerable. However, it can be argued with some plausibility – but here we leave the realm of mathematics and move into physics – that the number of possible scribbles is uncountable, and that the probability that a scribble you produce fits any of this countable set of descriptions is zero. Similarly, an actual nail would never way exactly π gram; it may be close, but almost certainly at least a fraction of 10⁻⁴⁰ off (or more precisely, such precision has no physical meaning). So this is a negative result.

More positively, if the scribble can be viewed, abstractly, as an open bounded and connected subset of the plane produced by the ink of a pen with finite (meaning non-zero) thickness, then it can be proved that for any positive value ε, however small, there is a curve described by one of these mathematical descriptions such that it is wholly contained within your scribble, and comes within a distance ε of any point in the scribble. If S is the scribble and C the curve, this can be formulated mathematically as

C ⊆ S ⊆ C ⊕ D_ε(0,0),

in which the dilation operation denoted by ⊕ is Minkowski addition, and its second operand is the disk with radius ε centred on (0,0). To explain this fully would take up much space, but some mathematical background can be found in our article Approximation theory. --Lambiam ^Talk 17:24, 1 April 2007 (UTC)

Thanks for the help, that's cleared it up somewhat. I've always sort of assumed that anything I scribble has to be describable mathematically, since it, if drawn under "perfect" conditions" (ie a plane whose smallest division is zero, a pen that draws a line of zero width), physically exists - basically, I've always assumed for something to physically exist, it has to be mathematically describable. Is that a valid assumption? 81.102.34.92 18:27, 1 April 2007 (UTC)

That is a question about physics – properties of nature – and not about mathematics. I see no reason to assume this to be true, or even that it is a meaningful statement, and since "perfect" conditions can't exist, I consider the application to your scribbles highly dubious, even if we assume that physical reality admits of a mathematical description. If it is true, we have no way of knowing it. Of course, modelling physical reality by coercing it into a mathematical abstraction has been and continues to be fruitful, but on sufficiently close examination the models thus far always come apart. --Lambiam ^Talk 20:16, 1 April 2007 (UTC)

Generally when working with mathematically defined curves, a the curve is simplified as passing through a certain number of points, and a polynomial function is used to interpolate a curved line between them. Splines are a commonly used technique for making curves. Also, yes, where the curve is not smooth, you would add many more points to increase the accuracy. You don't necessarily need a piecewise function (like splines), but it is usually easier to work that way because otherwise your polynomial goes up a degree for every point you add (which could get very tedious to solve). - Rainwarrior 20:02, 1 April 2007 (UTC)

I don't see the above answers as being as strongly on topic as two other directly applicable topics: Parametric_equation and the issue of discontinuities (first order, second order, etc.) in the curve being drawn, see Classification_of_discontinuities.

Taken together, those two articles pretty much cover the theoretical issues of drawing curves on paper, and are in fact arguably the best starting point if one wanted to develop algorithms for drawing or modeling curves in real life. Dougmerritt 20:35, 1 April 2007 (UTC)

I'm getting bogged down in the physical reality of the situation - what I really want to know is, is any curve drawn in the (mathematically ideal 2D plane) describable by a function, no matter how messy? Surely it must be possible, even if it comes down to a gigantic string of piecewise, constant functions, each applicable only at a single point? 81.102.34.92 20:52, 1 April 2007 (UTC)

In short, the answer is "yes".

If one tried to do it non-parametrically, then even circles tend to end up being relations rather than functions, thus the first article I pointed to.

If one worries about whether piecewise sequences are an issue, well, they come up as a result of discontinuities between continuous sections of the curve/function, thus the second article.

But actually one can even have an infinite number of discontinuities, or even a function that consists of nothing but discontinuities; such things historically presented challenges to the theory of functions (and to the historical definitions of "function"). Poincare' called them "monsters". But such functions can exist in a suitably modern theory of functions. They don't really come up in regard to drawing on paper, so that's just to illustrate the utter lack of serious problems with drawing curves. Dougmerritt 21:27, 1 April 2007 (UTC)

(Below, Salix and Lambiam bring up some interesting mathematical issues, but it's not clear to me, as yet, that they involve issues that the original questioner would care about, even though they are important in abstract math. I guess we shall see, as the discussion proceeds. Dougmerritt 02:55, 2 April 2007 (UTC))

There a theorem of Hassler Whitney, stating that any close set of points in Rⁿ can be defined as a set

f - 1 (0)

, for a smooth function f. In particular any curve in the plane could be defined in this way. See Algebraic curve. --Salix alba (talk) 23:17, 1 April 2007 (UTC)

Hmm. I'm not sure what you're driving at (tell me to go read the article you cited, if it indeed suffices). Surely "close set of points" ends up being the same thing as "smooth" in this context? Dougmerritt 02:55, 2 April 2007 (UTC)

The answer to your last formulation really hinges on what you mean by "describable". By definition, the mathematical abstraction we call a curve is a function, or, if you will, an equivalence class of functions, but then any member represents the curve. However, we have no way of exactly describing these functions in any finite way, however gigantic the strings. We can give finite descriptions that approximate them arbitrarily close. For simplicity, say you put just two points on paper, which can be described by giving the distance between them, a single real number. But is that number "describable"? --Lambiam ^Talk 23:33, 1 April 2007 (UTC)

Yes indeed. However, the original question is about an idealized model of drawing on paper. It seems to me that all straightforward such models would pretty much inherently involve "describable" points and paths. How would undescribable points or paths arise in this sort of context, despite the fact that such things are in fact important in general in mathematics? (E.g. the ineffability of almost all transcendental numbers/functions, versus the inherent finite-hood (arguably, to a first approximation) of real-world methods, measurements, and entities.) Dougmerritt 02:55, 2 April 2007 (UTC)

If in this idealized model you can only draw describable curves, then the answer is: Yes, indeed, any curve you can draw is describable. It is not clear to me, however, that this corresponds to the intuitive notion of the questioner. It would seem to imply (depending on what you mean by "describable") that you can only mark a countable subset of points in the plane, which to me would be counterintuitive. In an idealized setting, is any weight in kg you obtain by weighing an object necessarily describable? "Undescribable" weights would arise just because they happen to be the weight of the object under consideration, which presumably can be any positive real, of which there are more than even the mathematician's pen can describe. --Lambiam ^Talk 03:24, 2 April 2007 (UTC)

That's what I thought you meant. You and I differ in what we find intuitive, then. To me, when a question is grounded in real world considerations, I take for granted that all inputs and methods are describable. After all, undescribable entities cannot, by definition, be described in the real world. That's not to say that your intuition is wrong, it's just to give some explanation for why my intuition is different than yours.

