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October 24

Complete list of Memoirs of the American Mathematical Society

As a member of our LOCAL Library committee, I am supposed to find out the complete list of the Memoirs of the American Mathematical Society, including the volume number, the author(s) and the title. It is less than 190 entries. From (MathSciNet, Math Reviews on the Web), I got the following message:

Mem. Amer. Math. Soc.

Memoirs of the American Mathematical Society.

Amer. Math. Soc. , P.O. Box 6248, Providence RI 02940.

Status: No longer indexed

I filled in the form at

http://www.ams.org/mathscinet/support_mail.html

asking for instructions to use their site for this specific purpose but so far nobody paid any attention to my call for about the last 5 days. This is why I come to here. I am sorry that this is NOT a mathematical question but a mathematical ADMINISTRATIVE question. If not appropriate for this site, please delete me and I have no hard feeling. However, I would be very grateful if somebody could help. Thanks in advance. Twma 03:24, 24 October 2006 (UTC)

Maybe you have already seen this, but the overview page about the Memoirs says "This journal is indexed in Mathematical Reviews, Zentralblatt MATH, Science Citation Index®, SciSearch, Research Alert, CompuMath Citation Index®, and Current Contents®, Physical, Chemical & Earth Sciences" - so that gives you some possible sources. Alternatively, if a printed list would meet your requirements, you could perhaps request one by snail mail. Gandalf61 11:57, 24 October 2006 (UTC)

I cannot find a complete list of their *** the volume#, author(s), title *** from the above overview page. My problem is not yet solved. Thanks. Twma 03:47, 25 October 2006 (UTC)

Hi: "MELVYL" catalog has 883 entries going back to 1954.I don't know when they started or if that is complete. My way was to google "uc davis",click on "library" on ucdavis main page then click on melvyl and search under titles. I hope this is helpful. Let us know how it's going & don't hesitate to ask again and again here. Good luck,Rich 04:36, 25 October 2006 (UTC)

Following your instructions faithfully, I got 21 lines probably because I am an outsider. Extending your idea, I tried a few other university libraries including our own Australian National University. No luck. Perhaps the best solutions is to accept the statement of my colleague that it is POSSIBLE but not a SIMPLE way to get the complete listing of merely three items: volume number, author(s), title although we are in an information technological era. Originally I would expect that clicking the name of the Journal from mathscinet would give me the answer with 100 records per page. Once that failed, I should have stopped immediately. Mission abort. Instead, I came here. Cannot afford the time to work any further. Thanks to both Gandalf61 and Rich. Twma 09:28, 26 October 2006 (UTC)

Google gives me this... Does it meet your needs? --TeaDrinker 20:40, 26 October 2006 (UTC)

This is EXACTLY what I am looking for. I did try Google but I missed it. Thank TeaDrinker for his/her help with appreciation. My problem is solved. My colleague said, POSSIBLE but NOT SIMPLE. From TeaDrinker, it is very simple. This case is close. Thanks again. Twma 01:07, 30 October 2006 (UTC)

Difference between Diminishing Rate of interest and Flat Rate of Interest

What is the difference between Diminishing Rate of interest and Flat Rate of Interest? Which is more advantageous?

Diminishing rate of interest is diminishing i.e. getting lower over time and flat rate of interest stays the same. If you take out a loan you'd want diminishing interest more but if you were earning interest then you'd prefer flat rate of interest. --WikiSlasher 08:11, 24 October 2006 (UTC)

Complex number dimensions

So we can have 1 dimension (line), 2 dimensions (square), 3 dimensions (cube)... but can we have i dimensions? I mean in the sense that if you "square" the line you add 1 dimension, but if you square the i you get -1 dimension. But if you square -1 dimension you get 1 dimension. So then i would be the 4th root of the 1st dimension... would that serve any useful purpose? Sorry it's 4:30am here this might be a garbled message. Thanks --Ķĩřβȳ♥ŤįɱéØ 11:38, 24 October 2006 (UTC)

When you square the imaginary number i you get the number −1, not some dimension –1, which is not a particularly meaningful notion. I would say, No, we cannot have i dimensions. The number of dimensions is the number of characterizing values you need to supply in some context, such as height and width. There can't be i of such any more than you can have 2-3i grandchildren.  --Lambiam^Talk 12:58, 24 October 2006 (UTC)

Imaginary numbers are neither ordinal numbers nor cardinal numbers, so it wouldn't make sense to describe a dimension (which are described either ordinally or cardinally, depending on the semantics used) using an imaginary number.

Well ... there are definitions of dimension which extend the concept to fractional dimensions (e.g. box-counting dimension, Hausdorff dimension, correlation dimension). But they all very roughly depend on calculating or measuring how some quantity related to the "size" of a set varies as a power of some other quantity, which can be interpreted as a "length". So it is difficult to see how any extension of these ideas could give a meaning to an imaginary number of dimensions. I think you would need to start by finding a complex-valued equivalent of the concept of measure. Gandalf61

It is almost meaningful to raise a (real or complex) number to the power i: the "almost" because it's multivalued. See Exponential_function#On the complex plane. This means that you can determine (up to multiple values) the volume of an i-dimensional cube of any given side length. Similarly, you can compute the Hypersphere#Hyperspherical volume of an i-dimensional sphere. —Blotwell 02:57, 26 October 2006 (UTC)

what is progressive matrix that are asked in aptitude test

what is progressive matrix that are asked in aptitude test

Raven's Progressive Matrices is the page your looking for. --Salix alba (talk) 14:22, 24 October 2006 (UTC)

Probability question

I am a maths teacher - in class this question came up - but I can't get my head around it - but I'm sure that someone here can help:

If we have two identical lines A and B next to each other

________ A 

________ B

if I randomly choose any point on each line then there is a probability of 1/2 that the point from line B will be further right than the point from line A (symmetry)

If the two line look like this

________ A
                    ________ B  

now there is a probability of 1 that the point from line B will be further right than the point from line A

My question:

What if the two line look like this:

________ A
    ________ B

This time line B starts 1/2 way along from A (or any other fraction) What is the probabilty that if I choose a random point from each line then the point from line B will be further right than the point on line A? --yakov Korer 18:21, 24 October 2006 (UTC)

How about we draw a square on a grid, with corners at (0,0.5), (0,1.5), (1,0.5) and (1,1.5). The x-values represent the possible values of point A. The y-values represent possible values of point B. As we assume that they are both distributed uniformly - any unit area in that square has the same probability of occurring as any other unit area. Now we need to look at what it means for point B to be further right than point A - it's the same as y - x > 0. Now draw the line y - x = 0 on the grid. Anything above that line has y-x > 0, so point B is further right than point A. Anything below the line has y-x<0, so point A is further right than point B. It's straightforward to see what the required probability is - 7/8. To satisfy you that this is right, you could try this method with the other examples you posted - to see that you get the correct answers i.e. 1/2 and 1. For a more general look, see our article on convolutions, in particular, the bit about half-way down that page. Richard B 20:10, 24 October 2006 (UTC)

gear ratio selection

If i have a block that fits on a ball screw with a pitch of 6 mm and a diameter of 6 mm, how can I determine a gear box, with an appropriate gear ratio, that will allow the block to move along the ball screw with an accuracy of +-1 micron?

