Wikipedia:Reference desk/Archives/Mathematics/2006 September 26

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uncountable set

x is uncountable set, y is subset of x, if y is countable, is that x\y uncountable.

Yes but you did not ask for proof. Twma 01:55, 26 September 2006 (UTC)

yes, the proof may be like this, assume x\y is countable, so N ~ x\y, and x\y is finite set, so every element in N, there is corresponding element in x\y, and y is subset of x, and y is countable, so x must be finite set, but this is a contradiction with x is uncountable set, so x\y is uncountable.

That doesn't make sense. An easy proof notes that x is the union of y and x\y, and the union of two countable sets is countable, so one of the two must be uncountable. You seem to have many questions-- perhaps a textbook like [1] would help? Melchoir 05:12, 26 September 2006 (UTC)

countable set

P is the set of polynomial function with rational coefficient, is P is countable set?

Yes but you did not ask for proof. Twma 01:56, 26 September 2006 (UTC)

The set of polynomials of degree 1 is clearly isomorphic to Q (the rationals). The set of polynomials of degree 2 is clearly isomorphic to QxQ. And similarly for higher degrees: Q³, Q⁴, Q⁵, ... The total number is therefore Q¹ + Q² + Q³ + ... This is clearly 1/(1-Q). Since Q isn't finite, this is -0.

Alternatively, for an actually helpful answer, you might want to start at Countable set. -- Fuzzyeric 02:43, 28 September 2006 (UTC)

The quadratic formula to solve x

For quadratic equation ax² + bx +c:

x = ( -b +- sqrt( b² - 4ac ) ) / 2a

I asked my math teacher the other day how this equation was formulated, and he told me that you must complete the square of ax² + bx + c ... he showed me, but I didn't quite understand. When I formed two squares, it was the sum of two squares to equal zero, which has no solution. Could somebody take me through it?

See quadratic equation#Derivation. Conscious 04:35, 26 September 2006 (UTC)

To have a quadratic equation, we must have an equality:

$0=ax^{2}+bx+c.\,\!$

Perhaps before studying the formal manipulations of "completing the square" we should look at simpler cases. We are looking for values of x that make the quadratic polynomial (the right-hand side of the equation) equal to zero. In more geometric terms, we wish to find where the parabola defined by

$y=ax^{2}+bx+c\,\!$

crosses the x-axis. This may happen zero, one, or two places. Suppose b is zero:

$ax^{2}+c=0.\,\!$

Then if a is not zero, we can divide both sides by a to give

$x^{2}+c/a=0.\,\!$

Now subtracting c/a from both sides, we see that the square of x must equal −c/a. So long as a and c have opposite signs, we can take a square root to get one answer and take its negation for another.

However, if b is non-zero the parabola is shifted left or right so that the zero-crossings do not occur symmetrically on each side of the y-axis. For example, suppose c is zero:

$ax^{2}+bx=0.\,\!$

Since both terms of the polynomial contain a factor of x, we can rewrite this as

$(ax+b)x=0.\,\!$

If x is zero, clearly the polynomial is zero. Otherwise we must solve

$ax+b=0.\,\!$

Divide by a and subtract to reveal the unique solution, −b/a. So in this case the parabola is centered halfway between 0 and −b/a, at −b/2a. If we shift the x values by this amount, the parabola and its zero-crossings will again be centered around zero.

Another way to describe "completing the square" is, "centering the parabola". Conveniently, we have just found the solution! For, if we add c back in to the polynomial, it bumps the parabola up or down, but does not change its center. That is, the parabola will always cross zero at −b/2a plus or minus a square root.

All that remains is to decide what should be inside the square root. We make the shifting substitution

$x=X-{\frac {b}{2a}},\,\!$

to produced the centered form

$aX^{2}-{\frac {b^{2}}{4a}}+c=0.\,\!$

This has no X term, so we can adapt the solution from above. Divide by a, subtract the constant, and simplify to produce:

$X=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}.\,\!$

Thus the solutions for x, rather than X, are

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.\,\!$

To recap, the quadratic formula combines centering with square roots.

The question then naturally pops into our minds: Can we do something similar for cubic polynomials or higher? The answer is not at all obvious, took centuries to resolve, and produced some remarkable new mathematics. We can, with a little more difficulty, use shifting, square roots, and cube roots to solve cubics. With a great deal more difficulty we can similarly solve all quartics as well, though the method is so complicated it finds little use. But we can go no further; a polynomial of degree five or higher generally has no such closed-form solution.