Aside from that, I don't see how you can introduce the (mathematically quite valid) undescribable transcendental number of the mathematician's pen into what is supposed to be a model of the real world. Dougmerritt 03:36, 2 April 2007 (UTC)

(Correcting myself: I accidentally used your phrase "mathematician's pen" as if it meant undescribables, whereas you very clearly used it to mean "describables, even the ones that require sophistication to describe". Mea culpa. I don't think anything in the discussion really changes as a result, though. Dougmerritt 03:56, 2 April 2007 (UTC))

A fascinating discussion ! Breaking the original question down into sub-questions, I think we have the following:

Can an arbitrary "scribble" be approximated as closely as we like by the curve of a mathematical function that can be described in a finite number of steps ? Answer: yes. You just need to find a family of "describable" functions whose curves are dense in the space of "scribbles". Splines will probably work.
Is there a "describable" function whose curve is exactly the same as a specific (but arbitrary) "scribble" - this one here on this piece of paper, let's call it scribble 1784032 ? Answer: yes. I describe the function as follows: "the function whose curve is scribble 1784032". If you think this is "cheating", this may be because your definition of a "describable" function is more restrictive than mine.
Does any and every "scribble" correspond to the curve of a "describable" function ? Answer: no. As Lambian has pointed out, the set of "describable" functions can only be countable (regardless of how we define "describable"), whereas the set of scribbles is uncountable. So we can apply a version of Cantor's diagonal argument to show that there must be some scribble that is not included in the curves of any set of "describable" functions. Of course, if we find a specific scribble that is not covered by our set of "describable" functions - say scribble 1784032 - we can always add "the function whose curve is scribble 1784032" to the set. But there will always be another "undescribable" scribble - in fact, uncountably many.

The difference between cases 2 and 3 is subtle and non-intuitive. But most mathematicians find it both interesting and important (your mileage may vary !). Gandalf61 10:58, 2 April 2007 (UTC)

—Q: I have drawn this scribble on this piece of paper here. Can you give a mathematical description of it?

—A: Sure thing, no sweat. Let's call your scribble "Bobbie". Then here is its mathematical description: "Bobbie"! Easy, no?

--Lambiam ^Talk 19:44, 2 April 2007 (UTC)

Almost, but not quite. "Bobbie" is a label or identifier for the scribble. A mathematical description of the curve would be "the set of points that lie on the scribble Bobbie", or even "the co-ordinates of the points that lie on the scribble Bobbie" if you don't mind a frame dependent description. Gandalf61 20:27, 2 April 2007 (UTC)

Well this was certainly interesting! Thanks for summing everything up Gandalf. One last question - from a practical point of view, how would I actually go about trying to describe a scribble? I'm assuming some of the articles mentioned would help me (such as Interpolation, perhaps?) but they're all a bit abstract. What are the branches of maths I should be searching for? Having signed in, Icthyos 17:59, 2 April 2007 (UTC)

(Thanks for the signature.) As a practical matter, you might have a look at Potrace. :-) --KSmrq^T 18:34, 2 April 2007 (UTC)

See curve fitting. StuRat 02:06, 3 April 2007 (UTC)

[edit] Numbers

I'm looking for a number that is the sum of two consecutive positive integers, of three consecutive positive integers, and of four consecutive positive integers. Any suggestions? --Carnildo 18:57, 1 April 2007 (UTC)

Elementary observations:

$\begin{align} (k) + (k+1) = 2k + 1 &\Rarr n \equiv 1 \pmod{2} \\ (k) + (k+1) + (k+2) = 3k + 3 &\Rarr n \equiv 0 \pmod{3} \\ (k) + (k+1) + (k+2) + (k+3) = 4k + 6 &\Rarr n \equiv 2 \pmod{4} \end{align}$

What does this tell you? --KSmrq^T 19:20, 1 April 2007 (UTC)

I find that there is no such number.

If n is your number, then n = a + (a+1) = b + (b+1) + (b+2) = c + (c+1) + (c+2) + (c+3)

Or n = 2a + 1 = 3b + 3 = 4c + 6

Then for example 2a+1 = 4c + 6 leads to c = (2a-5)/4

But (2a-5)/4 cannot be an integer for any integer value of a, because 2a is even, then 2a-5 is odd so 2a-5 cannot be a multiple of 4.

-Xedi 19:22, 1 April 2007 (UTC)

Oh well, KSmrq beat me to it. :)

It's April 1. I can always hope that someone provides me with a number that is both even and odd. --Carnildo 20:56, 1 April 2007 (UTC)

It's April 2 now. For penance, find a number that can be written in 23 different ways as the sum of two or more consecutive positive integers. --Lambiam ^Talk 03:38, 2 April 2007 (UTC)

[edit] Binomial expansion for complex powers

Possible?--Ķĩřβȳ ♥♥♥Ťįɱé Ø 21:11, 1 April 2007 (UTC)

See Exponentiation, particularly Exponentiation#Complex_powers_of_complex_numbers. --Ybbor^TSurvey! 21:16, 1 April 2007 (UTC)

Actually, I think the appropriate reference is the binomial theorem. --KSmrq^T 21:51, 1 April 2007 (UTC)

I'm sorry if I didn't ask my question correctly. I meant, what does it even mean to have a complex exponent? Applications? etc. --Ķĩřβȳ ♥♥♥Ťįɱé Ø 00:16, 2 April 2007 (UTC)

Here, the first response may be helpful. --Lambiam ^Talk 00:35, 2 April 2007 (UTC)

[edit] April 2

[edit] is a sum of f(eigenvalue) meaningful?

