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October 25

Math

What is the factors of 24

Half of me is saying that I shouldn't answer this because it looks like a homework question, while the other half says I should... But I guess I won't. I'd like to point you to the Divisor article if you don't know what a factor is.

Anyway, the factor of a number (X) are all whole numbers other than 0 which can be multiplied by another whole number (which is also a factor of X) to get X, such as y (8) and z (3) which, multiplied together, result in yz = x (8 x 3 = 24). There isn't a "formula" to find factors of numbers aside from dividing the number by all whole numbers from 1 to whatever until you get one that works - that's what supercomputers are for (see RSA). The number ends in 4, so you can assume that it's divisible by 2, possibly by 3, 4 or 6 but not by 5 or 7, etc. But that's not needed for a small number like this. Also, the smallest factor is always 1 (assuming it's a positive nonzero integer) and the largest factor is always the number itself (1 x 24 = 24).

Just out of curiosity, how old are you and what grade are you in? I'm just asking because this is middle-school stuff. Corporal 02:10, 25 October 2006 (UTC)[reply]

To answer your question, I suggest that you read these articles:

The fundamental theorem of arithmetic establishes the importance of prime numbers. Prime numbers are the basic building blocks of any positive integer, in the sense that each positive integer can be constructed from the product of primes with one unique construction. Finding the prime factorization of an integer allows derivation of all its divisors, both prime and non-prime.

202.168.50.40 05:07, 25 October 2006 (UTC)[reply]

Yo, any whole number you can divide it by to make a whole number is a factor. End of story.

ahem....since this is middle school stuff, i should point out that just because it ends in 4 doesn't necessarily mean you CAN'T have 7 as a factor. 14. Lrpelkey 11:39, 28 October 2006 (UTC)[reply]

Economic Production function to profit function:

I would like to make sure I am accurate in my technique, am I right?

Once you have the cost function, formulated from a production function, there are two ways to find maximum profit:

You can either plug what optimal Q is equal to w.r.t. the constraint, and then maximize with respect to inputs X and Y, and then replug these optimals into the constraint for the optimal Q...

Or, you can plug in the cost function, and merely derive w.r.t. Q, then solve for the optimal Q?

After both of these are solved for Q, just plug them into: pi = Q(PQ) - X(pX) - Y(pY).

Correct? Thanks so much, ChowderInopa 02:49, 25 October 2006 (UTC)[reply]

You are not making much sense to me. What is X , Y , P and Q ? What is pi (I don't think you mean 3.1416)? Is lower case p different from upper case P? What is a production function? Is it a function of quantity? 202.168.50.40 05:01, 25 October 2006 (UTC)[reply]

(after edit conflict) The method you propose is not entirely clear. What is the relationship between the cost function and the production function? What is the role of the constraint? Is it a constraint on the inputs (for example X+2Y < 100)?

In general you cannot do optimization unless you actually know the objective function. If it is given analytically, by an equation, you may hope to use methods such as that of Lagrange multipliers to find an optimum. Or perhaps you can use quadratic programming. Otherwise, you could resort to numerical techniques such as hill climbing, modified to obey the constraint. --Lambiam ^Talk 05:10, 25 October 2006 (UTC)[reply]

Are all real numbers with multiple decimal representations rational?

Today's featured article, besides drawing out an unusually broad array of cranks, got me to thinking. Obviously, 0.999... = 1, 0.34 = 0.3399999..., 330 = 329.999..., etc. But all of these numbers with multiple decimal represenations are constructed by taking a simple terminating decimal number and replacing a final digit n with the digit sequence (n-1)99999... Are there any other types of numbers that also have multiple representations? --Pyroclastic 07:32, 25 October 2006 (UTC)[reply]

e (number) = 1 + 1/1 + 1/2! + 1/3! + ... Which is a basis for Euler's formula, and actually quite elegant. Also look at pi. 169.229.89.53 09:25, 25 October 2006 (UTC)[reply]

