Wikipedia:Reference desk/Archives/Mathematics/2011 May 31

= May 31 =

Three equations, three unknown angles
I'm a physicist working with an acoustics problem and have the following three equations which must be analytically solved for three unknown angles θ1, θ2 and θ3. All other letters represent known quantities. In particular, I must determine θ3 as a function of the input constants A, B, C and D.

sin(θ1)=A*sin(θ1-θ3), sin(θ2)=B*sin(C+θ2-θ3), sin(θ1)=D*sin(θ2)

I greatly appreciate any insight.AO93 (talk) 00:09, 31 May 2011 (UTC)


 * Use the formula $$\sin (a + b) = \sin (a) \cos(b) + \cos (a) \sin (b)$$. Looie496 (talk) 00:36, 31 May 2011 (UTC)


 * Unfortunately, that alone is insufficient to isolate the variables (at least as far as I can tell).AO93 (talk) 00:50, 31 May 2011 (UTC)

You can formally treat the equations as linear equations for sin(θ1), sin(θ2), and sin(θ1 - θ3), by treating sin(C + θ2 - θ3) as a formal parameter. If you then solve these equations, you get sin(θ1), sin(θ2), and sin(θ1 - θ3), in terms of sin(C + θ2 - θ3). Then, because

sin(C + θ2 - θ3) = sin(C + θ2 - θ1 + θ1- θ3)

you can expand this in terms of the three variables you solved for. So, you get a single equation for sin(C + θ2 - θ3), which then immediately yields sin(θ2) and then you can extract θ3 from that. Count Iblis (talk) 01:15, 31 May 2011 (UTC)
 * Count Iblis, Forgive me for editing your contribution. I replaced 'theta' with 'θ' in order to enhance legibility. Bo Jacoby (talk) 16:21, 31 May 2011 (UTC).
 * That makes sense, and I think gets me a step closer, but I'm not aware of a way of expanding the last expression purely in terms of sine functions. If the expansion in terms of the three variables will incorporate cosine functions (which, when converted to to sine functions, introduce radicals throughout the expression), then, unless I am again missing something, the resulting ugliness looks awfully difficult to solve.AO93 (talk) 16:05, 31 May 2011 (UTC)

The first step in solving the equations is to isolate the trigonometrics. Let
 * x1 = cos(θ1); y1 = sin(θ1); x2 = cos(θ2); y2 = sin(θ2); x3 = cos(θ3); y3 = sin(θ3); xc = cos(C); yc = sin(C);

Then the three equations are (using the formula sin(a+b-c) = sin(a)sin(b)sin(c) + sin(a)cos(b)cos(c) + cos(a)sin(b)cos(c) - cos(a)cos(b)sin(c))
 * y1 = A(y1x3 - x1y3)
 * y2 = B(ycy2y3 + ycx2x3 + xcy2x3 - xcx2y3)
 * y1 = Dy2

supplemented by the trivial x12+y12 = x22+y22 = x32+y32 = xc2+yc2 = 1 giving 6 algebraic equations involving the 6 unknowns x1, y1, x2, y2, x3, y3 and the known constants A, B, xc, yc, D.

The equations are fi(x1,y1,x2,y2,x3,y3) = 0 for i=1,..,6, where
 * f1 = - x1y3A + y1(x3A - 1)
 * f2 = x2(x3yc - y3xc)B + y2(x3Bxc + y3Byc - 1)
 * f3 = - y1 + y2D
 * f4 = x12 + y12 - 1
 * f5 = x22 + y22 - 1
 * f6 = x32 + y32 - 1
 * f7 = xc2 + yc2 - 1

The uninteresting variables x1,y1,x2,y2,x3 are eliminated one by one.

To eliminate x1 compute
 * f1a = (x1y3A + y1(x3A-1))f1 = - x12y32A2 + y12(x3A-1)2 = - x12y32A2 + y12(x32A2-x3A2+1 )
 * f4a = y32A2f4 = y32A2(x12 + y12 - 1) = x12y32A2 + y12y32A2 - y32A2
 * f8 = f1a + f4a = y12(x32A2-x3A2 +1) + y12y32A2 - y32A2 =  y12((x32+y32)A2-x3A2+1) - y32A2
 * f6a = y12A2f6 = y12(x32 + y32 - 1)A2
 * f9 = f8 - f6a = y12(-x3A2+A2+1) - y32A2

The 5 equations f2=f3=f5=f6=f9=0 define the 5 variables y1,x2,y2,x3,y3.

Eliminate y1:
 * f3a = (y1+y2D)f3 = (y1+y2D)(-y1+y2D) = - y12 + y22D2
 * f10 = f9 + (-x3A2+A2+1)f3a = y22(-x3A2+A2+1)D2 - y32A2 = y22a10 - y32A2

where
 * a10 = (-x3A2+A2+1)D2 = - x3AD22 + (A2+1)D2

The 4 equations f2=f5=f6=f10=0 define the 4 variables x2,y2,x3,y3.