Spherical Pythagorean theorem

I had read once about the spherical Pythagorean theorem and discovered some, which appeared to be complex, explanations of the theorem. I can not find any reference to this anywhere on Wikipedia, and I have searched the Internet to no avail. Could someone please enlighten me on the spherical Pythagorean theorem? Thank you. —The preceding unsigned comment was added by 72.160.189.183 (talk • contribs) .

Look at the law of cosines for the case that C is a right angle.  --Lambiam^Talk 20:51, 24 October 2006 (UTC)

Radius/semi-axis vs. diameter/axis

With a circle, ellipse and ellipsoid, the radius/semi-axis is half the diameter/axis. When finding an elliptic parameter (eccentricity, flattening, etc.), "a" and "b" are used to calculate these parameters (ultimately trig functions of the angular eccentricity, $o\!\varepsilon \,\!$). Are "a" and "b" technically radii or diameters in calculating $o\!\varepsilon \,\!$? While this may seem just semantical, consider an ellipsoid where "b_north" ≠ "b_south":

Since the polar radii aren't equal, the four semi-axes don't intersect in the middle. Does this mean the calculation of $o\!\varepsilon \,\!$ (and the associated elliptic paramters) isn't as simple as $\arccos \!\left({\frac {b_{n}+b_{s}}{2a}}\right)\,\!$ (since $\arccos \!\left({\frac {b}{a}}\right)=\arccos \!\left({\frac {2b}{2a}}\right)\,\!$), or is it?  ~Kaimbridge~23:04, 24 October 2006 (UTC)

The image does not look symetric about any horizontal line, hence its not an ellipse. This means that talking about the usual description of an ellipse do not hold. I don't know much about angular eccentricity as it is a rather non standard presentation. --Salix alba (talk) 12:53, 25 October 2006 (UTC)

It might not be an ellipse, but it is an ellipsoid——consider Mars: (Mean) equatorial radius = 3396.200, north polar radius = 3376.189 (typo corrected) and south polar radius = 3382.580. Is there an equation for an oval's eccentricity?  ~Kaimbridge~14:24, 25 October 2006 (UTC)

An ellipsoid is a three-dimensional shape. The curve in the figure is neither an ellipse nor an ellipsoid, and it has no eccentricity because it's not a conic section. —Keenan Pepper 18:38, 25 October 2006 (UTC)

Okay then, forget that figure. Keeping in mind the above Mars example (so this example is certainly possible), let's say there is an oblate spheroid, where a = 10000, b_north = 7000 and b_south = 8000. The obvious, single value for b is 7500, but would that be technically correct for finding elliptic parameters (such as eccentricity or flattening), since (e.g.) ${\frac {(2a)^{2}-(2b)^{2}}{(2a)^{2}}}={\frac {a^{2}-b^{2}}{a^{2}}}\,\!$, or——like the radius of an ellipse——would the parameters vary from latitude to latitude, perhaps utilizing some integrand?  ~Kaimbridge~22:04, 25 October 2006 (UTC)

The keyword is ovoid (or, if you wish, oval). As a class, they are not defined precisely enough to make the original question meaningful. One could study a specific class of ovoids, for example those satisfying an equation like ax²+kx³+by² = 1, and try to define notions similar to minor axis and such. Technically, definitions developped for ellipses do not apply, because these shapes are not ellipses.  --Lambiam^Talk 19:35, 25 October 2006 (UTC)

Abstraction of Boolean Rings

I am wondering whether there is an abstraction of the concept of a Boolean ring. This would be some ring defined by the n-th power of each element being the element itself, for some fixed n. Are there any interesting results that can be found from these rings?

I have nothing useful to say about this, other than to suggest you think about associative algebras over finite fields. (Other articles that might interest you are modular arithmetic and roots of unity.) –Joke 05:09, 25 October 2006 (UTC)

October 25

Math

What is the factors of 24

Half of me is saying that I shouldn't answer this because it looks like a homework question, while the other half says I should... But I guess I won't. I'd like to point you to the Divisor article if you don't know what a factor is.

Anyway, the factor of a number (X) are all whole numbers other than 0 which can be multiplied by another whole number (which is also a factor of X) to get X, such as y (8) and z (3) which, multiplied together, result in yz = x (8 x 3 = 24). There isn't a "formula" to find factors of numbers aside from dividing the number by all whole numbers from 1 to whatever until you get one that works - that's what supercomputers are for (see RSA). The number ends in 4, so you can assume that it's divisible by 2, possibly by 3, 4 or 6 but not by 5 or 7, etc. But that's not needed for a small number like this. Also, the smallest factor is always 1 (assuming it's a positive nonzero integer) and the largest factor is always the number itself (1 x 24 = 24).

Just out of curiosity, how old are you and what grade are you in? I'm just asking because this is middle-school stuff. Corporal 02:10, 25 October 2006 (UTC)

To answer your question, I suggest that you read these articles:

• Fundamental theorem of arithmetic
• Prime factorization algorithm

The fundamental theorem of arithmetic establishes the importance of prime numbers. Prime numbers are the basic building blocks of any positive integer, in the sense that each positive integer can be constructed from the product of primes with one unique construction. Finding the prime factorization of an integer allows derivation of all its divisors, both prime and non-prime.

202.168.50.40 05:07, 25 October 2006 (UTC)

Yo, any whole number you can divide it by to make a whole number is a factor. End of story.

ahem....since this is middle school stuff, i should point out that just because it ends in 4 doesn't necessarily mean you CAN'T have 7 as a factor. 14. Lrpelkey 11:39, 28 October 2006 (UTC)

Economic Production function to profit function:

I would like to make sure I am accurate in my technique, am I right?

Once you have the cost function, formulated from a production function, there are two ways to find maximum profit:

You can either plug what optimal Q is equal to w.r.t. the constraint, and then maximize with respect to inputs X and Y, and then replug these optimals into the constraint for the optimal Q...

