Wikipedia:Reference desk/Archives/Mathematics/2006 November 24

= November 24 =

AREA
how do I find the area of a circle ? how do I find the area of a square?
 * See circle and square for answers and explanations. ☢  Ҡ i∊ ff   ⌇  ↯  06:35, 24 November 2006 (UTC)


 * (Tongue in cheek) First find the area of the square, then throw darts to find the area of the circle.
 * Millennia ago, Archimedes proved that the area of a circle equals the area of a right triangle with radius as altitude and circumference as base. He then showed how to bound that area from above and below by circumscribing and inscribing regular polygons, touching the circle either at edge midpoints or at vertices, respectively. Read here for how he did it, in his own words (translated). --KSmrqT 07:04, 24 November 2006 (UTC)

Set of modular equations
Hello. I need some help with solving a set of modular equations. I would know how to solve it if the equations were not modular, but since they are, and the solution I got is not in integers, I probably do something wrong.

The equations are:

9*x+0*y+12*z=19(mod26)

4*x+18*y+z=7(mod26)

14*x+13*y+3*z=23(mod26)

I have also this 2-equations set, and I have the solution to this one, so if anyone prefers to help me with this one instead of the previous one, it may be helpful too. The equations:

9*x+20*y=3(mod26)

11*x+24*y=11(mod26)

while the solution is:

x=11

y=3

But as I said before, I don't know how to solve it myself. thanks, Rachel Wa L.


 * It's been a while since I did modular arithmetic, but it might help to note that the expression z=3(mod 26) is equivalent to saying there exists some integer k such that z=26*k + 3. So your second problem, for example, is equivalent to saying that there exists integers a and b such that


 * 9*x+20*y=26*a + 3


 * 11*x+24*y=26*b + 11


 * Now if you solve those equations for x and y, you'll get formulas for them in terms of integers a and b. Then you can try and find specific integers a and b that produce integer values in those formulas.  Not sure if that helps, but good luck.  Dugwiki 17:40, 24 November 2006 (UTC)


 * If you know how to solve the equations without the "(mod 26)", then you know how to solve them with "(mod 26)". The beauty of modular arithmetic is that the normal approach works perfectly, with one modification. We have two equations, so the first thing we'll do is cancel the y terms:


 * 9x+20y=3 (mod 26)
 * 11x+24y=11 (mod 26)


 * Multiply the bottom equation by 3, then simplify the coefficients so the numbers don't get too big. (The reason I picked 3 is that 24*3=72=2*26+20 so 26*3=20 (mod 26). I didn't just guess this, though: I noted that 24 = -2 (mod 26) and 20 = -6 (mod 26), and seeing that 3*(-2)=-6 (mod 26) is easy.)


 * 3(11x+24y)=3(11) (mod 26)
 * 33x+72y=33 (mod 26)
 * 7x+20y=7 (mod 26)


 * Now we do what we always do, subtract one equation from the other. In this case, we'll subtract the second (modified) from the first equation.


 * . 9x+20y=3 (mod 26)
 * - 7x+20y=7 (mod 26)
 * .  2x    = -4 (mod 26)
 * .  2x    = -4 (mod 26)


 * Now there are actually 2 solutions to this equation, which may seem strange: obviously -2 satisfies 2(-2)=-4, so noting that -2 = 24 (mod 26) we have one solution. But -4 = 22 (mod 26), and it's also clear that 2*11=22. So both 11 and 24 satisfy 2x = 22 (mod 26). Anyway, the next step is always substitution, so let's plug one of our solutions back in: choose x=11, and plug it into 7x+20y=7 (mod 26).


 * 7(11)+20y=7 (mod 26)
 * 77+20y=7 (mod 26)
 * 20y=-70 (mod 26)
 * 20y = 8 (mod 26)


 * We're going to run into double solutions again: rewrite this as (-6)y = -18 (mod 26). We can divide by -3 to get 2y=6. This has the obvious solution of 3, but is also satisfied by 16. So we have two pairs of solutions: [x=11,y=3] and [x=11,y=16]. (If you look above, we choose x=11, but we also could have used x=24. If we had substituted x=24 in, we would have found the equation 20y = 21 (mod 26), which has no solutions. To see this, note that even numbers are only congruent to even numbers mod 26, and 20y will certainly be even, but 21 is odd.) So [x=11,y=3] and [x=11,y=16] are the final answers. You should check that these both really do work.


