Wikipedia:Reference desk/Archives/Mathematics/2009 October 1

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 1

The answer to this system of equations?

I would appreciate it if anyone could tell me the answer to this system of equations:

x + y = 50

x*y = 60

Thanks. 74.12.20.180 (talk) 04:43, 1 October 2009 (UTC)

Multiply the first equation by x giving xx+yx=50x, and subtract the second equation giving xx=50x−60. Now the variable y has been eliminated from the problem. You know how to solve this quadratic equation, right? Bo Jacoby (talk) 06:24, 1 October 2009 (UTC).

I might have done it like this: since

x + y = 50,

we get

y = 50 − x,

so we can put 50 − x in place of y in the second equation:

x(50 − x) = 60.

Then we have a quadratic equation in one variable. Michael Hardy (talk) 19:22, 4 October 2009 (UTC)

If s/he knows how to solve a quadratic equation, then maybe everyone should stop reading at this point. Michael Hardy (talk) 19:20, 4 October 2009 (UTC)

(Edit Conflict) There are a few aspects of this question which are unclear and thus I wish to clarify them here:

Precision - The domains of the variables x and y are not specified; that is, are you attempting to find integer solutions to the system of equations, or are you attempting to find rational or real solutions to the system of equations? Thus, embedded within your question are infinitely many other questions, each question associated to a particular ring, which one cannot hope to answer here.

Choice of wording - You seem to request "an answer to this system of equations". This remark is not only imprecise (the precise statement would be "the solutions to this system of equations"), but suggests that this is a homework problem. Usually, if this were not a homework problem, the questioner would make some remark regarding his/her attempts at the solution and interest in the question. I encourage you to do the same.

In fact, politely speaking, you should reword your question in a precise manner and present it again, so that it not only demonstrates your understanding of the question, but also the degree of effort you have put into it. This comment should not indicate that I am unwilling to help you, but rather that your expression conveys that you will not learn anything from a straightfoward answer. Nevertheless, I will show you the solution to a similar problem, which, after precisely stating the above problem, should enable you to derive its answer:

Problem - Suppose that the sum and difference of two numbers are given. How does one compute the numbers themselves? (Here, by "number" it is meant "real number")

Solution - Write x = a + b, and y = a - b, where a and b are the two numbers in question. Observe that x + y = a + a, and thus (x + y)/2 = a. Similarly (x - y)/2 = b (why?). This was what was to be demonstrated, since x and y are known.

With the ideas that above the underlying technique in mind, solve the slightly harder problem that you wish to solve. Note the identity (hint) - $x^{2}+2xy+y^{2}=(x+y)^{2}$ for all real numbers x and y (notice how I have incorporated precision into my remark). Also note that there may be more than one way to solve this problem.--PST 06:33, 1 October 2009 (UTC)

Come on, give your head a wobble! You know what the OP was trying to say, so there's no need to be condescending. Your little monologue helped whom? How?! To the OP: First, try drawing a picture. You'll see that there are two solutions. You can get these by substitution. Solve x + y = 50 to give y = 50 − x. Sub this into the second equation to give x² − 50x + 60 = 0. Solve this for x to give x = 25 ± √565. Sub this solution into x + y = 50 to give 25 ± √565 + y = 50; then solve for y. There are two solutions and they are

$(x,y)=\left(25\pm {\sqrt {565}},25\mp {\sqrt {565}}\right).$ ~~ Dr Dec (Talk) ~~ 11:13, 1 October 2009 (UTC)

If you did not understand the purpose of my comment, then do not comment upon it. The OP wanted an answer to a question which he did not state precisely. Although there is no problem with this since he surely would not do this intentionally, the fact that he did not give any indication of his attempts gives reason for one to not answer his question. In fact, it is against Wikipedia guidelines to do so. I do not imply by this that answering the question is an offence since I am sure that thus far everyone has answered his question with good intentions. However, I believe that a lesson should be learnt from this experience, which will be a useful one, should he pursue mathematics or in fact anything else. PST 12:47, 1 October 2009 (UTC)

