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Contents 1 December 11 1.1 Roots of unity in infinite field 1.2 Average Annual Increase?? 1.3 Maclaurin polynomial series and calculating error 1.4 Buying a new graphing calculator

[edit] December 11

[edit] Roots of unity in infinite field

Hey, In an infinite field is it possible to find, for every natural number n, an element u so that for all i = 1 ... n? Obviously false for finite fields, but is true for the reals. Can you have a crazy infinite field in which every element has uⁱ = 1 for some i < n? A slightly more strong question is is it possible for an infinite field to have the group of units be torsion? (stronger, because you might be able to have a torsion group of units, but have arbitrarily large orders of elements so as to always find a u for a given n). —Preceding unsigned comment added by 64.198.253.173 (talk) 01:36, 11 December 2007 (UTC)

For every natural number n and every field F, the polynomial f(x)=(x-1)*(x^2-1)*...*(x^n-1) has only finitely many roots in F. If F is infinite, then there must be an element u which is not a root of f(x), and so u^i is not equal to 1 for any i=1...n.

The algebraic closure of a finite field has a torsion group of units. Every element of the algebraic closure is algebraic over a finite field, so contained in some (possibly larger, but still) finite field, and so has finite order. There is no bound to the order, of course. JackSchmidt (talk) 05:06, 11 December 2007 (UTC)

[edit] Average Annual Increase??

Good morning everyone,

I’m looking at salary figures for my company. In 1997 a particular position made $19,679. By 2007, that same position earned $30,970. I want to say something like “this position had an average increase of X% per year over this 10-year period.” But I’m not sure what the average percent increase is???

The salary of $19,679 to $30,970 is a $11,291 increase, which equates to a 57% increase. So, over a 10-year period, 57% / 10 years = 5.7% annualy. But I ran the math, and this does not calculate the average annual percent increase. I’d appreciate some help here from someone smarter than I.

Much appreciated. Rangermike (talk) 16:41, 11 December 2007 (UTC)

If the salary starts at S and increases by X% every year, then after n years the salary will be (1 + X%)ⁿS. In your case you have 30970$ = (1 + X%)¹⁰19679$, so , and X% = 1.04639 − 1 = 0.04639 = 4.639%. -- Meni Rosenfeld (talk) 17:12, 11 December 2007 (UTC)

This is the geometric mean. Phaunt (talk) 17:55, 11 December 2007 (UTC)

(After edit conflict) Here Meni has calculated the geometric mean of the increases (well, here it's trivial because we have the total increase; hence to calculate the mean we just take the nth root). Notice that this is precisely the mathematical entity to work with when you want means of rates or ratios.

Calculating the arithmetic mean (what you intend in the first place) is a common (though not precise) procedure to obtain such a figure. The smaller the increase of the ratio, the smaller the difference will be between the means. But remember Meni's calculation is the precise answer to your case. Pallida  Mors 18:02, 11 December 2007 (UTC)

A few remarks. If we are given a list of the relative increases every year (or a least of values from which we can calculate them), we can decide what we want to do with them. One possibility is to calculate the arithmetic mean of the values. However, here we are only given the start and end values. Given only those, the arithmetic mean of increases is not well-defined. The question "what is the average increase" is only meaningful if by "average" we mean some quantity which is invariant given the start and end. The only plausible such quantity is the value which, if it was the relative increase every year, would lead to the observed total increase. This happens to be equal to the geometric mean of the ratios minus 1 (not the geometric mean of the increases), and it is what I have calculated. I'll leave it to the OP to decide if this is what he wanted, with the understanding that anything else would require additional data. -- Meni Rosenfeld (talk) 18:25, 11 December 2007 (UTC)

Thanks for everyone's reply. I always learn something new on Wikipedia.Rangermike (talk) 19:26, 11 December 2007 (UTC)

Meni is right in his remarks about obscurity of geometric/arithmetic means used above. Let me clarify my statement:

For ratios β_i,i=1...n, the geometric approach to an average increase is given by

...While the arithmetic [incorrect, but approximate] approach is just to compute

.

Yours The OP's was a question about final fractions, not intermediate ratios, but a similar discussion takes place in such queries. So, for instance, if someone says that the annual inflation rate is 6%, you will instantly get a figure of around 0.5% for a monthly rate. The exact figure is something like 0.487% though.Pallida  Mors 18:20, 12 December 2007 (UTC)

[edit] Maclaurin polynomial series and calculating error

Find the third-order Maclaurin polynomial for (1+x)^-1/2 and bound the error R₃(x) if -.05 ≤ x ≤ .05.

