Wikipedia:Reference desk/Archives/Mathematics/2012 February 12

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February 12

Preliminary tests for regression analysis in SPSS; ANCOVA in SPSS

How do I check the validity of the assumptions of normality and homoscedasticity for a simple linear regression, in SPSS (please give detailed instructions if possible)?

Another issue - when I run an ANCOVA in SPSS, why is it that in the output table "Test of between-subjects effects" sometimes the "corrected model" is mentioned in the first line, but sometimes the table does not mention this parameter?

Thank you in advance. Gidip (talk) 08:57, 12 February 2012 (UTC)

Division by zero

Is it right to say that division by zero is infinite sometimes, meaning it's undefined? 212.170.181.95 (talk) 13:43, 12 February 2012 (UTC)

See division by zero. Infinite is different from undefined. It is sometimes okay to say division by zero gives infinity, for instance when dealing with the real projective line. Also in floating point computing they have signed zero which enables them to give either a positive infinity or negative infinity. Normally though division by zero is considered as giving an undefined result. Dmcq (talk) 14:20, 12 February 2012 (UTC)

My theory is this.

The infinitesimal (which is the theoretically smallest possible number greater than 0) is 1/infinity. If you consider the following:

${\begin{aligned}&V&{\frac {1}{{\tfrac {1}{V}}+V}}\\&.1&.099009900990099\dots \\&.01&.009999000099990000999900\dots \\&.001&.000999999000000999999000\dots \\&.00001&.0000099999999990000000000999999999900000\cdots \\&\quad \dots {\color {white}_{\Big |}}&\quad \dots \end{aligned}}\,\!$

and, if the infinitesimal (“iota”), ${\color {white}|}\iota \,\!$, is the smallest possible number, then

${\frac {1}{{\tfrac {1}{\iota }}+\iota }}={\frac {1}{\infty +\iota }}=0\,\!$

but now you have to consider $-\iota \,\!$: As ${\sqrt {V{\cdot }V}}=V\,\!$, then ${\sqrt {V\cdot -V}}={\sqrt {-1}}{\cdot }V.\,\!$ Thus,

${\frac {1}{0}}:=\lim _{V\to \iota }{\frac {1}{{\sqrt {-1}}({\tfrac {1}{V}}+V)}}={\frac {1}{i(\infty +\iota )}}\,\!$

So, since it involves ${\sqrt {-1}}\,\!$, it is still not a “real” quantity!

Of course, this is just my personal, unsubstantiated, unprovable theory, so don’t even think of using it anywhere in Wikipedia! P=)  ~Kaimbridge~ (talk) 15:24, 12 February 2012 (UTC)

We already have infinitessimal plus articles on non-standard analysis. Dmcq (talk) 16:18, 12 February 2012 (UTC)

If you are considering the real numbers, then there are no such infinitesimals. See our article on the completeness of the the reals. If you want to include infinitesimals in a mathematically sound way then you need to consider the hyperreals, but these are different from the real numbers that most people think of when you use the word "number". — Fly by Night (talk) 22:24, 12 February 2012 (UTC)

Completeness of the reals doesn't sound quite so convincing when it is based on the Completeness axiom Dmcq (talk) 22:41, 12 February 2012 (UTC)

My view is that the reals, including completeness, correctly capture the informal geometric intuition of a line without holes. So from this POV it's not a question of axiomatics.

In any case, completeness has little to do with division by zero. Structures that are like the reals except for having infinitesimals and infinite nonstandard reals, still don't have division by zero. On the other hand, structures such as the real projective line and the interval [0,+∞] do have division by zero, but do not have infinitesimals (and do have completeness). --Trovatore (talk) 22:47, 12 February 2012 (UTC)

0/0=0^0?

