Wikipedia:Reference desk/Archives/Mathematics/2007 October 31

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October 31

Another Algebra Question

It is quite coincidental that I just came to this Math Ref. Desk to ask the following question ... and I noticed another Algebra question posted immediately above mine. I was wondering ... is there a proper term (such as mathematician, scientist, composer, astronomer, musician, pianist, actor, botanist, horticulturist, obstetrician, cardiologist, etc.) to describe the "occupation" (as it were) of one who does algebra? Algebraicians ... that's not a real word, I assume. Algebraist ... is that a "real" word? What would be the correct / recognizable term for this? Thanks. (Joseph A. Spadaro 06:19, 31 October 2007 (UTC))

That's a silly question. Nobody does algebra in a vacuum. The occupation of most people who does algebra are as you said engineers, scientists, technicians, teachers, lecturers, builders, accountants, economists etc etc.

It's like asking "What is the proper term for people who does arithmetic? Arithmeticians???"

58.109.93.128 07:48, 31 October 2007 (UTC)

Why is that a silly question? Yes, there is a word algebraist in English dictionaries[1]. For example, you can say that "Bartel Leendert van der Waerden was "a real professional algebraist".[2] We even have a User:Algebraist. Likewise, arithmetician is a normal English word.[3]  --Lambiam 08:01, 31 October 2007 (UTC)

See also Wiktionary, a Wiki dictionary: wikt:algebraist, wikt:arithmetician. --CiaPan 08:14, 31 October 2007 (UTC)

Ian Banks wrote a book called The Algebraist, a good read but no mathematics. --Salix alba (talk) 08:37, 31 October 2007 (UTC)

Note that the algebra in the mathematician specialization "Algebraist" is not quite the same as the algebra which is taught at school. -- Meni Rosenfeld (talk) 10:33, 31 October 2007 (UTC)

Algebraicist also seems to be in occasional use, though I can't find a dictionary that approves of it. Algebraist 21:00, 31 October 2007 (UTC)

The term "algebraist" is certainly used to refer to research mathematicians who do algebra (but see Meni Rosenfeld's comment). I've never heard the term arithmetician. The more appropriate term for what that might mean is number theorist, or possibly "arithmetic geometer" (both of which are quite prevalent). kfgauss 07:19, 1 November 2007 (UTC)

Thanks to all for the input -- much appreciated! Thanks. (Joseph A. Spadaro 04:40, 5 November 2007 (UTC))

Energy Resources Question

Here is a question related to the Global Warming (well sort of).

It's to do with the world's coal reserves.

If the world's coal reserve is R

and the current coal consumption (per year) is C

and the growth (per year) in annual coal consumption is G (where G > 1)

In how many years will the half the current reserve be consumed?

58.109.93.128 08:06, 31 October 2007 (UTC)

Assuming steady exponential growth, measuring time t in years, and setting t = 0 for the present, the coal consumption rate at time t is C exp(λt), where λ = log G (so that, for t = 1, we get C exp λ = CG). To get the amount of coal consumed from t = 0 to t = T, we integrate this, which gives us Cλ⁻¹(exp(λT) − 1). Half the reserve will be consumed when this equals .5R. Solving Cλ⁻¹(exp(λT) − 1) = .5R for T results in T = λ⁻¹log(.5RC⁻¹λ + 1).  --Lambiam 08:32, 31 October 2007 (UTC)

T = (1/log G) * log(((log G) * 0.5R/C)+1) . Hmmm Let me check the answer.

Using R=100 C=2 G=1.02

According to yours, I get T1=20.31

According to mine, I get T2=20.47

My formula is T = log[1-(0.5R (1-G)/C)]/log G

211.28.129.206 11:03, 31 October 2007 (UTC)

As G goes to infinity, how long will half the reserve last? With your formula more than a year. But if G > (R/C)², the rate of consumption will exceed R after half a year, so if there is still .5R left, it will certainly be consumed within the next half year. I showed how I derived my formula. Where did yours come from?  --Lambiam 11:57, 31 October 2007 (UTC)

The second formula T = log[1-(0.5R (1-G)/C)]/log G assumes a discrete model in which an amount C is consumed in year 1, CG in year2, CG² in year 3, so after T years the total amount consumed is C(1-G^T)/(1-G). Then equate this to 0.5R and solve for T.

Of course, a non-integer value for T breaks the discrete model - the correct way to use a discrete model would be to take the integer part of T, 20 years, find the proportion of R remaining at the beginning of year 21 and the end of year 21, and use linear interpolation to find the value of T corresponding to 0.5R. Gandalf61 12:06, 31 October 2007 (UTC)

I would have to say that assuming a discrete model is quiet flawed since consumption is an on going process, it would be very much continuous. A math-wiki 19:38, 31 October 2007 (UTC)

I agree. I didn't say it was a appropriate model - I just said that it seemed to be the source of the second formula. Gandalf61 22:56, 31 October 2007 (UTC)

According to this http://www.energybulletin.net/29919.html

According to the widely accepted view, at current production levels proven coal reserves will last 155 years (this according to the World Coal Institute). The US Department of Energy (USDoE) projects annual global coal consumption to grow 2.5 per cent a year through 2030, by which time world consumption will be nearly double that of today.

