Wikipedia:Reference desk/Archives/Mathematics/2007 January 28

Welcome to the Wikipedia Mathematics Reference Desk 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 on one of the current reference desk pages.

January 28

algebra 1

461=180t-16(t(t) what does is t —Preceding unsigned comment added by 71.142.60.19 (talk • contribs)

What's with the brackets here? At any rate, this isn't a place to ask homework questions. --h2g2bob 02:44, 28 January 2007 (UTC)

Just type it into your friendly neighborhood graphing calculator. Rya Min 05:44, 28 January 2007 (UTC)

It took me a while to find the "=" button on my TI-86, but it looks like the answer is 3910461. --Carnildo 09:55, 28 January 2007 (UTC)

That doesn't sound right at all. --Wirbelwindヴィルヴェルヴィント (talk) 18:33, 28 January 2007 (UTC)

I got approximately 7.31 and 3.94? That looks wrong too. =P --Wirbelwindヴィルヴェルヴィント (talk) 18:38, 28 January 2007 (UTC)

Thats what I got Rya Min 19:26, 28 January 2007 (UTC)

algebra 1

find two consecutive positive numbers such that the pruduct of the sum and difference of the number plus eight is the sum of thier squares —The preceding unsigned comment was added by 71.142.58.30 (talk) 04:22, 28 January 2007 (UTC).

Since they are consecutive numbers, let's call them n and n+1. Then their sum is 2n+1 and their difference is 1, so the product of their sum and their difference is 2n+1, and adding 8 to this gives 2n+9. So we have

$2n+9=n^{2}+(n+1)^{2}$

I'll let you take it from there. Gandalf61 12:23, 28 January 2007 (UTC)

Teaching yourself Calculus

I have a mathematical education up to geometry. Do you think, using a textbook I found in my basement, I could teach my self some calculus? I'm always reading math and physics pages on Wikipedia and the calculus looks like some ancient middle eastern symbology, but I want it to make at least some sense to me. I don't expect to be able to do advanced calculus, just be able to understand what those long S's (integrals?) are and what derivatives are, and be able to use them in a basic equation. Thanks for your help. :-) —The preceding unsigned comment was added by Imaninjapirate (talk • contribs) 06:23, 28 January 2007 (UTC).

If you put your mind to it, I don't see why not. Have a brief look at derivative and integral and see if anything makes sense. You can probably get by with asking a few questions here after you understand some concepts there. --Wirbelwindヴィルヴェルヴィント (talk) 06:43, 28 January 2007 (UTC)

It's ironic that you'd refer to calculus as an "ancient middle eastern symbology", since the real ancient Middle Eastern notation is the system of Arabic numerals that everyone uses. ;) Calculus is more like... fancy German baroque notation? - Rainwarrior 07:13, 28 January 2007 (UTC)

You could try looking at wikibooks http://en.wikibooks.org/wiki/Calculus Theresa Knott | Taste the Korn 09:08, 28 January 2007 (UTC)

• Try to think about it in familiar terms, and it will be easier for you. You've been doing derivatives all your life. You know miles per hour? It's velocity, or $ds/dt$, where s is a position, and t is time. The only difference is that you make t really tiny, until it becomes an infinitesimal. An integral is about the opposite thing; right now, if you have a constant velocity, and you have a known amount of time, you can figure out the distance you traveled. Using integration, you can figure out the distance you traveled, even if your velocity is not constant. Titoxd^(?!?) 09:21, 28 January 2007 (UTC)

Yes of course you could teach yourself calculus (provided you have a calculus textbook that starts at a level you understand). I'm sure someone here could recommend you one, if your basement book is no good, or just look for a maths textbook that starts at the level you finished at school.

I would though recommend getting at least one other book to learn from as learning maths from a textbook written in one style can be a bit tiring. (I'm not a bookseller honest).

