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

< September 30	<<Sep \| October \| Nov>>	October 2 >

Humanities

Science

Mathematics

Computing/IT

Language

Miscellaneous

Archives

The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions at one of the pages linked to above.

Missing posts[edit]

Has anyone else noticed posts going missing on these pages? Like you click on an item in your watch list by UserXXXX, and it aint there? Im posting this msg on all ref desks.--Light current 11:27, 1 October 2006 (UTC)[reply]

Now restored. See Wikipedia_talk:Reference_desk#Archive_dump. --hydnjo talk 14:11, 1 October 2006 (UTC)[reply]

Simplifying rational expressions[edit]

If i have the expression: x(x-2)/3x(x-1) Can I cancel the 2 x's out and turn the expression into: x-2/2x(x-1)? Why or why not? Jamesino 17:54, 1 October 2006 (UTC)[reply]

I assume that multiplication in your formula takes precedence over division, so 3x(x-1) is the denominator of this fraction. Presumably, you mean "(x-2)" in the last formula; what you wrote would normally be interpreted as x-(2/2x(x-1)). What you appear to be doing is divide by x in the numerator (replacing x by x/x = 1, which can be omitted), while you subtract x from a fragment of the denominator (replacing 3x by 3x−x = 2x). There's no way you can do that. One of the simplest sanity checks is to substitute some concrete value and see if things pan out. If we substitute 10 for x in the original fraction, we get (10×8)/(3×10×9) = 80/270 = 8/27. For the second fraction we get 8/(2×10×9) = 8/180. These are clearly different numbers.

What you can do is cancel equal factors against each other by dividing by them. The basis of this rule is that (p/q)×(r/s) = (p×r)/(q×s). (See the rules for multiplying fractions.) Switching the two sides, and replacing q by p, we get: (p×r)/(p×s) = (p/p)×(r/s). Now if p is not equal to 0, p/p = 1, and we get (p×r)/(p×s) = r/s. But if p = 0, the expression (p×r)/(p×s) stands for 0/0, a so-called indeterminate form that has no meaning. You are not allowed to state that that is equal to something that has a meaning, like r/s.

Applied to your original problem: You can simplify x(x-2)/3x(x-1) to (x-2)/3(x-1) under the condition that x ≠ 0. --Lambiam ^Talk 18:38, 1 October 2006 (UTC)[reply]

Ah ok thanks. Using your explanation i figured that X also cannot equal 1, -2, or 2. Jamesino 18:48, 1 October 2006 (UTC)[reply]

Both x = 2 and x = −2 are fine. For x = 2, both x(x-2)/3x(x-1) and (x-2)/3(x-1) can be simplified to 0. For x = −2, nothing special is going on, and both fractions can be simplified to 4/9. The case x = 1 is indeed problematic: it implies division by zero. In this case both fractions are undefined (meaningless) in the domain of real numbers. Before, for the case x = 0, one was undefined and the other well defined. --Lambiam ^Talk 21:02, 1 October 2006 (UTC)[reply]

To avoid potential confusion, write the expression as

{\frac {x(x-2)}{3x(x-1)}}.

The value of a fraction is unchanged if both numerator and denominator are multiplied by a nonzero quantity. Likewise, a nonzero quantity can be cancelled top and bottom. These are our only possibilities. For example:

{\frac {3}{4}}={\frac {3\times 5}{4\times 5}}={\frac {15}{20}}.

{\frac {6}{8}}={\frac {3\times 2}{4\times 2}}={\frac {3}{4}}.

When dealing with expressions involving unknowns, we usually do not know in advance what is nonzero, so any manipulations depending on that assumption must be explicitly so qualified.

In the following example, we see a factor of z in both numerator and denominator. If we stipulate that z is not zero, we are allowed to cancel it:

{\frac {3z}{4z}}={\frac {3}{4}}.

Polynomials are much like integers; we can write both in terms of "primitive factors". For both, the factors are essentially unique. We say "essentially" because we are not concerned about the order of the factors, nor pairs of sign changes. For example,

15=3\times 5=(-5)\times (-3).\,\!

z^{2}-1=(z+1)(z-1)=(1-z)(-1-z).\,\!

With a little practice we can cope with fractions of polynomials nearly as easily as common fractions. --KSmrq^T 21:11, 1 October 2006 (UTC)[reply]

Is my derivative correct?[edit]

I have been using the Calculus Wikibook to teach myself derivative calculus. I get the derivative of $f(x)=2x+3$ to be equal to $2x^{2}+4+3x$ . Is this correct? If not, what is the correct answer? Thanks. --80.229.152.246 21:00, 1 October 2006 (UTC)[reply]

Nevermind, I have found where I went horribly wrong... --80.229.152.246 21:05, 1 October 2006 (UTC)[reply]

BTW, the term commonly used by mathematicians is differential calculus. --Lambiam ^Talk 21:14, 1 October 2006 (UTC)[reply]

:::Thanks, I forgot about that. That's what you get for typing a question in a hurry. By the way, can anyone help me with finding the correct derivative of $f(x)=2x+3$ ? Can you tell me what is it and how you get it? I can't even get the first demonstration to equal to what I work out now! That's what you get for reading stuff like that late at night wen you should really be in bed. I've just realised now I've spent more words talking off the track than on it. I think I better stop now. Thanks in advance. --80.229.152.246 21:17, 1 October 2006 (UTC)[reply]

Nevermind, after looking at some articles on Wikipedia, I think I have got the hang of it. Would I be right in saying that the answer is 2? Thanks for your help. --80.229.152.246 21:30, 1 October 2006 (UTC)[reply]

Yes, that's more like it. --Lambiam ^Talk 21:31, 1 October 2006 (UTC)[reply]

N-Dimensional Geometry question[edit]

If I have some n-dimensional plane (a1*x1 + a2*x2 .... = 0) (a's are constants, x's are variables) that passes through the origin, how can I determine if the plane intersects with a corner of the n-dimensional cube defined by ( 0 <= x1,x2 ... <= 1). Thanks for any help with this problem, the only thing I've found is trying the co-orindate of each corner into the plane equation. I'm looking for something more effecient than that. Thanks again. AmitDeshwar 23:29, 1 October 2006 (UTC)[reply]

I think I can rephrase the question in a way that may be useful... you want to know if there is a solution to the equation

a_{1}x_{1}+\cdots +a_{n}x_{n}=0

for which all x's are 0s and 1s. This is the same as asking whether some subset of the coefficients (other than the empty subset) add up to zero. Does that help at all? -GTBacchus^(talk) 23:37, 1 October 2006 (UTC)[reply]

This is 0-1 Integer Programming, which is on the original list of Karp's 21 NP-complete problems. So if you find an efficient method, let us know. --Lambiam ^Talk 00:26, 2 October 2006 (UTC)[reply]

I see now that this is also the Subset sum problem. I'll let you know if I find an effecient solution ;). AmitDeshwar 03:43, 2 October 2006 (UTC)[reply]