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March 22

An algebra word problem

I have been trying to figure out this simple problem, but i think that there is some information either missing or hidden somewhere. Please guide me in this regard. Question: The sum of two digits of a number is 13. When the position of two digits is interchanged, a new number is formed that is 45 more than the first number. What are the numbers? —The preceding unsigned comment was added by 202.69.56.190 (talk) 04:03, 22 March 2007 (UTC).[reply]

A two-digit number can be expressed as

10x + y

and that number with the digits reversed can be expressed as

10y + x

Combined with the information you're given, you can create a system of two equations in two unknowns, which you then can solve for x and y. Plug those back into the first equation, and you've got your number. --Carnildo 04:17, 22 March 2007 (UTC)[reply]

The sum of two digits of a number is 13.

x + y = 13

When the position of two digits is interchanged, a new number is formed that is 45 more than the first number.

10y + x = (10x + y) + 45

Limitation of x and y

0 <= x < 10 and x is an integer

0 <= y < 10 and y is an integer

Now solve for x and y

210.49.121.183 09:10, 22 March 2007 (UTC)[reply]

Thanx a lot Carnildo and User:210.49.121.183 for your response. I shall be very grateful to you both. Thanx again man!

SET THEORY: Binary Connectives

I'm in a class in which one of the problems was given that NAND and NOR have truth values that can make them complete on their own, and to prove that nand and nor are both complete on their own, AND that the other 8 binary connectives ARE NOT complete on their own. I have the first part proven. My question to the reference desk is: Do I have to write a proof of induction for each binary connective to prove that each connective is not complete on its own? Does my professor want 10 pages of proofs for this one question? Please let me know if there is an easier way. Thanks, Panuelo24.210.245.242 05:19, 22 March 2007 (UTC)[reply]

Find counterexamples? —The preceding unsigned comment was added by 129.78.64.102 (talk) 05:52, 22 March 2007 (UTC).[reply]

What do you mean the other 8 binary connectives? I thought there were 16 two-variable connectives in total.

Anyway, for your original question, Urbán János, Matematikai logika (Typotex, Nemzeti tankönykvkiadó, 1998) contains a simple condition for a set of boolean functions to be complete (generate any boolean function). While it doesn't prove that these conditions are sufficent, and that proof wouldn't be easy, you don't need that direction, because you've already done that part of the task. It is, however, probably easy to prove that the conditions are neccessary, and that all of your functions break at least one of them.

The conditions are the following. There are five conditions given, and for each five there has to be a function in the set that fulfils it. (In your case, when you want a single function that is complete in itself, that function has to fulfill all five. Also, this theorem considers functions of arbitary number of variables, while you consider only two-variable connectives.) Here they are:

You need a function f such that f(1, ..., 1) = 0.
You need a function f such that f(0, ..., 0) = 1.
You need a function that is not monotonous, that is, for which there exist booleans x_1, ..., x_n, y_1, ..., y_n for which x_1 <= y_1 and ... and x_n <= y_n and f(x_1, ..., x_n) > f(y_1, ..., y_n). (Here <= and > means the natural ordering of booleans.)
There has to be a function that is not self-dual, that is, for which there exist booleans x_1, ..., x_n where f(x_1, ..., x_n) != not f(not x_1, ..., not x_n).
There has to be a function that is not linear, where we call linear the functions that can be written in form f(x_1, ..., x_n) = c_0 xor (c_1 and x_1) xor ... xor (c_n and x_n) for some constants c_0, c_1, ..., c_n.

– b_jonas 08:53, 22 March 2007 (UTC)[reply]

For the case of binary connectives, the first, second and fourth of the five conditions above already suffice to rule out everything except NAND and NOR. The induction proofs for establishing the necessity of each of these three conditions are very simple. If f(1,1), then the inductive hypothesis is that any function g built using only f also has g(1,1) = 1. Likewise for f(0,0) with 0 and 1 swapped. For a self-dual function like f(x,y) = NOT x, all functions you can build are self-dual. --Lambiam ^Talk 10:39, 22 March 2007 (UTC)[reply]

Probability

You have N objects, numbered 1 through N. You select one at random, note which it is, and put it back. On average, how long will it take before you draw object "A" followed immediately by object "B", or "B" followed immediately by "A"? --Carnildo 05:35, 22 March 2007 (UTC)[reply]

Uh, I thought I might be able to do this.

The probability that you pick A first is A/N.