I'd suggest working though a couple of similar books together - this should help prevent you getting stuck when one book doesn't give a clear explanation. Good luck87.102.33.144 13:11, 28 January 2007 (UTC)

Thank you all for your advice! I read the derivatives and integrals and it did make sense so I guess I'm going to start reading that book, and if I don't understand things Wikipedia or Wikibooks looks like it would be a good source to help clear things up for me. :-) Imaninjapirate^{talk to me} 15:08, 28 January 2007 (UTC)

I have never understood why calculus is viewed with such fear and awe. Archimedes was using the basic ideas thousands of years ago, while Newton and Leibniz began to formalize the essentials many hundreds of years ago. These people were living in a world of knowledge that was incredibly primitive by today's standards, yet they could manage calculus. Even if they were remarkable mathematicians, their ideas were embraced and used widely. Like much of mathematics, including integers, calculus can be tricky to detail formally, but the calculations (and intuition) are so elementary we now routinely do them with computer algebra systems, in both symbolic and numerical form.

Calculus is taught to students in many disciplines in universities throughout the world, so the web contains of a vast range of supporting material. Among these are free, quality textbooks. Some examples:

• Gilbert Strang. Calculus.
• Jerome Keisler. Elementary Calculus: An Approach Using Infinitesimals.

Others can be found through George Cain's site and elsewhere using web searches. A free, open-source, cross-platform computer algebra system to use in checking your work is

• SAGE, downloadable here.

Good luck! --KSmrq^T 18:35, 28 January 2007 (UTC)

Wow, those are all really useful! Thanks a lot. :-) Imaninjapirate^{talk to me} 00:26, 29 January 2007 (UTC)

Of course you can teach yourself, I taught myself calculus. And now I have a degree in mathematics and physics, along with graduate studies. Many famous people in the past taught themselves. Sometimes it is even the best way to learn! Mathmo ^Talk 00:36, 29 January 2007 (UTC)

Also take a look at the precalculus topics. At the bottom of the article you'll find links pointing to e-books. Mr.K. 19:57, 30 January 2007 (UTC)

Linear regression problem

I have some data that I tried analysing using a linear model with R (programming language). The dependent variable is Y. There is an explanatory variable X1. The data consists of two populations, represented by another variable X2. I did a regression analysis Y~X1 in each subpopulation (X2=1 and X2=2), as well as a regression Y~X1 for the combined data set. My problem is that the regression lines, when plotted, are quite different from what I would have expected from the scatterplot. I've put the data, plots and R code on an external page, and would be extremely grateful for feedback on whether I am making a silly mistake in the regression analysis or plotting, or if I might need to consult an opthalmologist.

The first figure shows the results of the regression analysis for each subpopulation (red and blue), as well as the result for the whole data set (green). The second figure repeats the regression line for the combined data set (green), and shows where I had expected to find the regression line(orange). Thanks in advance for any feedback. --NorwegianBlue ^talk 13:46, 28 January 2007 (UTC)

1.

Different lines can be obtained by using different weights to off mean points eg |x_n-x_av| , (x_n-x_av)² it depends on how you want to emphasise the shape of the scatter plot. You could do an analysis of the scatter plot to see if the spread of values seems to match any known distribution.87.102.33.144 14:23, 28 January 2007 (UTC)

2.

The slope of the line of the combined data set didn't match what you expected - here's what I would do.

There are n points. Take a pair of points and find the slope and second variable (constant) for this pair of points eg y=ax+c, y2-y1=(x2-x1)a. Repeat this for each combination of points - there are n(n-1)/2 pairs of points..

Now you should have n(n-1)/2 values for 'a' (the slope) and 'c' (the constant). You can analyse this data set for and average 'a' and 'c', and also find the variance of both.

This should give you not only average slope but a range (amount of accuracy) of values in which 'a' can be expected to lay based on your data set.

