Wikipedia:Reference desk/Archives/Mathematics/2006 December 1

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December 1

tough physics...

Question: During an Apollo lunar landing mission, the command module continued to orbit the Moon at an altitude of about 100 km. how long did it take to go around the moon once?

i have to use the formula periodic cycle = radical 4pisquared times the radius (1.84 x 10tothe6thpower)cubed divided by 6.67 x 10tothe-11 Nmeterssquared by kgsquared.

thanxs for the help!

I'm not sure if this is the formula you mean: $periodiccycle={\sqrt {{4(\pi )^{2}r*(1.84*10^{6})^{3}} \over {(6.67*10^{11})^{2}}}}$ But I can ask you this: what is the relationship between the altitude and the radius of your orbit? (hint: the altitude is measured from the surface)

Is the set of real numbers a vector space over the complex numbers?

The converse is clear, since we have z = a+bi with multiplication by a real and addition well defined in the normal sense. But is there a way to define multiplication of a real number by a complex one such that the additive group R is a vector space over C? 06:06, 1 December 2006 (UTC)

Here's a cheap attack: given the Axiom of Choice, both R and C are themselves rational vector spaces of equal dimension, so they are isomorphic groups in particular, and since the group C has a C-vector space structure, so does R. But I wouldn't be at all surprised if it were impossible to actually define such a structure. Melchoir 08:04, 1 December 2006 (UTC)

predict to 4D Singapore Pools Draw

A fenomena in my country, Indonesia there was a lottery draw called TOGEL (Toto Gelap as Indonesian lang) that adopted result from Singapore pool 4D. it's take on 1st prize Only but we could play/bet 2 dgits, 3 digits and 4 digits with different prize. 4D strike winning payd up Rp.3,000,000, 3D strike payd up Rp.300,000, and 2D strike we won Rp.70,000.

till now I'm still looking for a good software which can do predict analys for the next draw. whoever can help me, please do let me know. my alternate email: aryadewangga@plasa.com

thanks for all

Riza

Unless the lottery is fixed, there should be no way to predict it. If there was, don't you think a million people would select the winning numbers ? StuRat 06:45, 1 December 2006 (UTC)

Nothing a time machine can't fix. Just buy tomorrow's newspaper today. 211.28.131.37 11:37, 1 December 2006 (UTC)

Actually, I think knowing the future with absolute certainty poses several temporal paradoxes. Thus, you can't go back in time in your own time line, although you may be able to go back in a parallel time line. However, it's not 100% certain that the lottery results will be the same in that parallel universe. StuRat 12:18, 1 December 2006 (UTC)

Oh well, chances can't get any worse anyway, can they? ☢ Ҡi∊ff⌇↯ 12:28, 1 December 2006 (UTC)

Trig uses

I needed some info on the real life applications of trigonometry. I couldn't find much through google; I need near 20 pages of stuff. Links to info pages would be appreciated, or you could just tell me in a shortened and simplified form. Cheers, and thanks in advance. --May the Force be with you! Shreshth91 08:28, 1 December 2006 (UTC)

Did you look at our article on trigonometry and the link to uses of trigonometry? It's a start. You just need to look further inside those topics mentioned. ☢ Ҡi∊ff⌇↯ 08:42, 1 December 2006 (UTC)

Surveying, navigation, calculating the height of mountains, GPS, radar, astronomy, etc. StuRat 08:46, 1 December 2006 (UTC)

Electrical Engineering, see phasor. 211.28.131.37 11:38, 1 December 2006 (UTC)

Trig functions are also very important in circular and simple harmonic motion. 82.68.188.118 18:56, 1 December 2006 (UTC)

My carpenter friend wanted to put a sloping roof on an octagonal building, with eight sides sloping to a point at the top, and asked me to help him figure out the angles at which he would have to cut the wood. It was all about sines and cosines, it turns out. -GTBacchus^(talk) 19:01, 1 December 2006 (UTC)

Need an equation

I need an equation. The preliminary equation is k = a * b/c. It needs to fulfill the requirement that as a goes up and/or b/c goes down, k must go up. What constans can be added to the equation to satisfy these requirements? Thank you. Jack Daw 10:27, 1 December 2006 (UTC)