 * If you're running as fast as you can away from these "double solutions", don't. Let's see where the problem arises. It happened when we had equations of the form 2x=4 or 2x=6. The number 2 is special here, and the reason is that 2 divides 26. With numbers that don't divide 26, like 3 or 5, there's no problem: the equation 3x=4 or 5x=6 has a unique solution mod 26. With 13, however, which does divide 26, we would have had the same problem: 13x = 0, for example, has not just two but thirteen solutions! (Can you see why it's all the even numbers?) If you know what a prime number is, you can see that it would be nice to do modular arithmetic modulo 7, for example, or modulo 23, because then there are no numbers that divide the modulus.


 * In conclusion: 1) with modular arithmetic you can add, subtract, and multiply to your heart's content. 2) You can also "divide" by any number that does not divide the modulus (in this case, 26) without worrying. 3) When you "divide" by a number which does divide the modulus, you might have multiple solutions (2x = 4) or no solutions (2x = 3), so be careful. [When I say "divide", I mean taking an equation like 2x=4 and canceling out the 2 to solve for x.] Hope this helps. Tesseran 19:31, 24 November 2006 (UTC)


 * The second set of equations, for which you have a solution, can be approached using Bézout's identity (though Brahmagupta probably deserves the name), which is one way to solve linear Diophantine equations. Of course, here we want not just integer solutions (which is the Diophantine demand), but modular solutions. Still, one immediately yields the other. Both sets of equations are linear, which makes solution much easier. We have two tools of interest. The first was mentioned in another recent post, the Gröbner basis computation using Buchberger's algorithm, here restricted to integers. For the three-variable problem we can find the basis
 * $$\begin{align} &893 - 2031 z,\\ &52 - 2031 y,\\ &3097 - 2031 x . \end{align}$$
 * The second tool is an integer variation of the QR decomposition, which reduces integer equations to Hermite normal form. Writing the three-variable problem in matrix form as
 * $$\begin{bmatrix}9&0&12\\4&18&1\\14&13&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}19\\7\\23\end{bmatrix},$$
 * we can convert the matrix to upper triangular (Hermite normal) form by multiplying on the left by unimodular matrix U, where
 * $$U = \begin{bmatrix}89&52&-72\\58&34&-47\\200&117&-162\end{bmatrix}.$$
 * Note that U is an integer matrix (and so is its inverse), which we put to good use. Multiplying both sides of the equation by U gives
 * $$\begin{bmatrix}1&0&904\\0&1&589\\0&0&2031\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}399\\259\\893\end{bmatrix},$$
 * from which we can immediately read off 2031z = 893, a result we recognize from the Gröbner basis. Since this equation holds in the integers, it must also hold in the integers modulo 26; thus
 * $$ 3z \equiv 9 \pmod {26}, \,\!$$
 * and we deduce that z = 3 will work. As in Gaussian elimination, we can proceed to find y, then x.
 * Although I have not provided a detailed algorithm for finding a Gröbner basis nor one for computing a Hermite normal form, knowing what to look for should allow a web/wiki search to fill in the gaps. Thus linear equations using modular arithmetic permit systematic solution methods which are effective and efficient. (Nonlinear equations are not so easy!) --KSmrqT 21:19, 24 November 2006 (UTC)

axioms of mathematical logic
i'm a math student on third year. i took many courses in analysis and algebra but there won't be a logic course this year in my university. so i decided to learn logic from a book. now i have a problem/question, but i'm not sure how to phrase it exactly. so i'll have to give some background.

the way i learned math until now was:
 * first was taught the language of predicate logic, and the basic rules to manipulate expressions (i think they're called inference rules).
 * then given the list of ZFC axioms.
 * then we built everything in that setting (all analysis and algebra i learned). all objects i dealt with until now were sets defined with ZFC.
 * in set theory course the ordinals and cardinals were defined too, all based on ZFC. also proved the transfinite induction and zorn's lemma.

and for me it's really fine. i like math to be formal just like that.

now i started learning logic and in the book they talk about expressions as if they are themselves some objects they can manipulate. they chose a language and define the set of all expressions over the language. but then saying they can chose the axioms to use only after that. so my problem is more or less: how can the set of expression over a language be a set if i haven't yet chosen the axioms (be ZFC or others) to define it as a set? they use zorn's lemma on sets of sentences to prove a compactness theorem. if so, can it be done under other settings too?
 * also like the valuation functions. a function f:A->B must be a subset of AxB (or another set in some equivalent way, but a set nontheless) therefor a set by ZFC.
 * by what axioms the set of constant letters in the language is a set?
 * in what sense a model is a set?
 * is it all done already inside the setting of ZFC or some other list of axioms?

