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December 13[edit]

Partial (trapezoidal) section of sphere/ellipsoid surface[edit]

(Moved by suggestion from Talk:Polar coordinate system)

What is a trapezoidal shaped section of a sphere or ellipsoid's surface called? I think it is something like "quadratic angle" or "quadrilateral section"—? ~Kaimbridge~01:04, 13 December 2006 (UTC)[reply]

The thing you describe will bound a solid angle. It's called a "spherical polygon" provided the lines creating it are arcs of great circles (see great circle also geodesic) see:Spherical trigonometry.83.100.254.21 17:57, 13 December 2006 (UTC)[reply]

No, that's not the term (I came across the term and an image of it a while back, while looking something else up——I think either here in Wikipedia or MathWorld).

The closest description would be "spherical isosceles trapezoid" (but, like segments in the above image, not necessarily "upright"). ~Kaimbridge~20:11, 13 December 2006 (UTC)[reply]

Is it a specific term that means the same as "spherical isosceles trapezoid" you are looking for, or is it a slightly different thing? To be honest I can't find anything the same, http://mathworld.wolfram.com/SphericalTriangle.html http://mathworld.wolfram.com/SphericalPolygon.html gives no alternative names. The only 'specialised' term I can think of in terms of spherical geometry is spherical lune. Have you looked at 'Hosohedron'?83.100.254.21 20:39, 13 December 2006 (UTC)[reply]

Looking at the two diagrams you supply do you mean the area/volume produced by the intersection of a spherical wedge and spherical segment/sector?83.100.254.21 20:43, 13 December 2006 (UTC)[reply]

Question are the parallels great circles (or not), are the 'legs' definately great circles.? alternatively is the shape formed from two longitudes and two latitudes?83.100.254.21 20:47, 13 December 2006 (UTC)[reply]

It doesn't have to be: See image——any of those grid sectors qualify.

I have heard something like this called a "patch" on the sphere's surface in the context of computer graphics, but I've never heard of a mathematical term for such a thing. - Rainwarrior 22:42, 14 December 2006 (UTC)[reply]

No, like I said, I came across the term/diagram in some math source. Oh well, I'll likely see it again (when I least expect it! P=) ~Kaimbridge~ 17:29, 18 December 2006 (UTC)[reply]

How to distinguish between different local maxima points and numerical errors[edit]

Hi all! I think this is a simple question, but I cannot think a clever algorithm for this problem: We will work in standard x-y-z space. We have an objective (strictly positive and bounded) function f(x,y) in domain, says, [0,10]x[0,10], and we want to write a computer program to find the set of all points {(x,y) } in the domain which locally maximizes f(x,y). However, suppose f(x,y) contains many many local maximum points, says, more than hundred mountains in the domain. To simplify the problem now, I suppose there is no ridge in any mountains.

Let us suppose that we can calculate the gradient (and hessian) of f(x,y), my basic scheme is to try using standard gradient ascent (or conjugate gradient, etc.) many times with different random starting points to discover all mountains. However, suppose my 2 answers (x1,y1) and (x2,y2) are very closed, and also f(x1,y1) and f(x2,y2) are also very closed. How can we check that they come from the same mountain (but with small numerical errors), or they really come from two different mountains?? -- 131.111.164.218 12:55, 13 December 2006 (UTC)[reply]

Why random points? Can't you just find the 0's of the {df/dx,df/dy}? yandman 13:04, 13 December 2006 (UTC)[reply]

Sorry, I didn't make my problem precisely. Suppose I have to solve the problem by wrting the computer program, so I have to concern about numerical error. Now suppose I find two 0's of the {df/dx,df/dy} which you mentioned, and they are very closed. How can I know that these two points come from different mountains or they come from the same mountain with small numerical error? -- 131.111.164.218 13:32, 13 December 2006 (UTC)[reply]

If they're from different mountains, there'll be another 0 of {df/dx,df/dy} between them (the bottom of the "valley"). yandman 13:43, 13 December 2006 (UTC)[reply]