I just recently got interested in eigenvectors. If $x i$ are the eigenvalues of a given matrix, is $\sum x_i$ or $\sum x_i^2$ or $\prod x_i$ equal to anything interesting? —Tamfang 20:47, 2 April 2007 (UTC)

The sum and product of the eigenvalues (taken with appropriate multiplicities) are related to simple properties of the matrix. There is more information in our article on eigenvalues (although you have to look quite hard to find it). Gandalf61 21:00, 2 April 2007 (UTC)

Since the trace, or the sum of the elements on the main diagonal of a matrix, is preserved by unitary equivalence, the Jordan normal form tells us that it is equal to the sum of the eigenvalues;
Similarly, because the eigenvalues of a triangular matrix are the entries on the main diagonal, the determinant equals the product of the eigenvalues (counted according to algebraic multiplicity).

From eigenvalues. --Xedi 23:43, 2 April 2007 (UTC)

Thanks. Why I asked: the application that interests me at the moment is deriving topological coordinates from a graph's adjacency matrix. It occurred to me that the entropy function of the set of (some increasing function of) eigenvalues would give an objective, albeit fuzzy, value for the "natural" number of dimensions, possibly better than the subjective number given by the "scree test". —Tamfang 06:18, 3 April 2007 (UTC)

If your matrix has finite order, then the sums $\sum x_i^k$ over all k in fact determine all the eigenvalues. (A fact of basic algebra, not matrices or eigenvalues. This fact is related to the power of character theory in the group representation theory of finite groups. (If this interests you and you can't see how this is relevant, I can elaborate. Respond here or leave me a message.) Tesseran 01:39, 4 April 2007 (UTC)

[edit] Simple groups

Is it true that any group is expressible as the direct product of a number of simple groups? —The preceding unsigned comment was added by 129.78.208.4 (talk) 22:56, 2 April 2007 (UTC).

~~This is true for any finite group~~ No - any finite group can be constructed as a product of simple groups, but not necessarily as a direct product. A construction that creates a finite group as the ~~direct~~ product of a sequence of simple groups is called a composition series, and every finite group has a composition series (note that this is not true for infinite groups). As an analogy, we can construct the following composition series to show that 12 is the product of a sequence of prime numbers:

$1\xrightarrow{\times2}2\xrightarrow{\times2}4\xrightarrow{\times3}12$

Of course, this composition series for 12 is not unique - two alternative composition series are:

$1\xrightarrow{\times2}2\xrightarrow{\times3}6\xrightarrow{\times2}12$

$1\xrightarrow{\times3}3\xrightarrow{\times2}6\xrightarrow{\times2}12$

We can see that the prime numbers involved in each of these composition series are the same (2,2 and 3) although they appear in different orders. The Jordan-Hölder theorem says that the same property holds for groups - each composition series for a given group contains the same set of simple groups as composition factors, in different orders. In this sense, the "factorisation" of a finite group into simple groups is unique. Gandalf61 08:46, 3 April 2007 (UTC)

Thanks, I think I might have actually asked that before, but promptly forgot the answer ;) —The preceding unsigned comment was added by 149.135.62.165 (talk) 10:05, 3 April 2007 (UTC).

I think the answer is false, even for finite groups. – b_jonas 13:49, 3 April 2007 (UTC)

Ah - yes - because the group products involved in a composition series are not necessarily direct products ? Gandalf61 14:40, 3 April 2007 (UTC)

Sometimes they're semidirect, but it gets worse than that even. Incidentally, the simplest counterexample to the original question is the cyclic group of order 4. Algebraist 00:19, 4 April 2007 (UTC)

Right, that counterexample is helpful. Then does that mean that the necessary paragraph on Simple group is wrong? (or if not wrong, needs clarification?) Is there still some known notion of "irreducible" arising from a decomposition of a group into a direct product of smaller groups? (we would consider the cyclic group of order 4 as an "irreducible", then). There is a well-known notion for abelian groups. Restrict to finite groups if needed. 01:20, 4 April 2007 (UTC)

Okay, I have corrected my original response above, to avoid confusion. One more question - why don't the product operations in a composition series have to even be semidirect products ? Since a composition series is a normal series, each subgroup in the series is a normal subgroup of the one above it - so doesn't that mean that the product operations are at least semidirect, even if they are not direct ? Gandalf61 09:24, 4 April 2007 (UTC)

No. Consider C₄ again. It has a normal subgroup isomorphic to C₂, but no disjoint C₂ subgroup to form a semidirect product with. I think (haven't done enough algebra to be sure) that the Jordan-Hölder theorem is about the best you can do here. Algebraist 19:04, 4 April 2007 (UTC)

[edit] April 3

[edit] time for thrown baseball to reach height s

Heres my problem from homework: "If a baseball is thrown straight upward from a level ground with an initial velocity of 72 ft/s, its altitude s (in feet) after t seconds is given by s = −16t² + 72t. For what values of t will the ball be at least 32 feet above the ground?" So I figured you make an equation 32 ≤ −16t² + 72t and figure it out algebraically from there. However, in the back of the book it gives the answer as 1/2 ≤ t ≤ 4. Could someone guide me to how you would obtain this answer? Thanks 65.30.153.24

Solve the inequality. You don't know how to do that? Factor the quadratic. —The preceding unsigned comment was added by 129.78.208.4 (talk) 03:18, 3 April 2007 (UTC).

We know from physical experience, if not from mathematics, that the ball will rise up and then fall down again. If it goes above 32 feet, then for two values of time it will be exactly 32 feet high. Thus we seek solutions for the equation

$72t - 16t^2 = 32 . \,\!$

If you cannot solve this equation on your own, probably you should drop the course, as the challenges will only get harder.

But let's assume you can find two solutions; then one of the times will be less than the other, and on physical grounds we may expect times between these to satisfy the demand. (To verify, calculate the height at their average.) --KSmrq^T 04:01, 3 April 2007 (UTC)

=================================================================

32 <= X

Add (-32) to both sides

0 <= X - 32

Replace X with -16t^2 + 72t

0 <= -16t^2 + 72t - 32

From here on, the problem is very simple to solve.

If you are a lazy bastard then plot the curve -16t^2 + 72t - 32 on your graphical calculator.