Or, you can plug in the cost function, and merely derive w.r.t. Q, then solve for the optimal Q?

After both of these are solved for Q, just plug them into: pi = Q(PQ) - X(pX) - Y(pY).

Correct? Thanks so much, ChowderInopa 02:49, 25 October 2006 (UTC)

You are not making much sense to me. What is X , Y , P and Q ? What is pi (I don't think you mean 3.1416)? Is lower case p different from upper case P? What is a production function? Is it a function of quantity? 202.168.50.40 05:01, 25 October 2006 (UTC)

(after edit conflict) The method you propose is not entirely clear. What is the relationship between the cost function and the production function? What is the role of the constraint? Is it a constraint on the inputs (for example X+2Y < 100)?

In general you cannot do optimization unless you actually know the objective function. If it is given analytically, by an equation, you may hope to use methods such as that of Lagrange multipliers to find an optimum. Or perhaps you can use quadratic programming. Otherwise, you could resort to numerical techniques such as hill climbing, modified to obey the constraint.  --Lambiam^Talk 05:10, 25 October 2006 (UTC)

Are all real numbers with multiple decimal representations rational?

Today's featured article, besides drawing out an unusually broad array of cranks, got me to thinking. Obviously, 0.999... = 1, 0.34 = 0.3399999..., 330 = 329.999..., etc. But all of these numbers with multiple decimal represenations are constructed by taking a simple terminating decimal number and replacing a final digit n with the digit sequence (n-1)99999... Are there any other types of numbers that also have multiple representations? --Pyroclastic 07:32, 25 October 2006 (UTC)

e (number) = 1 + 1/1 + 1/2! + 1/3! + ... Which is a basis for Euler's formula, and actually quite elegant. Also look at pi. 169.229.89.53 09:25, 25 October 2006 (UTC)

To make it even more elegant, why not write it as e (number) = 1/0! + 1/1! + 1/2! + 1/3! + ... StuRat 17:29, 25 October 2006 (UTC)

Um, but that's not a definition in terms of an infinite decimal sequence?--140.180.157.26 07:18, 26 October 2006 (UTC)

Well, there's 0.000… = -0.000…, although in some authors' development of decimals, this is written 0+.000… = -1+.999…, so it may or may not fit the pattern. Anyway, every number with multiple decimal representations is rational, yes: in particular, a rational which can be represented by a fraction with a power of 10 in the denominator. Melchoir 09:31, 25 October 2006 (UTC)

Well, I've no proof but a strong hunch that there are an infinite number of series representations of any rational number, since there's an infinite number of series that converge. But the number is probably finite for the irrationals? --BluePlatypus 12:37, 25 October 2006 (UTC)

It draws cranks for two reasons. One, there will always be some for such a topic. Two, the article threw away the instructional care of earlier versions to throw up a façade of being “encyclopedic”. Far from representing the best of our mathematical writing, it is an object lesson in what not to do. (Pardon me while I dismount my soapbox.)

Do you mean to restrict attention to decimal expansions? If so, let's review the options. A number like 0.84375, the finite decimal expansion of ²⁷⁄₃₂, can be normalized to an infinite expansion by adding trailing zeros. Such numbers are always rational, and — if nonzero — their denominator contains only prime factors of two and five. Exclude zero, and we have the only real numbers that admit two decimal expansions, the second one having trailing nines.

If we do not restrict ourselves to decimal expansions, every real number has numerous representations. We have alternatives like expansions in a radix of our choosing, continued fractions, and so on. We can represent a real algebraic number as a select root of a polynomial with integer coefficients. Expanding our scope, we can write down the digits using a different alphabet, like Devanāgarī or Hangul. The idea of a representation is broad!

As far as the series question, no, every real number is the limit of an infinite number of different series. A proof is trivial: Take any given series and preface it with two new terms that cancel each other; the result is a new series with the same limit. --KSmrq^T 14:13, 25 October 2006 (UTC)

A trivial proof but for a trivial case. Not a very interesting or enlightening answer. --BluePlatypus 20:23, 25 October 2006 (UTC)

Nonsense. You speculated that any given irrational number was the limit of a finite number of series. I showed that is false. More elaborate variations could be given, but there is no need; false is false. If you claim every mammal has a placenta, and I show you a platypus (!), must I also point out a Virginia Opossum, and a Tasmanian Devil, and a Monito del Monte? --KSmrq^T 01:59, 26 October 2006 (UTC)

No one seems to have directly answered the original poster's question. The answer is yes, all numbers with two distinct decimal representations are rational, and in fact are of the form ${\frac {n}{10^{m}}}$, where n and m are integers. --Trovatore 17:33, 25 October 2006 (UTC)

Oh, actually I guess Melchoir did answer it; sorry about that. --Trovatore 17:41, 25 October 2006 (UTC)

Is there some reason you ignore my answer? Also, your form of the fraction is misleading, because n cannot always be relatively prime to 10, which we would expect from the presentation of a rational number. The very example I gave, 0.84375 = ²⁷⁄₃₂, illustrates that. Sheesh. --KSmrq^T 18:39, 25 October 2006 (UTC)

Well, you sort of went on a bit, KSmrq, and your answer to the original question was in the middle of it all. The OP's question had a direct answer—"yes"—which no one bothered to give directly. --Trovatore 18:45, 25 October 2006 (UTC)

My answer mirrored the question sequence: cranks, decimal expansions, other representations, series. And if you think the answer to the poster's question is "yes", maybe you should go back and decide which question you were answering, the one in the title or the one in the post itself. Oy vey! --KSmrq^T 01:24, 26 October 2006 (UTC)

Help with diff eq

Hey,

I'm looking for a solution for next differential equation:

(b'(t))²/2 = A*b - B*b^4/4 + U²/2 , with A,B en U positive constants.

I actually need an explicit formula for b(t)! I would be very happy if someone could help me with this problem.