$k=1+a\div {\frac {b}{c}}$ 211.28.131.37 11:43, 1 December 2006 (UTC)

k = a − b/c also does the job. But, if I may say so, this is a peculiar problem. Where does it come from?  --Lambiam^Talk 12:56, 1 December 2006 (UTC)

a is a food ingredient. The more the better. The product comes in boxes of c, which costs b. The cost for a single product is b/c. The less the better. That's where it comes from. Jack Daw 20:52, 1 December 2006 (UTC)

Suppose one box costs 100 crowns and contains 5 spettekaka, each of which weighs 200 g. I assume that then a = 200 g, b = 100 crowns and c = 5. So the net weight of a box is c × a = 5 × 200 g = 1000 g. For that you paid the amount b. The conventional way to compare prices is to look at the cost per net weight, which is then b / (c × a) = 100 crowns / 1000 g = 0.10  crowns per gram. You want this to be low. Alternatively, you can look at how much you get per the amount you pay. That is of course the (multiplicative) inverse c × a / b = 1000 g / 100 crowns = 10 g per crown.  --Lambiam^Talk 18:24, 2 December 2006 (UTC)

Well a is an ingredient. I have no idea what spettekaka is, never heard of it :) The products I'm checking out are meal replacement bars. You buy these products, usually, in boxes of 12 (c), and their cost varies (b). Thus, one bar costs b/c. A low price is great, but it also has to contain adequate carbs. So the equation I needed was for a comparison of these to get some sort of "index" where a higher price lowers it and the amount of carbs per serving increases it. Of course there are other factors that would determine the "index", such as protein/carb-ratio (not too low but not too high), fructose and sodium. However you see there's a list of 120 products of which half are no use to me (such 45g protein and 10g carbs), so I need a formula with which I can "disqualify" the majority of the products that are of little use to me, and so to save me time when I want to know exactly what's in a product (I could hardly well inspect all 120 of them thoroughly). This is the list http://www.bodybuilding.com/store/barscarb.htm I'm not sure if I'm making much sense, hence my explaining things over and over. Ahem. Jack Daw 16:49, 3 December 2006 (UTC)

What you're talking about (the "index") is known as a cost function — a function which describes the "goodness" or "badness" of something numerically. One thing that's helpful in designing such functions is that they should have consistent units of measurement, even if their output is just a number (instead of, say, a price or weight). So write your varables carbohydrates in a bar (mass), mass of a bar (mass), price (currency/box), number (which has units of "per box"). As is obvious, the thing to do is form $p/n$ and $c/m$; the insight is why: using those ratios leaves you with a currency value and a unitless number, which is simpler than what you had previously. But then to combine these quantities, you need some sort of reference value; let's say for simplicity that you really badly want your carbohydrates, and that (other concerns like taste aside) you would be willing to pay 15 kr for a bar that was all carbohydrates. Of course, bars with no carbohydrates are worthless. So a simple choice of cost function assigns 0 cost to 0 money or 100% carbohydrates, and assigns 15 kr the same badness as 0 carbohydrate ratio, so that a worthless bar for 0 cost is the same badness as a perfect bar at the cost you're willing to pay for that. Then we have $f:={p \over n(15{\mbox{ kr}})}-{\frac {c}{m}}$. I'm using f instead of your k because this is a badness function that should not be confused with your goodness function. (Obviously, if any real number is considered a valid value for such functions, they can be interconverted simply by changing their sign.)

But this isn't the whole story: this would rate bars at 10 kr each but that only massed one milligram better than the same bars at 11 kr each that weighed a kilogram each. So the mass of the bar should come into play outside of the carbohydrate proportion. In fact, really all you care about is the carbohydrate amount; the problem comes when the carbohydrates are so sparse that you would be full (or over-caloried) before you had the carbohydrates you wanted. So instead of dividing c by m, consider c a good quantity on its own and ascribe a separate penalty to large m values. (One could also use total calories or so as a penalty.) Separating these two makes them both need a reference quantity: if this replaces one of three meals that should together provide about 300g of carbohydrates and should each mass maybe 200g, we can use reference quantities of 100g carbohydrates and 200g mass. Then, continuing to make the cost function as simple as possible, we have something like $f:={p \over n(15{\mbox{ kr}})}-{\frac {c}{50{\mbox{ g}}}}+{\frac {m}{200{\mbox{ g}}}}$. We have to double the middle term because there is only the one subtractive term to counteract two normalized additive terms; of course, this is then just a different normalization constant.