Yes, you're right, but in order to find that point in between, it looks like I have to solve another problem (with the same difficulty) for every pair of mountains (and there are a lot of mountains). So, now I'm looking for more simple checking method such as this: if |f(x1,y1) - f(x2,y2)| < c1 & if |(x1,y1) - (x2,y2)| < c2, then just ignore the error. But I'm still strucking of how can we know these c1 and c2? Are there any good rule of thumbs, (or totally different methods?). 131.111.48.168 15:34, 13 December 2006 (UTC)[reply]

I'm no expert on numerical methods, but I guess the answer can depend heavily on what kind of assumptions we can make about f (how smooth it is, what order of derivatives can be assumed to be nonzero, etc.). My idea is to calculate f at random points in the vicinity of the two maxima, and to check the statistical correspondence between receiving high values and being close to the maxima. If there doesn't seem to be a difference between, say, points in the middle and points near the maxima, we can assume that these are actually the same maximum. The exact implementation can vary. -- Meni Rosenfeld (talk) 16:44, 13 December 2006 (UTC)[reply]

Thank Meni! But can you explain more about your sentence check the statistical correspondence between receiving high values and being close to the maxima? I think I do not clearly understand 131.111.164.218 17:04, 13 December 2006 (UTC)[reply]

Heh, that sentence was left vague in purpose, because the exact way you implement it depends on many things, and even if those things are given, I'm not sure what's best. But consider this simplified example: Suppose your function is defined on

\mathbb {R}

, and your analysis shows that the points +1 and -1 are suspected local maxima, with a value of 0. You are trying to figure out whether the function is something like

1-x^{2}

, so these are actually the same maximum at 0, or the function is something like

-(x-1)^{4}-(x+1)^{4}

, so there really are maxima at +1 and -1. Since there are numerical errors (which we can model as random noise), you can't check this directly. However, you can evaluate the function at many random points, and calculate Pearson's correlation coefficient between, on one hand, the resulting values and the first of these candidate functions, and on the other hand, between the resulting values and the second of these candidate values. Whichever gets a higher coefficient is more likely to resemble the true function, so this can estimate whether the two maxima are actually one. This idea can probably be generalized to a more useful setting (just find two or more functions which may resemble the true function, and go with the one that corresponds better with the data). -- Meni Rosenfeld (talk) 17:25, 13 December 2006 (UTC)[reply]

Thanks for your clarifying :). That's surely a clever (though requires some works) method. In fact, I did not understand that sentence, because of my English, not because of its vagueness ... 131.111.164.218 17:36, 13 December 2006 (UTC)[reply]

Firstly a remark: If you have Hessians (i.e. second derivatives) of f, why not use Newton to find zeros of the derivative, rather than conjugate gradients? On your specific question: if your (x1,y1), (x2,y2) are very close (say

10^{-7}

) and you cannot get more accurate solutions due to numerical errors (say your function values are only exact to 7 decimal places), then there is nothing you can do. Sampling intermediate points will just give you noise based on those numerical errors. You will just have to accept that up to 7-digits exact these ARE the same mountain. Some smoothness argument (Lipschitz continuity of the derivative) will help in establishing that in a vincinity of your (x1,y1)/(x2,y2) there are no points with large values of f' (which would make those two separate mountains).195.128.250.110 23:04, 13 December 2006 (UTC)[reply]

You may be able to achieve superresolution in this numerical problem by replacing your function with its Pade approximant. The Pade approximant is a rational polynomial approximation to the function. If your f() is sufficiently well-behaved, then its approximant will be a faithful approximation. Note that meromorphic is sufficient. The advantage of the approximant is that you can use symbolic methods (derivatives of rational polynomial functions can be computed by what amounts to elementary arithmetic on lists of pairs of coefficients and powers) to determine whether there's a zero between the two peaks and/or to determine whether the Hessian has the same sign at the two peaks.