To make it even more elegant, why not write it as e (number) = 1/0! + 1/1! + 1/2! + 1/3! + ... StuRat 17:29, 25 October 2006 (UTC)[reply]

Um, but that's not a definition in terms of an infinite decimal sequence?--140.180.157.26 07:18, 26 October 2006 (UTC)[reply]

Well, there's 0.000… = -0.000…, although in some authors' development of decimals, this is written 0+.000… = -1+.999…, so it may or may not fit the pattern. Anyway, every number with multiple decimal representations is rational, yes: in particular, a rational which can be represented by a fraction with a power of 10 in the denominator. Melchoir 09:31, 25 October 2006 (UTC)[reply]

Well, I've no proof but a strong hunch that there are an infinite number of series representations of any rational number, since there's an infinite number of series that converge. But the number is probably finite for the irrationals? --BluePlatypus 12:37, 25 October 2006 (UTC)[reply]

It draws cranks for two reasons. One, there will always be some for such a topic. Two, the article threw away the instructional care of earlier versions to throw up a façade of being “encyclopedic”. Far from representing the best of our mathematical writing, it is an object lesson in what not to do. (Pardon me while I dismount my soapbox.)

Do you mean to restrict attention to decimal expansions? If so, let's review the options. A number like 0.84375, the finite decimal expansion of ²⁷⁄₃₂, can be normalized to an infinite expansion by adding trailing zeros. Such numbers are always rational, and — if nonzero — their denominator contains only prime factors of two and five. Exclude zero, and we have the only real numbers that admit two decimal expansions, the second one having trailing nines.

If we do not restrict ourselves to decimal expansions, every real number has numerous representations. We have alternatives like expansions in a radix of our choosing, continued fractions, and so on. We can represent a real algebraic number as a select root of a polynomial with integer coefficients. Expanding our scope, we can write down the digits using a different alphabet, like Devanāgarī or Hangul. The idea of a representation is broad!

As far as the series question, no, every real number is the limit of an infinite number of different series. A proof is trivial: Take any given series and preface it with two new terms that cancel each other; the result is a new series with the same limit. --KSmrq^T 14:13, 25 October 2006 (UTC)[reply]

A trivial proof but for a trivial case. Not a very interesting or enlightening answer. --BluePlatypus 20:23, 25 October 2006 (UTC)[reply]

Nonsense. You speculated that any given irrational number was the limit of a finite number of series. I showed that is false. More elaborate variations could be given, but there is no need; false is false. If you claim every mammal has a placenta, and I show you a platypus (!), must I also point out a Virginia Opossum, and a Tasmanian Devil, and a Monito del Monte? --KSmrq^T 01:59, 26 October 2006 (UTC)[reply]

No one seems to have directly answered the original poster's question. The answer is yes, all numbers with two distinct decimal representations are rational, and in fact are of the form

{\frac {n}{10^{m}}}

, where n and m are integers. --Trovatore 17:33, 25 October 2006 (UTC)[reply]

Oh, actually I guess Melchoir did answer it; sorry about that. --Trovatore 17:41, 25 October 2006 (UTC)[reply]

Is there some reason you ignore my answer? Also, your form of the fraction is misleading, because n cannot always be relatively prime to 10, which we would expect from the presentation of a rational number. The very example I gave, 0.84375 = ²⁷⁄₃₂, illustrates that. Sheesh. --KSmrq^T 18:39, 25 October 2006 (UTC)[reply]

Well, you sort of went on a bit, KSmrq, and your answer to the original question was in the middle of it all. The OP's question had a direct answer—"yes"—which no one bothered to give directly. --Trovatore 18:45, 25 October 2006 (UTC)[reply]

My answer mirrored the question sequence: cranks, decimal expansions, other representations, series. And if you think the answer to the poster's question is "yes", maybe you should go back and decide which question you were answering, the one in the title or the one in the post itself. Oy vey! --KSmrq^T 01:24, 26 October 2006 (UTC)[reply]

Help with diff eq

Hey,

I'm looking for a solution for next differential equation:

(b'(t))²/2 = A*b - B*b^4/4 + U²/2 , with A,B en U positive constants.