220.239.107.54 13:33, 3 April 2007 (UTC)

[edit] how to generate prime numbers?

is there any way to generate an arbitrary prime number?or ageneral way to generate prime numbers? 80.255.40.168 08:04, 3 April 2007 (UTC)ARTHER

Take a random number, test if prime. If not, add one and repeat. There are reasonable primality testing methods so this method is valid to generate a random prime. —The preceding unsigned comment was added by 149.135.62.165 (talk) 10:03, 3 April 2007 (UTC).

Some minor points. First, "random" or "arbitrary" on its own is not well defined - you need to say what distribution you want your prime numbers to follow. But let's assume that "random" means that you want a uniform distribution, where each prime number is equaly likely to be picked. Then you need to put an upper limit on the range of your prime numbers - so you want a random prime number between 1 and 1,000, say, or between 1 and 1,000,000. Finally, if you pick a random number with a uniform distribution and the discard non-primes, this will not produce a uniform distribution of primes. There will be a small bias towards smaller primes because a large number is less likely to be prime than a small number (see prime number theorem). Gandalf61 10:59, 3 April 2007 (UTC)

See also Generating primes. PrimeHunter 11:53, 3 April 2007 (UTC)

To clarify a bit on something Gandalf said, you won't get a uniform random distribution of primes if you follow the "discard and add one" procedure. In fact, though, I think the bias in that procedure would be to make individual large primes more likely than individual small primes. The reason is that prime deserts, which are large sequence of consecutive composite numbers, will all result in the same prime number being chosen. If you start in a desert, and add one repeatedly until you get a prime, you'll get the prime at the high end of the desert. So the probability of getting a prime at the high end of a desert is proportional to the size of the desert of composite numbers preceding it. Thus the most likely individual primes in that distribution are the ones that are preceded by a large number of non-prime numbers.

You can modify your procedure, though, to get a uniform distribution by doing a new uniformally random selection over your range every time (ie randomly pick a number in the range, check if it's prime, and if not randomly pick a different number). Under that algorithm the probability of any particular prime number being eventually chosen is equal. Dugwiki 15:34, 4 April 2007 (UTC)

Aha, I should have read the anon contributor's answer more carefully. Yes, the "discard and add one" method introduces a variable bias because the probability of a prime being chosen is proportional to the length of the non-prime gap immediately below it. The upper prime in a twin prime pair, for example, will have a much smaller chance of being chosen than a prime at the high end of a "desert". But the "discard and choose a new random number" method - which is what I thought had been suggested - still has its own systematic bias towards smaller primes. For example, there are 15 primes below 50, and only 10 primes between 50 and 100, so if your upper limit of your range is 100, then 60% of your "random" primes will be in the bottom half of your range, and only 40% will be in the upper half. Gandalf61 15:59, 4 April 2007 (UTC)

[edit] why cannot we find ln(-1)?

we know that sqr(-1)=i,this complex number comes from finding roots or (x^2)+1=0.the function f(x)=(x^2)+1 has no graph when x<1.the question is,the function f(x)=lnx has no graph when x<0,so why cannot we find roots or sort of numbers like ln(-1)?

80.255.40.168 08:22, 3 April 2007 (UTC)ARTHER

The complex logarithm is defined for all non-zero complex numbers. It is also multi-valued. Since

e (2 k + 1)π i = - 1

(for any integer k),

ln( - 1) = (2 k + 1)π i

for any integer k. Usually people restrict the angles to be within

( - π,π]

, so they say

ln( - 1) = π i

. --Spoon! 09:40, 3 April 2007 (UTC)

[edit] Geography

1) What is the angle (degrees from North) from Lahore, Pakistan to Makkah, KSA? 2) What is the angle (degrees from North) from Lahore, Pakistan to Mashhad, Iran? —The preceding unsigned comment was added by 203.81.194.11 (talk) 09:14, 3 April 2007 (UTC).

How accurate do you want the answer? If a not very accurate answer will suffice then bring out a map and draw a line from Lahore to Mecca. Then read the angle off the map.

This is obviously wrong. This method gives you always direction of 90° or 270° for two places on the same circle of latitude, while the orthodrome is a circle of latitude only if it is the equator — but in most cases it is not a circle of latitude, so the direction sought is neither west nor east. For some pairs of points of the same latitude it can even be NORTH! (Consider two points at latitude 50°N, one at longitude 90°W, the other one at 90°E.) --CiaPan 14:19, 3 April 2007 (UTC)

If you want a very accurate answer then you need to use spherical coordinates. Using the dot product multiply the two vectors (from the centre of the earth to both cities) to get the angle between the two cities. Then use the Law_of_cosines_(spherical) rule to find the desired angle. Problem solved. 220.239.107.54 13:11, 3 April 2007 (UTC)

PS: You probably figure out by now that we will not give you the answer. We just tell you how to find/calc the answer for yourself. 220.239.107.54 13:44, 3 April 2007 (UTC)

Who is "we"? --Lambiam ^Talk 13:51, 3 April 2007 (UTC)

The OP actually was asking for the azimuth from one point to the other, not the angle subtended at the core by the great circle segment between them. Anyway, tools to calculate this for you are readily available online. --Tardis 16:16, 3 April 2007 (UTC)

.....and the 220.239.107.54 user's answer leads toward finding the azimuth value. Given the calculated (angular) distance from A to B you can construct a spherical triangle A-B-pole, which has all sides known. Then it is possible to calculate its angles (actually you need only one of them) and eventually get the described direction as the horizontal angle from North. --CiaPan 16:55, 3 April 2007 (UTC)

I found only one free online azimuth calculator: the Great circle azimuth bearing and range calculator (with magnetic north). From the results I get it appears that it uses a spherical model. For the locations that are relevant here, the resulting discrepancy with the standard ellipsoid model is about 0.1°. The data I used as geolocation coordinates, picking, if possible, a striking spot in the cities using Google Earth – but I didn't find anything particularly striking in Lahore:

Lahore: 31.549°N, 74.342°E
Makkah: 21.4225°N, 39.8261°E
Mashhad: 36.290°N, 59.598°E

The azimuths I found with respect to Lahore, using the oblate spheroid model with standard parameters, are:

Makkah: 99.60°
Mashhad: 64.89°

These are the respective deviations from true North, measured counterclockwise (in Westerly direction). --Lambiam ^Talk 20:53, 4 April 2007 (UTC)

[edit] Cox proportional hazards model

In survival analysis, I was wondering if it was possible to use a Cox PH model to calculate expected survival times. From what I see the cumulative hazard will never be infinite so the expectation will always diverge. Am I missing something? Thanks. Faisel Gulamhussein 23:35, 3 April 2007 (UTC)

[edit] April 4

[edit] Foot and Penis Size

Hi, I heard that if you take 3 quaters of the measurement of your foot then times it by 3, it shows your erect size. Is this true? Thank you —The preceding unsigned comment was added by 202.161.2.238 (talk) 06:54, 4 April 2007 (UTC).

