Wikipedia:Reference desk/Archives/Mathematics/2007 June 13

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June 13

Regression

How would one perform a linear regression of Y on X and X² (simultaneously) using Excel?
Zain Ebrahim 11:06, 13 June 2007 (UTC)

Be very careful if you are using Excel for something like this. Excel has numerous errors in its statistical functions and Microsoft has a poor track record when it comes to fixing them. See, for example, [1], [2] and [3]. A short quote from the latter: "In its Excel XP release, Microsoft attempted to fix some statistical problems, but it did not do a good job (McCullough and Wilson, 2002). This failure presaged Microsoft’s attempt at a major overhaul with Excel 2003. While it fixed many functions, it failed to fix many others."

If you need accurate results, you would be much better off investing time and perhaps even some money in serious statistical software such as R (free) or the SAS System (not free). If you prefer to use a spreadsheet, Gnumeric has a better track record on accuracy than Excel.[4] If you must use Excel, don't use versions earlier than Excel 2003, and read the relevant parts of this site first: Errors, Faults and Fixes for Excel Statistical Functions and Routines. The LINEST function is Excel's regression workhorse. -- Avenue 13:05, 13 June 2007 (UTC)

Don't be too put off: for the simple regression you describe I'm sure Excel will do an adequate job. The papers referred to above say that no problems have been found with the linear regression routines in Excel 2003 and onwards, and in previous ones it was only some of the peripheral output that was questionable. You don't need a sledgehammer to crack a peanut. To actually answer your question:

Set out your Y, X and X² data in three successive columns. Then go to the menu item Tools|Data Analysis. If Data Analysis doesn't appear under Tools, you will first have to go to Tools|Add Ins and tick the Analysis ToolPak box. Under Tools|Data Analysis select Regression. In the following dialog box, identify your column of Y values in the Input Y Range box, and your double column of X and X² data in the Input X Range. The Help button in the dialog box will explain the other options. Nominate an output location for the results, and click OK. Good luck. --Prophys 10:42, 14 June 2007 (UTC)

abstract algebra

if G/Z(G)is cyclic then show that G is abelian

Perhaps reading Inner automorphism#The inner and outer automorphism groups will give you a hint.  --Lambiam^Talk 15:52, 13 June 2007 (UTC)

Just think about what it means for a group to be abelian; you need to show that for all $x,y\in G$, $xy=yx$. Since you know that G/Z(G) is cyclic, you can express any element in G/Z(G) as a power of a generator for G/Z(G). Think about what that means for arbitrary elements of G... –King Bee ^{(τ • γ)} 17:41, 13 June 2007 (UTC)

It's funny. A few hours ago I was working on an exercise in which I analyze the characters of a non-abelian group of order $p^{3}$, and had to effectively prove this statement in order to conclude that $|Z(G)|=p$. Perhaps the OP is a classmate :-) -- Meni Rosenfeld (talk) 20:01, 13 June 2007 (UTC)

abstract algebra

prove that Z/143Z is cyclic

Do your own homework. The reference desk won't give you answers for your homework, although we will try to help you out if there's a specific part of your homework you don't understand. Make an effort to show that you've tried solving it first. Dep. Garcia ( Talk + | Help Desk | Complaints ) 15:54, 13 June 2007 (UTC)

This is almost by definition. Take the time to read and understand cyclic groups (or any other topic that builds along the way) and you will find abstract algebra much easier. iames 15:56, 13 June 2007 (UTC)

(after edit conflict) What about: prove that Z/nZ is cyclic if n ≠ 0. See also Modular arithmetic#The ring of congruence classes.  --Lambiam^Talk 15:59, 13 June 2007 (UTC)

I suggest the following strategy: Prove that Z/3Z is cyclic. Prove that Z/4Z is cyclic. (Brute force may suffice.) Combine what you have have learned to attack Z/143Z. Question: How many groups have order 143? Have fun! :-) --KSmrq^T 16:26, 13 June 2007 (UTC)

I don't know much about group theory, and even less about Galois field theory, so maybe I am missing something here, but doesn't 11 * 13 = 143? Baccyak4H (Yak!) 16:35, 13 June 2007 (UTC)

Good catch; thanks. (That's what comes from using Factor instead of FactorInteger.) The fact that 1−4+3 = 0 shows that 143 is a multiple of 11. Fixed. --KSmrq^T 18:28, 13 June 2007 (UTC)

Composite in all bases

Assuming good faith on the part of KSmrq, I wondered if there exists a base > 5 for which 143 (expressed as 143 in that base) was prime, and at least up to base 32, the answer is no. Are there any results about this sort of thing? iames 16:47, 13 June 2007 (UTC)

I guess if I thought about it for 5 seconds I'd see how $n^{2}+4n+3=(n+3)(n+1)$ ... iames 16:59, 13 June 2007 (UTC)

I just recognized it as 12² − 1², which is of course (12+1)*(12-1). But ignore my comment if "prime" means something different in Galois theory. And even if it does not, I would definitely assume good faith; it wasn't like he said "60". Baccyak4H (Yak!) 17:09, 13 June 2007 (UTC)

One can assume good faith without assuming good judgment (a common Wikipedia confusion)! No sophisticated theory of bases or Galois fields is involved, I just made a mundane mistake. <:-O But the number of groups of order 143 remains of interest (for those interested in such things). --KSmrq^T 18:38, 13 June 2007 (UTC)

It turns out to be just 1[5] $Z_{11}\times Z_{13}$. I got side tracked into looking at groups whos order is the product of two distinct primes, p, q say, with p>q. A brief test on a few orders seems to suggest that if p % q == 1 then there will be two groups of order pq, but if p%q!=1 there is only a single group. Does any one know if this is true in general if so why? Also there seems to be two group of order 21 whats the other one look like? --Salix alba (talk) 00:06, 14 June 2007 (UTC)