Solutions can be send to: #@#@#@#@#

Yours sincerely,

Sam

Please read the guidelines at the top of this page, and follow them. --KSmrq^T 17:06, 25 October 2006 (UTC)

On the look of it, this has no analytic solution. For that, we need an antiderivative of the function f(x) = (1 + λx − μx⁴)^−1/2, and that antiderivative has to have an explicit inverse to boot. But I'm already stuck on the first part. You could perhaps try looking at expansions of b (for example as a sum of exponentials near the roots of the polynomial Ax - Bx^4/4 + U²/2), using the method of equating coefficients (in a more general sense than treated in our article under that name).  --Lambiam^Talk 19:13, 25 October 2006 (UTC)

Basically the equation is ((b')²)/2 = Ab - (Bb^4)/4 + U²/2 (I'm just confused about b^4/4 because that could be (b^4)/4, or b^1), right? --AstoVidatu 22:32, 25 October 2006 (UTC)

I'm not done, but try rewriting the equation as b' = (k)^(1/2), and set k equal to all that junk in the middle. Then take the integral of that equation (easy as pie) and sub in k and k' using k = 2*(Ab - (Bb^4)/4 + U²/2)... And remember U is just a number, so it goes to 0 when you take the derivative. --AstoVidatu 22:37, 25 October 2006 (UTC)

Infinitesimal

Is an infinitesimal the recipricol of infinity? —Preceding unsigned comment added by 86.142.195.245 (talk • contribs)

No, I believe the reciprocal of infinity is 0. It is somewhat debated, however, because infinity can have more than one value, while 0 is only 0. PullToOpen ^talk 20:10, 25 October 2006 (UTC)

But couldn't a case be made for 0 to be equal to an infintesimal? If you define 1/0 as infinity, the reciprocal is 0. However, that's just the ramblings of a 15-year-old, so don't set too much store by it. —Daniel (‽) 20:30, 25 October 2006 (UTC)

No it is not debated within professional mathematicians. The reciprocal of infinity is undefined, because infinity is not a real number. There are any number of Math FAQs out there that explain this, as well as the article on Infinity. --BluePlatypus 20:33, 25 October 2006 (UTC)

Let's clarify things a bit. The answer depends on the context. In the real numbers, there is no infinity or infinite quantity. In the real projective line, the reciprocal of ∞ is 0. In the surreal numbers, there is no ∞, but there are infinite quantities like ω, the reciprocal of which is 1/ω, an infinitesimal. -- Meni Rosenfeld (talk) 20:55, 25 October 2006 (UTC)

For "infinitesimal" to be a useful concept, you want to retain some conventional algebraic properties, such as, letting ε stands for some infinitesimal, (2ε)/ε = (ε+ε)/ε = ε/ε + ε/ε = 1 + 1 = 2. If ε was equal to 0, this would not be possible; we would get the useless form 0/0.  --Lambiam^Talk 21:45, 25 October 2006 (UTC)

Lambiam, isn't there a problem there? Your example says that 2ε/ε = (ε+ε)/ε = ε/ε + ε/ε = 1 + 1 = 2, but if ε is infintesimal, surely adding multiplying it by any finite number will leave it unchanged, thus 2ε/ε = ε/ε = 1? Have I missed something?

Yes, you have missed something. The idea is related to "differential quotients" suggesting the ratio of two infinitesimal quantities, as in dy/dx = 2. Mathematicians find it useful to manipulate this like dy = 2dx, as if dy and dx are actual quantities. Actually that practice preceded the invention of the differential calculus. Think of dy and dx as very very (as in VERY) small changes. It took quite some time before someone invented a way to put that on a safe footing.  --Lambiam^Talk 22:45, 25 October 2006 (UTC)

0 would be the recipricol of infinity. Infinity is defined as (infinity/1). And the recipricol of that would be (1/infinity), which is 0. --AstoVidatu 22:28, 25 October 2006 (UTC)

Yes, the typical infinitesimal is the reciprocal of an infinity. By definition, a positive infinitesimal number is smaller than the reciprocal of any positive integer: ε < ¹⁄_n. (Formally, we would add "for all n ∈ Z⁺".) Inverting, ¹⁄_ε > n; this says the reciprocal of an infinitesimal is greater than any integer.

However, if we want to do mathematics, not just wave our hands, we must be careful and more precise. The ordered systems of arithmetic we use most often — the integers, the fractions, the real numbers — exclude both infinities and infinitesimals. There is no integer greater than every other integer; there is no positive fraction smaller than every other positive fraction. So, if we are to do mathematics with infinities and/or infinitesimals, we must define a different system of arithmetic.

We have choices. In many of the new systems what I said is true. (An example is hyperreal numbers.) In some, we have no infinities, and no infinitesimal has a reciprocal. (An example is dual numbers.) Or we can extend the real numbers with no infinitesimals, but with a single infinite value, ∞, whose reciprocal is zero. (An example is the real projective line.) We also have exotic possibilities like surreal numbers.

All of these systems are valid choices and allow us to do interesting mathematics. However, they can have strange new properties and lack familiar old properties. Choose wisely. --KSmrq^T 00:06, 26 October 2006 (UTC)

So what do you get if you multiply an infinity by an infinitesimal? --86.139.127.29 21:14, 26 October 2006 (UTC)

That depends which infinity and which infinitesimal. Suppose ε is a positive infinitesimal in an appropriate system. Then 1/ε is infinite, and in the following, the first multiplicand is an infinitesimal and the second is infinite (and the result can be pretty much anything):

• ε * (1/ε) = 1
• ε * (2/ε) = 2
• ε * (1/ε²) = 1/ε
• ε² * (1/ε) = ε

-- Meni Rosenfeld (talk) 23:03, 26 October 2006 (UTC)

If .999... = 1in a base 10 number system what about others?

Would it also be true that .777... = 1 in a base 8 system? If so, would .000... = 1 in base 2? I'm just curious and hope this is not an inappropriate post. Downhill.geezer 21:36, 25 October 2006 (UTC)

No worries! Yes, in base 8, .777... = 1. In base 2, however, .111... = 1, not .000.... There's some more information at 0.999...#Generalizations. Melchoir 21:44, 25 October 2006 (UTC)

Your post is welcome.

If d is an integer greater than 1, then we can use it as a radix (base) just like 10 or 8 or 2. When we do, the equivalent of 9 repeating is d−1 repeating. So 0.777…₈ = 1 for radix 8, as you propose; but 0.111…₂ = 1 for radix 2, contrary to your guess. Independent of which d we choose, 0.000… is always exactly zero.