Yet more choices of variables are possible: if, for instance, the bars were all very small so that they could be treated as a continuum, it might make sense to throw out the mass per bar as a variable, return to the notion of carbohydrates per mass, and normalize price against mass rather than against number. Then you would have something like $f:=\rho \left(20{\frac {\mbox{g}}{\mbox{kr}}}\right)-\kappa$ where $\rho :={\frac {p}{m}}$ and $\kappa :={\frac {c}{m}}$ so that the cost is defined in terms of specific price and carbohydrates (see specific heat or specific energy; specific here means "per mass"), which are variables appropriate to a continuous substance.

As is probably by now clear, creating these functions can be quite complicated, as it involves figuring out what things you really want and quantifying their desirabilities in a compatible manner. It's all heuristic and imprecise, but a well-chosen cost function is very powerful for making consistent, logical decisions. They're often used in such fields as image processing, design optimization, and other fields where there aren't even enough people to look at every single subject except through the application of a cost function to a set of hundreds of them. --Tardis 17:29, 4 December 2006 (UTC)

That's a lot of math in one sitting :D The main problem with such cost functions, then, I think is because of "reality" being too inexact. With the bars for example, the one at the bottom is the optimal one, but it would be "disqualified" by a cost function that would demand 2 times more carbs than protein - this one has a ratio of 1.97:1 I think. However, because of other superior qualities with that particular bar, that's not a problem, and hence my choice. Thank you very much for your help. It stands as yet another monument against those who bash mathematics for not being applicable to the real world. Jack Daw 15:13, 5 December 2006 (UTC)

There's no "disqualification" going on, at least not with the functions I described. It's actually almost always the case that you should specifically avoid discontinuities in your cost function, because then a trivial (yea, perhaps even unnoticable) change in the parameters of the problem may make a catastrophic change in your decision, which is illogical. The 1.97:1 bar should be penalized by an amount that is (to first order) proportional to its 0.03 failing, not merely assessed a fixed "insufficient ratio" cost. If you include enough relevant parameters in your cost function, weight them appropriately, and avoid large derivatives (remember that a discontinuity can be treated as the limit of large derivatives), you will find that such "minor failings" will be overlooked precisely when there are worthy mitigating factors. Finally, it's helpful to look at all of your cost outputs, not just whatever object "won"; if of 30 choices, 25 are bad and the other 5 are all but identical in cost, perhaps the ideal solution (for other reasons that are hard to quantify) is the third-best, or even a mix of the 5. --Tardis 15:43, 5 December 2006 (UTC)

Layout of trig

I've got a formula in the form $\sec x={\frac {g}{{\omega }^{2}}}$, where x should be determined as a function of ω. However, I'm not sure what the best way to lay this out would be; $\cos x={\frac {{\omega }^{2}}{g}}$, $x=\arccos {\frac {{\omega }^{2}}{g}}$ or $x=\operatorname {arcsec} {\frac {g}{{\omega }^{2}}}$

These should all be identical, but is there any layout which is prefered for trig functions. Is it better to have sin x = y or x = arcsin y for example? Laïka 13:18, 1 December 2006 (UTC)

Assuming that ω may be zero while g is known to be non-zero, I'd go for arccos(ω²/g), as it avoids a possible division by zero, or in any case an apparent (but removable) singularity. When asked to define x as a function of y, given y = sin x, the answer "sin x = y" might not be considered equally satisfactory by everyone (and in particukar not by the grader if this is a question on a test). However, in using the inverse trigonometric functions, there is alway the issue what principal values to use, or to consider all solutions, since these inverse functions are tricky indeed: properly speaking they are not functions but relations, or so-called multivalued functions. By leaving the answer as: "x such that cos x = ω²/g", you avoid (or evade) that issue. Further, in general, there are no rules here for which of several equivalent expressions to choose, and you can select one using your own esthetic judgement, such as what looks simplest to you, or most symmetric, or whatever pleases you.  --Lambiam^Talk 15:02, 1 December 2006 (UTC)

That makes good sense; thanks! Laïka 16:47, 1 December 2006 (UTC)

Prolate radii and radii of curvature?