i tried to look for this here on wikipedia, going through the list of mathematical logic articles, nothing seemed to answer that. it's like in the book i read.

if it's long or complicated at least direct me to where i can find answers. --itaj 16:40, 24 November 2006 (UTC)


 * In studying formal logic you use mathematics to describe formal systems. These systems are a bit like games such as peg solitaire: there are precise rules on what are possible "configurations", and what are the allowed "moves" to go from one configuration to the next, and there is a notion of some goal, a configuration having a precisely defined property. In a logic the rules on wffs determine the possible configurations, which (in simple kinds of logic) are sets of wffs. The "proof rules" define the moves you may do. Typically, proof rules have one or more "antecedents": schematic wffs (with dummies). Proof rules with zero antecedents are also called "axioms" or "axiom schemas". This is just a name. In reasoning about such logic games, you use whatever setting you are comfortable with, just as you would do when reasoning about peg solitaire. If you are of the ZFC persuasion, use ZFC. Ignore the fact that while there are certain axioms you believe in (the ZFC axioms), in these objects you are studying certain kinds of game rules have confusingly also been called "axioms", which you are not required to believe in, anymore than people studying comparative religion need to believe the dogmas of the religions they study. --Lambiam Talk  20:16, 24 November 2006 (UTC)


 * thanks for the reply. but, i don't think i quite understand what you went at here. i don't believe in any set of axioms at all. i think of math as a game that starts with chosing inference rules and axioms then proving theorems. what i said about ZFC is that these are the axioms chosen in my studies (say: my univesity teaches math of ZFC axioms, i don't suppose it should surprise you :) ). my problem is that all this was fine while i studied set theory, algebra and analysis, but now that i start studying logic, it's not clear. maybe i don't understand your answer, but still i can't see what are the rules of the reasoning that is done on the wffs? what do you mean by set as in "sets of wffs", is it the "ZFC set" or what? --itaj 20:49, 24 November 2006 (UTC)


 * Perhaps an analogy will help clarify the situation. Let us suppose you speak French, which you learned growing up in Quebec. In your youth the subjects about which you spoke were music and cinema. (You were precocious, of course.) Later in life you notice that the French spoken in Paris is not quite the same as that you speak, and then there is that strange babble called English. So now you study language, and speak about those studies. And what language do you use to speak about language? Why French, of course.
 * Just so, you learned to do mathematics using a system of logic. Now you wish to formally study systems of logic. Naturally you will use the tools you know, including logic and set theory. All is well.
 * Some fascinating results appear when mathematics talks about itself. This is the source of Gödel's incompleteness theorems, for example. We must, of course, keep our minds and our notations clear so that we do not confuse the mathematics we are studying with the mathematics we are using to conduct the study.
 * We can also choose to adopt foundations other than logic and set theory. One idea, which started in topology(!), is the formal object known as a topos, which is a kind of category (as in category theory). We can first define a category and a topos, and then use those tools to describe logic and sets. We have the freedom to use a topos in which the logic is not a classical logic; for example, it may be an intuitionistic logic. We can also explore variants of set theory.
 * For your reading pleasure, some texts available on the web are:
 * Stefan Bilaniuk, A Problem Course in Mathematical Logic.
 * Robert Goldblatt, Topoi, the categorial analysis of logic.
 * Michael Barr and Charles Wells, Toposes, Triples and Theories.
 * Deirdre Haskell, Anand Pillay, and Charles Steinhorn (eds.), Model Theory, Algebra, and Geometry.
 * These may serve to give a taste for the interplay between logic and other areas of mathematics. --KSmrqT 22:27, 24 November 2006 (UTC)


 * OK, we're playing a game with axioms and inference rules. Now, axioms are particular formulæ (or, for axiom schemas, rules for constructing formulæ) while inference rules describe how to transform formulæ into other formulæ.  If we want not just to prove particular theorems (i.e. generate particular formulæ) but to prove statements about what formulæ can be generated, we need a formal system to represent the formulæ in.  The easiest such system to use is one where formulæ are strings of symbols, and collections of formulæ are sets.  Now, the game we're playing might use axioms and formulæ that appear to be about ZFC - but that's just the game we're playing; the actual process of proving statements about the game will use an entirely separate logic.  The meta-logic can use ZFC, but it doesn't have to; most systems of logic and set theory will allow you to prove the same things about the game.  It's perfectly acceptable, in fact, for the meta-logic to be informal; since (as I'm sure you've heard) a system capable of proving itself to be consistent is also capable of proving itself to be inconsistent, any process whereby mathematicians convince each other that a particular logic is suitable for playing games with must use informal reasoning at some point. EdC 00:12, 25 November 2006 (UTC)