I found PST's post well written and helpful as usual (whereas Declan's little irony helped whom?) --pma (talk) 13:05, 1 October 2009 (UTC)

If we're taking sides, I'm going with Declan. I think it is very safe to assume the OP is talking about real numbers (could be complex numbers, but the answers are real, so it doesn't matter). If they were talking about anything else it would have been a key part of the question and they would have mentioned it. It is also very clear that "the answer" is a solution set. The OP may not have realised that there will be two answers (for 2 linear simultaneous equations in 2 variables, the solution would be unique, the OP could easily have thought the same applied here), but we don't need clarification on the question to know that we should give both (or rather, how to get both). I wouldn't have actually given the answer, though, since this looks very much like a homework question. I think the Bo Jacoby's answer was the best. --Tango (talk) 14:31, 1 October 2009 (UTC)

I agree that it is possible to interpret the OP's question as it stands and that Bo Jacoby's answer was the most appropriate one. However, in my post I did not solely aim to to answer the question (although I did attempt to give the OP a hint), but rather I expressed my concerns that at some point, precision will be of importance in his/her life. Succintly, I suggested that carefully expressing his/her question would help him/her more than it would help us in answering his question. Perhaps I should have framed my remark in a less confrontational manner, but at its present length, it will at least leave an indelible mark on the OP's mind, which, I hope, will help him/her to realize the meaning of my comment later on. --PST 14:52, 1 October 2009 (UTC)

The OP came on asking how to solve two relatively simple questions. You can therefore assume that s/he may lack the necessary mathematical sophistication to phrase the problem perfectly. My guess is that s/he's been solving simultanious equations and has asked themselves "What if one of the equations was not linear?" I'd be surpised if the OP knows very much about complex number, much less about rings! I'll assume good faith on your part and assume that it wasn't an attempt to show your mathematical and linguistic superiority. That given, you've most probably totally baffled the OP and left him/her with more questions than when s/he started. Ask yourself this: If someone comes on here asking how to solve x + y = 50 and xy = 60 then are they going to have the faintest idea of what you were talking about? Of course not! ~~ Dr Dec (Talk) ~~ 15:57, 1 October 2009 (UTC)

Clearly, I should defend my remark. First of all, I did not intend to display any "mathematical or linguistic superiority"; in fact, why either is demonstrated in my post is beyond my comprehension. Secondly, as I have already made clear, I attempted to illustrate the unclear aspects of the OP's post. In my opinion, this serves to convey the importance of mathematical precision, which, at high school level, few students appreciate. Surely, my post will be in the forefront of the OP's mind when he does hear a mathematician speaking of precision. In fact, it will be important in his life later on, since no matter what you do, clarity of expression is crucial. At the reference desk, we work as a group of people who wish to help. Some people answer all sorts of questions that are posed (independent of what subject matter the question embodies) whereas some choose to discuss the aspects of the question other than the answer. If you would like to think of it in that manner, I am the latter for the simple questions. I know already that many dedicated people here would respond to the OP, and thus, in order to add another dimension to the OP's algebra skills, I explained the importance of precision. Although I do not think that your intentions were malacious, this discussion was started because of your unnecessary remark; not only the content, but the way in which you phrased it. You are certainly an important contributor to the reference desk, but at times your attitude is not. In effect, you must assume good faith and at least attempt to see the importance of others' remarks. --PST 01:28, 2 October 2009 (UTC)

type 1 and 2 errors of statistical hypothesis testing

There seems to be a bit of confusion here regarding the definitions of α, 1-α and 1-(1-α), and similarly, β, 1-β and 1-(1-β). Is the following true:

• α = FP
• 1-α = specificity = significance
• 1-(1-α) = rate of FP
• β = FN
• 1-β = sensitivity = power
• 1-(1-β) = rate of FN

Thanx! DRosenbach ^{(Talk | Contribs)} 12:55, 1 October 2009 (UTC)

α is the probability of a false negative and is called the "level of significance." I believe that β is the probability of a false positive. Wikiant (talk) 14:28, 1 October 2009 (UTC)

