Wikipedia:Reference desk/Archives/Mathematics/2006 July 1

Humanities Science Mathematics Computing/IT Language Miscellaneous 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 at one of the pages linked to above.

< June 30 Mathematics desk archive July 2 >

Mathcad to Wiki math markup

Is there an engine to covert mathcad format to wiki math markup format. I have a bunch of derived equations for solved variables that will take about two weeks to convert if done by hand? ...IMHO (Talk) 10:44, 1 July 2006 (UTC)

Wikipedia uses TeX, so that might help you in your search (i.e. don't search for "mathcad to wiki", search for "mathcad to tex"). —Keenan Pepper 18:38, 1 July 2006 (UTC)

Number speak

I realize this question is elementary to mathematicians but how do you speak the number, 1,072,915,200,000,000,000.

Thank You

See Names of large numbers. -- Meni Rosenfeld (talk) 16:51, 1 July 2006 (UTC)

I'd say "one point zero seven two nine one five two times ten to the eighteenth", but if you wanted to, you could say "one quintillion, seventy-two quadrillion, nine hundred fifteen trillion, two hundred billion". —Keenan Pepper 18:37, 1 July 2006 (UTC)

The honest answer is that we don't. For a number with that many digits it is not particularly helpful to say it. If the precise digits are meaningful, it is more helpful to see the number written down. In the common case where we don't care about those details, it is more helpful to use rounded scientific notation ("one point oh seven times ten to the eighteenth", 1.07×10¹⁸). Use of scientific notation is one effective way to combat the "size stun" effect of large numbers. ("A billion here, a billion there, and pretty soon you're talking real money!" — attributed to US Senator Everett Dirksen) --KSmrq^T 19:24, 1 July 2006 (UTC)

Number System

Why everything to the power of zero is equal to one ?

Because it follows from the rules of exponents. x^n/x^n = x^(n-n) = x^0 = 1. Splintercellguy 17:20, 1 July 2006 (UTC)

The question is almost true, but not quite. Zero to the power of zero is commonly considered undefined, not one.

The core of the answer is consistency, to make a system where everything works nicely together. Consider a sequence of powers of 10.

…	10−2	10−1	100	101	102	103	…
	1⁄100	1⁄10	?	10	100	1000

As we step right through the powers, each number is 10 times more than the one before. As we step left, each is 1⁄10 of its neighbor. The only way to fill in the entry for the power zero and be consistent is to use 1. This is true whether we use powers of 10 or of any other number besides zero. (Try it for zero!) --KSmrq^T 19:44, 1 July 2006 (UTC)

Empty product provides a good overview, and a nice quote to boot. RandomP 20:15, 1 July 2006 (UTC)

Much as I endorse the choice of Knuth and colleagues to define 0⁰ as 1, it is important to understand that they are advocating a position, not stating a commonly agreed convention. The argument for that choice is again consistency, in fact a consistency beyond that shown in my previous post. However, every positive power of zero is zero, and every negative power is undefined, so this is a qualitatively different sort of choice, not so easily made. --KSmrq^T 02:21, 2 July 2006 (UTC)

I'm not sure what your definition of "common" is, but I sincerely doubt there is even one reasonably recent textbook of analysis which hasn't adopted that convention when talking about power series.

(Of course, 0×∞ is now commonly defined to be 0, too. Measure theory just wouldn't work without it).

Another thing to keep in mind is that when 0 is considered a set, 0⁰ very plainly does have cardinality 1: there is exactly one empty function (by contrast, 0¹ is empty.) At the very least, if you think 0⁰ is undefined, you must go to the trouble of saying which 0 you use.

RandomP 12:20, 2 July 2006 (UTC)

Three points:

1. On "common": I have not done a survey of contemporary literature in research and teaching, but I'll wager the "undefined" version is still to be found, and surely literature predating Concrete Mathematics (which is a lot) will be full of it.
2. The qualitative difference of a nonzero base is clear, both in the arguments given and in the fact that essentially all the literature, current or past, agrees that a nonzero value raised to a zero power should equal 1.
3. Assuming we're still interested in helping the original poster understand, it's important to highlight the difference between the two cases.