The probability that you draw A, and then B, is A/N * B/N-1 = AB/N(N-1)

So if it takes you t time to draw N objects, and you draw objects subsequently at a uniform rate, then it will take you:

AB/N(N-1) units of time

If that's wrong, sorry. Tried my best, heh. --144.132.71.47 08:22, 22 March 2007 (UTC)[reply]

Probability of picking object A is 1/N

Probability of picking object A follow by object B is (1/N)*(1/(N-1)) = 1/(N*(N-1))

210.49.121.183 09:16, 22 March 2007 (UTC)[reply]

It is not clear from the question whether you could draw object A twice in a row, which I'll assume is indeed possible. It is also not quite clear whether the "success event" is ((A followed by B) OR (B followed by A))", or whether these two combinations constitute distinct success events. I'll assume the first, and also that A and B are different, which implies that N is at least 2. Consider a very long string of draws. For an adjacent pair, there are N² possibilities, of which 2 (AB and BA) are success. If all possibilities are equally likely, the chance of success is then 2/N². The average distance between successes is the inverse: N²/2. --Lambiam ^Talk 09:42, 22 March 2007 (UTC)[reply]

Nope, the average frequency of successes is indeed 2/N², but the expected time to first success is not the reciprocal of this. That's because successes are not independent: if you succeed at time t, then you clearly drew A or B last (rather than Z, say) so your chance of succeeding again at time t + 1 is greater. I get the answer N (N + 1)/2, by standard tedious Markov chain methods. There's probably a slick way too. —Blotwell 17:30, 23 March 2007 (UTC)[reply]

What I wrote is correct. I did not give a formula for the expected time to first success, but for the average distance between successes, which is the reciprocal of the frequency of successes. For example, in CABA, you have 3 pairs of consecutive draws (CA, AB, and BA), two of which are successes, at a distance of 1 apart. --Lambiam ^Talk 20:54, 28 March 2007 (UTC)[reply]

Math Reference Book

Is there an equivalent math reference book to the CRC Handbook of Chemistry and Physics? Thepatriots 06:35, 22 March 2007 (UTC)[reply]

Mathematics is a very broad field, but if you focus your attention on real and complex analysis with an emphasis on functions and computation, perhaps Abramowitz and Stegun qualifies. --Lambiam ^Talk 09:27, 22 March 2007 (UTC)[reply]

CRC also used to publish CRC Handbook of Mathematical Tables, although I am not sure if they still do. Should be easy to check though. Baccyak4H (Yak!) 14:34, 22 March 2007 (UTC)[reply]

A much broader reference work, though less for the layman, is the Encyclopedic Dictionary of Mathematics, 2/e, published by MIT Press, ISBN 978-0-262-09026-1. They also published the three-volume series Fundamentals of Mathematics, ISBN 978-0-262-52093-5, and so on (now out of print), which is kinder. An excellent Web source is the Springer Encyclopaedia of Mathematics. With a little luck and a lot of labor, in a few years Wikipedia itself will be the first place to turn. --KSmrq^T 22:07, 22 March 2007 (UTC)[reply]

A general math reference book I still use sometimes is Schaum's Mathematical Handbook of Formulas and Tables. However, if you have Mathematica, as I do now, about half of Schaum's is now obsolete. For example, using Mathematica to solve integrals works a lot better than flipping though old-fashioned tables of integrals. MrRedact 05:32, 23 March 2007 (UTC)[reply]

Question..

Is the logarithm to any real base of i a complex number (ie: ln(i))? This has greatly intrigued me. I can't see a way to express it in complex number format (a + bi), so is it not in the complex number system? What sort of number is this expression? Is f(i), where f is a transcedental function, a complex number for that matter? I have a feeling these are all "indeterminate forms." —The preceding unsigned comment was added by 144.132.71.47 (talk) 08:35, 22 March 2007 (UTC).[reply]

In our article Complex logarithm you can find that ln(i) = i^π/₂. Generalized to the multi-valued Log function, Log(i) = i(2k+¹/₂)π. So indeed, this value is a complex number and in fact a purely imaginary number – as is the case for the logarithm of any number whose absolute value is 1. If you pick another real base b, the effect is of dividing the natural logarithm by ln(b), so that does not change the imaginariness. I'm not sure I understand the question about transcendental functions. The function defined by f(x) = (x−i)exp(x) is transcendental, but f(i) = 0. --Lambiam ^Talk 09:23, 22 March 2007 (UTC)[reply]

