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December 23

lowest altitude of sky visible from this scenario?

Hi. If the edge of a roof is about three metres higher than a deck, and the centre of tripod on the ground of a telescope is located about 2 metres from the line where the edge of the roof extending down to the deck would be, and the pivital point, the point inside the mount that doesn't move when the telescope is moved (it's an equatorial mount) is about 115 cm off the ground, and the part of the telescope that doesn't move when only the declination axis is moved when the telescope is pointing due north (I live at about 45 degrees latitude) and the counterweight rod is perpendicular when viewed from the front to the right asenciation slow-motion rod underneath the RA setting circle, is about 10 cm higher than the RA pivital point, and the roof is probably at about a 45-degree angle, and the distance between the roof edge and the top of the roof (indicating the hypotenuse, not the other dimensions) is about 8 metres long, then about how low in terms of degrees from the ground, which is 0, would the lowest part of sky be viewable from the actual telescope be, and not be obscured by the roof even when the telescope is set to its lowest magnification, which gives a FOV of approximately 100 arcminutes, and if you require more specific information and measurements or parameters, which will probably be approximate, feel free to ask? Thanks. ~A H 1^(TCU) 19:24, 23 December 2007 (UTC)[reply]

I think I require a diagram... mattbuck (talk) 20:25, 23 December 2007 (UTC)[reply]

Let me try and work out what he said. A telescope is 3meters below and 2 meters away from a roof which slopes up 45 degrees for a length of 8 meters. The telescope is 115cm off the ground. So what angle can it look at and still make it over the roof? It ends up 1.85 meters below and 2 meters away from the corner of the roof, which is obviously less than 45 degrees, so you could see the roof and be blocked by it. The height of the roof is sin(45)*8= 5.656m, so the final view cut-off point is 5.656+2m wide and 5.652+1.85 meters high, which results in an angle of invtan(h/w)=44.4 degrees.--58.111.143.164 (talk) 22:27, 23 December 2007 (UTC)[reply]

(after edit conflict) You wrote a very long sentence that I find difficult to parse. Is the following a correct rephrasing?

When the telescope is pointing due north, and the counterweight rod is perpendicular (viewed from the front) to the right ascension slow-motion rod underneath the RA setting circle, the part of the telescope that doesn't move when only the declination axis is moved is about 10 cm higher than the pivotal point.

I don't know enough about telescopes in general and yours in particular to understand the role of the "right ascension slow-motion rod underneath the RA setting circle". Our Equatorial mount article mentions two perpendicular axes of motion known as right ascension and declination. These are (as I understand it) always perpendicular; that does not depend on whether you view things from the front. How is the slow-motion rod related to the right-ascension axis of motion?

By "higher", do you mean higher in the vertical direction? What is the distance between the central line through the telescope and the pivotal point?

Doesn't it matter in what direction the view-blocking roof edge is (to the north, east, south, west, ...) with respect to the telescope? And what is the direction you are thinking of when asking for the lowest viewable part of the sky? What keeps you from just pointing the telescope away from the roof, so that it doesn't block the view? Are you surrounded by a 3 metre high roof on all sides? Or do you mean: the lowest angle from the ground so that you have an unobstructed view when viewing at that angle for a full circle of 360° all around? That will be slightly less than 45°, say 44°; I don't know what precision you're aiming at.

I don't understand how the magnification and field of view is relevant; what is blocked from view by the roof should be independent of that.

--Lambiam 23:06, 23 December 2007 (UTC)[reply]

It's nice to be told we should feel free to ask, but what is the point if our questions are not answered? --Lambiam 20:51, 29 December 2007 (UTC)[reply]

Despite anything you may have heard to the contrary, it is the question that matters, not the answer... -- Meni Rosenfeld (talk) 23:47, 29 December 2007 (UTC)[reply]

Hi. I don't know if anyone will answer or if I will be able to read any replies now that it's in the archives, but I'll try to answer your confusidments. In my telescope there are two points which the telescope rotates around. When the right asenciation is rotated, the rotational point is about 10 cm lower than the rotation centre point when the declination is rotated, when the telescope is postitioned as described. I described it this way as to describe how the telescope is positioned when this is true. The thing is, if the telescope is rotated at the right asenciation axis so that it is lower than normal, then when the telescope is pointing in a particular direction it will be lower than normal, which means the roof is able to block higher areas of sky from that viewpoint. The thing is, when the telescope is pointed higher, its viewpoint is lower, and when it is pointed lower, the angle makes its viewpoint higher, so it is able to see lower areas of sky when it is pointed lower, and the lower it is pointed, the lower area of sky it can see. The roof is only on one side, and the reason I can't point in a different direction is because if there is a star that is obstructed by the roof in that direction, I can't view the star by pointing to another direction. I'm not even sure that the roof is at a 45-degree angle, what is the standard angle for this? remember that when the roof reaches its top, it slopes down on the other sides as well. If you extended a line from the top vertex of the roof to a point directly below it, deck-level, then extended a line parallel to the way a line directly from the observer towards the wall of the building so that it is perpendicular to both the edge of the roof and the wall, and extended the line so it is the minimum distance at a point from the observer, the distance would be about 5 m, and the roof extends in four directions to form a square-based pyramid, and each side would be the same, so what would be the minimum observable angle viewable from a direction so that the observer is facing the wall in a directional line that is perpendicular when viewed from above to the roof edge? The magnification and FOV might matter, because at a lower magnification the FOV of the telescope is more likely to be obstructed by the roof on part of the sky it is pointing at. Thanks. ~A H 1^(TCU) 18:57, 30 December 2007 (UTC)[reply]