It is indisputable that when Knuth et al. argued for 0⁰ = 1 the convention was not commonly agreed; so if you like, change "they are advocating" to "they were advocating". Either way, we should not ignore the issue. --KSmrq^T 19:20, 2 July 2006 (UTC)

My understanding is this is currently a high-school vs real mathematics issue: in high school classes, 0⁰ is left undefined (just as ${\sqrt {-1}}$ is, commonly), but for someone with more mathematical knowledge, 0⁰ is defined at least when 0 is considered as a cardinal number; it might yet be undefined when 0 is considered a complex number, and I have seen people go so far as to claim that a^b is undefined (or multi-valued, at least) for b a non-integer real number and a a positive real number, unless it is explicitly mentioned that real exponentiation, not (multi-valued) complex multiplication, is meant.

Sorry if that got lost. The important points are:

• it's a convention. conventions are chosen for convenience, not for truth.
• overwhelmingly, people consider 0⁰ = 1 to be the implied convention when the expression is used by someone else, though some may choose not to use the expression at all, leaving it undefined.
• there are situations where the expression 0⁰ is defined, and is 1.

RandomP 20:33, 2 July 2006 (UTC)

Of course, 0×∞ is now commonly defined to be 0, too. Measure theory just wouldn't work without it.

Can you explain why?--72.78.101.61 12:32, 2 July 2006 (UTC)

Well, I might have been a bit .. informal there. Of course you could reword all references to that definition, but it just so happens that when you define 0×∞ to be 0, things like the product measure Just Work. The 2-dimensional volume of the x axis, for example, is 0; however, it should also be 0×∞, because it's the cartesian product of a single point (measure 0) and the entire real line (measure ∞).

I believe that it used to be perfectly acceptable to write f(∞) = 7, for example, where today we would write $\lim _{x\rightarrow \infty }f(x)=7$. And there certainly are real functions f, g such that, using the sloppy notation, f(∞) = ∞, g(∞) = 0, fg(∞) = 17, or whatever number you want. However, now that we're standing on the shoulders of those giants (in patience, at least) who've eradicated the sloppy notation, we can easily define 0×∞, and be done with it (though we must then remember that considered as a map from [0,∞] × [0,∞] to [0,∞] (with the usual topologies), multiplication is no longer continuous).

Put yet another way, {0,∞} is the simplest semiring. If you so wish, you can think of that as a model for binary logic, and the only way for that to work is to define 0∞=0.

RandomP 13:31, 2 July 2006 (UTC)

There is yet another reason to adopt 0⁰=1 — you simply need this for the binomial theorem to work. Put y=0 and try to calculate (x+0)¹. You'll get: (x+0)¹ = x⁰·0¹ + x¹·0⁰. This will give the desired result of x only if 0⁰=1.
:-) CiaPan 17:46, 7 July 2006 (UTC)

I should point out that not everything to the power of zero equals zero. $0^{0}$, which is the only exception in the entire number set C, has two definite possible values, and one undefined. Here's the proof:

$0^{0}=e^{0\log 0}=$undefined

Since $0^{n}=0$
therefore $0^{0}=0$

Since $n^{0}=1$
therefore $0^{0}=1$

--JB Adder | Talk 14:12, 2 July 2006 (UTC)

That $0^{n}=0$ only holds for natural numbers n > 0 (with xⁿ interpreted as x multiplied by itself n times), so this isn't a "proof" at all, just incorrect appeal to a definition. Fredrik Johansson 22:43, 2 July 2006 (UTC)

No, it holds for every real n>0. See Exponentiation article and read how Exponentiation#Real powers of positive real numbers are defined from rational powers by continuity. In the same manner (i.e. by continuity) Pⁿ for positive P and positive n extends to positive powers of zero. Check Image:Root_graphs.png to see how Pⁿ approaches zero when P approaches zero, for different n. This definition is correct. --CiaPan 20:12, 11 July 2006 (UTC)

what is this function called?