Mathematicians have a habit of seeing how far an idea can go. Donald Knuth, in volume 2 of his series The Art of Computer Programming (ISBN 978-0-201-89683-1), presents a wide assortment of positional notation systems. For example, in balanced ternary, with digits 0, 1, and −1 (written 1), the notation 0.111… represents ¹⁄₂. With a radix of −10, digits 0 through 9, every real number (positive, zero, and negative) can be written without a sign. With a radix of i−1 (where i² = −1), digits 0 and 1, every complex number can be written. It might be fun and educational to explore some of these, seeking the equivalent of repeating 9s. --KSmrq^T 00:58, 26 October 2006 (UTC)

October 26

Terms in polynomial expansion

How can you determine the number of terms in a simplified (like terms have been combined) polynomial expansion? I am particularly interested in ${(a+b+c+d)}^{17}$, as that was on my math contest today, but any general help would be appreciated. Thanks. —Mets501 (talk) 01:49, 26 October 2006 (UTC)

Well for starters the powers of any of the terms should add up to 17. So the question is how many different ways can you make $a^{x}b^{y}c^{z}d^{w}$ where $x+y+z+w=17$? I'd have to think about this for a moment to give you an answer to that question though... - Rainwarrior 02:50, 26 October 2006 (UTC)

For just three variables ${(a+b+c)^{n}}$ should have $\sum _{i=1}^{n+1}{i}$, which is an easily computed arithmetic series. To do a fourth as well, it would compound, I think, as something like $\sum _{j=0}^{n}{\sum _{i=1}^{j+1}{i}}$? Since 17 is a fairly small number you could calculate the 18 values (for j=0 to n) and sum them yourself, I suppose, but I don't know of an easier way to compute the sum of this series. - Rainwarrior 03:07, 26 October 2006 (UTC)

Ah! It's a Tetrahedral number! (The 3 term polynomial problem was a Triangular number.) So... according to that formula, I'd say, 1140? - Rainwarrior 03:35, 26 October 2006 (UTC)

Tetrahedral numbers also turn up as a diagonal of Pascal's triangle. So, ${{n+3} \choose {3}}$ or ${{17+3} \choose {3}}=1140$. (I suspected there would be a solution related to the choose operation.) - Rainwarrior 04:08, 26 October 2006 (UTC)

The general question asks for the number of terms in $(x_{1}+x_{2}+\cdots +x_{k})^{n}$. A good way to tackle this type of question is to find the answers for specific cases and look for a pattern. For instance, if k = 2 then you're looking at $(a+b)^{n}$ which has $n+1$ terms according to the binomial theorem. The trivial case k = 1 is also interesting here; there is only one term if k = 1. Look at some more cases: n = 0, n = 1, perhaps n = 2, etc., and put all the results in a table. I think you will recognize the pattern in the table once you have enough entries. -- Jitse Niesen (talk) 03:06, 26 October 2006 (UTC)

OK, I'll try it out tomorrow. Thanks for the help! —Mets501 (talk) 03:10, 26 October 2006 (UTC)

See the multinomial theorem, which deals with this in generality, and may be useful to the problem you are considering. Dysprosia 08:42, 26 October 2006 (UTC)

In general for a k-termed polynomial raised to the power of n, you should have ${k+n-1} \choose {k-1}$, I think. - Rainwarrior 15:48, 26 October 2006 (UTC)

And there is an easy explanation: Each term in the expansion of $(x_{1}+x_{2}+\cdots +x_{k})^{n}$, apart from its multiplicity, has the form $x_{1}^{p_{1}}x_{2}^{p_{k}}\cdots x_{k}^{p_{k}}$ in which the exponents $p_{i}\,\!$ sum up to $n\,$. There is a one-to-one correspondence between the sequences $p_{1},p_{k},\ldots ,p_{k}\,\!$ of non-negative integers with sum equal to $n\,$, and the combinations of $k{-}1\,\!$ different choices from the set $\{1,2,\ldots ,n+k{-}1\}\,\!$. The correspondence is given by the mapping that sends the sequence $p_{1},p_{k},\ldots ,p_{k}\,\!$ to the set $\{p_{1}+1,p_{1}{+}p_{2}+2,\ldots ,p_{1}{+}p_{2}{+}\cdots {+}p_{k-1}+k{-}1\}\,\!$.  --Lambiam^Talk 15:09, 27 October 2006 (UTC)

"measureomorphisms"

Homeomorphisms are bijective & cts both ways. Has anything been done w/relations that're bijective outside of sets of measure zero and measurable both ways, say on [0,1]? Thanks,Rich 10:40, 26 October 2006 (UTC)

If you "forget" the topological structure, a homeomorphism is a function between the sets of the two spaces involved. What you write suggests that you are thinking of functions between power sets. Is that correct? For that to be a useful notion, you'd need a lot more in the way of restrictions. Did you have any particular function or application in mind?  --Lambiam^Talk 12:24, 26 October 2006 (UTC)

Measurable functions are perfectly well-defined set functions from one measure space to another. Except for the "sets of measure 0" part, you just have (I believe) an morphism-with-inverse in the category of measurable spaces. I don't know enough category theory to say whether that can be called an "isomorphism" in that category or not. I don't know whether this has been studied specifically in the category of measurable spaces. 128.135.72.214 04:23, 27 October 2006 (UTC)

So I'm not too sure what a "measurable space" (as opposed to a "measure space") is. Measurable functions don't naturally form a collection of morphisms, because they're not closed under composition. Maybe the condition you really want is that the preimage of a null set is null? In that case, I think your "measuremorphism" would be a Rudin–Keisler isomorphism between the Boolean algebras created by modding out by null sets. --Trovatore 04:42, 27 October 2006 (UTC)

I wouldn't be surprised if the problems with composability went away in most interesting cases. Note that a bounded measurable function on [0,1] (in particular, your measureomorphism) defines an element of the C-star algebra L^∞([0,1]), which is isomorphic to a continuous [0,1]-valued function on a space X. I wouldn't be surprised if it turned out that your other requirements then forced that [0,1]-valued function to extend to a topological automorphism of X, and that simple case, at least, could be subsumed in the study of the spectrum of commutative C*-algebras.

Put slightly simpler, your auto-measureomorphisms of a measure space S might just be C*-automorphisms of the C*-algebra L^∞(S), at least in some nice cases.

RandomP 20:23, 28 October 2006 (UTC)

Equation of midpoint

A line segment of length 6 moves so that A remain on the x-axis and B remains on the line y=2x. Find the equation of the locus of the midpoint M of segment AB.--Patchouli 10:53, 26 October 2006 (UTC)

Do your own homework: if you need help with a specific part or concept of your homework, feel free to ask, but please do not post entire homework questions and expect us to give you the answers. Letting someone else do your homework makes you learn nothing in the process, nor does it allow us Wikipedians to fulfill our mission of ensuring that every person on Earth, such as you, has access to the total sum of human knowledge. -- Meni Rosenfeld (talk) 11:05, 26 October 2006 (UTC)

If B has y = 0, where is A? What if y = 6? Find a general equation for A given B. From this it is straightforward to determine the behavior of the midpoint. --Pyroclastic 15:06, 26 October 2006 (UTC)

If B(0,0)→A(6,0)→M(3,0).

If B(3,6)→A(3,0)→M(3,3).