I actually need an explicit formula for b(t)! I would be very happy if someone could help me with this problem.

Solutions can be send to: #@#@#@#@#

Yours sincerely,

Sam

Please read the guidelines at the top of this page, and follow them. --KSmrq^T 17:06, 25 October 2006 (UTC)[reply]

On the look of it, this has no analytic solution. For that, we need an antiderivative of the function f(x) = (1 + λx − μx⁴)^−1/2, and that antiderivative has to have an explicit inverse to boot. But I'm already stuck on the first part. You could perhaps try looking at expansions of b (for example as a sum of exponentials near the roots of the polynomial Ax - Bx^4/4 + U²/2), using the method of equating coefficients (in a more general sense than treated in our article under that name). --Lambiam ^Talk 19:13, 25 October 2006 (UTC)[reply]

Basically the equation is ((b')²)/2 = Ab - (Bb^4)/4 + U²/2 (I'm just confused about b^4/4 because that could be (b^4)/4, or b^1), right? --AstoVidatu 22:32, 25 October 2006 (UTC)[reply]

I'm not done, but try rewriting the equation as b' = (k)^(1/2), and set k equal to all that junk in the middle. Then take the integral of that equation (easy as pie) and sub in k and k' using k = 2*(Ab - (Bb^4)/4 + U²/2)... And remember U is just a number, so it goes to 0 when you take the derivative. --AstoVidatu 22:37, 25 October 2006 (UTC)[reply]

Infinitesimal

Is an infinitesimal the recipricol of infinity? —Preceding unsigned comment added by 86.142.195.245 (talk • contribs)

No, I believe the reciprocal of infinity is 0. It is somewhat debated, however, because infinity can have more than one value, while 0 is only 0. PullToOpen ^talk 20:10, 25 October 2006 (UTC)[reply]

But couldn't a case be made for 0 to be equal to an infintesimal? If you define 1/0 as infinity, the reciprocal is 0. However, that's just the ramblings of a 15-year-old, so don't set too much store by it. —D a niel (‽) 20:30, 25 October 2006 (UTC)[reply]

No it is not debated within professional mathematicians. The reciprocal of infinity is undefined, because infinity is not a real number. There are any number of Math FAQs out there that explain this, as well as the article on Infinity. --BluePlatypus 20:33, 25 October 2006 (UTC)[reply]

Let's clarify things a bit. The answer depends on the context. In the real numbers, there is no infinity or infinite quantity. In the real projective line, the reciprocal of ∞ is 0. In the surreal numbers, there is no ∞, but there are infinite quantities like ω, the reciprocal of which is 1/ω, an infinitesimal. -- Meni Rosenfeld (talk) 20:55, 25 October 2006 (UTC)[reply]

For "infinitesimal" to be a useful concept, you want to retain some conventional algebraic properties, such as, letting ε stands for some infinitesimal, (2ε)/ε = (ε+ε)/ε = ε/ε + ε/ε = 1 + 1 = 2. If ε was equal to 0, this would not be possible; we would get the useless form 0/0. --Lambiam ^Talk 21:45, 25 October 2006 (UTC)[reply]

Lambiam, isn't there a problem there? Your example says that 2ε/ε = (ε+ε)/ε = ε/ε + ε/ε = 1 + 1 = 2, but if ε is infintesimal, surely adding multiplying it by any finite number will leave it unchanged, thus 2ε/ε = ε/ε = 1? Have I missed something?