For many "natural" problems, the Pade approximant is a surprisingly rapidly faithfully converging approximation. The simplicity of the arithmetic on rational polynomials is another benefit.

In this application, you'd sample a bunch of points scattered around your two nearly coincident peaks and use those pairs of inputs and output to build your approximant. There are canned codes for this and the Pade approximant article references "Numerical Recipes". Note that this is a variety of superresolution/extrapolation, so your result could be wrong, but you're making the best guess you can with the data you've got.

Fuzzyeric 04:17, 14 December 2006 (UTC)[reply]

I just found the two recent answers today. Perhaps it is too late, but "a lot of thanks" to Fuzzyeric and 195.128.250.110! You make me better understand the problem, and open some unknown wisdom to me! -- 131.111.164.227 12:56, 20 December 2006 (UTC)[reply]

Rate of Return with Multiple Capital Injections[edit]

I need assistance determining the the TRUE ROI for a serious of investments with differing profit/loss and differing capital injections.

For example:

Jan 06: Purchase $10000 in shares in X Jan 06: Sell all shares in X and make profit of $1000 & spend this $1000 Feb 06: Take $10000 & purchase share in Y Feb 06: Stock drops and purchase $5000 more share sin Y Mar 06: Sell all shares in Y & make profit of $1500.

Individually, the SIMPLE ROI for each investment is 10%

However, the total profit is $2500, with a maximum invesment of $15000 = 16.7% return. I think this is wrong.. I need validation

OR

Do i just weight each investment accordingly to better understand the correct SIMPLE ROI.

OR

Is there some other method to best figure out ROI?

Thanks Trevor Gartner trevor.gartner@telus.com

I'd say the ROI is actually closer to 18%, because you only actually invest 14k (the 1k you gain after the first "session" can be deducted from the 5k extra you put in). yandman 20:50, 13 December 2006 (UTC)[reply]

You shouldn't be surprised that your ROI for a set of consecutive investments exceeds that for the investments separately; if you buy $1 and sell at $1.10, but then immediately buy at that $1.10 again (neglecting fees) and sell at $1.21, then repeat and get $1.33, then $1.46, $1.60, $1.76, and $1.94, obviously you have +94% return even though each separate investment was +10%. The quantity that should be "conserved" like you want it to be is the (effective, average) interest rate for a simple or subdivided investment; it's important to consider it as having dimensions of (inverse) time. --Tardis 16:13, 15 December 2006 (UTC)[reply]

Non-Euclidian geometry[edit]

The five postulates of classical Euclidian geometry are

Any two points can be joined by a straight line.
Any straight line segment can be extended indefinitely in a straight line.
Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.
All right angles are congruent.
If two lines intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough.

By changing the fifth postulate, you get the hyperbolic and elliptical geometries. But what sort of geometry do you get if you change the fourth postulate? --Carnildo 23:49, 13 December 2006 (UTC)[reply]

None. People at that time were hypnotised on right angles, but the postulate just seems to introduce a definition of congruence e.g. "if it's right it's right and if the other one is right too they are identical" seems quite tautological. What do you think ? -- DLL ^{.. T} 18:07, 14 December 2006 (UTC)[reply]

I think it's hard to tell because we can't consider Euclid's postulates as an axiom system in the modern sense. – b_jonas 19:19, 14 December 2006 (UTC)[reply]

As a bit of a cheat - how about if the three orthogonal vectors ie i,j,k are replaced by sheared versions eg i'=i,j'=0.9j+0.1i,k'=0.9k+0.1i+0.1j ? (needs normalising) so then the new 'right angles' would be angle between vectors i'j', i'k', and j'k' - so the new angle bewteen two vectors would be calculated from the 'new right angle' would vary depending on the orientation of the two vectors - not really what you are looking for I suspect..(Though it does make the new 'right angles' non congruent in euclidean geom)83.100.174.70 19:53, 14 December 2006 (UTC)[reply]