What do you call the inverse to the real function x |-> xⁿ + x + 1. (Maybe only for odd n?) For some reason, I thought this was called the Eisenstein function, but now I can't find that name in any of my references, so I'm wondering if I just dreamed that up. -lethe ^{talk +} 17:11, 1 July 2006 (UTC)

It isn't one-to-one for even n—f(0) = f(-1) = 1, for example—so you'd better restrict to odd n. Tesseran 07:12, 2 July 2006 (UTC)

This is not an insurmountable barrier; for example, x² suffers from that defect as well, and that hasn't stopped us from giving it an inverse, the square root. But I think I mucked up my question. What I really want is the function that gives you the root of the polynomial xⁿ + x + a. I guess this function is actually the inverse of f(x) = xⁿ + x, rather than what I said above, whoops. It's the function that you need to write the solution of the quintic in closed form (does it work for polynomials of any degree?) -lethe ^{talk +} 07:33, 2 July 2006 (UTC)

A somewhat related article is Artin-Schreier theory.--gwaihir 08:24, 2 July 2006 (UTC)

Simplification

If

${\left({\frac {P+D}{P}}\right)}$

can be simplified as

${\left(1+{\frac {D}{P}}\right)}$

then what is the simplification of

${\left({\frac {P}{P+D}}\right)}$

...IMHO (Talk) 22:50, 1 July 2006 (UTC)

It can't be simplified further. The top equation only works because P/P = 1, but you can't split a denominator this way. --ColourBurst 23:07, 1 July 2006 (UTC)

I'd say that

${\frac {1}{1+D/P}}=\left(1+{\frac {D}{P}}\right)^{-1}$

is a simpler form, whichever way you prefer to write it. Melchoir 23:59, 1 July 2006 (UTC)

Oh yes it can!

${\left(1-{\frac {D}{P+D}}\right)}$

-- Jokes Free4Me 13:08, 4 July 2006 (UTC)

ColourBurst is correct. Melchoir is less correct, if not strictly wrong. Computer algebra systems explain. To some extent the criteria for which of two equivalent expressions is "simpler" than the other are in the eye of the beholder. When a computer program is asked to simplify an expression, it may not choose the form the user would prefer. That said, for most purposes, and with most computer algebra systems, the original expression is as simple as possible (assuming P does not divide D). The alternative expression has a fraction in the denominator, which is usually something we prefer to eliminate, not create. However, on some occasions and for some purposes the double division may nevertheless be what we want. It may be easier to appreciate the issues with a different example: Which is simpler, 1−x² or (1−x)(1+x)? The first form has fewer "pieces", but the second reveals the factors. Still, I think the original question is less ambiguous; if it comes up on an mathematics test you are likely to get a better grade with "no simplification". --KSmrq^T 01:48, 2 July 2006 (UTC)

Going a bit further without creating another section now let me ask whether in either case P can equal zero without resulting in a divide by zero error? ...IMHO (Talk) 03:40, 2 July 2006 (UTC)

In the first case, no, and in the second case, yes. The second case doesn't have a nonzero denominator (assuming D is nonzero), and in the first case either form has a zero denominator so it would be undefined. If you wanted to ask whether what the limit approaches as P approaches zero in the first case (which different from what you asked), the answer would be that it approaches positive infinity. --ColourBurst 05:18, 2 July 2006 (UTC)

Without jumping to any conclusion and avoiding the assumption that the answer is "no" allow me to ask then if in a situation where the value of P represents the number of grains of sand in the top portion of an hour glass and the value of D represents the number of grains of sand in the bottom portion of the same hour glass is it possible for the all of the states of the hour glass (upper chamber filled with sand and lower chamber empty and lower chamber filled with sand and upper chamber empty, plus any and all states in between) to be represented by the ratio of D to P or by the ratio of P to D corresponding to all of the states in terms of the location of each grain of sand in the upper chamber or the lower chamber of the hour glass specifically including the state of P where P is equal to zero? If not what equation would allow all states of the hour glass to be represented without incurring a divide by zero error for P. ...IMHO (Talk) 04:50, 2 July 2006 (UTC)