See "example of groups of order pq". PlanetMath..  --Lambiam^Talk 00:23, 14 June 2007 (UTC)

That example generalises, of course: if p and q are primes, p < q, and p does not divide q-1, then there is a unique group of order pq. If p does divide q-1, then there is another group (a semidirect product of C_p and C_q). I'm pretty sure there's only one nontrivial semidirect product, but I can't remember if I've proved it or not and don't feel like doing it now. Exercise: for which n is there a unique group of order n? (this was set to me two years ago, and I have the answer, but my only proof uses the odd order theorem, so I'm still working on it) Algebraist 10:59, 15 June 2007 (UTC)

As for numbers that are composite in all bases, the simplest example would presumably be 100_n = 10_n × 10_n for any base n. —Ilmari Karonen (talk) 18:55, 16 June 2007 (UTC)

Analysis qualifying exam question

Hi all, I'm studying for a pesky PhD qualifying examination, and one of these complex analysis questions has me stumped. I was hoping to get some help. Here's the statement:

Let f be analytic and nonconstant in the unit disk and let 0 < r < 1. Suppose that on the circle {z : |z| = r}, |f| assumes its maximum at the point z₀. Prove that $z_{0}{\frac {f'(z_{0})}{f(z_{0})}}$ is real.

I would like to tell you what I've tried on this problem, but to be honest, I don't even know where to begin. A couple of my fellow graduate students are the same way. Can you help us out? Hints over solutions are preferred. Thanks! –King Bee ^{(τ • γ)} 17:56, 13 June 2007 (UTC)

I'll close my eyes and point to Maximum modulus principle and Borel–Carathéodory theorem (and then to sci.math when you find I wasn't helpful.) iames 19:22, 13 June 2007 (UTC)

I haven't tried a proof, but the first thing that occurs to me is that at a maximum the derivative (think "gradient") is perpendicular to the circle. Does that help? --KSmrq^T 19:24, 13 June 2007 (UTC)

A small correction, though - The derivative is not perpendicular to the circle. For example, if $z_{0}=1+i$ and $f(z_{0})=1$, then $f'(z_{0})\;\!$ is in the direction of $1-i$. There is no need for the Borel–Carathéodory theorem, but you do need to consider the directions and the Maximum modulus principle. Expressing $f(z)$ and $f'(z)\;\!$ in terms of real and imaginary parts may help clarify things. -- Meni Rosenfeld (talk) 19:50, 13 June 2007 (UTC)

I like these ideas, and I should admit that I knew that the Maximum Modulus Principle would be an important thing, but I'm having trouble trying to figure out how/where to use it.

Meni - I decomposed the expression $z_{0}{\frac {f'(z_{0})}{f(z_{0})}}$ into its real and imaginary parts, and it would suffice to show that the imaginary part is 0. For the imaginary part, I got (if z₀ = x₀ + iy₀)

${\frac {1}{u(z_{0})^{2}+v(z_{0})^{2}}}\cdot (y_{0}u(z_{0})u_{x}(z_{0})+y_{0}v(z_{0})v_{x}(z_{0})+x_{0}u(z_{0})v_{x}(z_{0})-x_{0}u_{x}(z_{0})v(z_{0}))$

Is this what you were thinking, or am I way off? –King Bee ^{(τ • γ)} 21:13, 13 June 2007 (UTC)

Maybe I've drunk too much, but if you take f = exp, isn't |exp z| = exp |z|, so z₀ can be any point on the circle |z| = r. But then z₀f'(z₀)/f(z₀) = z₀, which need not be real.  --Lambiam^Talk 21:35, 13 June 2007 (UTC)

Postscriptum. Simpler counterexample: f(z) = iz. Do I misunderstand the problem statement? 21:40, 13 June 2007 (UTC)

Lambiam : You have definitely drunk too much if you say |exp z| = exp |z|. Also, for $f(z)=iz$, the value in question is 1, which is real.

King Bee: I guess that should work. Now take a small real $\epsilon$ and see what the maximality of $f(z_{0})$ has to say about $f(z_{0}(1+i\epsilon ))\;\!$, and then use the Cauchy-Riemann equations. I think this will work even without the maximum modulus principle (you could use it to learn about $f(z_{0}(1-\epsilon ))$ for small $\epsilon >0$, but this doesn't seem necessary). -- Meni Rosenfeld (talk) 22:07, 13 June 2007 (UTC)

I don't think it makes the problem any easier, but it may simplify bookkeeping to assume z₀ = 1. (As far as I can tell, replacing f with g(z)=f(z₀ z) immediately reduces this to the case where 1 is a maximum. Of course g is then analytic in a disk strictly bigger than the unit disc, so there's no problem there.) Tesseran 01:22, 14 June 2007 (UTC)

Yes, this is certainly a good transformation. Building up on that, letting $h(z)={\frac {f(z_{0}z)}{f(z_{0})}}$ is even neater. -- Meni Rosenfeld (talk) 13:29, 14 June 2007 (UTC)

For an alternative (and in my opinion, more elegant) approach, read KSmrq's comment: the gradient of |f| is perpendicular to the circle. In other words, the derivative of |f| along the circle vanishes. You should now recognize f' / f as the logarithmic derivative and remember that the real part of the logarithm is the modulus, and a proof will follow. No need to use maximum modulus or Cauchy-Riemann or similarly advanced results. -- Jitse Niesen (talk) 00:26, 15 June 2007 (UTC)

• Thanks for all your help. There are some pretty clever folks here at the ref desk! –King Bee ^{(τ • γ)} 15:15, 16 June 2007 (UTC)