This reference desk is for questions about Mathematics, Calculus, and Accounting. --Lambiam ^Talk 08:05, 4 April 2007 (UTC)

No, it's not true, see this mention on Snopes.com for further information. - Mgm|^(talk) 09:30, 4 April 2007 (UTC)

If you take 3/4 of the length of a typical adult male foot, say 10 inches, then multiply by 3, you get 9/4 of 10 or 22.5 inches. Does that sound like the proper length to you ? StuRat 20:51, 4 April 2007 (UTC)

BTW, doesn't this question sound like it would also work under the title of the previous question ? :-) StuRat 20:53, 4 April 2007 (UTC)

Sounds proper to me! :D Splintercellguy 22:50, 4 April 2007 (UTC)

[edit] Differentation

$\cfrac{(x + \delta x)^2 - x^2}{\delta x}$

Then...

$\cfrac{x^2 + 2x\delta x + \delta x^2 - x^2}{\delta x}$

Then...

$\cfrac{2x\delta x + \delta x^2}{\delta x}$

According to my book, by simplification, we get:

$2x + \delta x \,$

But I notice that two $δ x$ have been removed from the numerator when there is only one in the denominator. What have I missed here? —The preceding unsigned comment was added by 164.11.204.51 (talk • contribs).

A single factor of

δ x

has been removed from each of the terms in the numerator. Filling in the missing step, we have:

$\cfrac{2x\delta x + \delta x^2}{\delta x}=\cfrac{\delta x(2x + \delta x)}{\delta x}=2x + \delta x$ ... Gandalf61 10:16, 4 April 2007 (UTC)

[edit] largest prime number

i read in wikipedia that there are prizes to find the largest prime number. can we put (10^n)+1,n,is integer.then we start to pick up n`s as large as possible to find the lrgest prime number???is`nt 100001,10000000000001 are prime numbers??? 80.255.40.168 12:39, 4 April 2007 (UTC)fwfabii

Prime number records are typically created in this fashion. Take a look at Mersenne prime for example. (note that not all numbers of the form 10^n+1 are prime, for example 1001 is not. Sander123 12:45, 4 April 2007 (UTC)

To be precise, there is no such thing as the largest prime number. Euclid already proved that there are infinitely many prime numbers. The prizes are for very large prime numbers. The example numbers you give are not primes: 100001 = 11 × 9091 and 10000000000001 = 11 × 859 × 1058313049. They are also not very large; the largest of these two has 14 digits. To win the least of the prizes, the prime needs to have at least 10000000 digits; printed in a book that would take up something like one thousand pages. Although we have "fast" primality tests, they are not nearly fast enough to cope with primes that large. For that we need some mathematical breakthrough. --Lambiam ^Talk 13:05, 4 April 2007 (UTC)

Actually, there are known primality tests that can test numbers of special forms above 10,000,000 digits, for example the Lucas–Lehmer test for Mersenne numbers. A test takes a long time and a lot of tests are probably needed. Great Internet Mersenne Prime Search (GIMPS) has tested many candidates with it and found the largest known prime which currently has 9,808,358 digits. Note that 10ⁿ+1 for a positive integer n is known to be composite when n is not a power of 2. And $10^{2^m}+1$ grows so fast that it appears likely that 11 and 101 (for m = 1 and 2) are the only primes of that form. PrimeHunter 02:03, 5 April 2007 (UTC)

[edit] Angle problem

I was e-mailed this angle problem and I'm stuck trying to figure it out. It consists of a regular hexagon inside a circle. Below the image is what I've been able to figure out. Any help with this brainteaser would be greatly appreciated. Cheers. --MZMcBride 22:19, 4 April 2007 (UTC)

$\angle{1}$ = 30˚
$\angle{2}$ = 70˚
$\angle{3}$ = 50˚
$\angle{4}$ = 30˚
$\angle{5}$ = 45˚
$\angle{6}$ = 45˚

$\angle{7}$ = 120˚
$\angle{8}$ = 60˚
$\angle{9}$ = 90˚
$\angle{10}$ = 100˚
$\angle{11}$ = 50˚
$\angle{12}$ = 30˚

$\angle{13}$ = 60˚
$\angle{14}$ = 120˚
$\angle{15}$ = 45˚
$\angle{16}$ = 45˚
$\angle{17}$ = 60˚
$\angle{18}$ = 120˚

$\angle{19}$ = 90˚
$\angle{20}$ = 105˚
$\angle{21}$ = 75˚
$\angle{22}$ = 55˚
$\angle{23}$ = 125˚
$\angle{24}$ = 110˚

$\angle{25}$ = 70˚
$\angle{26}$ = 100˚
$\angle{27}$ = 80˚

Once you have found the answer for any of the blank slots, the remaining ones should be easy. Let us give a name to the point on the side AD of the squares serving as the vertex of angles 10, 11 and 12, and call it J. Also, for simplicity, set the length of the sides of the square to 1, and think of the diagram as embedded in the Euclidean plane with Cartesian coordinates, with B = (0, 0), A = (1, 0) and C = (0, 1). First, work out the length of the sides of the hexagon. That gives you the coordinates of H, and therefore the length of AH. Then find the length of AJ, and thereby the coordinates of J. Also find the coordinates of E. You then have the slope of the line EJ, and thereby the angle it makes with the horizontal. I performed these calculations, and unless I made a mistake, the answers are not nice round numbers. Once you've solved it, you can return the favour by posing essentially the same question with a pentagon. --Lambiam ^Talk 23:01, 4 April 2007 (UTC)

Just glancing over the data filled in, clearly angle 20 is wrong, and angle 13 is obvious (both for the same reason). When two lines cross, we have two pairs of equal angles, which will help. My suggestion is to use the theorem on external angles of a polygon. --KSmrq^T 00:29, 5 April 2007 (UTC)

