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December 31

Sinusoidal phase angle

In the article "Phase (waves)", it states that a phase is "the initial angle of a sinusoidal function at its origin". Does this mean gradient at x=0 or something different? — Preceding unsigned comment added by Demiseofanalog (talk • contribs) 22:24, 31 December 2014 (UTC)

I added a title. StuRat (talk) 23:05, 31 December 2014 (UTC)

Something different. In $f(x)=\sin(\omega x+\varphi )$, the phase is $\varphi$ - that is, when $x=0$, the sine function is taken on an angle of $\varphi$. -- Meni Rosenfeld (talk) 00:39, 1 January 2015 (UTC)

"angle" in this context means the argument of the trigonometric function, not the slope of the graph. -- ToE 14:24, 1 January 2015 (UTC)

January 2

Dimension of C(R) as a vector space over R

What is the dimension of the space of all continuous functions from R (the real numbers) to R, as a vector space over R? Is it a named cardinal invariant, or is it just $2^{\aleph _{0}}$, for some probably embarrassingly trivial reason I haven't noticed? (I'm pretty sure I have a proof that it's at least ${\mathfrak {d}}$, so in particular it's uncountable). --Trovatore (talk) 07:36, 2 January 2015 (UTC)

It is the same as the dimension of the space of all continuous functions from ℚ (the rational numbers) to ℝ, as a vector space over ℝ. Because ℚ is dense in ℝ. Bo Jacoby (talk) 11:56, 2 January 2015 (UTC).

The answer thus is "interesting" only of CH fails (while a proof may be interesting in either case)? YohanN7 (talk) 13:42, 2 January 2015 (UTC)

It's the continuum. Consider $f_{r}(x)=\max\{0,1-|x-r|\}$, a "spike" with support $[r-1,r+1]$. Then $\{f_{r}:r\in \mathbb {R} \}$ is an independent set. For suppose you had some finite dependence relation, and consider the smallest $r$ with $f_{r}$ having non-zero coefficient in the dependence relation. Then there is some $x$ in the interior of the support of $f_{r}$ but not in the support of any other function in the dependence relation. This $x$ witnesses that the linear combination is not the 0 function, so it's not really a dependence relation.--173.49.37.187 (talk) 15:54, 2 January 2015 (UTC)

Yes, that does it. Thanks. --Trovatore (talk) 16:19, 2 January 2015 (UTC)

Group theory question

If G is a finite group, and is not Cp^n, n>1, p prime, and has a normal proper nontrivial subgroup N, then is it true that G can always be written as NH, where H is a subgroup of G and N and H have trivial intersection? thanks.2601:7:6580:5E3:69F0:C6A7:1CEB:D852 (talk) 17:36, 2 January 2015 (UTC)

Basically you're looking for a normal subgroup which is not a factor in a semidirect product. I think technically G=C₁₂=<a> meets the criteria, since it has a subgroup H=<a²> of index 2 and only one subgroup <a⁶> of order 2 and this is contained in H. Not really in the spirit of the question though so perhaps the generalized quaternion groups Q_4n are a better example since they have a normal subgroup of order 2n but this contains the only element of order 2. In general, except for groups of small order, the semidirect products are a special case and more general types of group extensions are are more usual. --RDBury (talk) 18:33, 2 January 2015 (UTC)

Intersection of triangles in higher dimensions

The intersection of line segments in 1-D always includes some extreme points. This is not true in the plain. I would like to know if it is possible in d-dimensional space (d>3) to intersect 2 triangles in a way that only interior points are in the intersection. By interior points I mean points of the topological interior by the subspace topology induced via the plane of the respective triangles. Thus, no vertices nor edges would be allowed in the intersection. Is this possible, and if so, starting at which dimension? Is there a generalization to higher dimensional simplexes (tetrahedrons etc.)?

95.115.186.77 (talk) 20:03, 2 January 2015 (UTC)

No. Two intersecting planes always describe a line (unless the planes are identical). Doesn't matter how many dimensions you use. If the intersection is restricted to the portion of a plane described by a triangle, then the resulting line segment must terminate on an edge / vertex of a triangle. In general, two triangle can either A) not intersect, B) intersect at a single point at the boundary of both triangles (e.g. they share a vertex), or C) intersect with a line segment that runs from one boundary point to another boundary point. Dragons flight (talk) 20:21, 2 January 2015 (UTC)

Sorry if my question was not crystal clear (I kind of suspected incoming misunderstandings). Triangles in higher dimensions (d>3) don't fix a plane, but a hyper-plane, of dimension d minus one. 95.115.186.77 (talk) 20:32, 2 January 2015 (UTC)

Well, maybe you got a point that I can't see in my mind yet. Let's see who's faster, me or you, to pinpoint it out ;-). 95.115.186.77 (talk) 20:41, 2 January 2015 (UTC)