Wikipedia:Reference desk/Archives/Mathematics/2008 September 22

= September 22 =

Dedekind cuts
From the article:

A typical Dedekind cut of the rational numbers is given by


 * $$A = \{ a\in\mathbb{Q} : a^2 < 2 \lor a\le 0 \},$$
 * $$B = \{ b\in\mathbb{Q} : b^2 \ge 2 \land b > 0 \}.$$

This cut represents the irrational number $$\sqrt{2}$$ in Dedekind's construction.

Would one need to use both sets to represent $$\sqrt{2}$$, or just one? Defining it as A would imply the existence of B, and vice versa; i.e. if I wanted to define $$\sqrt{2}$$, I would just have to describe A or B, but is there any need for both outside of being totally rigorous? —Preceding unsigned comment added by 81.187.252.174 (talk) 07:23, 22 September 2008 (UTC)
 * See Dedekind_cut: "each of A and B does determine the other. It can be a simplification, in terms of notation if nothing more, to concentrate on one 'half' ". Bo Jacoby (talk) 09:30, 22 September 2008 (UTC).

Find the natural domain and correspong range of this function...
Note: if someone can and doesn't mind doing so, please fomat the notation properly so that it is easier to read.

$$x \rightarrow \sqrt{x(x-3)}$$

Im not really interested in the working or anything. I just want to know if someone can verify that:
 * the domain is the set real numbers "greater than or equal to 3" or "greater than or equal to 0"
 * the range is the set of real numbers "less than or equal to 0".

Thats what the book and my teacher say. I agree with the domain because you cant end up with a negative inside the square root. But I dont agree with the range because the square root of a number can be positive or negative. Therefore the range should just be the set of real numbers.

--RMFan1 (talk) 13:08, 22 September 2008 (UTC)
 * The square root function returns only one result. Yes, it's true that there are numbers a and -a square roots for some c, but the square root function returns only the positive number.  x42bn6 Talk Mess  13:16, 22 September 2008 (UTC)
 * Haven't you got the domain the wrong way round? x(x-3) is negative for x between 0 and 3, so the domain is *greater* than or equal to 3 or *less* then or equal to 0. --Tango (talk) 13:22, 22 September 2008 (UTC)


 * Also, your book should have specified that you are expected to find the domain and range on the basis that this is a function from real numbers to real numbers. If it is regarded as a function from complex numbers to complex numbers then its domain and range are the whole of the complex plane - in fact, it covers its range twice, with the exception of one critical value. (although this is sort-of-implied by the use of the square root function). Gandalf61 (talk) 13:38, 22 September 2008 (UTC)


 * Ok thanks. The error then is I, along with pretty much everyone else in the class take the square root function as returning the positive and negative values. Even the teacher so I dont know why she didnt think the book was wrong too --RMFan1 (talk) 13:51, 22 September 2008 (UTC)
 * Well, it depends on what you mean by "square root function". The square root symbol means "positive square root", that's almost universal, the words "square root", though, mean a multi-valued function, as does $$x^\frac12$$. --Tango (talk) 14:26, 22 September 2008 (UTC)
 * I don't know what you were taught, but both "square root" and $$x^\frac12$$ imply only the positive branch in my experience. The square root of 25 is 5, not +/- 5.  Dragons flight (talk) 21:16, 22 September 2008 (UTC)
 * In my experience, x1/2 is only required to be positive if x is implicited constrained to be a positive real. Otherwise we're using the complex exponential function, and it's inherently two-valued. Algebraist 21:19, 22 September 2008 (UTC)
 * So you'd say (-25)1/2 is +/- 5i? I'd still say this is only +5i, i.e. the positive branch.  What to do with complex x might be more ambiguous, but in my experience all real valued x would be understood as implying the positive branch.  Dragons flight (talk) 21:29, 22 September 2008 (UTC)
 * What do you consider to be the positive branch of the complex square root function? In my experience, there's no clearcut answer, and at least one standard answer is undefined precisely on the negative reals. Algebraist 21:40, 22 September 2008 (UTC)
 * If i is understood as "positive" and -i as "negative" then a*i is on the positive branch of the pure complex numbers iff a > 0. Mixed numbers, a+bi, give rise to ambiguities, but I was always taught that x1/2 for real valued x gives rise to either positive real numbers or positive pure imaginary numbers.  Obviously we are quibbling about definitions, but that's how it was in my education.  Dragons flight (talk) 04:52, 23 September 2008 (UTC)
 * This little subtlety arises when you are solving quadratic equations and no teacher I've ever known has explained it clearly. The square root function, $$ \textstyle \sqrt{} : \mathbb{R} \to \mathbb{R} $$, does indeed only return a positive result. It is a genuine honest-to-god function, not one of those icky multivalued things. HOWEVER, when you have an equation of the sort $$ \textstyle x^2 = a $$ and you try solve for x, this gives us two distinct possiblities: Either $$ \textstyle x = \sqrt{a} $$ or $$ \textstyle x = -\sqrt{a} $$. One of these two must be true if the original equation was true, but without further information (such as "x is positive"), we can't tell which one of the possibilities is the case. So the square root of a positive number is positive, but for any positive number there are TWO numbers whose squares are that number - one is the square root, and the other is the negative of the square root. Maelin (Talk | Contribs) 04:32, 23 September 2008 (UTC)
 * So unless I'm missing something, the foregoing discussion means that the range may be the set of real numbers, or it may be the set of real numbers greater than or equal to 0, but it should not be, as originally posted, the set of real numbers less than or equal to 0?…81.154.106.183 (talk) 08:22, 23 September 2008 (UTC)
 * You're missing something. The range is quite definitely the set of nonnegative reals. There's some disagreement as to what 'square root' means, and some disagreement as to what '$$x^\frac12$$' means, but everyone agrees that $$\sqrt x$$ is positive. Algebraist 10:39, 23 September 2008 (UTC)


