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August 28

Kernel of a transformation on the set of polynomials

Given integers $n,m\geq 1$ Let $x_{1},...,x_{n}$ be distinct points in $\mathbb {R}$ and $\mathbb {P} _{m}$ denote the set of polynomials of degree at most $m$ . Define a transformation $T:\mathbb {P} _{m}\rightarrow \mathbb {R} ^{n}$ by $T(f)={\begin{bmatrix}f(x_{1})\\\vdots \\f(x_{n})\end{bmatrix}}$ where $f\in \mathbb {P} _{m}$ . What is the dimension of the kernel of T?

The case m<n is trivial. Otherwise any such polynomial in the kernel of T can be written as $a\left(\prod _{i=1}^{n}(x-x_{i})\right)\left(\prod _{i=n+1}^{m}(x-c_{i})\right)$ where $c_{i}\in \mathbb {C} ,a\in \mathbb {R}$ are arbitrary (free?) variables. Is it sufficient to claim that the answer is $m-n+1$ because that is the number of free variables, or do I need to show something more (like find a basis for the kernel) ? Widener (talk) 11:57, 28 August 2011 (UTC)[reply]

You know the dimension of the image from the question on Aug. 21. Use dim(ker)+dim(im)=dim(domain).--RDBury (talk) 15:27, 28 August 2011 (UTC)[reply]

Yes, I know that, but I would like to know if it can be done without knowing dim(im) beforehand. Widener (talk) 17:38, 28 August 2011 (UTC)[reply]

You've more or less already answered your own question: any polynomial in the kernel can be written (uniquely) as

\left(\prod _{i=1}^{n}(x-x_{i})\right)p(x)

for some

p\in P_{m-n}

. Sławomir Biały (talk) 17:49, 28 August 2011 (UTC)[reply]

Partial Derivative of Volume of Cone

Hello. Why is the partial derivative of the volume of a cone not its surface area unlike the ball (solid sphere)? Thanks in advance. --Mayfare (talk) 15:24, 28 August 2011 (UTC)[reply]

The reason it works for a ball is that a small change in radius radius results in an shell with uniform thickness; it has approximately the same volume as area times the thickness. That reasoning fails for a cone. It also depends on the how you measure the size of the sphere. For example the formulas for area and volume in terms of diameter are V=1/6 π d³ and A = π d²; the derivative is off by a factor of 2 in that case. There are reasons that the derivative of volume formula is the area formula for a sphere, but it's kind of a one-off.--RDBury (talk) 15:39, 28 August 2011 (UTC)[reply]

Even if you maintain the height-to-radius ratio? --Mayfare (talk) 14:12, 29 August 2011 (UTC)[reply]

Ask yourself how the height

H

and the radius

R

vary if you increase the thickness of the cone by

x

units. Reasoning from the triangular cross-section may help. I get

H(x)=H+x+x{\sqrt {{\frac {H^{2}}{R^{2}}}+1}}

and

R(x)={\frac {R}{H}}H(x)

. Plug

H(x)

and

R(x)

into the formula for the volume of a cone, differentiate with respect to

x

, and evaluate at

x=0

. You should get the formula for the area. 98.248.42.252 (talk) 06:38, 30 August 2011 (UTC)[reply]

Cone terminology

There are four different things I think of as a "cone":

1) What most people call a "cone", that is, the portion trimmed between a plane and a vertex point.

2) A "semi-infinite cone", trimmed only to the vertex point.

3) An "infinite cone", with two lobes, unbounded at both ends.

4) A "conic frustrum", with a single lobe trimmed at both ends by planes.

So, am I using the proper terminology to describe these ? Is there a way to specify the first one that makes it clear I'm not talking about the others ? StuRat (talk) 17:49, 28 August 2011 (UTC)[reply]

See the disambiguation page cone. Those aren't too different and are only a small subset of its meanings. Dmcq (talk) 18:13, 28 August 2011 (UTC)[reply]

By the way, the word is frustum—there's only one R. —Bkell (talk) 14:16, 29 August 2011 (UTC)[reply]

In general, you can have cones of any dimension. A very nice class of cones are given by the zero sets of non-degenerate homogeneous polynomials (of any degree, and in any number of variables). Although it can sometimes easier to think of them over C instead of R. For example, the set of points (x₁, x₂, x₃, x₄) in R⁴ such that x₁² + x₂² + x₃² – x₄² = 0 is a cone, whose x₄–constant hyperplane sections give spheres. (Note that x₄ = 0 gives a sphere or radius zero.) You can, of course, construct much more exotic examples. — Fly by Night (talk) 01:59, 30 August 2011 (UTC)[reply]

Everyone's going a bit off track on this one. If I say "cone" to an average person (not a mathematician), there's no question about it being a 3D cone (versus other dimensions or any of the other things listed on the disamb page). There is, however, the question of it being infinite or finite. That's what I need terms for. Is there no shorter term for a "3D, finite, right, circular cone" ? StuRat (talk) 23:48, 30 August 2011 (UTC)[reply]

Are you saying that the word "cone", in ordinary non-mathematical usage, is confusing because it may refer to an infinite cone? I cannot imagine a non-mathematician hearing the word "cone" in a non-mathematical context and getting the impression of an infinite cone. Infinite objects do not arise in non-mathematical contexts. In ordinary usage, the word "cone" means "a solid or hollow object that tapers from a circular or roughly circular base to a point" (New Oxford American Dictionary). That description cannot possibly refer to an infinite cone, which has no base. Now, I don't know of a single word for the very precise concept of a "three-dimensional finite right circular cone", but you could probably remove the word "finite" from that description, since the concept of a "right circular cone" really makes sense only in the finite case. —Bkell (talk) 17:52, 31 August 2011 (UTC)[reply]

Romanian math

I noticed that in several international math journals, a disproportionally large number of problems were submitted by Romanians. Why is this? Are Romanians naturally better at math or something? --76.211.88.37 (talk) 19:41, 28 August 2011 (UTC)[reply]

No they're terrible at maths and can't solve any problems so they're trying to get other people to help them ;-) They've done a lot of fine maths over the years. Also mathematics has a good public reputation there, the original maths olympiad was held by Romania for instance. Dmcq (talk) 21:20, 28 August 2011 (UTC)[reply]

You find that a lot of the old Eastern Block countries are the same. They weren't touched by commercialism like the West was. Things like a good education still come before being popular and famous. In the West, being clever isn't cool. — Fly by Night (talk) 22:55, 28 August 2011 (UTC)[reply]

A question about $\zeta (0)$

I know that when evaluated this produces $-{\frac {1}{2}}$

I'm not quite sure why this is the answer, but I was wondering if there was any significance of this being -.5 and whether or not there are any other values z such that $\zeta (z)=-{\frac {1}{2}}$ And if there were, what the significance of THAT would be. As you can probably tell, I'm new to this function :-) Aacehm (talk) 21:14, 28 August 2011 (UTC)[reply]

I have literally no idea how the function works, but I do know a website that does. WolframAlpha thinks there are many more solutions z<-15. They aren't integers or anything like that, I guess if they are something special it would take more than me to recognise it. Grandiose (me, talk, contribs) 14:57, 29 August 2011 (UTC)[reply]

The article 1 + 1 + 1 + 1 + … is what you're looking for. — Fly by Night (talk) 20:13, 29 August 2011 (UTC)[reply]