I understand how to find the 3rd order Maclaurin polynomial, but I don't understand what it means, or how to bound the error between -.05 and .05. They give me an equation to use, which is (f⁽ⁿ⁺¹⁾(c)(x⁽ⁿ⁺¹⁾))/(n+1)!

The Maclaurin polynomial is 1 - (1/2)x + (3/8)x² - (5/16)x³...... —Preceding unsigned comment added by 64.113.85.178 (talk) 19:10, 11 December 2007 (UTC)

Perhaps Taylor's theorem will help.

The 3rd order Maclaurin polynomial of the function is the best polynomial of degree 3 for calculating the function approximately when . If you take some , you will see that is very close to . However, there will still be some error, and we want to know how big the error can be, that is, to say something like "If , then the error can be at most 0.001".

Taylor's theorem gives several ways to find such a bound for the error. One of them is that for every x, there is some c between 0 and x such that the error (here n=3) is . After calculating the 4th derivative of the function, you can use the fact that both c and x have an absolute value no greater than 0.05 to find an upper bound on the magnitude of the error. -- Meni Rosenfeld (talk) 19:43, 11 December 2007 (UTC)

[edit] Buying a new graphing calculator

My TI-89 Titanium was stolen recently and thus I need to buy a new graphing calculator. I'm a high school student and plan on using whatever I choose to purchase throughout college (or maybe even grad school), if at all possible. Texas Instruments recently created a new calculator, called the TI-Nspire. However, I don't know whether this is actually an improvement over the Titanium or whether I should just get an HP 50G. Which of the three should I buy? Bobguy7 (talk) 23:18, 11 December 2007 (UTC)

I personally never saw the point of a graphics calculator. Every exam I've ever done at school & universoty has explicitly forbidden calculators with more than 2 lines of display. mattbuck (talk) 00:44, 12 December 2007 (UTC)

I'm a college senior going to math grad school next year. The only calculator I have is a TI-83+, and I haven't really used it since freshman year. There's always Mathematica if you need to do something heavy duty, so for a graphing calculator I'd stick with something cheap and easy to use. 134.173.93.150 (talk) 00:54, 12 December 2007 (UTC)

Personally, I've found my TI-89 Titanium to be incredibly useful in my mathematics and engineering courses. I would never downgrade to anything less, despite the higher price. I suppose it also depends on what your professors will allow. My professors were quite lenient. Mattbuck's professors were (are) more strict. Yours could fall anywhere in between. - SigmaEpsilon → ΣΕ 03:07, 12 December 2007 (UTC)

I do maths at Nottingham, and the school's standard exam blurb states no more than 2 lines of display, presumably to stop people looking up answers with it, or using calculators like that to do calculus and stuff. I've never had need of more than a standard 2-line display anyway. mattbuck (talk) 03:36, 12 December 2007 (UTC)

Personally, my TI-83+ has been perfectly capable of handling all my engineering courses. I've never seen a restriction on the complexity of a calculator, either. sh¤y 15:57, 12 December 2007 (UTC)

The OP hasn't said what he intends to study at college, but as a mathematics student myself I have only very occasionally found a calculator useful (I didn't have a computer in those days), and have not sat an exam in which calculators were allowed since high school. Engineers and such need calculators more, of course. Algebraist 17:46, 12 December 2007 (UTC)

As long as we are sharing our personal experiences, I, currently a mathematics student, use calculators frequently for various purposes, and I don't recall ever doing an exam in which calculators were forbidden. I have never used a graphing calculator in my life, though - for anything beyond simple calculations, I use my computer with a CAS. -- Meni Rosenfeld (talk) 17:59, 12 December 2007 (UTC)

I used a graphics calcultor a few times at school, but only to learn how to as a class exercise. It was never useful to actually have one, and they cost too much. Maths students don't need calculators like that, heck, we barely need calcultors at all, unless you do statistics, but who wants to do that? mattbuck (talk) 18:20, 12 December 2007 (UTC)

I used a TI-83+ provided by my school, a fair amount during my Sophomore and Junior Years of High School, it was mainly used for Statistics but was also used a fair amount for Calculus. The only reason we really 'needed' the graphing calculator in Algebra 2 was for finding determinants of 3x3 or larger matrices, since they didn't bother to teach Minors and Cofactors(??) I currently own a TI-Voyage 200, which will likely never be allowed to be used during a test since it has a qwerty keyboard, but I do like using it to aid in my mathematical derivations. A math-wiki (talk) 19:57, 14 December 2007 (UTC)