1-(1-α) is just α of course. As far as I can recall (from many years ago) α is the probability of a false positive, as stated in the article, and is the significance level of the test, with 1-α being the specificity. Similarly, β (called the sensitivity) is the probability of false negative and 1-β is called the power or sensitivity of the test. The article does seem slightly confusing at first reading, but I think it is correct. Dbfirs 16:37, 1 October 2009 (UTC)

I believe that is incorrect. The verbal definition of α is, "the probability of rejecting the null hypothesis when, in fact, the null hypothesis is true." That's a false negative. Wikiant (talk) 18:33, 1 October 2009 (UTC)

No, it's a false positive. Rejection of the null hypothesis is the positive result. Algebraist 21:25, 1 October 2009 (UTC)

Can no one help me? -- the aforementioned has been culled from the confusing, double-talk of this section. DRosenbach ^{(Talk | Contribs)} 01:01, 2 October 2009 (UTC)

The terminology can be confusing, but the article is correct (I've just fished out a very old text book to check).

A Type I (α) error occurs when we reject the null hypothesis even though the null hypothesis is true. This is called a "false positive" and α is the significance level of the test. 1 - α is the specificity.

A Type II (β) error occurs when we fail to reject the null hypothesis (i.e. accept the null) even though the null hypothesis is false. This is called a "false negative" and β is the sensitivity of the test. 1 - β is called the power or sensitivity of the test. When I have time, I'll have a go at setting out the article more clearly, unless someone else beats me to it (I hope they do!) Dbfirs 12:47, 2 October 2009 (UTC)

Dbfirs -- I'm sort of glad that I have at least one person engaging my question, but I'm also sort of frustrated that you make conflicting seemingly conflicting statements of the article is correct and α is the significance level of the test -- as significance is listed above as being equivalent to not α, but 1-α. Perhaps I'm either misunderstanding your comments or the logic, but to me, x ≠ 1-x. DRosenbach ^{(Talk | Contribs)} 13:27, 2 October 2009 (UTC)

No, the article says that 1 - α is the specificity. α is the significance level of the test. The article does say this, but not expressed very clearly. Dbfirs 17:51, 2 October 2009 (UTC)

If α is the probability of a false positive, then clearly 1-α is the significance of the test. A test at 95% significance does not have a 95% chance of a false positive. Taemyr (talk) 17:56, 2 October 2009 (UTC)

No, that's wrong. The significance level is the rate of false positives, i.e. it is the rate of falsely rejecting the null hypothesis. Michael Hardy (talk) 23:17, 2 October 2009 (UTC)

Taemyr, you're confused. Nobody ever tests at the 95% significance level. 5% is commonplace. Michael Hardy (talk) 23:18, 2 October 2009 (UTC)

( ... sorry about the delay in replying, I 've just returned home ...) People often confuse specificity with significance (it probably doesn't really matter). A significance test is normally done at the 5% significance level (and sometimes 1%). This corresponds to a specificity of 95% (or 99%), but, informally, we often speak of testing at the 95% level of significance when we mean the 5% level of significance i.e. the 95% level of specificity. Hence the confusion. Dbfirs 23:08, 2 October 2009 (UTC)

Maybe the section's been edited since the time it was commented on above. Here's what I find there:

The false positive rate is the proportion of negative instances that were erroneously reported as being positive.

It is equal to 1 minus the specificity of the test. This is equivalent to saying the false positive rate is equal to the significance level.

${\text{false positive rate}}={\frac {\text{number of false positives}}{\text{total number of negative instances}}}$

In statistical hypothesis testing, this fraction is given the Greek letter α, and 1 − α is defined as the specificity of the test. Increasing the specificity of the test lowers the probability of type I errors, but raises the probability of type II errors (false negatives that reject the alternative hypothesis when it is true).

