Wikipedia:Reference desk/Archives/Mathematics/2006 December 29

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December 29

Integrals with Mathematica

I'm trying to figure out some output from Mathematica. Here's the input:

$\int _{0}^{1}{\frac {\log ^{16}{t}}{\sqrt {1-t^{2}}}}dt$

And this is what I get for output

$2092278988000\,$

$_{18}F_{17}({\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},$

${\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {1}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},$

${\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}},{\frac {3}{2}};1)$

Anyone know what the $_{n}F_{r}$ notation is? —Mets501 (talk) 03:18, 29 December 2006 (UTC)

See “hypergeometric series”. --KSmrq^T 05:09, 29 December 2006 (UTC)

Thanks! —Mets501 (talk) 13:47, 29 December 2006 (UTC)

Integral notation

Suppose I want to integrate the function $f$ over the set $S$:

$\int \limits _{r\in S}f(r)\,\mathrm {d}$

As you can see, there is something missing after the "$\mathrm {d}$". What should I really put there? Is "$r$" the right thing, or perhaps "$S$"? Also, when should there be a "$\mathrm {d}$" in the first place? In the area of Lebesgue integrals, I see you put a measure (often denoted by "$\mu$") there, but the "$\mathrm {d}$" seems to be something else. Thanks in advance! —Bromskloss 10:35, 29 December 2006 (UTC)

It should probably be an r. Strictly speaking, any variable would be valid, but any variable that isn't r would result in an uninteresting result - specifically, the "area" of S. So it's safe to assume the integral should be with respect to r - which means (intuitively) that we will vary r within the set S, and then sum up all the values we get from f(r). Maelin (Talk | Contribs) 10:56, 29 December 2006 (UTC)

You could use the notation

$\int \limits _{r\in S}f(r)\,\mathrm {d} S$

with the understanding that it is clear from the context what measure you are using on S. If, for example, S is a subset of $\mathbb {R} ^{2}$, r could be a point in Euclidean space but with polar coordinates, which gives a different measure than if you have Cartesian coordinates. See also Multiple integral and Differential form.  --Lambiam^Talk 11:34, 29 December 2006 (UTC)

Hmm, are you saying that

$\int \limits _{r\in S}f(r)\,\mathrm {d} r$ and $\int \limits _{r\in S}f(r)\,\mathrm {d} S$

represent the same thing? That sounds strange, don't you think? —Bromskloss 20:08, 29 December 2006 (UTC)

An example with the equivalent of $\mathrm {d} r$ can be seen in the third equation of Moment of inertia. Some example with the equivalent of $\mathrm {d} S$ are found in the first two equations of these lecture notes, several equations (e.g, (3.11)) of these lecture notes, and the left-hand side of the Kelvin-Stokes theorem. Note that the notation is not consistent across these examples. For yet another form, see Green's theorem, in which the A in dA stands for the concept "Area". Clearly, the notation is not de facto standardized and not always equally rational.  --Lambiam^Talk 21:47, 29 December 2006 (UTC)

The question omits necessary information. We cannot integrate over a set without additional structure either stated or implied. For example, suppose the set is {a,b,c} and the function assigns 1, 2, and 3 to each element, respectively; or suppose the set is the rational numbers, and the function takes the constant value 1. If we have a σ-algebra of subsets of S and a measure μ defined on it, and if f is a measurable function, then we could write

$\int _{S}f\,d\mu .$

(Please note that the d is usually set in italics, not roman, in this context.) But let's not speculate; suppose you just tell us the circumstances you have in mind. --KSmrq^T 00:56, 30 December 2006 (UTC)

I wasn't imagining anything incredibly fancy. Let's just say $S$ is the unit disc:

$S=\left\{(x,y)\in \mathbb {R} ^{2}\,:\,x^{2}+y^{2}\leq 1\right\}$

As for algebra and measure (if needed), can we assume something usual? ;-) I'm afraid I don't know enough about them to say anything useful. —Bromskloss 21:07, 30 December 2006 (UTC)

The usual area form on the disc might be written in a few ways when an author thinks the meaning is clear:

$d^{2}r=dA=dx\,dy=dx\wedge dy=d(x\,dy)=\cdots$

I'd throw in a few using polar coordinates, except that I usually get those wrong. For information on what the "d" means as an independent mathematical object, see Exterior derivative.

For most purposes, "dA" shouldn't be taken too literally as d of some well-defined mathematical object A. Certainly d²r doesn't mean what it appears to. Melchoir 23:55, 3 January 2006 (UTC)

A unit disk in the real plane allows us to assume enough structure to give a meaningful answer. We don't even need to formally discuss measure, though we have a standard natural definition.