So R=155 C=1 and G=1.025

So long long until half the current coal reserves is consumed? 202.168.50.40 00:31, 1 November 2007 (UTC)

Parameterizing a line by its length

For some reason I cannot seem to correctly parameterize a line by the length along the line. That is, I want to write: $x=m_{1}s+b_{1}$, $y=m_{2}s+b_{2}$ where $s$ goes from zero to the total length of the line. I did a page full of math and failed miserably at getting it right. So that we're communicating in the same language, my line begins at the point $(0,a)$ and ends at $(c,w/2)$. (There are reasons for this notation, this is part of a larger geometric figure I'm trying to mesh). I think my brain is just sore today from too much geometry... can someone help me out? Thanks, moink 18:20, 31 October 2007 (UTC)

I find it easier to use a parameter that goes from 0 to 1, so $x=x_{initial}+(x_{final}-x_{initial})t$ and $y=y_{initial}+(y_{final}-y_{initial})t$. With your initial and final points, $x=ct$ and $y=a+(w/2-a)t$. My parameter $t$ is your parameter $s$ divided by the length of the line, $t={\frac {s}{\sqrt {c^{2}+(w/2-a)^{2}}}}$ --tcsetattr (talk / contribs) 19:00, 31 October 2007 (UTC)

((edit conflict))

You need $x(0)=0,\ y(0)=a$ and $x(1)=c,\ y(1)=w/2.$ This constitutes a system of four linear equations. The solution is quite obvious: plug $s=0$ into your equations, and you get $b_{1}=0$ from x-equation and $b_{2}=a$ from y. Then plug $b_{1},\ b_{2}$ and $s=1$, and you get $m_{1}=c-0$ and $m_{2}=w/2-a.$ --CiaPan 19:07, 31 October 2007 (UTC)

Yeah, I figured it out just now and was going to come here and say "never mind" and give the answer. Anyway, it's important that it be the length of the line because I'm doing this for surface meshing purposes, and the triangles have to be approximately right. I just used:

$m={\frac {W/2-a}{c}}$

$l=c{\sqrt {m^{2}+1}}$ (line length, so s will go from 0 to l)

$x={\frac {s}{m^{2}+1}}$

$y=mx+a$.

Thanks! moink 19:10, 31 October 2007 (UTC)

For future reference, affine geometry and barycentric coordinates are helpful. To go between point p₀ and point p₁ we may use the affine combination (1-t)p₀+tp₁, with 0 ≤ t ≤ 1. To use a parameter a going from a₀ to a₁, set t = (a₁−s)/(a₁−a₀).

This generalizes in a variety of nice ways. For example, consider interpolation through three points, p₀, p₁, and p₂, with associated parameter values a₀, a₁, and a₂ (with a₀ < a₁ < a₂). Compute p₀₁ between the first two points using the first two parameter stops, p₁₂ between the last two points using the last two parameter stops, and then p between these computed points using the first and last parameter stops. Interpolation through any number of successive points is just more of the same; but it is not recommended for more than a few points because of Runge's phenomenon.

We can approach Bézier curves and B-splines in similar fashion. Take those same three points and use them to create a quadratic Bézier curve parameterized from 0 to 1 as follows. Use the first two points with unit parameterization, the last two points likewise, and then the two computed points. Thus

${\begin{aligned}\mathbf {p} &=(1-t){\big (}(1-t)\mathbf {p} _{0}+t\mathbf {p} _{1}{\big )}+t{\big (}(1-t)\mathbf {p} _{1}+t\mathbf {p} _{2}{\big )}\\&=(1-t)^{2}\mathbf {p} _{0}+2(1-t)t\mathbf {p} _{1}+t^{2}\mathbf {p} _{2}.\end{aligned}}$

With four points we get a cubic Bézier curve, and so on for any degree we like. The polynomials weighting each point in the expanded form are called Bernstein polynomials. Higher degrees get rather expensive, and also tend to do far more smoothing than we need, so we prefer splines — piecewise polynomial curves. Happily, B-splines are just as easy to compute (with non-uniform knot vectors, if we wish). We can even create triangular Bézier surface patches with a trivial variation of the same ideas. --KSmrq^T 23:14, 31 October 2007 (UTC)

Well, thanks, but that still wouldn't be parametrized by length (line length, arc length). It's really easy to parameterize something, it's much harder to do it by length! For instance, I've also had to mesh up the surface of an elliptical cylinder. I can tell I did it wrong because the triangles at the end of the major axis are much smaller/farther-from-equilateral than the triangles at the end of the minor axis. It's really easy to parameterize an ellipse, but doing it by arc length is a serious challenge. moink 23:52, 31 October 2007 (UTC)

I'm assuming you know how to compute the length of a line from its two end points!

Curves are usually expensive to parameterize by arclength. To find the arclength we must differentiate the position with respect to the existing parameter, do the Pythagoras thing to convert the resulting vector to a length, then integrate the resulting square root of sum of squares. For an ellipse, which is already simpler than an ellipsoid, this leads to elliptic integrals. Since we're generally lacking closed form solutions we are forced to use numerical approximations. If you then want to get approximately equal-area triangles on an ellipsoid, the challenge is greater still. --KSmrq^T 01:27, 1 November 2007 (UTC)