If 0/0 is indeterminate, Why does 0^0 = 1?--92.28.91.245 (talk) 22:57, 12 February 2012 (UTC)

Oh, you've stepped in it now :-) --Trovatore (talk) 23:01, 12 February 2012 (UTC)

See Exponentiation#Zero to the zero power. --jpgordon^{::==( o )} 23:01, 12 February 2012 (UTC)

(ec) To be less mysterious, the status of 0^0 is controversial. Some mathematicians like to insist on defining 0^0 to equal 1 in virtually every context in which the expression might come up. Others (I'm in this latter group) prefer to distinguish based on context. Roughly speaking, when exponentiation is defined as repeated multiplication, then it makes sense to say 0^0=1. However exponentiation in which both the base and exponent are real (or complex) numbers is actually not defined as repeated multiplication, and in this context it may make sense to leave 0^0 undefined. --Trovatore (talk) 23:06, 12 February 2012 (UTC)

In the latter context, you can say that 0^0 is "almost always" 1, which is to say that if you define it as a limit of f(x)^g(x) for real x->0 such that f(x)->0 and g(x)->0 as x->0, then the limit will be equal to 1 for almost any choice of f(x) and g(x). For example, if you restrict f(x) and g(x) to be polynomials of finite degree N, the limit of f(x)^g(x) will be equal to 1 regardless of the choice of coefficients. You could say that examples where f(x)^g(x) converges to anything other than 1 are pathological. --Itinerant1 (talk) 00:48, 13 February 2012 (UTC)

Well, this is an argument that is made. Personally I don't buy it. To me the most natural category in which to define basic operations on the reals is just continuous, not analytic. --Trovatore (talk) 01:55, 13 February 2012 (UTC)

That's not always true though, and there are fairly easy counterexamples: $f(x)=0,g(x)=x^{2}$ for example. Widener (talk) 01:05, 13 February 2012 (UTC)

Yes, I stand corrected. f(x) exactly equal to zero is the only counterexample. More generally, the article on exponentiation says (and it seems to be correct) that, as long as f(x) is nonnegative and it's not exactly zero and f(x) and g(x) are both equal to their Taylor series in some neighborhood of 0, f(x)^g(x) converges to 1. (This includes my statement about polynomials of finite degree as a special case.) Counterexamples shown in the article are pathological: e^{-1/x} is discontinuous at zero; e^{-1/x^2} is a classic example of a non-perturbative function (one which is infinitely differentiable but not analytic at x=0, because its Taylor series is equal to zero).--Itinerant1 (talk) 01:30, 13 February 2012 (UTC)

See our section: Indeterminate form#The form 0⁰. 0⁰ is one of the seven commonly listed indeterminate form; it's just not as indeterminate as the other six. -- ToE 01:56, 13 February 2012 (UTC)

Yeah in engineering it might always be 1, but this is the maths reference desk. Dmcq (talk) 02:01, 13 February 2012 (UTC)

Hmm. I would have thought that most engineers would be happy considering it to be undefined, and that it would be many mathematicians, particularly those working with discrete math, who would define it to be one. -- ToE 04:12, 13 February 2012 (UTC)

"Some textbooks leave the quantity 0⁰ undefined, because the functions x⁰ and 0^x have different limiting values when x decreases to 0. But this is a mistake. We must define x⁰ = 1 for all x, if the binomial theorem is to be valid when x=0, y=0, and/or x= -y. The theorem is too important to be artitrarily restricted! By contrast, the function 0^x is quite unimportant.

- Concrete Mathematics, by Graham, Knuth, and Patashinik, 1989, p. 162. Bubba73 ^{You talkin' to me?} 02:34, 13 February 2012 (UTC)

Knuth's argument is good if the exponent is considered to be an integer; not so good if it's considered to be a real number. This applies more generally to the arguments involving power series (the binomial theorem, in general, gives you a power series) because power series have integer exponents. --Trovatore (talk) 05:44, 13 February 2012 (UTC)

There seems to be invalid conflation of the concepts undefined and indeterminate. That 0⁰ is an indeterminate form follows from the definition of the latter. This in no way prevents it from being defined as 1 (or even makes defining it "not sensible"). Given this context, is there any reason to leave it undefined in any context? — Quondum^☏✎ 08:03, 13 February 2012 (UTC)

Yes. The standard definition of real-to-real exponentiation is

$x^{y}=e^{y\log x}$

Now, the log of 0 is undefined, so that makes the whole expression undefined. --Trovatore (talk) 08:28, 13 February 2012 (UTC)

I've always wondered about that - Using that definition, wouldn't $0^{1}$ also be undefined, or indeed zero to any power at all (even positive powers)? Widener (talk) 08:36, 13 February 2012 (UTC)