Hmmm, I think you have the wrong sense of ratio here. A "zero" ratio for D:P means that D is zero, and you couldn't infer anything about P from D. "Infinite ratio" means the opposite (0 for P). But ratios never infer anything about how many particles are really in the glass - just the proportion. --ColourBurst 05:18, 2 July 2006 (UTC)

Okay then what you are saying appears to be that the proportion of P to D or D to P can not be represented by the division of P by D or of D by P when the denominator is zero and therefore that the relationship expressed by such division is invalid when applied to the situation represented by an hour glass where P or D may have a value of zero? ...IMHO (Talk) 05:25, 2 July 2006 (UTC)

Well, the ratios in those two cases have a "meaning" - one of those elements is zero, depending on case. It's not invalid - you just can't infer anything about the other quantity without additional information. See elasticity (economics) for an example of what a zero ratio and an infinite ratio means. --ColourBurst 05:41, 2 July 2006 (UTC)

I'm not quite sure of your objective, but I think you had the right idea with P/(P+D). Given the extra information you have now provided, we know that P+D gives the total number of grains of sand, which should be constant. So this ratio will equal 1 when the top is full (bottom empty) and 0 when the top is empty (bottom full). It's a "time remaining" meter, or more accurately, a "proportion of grains remaining" meter. Assuming there is sand in the hourglass, this expression will never cause a divide by zero. --KSmrq^T 06:03, 2 July 2006 (UTC)

Such that we might make a direct substitution in the equation below? (Forgive me in the interest of time for not first working the answer out for myself.)

$t={\frac {1}{\lambda }}{\times }{\ln \left(1+{\frac {D}{P}}\right)}$

...IMHO (Talk) 06:11, 2 July 2006 (UTC)

$t={\frac {1}{\lambda }}{\times }{\ln \left({\frac {1}{\frac {P}{P+D}}}\right)}$

...IMHO (Talk) 06:36, 2 July 2006 (UTC)

What I am looking for is an equation that will render the same proportional values for t yet tolerate a value of zero for P. ...IMHO (Talk) 07:05, 2 July 2006 (UTC)

table

Hour Glass

Grains of Sand
Chambers		Proportion			t
Upper	Lower	form	inverse		form	inverse
P	D	1+(D/P)	P/(P+D)	λ	(1/λ)*ln(1+(D/P))	(1/λ)*ln(P/(P+D))
20	0	1	1	0.0121	0	0
19	1	1.05263157894737	0.95	0.0121	4.23911523864054	-4.23911523864054
18	2	1.11111111111111	0.9	0.0121	8.70748063287821	-8.70748063287821
17	3	1.17647058823529	0.85	0.0121	13.4313164874194	-13.4313164874194
16	4	1.25	0.8	0.0121	18.4416158110917	-18.4416158110917
15	5	1.33333333333333	0.75	0.0121	23.7753778885769	-23.7753778885769
14	6	1.42857142857143	0.7	0.0121	29.4772680941101	-29.4772680941101
13	7	1.53846153846154	0.65	0.0121	35.6018938919384	-35.6018938919384
12	8	1.66666666666667	0.6	0.0121	42.2169936996686	-42.2169936996687
11	9	1.81818181818182	0.55	0.0121	49.4080165913736	-49.4080165913736
10	10	2	0.5	0.0121	57.2848909553674	-57.2848909553674
9	11	2.22222222222222	0.45	0.0121	65.9923715882456	-65.9923715882456
8	12	2.5	0.4	0.0121	75.7265067664591	-75.7265067664591
7	13	2.85714285714286	0.35	0.0121	86.7621590494775	-86.7621590494775
6	14	3.33333333333333	0.3	0.0121	99.5018846550361	-99.5018846550361
5	15	4	0.25	0.0121	114.569781910735	-114.569781910735
4	16	5	0.2	0.0121	133.011397721826	-133.011397721826
3	17	6.66666666666667	0.15	0.0121	156.786775610403	-156.786775610403
2	18	10	0.1	0.0121	190.296288677194	-190.296288677194
1	19	20	0.05	0.0121	247.581179632561	-247.581179632561
0	20	#DIV/0!	0	0.0121	#DIV/0!	#NUM!

