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March 8

Not HW, studying. I swear

Choose to believe me or not, but I'm studying for a Calculus test, and I did a problem in my book for which the answer is not given. I just want to make sure my answer is correct. The question was whether for the equation x^3 - 7.8x^2 + 20.25x - 13 there is or is not a relative minimum and maximum where the graph seems to level off around x=2.6. I said yes. At 2.5 and 2.7, where there is a maximum and then a minimum, respectively. Therefore, I said, there is no horizontal tangent line at the point of inflection. Can anyone just assure me that I didn't screw it up? If you want my thought process, I derivated the equation to get 3x^2 - 15.6x + 20.25 , which i then found the zeroes for. Those, I think, should be the critical points, right? Then, because the graph has a very slight dip in between 2.5 and 2.7 before heading back up, there can be no tangent line at this point of inflection (that is the point of inflection, right, where the graph changes concavity?) Thanks, Sasha 70.108.199.130 03:31, 8 March 2007 (UTC) Maybe I should mention that the question focused in on the graph in the domain from 0 to 4[reply]

Yes, that sounds correct to me. Incidentally, we have no problem checking your homework for you, or even pointing out how to do your homework, it's just doing your homework for you that we refuse. StuRat 03:45, 8 March 2007 (UTC)[reply]

Thanks. That makes me feel better. I hate figuring something out and then hearing I've figured it out wrong. 70.108.199.130 05:24, 8 March 2007 (UTC)[reply]

I'm not so enthusiastic about your method. In this context, an inflection point also has a horizontal tangent, and therefore the derivative must be zero there as well. It is not; done. Formally, we have a two-step process. First, we look for points where the derivative is zero, so that the graph is locally (to the first order, infinitesimally) neither rising nor falling. Then, we inspect the second derivative at those points. If negative, then the first derivative is positive (so the function is rising) on the left, decreasing through zero to negative (so the function is falling) on the right; this is a local maximum. If positive, a similar argument says we have a local minimum. If zero, we have an inflection. Or, the second derivative does not exist, and we must use other means.

For an example of this last possibility (unlikely on most exams), consider

3x{\sqrt[{3}]{x^{2}}}

between −1 and +1. The first derivative exists everywhere, and is zero at x = 0. We cannot assign a value to the second derivative at that point; yet we know the function is strictly negative to the left and strictly positive to the right, so a minimum or maximum is impossible. --KSmrq^T 07:25, 8 March 2007 (UTC)[reply]

Inflection point need not have a horizontal tangent. Check sine or tan graphs. --CiaPan 07:44, 8 March 2007 (UTC)[reply]

Not so fast; I accepted the poster's definition and explicitly said in this context for a reason! If we consider a plane curve, rather than the graph of a function, meanings change. So do possibilities. For example, every point on a circle is an extremum, in one sense of the term; but if we consider the half circle which is the graph of (1−x²)^1/2, we could also say that the only extremum is at x = 0. For the differential geometry sense of inflection, we would look for a curvature reversal; that is, we ask where the curvature vanishes. But this is not that context. --KSmrq^T 11:42, 8 March 2007 (UTC)[reply]

(edit conflict)
First, please note that is an expression what you wrote, but NOT an equation (equation needs an equals sign, see equation article for the definition). It is not even a function (that needs explicit definition of its input variable). Next, you are wrong about the tangent line: of course there is a line, which is tangent to the function

f(x)=x^{3}-7.8x^{2}+20.25x-13

graph at x=2.6.
However you're right that the line is not horizontal (derivative value

f'(2.6)=-0.03

is non-zero), so the function has no local (relative) extremum at that point.
CiaPan 07:41, 8 March 2007 (UTC)[reply]

I assumed the poster meant there is no "horizontal" tangent line, the second time the tangent line was mentioned, because they were talking about a horizontal tangent line the first time. StuRat 07:47, 8 March 2007 (UTC)[reply]

Pi transcendental

Is there a proof that pi is transcendental that doesn't use complex numbers? —The preceding unsigned comment was added by 130.159.248.36 (talk) 10:36, 8 March 2007 (UTC).[reply]

Every mathematical statement involving complex numbers can be rephrased as a statement involving only real numbers. By applying this systematically to the statements in a proof of the transcendence of π, including the proofs of any other theorems and lemmas invoked directly or indirectly, you should eventually reach a proof devoid of any reference to complex numbers. The result of such a mind-numbing exercise will not be pretty and even less enlightening, but if the existence question encompasses the Book of Ugly Proofs in Platonic hell, then the answer is yes. I suspect that no-one bound to this orb has ever given a "non-complex" proof. --Lambiam ^Talk 15:27, 8 March 2007 (UTC)[reply]