That is correct in saying that α is the false-positive rate, i.e. α is the probability of rejecting a null hypothesis, given that the null hypothesis is true, i.e. α is the significance level. Michael Hardy (talk) 23:31, 2 October 2009 (UTC)

I've just realised that I've made a serious error above. The alpha parts are correct, but beta is not the sensitivity. This may be why we were getting confused. I had assumed a symmetry when there isn't one (at least not in the terms used). Dbfirs 19:27, 3 October 2009 (UTC)

Limits on discrete sets

What is $\lim _{n\to 0^{+}}{\frac {1}{{\frac {1}{n}}!}}$? And $\lim _{n\to {\frac {1}{4}}}{\frac {1}{{\frac {1}{n}}!}}$? --88.78.224.81 (talk) 15:23, 1 October 2009 (UTC)

What do you mean by x! for non-integer x? — Emil J. 15:30, 1 October 2009 (UTC)

Good point. But we can generalise the factorial function with the Gamma function. For z ∈ C with Re(z) > 0 we have

$\Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,dt\;$

It has the nice property that for n ∈ N we have Γ(n) = (n+1)! ~~ Dr Dec (Talk) ~~ 16:02, 1 October 2009 (UTC)

We need to use the Gamma function. So we can define ƒ : C → C given by

$f(z):={\frac {1}{\Gamma (z^{-1}-1)}}\ .$

This generalises the OP's formula because Γ(n−1) = n! for n ∈ N. The second limit is easier than the first, we have

$\lim _{z\to 1/4}f(z)={\frac {1}{2}}\ .$

As for the limit as z → 0, well, I'm still working on that one... ~~ Dr Dec (Talk) ~~ 16:17, 1 October 2009 (UTC)

...et ecce trabis est in oculo tuo! ( ↑ ) --pma (talk) 17:26, 1 October 2009 (UTC) ;-)

Couple of slips there, I think. For a start, Γ(n) = (n − 1)!, not (n+1)!. Gandalf61 (talk) 16:38, 1 October 2009 (UTC)

Ah, yes, quite right. But hey, what's a sign error between friends? ~~ Dr Dec (Talk) ~~ 16:49, 1 October 2009 (UTC)

The values of the limits are trivial to compute, once you manage to make any sense of the original notation by plugging in the Γ's (I know about this function, but I was not sure the OP does, or even is aware that there is a problem). The first limit is 0, as $\lim _{n\to 0^{+}}{\frac {1}{n}}!=\lim _{x\to +\infty }x!=+\infty$. The second limit, restated using Γ, is a limit of a continuous function, hence its value is simply the value of the function at 1/4, i.e., $\lim _{n\to {\frac {1}{4}}}{\frac {1}{{\frac {1}{n}}!}}={\frac {1}{4!}}=1/24$ (I'm not sure where you got the 1/2). — Emil J. 16:43, 1 October 2009 (UTC)

As Gandalf pointed out: I had a sign error. I had Γ(n) = (n+1)! instead of Γ(n) = (n−1)!. And in fact ƒ(z) → 1/2 as z → 1/4. The problem is that I should have had a plus in the denominator instead of a minus. When you replace it with a plus the limit is 1/24 as you say. I was having trouble with the case of z → 0 at the time of writing the post because

$\Gamma \left({\frac {1}{z}}+1\right)=\int _{0}^{\infty }t^{1/z}e^{-t}\,dt\;$

and I wasn't sure whether the limit of the integral on the right was well defined as z → 0 in the integrand. But it does seem that Γ(z⁻¹+1)⁻¹ → 0 as z → 0. ~~ Dr Dec (Talk) ~~ 17:07, 1 October 2009 (UTC)

Of course not. That limit doesn't exist, even for z in the real numbers... --pma (talk) 17:52, 1 October 2009 (UTC)

However, the original limit calls for z → 0+, and that one does exist. — Emil J. 18:01, 1 October 2009 (UTC)

The assumption that we consider the limit z → 0+ is tacit, since we assume that Re(z) > 0. I didn't see much point writing z → 0+, when it's clear that Re(z) > 0, and so z → 0− isn't well defined. ~~ Dr Dec (Talk) ~~ 18:15, 1 October 2009 (UTC)

