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April 2

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)

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:

$1{\xrightarrow {\times 2}}2{\xrightarrow {\times 2}}4{\xrightarrow {\times 3}}12$

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

$1{\xrightarrow {\times 2}}2{\xrightarrow {\times 3}}6{\xrightarrow {\times 2}}12$

$1{\xrightarrow {\times 3}}3{\xrightarrow {\times 2}}6{\xrightarrow {\times 2}}12$

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)

April 3

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)

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)

I think Gandalf is wrong- "discard and choose a new random number" does in fact give a uniform distribution on primes in the relevant set. The bias towards smaller numbers is because there are more small numbers in the set. The probability of picking any one element from that set is equal for all elements86.27.93.130 07:28, 10 April 2007 (UTC)

What about pick an arbitray number, if its even, and not 2, then add 1, and see if its prime. Oh. That again will only make the search more efficent.

You can use this link:

[Prime Number Test]

Another way to generate a prime number is the sieve of erostathnies.

Another way is to generate a list of prime numbers in increasing order, and pick one off the end, when you get tired. ::)) Artoftransformation 13:52, 9 November 2007 (UTC)

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 $e^{(2k+1)\pi i}=-1$ (for any integer k), $\ln(-1)=(2k+1)\pi i$ for any integer k. Usually people restrict the angles to be within $(-\pi ,\pi ]$, so they say $\ln(-1)=\pi i$. --Spoon! 09:40, 3 April 2007 (UTC)

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)

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. 23:35, 3 April 2007 (UTC)

April 4

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)

HAHA. --JianLi 01:24, 8 April 2007 (UTC)

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 $\delta x$ have been removed from the numerator when there is only one in the denominator. What have I missed here? —Preceding unsigned comment added by 164.11.204.51 (talk • contribs)

A single factor of $\delta x$ 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)

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 primality test. 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)

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)

April 5

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♥♥♥ŤįɱéØ 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)

$\int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}$ Gaussian integral --Spoon! 19:32, 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)

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♥♥♥ŤįɱéØ 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♥♥♥ŤįɱéØ 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 $(r_{i},\theta _{i})$ and $(r_{i+1},\theta _{i+1})$ 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♥♥♥ŤįɱéØ 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♥♥♥ŤįɱéØ 09:26, 5 April 2007 (UTC)

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)

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((tj), y(tj)) - f((tj), zj)] | ≤ L |y((tj) - (zj) |.

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

|ej+1|≤ (1+Lh)|ej| +½h2k

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

|ej|≤ (1+Lh)|ej-1| +½h2k  (*)

she then says that (*) is less than

(1+Lh){(1+Lh)|ej-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)

ok but where does the extra 1+lh come from?130.88.52.11 17:04, 10 April 2007 (UTC)

The same place the extra $h^{2}k \over 2$ comes from? You're nesting two "calls" to (*). --Tardis 21:20, 10 April 2007 (UTC)

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)

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)

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 $f(x)=Ke^{x}$ 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 $y(t)=Ke^{t}$ 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 $dy/dx=y$. Basically you get $dy/y=dx$, 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 $y(t)=Ke^{t}$ 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 "$(Ke)^{x}$" (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, $dy/y=dx$ 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)

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)

April 6

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)

One reasonable interpretation of the question is, "I have to sweep a floor. I'll start at a corner. I'll clear the floor in parallel lines, starting from one wall and moving to the opposite wall. Doing this, I'm guaranteed to clean the whole floor. However, if I swept in a less systematic way, I'd risk missing a spot. Am I right?" It's hard to answer that fully without knowing why you're asking the question, but I'd say, yes and no. It's true that that would clean every part of a normal floor, but there are other systematic ways of doing it. For instance, you could spiral inward and sweep the dust into a pile, then sweep that pile away. Or you could start at the center and sweep outwards in radiating lines, then sweep around the edge. Or (to steal a maids' trick) you could sprinkle coffee grounds over the whole floor, then sweep it any way you please until all the grounds are gone. You're right, however, that if you go at it blind, with no plan and no way of telling when you're done, you'll probably miss a spot. Black Carrot 08:14, 9 April 2007 (UTC)

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)

April 7

Poor Little Suzy

One day Little Suzy found a circular spot of bacterial growing in a pond. Deciding to include it in her Science Report, she measured the size of the bacterial everyday. The first day the bacterial patch had a radius of exactly 10cm. The next day it was exactly 11cm. For the next 8 days, she got the following results.