If you have "Add pages I edit to my watchlist" in your preferences checked, this archive page is now in your watchlist and you will see if anyone edits it. It is likely that Lambiam watches this, too. -- Meni Rosenfeld (talk) 19:04, 30 December 2007 (UTC)[reply]

Hi. Yes, I have that enabled. Anyway, when I stand where the telescope would be positioned, I can't see the roof, but only the edge. When I lean back, tilt my head, and stand up taller, I could just see the snow on the roof, but not the roof itself, and of course where the telescope is positioned, I even more wouldn't be able to see the roof, although the roof would be higher in the sky. Thanks. ~A H 1^(TCU) 16:06, 31 December 2007 (UTC)[reply]

For all practical purposes, light travels in a straight line, so if you can't see the roof when you hold your eye where the lens of the telescope is, then neither can you see it through the telescope. So the edge of the roof is the limiting factor. --Lambiam 21:53, 31 December 2007 (UTC)[reply]

Since you don't accurately know the dimensions or angle of the roof, no amount of trigonometric calculation is going to get you an accurate answer, even when we fully understand the mechanics of your telescope mount. A much simpler solution is to take a practical approach. Point your telescope at the roofline and read off the angle from mount adjsutment scale. Or if there is no scale, use a protractor. Spinningspark (talk) 01:14, 1 January 2008 (UTC)[reply]

Zero divided by zero = Error 2

Why does zero divided by zero = Error 2 and not zero? --Seans Potato Business 22:08, 23 December 2007 (UTC)[reply]

See Division by zero. 78.145.159.201 (talk) 22:19, 23 December 2007 (UTC)[reply]

This wording isn't precise. Error 2 being displayed doesn't mean 0/0 is equal to Error 2. According to Calculator#Scientific and financial calculators, Error 2 is used to distinguish from Error 1, an out of bounds error. –Pomte

As I understand your question, it is: Why does 0/0 result in an error? Why is it not just 0? The immediate reason is that division by 0 has been left undefined by mathematicians on purpose, also for the case 0/0. For other cases like 1/0 the reason is obvious: when a/b = c, then a = bc; this is the defining characteristic of division. So if 1/0 had been defined to be some number c, we would get 1 = 0c = 0, a contradiction.

For the case 0/0 we do not get such a contradiction. We run instead into another problem. Look at the defining characteristic of a/b = c for the case a = b = 0, so we are asking: what is c in 0/0 = c? The defining characteristic says then: 0 = 0c. But that is true for all values of c, it is a tautology that does not tell us anything about c. So we can't use this, and any possible choice would be arbitrary. You might argue: 0/b is 0 for all other values for b then 0, so why not extend this to the case 0/0? But then, consider this: a/a = 1 for all other values for a then 0, so why not extend that to the case 0/0? --Lambiam 23:25, 23 December 2007 (UTC)[reply]

If division by zero isn't allowed, doesn't that mean that, for example,

{\frac {x}{x}}

does not always equal one? asyndeton talk 23:29, 23 December 2007 (UTC)[reply]

Yes. That equals one iff x is not equal to 0. –Pomte 00:30, 24 December 2007 (UTC)[reply]

In complex analysis, 0/0 can have many different values: 0, infinity or any other complex number. It all depends on the multiplicity of zeroes. We define the multiplicty of the zero of a function f by the lowest power term in the Taylor series of the function about that zero. Consider the following function:

g:={\frac {(f')^{2}}{(f-a)(f-b)}}

where f is some function where all zeroes have multiplicity of at least 2, and a and b are complex constants. If we differentiate the taylor series to get f', then it is clear that if (f-a) has a zero of multiplicity X at z, then f' has a zero of multiplicity X-1 at that point. Since, f cannot take the values a and b simultaneously, at a zero of f - a, call it z, we have the bottom having multiplicity of X, and the top having

2(X-1)\geq X

, therefore if the bottom has a zero, the top will have a zero of at least that order. If X > 2, we have a zero of g, and if X=2, then we get top and bottom having the same multiplicity.