Thinking about the exponential function and its relationship to complex numbers will make things blindingly clear to you, and if not, will elucidate Lambiam's comments above. —The preceding unsigned comment was added by 203.49.190.52 (talk) 09:29, 22 March 2007 (UTC).[reply]

Okay, thanks for your answers. Is there such a number involving i that can't be expressed in complex-number form? For example, the sine of the secant of the tangent of the logarithm of i? What would that sort of crazy number be? I read up on the relationship between the exponential function and complex numbers and I'm assuming you're addressing Euler's formula? —Preceding unsigned comment added by 144.132.71.47 (talk • contribs)

That's a good question. The complex numbers have some special properties. They are an algebraically closed field, which means that if a polynomial has coefficients which are complex numbers then all of its roots are also complex numbers - this doesn't work for the field of real numbers, for example. They are also a complete metric space, which means that every convergent sequence of complex numbers will converge to some complex number, not to something different. Rational numbers, for example, don't have this property - you can create a sequence of rational numbers that converges to an irrational number, such as sqrt(2).

Functions like sine, secant, tangent and logarithm can all be defined as infinite Taylor series. If we plug in a complex number at one end, then any value we get out at the other end will be some other complex number (sometimes we get nothing out at the other end - we can't find a value for the logarithm of 0, for example, because the series does not converge). So we can be sure that the sine of the secant of the tangent of the logarithm of i, if it exists at all, will be a complex number.

Bottom line is that you can't define a function using only complex numbers that take you outside of the complex numbers. Of course, you can construct other types of numbers by starting with stuff outside of the complex numbers in the first place - see quaternions for one example, and hypercomplex numbers for other examples. Gandalf61 11:20, 22 March 2007 (UTC)[reply]

Even weirder than

\log i

is

i^{i}=e^{i\log i}=e^{i\left(i{\frac {\pi }{2}}\right)}=e^{-{\frac {\pi }{2}}}\approx 0.208

, a real number! --Tardis 16:44, 23 March 2007 (UTC)[reply]

Just wanted to agree with Gandalf that this is an excellent question to be asking at this point. Tesseran 14:15, 27 March 2007 (UTC)[reply]

Very simple equation

I was messing around with squares and square roots, and came up with the equation $x^{2}-{\sqrt {x}}=14$ . Trial and error quickly proves the answer to be 4, but I cannot actually solve it. I have come to $0=x^{4}-28x^{2}-x+196$ , but I don't know how to continue. Could anyone offer a hint as to how to continue? Hyenaste ^(tell) 18:33, 22 March 2007 (UTC)[reply]

A place to start would be Quartic equation#Solving a quartic equation. Dugwiki 19:02, 22 March 2007 (UTC)[reply]

I'm not sure what you think it means to solve an equation, but it simply means to find the values for the unknowns that make the equation hold, which you did, since indeed

4^{2}-{\sqrt {4}}=16-2=14

. If you're looking for a systematic approach, then you did the right thing by bringing the equation into polynomial form. Basically, if

x^{2}-{\sqrt {x}}-14=0

, then whatever you multiply the l.h.s. by, the result will still be 0. So

(x^{2}-{\sqrt {x}}-14)(x^{2}+{\sqrt {x}}-14)=x^{4}-28x^{2}-x+196=0

. Maybe this is cheating, but if you have a polynomial

P(x)

and you know that

P(a)=0

, then you know that

x-a

is a factor of

P(x)

. Indeed,

x^{4}-28x^{2}-x+196=(x-4)(x^{3}+4x^{2}-12x-49)

. This gives us a root

x=4

, which we knew already, and the roots of the cubic polynomial

x^{3}+4x^{2}-12x-49

. A little exploration shows that this has only one real root for

x

about equal to 3.483336. However, this is not a root of the original equation. It is an adjoint root: a root of

x^{2}+{\sqrt {x}}-14=0

, brought in by the multiplication to get rid of the square root. Of course, if the original equation had been

x^{2}-{\sqrt {x}}=13

, which does not have such a nice root, you have to fall back on either solving the quartic equation algebraically, or solving it numerically. --Lambiam ^Talk 20:00, 22 March 2007 (UTC)[reply]

Following an idea put forward by an anon who since removed it, the substitution

x:=y^{2}

turns the original equation into a quartic

y^{4}-y-14=0

, with an additional condition

y\leq 0

. Solving this algebraically should be more pleasant. The quartic has an additional root of about −1.86637, but that violates the condition. (Not coincidentally, its square is about 3.483336.) The root 2 gives us, after squaring, the true solution 4. --Lambiam ^Talk 22:12, 22 March 2007 (UTC)[reply]

Guessing is allowed in finding solutions! In examples like this, there are systematic ways to guess. We also want to verify our guess (trivial here), and perhaps find or exclude other solutions.