 * So yes, it should not be, as originally posted, "the set of real numbers less than or equal to 0". It was this bit I was querying, as the specific direction of single-sidedness in the original post hadn't been mentioned by anyone above.…81.154.106.183 (talk) 19:45, 23 September 2008 (UTC)


 * Indeed, after consulting our own article on the matter, I retract my prior comment to the effect that a number only has one square root. Like "roots of a polynomial", a number may have many square roots. Nonetheless, as Algebraist said, the standard square root function on the real numbers, by convention always takes the positive square root. Whether the noun phrase "square root of x" refers only to the principal square root or encompasses all (both) square roots is an unfortunate ambiguity of language that must be assessed on a case-by-case basis. Maelin (Talk | Contribs) 14:12, 23 September 2008 (UTC)

Rearranging equations
I have been given the function $$f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}$$ and am told that $$f'(x)=\frac{(x-1)}{2x\sqrt{x}}$$.

I get that $$f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}=x^\frac{1}{2}+x^{-\frac{1}{2}}$$ and that $$f'(x)=\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{\frac{3}{2}}$$ from this.

How do I go from here to the final answer of $$f'(x)=\frac{(x-1)}{2x\sqrt{x}}$$? No matter what I do, I keep on getting stuck. Thanks. --AFairfax (talk) 16:58, 22 September 2008 (UTC)


 * Surely you mean $$f'(x)=\frac{1}{2}x^{-\frac{1}{2}}+-\frac{1}{2}x^{-\frac{3}{2}}$$ (note the negative exponent on the second term)? -- Jao (talk) 17:03, 22 September 2008 (UTC)


 * (Assuming the missing minus sign was just a typo) Factorise out an $$x^\frac{-3}2$$ term. --Tango (talk) 17:14, 22 September 2008 (UTC)
 * Indeed it was, thanks for pointing that out. Still, I'm not getting anywhere factorising out an $$x^\frac{-3}2$$ term. I manage to get: $$x^{-\frac{3}{2}}\left(\frac{1}{2}x^{-\frac{4}{2}}+\frac{1}{2}x^{-\frac{1}{2}}\right)$$ —Preceding unsigned comment added by AFairfax (talk • contribs) 20:20, 22 September 2008 (UTC)
 * I've no idea how you got that. Factor out the x-3/2 term correctly, and you have the given answer. Algebraist 21:00, 22 September 2008 (UTC)
 * I thought that might be the problem, but I have no idea how to do it correctly. Nor can I get the suggestion below to work (it's too late, and my Maths is too bad...) --AFairfax (talk) 21:35, 22 September 2008 (UTC)
 * To work it backwards, split the fraction into two parts: $$\frac{x}{2x\sqrt{x}}-\frac{1}{2x\sqrt{x}}$$. Then just simplify it. To work it forwards (which is always the better way to learn - you can't rely on always having the answer in front of you!), "factorise out by X" means "divide each term by X, put brackets around the whole thing, and write 'X' at the front". --Tango (talk) 21:54, 22 September 2008 (UTC)
 * I understand the concept of factorising, it's just that in this case, I don't appear to be able to factorise out x-3/2 correctly. Thanks for all the help guys. --AFairfax (talk) 22:34, 22 September 2008 (UTC)
 * It's easy to work backwards from the known answer. Just make the division in $$f'(x)=\frac{(x-1)}{2x\sqrt{x}}$$ and compare. PrimeHunter (talk) 21:09, 22 September 2008 (UTC)


 * Following up on Tango's explanation of factorization, here we go step-by-step. We wish to factor $$x^{-3/2}$$ out of the two terms $$x^{-1/2} - x^{-3/2}$$ (I'll forget about multiplying by 1/2, it doesn't matter).  First we divide each term by $$x^{-3/2}$$:
 * $$\frac {x^{-1/2}}{x^{-3/2}} = x^{-1/2} \cdot x^{3/2} = x^{-\frac 12 + \frac 32} = x^1 = x $$
 * $$\frac {-x^{-3/2}}{x^{-3/2}} = -x^{-3/2} \cdot x^{3/2} = -x^{-\frac 32 + \frac 32} = -x^0 = -1 $$
 * Then we put brackets about and multiply back by $$x^{-3/2}$$:
 * $$x^{-3/2} \cdot (x - 1) $$
 * Which is the same thing as:
 * $$\frac {x - 1}{x^{3/2}} = \frac {x - 1}{x\sqrt x} $$
 * Since we have first divided by $$x^{-3/2}$$ and then multiplied by $$x^{-3/2}$$, the resulting expression is equal to the expression we started with. I've written the steps out in some detail to help you find where you might have made a mistake in your own work.  Eric.  65.96.172.100 (talk) 04:29, 23 September 2008 (UTC)

Mean of circular quantities
The article says:

"Convert all angles to corresponding points on the unit circle, e.g. α to (cosα,sinα). That is convert polar coordinates to Cartesian coordinates. Then compute the arithmetic mean of these points. The resulting point will lie in the unit circle.'"

Shouldn't it be unit disk? 83.23.221.221 (talk) 22:16, 22 September 2008 (UTC)


 * I think either "within the unit circle" or "on the unit disk" would work, "in the unit circle" is ambiguous. I've changed it to "on the unit disk", which seems the better of the two options to me (since a point on the unit circle might not be considered to be within, but it is on the (closed) unit disk). --Tango (talk) 22:28, 22 September 2008 (UTC)