Wikipedia:Reference desk/Archives/Mathematics/2009 December 29

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December 29

Hypervolume of an 8-sphere

Does anyone know the formula for the hypervolume of an 8-sphere? This is so I can remove all those question marks from Nine-dimensional space... 4 = 2 + 2 04:23, 29 December 2009 (UTC)

See n-sphere. PrimeHunter (talk) 04:32, 29 December 2009 (UTC)

transcendental functions

The article Lockhart's Lament, linked above has the sentence "The measurement of triangles will be discussed without mention of the transcendental nature of the trigonometric functions." Can someone please explain to me what the transcendental nature of trigonometric functions has to do with measurement of triangles. Thanks-Shahab (talk) 07:09, 29 December 2009 (UTC)

Could you please provide the context for that remark (the exact number of the page on which it was made)? I read that article earlier this year and do not recall such a statement being made. But it is quite possible that I have forgotten its presence. --PST 07:36, 29 December 2009 (UTC)

It's on the last page (Pg 25) of the article. The section deals with an "honest catalog" of the curriculum(The trigonometry course). The complete paragraph runs as follows:

Two weeks of content are stretched to semester length by masturbatory

definitional runarounds. Truly interesting and beautiful phenomena, such as the way the sides of a triangle depend on its angles, will be given the same emphasis as irrelevant abbreviations and obsolete notational conventions, in order to prevent students from forming any clear idea as to what the subject is about. Students will learn such mnemonic devices as “SohCahToa” and “All Students Take Calculus” in lieu of developing a natural intuitive feeling for orientation and symmetry. The measurement of triangles will be discussed without mention of the transcendental nature of the trigonometric functions, or the consequent linguistic and philosophical problems inherent in making such measurements. Calculator required, so as to

further blur these issues.

Regards-Shahab (talk) 07:52, 29 December 2009 (UTC)

Not sure what it's about. Trigonometric functions of any rational number besides 0 are always transcendental when you use radians, perhaps he was just referring to the problems of incommensurables the Greeks had with things like the square root of 2? Dmcq (talk) 08:35, 29 December 2009 (UTC)

I think he means that in many high school problems given, the lengths of the sides of the triangles involved are usually assumed to be simple values such as 1 or ${\sqrt {2}}$, and this convolutes one's understanding of more general triangles. --PST 09:49, 29 December 2009 (UTC)

I think he means that learning how to calculate the length of a certain side given an angle and the length of another side is not interesting maths (it's just computation) whereas learning (or, even better, deriving) that the trigonometric functions are transcendental is interesting maths. --Tango (talk) 14:56, 29 December 2009 (UTC)

No math after 12th grade?

A question about let me thinking. Is it too awkward to say that there is no math before 12th grade? Until 12th grade you are only performing calculations, you are more a human calculator than a real mathematician. ProteanEd (talk) 12:52, 29 December 2009 (UTC)

The word "mathematics" refers to a wide spectrum of activities. It includes things such as addition and multiplication, learned in grade school, and also includes esoteric facts about more complicated objects that are learned in graduate school. These are not fundamentally different topics, but are simply different expressions of the same subject. You have to crawl before you can walk, and later you learn to drive; all of these are forms of locomotion. — Carl (CBM · talk) 12:56, 29 December 2009 (UTC)

I disagree (in fact, I believe that there is not mathematics even in the 12th grade; at least in the current American education system). Mathematics is, after all, creativity. It is not about attaining tools such as computational ability, but rather about using the tools for a particular purpose. Thus mindless computation, on its own, is not mathematics, unless it is done in a creative manner. Even calculus, in the mechanical sense of solving problems, is not mathematical unless one does it creatively. Thus, while the routine technique of finding the maximum values of polynomials of degree two on a compact interval loses its mathematical value at a certain stage, inventing new techniques in numerical analysis can be safely referred to as mathematics. In another context, one could refer to painting blue dots on a white sheet of paper as "painting", but such would not be considered to be of true artistic value. Combining a wide range of colors with different tone, depth and emotion to create a painting is true art. --PST 13:10, 29 December 2009 (UTC)

You're free to argue that the English language should be changed. At the present, "mathematics" clearly refers both to the things we learn in kindergarten and the things we learn in graduate school. Similarly, finger-painting is indeed painting. Maybe you're talking about real mathematics, but that sort of discussion will always devolve into polemics. — Carl (CBM · talk) 13:16, 29 December 2009 (UTC)