 * I think what Itaj has run into here is one of the central problems of the formalist approach. It simply doesn't work for human reasoning. There are lots of other problems with the approach, but this is the one drawn into sharpest relief here. You have to be able to distinguish between object language and metalanguage, between the things you're talking about (such as sets) and the tools you use to talk about them (formal language) or draw conclusions about them (axiomatics, for example).
 * So my advice is, for the moment, think about sets not as "defined by ZFC" (that's a category error anyway; axioms don't define things), but rather as real objects whose behavior conforms to the ZFC axioms. You don't necessarily have to believe it; think of it as a convenient fiction, if you like. Over time you can ask yourself whether mere fictional reference is a sufficient explanation for the observed facts they predict, but that's a discussion for another day.
 * Then, if it makes you feel better, you can go back and reinterpret everything you said about sets, while thinking of them in real objects, in terms of formal sentences in a formal metalanguage, and use a formal theory to derive those sentences. --Trovatore 04:15, 25 November 2006 (UTC)


 * thanks for the help. i must say it's still not exactly clear, but you all gave me some base of what are the things i should put in order. --itaj 21:38, 28 November 2006 (UTC)

Second derivative
Why is the second differential $$ (\frac)$$ pronounced 'd two y, by dx squared', not 'd squared y, by dx squared'? Eŋlishnerd ( Suggestion? | wanna chat? ) 23:44, 24 November 2006 (UTC)


 * It's probably a subconcious recognition that the letter "d" isn't actually a variable that is getting squared, but is simply shorthand for "derivate" or "differential". On the other hand, the letter "x" actually represents a variable that could potentially be squared.  Just a guess though. Dugwiki 23:54, 24 November 2006 (UTC)


 * A guess: People often manipulate "dx" as if it were a variable (I don't understand it well enough to know whether such manipulation is really valid), so I guess you can square "dx" if you treat it as a variable. "d" seems to be considered an operation, and squaring an operation doesn't make sense.
 * Personally, I've only ever heard it read "the second derivative of y with respect to x", but maybe I haven't been exposed to it enough yet. —AySz88\ ^ - ^ 03:02, 25 November 2006 (UTC)


 * I've been exposed, but I would prefer "second derivative" as well. Taking the derivative with respect to x is an operator; its input is a function of x, say f, and its output is another function of x, d&frasl;dxf. Applying the operator twice produces

\begin{align} \frac{d}{dx} \left(\frac{d}{dx} f\right) &{}= \frac{d^2}{dx^2} f\\ &{}= \frac{d^2 f}{dx^2} \end{align} .$$
 * We do also see differentials like
 * $$ du^n = n u^{n-1}\,du, \,\!$$
 * which make sense, but take separate explaining. --KSmrqT 03:38, 25 November 2006 (UTC)


 * I've certainly heard and repeated "d squared y, d x squared", and never heard "d two y, d x squared". "Squaring" an operator can make sense if we think of composition as our operation, which combines two operators (or mappings or functions of whatever kind) into one.  This is a common notation for iterating functions, which is consistent with the notation $$f^{-1}$$ for the inverse function of $$f$$.  If $$f^{-1}(x)$$ means to apply $$f$$ "backwards", as it were, then $$f^2(x)$$ can certainly mean to apply $$f$$ twice - so $$f^2(x)=f(f(x))$$.  Similarly, $$\frac{d^2}{dx^2}\left[ f\right] = \frac{d}{dx}\left[ \frac{d}{dx}\left[ f\right] \right]$$. -GTBacchus(talk) 04:01, 25 November 2006 (UTC)


 * Its 'mispronounced' d 2 squared by people who dont know the history of the notations. Newton used d' ( or D-Prime ) It was really messy. I think that Lebinowitz used the d^2 notation, and had clearer writings, but didnt get as far, ( althought neither got very far with the infintessimals ). The notation is really the second diretive with respect to y of x. That is how the professor who explained the question on notation would say it. (yea, Walters at Contra Costa College. I should give him a call...) Artoftransformation 07:10, 25 November 2006 (UTC)


 * Do you mean the second derivative "of y, with respect to x"? -GTBacchus(talk) 08:46, 25 November 2006 (UTC)