To give a simple example, consider f to be

\cos ^{2}z

, with a = 0, b = 1. Then we have the numerator as

4\cos ^{2}z\sin ^{2}z

and the denominator as

(\cos ^{2}z-1)\cos ^{2}z

. If we take z = π,

\cos ^{2}z-1=0

, so we have a zero (of multiplicity 2) on the bottom. In the numerator, we have

\sin ^{2}z=0

, and thus a zero, ie

g(z)={\frac {0}{0}}

. BUT,

\cos ^{2}z-1=-\sin ^{2}z

, so

g={\dfrac {4\cos ^{2}z\sin ^{2}z}{-\sin ^{2}z\cos ^{2}z}}=-4

. Thus here 0/0 = -4. mattbuck (talk) 00:28, 24 December 2007 (UTC)[reply]

I'm not sure what you mean by "can have many different values". Also in complex analysis, 0/0 is undefined and has no value. Perhaps you are thinking of the indeterminate form 0/0 that may be obtained in the process of attempting to determine the limit of some expression that is a quotient. Although this has the form of an algebraic expression, it is just a formal expression that is not supposed to have a meaning, just like ∞ − ∞ has no meaning. If you get an indeterminate form, it signals that your attempt to find the limit you were looking for ran into a problem, and that in fact still everything is possible. --Lambiam 01:16, 24 December 2007 (UTC)[reply]

I mean that you can have a quotient where both numerator and denominator are both zero yet the quotient itself still has a calculable value. It's just the way I was taught it... mattbuck (talk) 11:55, 24 December 2007 (UTC)[reply]

In that case, I'm afraid, you were taught incorrectly. If a/b is defined (which is what I guess you mean by "has a calculable value"), then its value only depends on the values of a and b, and not on some history of how those values arose. If a = c and b = d, and a/b is defined, then so is c/d, and a/b = c/d. --Lambiam 12:54, 24 December 2007 (UTC)[reply]

We can put it this way. Take the function

f(z)={\frac {z}{z}}

in its natural domain. It is defined for every

z\neq 0

and has a removable singularity at 0. Since removable singularities want to be removed, it is natural for us to consider the function obtained by removing this singularity. We might even abuse notation and call the extended function f as well. We would then have

f(0)=1

. Since the original definition is

f(z)={\frac {z}{z}}

, we might even go so far as saying

1=f(0)={\tfrac {0}{0}}

and thus

0/0=1

"here". But this is just abuse (misuse, even) of notation. It has nothing to do with the actual operation 0/0. -- Meni Rosenfeld (talk) 14:47, 24 December 2007 (UTC)[reply]

Your original question also suggests a misunderstanding of what calculators really are. A calculator is not an oracle, a device that provides a glimpse into the world of absolute truth with answers to life, the universe, and everything. It is just a pile of silicon and plastic, where pressing buttons causes electric currents ultimately leading to something being displayed on an LCD screen, in a way designed by some guy you have never met in your life. The fact that you type "0/0" into your calculator and get "Error 2" doesn't mean that "0/0 = Error 2", it just means that whoever designed the calculator wanted it to display this message. Why? Because there isn't any suitable alternative, because the mathematical operation "0/0" is undefined (i.e., the definition of division doesn't give a value for 0/0). So the question you should have asked is "why is 0/0 undefined?". Hopefully you will find the answers given here to this question satisfying. -- Meni Rosenfeld (talk) 11:39, 24 December 2007 (UTC)[reply]

Why does x/x = 1 iff x <> 0? I would submit x/x = 1 for all numbers even if x = 0. This is obviously not the same as 0/0 = 1. Isn't x/x = 1 by definition just like 0! = 1 by definition? NYCDA (talk) 19:50, 24 December 2007 (UTC)[reply]
- What??? If ${\frac {x}{x}}=c$ then $x=cx$ , which has no unique solution when $x=0$ , per Lambiam's comment above.. And if I'm not mistaken, $0!=1$ not by definition, but because multiplication over an empty set yields the multiplicative identity (1), just like addition over an empty set yields the additive identity (0). --Kinu ^t/_c 20:56, 24 December 2007 (UTC)[reply]

NYCDA, I think you are a little confused. If x = 0 then x / x is 0 / 0. Thus if 0/0 is undefined, so is x/x when x=0.