Radius of the circular bacterial patch

r[0] = 10.00 exact

r[1] = 11.00 exact

r[2] = 12.21

r[3] = 13.70

r[4] = 15.58

r[5] = 18.00

r[6] = 21.25

r[7] = 25.76

It soon became clear to her that the radius of the bacterial patch follows the following rules. The first 3 days are a dead give away.

r[0] = 10

r[t] = r[t-1] + ( r[t-1] * r[t-1] ) / 100

For her Science report, she decides to use a differential equation.

She wrote ${\frac {\text{dr}}{\text{dt}}}=k\cdot r^{2}\,$

with the solution $r(t)={\frac {R_{0}}{1-R_{0}\cdot k\cdot t}}$ where $R_{0}$ is r(t=0) which is 10.00

However no matter how hard she tried, she could not find the values of k that will give the results consistent with her measurements.

She tried k=1/100 which gives r(7)=33.33

She tried k=0.00874 which gives r(7)=25.76 but r(3)=13.55

Why can't Little Suzy find a value of k that works? 220.239.107.54 06:31, 7 April 2007 (UTC)

Is this a serious question? The function r, at least for the observed experimental data, satisfies

${\frac {\Delta r}{\Delta t}}=0.01\cdot r^{2}~\mathrm {for} ~\Delta t=1\,.$

In general, the difference quotient will not be the same as the derivative.  --Lambiam^Talk 09:12, 7 April 2007 (UTC)

Indeed. Suzy is approximating a difference equation by a differential equation. In the general case, the solutions to the two equations may have quite different behaviours. In this case, for example, the solution to the differential equation becomes unbounded ("infinitely large") as t approaches 1/kR₀, whereas the solution to the difference equation is clearly bounded for all t. Gandalf61 10:37, 7 April 2007 (UTC)

Tell me. If you are Suzy, what differential equation would you use for your Science Report? It has to be consistent with the actual measurements. 220.239.107.54 12:55, 7 April 2007 (UTC)

If I was Suzy, I would disguise my homework question as a cute little story. − Twas Now ( talk • contribs • e-mail ) 23:06, 7 April 2007 (UTC)

If this is homework, it is an impossibly difficult assignment. Compare the related sequence 1, 2, 6, 42, 1806, 3263442, 10650056950806, ... (sequence A007018 in the OEIS). Although a closed-form expression for the n'th term a_n is given, it cannot in any obvious way be extended to a differentiable function f on the reals such that f(n) = a_n. The relationship becomes obvious if you define a_n = r_n/100, where r_n is Little Suzy's sequence. The recurrence relation is then the same as for A007018: a_n+1 = a_n + a_n². The only difference is the starting value.  --Lambiam^Talk 23:48, 7 April 2007 (UTC)

New exciting developments ?

I was just wondering if any of you helpers here have written any noteworthy papers on mathematics or made any contributions in general to math. I know there are some brilliant minds here like Lambiam and Ksmrq (and many more). Or are you noticing any new exciting developments in the field of mathematics? — Preceding unsigned comment added by 144.132.64.209 (talk • contribs)

Thanks for the compliment. I choose not to reveal my real-world identity, but some publications of which I'm an author or co-author are cited in Wikipedia articles (not added by me). I think some noteworthy papers by KSmrq are also cited, if I'm not mistaken about his/her identity. There are many new and exciting developments (for instance the novel techniques and insights in the proofs of Fermat's Last Theorem and the Poincaré conjecture, or the advances in number theory and other fields of mathematics made possible by fast computers), most of which is regrettably too technical for me to really understand it.  --Lambiam^Talk 13:26, 7 April 2007 (UTC)