So enjoy the quadratic formula. It's remarkably useful, and rather special. --KSmrq^T 16:51, 26 September 2006 (UTC)

equivalent class

if S is any infinite set, and x(- S, is that S is equivalent class to S\{x}.

I'm not quite sure what you're trying to say by "is that S is equivalent class to S\{x}", but if you're trying to ask whether S is in the same equivalence class (or, more succintly, "is equivalent to") S\{x}, you need to state what your equivalence relation is - for example, S and S\{x} are equivalent if the relation is "has the same cardinality" (since S is infinite), but they are not equivalent if the relation is "contains x". Confusing Manifestation 07:50, 26 September 2006 (UTC)

solve equations simultaniously

Solve equations 2x+3y=2 and 6x-y=-4 simultaneously.

What does this mean? How do I do this? I've probably been doing problems like this for about 5 years, but never got it well enough to remember anything. Thanks. 71.231.150.146 05:18, 26 September 2006 (UTC)

It means the value for X is the same in both equations and the value for Y is the same in both equations. There are many ways to solve them, the substitution method, addition method, graphing, using matrices, etc. StuRat 10:08, 26 September 2006 (UTC)

See Simultaneous equations. Conscious 05:38, 26 September 2006 (UTC)

In this case, I recommend you to read System of linear equations in the first place. JoergenB 10:00, 26 September 2006 (UTC)

Each separate equation defines a straight line in the plane; its solutions are precisely the points (x, y) lying on that line. The solution to both equations simultaneously is the coordinate where the lines intersect. Fredrik Johansson 11:26, 26 September 2006 (UTC)

It's hard to know how best to answer your question, since you do not make it clear what you already understand and what you want. The specific example you give is two equations in two variables, where each equation has at most one variable in a term. Is your question about this limited class of problems? Or are you also interested in more equations with more variables? Or perhaps more complicated equations such as quadratics? If you can "do them", what is it you'd like to remember? Are you saying you have to look up the solution technique each time, so would like help retaining the procedure? Help us help you. --KSmrq^T 17:17, 26 September 2006 (UTC)

Square Root?

What is the square root of France?

${\sqrt {(}}France)$

Thank you in advance for your assistance in solving this highly complex mathematical expression?

24.39.182.101 17:22, 26 September 2006 (UTC)

The usual definition of France is not mathematically rigorous, so we cannot use it to determine if √France is meaningful and its value. If seen as a variable which can take positive real values, then there is no way to simplify √France without knowing more about the value it represents. -- Meni Rosenfeld (talk) 17:34, 26 September 2006 (UTC)

France does not represent much of value. For most practical purposes, its square root may be considered zero. Fredrik Johansson 18:20, 26 September 2006 (UTC)

Think carefully before answering, as the verbal question would properly be notated

${\sqrt {\operatorname {France} }},\,\!$

which is not what we see. Perhaps what is meant is, "What country, if it applied itself twice over, could equal France?" But if so, it's a trick question, because France is, of course, incomparable. --KSmrq^T 19:07, 26 September 2006 (UTC)

Approximately $1.64(evil)^{\frac {1}{4}}$, of course. Factor out the root e, then apply the principle that money is the root of (all) evil. —AySz88\^-^ 22:24, 26 September 2006 (UTC)

I belive in this case, evil is the root of all money. Which, considering the widespread notion that money is the root of all evil, poses an interesting pair of questions: what is money? and what is evil? so that such an interesing pair of relations can occur.

There exist only complex numbers that, when squared, equal each other. So, economics and morality must both be complex. Black Carrot 06:44, 27 September 2006 (UTC)

Not exactly. The equations a²=b, b²=a have solutions in reals, (0,0) and (1,1). We can thus safely conclude that money = evil. -- Meni Rosenfeld (talk) 15:04, 27 September 2006 (UTC)

... or that neither money nor evil exist (or at least have no magnitude). -- Fuzzyeric 02:53, 28 September 2006 (UTC)

But we know that evil does exist. This leaves us with evil = 1, which means that evils is the fundumental unit. Therefore, everything in the world is made of evil. This explains a lot. -- Meni Rosenfeld (talk) 14:12, 28 September 2006 (UTC)

Pi

For some reason, I just can't comprehend how the Bailey-Borwein-Plouffe_formula calculates Pi. Does it really calculate individual digits? Could I easily calculate the, say, 84th digit? The article isn't too clear to me. Maybe it's not possible to explain in simple terms? --Russoc4 20:18, 26 September 2006 (UTC)