Further discussion

When a function diverges, you can't fix it by rearranging the way it's written. Probably the math is trying to tell you something. What is the application for all this? Melchoir 02:42, 3 July 2006 (UTC)

The inverse column does at least show that the divide by zero error is a number too large (probably infinity) to be calculated which tells me that P has attained a value of zero. (The issue is important because others have disputed this and think that P can never reach zero which is obviously not true for grains of sand or any microscopic, but whole nonetheless objects, that are similarly undivisible). What the math is telling me is it is the value of t that is indeterminate when P equals zero rather than the value of P being indeterminate. ...IMHO (Talk) 04:32, 3 July 2006 (UTC)

Okay... have you considered the possibility that these equations don't appropriately model the behavior of sand grains; and that nothing should be concluded about sand from examining them, anyway? Melchoir 05:01, 3 July 2006 (UTC)

Absolutely and in fact my reason for bring this matter to the attention of the mathematics desk. I am looking for a mathematical model of an hour glass. ...IMHO (Talk) 05:24, 3 July 2006 (UTC)

Oh! Well, the underlying microscopic dynamics aren't well-understood,[1] but on a large scale, the flow of sand through an hourglass proceeds at a constant rate, regardless of the amount of sand in the upper bulb.[2] Melchoir 06:07, 3 July 2006 (UTC)

That is true in terms of the rate of flow as the equations and value of t would suggest. In the formula I am looking for the proportional difference between the upper and the lower bulbs would have significance and be exponential but not related to or based upon a rate of flow but rather only on the proportional difference in the number of grains of sand in each bulb even if the hour glass fell on its side for a couple of hundred years. ...IMHO (Talk) 08:32, 3 July 2006 (UTC)

What are you talking about? Here's a formula:

$D=rt.\,\!$

Here's another formula for the uptight:

$D(t)={\mbox{mid}}(0,rt,P(0)).\,\!$

Given a sufficient reserve of creativity and time, I'm sure you could also work some proportional differences and exponents in there, but why would you even want to? Melchoir 08:49, 3 July 2006 (UTC)

Actually having now just plotted the values for t using a line graph I may have found the actual solution to the problem. Instead of needing a new equation what it appears that I need (and which I speculated about in an earlier discussion) is simply the need for integer versus decimal calculations or presentations of the results of the calculations. In other words if you do a line graph then you have a nasty line for the plot of t that is returning to zero from the value of 247.58118. If instead you use a bar graph instead of a line graph then the aesthetics prevent the viewer from having a misconception. ...IMHO (Talk) 10:15, 3 July 2006 (UTC)

Okay, I've had a look at your contributions to see what you're really up to, and it has nothing to do with sand. In fact, it has little to do with even its stated topic. Just don't put it into a Wikipedia article. Melchoir 18:46, 3 July 2006 (UTC)

LOLFDAC! Well after another effort to manipulate the equation so that it would not end up with a divide by zero error or an "infinite ratio" (by adding 1 to P and other compensating measures) I again came to the conclusion that the problem is not the equation but rather the misinterpretation of the results that P can never reach zero. In fact I believe now that the occurrence of a divide by zero error or an "infinite ratio" is necessary to provide a reliable indication that the value of P has in fact reached zero, i.e., that there are no grains of sand left in the upper bulb if or so long as these results are not misinterpreted.

I now realize that the basis for misinterpretation is that if P reaches zero then the value of t would be infinite (or at least greater than the time since the Big Bang) which would suggest a lower limit of $P=1.25*10^{-9}\$ where $P={\frac {D}{\exp {\left({t*\lambda }\right)}-1}}$ such that the value of t in the equation $t={\frac {1}{\lambda }}{\times }{\ln \left(1+{\frac {D}{P}}\right)}$ does not exceed the amount of time since the Big Bang or approximately 16 billion years. Otherwise we should prehaps question the validity of all equations that produce a historical date which is or can be farther back in time than the date the universe is determined to have been created.

Rather than serving as the basis for a Wikipedia article this realization will instead allow me to determine how large the grain of salt must be that is taken with the Wikipedia articles I read. ...IMHO (Talk) 22:20, 3 July 2006 (UTC)