The first thing to deal with is the set inclusion notation: Get rid of it; it is highly nonstandard and semantically wrong. We are integrating over an area, so conceptually we need to sum over an infinite number of infinitesimal areas, not over points.

The next thing to discard is the roman typeface on the “d”, which is also semantically wrong here, at least usually.

Now, for a concrete example, take f(x,y) to be

$f(x,y)=\rho \,x,\,\!$

as when finding the x coordinate of the center of mass of a non-uniform thin plate with density ρ(x,y). One common notation would be

$\int _{S}\rho \,x\,dA,$

where the dA denotes a differential area, canonically an infinitesimal box in x and y. If we wish to be more explicit we could write a double integral

$\iint \limits _{S}\rho \,x\,dx\,dy,$

which we might then convert to nested integrals. However, for a disk it is often more natural to use polar coordinates, r and θ, rather than cartesian x and y. The dA notation allows us to suppress such detail, but if we want explicit polar coordinates we would convert x to r cos(θ), convert dA to r dθ dr, and write

$\iint _{S}\rho \,r\cos(\theta )\,r\,d\theta \,dr=\int _{0}^{1}r^{2}\left(\int _{0}^{2\pi }\rho \cos(\theta )\,d\theta \right)\,dr.$

We have implicitly converted ρ to polar coordinates as well; and if ρ happens to be radially symmetric (so that it does not depend on θ), we can pull it out of the inner integral.

Suppose we wish to go one step further and apply Stokes' theorem (or its specialization, Green's theorem). This requires us to view the thing we are integrating as the exterior derivative, dω, of a differential form, ω (for suitable ω), in which case we may replace the integral over the disk by an integral around its bounding circle.

$\int _{S}\mathrm {d} \omega =\int _{\partial S}\omega .\,\!$

(This equation would be an example of an acceptable use of a roman “d”, though even here italic is common.) To illustrate, let us integrate an implicit polynomial for a trifolium using the machinery of differential forms explicitly. We can write

${\begin{aligned}\omega &{}=-\left(3x^{4}y-3x^{3}y+x^{2}y^{3}+3xy^{3}\right)\,dx+\left(3xy^{4}+x^{3}y^{2}\right)\,dy,\\\mathrm {d} \omega &{}=3\left((x^{2}+y^{2})^{2}-x^{3}+3xy^{2}\right)\,dx\wedge dy.\end{aligned}}\,\!$

Since the boundary of the disk is a circle, we may choose to integrate ω using x = cos θ, y = sin θ. Alternatively, we could use y = ±(1−x²)^1/2, and integrate from right to left along the top half of the circle and from left to right along the bottom half (thus always keeping the area to our left).

${\begin{aligned}\int _{S}\mathrm {d} \omega &{}=\int _{S}3\left((x^{2}+y^{2})^{2}-x^{3}+3xy^{2}\right)\,dx\wedge dy\\&{}=\int _{\partial S}\omega \\&{}=\int _{\partial S}\left[-\left(3x^{4}y-3x^{3}y+x^{2}y^{3}+3xy^{3}\right)\,dx+\left(3xy^{4}+x^{3}y^{2}\right)\,dy\right]\\&{}=\int _{1}^{-1}\left(6x^{3}-4x^{2}-3x\right){\sqrt {1-x^{2}}}\,dx-\int _{-1}^{1}\left(6x^{3}-4x^{2}-3x\right){\sqrt {1-x^{2}}}\,dx\\&{}=\pi \end{aligned}}$

These are not the only possibilities, but perhaps they will serve to give some idea of the diversity. --KSmrq^T 12:59, 6 January 2006 (UTC)

Indefinite multiple integrals

Taken from the Multiple integral article:

"Because it is impossible to calculate the antiderivative of a function of more than one variable, indefinite multiple integrals do not exist so they are all definite integrals."

Why is it impossible? Thanks. --Ķĩřβȳ♥ŤįɱéØ 11:18, 29 December 2006 (UTC)

It is formulated rather absolutely, but the essential problem is in defining what it would mean in the first place. Without a definition we have a meaningless concept. We would need some kind of mathematical object, and a notion of derivative defined on that kind of object, such that applying it gives you the "body" of the "indefinite multiple" integral. Actually, such a thing does exist: differential forms may have an "exterior derivative" that is again a differential form, and in some sense the body of a multiple integral is a differential form, so you could look for an "exterior antiderivative". But this is not a direct generalization of conventional indefinite integrals for a single variable.  --Lambiam^Talk 11:50, 29 December 2006 (UTC)

