August 27

Is there a term for quantities, such as absolute temperature and P/E ratio, that are effectively higher when negative than when positive? NeonMerlin 00:13, 27 August 2007 (UTC)[reply]

The original Celsius scale of temperatures—see History of the Celsius temperature scale on The Uppsala Astronomical Observatory's web server (one of external links from Celsius article.) --CiaPan 08:30, 27 August 2007 (UTC)[reply]

I'm not sure this is really relevant. The OP's intention was for quantities where high positive real numbers are between low positive numbers and negative numbers. Also, he asked for a name for such quantities, not examples. Anyway, I'm not sure there is a general name, but this is mostly because the phenomenon isn't that deep - it is simply the result of taking the inverse of an "ordinary" measure (for example, the "true" measure of "hotness" can be said to be the exponent in the Boltzmann distribution, which is proportional to ${\displaystyle -{\tfrac {1}{T))}$). -- Meni Rosenfeld (talk) 16:22, 27 August 2007 (UTC)[reply]

Stellar magnitude is an example where a ranking system ("stars of the first magnitude") evolved into a continuous scale; thus the planet Jupiter currently has an apparent magnitude of approximately −2, brighter than the North star Polaris at approximately +2. We also have electrons with a negative electrical charge. I don't think there is a general name for such situations. --KSmrq^T 01:00, 28 August 2007 (UTC)[reply]

Would not golf scores fall into this category? A golfer with a score of −3 (in relation to par) would be five strokes ahead of a golfer at +2. And if you are counting actual strokes, on most courses that would equate to a score of 69 beating a round of 74. — Michael J 22:39, 29 August 2007 (UTC)[reply]

A few months ago, I've read a very interesting mathematical article I found at Digg. The author, Robert Palais, explains why our definition of pi based on the circle's diameter is bad, and that the value that should've got that name is 6.283185..., that is, ${\displaystyle 2\pi }$. The idea wasn't new to me, but it never occurred me how deep this bad definition goes before reading the article. The most convincing argument is probably a better Euler's identity: ${\displaystyle e^{\pi i}=1}$.

So I've been wondering, what else has been written on the subject? Is there any relevant fuzz about this in the mathematical community? Is there any standard name or symbol for ${\displaystyle 2\pi }$? The author proposes ${\displaystyle \pi \!\!\!\pi }$, but it's kind of patchy. Can you guys think of examples where 3.14... seems like a better choice than 6.28...? — Kieff | Talk 05:11, 27 August 2007 (UTC)[reply]

Perhaps you will be interested in Talk:Pi/Archive 3#2pi?, Talk:Pi/Archive 4#π vs 2π and the unit circle and Talk:Pi/Archive 5#2 pi. -- Meni Rosenfeld (talk) 08:08, 27 August 2007 (UTC)[reply]

To address your last question more directly: I don't think there is any situation where ${\displaystyle \pi }$ is certainly better than ${\displaystyle \pi \!\!\!\pi }$, but two examples where at least superficially it appears so is ${\displaystyle e^{\pi i}+1=0}$ (which in fact conveys more information than ${\displaystyle e^{\pi \!\!\!\pi i}=1}$), and ${\displaystyle A=\pi r^{2))$ (though, of course, it can be argued that ${\displaystyle A={\frac {1}{2))\pi \!\!\!\pi r^{2))$ is more natural with the half coming from integration, from the formula for the area of a triangle or whatever). -- Meni Rosenfeld (talk) 08:26, 27 August 2007 (UTC)[reply]

It bears repeating though that Euler's formula, both geometrically (one-half turn gives -1) and algebraically (-1 is a square root of 1), makes more sense with the three-legged pi. And it introduces the constant 2, which is extremely important throughout math. nadav (talk) 08:45, 27 August 2007 (UTC)[reply]

Thanks for that - I think I've seen that article before. What I'd like to ask is why did 'we' pick pi=3.14.. and not 6.2.. both seem equally as likely (especially pre Euler) - anyone got a historical perspective on that (long memories?)87.102.45.106 12:24, 27 August 2007 (UTC)[reply]

scrub that found the answer at History_of_π#History_of_the_notation87.102.45.106 12:51, 27 August 2007 (UTC)[reply]

Apart from the notation π coming to mean the ratio between a circle's circumference and its diameter – rather than radius – the number thereby represented, approximately 3.14, had been known since long as Archimedes' constant. Why Archimedes chose to give a numerical approximation of π rather than 2π is unknown, as far as I know.  --Lambiam 14:17, 27 August 2007 (UTC)[reply]

So we should be blaming Archimedes for this? 189.15.89.220 15:04, 27 August 2007 (UTC)[reply]

It looks like there is a natural human tendency to pick pi and not 2pi as a standard see History_of_π#Computation - of the ancient estimates of pi only one has gone for 2pi. (note I don't have access to the original documents so I've taken what is written there literally). Why this would be I've no idea.87.102.85.15 15:12, 27 August 2007 (UTC)[reply]