 * That's how I say it, too. StuRat 06:33, 29 November 2006 (UTC)


 * As a teach of an AP Calculus class, I will tell you that pronouncing it "squared" is extremely wrong. Not only is it just incorrect, it actually totally misrepresentss the meaning of notation.  Please read the previous three comments for clarification.  Finally, let me add, that I would be very upset with my students for such a mistake. Justin Custer 16:26, 30 November 2006 (UTC)

Implicit Function
How would I graph an equation in the form: $$a^2 x^3 + axy^4 + a^2 y^2 = 0$$? Could someone point me in the right direction or perhaps walk me through it? I've spent hours on this problem and looked all over the internet. Thanks. -- Sturgeonman 00:24, 25 November 2006 (UTC)
 * Are you sure that's the equation in question? Seems a bit unfriendly.
 * Anyway, here's how I plot that (thanks to the Gnuplot NSFAQ):
 * First, solve for a:

gnuplot> f(x, y) = -(x*y**4)/(x**3 + y**2)
 * Then plot for values of a from -1 to +1:

gnuplot> set xrange [-2:2] gnuplot> set yrange [-2:2] gnuplot> set view 0,0 gnuplot> set isosample 100,100 gnuplot> set size square gnuplot> set cont base gnuplot> set cntrparam levels incre -1.0,0.25,1.0 gnuplot> unset surface gnuplot> splot f(x,y)
 * HTH. EdC 02:35, 25 November 2006 (UTC)

It is the correct equation, and it's a nasty one. However, I need to know how to graph it by hand. A better phrased question would have been, is this equation able to be put in explicit form? Can it be solved for x or y? What if a=16 for an intial value? Can it be graphed if the equation = 0 ? -- Sturgeonman 02:57, 25 November 2006 (UTC)


 * Since y occurs in even powers, the graph must be symmetric about the x axis (if y0 is a solution, so is −y0). Using the quadratic equation we can solve for y2 as a function of x,
 * $$ y^2 = \frac{-a \pm \sqrt{a^2 - 4 a x^4}}{2x} . $$
 * We may then plot two values of y for each right-hand side. For a positive value of a, we must bound x2 by 1&frasl;2&radic;a to produce real results. We must also constrain the right-hand side to be positive. (Note if a is zero, then all values of x and y give solutions.)
 * The program GrafEq is a robust tool for plotting implicit functions; you might want to compare its results to yours. --KSmrqT 05:08, 25 November 2006 (UTC)


 * Another approach is to consider asymptotic behaviour. For example, (0,0) is clearly a solution; as x and y tend to 0, the $$axy^4$$ term disappears fastest, so the graph approaches that of $$a^2 x^3 + a^2 y^2 = 0$$ i.e. $$y^2 = -x^3$$. Also consider e.g. x tending to 0 with y large: asymptotic to $$x = -a/y^2$$; and other asymptotes e.g. x tending to plus or minus infinity. With that and the bound on $$x^2$$ KSmrq demonstrated, we can draw a load of asymptotes and know that the graph runs between them. EdC 12:01, 25 November 2006 (UTC)


 * Indeed, if a is positive, the only values of x that produce solutions are $$[-(a/4)^{1/4}, 0]$$. Each x-value inside that range has 4 values of y; two on $$y^2 = -x^3$$ and two on $$x = -a/y^2$$. So, just draw those graphs plus a vertical tangent point at $$x = -(a/4)^{1/4}$$, and run curves between them. You get something that looks like a moustache turned on its side. EdC 12:07, 25 November 2006 (UTC)


 * [[Image:X3+xy4+y2.gif|thumb|right|200px]]There are algorithms which can plot implicit equations directly, I've developed one as part of my SingSurf package. Basically you use a recursive algorithm based on quadtree subdivision, you can use a bernstein polynomial representation of the curve which allows a test for whether the polynomial has a zero in a given domain. The tricky part comes if the curve has a singularity. Raytracing approaches can also be used. --Salix alba (talk) 14:56, 25 November 2006 (UTC)

I tried the quadratic technique shown above, but I didn't get the same graph. Basically, how would I solve this simplified version for x or y: $$ax^3 + xy^4 +ay^2$$


 * Wait - forget it. I got the correct graph on a different calculator; my Ti-84 apparently sucks. Immense thanks to everyone who helped; I think I'll write an article as repayment. -- Sturgeonman 16:07, 25 November 2006 (UTC)
 * For negative values of a the graph has three components. --Lambiam Talk  16:28, 25 November 2006 (UTC)