Yes, you have missed something. The idea is related to "differential quotients" suggesting the ratio of two infinitesimal quantities, as in dy/dx = 2. Mathematicians find it useful to manipulate this like dy = 2dx, as if dy and dx are actual quantities. Actually that practice preceded the invention of the differential calculus. Think of dy and dx as very very (as in VERY) small changes. It took quite some time before someone invented a way to put that on a safe footing. --Lambiam ^Talk 22:45, 25 October 2006 (UTC)[reply]

0 would be the recipricol of infinity. Infinity is defined as (infinity/1). And the recipricol of that would be (1/infinity), which is 0. --AstoVidatu 22:28, 25 October 2006 (UTC)[reply]

Yes, the typical infinitesimal is the reciprocal of an infinity. By definition, a positive infinitesimal number is smaller than the reciprocal of any positive integer: ε < ¹⁄_n. (Formally, we would add "for all n ∈ Z⁺".) Inverting, ¹⁄_ε > n; this says the reciprocal of an infinitesimal is greater than any integer.

However, if we want to do mathematics, not just wave our hands, we must be careful and more precise. The ordered systems of arithmetic we use most often — the integers, the fractions, the real numbers — exclude both infinities and infinitesimals. There is no integer greater than every other integer; there is no positive fraction smaller than every other positive fraction. So, if we are to do mathematics with infinities and/or infinitesimals, we must define a different system of arithmetic.

We have choices. In many of the new systems what I said is true. (An example is hyperreal numbers.) In some, we have no infinities, and no infinitesimal has a reciprocal. (An example is dual numbers.) Or we can extend the real numbers with no infinitesimals, but with a single infinite value, ∞, whose reciprocal is zero. (An example is the real projective line.) We also have exotic possibilities like surreal numbers.

All of these systems are valid choices and allow us to do interesting mathematics. However, they can have strange new properties and lack familiar old properties. Choose wisely. --KSmrq^T 00:06, 26 October 2006 (UTC)[reply]

So what do you get if you multiply an infinity by an infinitesimal? --86.139.127.29 21:14, 26 October 2006 (UTC)[reply]

That depends which infinity and which infinitesimal. Suppose ε is a positive infinitesimal in an appropriate system. Then 1/ε is infinite, and in the following, the first multiplicand is an infinitesimal and the second is infinite (and the result can be pretty much anything):

ε * (1/ε) = 1
ε * (2/ε) = 2
ε * (1/ε²) = 1/ε
ε² * (1/ε) = ε

-- Meni Rosenfeld (talk) 23:03, 26 October 2006 (UTC)[reply]

If .999... = 1in a base 10 number system what about others?

Would it also be true that .777... = 1 in a base 8 system? If so, would .000... = 1 in base 2? I'm just curious and hope this is not an inappropriate post. Downhill.geezer 21:36, 25 October 2006 (UTC)[reply]

No worries! Yes, in base 8, .777... = 1. In base 2, however, .111... = 1, not .000.... There's some more information at 0.999...#Generalizations. Melchoir 21:44, 25 October 2006 (UTC)[reply]

Your post is welcome.

If d is an integer greater than 1, then we can use it as a radix (base) just like 10 or 8 or 2. When we do, the equivalent of 9 repeating is d−1 repeating. So 0.777…₈ = 1 for radix 8, as you propose; but 0.111…₂ = 1 for radix 2, contrary to your guess. Independent of which d we choose, 0.000… is always exactly zero.

Mathematicians have a habit of seeing how far an idea can go. Donald Knuth, in volume 2 of his series The Art of Computer Programming (ISBN 978-0-201-89683-1), presents a wide assortment of positional notation systems. For example, in balanced ternary, with digits 0, 1, and −1 (written 1), the notation 0.111… represents ¹⁄₂. With a radix of −10, digits 0 through 9, every real number (positive, zero, and negative) can be written without a sign. With a radix of i−1 (where i² = −1), digits 0 and 1, every complex number can be written. It might be fun and educational to explore some of these, seeking the equivalent of repeating 9s. --KSmrq^T 00:58, 26 October 2006 (UTC)[reply]

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