Flattery, eh? Mathematics is a field in which humility is recommended; for, the really good problems are so difficult, and the accomplishments of Euler, say, are so overwhelming. (Euler could do great mathematics late in life, blind, with children playing in his lap, across many topics, more prolifically than any other mathematician that ever lived.) Even if you prove Fermat's Last Theorem or the Poincaré conjecture, the world takes only fleeting notice and few will ever understand (or care) what you have done. For Wikipedia, I just try to explain a few things as best I can.

In a field as old as mathematics, most researchers dig deeper and deeper into very specialized topics trying to extract some gem. Broadly speaking, many fascinating recent events have the opposite character, revealing structure connecting specialties. More subtly, I see indications of a renaissance in attitude towards teaching, shaking off some of the bad side effects of the Bourbaki approach, which left students bewildered by seeing the general without the prior motivation of the specific. Also, connections to physics and other applications are no longer quite so unsavory, and we see inspiration and insights flow both directions.

Computers have helped, in ways we tend to overlook. Especially, the web has facilitated world-wide conversation and collaboration, and allowed papers to be seen much more quickly. Computer typesetting, the ubiquitous TeX, has also made publication of results considerably easier than in the days of typewriters (with hand-lettered symbols) and "penalty copy" burdens on publishers. Explorations and experiments can also use computers (sometimes), and a computer algebra system can help us keep our signs straight.

If you would like to keep an eye on advanced developments, try the AMS web site. --KSmrq^T 23:23, 7 April 2007 (UTC)

April 8

Advantages!

I am trying to find out in which instances would be advantageous to use a mixed number and in which an improper fraction. 72.40.60.1 00:01, 8 April 2007 (UTC)jc

If what you care about is the magnitude, and you require more precision than offered by rounding to a whole number, then a mixed number is probably more reasonable than an improper fraction. Imagine asking for a ³¹⁵/₈th inch joist, rather than a 39³/₈" joist. (Of course, 100 cm is even easier.) In this case mixed notation is also traditional, and using traditional notation, even if irrational, has certain advantages when attempting to communicate with people who are used to traditional notation. In mathematics, the convention is to use vulgar fractions for rational numbers (except of course integers), improper or not. So the fourth convergent of the continued fraction for π will be given as 355/113, not as 3¹⁶/₁₁₃. Doing arithmetic is easier with vulgar fractions: try to compute 2⁵/₁₄ ÷ 2¹³/₂₁.  --Lambiam^Talk 01:15, 8 April 2007 (UTC)

Big pi notation

What is the origin of the "big pi" notation for products? Who introduced it and when? Is it related to Gauss's notation for the gamma function (was one derived from the other)? Fredrik Johansson 00:30, 8 April 2007 (UTC)

I don't know when the notation was introduced, but I could not find a use by Euler and Leibniz. They still wrote a bunch of multiplied factors (or summed terms for a sum), followed by "etc." (in present-day notation "···"). But the introduction of the integral sign by Leibniz clearly paved the way for a summation sign. Since the Latin letter "S" for Summa had already been taken (in a shape nowadays no longer recognized as a letter), the Greek letter "S", or Sigma, was used instead. From there the step to using the Greek letter "P", or Pi, for Productum, is obvious. The large size gives more horizontal room for denoting the limits of summation (subscript and superscript) and matches the traditionally long shape of the long s.  --Lambiam^Talk 01:50, 8 April 2007 (UTC)

A list of notational "firsts" can be found here. There seems to be a conflict of claims:

The product symbol (∏) was introduced by Rene Descartes, according to Gullberg.

Cajori says this symbol was introduced by Gauss in 1812 (vol. 2, page 78).