As for the factorial - that depends on the definition. You could use a general definition from which it follows immediately that 0!=1, or you could use a simpler definition, which only works for positive numbers, and then add 0!=1 as a special case. -- Meni Rosenfeld (talk) 21:16, 24 December 2007 (UTC)[reply]

That's the distinct I'm trying to make, x/x is not the same as 0/0 when x = 0. One proof is x^n/x^m = x^(n-m) such that x^3/x^2=x^(3-2) and x^3/x^3=x^(3-3)=x^0=x/x=1. But x^0=1 for all x right? So x/x=1 even if x=0 but 0/0=undefined. In calculus, x = lim n->infinity 1/n = 0 but x/x = lim n->infinity 1/n divided by lim n->infinity 1/n = lim n->infinity (1/n)/(1/n) = 0/0 = 1. If you said x/y=1 when x = 0 and y = 0, I would say it's undefined. But x/y=1 when x = y and x = 0, then I would say it's correct. x/x=c was not what I said, it was x/x=1 for all x which does not violate x=1x. I'm glad you brought up multiplication over an empty set since that defined 0^0 = 0/0 = 1. See http://wiki.riteme.site/wiki/Exponentiation#Zero_to_the_zero_power NYCDA (talk) 18:01, 26 December 2007 (UTC)[reply]

You're thinking completely backwards. Mathematical theorems conform to mathematical principles, not the other way around. And mathematical principles are remarkably simple (it's the way we combine those principles to create useful statements that can be complicated). If x = 0 then x/x is 0/0. Period. Nothing you can ever do or say will change that, because this is more fundamental than pretty much anything. I don't understand why you insist on making this more complicated than it really is.

All those formulae you mention obviously have limitations. None of them were given on the stone tablets, they were derived from first principles and their validity is limited by what those will allow. For example,

x^{n-m}={\frac {x^{n}}{x^{m}}}

is clearly valid only for

x\neq 0

(always read the fine print). Your suggestion that we break mathematics in order to make this formula a little more general is ridicuolous.

As for 0^0, it is sometimes defined as 1, depending on the context. It is certainly not the same as 0/0. -- Meni Rosenfeld (talk) 18:17, 26 December 2007 (UTC)[reply]

I'm trying to present a different view point. x/x when x=0 must be analyzed instead of just declaring it undefined. There are those who define 0⁰=1, whereas others leave it undefined. NYCDA (talk) 23:44, 26 December 2007 (UTC)[reply]

You are free to define anything in any way you like, as long as you don't break any fundamental principle. There is no problem with defining

0^{0}=1

. There is no problem with defining

0^{0}=2

, except for it being a useless and motivation-less definition. But there is a problem with saying that

x^{x}

when

x=0

is not

0^{0}

, because that's just wrong. It contradicts the meaning of equality - if

x=0

then x and 0 are the same thing, and whenever you see x you can write 0 instead.

Back to the original point, there is no problem with defining

0/0=1

(except that some arithmetic theorems no longer hold). There is no problem with defining

0/0=2

, or with defining

0/0=0

(which makes more sense than those other options). But there is a problem with saying that

x/x

when

x=0

is not

0/0

.

Again, this is completely beyond debate and beyond "different view point", unless you are willing to create an entirely new kind of mathematics, one where equality means something completely different. But then WP:RD/math would not be the correct place to discuss it, you'd need a new section - WP:RD/NYCDA'sNewKindOfMath. -- Meni Rosenfeld (talk) 11:53, 27 December 2007 (UTC)[reply]

Fat tails in stock market index movements

It is said that stock market index changes have fatter tails than those of a gaussian distribution. Is there any consensus about what statistical distribution they do correspond to please? 80.0.124.1 (talk) 23:10, 23 December 2007 (UTC)[reply]

The changes in stock market index cannot have a gaussian distribution, which would allow the index to obtain negative values with positive probability. The logarithm of the index behaves better. Actually the ratios between index values rather than the difference between index values have interpretations. Do the distribution of changes of the logarithm of the index also have fatter tails? Bo Jacoby (talk) 11:08, 24 December 2007 (UTC).[reply]

I bet they do. Most of the time for most stocks the trading that goes on is kind of random, with a normal volume and about equal amounts of sellers and buyers. But sometimes, because of some rumour or news story, large amounts of people think this is a good time to sell or buy some stock, with most going in the same direction. How many will depend on how compelling the story is and how much news coverage it gets. All this does not easily suggest a mathematical model. --Lambiam 11:22, 24 December 2007 (UTC)[reply]

The condition, that all the contributions to the total change are independent, is sufficient, but not necessary, for the resulting distribution to be gaussian. The gaussian distribution accounts for that most changes are rather small, but exceptionally they are big. In the long run exceptional events happen. I am not a betting person, and the outcome of the bet is hard to decide, as it takes a big number of observations to conclude that the tails are significantly fatter. Even when there is lots of handwaving why some mathematical model will not necessarily work, it may work anyway. Have a merry christmas! Bo Jacoby (talk) 12:30, 24 December 2007 (UTC).[reply]