The pairs of M do not help me to find an equation. I can't use the 2 point to find the equation of a line. I need to be able to impose the constraint that the total distance between A and B always remains 6. That would yield a conic. I don't know how to proceed.--Patchouli 16:25, 26 October 2006 (UTC)

• It cannot be a rotated parabola with A as the focus and y=2x as the directrix because AB is a (straight) line segment. Furthermore, the conic that I am interested in cannot extend indefintely in any direction forasmuch as AB would then exceed 6. Therefore, it must be a rotated ellipse. --Patchouli 16:28, 26 October 2006 (UTC)

Let's call the coordinates of A (x_A, y_A), and the coordinates of B (x_B, y_B). I would start by finding equations for x_B and y_B in terms of x_A. Chuck 17:10, 26 October 2006 (UTC)

I did find an equation. It was an interesting exercise. You should try it. :-)

As you surmise, the result is a rotated ellipse. It is centered at the origin, and can be given by an implicit algebraic equation whose coefficients are small integers (less than 10). The semimajor axis is three times the golden ratio.

This should be enough to help you check your work, and even mount a brute force assault.

Better still, an ellipse is determined by five points. You already have four: the two above plus their symmetric mates. It's easy to find another. If A is at the origin, where's B?

Or, consider the following line of attack.

Since A is on the x axis, it has coordinates (a,0). Let the y coordinate of the midpoint be y; then the y coordinate of B must be 2y, and its x coordinate is immediate. From the coordinates of A and B we can use Euclidean distance to write a quadratic equation in a and y constraining the (squared) length. Implicitly, a is a function of y.

See where these hints take you, and ask again with more details if you get stuck. --KSmrq^T 18:16, 26 October 2006 (UTC)

the y coordinate of the midpoint be y; then the y coordinate of B must be 2y

You are assuming that AB is vertical, but it isn't always so.

Also, what are the x coordinates of B and M?--Patchouli 18:37, 26 October 2006 (UTC)

No, I am not assuming AB is vertical. Nor will I hand-feed you everything; I expect you to find the x coordinates of B and M yourself. And ethically, I expect you to inform your instructor that you received assistance. --KSmrq^T 20:34, 26 October 2006 (UTC)

Author of the Kramer Theorem

Who is the the author of the Kramer Theorem? I have heard an opinion that it is the Gaon of Vilna (Rabbi Eliyahu Kramer). He lived at the end of the 17th century and is known to have mastered many areas of knowledge, such as algebra, geometry, etc.

Thank you

Olga Yorish

They had no electricity in Vilna at that time. -- DLL ^{.. T} 21:06, 26 October 2006 (UTC)

If you mean Cramer's rule, that would be Gabriel Cramer. Or perhaps you meant the Cramér-Wold theorem ? In that case, the authors are Herman Ole Andreas Wold and Harald Cramér. StuRat 00:55, 27 October 2006 (UTC)

I see there is also a Kramers' degeneracy theorem in the physics field. Is that the one you mean ? It appears to be named after Hendrik Anthony Kramers. StuRat 03:59, 27 October 2006 (UTC)

October 27

Conundrum

I am as old as Casey will be when I am twice as old as Casey

was when my age was half the sum of our present ages.

Casey is as old as I was when Casey was

half the age he will be 10 years from now.

How old am I?

Can anyone help? MusicMan1015 02:13, 27 October 2006 (UTC)

You don't know how old you are? Sorry, I can't help. (Check your birth certificate.) You want someone else to do your homework for you? Sorry, the instructions at the top of this page say don't ask here. Want help with algebra? Ask something specific, showing you've put in some effort of your own. Want a hint? Let a be your age and let c be Casey's age. Translate the two sentences into equations in a and c; standardize as a system of linear equations. --KSmrq^T 02:56, 27 October 2006 (UTC)

It's not HW. I'm well aware of the rules. MusicMan1015 03:21, 27 October 2006 (UTC)

I'll get you started:

 C      = Casey's current age               C+10    = Caseys age 10 years from now
   M    = My current age                   (C+10)/2 = Half Caseys age 10 years from now
 C+M    = Sum of our current ages
(C+M)/2 = Half the sum of our current ages

Continue like this and show us all your assignments, then we may help some more. StuRat 03:36, 27 October 2006 (UTC)

To be fair, the wording is a bit tricky. Maybe this will help:

I am as old (as Casey will be when I am (twice as old (as Casey was when my age was (half the sum of our present ages))))).

Casey is as old (as I was when Casey was (half the age Casey will be 10 years from now)).

Start by translating what's in parentheses, then work your way out in stages. There are three numbers worth keeping in mind: your age now, Casey's age now, and the difference between them (which is constant as you age). You'll wind up with two pretty simple formulas in two variables (C and M), which you can solve if you've learned Algebra. Black Carrot 21:25, 27 October 2006 (UTC)

When you've worked that one out, you can try your new-found skills on this one:

• Ten years from now, Tim will be twice as old as Jane was when Mary was nine times as old as Tim. Eight years ago, Mary was half as old as Jane will be when Jane is one year older than Tim will be at the time when Mary will be five times as old as Tim will be two years from now. When Tim was one year old, Mary was three times as old as Tim will be when Jane is three times as old as Mary was six years before the time when Jane was half as old as Tim will be when Mary will be ten years older than Mary was when Jane was one third as old as Tim will be when Mary wil be three times as old as she was when Jane was born. How old are they all now? JackofOz 11:57, 30 October 2006 (UTC)

ultra-elliptical functions

I really need to know what ultra-elliptical functions is. Can somebody please explain this to me? —The preceding unsigned comment was added by 213.64.69.132 (talk • contribs) .

They seem to be due to Sofia Kovalevskaya (Sophia Kovalevsky}; our article mentions the publication Mémoire sur un cas particulier du problème de le rotation d'un corps pesant autour d'un point fixe, ou l'intégration s'effectue à l'aide des fonctions ultraelliptiques du temps (1886). Another source gives this reference: "Memoirs présentés par divers savants étrangers à l'Académie des Sciences de l'Institut National de France". Paris. 1890. V.31. p.1–62. The term does not appear to have gained much currency.  --Lambiam^Talk 21:08, 27 October 2006 (UTC)

MU key in calculator

MU stands for Mark Up, which is for calculating Gross margin used in business environments. There are different methods of calculating this, depending on the model of the calculator.
Below some examples that works with my calculator:

Example 1: Enter 100 then press the divide key, then enter 15 then press the MU key, which displays 117.65 (15% of this total equates to 17.65)
Example 2: Enter 125 (selling price) then press the minus key, then enter 100 (seller's costs) then press the MU key, which displays 25 (25% Gross margin)

