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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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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 ♥Ťįɱé Ø 11:38, 24 October 2006 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]