I guess you are right. --PST 13:22, 29 December 2009 (UTC)

That definition is entirely subjective. In particular, at which point mathematics becomes creative is not properly ascertainable. If "creative" means "original", many mathematicians never end up doing mathematics. In the same way that the composition of a short letter and the composition of a classic piece of fiction are both termed "writing", everything from basic addition to the discovery of new theorems in esoteric fields is mathematics. So, yes, mathematics is done in high school and before. —Anonymous Dissident^Talk 13:30, 29 December 2009 (UTC)

As I mean it, doing mathematics is not necessarily creating original maths, nor one needs to discover original results to enjoy it greatly, in the same way that doing music is not necessarily composing. When I was 6 I had a wonderful teacher, which was about 19, so sweet. She made us solve nice addition problems drawing colored balls... when she would come to me and lean over my little desk, how happy I was! I scarcely listened towhat she would tell me, but only hear the sound of her sweet voice and sniff the perfume of her black hairs. Maybe that was no original maths, but how exciting! ;)--pma (talk) 15:39, 29 December 2009 (UTC)

Symmetries and perturbations

What do you think about this statement:

Suppose f is a continuous function and h is a (continuous) symmetry for f i.e.

$f\circ h=h\circ f$

Then for any ε there exists δ such that if d(f,f ')<δ then there exists h ' such that h ' is a simmetry for f ' and d(h,h ')<ε.

assuming d is the standard distance given by the uniform norm.

Is it true or false?--Pokipsy76 (talk) 13:57, 29 December 2009 (UTC)

I see it more false than true in general, though I don't have a counterexample. Could you specify the domain? If f₀ is an Anosov diffeomorphism and f₁ is (sufficiently) C¹ close to f₀ then f₁ is Anosov and conjugated to f₀ with a conjugation H which is C⁰ close to the identity, together with its inverse H^-1. So by conjugation with the same H you do have a h₁ C⁰ close to h₀ doing the job. --pma (talk) 14:36, 29 December 2009 (UTC)

I didn't want to be specific about the domain, I wanted to ask the question in general, however I'm curious in particolar for functions in R and R². I supopose that a first step to find a counterexample could be to find a function which doesn't have any nontrivial simmetry... are there such functions?--Pokipsy76 (talk) 15:26, 29 December 2009 (UTC)

Consider the following counter-example on $\mathbb {R} .$ Let $f_{0}:=\mathrm {id}$ and $h_{0}\in C^{0}(\mathbb {R} ,\mathbb {R} ),$ any function, with $h_{0}\neq \mathrm {id} .$ Say $\|h_{0}-\mathrm {id} \|_{\infty }>r>0.$ Then $f_{0}$ and $h_{0},$ of course, do commute, and I claim that there exists a function $f$ as close in the uniform distance to $f_{0}$ as we want (in fact, even in the $C^{\infty }$ distance if you wish), with the property that any $h\in C^{0}(\mathbb {R} ,\mathbb {R} )$ that commutes with $f$ necessarily has a uniform distance $\|h_{0}-h\|_{\infty }\geq r.$

Indeed, let $a\in \mathbb {R}$ such that $|a-h_{0}(a)|\geq r.$ There exists a function $f$ such that $a$ is a globally attractive fixed point of $f$, meaning that $f(a)=a$ and $f^{\,n}(x)\to a$ as $n\to \infty$ for all $x\in \mathbb {R} ;$ such an $f$ may be chosen arbitrarily close to $f_{0}$ (take e.g. $f(x):=x-\epsilon \,\arctan(x-a)$ for a small $\epsilon >0$). Let $h\in C^{0}(\mathbb {R} ,\mathbb {R} )$ a function that commutes with $f$. Then we have $h(a)=h(f^{n}(a))=f^{n}(h(a))\to a$ as $n\to \infty$ so $h(a)=a$ and $\|h-h_{0}\|_{\infty }\geq |h(a)-h_{0}(a)|=|a-h_{0}(a)|\geq r.$ Is that clear? Actually this argument could be adapted in order to show that the original property fails to hold with $f_{0}:=\mathrm {id}$ and $h_{0}\neq \mathrm {id}$ with any topological manifold as a domain. --pma (talk) 20:28, 29 December 2009 (UTC)

Very nice proof!--Pokipsy76 (talk) 19:37, 5 January 2010 (UTC)