The references in question are

• Cajori, Florian. A History of Mathematical Notations. 2 volumes. Lasalle, Illinois: The Open Court Publishing Co., 1928–1929.
• Gullberg, Jan. Mathematics: From the Birth of Numbers. New York: W. W. Norton & Co., 1997.

Since Descartes flourished long before Gauss, he would have priority; however, since one would expect numerous intervening uses, Gauss seems more likely. --KSmrq^T 03:07, 8 April 2007 (UTC)

In Gauss 1812, Disquisitiones generales circa seriem infinitam ... §18, the notation Π (k, z) is introduced for a two-argument function, and in §20 the notation Π z is defined as "the limit of Π (k, z) for k = ∞". This is the same as the Gamma function shifted one over. Although the two-argument function could have been defined as a product, it is not; the expression used by Gauss is

$\Pi (k,z)={\frac {~1~~~\,.\,~~~2~~~\,.\,~~~3\,\,.\,.\,.\,.\,.\,\,k~~~}{(z+1)(z+2)(z+3)\,.\,.\,.\,.\,(z+k)}}k^{z}\,.$

This should, in my opinion, not be considered an introduction of a product symbol. It is not in any way different from the introduction, a few sections later, of Ψ for denoting what we now call the Digamma function. The case is not closed. What we really want to see is a use with a running index.

I also looked at Euler's use of Σ for summation, said to be the first (Institutiones calculi differentialis, 1755). Interestingly, what Euler defines is an "indefinite sum" or "antidifference" operator, just like the integration symbol without bounds functions works as an indefinite integral or antiderivative operator. When the notation is introduced, Euler writes:

${\mbox{For if thus}}~\Delta ~x=\omega ;~{\mbox{then it will be the case that}}~\Sigma ~\omega =x+C\,,$

complete with an "indefinite constant of summation". Again no running index. It is funny to see how the printer, at some point apparently having run out of $\Sigma \,\!$'s, resorts to using ${\mbox{M}}\,\!$'s rotated by a quarter turn.  --Lambiam^Talk 10:27, 8 April 2007 (UTC)

For those who would like to examine Gauss for themselves, here is a link. (Euler's works are also online.) --KSmrq^T 11:07, 8 April 2007 (UTC)

Did a bit of digging myself, and found one use of $\Sigma$ for summation in Fourier's 1822 Théorie analytique de la chaleur and one in Cauchy's 1826 Leçons sur les applications de calcul infinitésimal. Both authors assume the reader is unfamiliar with the notation and provide an explanation. Some thirty years later, Riemann in his 1859 Über die Anzahl der Primzahlen unter einer gegebenen Größe and Dedekind in his 1863 Vorlesungen über Zahlentheorie are both using $\Sigma$ and$\Pi$ without further explanation. This indicates that the summation and product notations passed into general usage sometime between 1820 and 1860. Gandalf61 11:34, 8 April 2007 (UTC)

Carpet cleaning

I s u p p o s e this counts under "Accounting"...I spilled some Pepsi on white carpet, and Mum's making me pay for cleaning it. How much do you think it would cost to have the carpet cleaned?--the ninth bright shiner ^talk 02:30, 8 April 2007 (UTC)

The Misc Desk might be more appropriate. If you give us your city and country we might be able to find a price to clean the carpet in one room in your area, though. (They will likely need to clean the entire room so the cleaned spot matches the rest of the carpet.) The minimum charge is $75 at one company in northern Kentucky, just to give you an idea of the prices you are dealing with: [1]. StuRat 04:00, 8 April 2007 (UTC)

I'm in Kernersville, North Carolina (27284-7844). Something to consider is that the spot in question is in the room of my grandmother, who is crippled and therefore bedridden, and I doubt we'll be able to move her, or the computer desk in the room. This carpet is relatively new, less than half a year old, so maybe cleaning all of the carpet won't be necessary.--the ninth bright shiner ^talk 04:40, 8 April 2007 (UTC)