October 28

Finding a formula based on known numbers

Fixed Number = 00000000000 / Outcome = 0
FN = 11111111111 / O = 3
FN = 22222222222 / O = 6
FN = 33333333333 / O = 9
FN = 44444444444 / O = 2
FN = 55555555555 / O = 9
FN = 66666666666 / O = 2
FN = 77777777777 / O = 5
FN = 99999999999 / O = 1
FN = 12345678901 / O = 5
FN = 11223344556 / O = 1
FN = 10987654321 / O = 5
FN = 22446688001 / O = 6
FN = 11335577990 / O = 2
FN = 53216854234 / O = 0
FN = 43583209432 / O = 8
FN = 12309124097 / O = 8

Is there a way to find out the formula that reaches these conclusions, or is this some known calculative formula i am missing?Avalean 16:50, 28 October 2006 (UTC)

I think it may have something to do with sum of the number's digits. Try making a table of sums and results, and see where that gets you. -- Meni Rosenfeld (talk) 16:57, 28 October 2006 (UTC)

53216854234 and 43583209432 both have 43 for the sum of digits but have different outcomes. Check digits are often formed by taking a weighted sum of the digits modulo some number, which in the beginning looks promising but breaks down at 55...5. The number of possibilities is endless, and without some inspired guess you aren't going to get anywhere. What is the source of these numbers? Would it have been possible to vary just one number at a time?  --Lambiam^Talk 20:05, 28 October 2006 (UTC)

The use of these numbers is similar to a bar code check number. The final digit to a bar code. I have the program that generates these codes but I'm trying to figure out the formula to it. As you said above lambiam, its not mod or the sum of some numbers. I was trying to get the formula to be able to calculate the check digit for many bar codes in one swoop.Avalean 01:50, 29 October 2006 (UTC)

If you have a large database of valid numbers with their codes, try comparing ones that differ only by one number (or as few as you can find). - Rainwarrior 04:08, 29 October 2006 (UTC)

Hello! Did you try any research for barcode check number ? this looks interesting : [1] -- DLL ^{.. T} 22:04, 29 October 2006 (UTC)

If by "barcode" you mean Universal Product Code, its check digit formula fits the pattern I gave above (weighted sum of digits modulo some number). This is ruled out by the actual outcomes: 55...5 and 33...3 both give 9, but the difference 22...2 does not give 0.  --Lambiam^Talk 04:58, 30 October 2006 (UTC)

But... the barcode generator servlet linked to by DLL does turn out to give the same outcomes if for "Barcode symbology" you select "MSI". Changing a few select digits one at a time shows patterns suggestive of the general weighted-sum pattern modulo 10. So when/where/why does this break down? Or we could try to cheat and find the MSI algorithm somewhere on the Internet.  --Lambiam^Talk 05:16, 30 October 2006 (UTC)

October 29

Learning update schedules

This is sort of also a computer science question.

I want to create an RSS reader that "learns" approximately the times of day that an RSS feed is likely to be updated since the last check, and chooses to check the feed according to the probability that it has been updated since it was last checked.

I'm sort of at a loss as to how to actually implement this, especially because (I think) the probability is based on the time of day, weekday, AND the amount of time since the feed was last checked.

Does anyone have any ideas here?

You could use Unix timestamp (the unix seconds) to average out when one update happened and the last one before that also to create an estimate based on the prior updates.

I.e One update was at 09:00 one day and at 09:30 the next. Therefor the program can "predict" the next update is going to be at 9:15, but if you are trying to make a perfect AI that thinks ie that the guy posting is always late on wednesdays, thats a whole other can of wormsAvalean 01:53, 29 October 2006 (UTC)

This looks like a lot of work; is checking for updates costly? There are (at least) three aspects to this: (1) Selection of a model; (2) Estimating the model parameters; (3) Computing the probability. You need a model for the random variable of the update times. Depending on what you observe, this may be like: typically each next update is about 2 hours after the last, except in the wee hours it's more like 4 hours between updates; or: every day around 10am, noon, 15am and 22am. In the first case the times are much more irregular, of course. These possibilities require very different models. The second class is easier: use separate r.v.s for the update time at each part of the day, such as "the afternoon update", which maybe is on the average at 14:52am with a spread of 12 minutes. A normal distribution may be good for this, but probably you also want to add a probability that the whole update is skipped. The parameters (μ and σ) may be made to be different not only depending on the part of day, but also on the weekday. For the first class things are more complicated. You may consider a log normal distribution for the time between updates, but now we are in a situation where the parameters of the distribution change continually. One way of handling this might be to change instead the time scale, as if the clock ticks much slower in the wee hours than in busy periods. Estimating the model parameters can be done if you have enough data using maximum likelihood methods. The more parameters you have, the more data you will need. This can almost certainly not be done analytically; you will need to do numerical optimization, which is tricky. Now assuming you have all parameters, and the data on recent observations (last known update at 10:05, no new update yet at 12:22), you should be able to compute the conditional probability of an update having happened given these recent observations.  --Lambiam^Talk 06:44, 29 October 2006 (UTC)

Fractions whose value is one

what is the name for a number of fractions whos value is one? —The preceding unsigned comment was added by 124.206.195.150 (talk • contribs) 17:59, October 29, 2006 (UTC).

Are you talking about these?

${\frac {1}{1}}={\frac {2}{2}}={\frac {3}{3}}={\frac {4}{4}}$

—The preceding unsigned comment was added by 202.168.50.40 (talk • contribs) 21:44, October 29, 2006 (UTC).

Last time I checked, they were called equivalent fractions... Titoxd^(?!?) 16:22, 30 October 2006 (UTC)

cartesian product of unit circle with itself

It's indeed 'just' a torus as a topologist dismissed the following question with(I'm interested in geometry as well as topology, and right now, in differential geometry): Since not even the tiniest region of the surface lives in a 3 dimensional space, what curvature can be assigned to it?-Is it a tensor thing? Thanks,Rich 22:21, 29 October 2006 (UTC)

It embeds into 4-dimensional space (a two-dimensional subset of the three-dimensional sphere in R⁴) and has zero curvature everywhere. —David Eppstein 00:46, 30 October 2006 (UTC)

(after edit conflict) If by "circle" you mean the topological concept of a circle "up to homeomorphism", also known as the 1-sphere S¹, and by "product" you mean the topological product, the concept of curvature simply does not apply. You get indeed the topological torus, which by definition is the topological product of two topological circles: S¹ × S¹. If your circles are embedded in the Euclidean plane R², and you take the square into four-dimensional Euclidean geometry R⁴, it is possible to assign a meaning to the curvature (which then for reasons of symmetry must be the same everywhere).  --Lambiam^Talk 01:00, 30 October 2006 (UTC)