Well, actually, now I finally noticed the title of this section, and it seems that the OP is after all not interested in computing the limits, but in the question of how to make sense of them, as the functions are not defined on a neighbourhood of the offending point. The answer is that in this particular case, one could interpret x! as a shorthand for Γ(x + 1) as above, which makes the limits well-defined. In general, it is possible to define a limit relative to a set $\lim _{\begin{smallmatrix}x\to a\\x\in A\end{smallmatrix}}f(x)$ provided a is a limit point of A. This would take care of $\lim _{n\to 0^{+}}{\frac {1}{{\frac {1}{n}}!}}$, except that one should use a different notation to make it clear that the limit is only taken relative to $\{1/n\mid 0<n\in \mathbb {N} \}$. The second limit is still not well-defined using this approach, as 1/4 is isolated point of the domain of the function. — Emil J. 17:00, 1 October 2009 (UTC)

P value

Please see here for question. DRosenbach ^{(Talk | Contribs)} 17:45, 1 October 2009 (UTC)

The answer is that your word "still" is wrong. The p-value goes from above 5% to below 5%. Taemyr (talk) 21:18, 1 October 2009 (UTC)

Absolute value has nothing to do with this -- if a p value is .01%, .1% or even 1%, that's acceptable, because it's below 5%. In the example given (copied from the p value article), the p value is 4.41% -- still below 5%, yet is said to allow the null hypothesis to be rejected. Why? DRosenbach ^{(Talk | Contribs)} 00:03, 2 October 2009 (UTC)

The null hypothesis is accepted failed to be rejected when the p-value is greater than 5%. That means the result is within the range of what we'd expect to happen 95% of the time if the null hypothesis were true. If the p-value is less than 5% that's when we suspect something fishy is happening and reject the null hypothesis. The lower the p-value, the more extreme the result, and the less likely it is that the null hypothesis would lead to that result. Rckrone (talk) 05:32, 2 October 2009 (UTC)

You are right, of course, but I find your terminology misleading. When the p-value is greater than 0.05, I find it more suggestive to say "we failed to reject the null hypothesis" or "we cannot reject the null hypothesis"; under no conditions do we "accept" the null hypothesis, because even a high p-value does not lend support for the null hypothesis -- it is merely consistent with the null hypothesis being true. Eric. 131.215.159.109 (talk) 11:21, 2 October 2009 (UTC)

Thanks for that correction. I'll admit I'm no good at statistics. Rckrone (talk) 16:58, 2 October 2009 (UTC)

I make no statement as to the acceptance of a null hypothesis. And if I am correct, then why does the article say we use the p value to reject the null hypothesis if it is below 5%? DRosenbach ^{(Talk | Contribs)} 13:29, 2 October 2009 (UTC)

The p-value tells us the probabilty that random chance could account for the observed outcome given the null hypothesis. So if the p-value is low we can reject the null hypothesis because the observed outcome would have been too unlikely. In short, below 5%, then we can reject the null hypothesis. Above 5% and we can not reject the null hypothesis. Taemyr (talk) 14:56, 2 October 2009 (UTC)

Not to nitpick, but this explanation needs some clarifying. The p-value is the probability that random chance would obtain an outcome at least as inconsistent, if not more so, as the observed outcome is, with respect to the null. The idea is that if one rejects the null with this particular outcome, then any more anomolous outcomes would have drawn the same conclusion. In rigorous foundational terms, the probability of this particular outcome tends to become a measure-theoretical morass. Thus as a decision procedure, determined before any outcome was obtained, one can still speak of not rejecting 95% of the time if the null is true. Baccyak4H (Yak!) 16:49, 2 October 2009 (UTC)

@DRosenbach: Sorry if I wasn't clear, my comment @11:21 was directed towards Rckrone, not towards you. Eric. 131.215.159.109 (talk) 18:48, 2 October 2009 (UTC)

Great -- got it now. Thanx everyone! DRosenbach ^{(Talk | Contribs)} 02:05, 5 October 2009 (UTC)