Can every mathematical statement that uses complex numbers really be rephrased using only real numbers? Is this just an intuition or is it provable? —The preceding unsigned comment was added by AbcXyz (talk • contribs) 19:11, 8 March 2007 (UTC).[reply]

In a limited sense this is true because C is a two-dimensional vector space over R. A more interesting question is whether certain mathematical facts can be proved without using complex analysis. A proof in number theory that does not use complex analysis is called an "elementary proof" - although it may be far from elementary in the usual sense of the word. Gandalf61 22:09, 8 March 2007 (UTC)[reply]

It would suffice to use only zero and one! The key insight is that we can construct complex numbers and their theory from real numbers. Two examples:

Consider polynomials in one variable, x, with real coefficients, and reduce modulo the polynomial x²+1.
Consider all 2×2 real matrices formed as linear combinations of
$I={\begin{bmatrix}1&0\\0&1\end{bmatrix}},\qquad J={\begin{bmatrix}0&-1\\1&0\end{bmatrix}}.$

Either construction gives an algebra equivalent to complex numbers. Of course, we can construct real numbers from rational numbers, rational numbers from integers, and so on. We can also state what it means for a complex function to be holomorphic in terms of its component real and imaginary functions, using the Cauchy-Riemann equations. --KSmrq^T 23:20, 8 March 2007 (UTC)[reply]

I am sorry, but I do not get your point: what are you trying to say that has not been said before?--80.136.129.213 23:24, 8 March 2007 (UTC)[reply]

First, although we may view C as R², that does not capture the algebra. The two constructions I give support multiplication and reciprocals, for example. Second, Gandalf61 raised the question of complex analysis, so I pointed out the relevance of the Cauchy-Riemann equations, which are purely real. Third, I affirm and extend Lambiam's assertion; we don't even need real numbers, because they in turn can be constructed from more primitive raw material (if you like, an allusion to the systematic foundation work of Nicolas Bourbaki). Overall, I hope the concrete examples of constructions and methods help make the abstract claims more plausible and intuitive.

Incidentally, Lambiam's Book of Ugly Proofs has a positive counterpart, which Paul Erdős called simply The Book. --KSmrq^T 08:10, 9 March 2007 (UTC)[reply]

4 coloured map of Europe

HI, does anyone know where you can get a 4 coloured map of Europe? I've found one for the US but, considering the audience I have to present to, I'd prefer one of Europe. Any ideas? thanks 130.88.137.220 12:29, 8 March 2007 (UTC)[reply]

Try commons:Category:Maps_of_Europe there are some blank maps in their which you could colour yourself. --Salix alba (talk) 13:15, 8 March 2007 (UTC)[reply]

Try commons:Category:Blank maps of Europe, and commons:Image:Europa_zemljevid.png --CiaPan 14:36, 8 March 2007 (UTC)[reply]

Equation of artillery fire?

I searched wikipedia with the search function but couldn't turn up anything on the path an artillery shell takes as a function of the angle and the energy put into firing the shell. Is there an equation for that? (and should this question be here or in the physics page)Coolotter88 23:20, 8 March 2007 (UTC)[reply]

It should be on the physics page. Ballistics calculations in the real world are not simple, and were a major impetus for the development of electronic digital computers. In the mathematical world we often idealize the trajectory as a parabola. Caveat emptor. --KSmrq^T 23:32, 8 March 2007 (UTC)[reply]

Please see Projectile motion.--Ķĩřβȳ ♥Ťįɱé Ø 00:01, 9 March 2007 (UTC)[reply]

The limited discussion in that article cannot explain how David Beckham can "bend" a kick, or how a boomerang can return, nor why a rifle is designed to spin a bullet, nor why the shape of projectiles evolved from spherical to tapered. One would hardly need a digital computer to determine a parabolic path. The most important neglected influence is fluid dynamics, especially viscosity and Bernoulli effects. These depend on altitude, temperature, and humidity, and also on the shape, speed, and motion of the projectile. Applied mathematicians with an interest in numerical analysis may be intrigued to learn that I. J. Schoenberg was working on ballistics at Aberdeen when he first developed splines. The "real world" can be complicated, but nevertheless may inspire some nice mathematics. --KSmrq^T 08:56, 9 March 2007 (UTC)[reply]