Why not just get a good spot carpet cleaner? Althought the sooner after the spill you do it, the better / easier it is to remove it, in general. --Wirbelwindヴィルヴェルヴィント (talk) 18:20, 8 April 2007 (UTC)

I agree. Get some good carpet cleaner and do as good a job as possible yourself. Show it to your mum and hopefully she will accept that. You could also apologize and promise to help her out in some other ways to make it up to her (like washing dishes, etc.). StuRat 04:53, 9 April 2007 (UTC)

Or try (diluted) white vinegar with a dash of mild detergent; spray, let stand, blot gently; repeat as needed; spray with clean water and blot. Do not wet too much or the carpet may wrinkle.  --Lambiam^Talk 05:55, 9 April 2007 (UTC)

What on earth has this got to do with mathematics?

It is like integration over an area.  --Lambiam^Talk 05:55, 9 April 2007 (UTC)

Factorial Problem

1! + 2! + 3! + 4! ... 97! + 98! + 99! + 100! is divided by 18. What is the remainder?

How would one go around solving such a problem? What concepts are involved? Are there any online calculators that can brute force it? But more interestingly, how can it be solved without doing so? --Proficient 06:07, 8 April 2007 (UTC)

This is much easier than you might think. Factor 18 as 3×6. Now recall that 6! is 6×5×4×3×2×1, and every higher factorial is a multiple of this one; what major simplification does this allow? --KSmrq^T 06:23, 8 April 2007 (UTC)

Easiest way I can think of for solving it is that it's equal to 100 * 1 + 99 * 2 + 98 * 3 + 97 * 4 + ... + 2 * 99. But that doesn't seem any easier to put into a calculator. --Wirbelwindヴィルヴェルヴィント (talk) 18:19, 8 April 2007 (UTC)

Perhaps you wanted to say that it's equal to 1*(1 + 2*(1 + 3*(1 + 4*(1 + ... 98*(1 + 99*(1 + 100))...)))) – b_jonas 19:41, 8 April 2007 (UTC)

KSmrq's point is that anything that has a factor of 18 does not contribute to the problem. You can dramatically reduce what you have to calculate. Root⁴(one) 18:42, 8 April 2007 (UTC)

To answer your less interesting question, yes, it can be brute forced, after all 100! is only 158 digits long. 1! + 2! + 3! + ... + 100! =

94 269001 683709 979260 859834 124473 539872 070722 613982 672442 938359 305624 678223 479506 023400 294093 599136 466986 609124 347432 647622 826870 038220 556442 336528 920420 940313

Dividing that by 18, the remainder you get is 9. – b_jonas 19:41, 8 April 2007 (UTC)

I see now. Thanks for your help. --Proficient 22:44, 8 April 2007 (UTC)

significant or nonsignificant

The study examined moderators of the relationship b/w trait anxiety and information received by patients postmyocardial infraction. When information received was regressed on the product of trait anxiety and gender, the beta was -.10 (p=.42) and the R squared change was .009 F= .66 and p= .42. —The preceding unsigned comment was added by Rtuscan (talk • contribs) 23:28, 8 April 2007 (UTC).

Besides for not knowing what you're talking about, I have no idea what you're asking. --Wirbelwindヴィルヴェルヴィント (talk) 23:43, 8 April 2007 (UTC)

Maybe the question is: what does this mean? It should be postmyocardial infarction, not infraction, but even then we to are lost when it comes to explaining what it means to multiply anxiety by gender.  --Lambiam^Talk 02:27, 9 April 2007 (UTC)

The variable "F" above presumably stands for the value of a test statistic that has an F-distribution. Here is an online a table of upper critical values.  --Lambiam^Talk 02:55, 9 April 2007 (UTC)

Can your heart be arrested for commiting a serious infraction ? :-) StuRat 04:48, 9 April 2007 (UTC)

y = x^2

What's the curve called if y = x²? Thanks very much. 208.72.125.179 02:20, 9 April 2007 (UTC)

It's a parabola. -- JackofOz 02:24, 9 April 2007 (UTC)