Much has been written about oblate spheroids (a>b), but very little regarding the prolate case.
There are 2 principal elliptic integrands: Where $o\!\varepsilon \,\!$ is the angular eccentricity, equaling (for the oblate) $\arccos \left({\frac {b}{a}}\right)\,\!$,

• $E'(\theta )={\sqrt {1-(\sin(\theta )\sin(o\!\varepsilon ))^{2}}};\quad C'(\theta )={\sqrt {1-(\cos(\theta )\sin(o\!\varepsilon ))^{2}}};\,\!$

Now let

•  $n'(\theta )={\frac {1}{E'(\theta )}};\quad m'(\theta )=\cos(o\!\varepsilon )^{2}n'(\theta )^{3};\,\!$

$r'(\theta )=n'(\theta ){\sqrt {\cos(\theta )^{2}+\sin(\theta )^{2}\cos(o\!\varepsilon )^{4}}};\,\!$

For the oblate spheroid, the parametric latitude, $\beta \,\!$, is less than the planetographic, $\phi \,\!$ (since it "squashes" towards the equator):

$\beta =\arctan(\cos(o\!\varepsilon )\tan(\phi ));\,\!$

For the meridional radius of curvature, M, its perpendicular/"normal" counterpart, N, and the radius, itself, "R":

$M=M(\phi )=a\cdot m'(\phi );\quad N=N(\phi )=a\cdot n'(\phi );\,\!$

$R=R(\phi )=a\cdot r'(\phi )=a\cdot E'(\beta )\,\!$

At the equator, $M={\frac {b^{2}}{a}}\,\!$ and $N=a\,\!$; at the poles, they both converge at ${\frac {a^{2}}{b}}\,\!$.
Okay, now what about the prolate (b>a)? Am I right in presuming that,

$o\!\varepsilon _{p}=\arccos \left({\frac {a}{b}}\right);\,\!$

$\beta =\arctan(\sec(o\!\varepsilon _{p})\tan(\phi ));\,\!$

(since it is "elongated", rather than "squashed")

Since, at the equator, N still equals a, and (it would seem) $M={\frac {b^{2}}{a}}\,\!$ (since, here, the arc is greater than the axis——or would it just be b?). Likewise, at the poles, they would still seem to converge at ${\frac {a^{2}}{b}}\,\!$. Experimenting with different combinations at the equator and poles, it would seem that,

$M=M(\phi )={\frac {a\cos(o\!\varepsilon _{p})}{C'(\phi )^{3}}}={\frac {b\cos(o\!\varepsilon _{p})^{2}}{C'(\phi )^{3}}};\,\!$

$N=N(\phi )={\frac {a\cos(o\!\varepsilon _{p})}{C'(\phi )}}={\frac {b\cos(o\!\varepsilon _{p})^{2}}{C'(\phi )}};\,\!$

Therefore, I would say,

$m'(\theta )_{p}={\frac {\cos(o\!\varepsilon _{p})^{2}}{C'(\theta )^{3}}};\quad n'(\theta )_{p}=m'(\theta )_{p}C'(\theta )^{2};\,\!$

$M=M(\phi )=b\cdot m'(\phi )_{p};\quad N=N(\phi )=b\cdot n'(\phi )_{p};\,\!$

Are there any sources to confirm this?
Also, what about R? Given the change with M and N, would the base value of R be $b\cdot C'(\beta ),\!$, with the $\phi \,\!$ version adjusted accordingly?
Again, any confirming sources?  ~Kaimbridge~14:20, 1 December 2006 (UTC)