I don't really have time to answer this now, but if nobody gets back to you I'll try to explain this evening. But I have a question: is there a known, simple formula for calculating the decimal digits of π? It seems that to get an individual digit with these binary formulae you need to calculate all the preceding digits. –Joke 20:41, 26 September 2006 (UTC)

No, I know of no cheat for decimal digits. For hexadecimal digits, we can — as requested — calculate the 84th digit, say, without first accumulating all the preceeding hex digits. The reason this works is that the summation formula has two lucky properties: (1) it expands in powers of 1/16, and (2) it limits the contributions to any selected hex digit. Read carefully about the use of the modulo operator, which is key to the latter. --KSmrq^T 21:12, 26 September 2006 (UTC)

Why can't the hex values be converted to decimal? --Russoc4 23:24, 26 September 2006 (UTC)

They can, but to convert a hexadecimal expansion to decimal you need to have all its hex digits to find the last decimal digit. If I tell you the number is 3.???????????????????????DEADBABE, where each ? is a hex digit I chose not to reveal, there's nothing you can do with the deceased infant at the end. --Lambiam^Talk 23:45, 26 September 2006 (UTC)

At the mathworld article on BBP formulas, it says that it has been proven that the BBP method for PI cannot be adapted to any base that is non-binary (power of two)m, though there may still be a different method, as yet undiscovered, for finding decimal digits. - Rainwarrior 00:21, 27 September 2006 (UTC)

Shortest algorithmn for calculating infinite digits of pi (written in python programming language)

 #!/usr/bin/python
 from sys import stdout
 k, a, b, a1, b1 = 2, 4, 1, 12, 4
 while 1:
     p, q, k = k*k, 2*k+1, k+1
     a, b, a1, b1 = a1, b1, p*a+q*a1, p*b+q*b1
     d, d1 = a/b, a1/b1
     while d == d1:
         stdout.write('%d' % d)
         a, a1 = 10*(a%b), 10*(a1%b1)
         d, d1 = a/b, a1/b1

Sample run

 $ python short_pi.py 
 3141592653589793238462643383279502884197169399375105820974944592307816406286208998
 6280348253421170679821480865132823066470938446095505822317253594081284811174502841
 0270193852110555964462294895493038196442881097566593344612847564823378678316 and so on

202.168.50.40 00:05, 27 September 2006 (UTC)

It's easy to reduce this by one line. Here's a short astounding obfuscated C program by Dik T. Winter and Achim Flammenkamp:

 a[52514],b,c=52514,d,e,f=1e4,g,h;main(){for(;b=c-=14;h=printf("%04d",
 e+d/f))for(e=d%=f;g=--b*2;d/=g)d=d*b+f*(h?a[b]:f/5),a[b]=d%--g;}

However, it gives only 15000 decimals instead of an infinitude like in the Python program. --Lambiam^Talk 01:50, 27 September 2006 (UTC)

If hex digits of π are fine, I've written a Python program for calculating them with the BBP formula. The page explains how you rewrite the formula to calculate isolated digits. If a base 10 BBP-type formula existed for π, the same could be done for decimals, but as was pointed out above, there is no (arctangent-based) BBP formula for π in any non-binary base. By the way, you can calculate isolated decimal digits of ln(9/10). - Fredrik Johansson 04:41, 27 September 2006 (UTC)

That last note is interesting...can you elaborate? :-) --HappyCamper 17:09, 27 September 2006 (UTC)

Well, $\log(1-{\frac {1}{10}})\ =\ \sum _{k=1}^{\infty }{\frac {-1}{k\,10^{k}}}$. -- Fuzzyeric 03:00, 28 September 2006 (UTC)

Straight Angle's interior

My math teacher doesn't seem to know. Does a straight angle have an interior or exterior? Reywas92 21:41, 26 September 2006 (UTC)

Okay, a straight angle is 180 degrees, or π radians. A 120 degree angle has an interior angle of 120 and an exterior angle of 60. A 150 degree angle has an interior angle of 150 and an exterior angle of 30. Therefore, it would make sense for a 180 degree angle to have an interior angle of 180 degrees, and an exterior angle of 0 degrees. Makes sense to me at least. --AstoVidatu 22:42, 26 September 2006 (UTC)

Huh? Wouldn't a 150 degree angle have an exterior angle of 210 deg, not 30? Reywas92 00:20, 27 September 2006 (UTC)

Check out the Internal angle page. The interior/exterior angle diagram was the type of thing that I was thinking of. I'm pretty sure interior angles and exterior angles add up to 180 degrees. --AstoVidatu 00:51, 27 September 2006 (UTC)