Moreover, if your circle had radius r, and hence arc curvature r^-1 at each point, then the curvature tensor naturally associated with the topological product would be ${\begin{pmatrix}r^{-1}&0\\0&r^{-1}\end{pmatrix}}$. (More generally, one would take the direct sum of curvature tensors of the multiplicands.) So the mean curvature of the product is r^-1 and the Gaussian curvature is r^-2. —Blotwell 01:12, 30 October 2006 (UTC)

Blotwell, you're sure that calculation is valid when dealing with immersions into R⁴ too? Because it looks way too similar to what you'd get by immersing it in R³, but without taking into account that one of the eigenvalues of the Gauss map does change sign at some points. (Similar for mean curvature). And by Gauss-Bonnet (as mentioned below), curvature cannot be everywhere positive on the torus, as the integral of the Gaussian curvature over the whole torus is zero. \Mike(z) 14:51, 30 October 2006 (UTC)

I did mean of course the product of two planar "unit" circles and it lives in 4 dimensional Euclidean space as for example (cost,sint,cosu,sinu) 0<=t,u<2pi.I agree that the curvature is of course everywhere the same. I am a bit nonplussed to hear the curvature is zero everywhere, David Eppstein. Are you certain? Regards,Rich 01:21, 30 October 2006 (UTC)

Gaussian curvature is intrinsic, locally a circle is (in an intrinsic view) the same as a line, and so locally the product of two circles is the product of two lines, a plane, which has no curvature. For other kinds of curvature I imagine it would matter how it was embedded into R⁴, though. —David Eppstein 01:36, 30 October 2006 (UTC)

PS if you want some more verifiability than my word about the curvature being zero, try Googling for "flat torus". —David Eppstein 03:29, 30 October 2006 (UTC)

Cf. also Gauss-Bonnet theorem: if the Gaussian curvature is everywhere the same, it must be zero.--gwaihir 11:35, 30 October 2006 (UTC)

Just to clarify: if the Gaussian curvature is everywhere the same on the torus, it must be zero. Tesseran 13:37, 30 October 2006 (UTC)

I was figuring the types of curvature I had learned about for lines and curves in 3D space would be at best "incomplete" and perhaps somehow incorrect for a surface that doesn't fit into 3D space at all. Rich 02:32, 30 October 2006 (UTC)

If you look at the torus as a quotient of R² (by the integer lattice, or by the group of integer translations), it inherits the Euclidean metric from R². In this view, it is clear that the metric is flat. This is true of the cylinder too, for example, because it also can be seen as a quotient of the plane. (It is also S¹ × R, in case you worry about the flatness of the cylinder; it can also be seen as a subsurface of R² where one of the principal curvatures is flat.)

Your question hits upon one of the fundamental issues in the development of mathematics: the distinction between intrinsic and extrinsic properties of objects. In group theory, groups were once thought of primarily as subgroups of permutation groups or matrix groups, and it was the step towards abstraction in defining groups axiomatically that allowed a lot of the advances in modern algebra. Furthermore, now we can study all the different ways that one abstract group can be realized as a subgroup of a matrix group—this is representation theory. In topology and geometry, surfaces and curves were thought of for a long time as subsets of some Euclidean space; moving to an abstract definition of a manifold allows us to distinguish between intrinsic properties of a manifold (like dimension, lengths, or angles) and extrinsic properties that depend upon the way the manifold is embedded into Euclidean space. One of my favorite results is that Gaussian curvature is an intrinsic property; though the principal curvatures may vary based on the embedding of a manifold, their product depends only on the metric. So you can, for example, calculate the curvature of the metric in any model that you like. Tesseran 13:37, 30 October 2006 (UTC)

Is it meaningful to talk of the Gaussian curvature of a surface in 4D? Our article only mentions the 3D case and does not talk of extensions. So I'm a little confused as to which notion of curvature we are refering to. --Salix alba (talk) 16:07, 30 October 2006 (UTC)

"intrinsic". It's a 2D thing and does not depend on ambient space.--gwaihir 16:10, 30 October 2006 (UTC)

Yes, 'sectional curvature' is the more general notion in this case, and for two-dimensional surfaces they coincide. \Mike(z) 16:25, 30 October 2006 (UTC)

history of the compass

I am trying to find information on (geometry) compasses for a school project for my son and am having trouble finding info on the history of them. Ie: where they come from, who made them etc. Any links or info would be appreciated. Kerry. --203.58.44.15 03:45, 30 October 2006 (UTC)

The third postulate of Euclid's Elements: "Given a straight line segment, a circle can be drawn using the segment as radius and one endpoint as center", shows that the compass was known circa 300 BC. Presumably it is much and much older as an instrument to transfer a reference length. This web site states: "[The Egyptian god] Ptah's major symbols are the Masonic Square and Compass, encircled by a knotted length of rope." I don't know how reliable that site is, and could not find further corroboration of this statement, but by itself it is plausible that the ancient Egyptians knew the compass. If made of wood, you don't expect much of these to have survived 50 centuries or so, even in a desert climate. Since it does not require particularly advanced skills to make a compass, it would not be strange if not much is known about their early history.  --Lambiam^Talk 04:44, 30 October 2006 (UTC)

Value of a Life Estate

When my uncle dies he is leaving his $350,000 home to his 65 year old friend.

When his friend dies or moves out, the home will go to me, according to my Uncle's Will.

What might I have to pay his friend to encourage her to move now (the value of the Life Estate)?

Thank you.

It sounds like all the friend gets is the use of the home (as opposed to ownership), for the remainder of their life. If we figure a life expectancy of 85, and a value of US$15,000 per year, that would be US$300,000. Unfortunately, with taxes and other transaction costs, this would make that too much money to pay. Of course, you would know more about this person's health and the rental prices for a comparable house in your area, so could come up with a more accurate estimate. We could have attempted to take inflation during the 20 years into account, as well, but that would more or less even out if it was invested such that it returned the rate of inflation, after taxes. Also note that many people are rather short-sighted, and would take a much smaller amount of cash NOW, rather than a larger sum later. StuRat 14:22, 30 October 2006 (UTC)

Note that I was assuming your uncle dies now. If we have to take his life expectancy into account relative to his friends, then things get complicated. The value to his friend might be almost zero, if your uncle is likely to outlive the friend. Another thought, perhaps you could pay the friend somewhat less, but still allow him to live in a portion of the house, with the rest available for you. This might work out better financially for both of you. StuRat 14:29, 30 October 2006 (UTC)