Question - does this range of values include your expected (orange) line.. Hope that helps. I'm suggesting you find the 'amount of error' in the line.87.102.33.144 14:40, 28 January 2007 (UTC)

Thanks! Re 1: I was thinking plain old least squares linear regression. My problem isn't that I had some preconceived expectations about the regression lines, but that they are less steep than what my eyes are telling me that they should be when I look at the scatter plot. I have plotted the residuals against quantiles of the normal distribution (qqnorm in R). Except for two outliers at the lower end, and four at the upper end, the residuals appeared to be reasonably normally distributed. And with approx. 1500 data points, I wouldn't expect 6 outliers to have a tremendous effect.

Re 2. That's interesting, I'll try to do what you suggest. Should the average slope and intercept using this procedure be identical to those obtained using least squares linear regression? --NorwegianBlue ^talk 15:11, 28 January 2007 (UTC)

As I understand it - not identical - method of least squares emphasises 'erratic' points more than a magnitude (absolute value) method. So there will be a minor difference (except of course when all the points are exactly on a line - not the case here).

I've just 'thought' of something else - is your method using - vertical offsets (my guess is it is - as it is simpler and more common) - if so you could try perpendicular offsets (much better) see http://mathworld.wolfram.com/LeastSquaresFitting.html (second set of diagrams down) - I think the picture speaks for itself. If your not using perpendicular offsets you really should try this - it does give better results (especially when the slope of the line is high) - you might find it gives a result closer to the one you expected - the eye is usually a good measurer..87.102.33.144 16:19, 28 January 2007 (UTC)

You also might want to look at http://www.tufts.edu/~gdallal/slr.htm which gives "confidence bands from the regression line " - I haven't checked the maths on this - what you need to do is find "the confidence bands from the regression line" - a google search might help or you could ask here - I'd be very suprised if your orange line didn't lie within the confidence bands.

The confidence bands effectively give a measure (statistical) of where the actual line can be expected to lie (within certain probabilities) - like a broad thick line that the actual line must lie within.87.102.33.144 16:19, 28 January 2007 (UTC)

I think you are right in that the lm() function of R minimizes the squared vertical offsets. Browsing the help files, I found that R has a procedure called line() which does robust line fitting. The line thus obtained was slightly "better" than the green line obtained using lm(), but still far from my orange line. --NorwegianBlue ^talk 18:48, 28 January 2007 (UTC)

Looking at your picture, the orange line is more or less the major axis of an ellipse containing most of the points of the joint scatter plot. In general, the line obtained by linear regression has a slope that is less steep. Let U and V be two independent normally distributed (μ = 0, σ = 1) random variables. If X = aU + b, Y = mX + c + dV, for sufficiently large samples the least squares linear regression fit will approximate the line y = mx + d. The lines of equal density in a scatter plot are then ellipses of the form u² + v² = r², for various values of r, where u = (x−b)/a and v = (y-(mx+c))/d. Expressed in x-y coordinates: ((x−b)/a)² + ((y-(mx+c))/d)² = r². In general, the slope of the major axis of these ellipses is not equal to m. As you change the scale of Y, the value of m changes accordingly. But the major/minor axes of an ellipse are not projective invariants. In particular, they are perpendicular, but linear scale transformations do not preserve angles and consequently do not preserve these axes.  --Lambiam^Talk 12:36, 29 January 2007 (UTC)

Thanks, Lambiam. It's correct that I expected the regression line to more or less follow the major axis of that ellipse, and I didn't know that a less steep line was to be expected. Can I infer from your comment, then, that there is nothing obviously wrong with the regression lines as plotted? --NorwegianBlue ^talk 18:58, 29 January 2007 (UTC)

You can infer from my comment that I see nothing obviously wrong with these lines. For a simple visual check, if you have a nicely ellipse-shaped cluster, take the vertical tangents at the left and right sides of the ellipse. The regression line should pass through the points where these lines touch the ellipse. For each point on the regression line, the vertical extents above and below should be (roughly) equal. As far as I can see without a visit to the ophthalmologist, your red, blue and green lines pass this visual check; the orange line does not.  --Lambiam^Talk 13:36, 30 January 2007 (UTC)

Thanks a lot! --NorwegianBlue ^talk 22:11, 30 January 2007 (UTC)

???