I suppose the ratio of the circumference to the diameter was chosen simply because a diameter is easier to measure empirically -- for example, measuring the distance across a hole rather than to the center. --TrippingTroubadour 19:34, 15 September 2007 (UTC)[reply]

I have just noticed that if I calculate the integral of 1/x between a and -a I get iπ - this never happened before

ʃ(1/x)dx = ln(x) + C

So between a and -a I get

ln(a) + C - ( ln(-a) + C ) = ln(a) - ln(-a) = ln(a) - ln(-1a) = ln(a) - ln(a) - ln(-1)

ln(a) - ln(a) - ln(-1) = ln(-1) = iπ (+2inπ n=integer)

But the graph is of 1/x is symmetrical about x=0 so where does the imaginary bit come from?

I must have made a mistake or is it more complex (excuse the pun)?87.102.85.15 14:57, 27 August 2007 (UTC)[reply]

The anti-derivative of 1/x is ln(|x|), not ln(x). Otherwise, as you see, you get strange imaginary problems. The integral of the stuff on the left must be equal to the integral of the stuff on the right. Gscshoyru 15:31, 27 August 2007 (UTC)[reply]

The standard definitions for integrals of real functions simply do not apply to the function ${\displaystyle f(x)={\tfrac {1}{x))}$ in an interval containing 0, and neither do the fundumental theorems of calculus. At the most basic level, the mistake was trying to apply them in the first place.

Instead, you may want to consider a path integral in the complex plane; however, since the function has a simple pole, the integral is path-dependent and cannot be expressed simply as the difference of an antiderivative. The antiderivative is multi-valued. -- Meni Rosenfeld (talk) 15:43, 27 August 2007 (UTC)[reply]

However, the path-integral does evaluate to the given value (2n+1)iπ, with the n path-dependent Algebraist 15:55, 27 August 2007 (UTC)[reply]

it seems to me that when x>0 there are a number of curved lines I can move along for ln(x) ie ln(x) , ln(x)+2iπ , ln(x)+4iπ etc and also another set of lines when x<0 this time ln(x) + iπ (odd numbers).. (is this is what you mean by 'path integral'?)

Is it right to assume that there is no connection of this lines across x=0?87.102.85.15 17:40, 27 August 2007 (UTC)[reply]

This is hard to answer unless you have a background in complex analysis, a fascinating and fundamental subject which you would do well to study in more depth (after which you will have no need for our answers). Within this framework, we would say that the logarithm function has a branch point at 0, which is the cause of all the strange phenomena we see here. Also, you may want to take a look at path integral. -- Meni Rosenfeld (talk) 18:26, 27 August 2007 (UTC)[reply]

Thanks.87.102.85.15 18:37, 27 August 2007 (UTC)[reply]

If you just think about it trivially, the integral is the area under the curve. If you integrate from some positive constant to some other positive constant, or some negative to some negative constant, you should be fine. But if you are integrating from 0 to something you will run into a problem, because the graph runs off to infinity at 0. So you will have a problem there! —Preceding unsigned comment added by 129.78.64.102 (talk • contribs) 22:47, 27 August 2007

You can split the integral into two improper integrals, the part before 0 and the part after 0: ${\displaystyle \int _{-a}^{a}{\frac {1}{x))\,dx=\int _{-a}^{0}{\frac {1}{x))\,dx+\int _{0}^{a}{\frac {1}{x))\,dx}$

but the problem is that both the parts diverge (the first to negative infinity, the second to positive infinity) so the integral is not defined. However, the Cauchy principal value evaluates to 0. --Spoon! 12:40, 28 August 2007 (UTC)[reply]

Is there any free Matlab-like program that can read Matlab scripts please? The List of numerical analysis software lists a few free programs similar to Matlab, but can any of these interperate a Matlab script? Thanks 80.0.121.94 16:20, 27 August 2007 (UTC)[reply]

GNU Octave can generally run scripts written for recent versions of MATLAB. —Bromskloss 19:14, 27 August 2007 (UTC)[reply]

for the Point Isabel Regional Shoreline article i need the equivalents for:

• 23 acres in sq.km and sq.mi.
• 16 to 25 feet in m
• 3.2mi in km

thanksCholga^talK! 21:46, 27 August 2007 (UTC)[reply]

Google can do these calculations: e.g. 23 acres in square kilometers, 23 acres in square miles nadav (talk) 22:06, 27 August 2007 (UTC)[reply]

Also, if you enjoy knowing you can do problems like this yourself, our various articles have the conversion factors you'd need. For example, Acre notes that 1 square mile is 640 acres. See also Conversion of units#Tables of conversion factors, which lists the SI equivalent of any unit you can think of. Tesseran 00:05, 28 August 2007 (UTC)[reply]