Wikipedia:Reference desk/Archives/Mathematics/2007 November 29

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November 29

Different forms of the Rayleigh quotient

My PDE book has a formula for the Rayleigh quotient that's equivalent to the one at the bottom of this section: Sturm-Liouville theory#Sturm-Liouville theory, but the Rayleigh quotient article has something that looks much simpler. Are these expressions actually related, and if so, how? —Keenan Pepper 00:19, 29 November 2007 (UTC)

The Sturm-Liouville version is a special case of the general definition. The vector space is the space of nice functions satisfying appropriate boundary conditions, the inner product is $\langle y_{1},y_{2}\rangle =\int _{a}^{b}y_{1}(x)y_{2}(x)w(x)\,dx$ and the selfadjoint (=Hermitian since we're real) operator is $(w(x))^{-1}\left(-{\frac {d}{dx}}\left[p(x){\frac {d\,}{dx}}\right]+q(x)\right)$, if I remember my applied correctly and have made no τεχnical errors. Algebraist 01:39, 29 November 2007 (UTC)

I was about to complain that they still didn't look the same, but then I just integrated by parts and all was well. Thanks! I might add a section to one of those articles later. —Keenan Pepper 04:06, 29 November 2007 (UTC)

ANOVA/regression analysis

I was running an ANOVA as well as Regression Analysis with one independent variable or one predictor variable. I was surprised to see that the F value as well as the significance level and R square value were identical for ANOVA and Regression Analysis. Could anyone help me to understand what that means or tells me? I appreciate any comments or insights. Thanks much. -- Sophiepsych (talk) 02:39, 29 November 2007 (UTC)

First, are you sure you spoke rightly? A predictor variable typically is an independent variable. Did you mean predictor and dependent?

The situation you describe can happen when you have a continuous X variable but which happens to take on only two unique values in your experiment. The predictions in both models are simply the means of Y for each subset with the unique X values. The main difference is that for the ANOVA, it makes no sense to as what the prediction is for an X value halfway (say) between the two in your experiment, while that does make sense for a regression. In other words, one or the other of these analyses is preferred for your experimental situation, even though both give the same results. Baccyak4H (Yak!) 14:31, 29 November 2007 (UTC)

Yes, I meant predictor variables (IV), in both cases, ANOVA and Regression Analysis. Thank you for your help --Sophiepsych (talk) 03:10, 30 November 2007 (UTC)

Why is 0 undefined?

Just out of curiousity, if 0 * 0 = 0 then therefore 0² = 0. If 0² equals to zero then why is square root of 0 undefined and why does the calculator value of $0$ equate to zero?

And another question, if the factorial of 5 is given by 5*4*3*2*1 then why is the factorial of zero, one? If to find the factorial of 0, you must evaluate 0*1? —Preceding unsigned comment added by 122.105.132.73 (talk) 07:02, 29 November 2007 (UTC)

There seems to be some misunderstanding. The number 0 is not undefined. The word zero is the English word that is the name of the number (as well as the digit) 0, just like one is the name of the number 1, and two is the name of the number 2.

There are several explanations for why 0! = 1. The most direct one is that this is so, because that is how the factorial function has been defined. It is true by definition. You might ask why it has been thus defined. Then the reason is that with this definition many mathematical formulas using the factorial come out nicer. For example, for all non-negative integers n, it is the case that (n+1)! = (n+1) × n!, so for example 5! = 5 × 4!. Then 1! = 1 × 0!. If you work this out, you will find that this implies 0! = 1.  --Lambiam 07:49, 29 November 2007 (UTC)

Sorry I think I mistyped the question. I meant if 0² equals to 0 then why is the square root of 0 undefined?—Preceding unsigned comment added by 122.105.132.73 (talk) 10:10, 29 November 2007 (UTC)

The square root of 0 isn't undefined. The square root of 0 is 0, just as you expect. Which make and model of calculator are you using ? Gandalf61 (talk) 09:52, 29 November 2007 (UTC)

I don't think there is a problem with the calculator, if by "the calculator value of 0 equate to zero" the OP meant that the calculator gives ${\sqrt {0}}=0$ (which is correct). Perhaps the OP was thinking about $0^{0}$, which is sometimes left undefined (in particular, if the exponent is taken to be a real number rather than an integer).

As for 0! - the factorial of a nonnegative integer n can be defined as "the product of all integers which are at least 1 and no more than n". With this definition, 0! is not 0*1 but rather the product of all intgers which are at least 1 and at most 0. Of course, no such integer exists, so 0! is the product of nothing, or an empty product. It makes a lot of sense to consider the empty product to be the multiplicative identity, 1. One way to think about it - 1 can always be considered to be a part of the product, since it doesn't change the result. So it doesn't matter if you think about 5! as 1*2*3*4*5 or 1*1*2*3*4*5. For 0!, there are no additional factors that need to be included by the definition, so all that is left is the default 1. -- Meni Rosenfeld (talk) 11:53, 29 November 2007 (UTC)

Excellent answer. I never though about that. With this definition, 0!=1 becomes a theorem. nadav (talk) 12:34, 29 November 2007 (UTC)

One point: it's not that it just makes sense to define the empty product as 1. Rather, if it is to equal some number, then that number has to equal 1. nadav (talk) 12:39, 29 November 2007 (UTC)

These definitions unfortunately don't work for negative numbers, which can also be defined. The most versatile definition of the factorial is the Gamma function, which when evaluated for 0! will give an answer of 1. SamuelRiv (talk) 15:01, 29 November 2007 (UTC)

(ec) One can also define n! as the number of permutations of n objects. Since the empty function is the unique permutation of no onjects, we have 0!=1. In an attempt to understand the OP's question, note that when defining complex exponentiation it's customary to rule out zero entirely, so in some rather silly sense the complex square root of zero is undefined. Algebraist 15:04, 29 November 2007 (UTC)