Great catch on angle 20! I put the answers that I got up. The slope method seemed to work really well, the answer I got for the slope was 10˚. I wasn't really sure what you meant about using a pentagon instead, but I made one below anyway. Thanks for all the help. --MZMcBride 01:00, 5 April 2007 (UTC)

I meant sending it as a problem to whoever sent the first problem to you. Another variant would be one in which the upper vertex of the pentagon lies on the upper side of the square. --Lambiam ^Talk 11:00, 5 April 2007 (UTC)

[edit] April 5

[edit] square root of pi

My siblings sometimes mention the square root of pi to make fun of me when I get too technical in my explanation of something. I was wondering if the square root of pi actuall has any use anywhere in mathematics.J.delanoy 01:31, 5 April 2007 (UTC)

One in in the probability density function of the normal distribution. (Note you can factor out the root of 2.) There are many others if you look around... Baccyak4H (Yak!) 01:55, 5 April 2007 (UTC)

Error_function, although this is the reciprocal of $\sqrt{\pi}$ --Ķĩřβȳ ♥♥♥Ťįɱé Ø 02:45, 5 April 2007 (UTC)

"square root of pi" gives 15100 Google hits. I have no idea how many have use in mathematics. PrimeHunter 02:55, 5 April 2007 (UTC)

Γ(1/2), and Stirling's approximation. —David Eppstein 03:14, 5 April 2007 (UTC)

It shows up when you try to square the circle. It's the width of a square with the same area as a circle of radius 1. In fact, that's why squaring a circle is impossible using ruler and compass - Pi is transcendental. Black Carrot 04:03, 5 April 2007 (UTC)

I don't know much about $\sqrt{\pi}$ , but apparently $\left( \sqrt{\pi} \right)^2$ has a lot of use. --Hirak 99 15:16, 5 April 2007 (UTC)

That's hilarious, NOT.J.delanoy 23:47, 5 April 2007 (UTC)

Thanks for answering the question!!!! You guys ROCK!!!!!J.delanoy 23:49, 5 April 2007 (UTC)

And gamma(1/2) = (-1/2)! = root pi→81.153.220.170 11:52, 6 April 2007 (UTC)

[edit] Triangle integrals?

The Riemann, Darboux, Lebesgue, etc. integrals use rectangles. By are there any integral definitions which use triangles? After all, triangles are very "special" and have their own study (trigonometry). Couldn't the area under a curve be calculated using triangle approximations? All polygons can be broken up into triangles, but not all polygons can be broken up into rectangles. So wouldn't triangle integration converge towards the actual value much quicker than Riemann integration? Thanks.--Ķĩřβȳ ♥♥♥Ťįɱé Ø 04:28, 5 April 2007 (UTC)

Sure. Take any rectangular integral, and divide each term by two. Then multiply by two. This corresponds to slicing each rectangle from one corner to another, and reassembling the triangles. Or, measure it with trapezoids (which are triangle-like) or, for something even more fun, try wiggly segments of polynomials. There are loads of choices, and I wouldn't be surprised if it was possible to define integrals using pretty much any shape you want. For instance, applying your idea about polygons, it's possible to find the area of a circle by essentially integrating over triangles. Probably other blobby shapes too. It wouldn't be all that different from polar integrals, which use wedges of circles. Then there's integrating solids of revolution, which use washers and disks. The big question in each case is, is this really the easiest way to do this? Rectangles are useful because, in a rectangular coordinate system, they're really really easy to keep track of. If you'll think back to trigonometry, you may recall that powerful as the system is, it's anything but convenient. Black Carrot 05:59, 5 April 2007 (UTC)

So is it because of the orthogonal coordinate system that makes rectangles simpler? If, say, we wanted to find integrals in non-orthogonal coordinate systems, would triangles be better(Like, if instead of the x and y axes being perpendicular, if they made a 45° angle)?--Ķĩřβȳ ♥♥♥Ťįɱé Ø 06:12, 5 April 2007 (UTC)

That's right. For example in polar coordinate system the area would be easily integrated by triangles, which approximate sectors between some points

(r i,θ i)

and

(r i + 1,θ i + 1)

with common third vertex at the system's origin

(0,0)

. --CiaPan 06:19, 5 April 2007 (UTC)

You can dissect a rectangle into a trapezoid and a triangle. Remove the triangle and you get a trapezoidal approximation. For an example see Riemann sum#Trapezoidal rule. This in fact is almost (i.e. for partitioning into small sub-intervals) equivalent to Riemann sum#Middle sum. For some cases the Simpson's approximation with parabolas is even better than trapezium method. --CiaPan 06:15, 5 April 2007 (UTC)

Thanks for all the information. I thought up triangle integrals in the shower today... yes, I am that "weird". In any case:

This picture answers my question.--Ķĩřβȳ ♥♥♥Ťįɱé Ø 06:31, 5 April 2007 (UTC)

Now keep in mind, those aren't triangles in the picture, they're wedges from circles. It's just that triangles give almost the same answer if they're narrow and exactly the same answer in the limit. Black Carrot 07:59, 5 April 2007 (UTC)

yes, those are sectors, but their similarity with a triangle shape makes triangles a possible surrogate.--Ķĩřβȳ ♥♥♥Ťįɱé Ø 09:26, 5 April 2007 (UTC)

[edit] how to undestand this?

suppose we want to devide($17)up on 3 shares,the first share is(1/9),second share is (1/3)and the third share is(1/2).abviously we cannot get integer shares.NOW,add (1+17=18),the three shares of(18)would be(2,6,9)respectively.we can see(9+6+2=17)which means that we get fully three shares without losing the additional($1).how can we understand this in mathemetical way?80.255.40.168 11:59, 5 April 2007 (UTC)ARTHER

Suppose you were asked to split $100, giving 1/3 to one person and 1/3 to another person. You could borrow $50 so that you have $150, then give 1/3 of this to each person ($50 each), and still have $50 left to repay the loan. Of course, you haven't really split the $100 into 1/3 and 1/3, because 1/3+1/3 is not equal to 1 - it is only 2/3. Your problem is just a slightly more complicated version of this $100 split. Notice that 1/9 + 1/3 + 1/2 is not equal to 1 - it is only 17/18. Gandalf61 13:31, 5 April 2007 (UTC)