94.874^2.3 to 4d.p

(added by anon)

94.874^2.3 = 94.874^23/10 = (94.874²³)^0.1

94.974^23 can be easily calculated - from this what is needed is the tenth root of the result - there are 10 tenth roots , 5 pairs of |r_n| and -|r_n|, 4 pairs of these are imaginary, one pair is real. 87.102.33.144 16:28, 28 January 2007 (UTC)

Complex measures

I'm working through a pretty tough exercise (I think so, anyway) in Royden's Real Analysis, and I was hoping to get some insight. I'll try to keep it as short and sweet as possible.

If you have a set function $\nu$ that maps sets in a sigma-algebra to the complex numbers that maps the empty set to 0 and is countably additive over sequences of disjoint sets, then we call this set function a complex measure. We can write $\nu =\mu _{1}-\mu _{2}+i\mu _{3}-i\mu _{4}$ by first splitting $\nu$ into its real and imaginary parts (which are signed measures), and then applying the Jordan decomposition theorem to each part. This is fine. Next, we need to show that there is a measure $\mu$ and a complex-valued measurable function $\varphi$ with $|\varphi |=1$ such that

$\nu E=\displaystyle \int _{E}\varphi \ d\mu .$

This is where I'm stuck. My guess is that $\mu =\mu _{1}+\mu _{2}+\mu _{3}+\mu _{4}$, since that's what the hint seems to suggest, and then we can apply the Radon-Nikodym theorem to each $\mu _{j}$ to get $\varphi _{j}$, and it seems it should just follow from there. However, I see no reason why $\varphi =\varphi _{1}-\varphi _{2}+i\varphi _{3}-i\varphi _{4}$ should have modulus 1. Am I missing something obvious? It's clear to me that $|\varphi |>0$ $\nu$-a.e., but not that it is precisely equal to 1. Any help would be appreciated, and thanks for taking the time to read this lengthy post. –King Bee ^{(T • C)} 20:03, 28 January 2007 (UTC)

I hereby retract this question. I figured it out. For those who are wondering, it hinges on the fact that the $\varphi _{j}$ have disjoint support, and uses the fact that $\mu _{1}\perp \mu _{2}$ and $\mu _{3}\perp \mu _{4}$. –King Bee ^{(T • C)} 16:41, 29 January 2007 (UTC)

Differentiation question

How do you differentiate a square root? --Lazar Taxon 20:30, 28 January 2007 (UTC)

Do you mean from the definition of the derivative? You'll want to multiply the numerator and denominator of the difference quotient by the conjugate of the numerator. If you mean just by using the rule

${\frac {d}{dx}}(x)^{n}=nx^{n-1},$

begin by noting that ${\sqrt {x}}=x^{1/2}$. –King Bee ^{(T • C)} 20:44, 28 January 2007 (UTC)

We can approach this in a variety of ways. One is, as King Bee (implicitly) suggests, to take advantage of Newton's general form of the binomial theorem, where we can expand (a+b)^c for values of c other than integers. But I think in this case we ought to use the opportunity to exploit a fact about inverse functions.

Suppose y is a function of x, namely x². Then at any given value of x we find that the ratio of change in y to change in x, ^dy⁄_dx, is 2x. We call this the derivative of y with respect to x, right? But then the ratio of change in x to change in y is the reciprocal, ^dx⁄_dy. That is, consider x as a function of y, namely √y. The derivative at a given value of y is ¹⁄_2x, which is ¹⁄_2√y. We may also write the result as ¹⁄₂y^−1/2, for comparison with the binomial version. (Note we have switched the roles of x and y, for purposes of explanation.) Beyond the immediate application, this concept is worth remembering. --KSmrq^T 02:35, 29 January 2007 (UTC)