Algebra resources

As some people have pointed out in my calculus question posted on here a few days ago, my algebra is not too good. Unfortunately, my GCSE textbook that I have doesn't cover anything like the binomial theorem, limits, or anything else that is involved in some of the more 'interesting' areas of maths, such as calculus. Can anyone point me to some free online resources for learning algebra that include these topics and explain them reasonably well and correctly? (I have found some sites that tell you to do the wrong thing, and plenty of ones that skip out steps that they think are obvious, which they are to someone who knows what to do but not to someone who doesn't) Thanks very much. As an aside, when people mention how intelligent Newton was, they normally talk about his theories of gravitation. Surely his invention/discovery of calculus and other mathematical tools is more important, especially seeing as he was wrong about gravity, and the fact that most new physical ideas need calculus to be worked with? --80.229.152.246 14:21, 29 December 2006 (UTC)

You can use our site! See for examples Binomial theorem, Limit (mathematics), and Calculus. You could also try navigating Category:Mathematics, which organizes all of our mathematics articles by topic. —Mets501 (talk) 16:44, 29 December 2006 (UTC)

Wikipedia isn't really well suited to teaching subjects, but Wikibooks is, provided that they have books available on your subject. Also, Newton didn't single-handedly invent calculus, many others contributed, too. StuRat 19:13, 29 December 2006 (UTC)

I am delighted by your question. It takes uncommon wisdom and humility to acknowledge a need to learn more and to seek assistance. In this case, your need is a shared one, and there are quality sources on the web.

One place to start is The Math Forum at Drexel, where you will find a page of resources. For online mathematics texts, the AMS has links, though many are on more advanced topics. Still, you can find jewels like Calculus, by Gilbert Strang, a Professor of Mathematics at MIT. You might also browse topics at The Mathematical Atlas.

If I have a chance, I'll see if I can dig up a good elementary discussion of the binomial theorem. Meanwhile, these sources should keep you in good reading. --KSmrq^T 22:07, 29 December 2006 (UTC)

Thanks for all your answers everyone. The sites look to be quite good and useful and I'll make sure I check them all out. The problem with using Wikipedia and Wikibooks for learning this kind of thing is not that the articles are bad or hard to understand, but that it is hard to find them without knowing what you are looking for. For example, I now know the binomial theorem thanks to a link posted on here. However, I had never even heard of it before and so could not learn it without the help given here. Although the Category:Mathematics is a nice list of topics, I have no idea which ones to look at first in order to be able to understand them. The only reason I went onto looking at calculus before I looked at things like the binomial theorem (which I should really know before attempting calculus) is because I had actually heard of it and could therefore search for it. Looking at my GCSE textbook for topics is no good as I already know the ones that are included (which in my opinion isn't very many), and if I don't then the textbook explains them reasonably well. Looking at the website for the National Curriculum is a waste of time as well as it doesn't really say anything about what you should learn at GCSE and A level. Because of this, I have to take my chances looking at random articles, hoping that I can understand them and reading articles in magazines and newspapers that contain little bits of information. The problem with doing this is that the articles in the newspapers often do not have enough room to explain things clearly and thoroughly and only really serve as a pointer that such a thing exists. In some cases, the magazines and newspapers get things wrong and completely confuse one subject with another. This is why I think that your answers are so good and that you are all very nice people, encouraging me to learn more rather than laughing at my ineptitude. Thank you very much for your kindness and answers. --80.229.152.246 14:13, 30 December 2006 (UTC)

You might want to get yourself an algebra textbook (you could buy one, but hopefully a friend has an old one or you can check one out of the library). While the textbook may or may not be a good learning aide, it's a safe bet that it will at least list the names of all the relevant concepts. Once you have the names, you can look them up at Wikipedia or elsewhere. StuRat 16:46, 30 December 2006 (UTC)

Equation of a Line

Hi, I was reading an explanation on the equation of a line and I got stuck on something:

How do we get

${\frac {y-y_{1}}{y_{2}-y_{1}}}={\frac {x-x_{1}}{x_{2}-x_{1}}}$

from

$y=mx+c\,$ ?