Yes you do get that. The limit is always 0 though as you approach 0 from a positive number. Of course that is if y is a real. If y is the integer 1 you get normal integer exponentiation. Dmcq (talk) 08:50, 13 February 2012 (UTC)

Not only from a positive number. Whatever branch of the logarithm you choose, you get $\lim _{x\to 0}\exp(1\ln x)=0$, so the discontinuity is removable and should be removed by setting $0^{1}=0$. To abuse notation, ln 0 isn't undefined, it's $-\infty +ki$, and for any value of k, $\exp(1(-\infty +ki))=0$. But $0\cdot (-\infty )$ is indeterminate, and thus so is $0^{0}$. -- Meni Rosenfeld (talk) 14:00, 13 February 2012 (UTC)

Defining 0⁰ as 1 doesn't work with indeterminate forms, it breaks the mechanism. The definition of an indeterminate forms is bound with not only the limit but the value at a point. FDor instance at indeterminate form we have "Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form". If 0⁰ was defined to be 1 then the expression x^y would give enough information after the limits 0 were substituted for x and y to determine the original limit. This simply does not lend itself to automating in any sort of straightforward way. You've either got to treat exponentiation as a black box function and not one of the straightforward arithmetic operations or else you've got to leave 0⁰ undefined. Dmcq (talk) 08:32, 13 February 2012 (UTC)

Very nicely explained, Dmcq. Nevertheless, such a mechanism should surely inherently involve caveats, just as for the process of cancelling factors from the numerator and denominator. And the mechanism breaks for many functions anyway (what you call "black box functions"). It would seem that being able to treat it as a "straightforward arithmetic operation" without checks during the replacement process described (and then only, it seems to me, in the real case) exacts too steep a price in numerous other areas, such as with the sigma expression for a Taylor's series. The power function needs to be treated with extreme care in any event: the usual identities have massive pitfalls. So it would really make sense (IMHO) to classify this function as a "black box" function rather than a "straightforward arithmetic operation". What a minefield. — Quondum^☏✎ 10:21, 13 February 2012 (UTC)

There's no big price to pay. You simply treat real powers differently from integer powers. Taylor series only have integer powers. This is actually something computers find extremely easy but people seem to have problems with, saying the real number 0.0 is not the same as the integer 0. This way of dealing with things also helps with problems like Clausen's paradox in exponentiation#Failure of power and logarithm identities where e to a complex power is a complex power of a real whereas e^1+2πin is a complex number with the value e+i0. A computer would never fall into that trap. Dmcq (talk) 13:03, 13 February 2012 (UTC)

What you are saying implies that 0⁰ is defined in many cases (in your example, the function defined as a map ℝ×ℤ→ℝ, so we really have distinct functions for each domain, with a heuristic for determining which domain we're dealing with). I have little problem with this resolution of the problem (it seems quite natural to me), though it carries its own price (not too big, as you say): when one domain embeds into another domain (ℝ×ℤ into ℝ×ℝ, say), we can lose some points of the first domain. I suspect that there is not a consensus on this amongst mathematicians. If I'm wrong and what you are saying is the consensus view, perhaps we have some articles to clarify. — Quondum^☏✎ 13:52, 13 February 2012 (UTC)

I'm surprised to hear someone on this desk talk about computers "finding things easy" or "not falling into traps". That computers can do things that humans find difficult is fundamental to why we have them in the first place. But it's not a question of computers experiencing ease or difficulty. They are machines that either can or cannot be programmed to perform a certain task; if they can be so programmed, they do exactly and only what they're programmed to do. -- Jack of Oz ^{[your turn]} 20:21, 14 February 2012 (UTC)

You've obviously not had much experience with computers. What you describe as doing only what they are programmed to is really them working to rule, they are malicious, cranky, peevish, petty and perverse and will exploit the slightest mistake in anything you say. Have you never fought your pc for possession of a a CD for instance? Or never had it quietly stick in some directory you'd never think of into a save menu so your file will be effectively lost? Or perhaps you've been indoctrinated by them into blaming the victim for any problems rather than the perpetrator? ;-) Dmcq (talk) 00:30, 15 February 2012 (UTC)

I agree with Quondum and disagree with Trovatore and Dmcq.