On the other hand, in the example above, you are not really splitting $100 into thirds, you are splitting $150.J.delanoy 23:51, 5 April 2007 (UTC)

Perhaps David Eppstein can clarify the history of this classic use of Egyptian fractions, but some version of this puzzle has been challenging and entertaining minds for a very long time. I believe I first saw it as a tale of Nasrudin, but it was already old by then. --KSmrq^T 20:10, 5 April 2007 (UTC)

[edit] Initial Value Problems convergence etc

hi, i'm going through my lecture notes (on Numberical solutions of ODEs btw) and i cant get what my lecturer has done. I wont write out the whole thing becuase a) its 3 pages long and b) i dont have latex. but i'm guessing what i'm stuck on isn't IVP specific but maybe it is, i dont know... anyway, heres an outline... we're dealing with Eulers method which we've found the local truncation error for, we're also told that f(t,y) satisfies a lipschitz condition so

|[f((t_j), y(t_j)) - f((t_j), z_j)] | ≤ L |y((t_j) - (z_j) |.

So we keep chunking through, doing our thing until we get to:

|e_j+1|≤ (1+Lh)|e_j| +½h²k

so we're then supposed to replace j with j-1 sp predictably enough we get:

|e_j|≤ (1+Lh)|e_j-1| +½h²k  (*)

she then says that (*) is less than

(1+Lh){(1+Lh)|e_j-2|+½h²k} +½h²k

wtf?

If anyone knows whats going on i'd be so appreciative. (Apologies if i'm not being very clear or its not readable). thanks 130.88.52.26 12:32, 5 April 2007 (UTC)

That's the result of writing your (*) expression again — expressing

| e j - 1 |

in terms of

| e j - 2 |

— and substituting into (*). --Tardis 14:55, 5 April 2007 (UTC)

[edit] Famous function?

Does the graph of $x / (1 + x 2)$ have any specific name or function?Wbchilds 13:25, 5 April 2007 (UTC)

It's related to the Lorentzian function; in fact, integrating it over all reals would yield the mean of that probability distribution if it existed. --Tardis 15:15, 5 April 2007 (UTC)

[edit] Volume of sphere

How to calculate the volume of a 'perfect' sphere besides putting it into a jar and see the rise in the water volume? If so, what is the formula? —The preceding unsigned comment was added by Invisiblebug590 (talk • contribs) 13:51, 5 April 2007 (UTC).

$\frac{4}{3} \pi r^3$ where r is the radius of the sphere. See sphere. x42bn6 Talk 14:02, 5 April 2007 (UTC)

And where π is tasty. − Twas Now ( talk • contribs • e-mail ) 17:21, 5 April 2007 (UTC)

[edit] Proof that Ke^x for contants K are the only functions that are their own derivative

The article exponential function correctly says that functions of the form $f (x) = K e x$ for contant K are the only functions that are their own derivative. Any suggestions on a link to a proof of that uniqueness? Dugwiki 16:11, 5 April 2007 (UTC)

Assume that f is a differentiable function with f(0) = 1, and put g(x) = log f(x). Then by the chain rule g'(x) = f'(x)/f(x), which equals 1 if (and only if) the functions f and f' are the same. Function g'(x) is then Lipschitz continuous. Now apply the Picard-Lindelöf theorem to obtain g(x) = x, and use f(x) = exp g(x). The extension to other values of K is not hard. You don't actually need this sledgehammer to swat this gnat; all you need is that only constant functions have a derivative that is 0 everywhere. --Lambiam ^Talk 16:28, 5 April 2007 (UTC)

In fact, as I mentioned below, the solution using the Picard-Lindelöf theorem appears to be even simpler. Set

f (t, y (t)) = y (t)

(the identity function) and y(0)=K and y'(t)=f(t,y(t))=y(t). Then PLT says that

y (t) = K e t

is the unique solution. No need to even mess with the chain rule there. Dugwiki 17:52, 5 April 2007 (UTC)

It's the only solution of the differential equation $d y / d x = y$ . Basically you get $d y / y = d x$ , and integrating $ln(y) = x + C$ . From there your equation follows. For more detailed proof, see Lambiam's links above. --Hirak 99 16:35, 5 April 2007 (UTC)

PS. Unless you allow complex values for the constant of integration $C$ , you won't explain why negative (and for that matter, complex) $K$ works in your equation. --Hirak 99 16:40, 5 April 2007 (UTC)

Yeah, someone else pointed out my error on the article discussion page. Basically the PLT says that

y (t) = K e t

is the unique solution to the differential equation

y'(t) = y (t), y (0) = K

with

f (t, y (t)) = y (t)

. Dugwiki 17:41, 5 April 2007 (UTC)

P.S. I almost overlooked Hirak's comment about negative K. I think you are reading what I typed as "

(K e) x

" (which doesn't make sense in the reals if K is negative). That's not what I mean, though. I'm talking about "

K (e x)

", which is well defined for all real K. Dugwiki 20:05, 5 April 2007 (UTC)

I think Hirak referred to the relationship C = ln(K), which, if C is real, implies K > 0. To avoid the issue I assumed f(0) = 1 in my first reply. --Lambiam ^Talk 21:01, 5 April 2007 (UTC)

Ok, I thought he was replying to my post. Thanks for clearing that up. Dugwiki 21:06, 5 April 2007 (UTC)

Actually,

d y / y = d x

integrates to

ln | y | = x + C

(

ln | y |

is a better antiderivative of

1 / y

because it works for negative numbers). So the absolute value around

y

ensures that for every solution

y = f (x)

y = - f (x)

is also a solution. Which accounts for the negative K's.

Also, when you divided by

y

initially (like when you divide by any expression), you need to separately consider what happens when the thing you are dividing by is zero. When

y = 0

, that is indeed a (trivial) solution to the equation, so that accounts for K=0. --Spoon! 10:48, 6 April 2007 (UTC)

[edit] Which Statistical Test?