Thanks, Alex Ng 23:08, 29 December 2006 (UTC)

Who says you're supposed to get the first one from the second one? They use different parameters to describe the line: the first uses two points, and the second uses the slope and the y-intercept. —Keenan Pepper 23:12, 29 December 2006 (UTC)

Thanks for the quick reply. Sorry, I made a wrong assumption that the first was derived somehow from the second. I should restate my question: Where does the first equation come from (if from anywhere at all)? Alex Ng 23:30, 29 December 2006 (UTC)

Start with the slope equal to the change in y over the change in x:

$m={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}$

Next, can you define c in terms of ($x_{1},x_{2}$) and ($y_{1},y_{2}$) ? If so, you can substitute both into $y=mx+c$ and solve from there. StuRat 23:37, 29 December 2006 (UTC)

I don't know if this is of any help, but I think you may need the standard equation for a line in the form $Ax+By=C$ --Ķĩřβȳ♥ŤįɱéØ 01:02, 30 December 2006 (UTC)

The assumption is that the points (x₁, y₁) and (x₂, y₂) are points on the line and so satisfy the equation for points on the line:

$(1)~y=mx+c\,\!.$

Since (x₁, y₁) is on the line, we then have:

$(2)~y_{1}=mx_{1}+c\,\!.$

Substract (2) from (1) to give

$(3)~y-y_{1}=m(x-x_{1})\,\!.$

This equation is again an equation for the whole line, and is a suitable form if the slope of the line and one point on it are given. Using the fact that (x₂, y₂) is also on the line, we now have:

$(4)~y_{2}-y_{1}=m(x_{2}-x_{1})\,\!.$

Divide (3) by (4) – only possible if y₁ ≠ y₂ – to get the final equation:

${\frac {y-y_{1}}{y_{2}-y_{1}}}={\frac {x-x_{1}}{x_{2}-x_{1}}}.$

 --Lambiam^Talk 04:35, 30 December 2006 (UTC)

(Edit conflict) We can write equations for a line in a number of ways, each useful for certain purposes. The choice may depend on what data we are given, or perhaps on what information we want to reveal. One common form of data would be a pair of distinct points on the line,

$P_{1}=(x_{1},y_{1}),\qquad P_{2}=(x_{2},y_{2}).\,\!$

The first equation you propose is a natural form in this case. If we substitute x₁ for x and y₁ for y, both sides are zero, so point P₁ satisfies the equation. If we substitute x₂ for x and y₂ for y, both sides are one, so point P₂ satisfies the equation. In fact, if x equals (1−t)x₁+tx₂ and y equals (1−t)y₁+ty₂, the equation is satisfied.

This leads to another form of line equation,

$P=(1-t)P_{1}+tP_{2}.\,\!$

Where the previous equation allows us to test if a point is on the line, this one allows us to generate points guaranteed to be on the line.

So what about y = mx+c? This form does not allow us to insert two known points to get an equation; in fact, if x₁ = x₂ (so the line is vertical), we cannot use this form. However, if the line is not vertical it allows us to both test and generate points. Any value of x substituted on the right gives the associated value for y. In particular, when x equals zero, the point lies on the y axis, at position (0,c). This is called the “y intercept” of the line. When x equals −^c⁄_m, the right side is zero, so the line crosses the x axis at (−^c⁄_m,0). (We cannot do this if m is zero, indicating a horizontal line.) The value c shifts the line up or down, while the value of m (the slope) tilts it.

Since the slope-intercept form does allow us to generate points, we can convert to the first form by doing so. Converting in the reverse direction merely requires a little algebraic manipulation, to solve for y, as StuRat has indicated.

$y={\frac {y_{1}-y_{2}}{x_{1}-x_{2}}}x+{\frac {x_{1}y_{2}-x_{2}y_{1}}{x_{1}-x_{2}}}$

For some uses of lines the slope-intercept information is what is provided. For example, if we are at an elevation of 200 m and climbing a 20% grade, we can take the starting elevation as the intercept and the grade (written 0.20) as the slope. --KSmrq^T 05:18, 30 December 2006 (UTC)

Thanks everyone for taking the time to help me understand this equation. StuRat KSmrq *correction* (sorry!), I have a question: I don't understand where (1−t)x1+tx2 comes from and what the t means. Could you maybe show me a quick example of it in use? Alex Ng 08:07, 30 December 2006 (UTC)

That was me, KSmrq, not StuRat, who introduced t. A weighted sum of (the coordinates of) points in which the weights themselves sum to 1 is called an affine combination. (We have an article, but it needs work.) Such combinations have nice geometric properties, which I used to advantage. Here t is an arbitrary real number (and t plus 1−t clearly sum to 1). When 0 ≤ t ≤ 1, the equation I gave produces points on the line segment from P₁ to P₂, which is often handy in applications. We call t a parameter, and say we have a parametric equation for the line.

Remember I said that the slope-intercept form is not available for vertical lines? The parametric equation is always available. For example, let P₁ = (3,0), and let P₂ = (3,4). If t is ¹⁄₂, we get the midpoint of the line segment, (3,2). The parameter t allows us to select different points; but since all points have the same x value, the slope-intercept form does not. --KSmrq^T 09:20, 30 December 2006 (UTC)

I see how it works now! Thanks KSmrq! Alex Ng 20:00, 30 December 2006 (UTC)