1. The power aⁿ is 1×a×···×a (n multiplications by a). This means that a⁰=1 for all values of a, also for a=0. So 0⁰=1.
2. The real number 0 is an also integer. Computers have different internal representations between real 0.0 and integer 0, but mathematically 0.0=0. (And, Dmcq, e+i0=e).
3. The values of mathematical expressions do not depend on context. Context is not a mathematically defined concept.
4. The formula $x^{y}=e^{y\log x}$ works for x>0, but for x=0 and y≥0 the left hand side is defined and the right hand side is undefined. For example: $0^{1}=0$ even if $e^{\log 0}$ is undefined.
5. "Limits involving algebraic operations are often performed by replacing subexpressions by their limits", but this procedure is sometimes not justified. lim(f(x))=f(lim(x)) requires continuity. The function $0^{x}$ (for x≥0) is not continuous. $0^{x}=0$ for x>0, and $0^{x}=1$ for x=0.

Bo Jacoby (talk) 11:06, 14 February 2012 (UTC).

We all know you disagree with me. However, if you think you agree with Quondum, you may not have carefully read what he/she wrote. Is it possible that you saw I have little problem and interpolated I have a little problem? --Trovatore (talk) 16:26, 14 February 2012 (UTC)

I am finding this immensely instructive. I am comfortable with any of these approaches, but it is clear that the premises/definitions must be identified and consistent. Different people regard different properties (e.g. continuity or universality) as more important/convenient, but one can't have it all. Generalize to complex numbers and a new type of problem occurs. I wonder about Clifford numbers... (heh-heh) — Quondum^☏✎ 19:49, 14 February 2012 (UTC)

Well, most of the time it doesn't matter, so in those cases, just pragmatically, it isn't really necessary to identify all the premises and definitions and make them consistent. When it does matter, people generally do it; I don't ever recall seeing any serious piece of mathematics that ran into any serious trouble on this point. If you want to come up with a perfectly well-specified approach just for your own sense of mental tidiness, of course you're free to do that, but you can't count on anyone else adopting it. --Trovatore (talk) 22:41, 14 February 2012 (UTC)

Considering that Trovatore says that "most of the time it doesn't matter" it is strange that he still insists that the text in Exponentiation#Integer_exponents

Any nonzero number raised to the exponent 0 is 1; one interpretation of these powers is as empty products.

These equations do not decide the value of 0⁰. This is discussed below.

is not simplified to

Any number raised to the exponent 0 is 1.

Bo Jacoby (talk) 12:02, 16 February 2012 (UTC).

I said "most of the time it doesn't matter", not "it never matters". As Einstein put it, "things should be as simple as possible, but no simpler". By the way, you left out a subjunctive. --Trovatore (talk) 23:52, 16 February 2012 (UTC)

You let the unimportant and irrelevant fact that "some authors do not define 0⁰" pollute the article with stuff that is incomprensible to the uninitiated reader. It is unforgivable. Bo Jacoby (talk) 07:01, 17 February 2012 (UTC).

Bullshit. You're trying to push your particular POV as though it were the standard one. It is one that is represented in the mathematical community, but it is not by any means the consensus. --Trovatore (talk) 09:10, 17 February 2012 (UTC)

You are rude because you are mistaken. Bo Jacoby (talk) 15:10, 17 February 2012 (UTC).

Aside

Just as an aside, I've been looking at how my calculator and various pieces of software cope with the problem. My calculator (a Casio scientific) gives an error, as does Microsoft's Excel, but Window's Calculator tool returns 1. The Sinclair Spectrum also returned 1 (at least, an emulator does, so I assume the original did), as did the ZX81. (Now I'm showing my age!) Wikipedia's #EXPR function ({{#expr:0^0}}) gives 1. Anyone else got other software etc. to try it on?  An optimist on the run! 21:53, 14 February 2012 (UTC)

There's bit about this in exponentiation#Treatment on computers. In the 20008 IEEE floating point standard there are three different functions, pown which takes integer powers and a power 0 always gives 1, powr which gives undefined and pow which gives the combined result of pown and powr and is usual one when one wants to assume that a person normally does things right. It also has rootn which gives the nth root of a number where n is an integer, this so one can do the business of for instance returning -3 as the cube root of -27 which doesn't work with real powers. Dmcq (talk) 23:18, 14 February 2012 (UTC)