I've done a test with 3 groups, one control (in this case water) and two experimental (two different kinds of drinks) and measured final time for a distance covered. I know how to standard deviations and all that but after that i am slightly confused, what do i have to do after this to find any statistical difference between all the groups, will i need a t-test, ANOVA or some other test? If it is an ANOVA could someone explain how to do it in a laymans way or put a link up to somewhere that does, i can never make sense of the mathmatical stuff on wikipedia.

Any help would be appreicated. Rickystrapp 19:45, 5 April 2007 (UTC)

I won't comment on the math involved, but will comment on your control. The subjects presumably can tell if they are drinking water or one of the drinks. This makes the control not very valuable, as the placebo effect can come into play. Ideally, the study should be double-blind, where neither the experimenter nor subjects can identify whether they have been given the control until the results are in. In the case of water, at very least you could add dye to make it look like it might be something else. If you can find an additive you're sure will change the taste and smell without altering performance (perhaps you know this from a previous study), you could also add that to the control. StuRat 20:05, 5 April 2007 (UTC)

I did take that into consideration when designing the study, however i decided it didnt matter enough, i wont go into my experiment but it was agreed not to do that but mention it in the limitations of the study. Rickystrapp 20:26, 5 April 2007 (UTC)

If the only variable in the experimental set-up under your control was the type of potion quaffed, then I see no use for analysis of variance. If the data in each group looks like it follows a normal distribution – you can use Shapiro-Wilk or Kolmogorov-Smirnov to test for that – then indeed Student's t-test is a reasonable test for significant differences. In view of the origin of the test it would be quite fitting if one of the experimental elixirs was Guinness. You should decide on the confidence interval before you compute the test statistic. --Lambiam ^Talk 21:16, 5 April 2007 (UTC)

[edit] April 6

[edit] Sweeping the floor

A child is instructed to sweep a floor. He concludes that to sweep every single bit of the floor- he must sweep it in a systematical way, going up and down the floor, then moving a broom's length to the left or right. By doing this, it is deduced that the chance of sweeping every part of the floor is 100%. Therefore, it is deduced that if the child did not sweep it systematically, there would be a much less chance of sweeping 100% of the floor.

Would this be true, or is it a flaw in our logic?

p.s i have tried the humanities desk but have been referred here.--Howzat11 01:38, 6 April 2007 (UTC)

Not a great question. "He concludes that … he must …." (Emphasis added.) If the conclusion is correct, then there is no choice. If the conclusion is incorrect, it is rather confusing to state it. Without the "conclusion", we have the following parallel example: "A child has a piece of cake, and finds it a satisfying dessert. If ice cream is substituted for the cake, dessert will not be satisfying. True or false?" --KSmrq^T 01:52, 6 April 2007 (UTC)

If it is true that withered flowers come only from not watering them, then not not watering them cannot result in (as opposed to give "a much less chance of") withering them. If on the other hand withering can come by several means (e.g. age), (not not) watering them gives no guarantee of unwitheredness.

Reverting to the floor, it seems reasonable anyway that a systematic way is not necessary for 100% sweeping - a sufficiently long random approach would eventually cover every part.81.153.220.170 12:15, 6 April 2007 (UTC)

Wouldn't the area not sweeped just become infinitely small? --Ybbor^TalkSurvey! 13:30, 6 April 2007 (UTC)

No, the expected area sweeped would never reach zero, but the actual area sweeped will. The probability of the floor being completely sweeped will approach, but never reach, 100%. As to answer the question, assuming not sweeping systematically makes sweeping the whole floor highly probable, but not certain, the probability will be slightly less, but the odds will be infinitely less. For example, if there is a 99% (0.99) probability of sweeping the whole floor randomly, then the probability is 1% (0.01) less, but the odds of doing it randomly are 99:1 (99) and the odds of doing it systematically are 1:0 (infinite) making the odds infinitely less. — Daniel 17:32, 6 April 2007 (UTC)

These are sweeping statements. For a mathematically and logically rigorous treatment of the original question, we need a definition of "systematic". The method of sweeping only along the edges of the area is a system and so could be considered systematic under at least one reasonable definition of the term. It is not clear why the age or developmental phase of the sweeper is introduced; can we abstract from that? --Lambiam ^Talk 20:06, 6 April 2007 (UTC)

[edit] Numbers not definable

Is there any name for numbers not definable in a finite amount of data? e would not be one because in could be defined as $e = \sum_{n = 0}^\infty \frac{1}{n!}$ . They can be proven to exist because each definition can be given a unique Gödel number which will be a natural number and therefore there will be at most $\aleph_0$ numbers that can be defined and there are Cardinality of the continuum real numbers, which has been proven to be more. An interesting paradox arises because, although they exist, it is impossible to find an example. — Daniel 21:18, 6 April 2007 (UTC)

You might call them "undefinable numbers". Precisely because a counterexample cannot be constructed, constructivists will deny that such numbers exist – and escape a contradiction because they don't believe either that a bijection can be constructed between the natural numbers and those Gödel numbers that effectively define a real number. --Lambiam ^Talk 22:06, 6 April 2007 (UTC)

Excuse my ignorance, but in the original question isn't Euler's number being defined by a "non-finite" amount of data, in that n has to go to infinity? Icthyos 22:11, 6 April 2007 (UTC)

Depends on what you mean (BTW, no one calls it "Euler's number"; took me a minute to figure out what you were talking about -- it's just e; that's the standard name). You have kind of hit on the key issue, though. For a given fixed, precise notion of definability, we can define a specific real that isn't definable according to that fixed scheme. "Definable" itself, it seems, is not definable. --Trovatore 22:36, 6 April 2007 (UTC)

I'll just edit the phrase "Euler's number" out of the second sentence of its article, shall I? :P~ So what you're saying is, for any interval of real numbers there will always be an irrational number in there? I know irrational numbers aren't undefined, but I think I've come across a proof of that (my case) before. What precisely do you mean by "notion of definability"? Icthyos 22:44, 6 April 2007 (UTC)

I think the "specific real" refers to Cantor's diagonal argument; look for s₀ there. --Lambiam ^Talk 23:02, 6 April 2007 (UTC)

See also Definable real number. PrimeHunter 23:43, 6 April 2007 (UTC)

But take it with a grain of salt. It's not as bad an article as it once was, but it's still a bit shaky. --Trovatore 23:47, 6 April 2007 (UTC)

[edit] April 7

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