Brackets in equations

I've got a maths book (part of the Murderous Maths series) which explains algebra. Most of it is pretty basic, but there's one bit that doesn't make any sense. It states that although $(f+g)=x$ can be made into $f+g=x$, some mathematicians consider the reverse to not be true; the brackets cannot be put back in to make $(f+g)=x$ again. Is this true, and if so, why? Laïka 16:53, 1 December 2006 (UTC)

Parentheses are used in mathematical notation with a variety of meanings, but the only customary meaning that makes sense here is grouping, which is then done to avoid unintended interpretations because the precedence of the operations would require that, or is not clearly defined, or possibly for clarity. Now it is the case that under the standard rules $f+g=x$ has only one possible interpretation. Therefore, inserting parentheses here serves no purpose and is silly, and in any case quite unconventional. No working mathematician would write that, and if a student did, they'd probably correct it. What would you think of the formula $((2)x+3((y))+1)=(((z)^{(2)}))$? I would rewrite is as $2x+3y+1=z^{2}$. Perhaps that is what the book is trying to say: Don't use parentheses that serve no purpose.  --Lambiam^Talk 17:54, 1 December 2006 (UTC)

The only other possible meaning I could see here is pointing out some sort of ambiguity between (f+g) = x and f + (g=x), but that's rather silly if you're working in the kind of mathematical contexts you seem to be talking about (as opposed to computer science, where you would need to tell the computer these sorts of things). —AySz88\^-^ 04:57, 4 December 2006 (UTC)

e=mc to a layman

Please explain this to me so i can understand it- energy = mass squared?—The preceding unsigned comment was added by 12.161.241.50 (talk • contribs).

A good place to start is the article E=mc². Cheers, --TeaDrinker 18:35, 1 December 2006 (UTC)

Let's get the equation right first, shall we? It's

$E=mc^{2},\,\!$

where E is energy, m is mass, and c is the speed of light (in a vacuum). Each quantity must be expressed in suitable units for this to be meaningful. In theoretical physics it is often convenient to use units in which the numerical value of c is 1. Either way, the essence of the equation is that energy and mass are two views of the same thing. The definition of (inertial) mass is fairly tangible. Float a body in space, and push on it. The harder it is to start or stop or redirect, the greater its mass. The definition of energy is more intangible; in many ways it is a bookkeeping trick. For example, a swinging pendulum has potential energy but no kinetic energy at the height of a swing, and the reverse at the bottom. A fundamental rule of physics is that the books must balance: the total energy never changes. Einstein's equation is necessary to balance the books in special relativity.

Here's a basic example. An electron picks up (kenetic) energy as it travels the length of a linear accelerator (which is just a high-tech version of the CRT in a television set). It also becomes harder to push around, harder to speed up. That is, the increase in energy is visible as an increase in mass. The more energy we give the electron, the more noticeable the effect becomes. A really powerful accelerator will not work properly unless we compensate.

We find a kindred phenomenon in the relationship between lightning and a lodestone (a natural magnet), an equivalence between electricity and magnetism. When we move a current of electricity through a wire, the wire acts as a magnet; and when we move a magnet near the wire, we generate electrical current. As a practical consequence we get electric motors and electrical generators. In both energy/mass and electricity/magnetism, superficial differences conceal a deeper unity of practical importance.

But instead of explaining more, I must state the obvious: This question should be posted on the reference desk for science, not the desk for mathematics. --KSmrq^T 02:46, 3 December 2006 (UTC)

I don't agree with you on that certain units must be used for the expression to make sense. You can have any units you want (or no specific units at all) in mind, and it is still perfectly fine. —Bromskloss 15:20, 3 December 2006 (UTC)

As long as the units are applied such that units cancel out correctly within the equation it should be fine. The relation should hold irrespective of units. --AstoVidatu 22:37, 3 December 2006 (UTC)

AstoVidatu has captured my intended meaning. Perhaps it was more confusing for me to bring up the issue, as it applies to all equations in physics. However, it's an issue that gets raised on the article talk page, and this questioner may not be aware of such conventions, so I thought it worth a remark. --KSmrq^T 05:46, 4 December 2006 (UTC)