Because of the square root, x cannot be negative. Both x² and √x strictly increase as x increases beyond 1, so we can easily show we have at most one solution.

Let us rewrite the question as, "What value of x makes x²−14+√x equal to zero?" If we guess 1, we get −14; and if we guess 9, we get 64. This function has no abrupt jumps (it is continuous), so somewhere between the negative and the positive must be a zero. This simple statement is very powerful!

Once we have bracketed a zero, one strategy is bisection. That is, look midway between the ends, which at this step would be 5. When we try that, we get another positive result, approximately 8.76. But that means we have cut our range of possibilities in half, from somewhere between 1 and 9, to somewhere between 1 and 5. If we do this repeatedly, we should get a sufficiently precise answer.

Often we have faster methods. For example, this function is not only continuous, it is also differentiable. Thus at any point of interest we can find a straight-line approximation, and ask where that line crosses zero. Since the derivative is 2x−¹⁄_2√x, at x = 5 the slope of the line is approximately 9.77. So we expect that 5−^8.76⁄_9.77, approximately 4.1, will be a better guess. And, as we know in this case, it is. This is the essence of Newton's method for finding the zeros of a function. If we take three more steps like this, we get x = 4.00000000000000787, which is an excellent approximation. (The bisection method would still have a long way to go.)

As it happens, even when our function is merely a polynomial, we usually have no choice but to use systematic guessing when the degree is greater than 4. In fact, it is what we use to find an approximate value for √2, a zero of x²−2. --KSmrq^T 23:02, 22 March 2007 (UTC)[reply]

Thank you Dugwiki for pointing me to quartic equation, and especially thank you Lambiam and KSmrq for showing and explaining some methods of solution. I am quite pleased with my math here which I could not have done without your guidance. Hyenaste ^(tell) 18:25, 23 March 2007 (UTC)[reply]

Another system they haven't mentioned is to assume the answer will be easy to get, and explore that option first. For instance, if you assume the answer will be a whole number, it's not hard to see what it would have to be. It would have to be square, that is, x=z^2 for some whole z, because of the square root, and it would have to be positive. That gives z^4+z=14. If you factor that, you get z(z^3+1)=14. That's not useful if z could be any real number, but to be a whole number it would have to be a factor of 14. Tossing the four options in, only z=2 works, which gives x=4. If you start with the wholes, then move to the rationals, then to the reals, and finally to the complexes, you'll never risk solving the entire quartic formula just to get x=4. Black Carrot 16:25, 24 March 2007 (UTC)[reply]

Incidentally, that also demonstrates that there is no fractional answer. If z was a fraction, the denominator would have to be a whole factor of 1, in other words equal to 1, making it a whole number. Black Carrot 16:35, 24 March 2007 (UTC)[reply]

Black Carrot is referring to the rational roots test (though if you don't understand why BC's assertion is true, you might think about it on your own before clicking that link). Tesseran 14:06, 27 March 2007 (UTC)[reply]

Terminology question: what is your preferred way of saying this?

Let $R$ and $S$ be two $n$ -ary relations. What is your preferred way of saying

$R(x_{1},x_{2},\ldots ,x_{n})\Rightarrow S(x_{1},x_{2},\ldots ,x_{n})$ , for all $x_{1},x_{2},\ldots ,x_{n}$

in English? (Do you say, for example, " $S$ subsumes $R$ "? " $R$ is a stronger relation than $S$ "? " $R$ is a subrelation of $S$ "?) --64.236.170.228 19:32, 22 March 2007 (UTC)[reply]

I would avoid "subsumes" because I'd be afraid that many readers wouldn't be able to keep straight which subsumes which. "Subrelation" is fine if the readers, in the context, think of the relations as subsets of some Cartesian product, but then why not write R ⊆ S?. If they have been encouraged to think of them as predicates, "stronger" seems better. Personally I might settle for "universally implies". --Lambiam ^Talk 20:17, 22 March 2007 (UTC)[reply]