Eliminate x2:
 * f2a = (-x2(x3yc-y3xc)B + y2(x3Bxc+y3Byc-1))f2 = - x22(x3yc-y3xc)2B2 + y22(x3Bxc+y3Byc-1)2
 * f5a = (x3yc-y3xc)2B2f5 = (x3yc-y3xc)2B2(x22+y22-1) = x22(x3yc-y3xc)2B2 + y22(x3yc-y3xc)2B2 - (x3yc-y3xc)2B2
 * f11 = f2a + f5a = y22((x3Bxc+y3Byc-1)2 + (x3yc-y3xc)2B2) - (x3yc-y3xc)2B2 = y22a11 - (x3yc-y3xc)2B2

where, using f6=f7=0,


 * a11 = (x3Bxc+y3Byc-1)2 + (x3yc-y3xc)2B2 = - x3Bxc2 + y3Byc2 + B2+1

The 3 equations f11=f10=f6=0 define the 3 variables y2,x3,y3.

Eliminate y2:
 * f10a = a11f10 = a11(y22a10 - y32A2) = y22a10a11 - y32a11A2
 * f11a = a10f11 = a10(y22a11 - (x3yc-y3xc)2B2) = y22a10a11 - a10(x3yc-y3xc)2B2
 * f12 = - f10a + f11a = y32a11A2 - a10(x3yc-y3xc)2B2

The 2 equations f12=f6=0 define the 2 variables x3,y3.

Afeter eliminating x3 you get the analytical answer to your problem. The equation is solved by numerical approximations, see root-finding algorithm. If some of the six solutions y3 satisfy -1&le;y3&le;1, then sin(&theta;3)=y3 define the angle &theta;3. Bo Jacoby (talk) 00:43, 1 June 2011 (UTC). I have now simplified the above somewhat. It is a tedious calculation to do by hand. I should have used Maple or Mathematica. Bo Jacoby (talk) 10:36, 3 June 2011 (UTC). Bo Jacoby (talk) 16:17, 6 June 2011 (UTC).

Confirm whether I've proved this right: normed spaces and linear subspaces
Hello,

I want to know if my argument is fine for this problem: it seems to be to me, but I thought I remembered seeing a much lengthier proof of the same thing a while back and fear I may have omitted or ignored something important.

I wish to prove that if X is a normed space, then every proper linear subspace V of X has empty interior. I argue suppose not: then it must contain a ball B at some point x in V. Then since v is linear, the ball B - v must also be in V: but this is a ball around the origin, and by linearity we can scale this ball up to any size to show that V must contain everything of norm less than any given R, which shows V must be X and thus not proper.

Is my argument valid or have I neglected some important point? Many thanks 131.111.185.74 (talk) 16:55, 31 May 2011 (UTC)
 * Sounds good to me.--RDBury (talk) 02:03, 1 June 2011 (UTC)

what is small v here .what is B-v

Matrices
Hello all, me again. I'm compiling a summer reading list and I need a good text about matrices and matrix theory, again that begins with the very basics (definition of a matrix, definition of matrix addition/multiplication, etc) but is fairly rigorous and covers a good amount of ground. Thanks. 72.128.95.0 (talk) 17:33, 31 May 2011 (UTC)
 * Just about any book titled Linear algebra is going to do that. Is there a specific application area you're wanting to focus on?  i kan reed (talk) 21:28, 31 May 2011 (UTC)


 * A quick google lead to this free online textbook, which appears to have good reviews. 130.88.73.71 (talk) 09:11, 1 June 2011 (UTC)


 * I enjoyed Smale's "Differential Equations, Dynamical Systems and Linear Algebra". It has a much wider scope than just matrices, but the matrix algebra intro is very accessible. Overall I think it is a nice treatment, starting with basics and with decent rigor. It will give you a very different (IMO, better) feel of the subject than some random current college text. SemanticMantis (talk) 13:46, 1 June 2011 (UTC)

Particular graph example: Turan graph related
Hello: I am looking for a graph G on n vertices with fewer edges than the Turan graph $$T_{r-1}(n)$$ where r < n, but for which adding any edge to G produces a $$K_r$$, i.e. a maximal $$K_r$$-free graph, for each value of n and r with n > r. Could anyone help me? Nothing obvious comes to mind. Thanks Typeships17 (talk) 22:52, 31 May 2011 (UTC)


 * The Kneser graph $$KG_{2r-1,2}$$ contains no $$K_r$$ (because it is impossible to have r disjoint 2-subsets of a set containing $$2r-1$$ elements). Moreover, adding any edge produces a $$K_r$$. For, consider any two nonadjacent vertices in $$KG_{2r-1,2}$$; without loss of generality, these vertices are the subsets $$\{1,2\}$$ and $$\{2,3\}$$. Once these vertices are joined by an edge, the vertices $$\bigl\{\,\{1,2\},\,\{2,3\},\,\{4,5\},\,\{6,7\},\,\ldots,\,\{2r-2,2r-1\}\,\bigr\}$$ form a $$K_r$$. So $$KG_{2r-1,2}$$ is a maximal $$K_r$$-free graph.
 * The Kneser graph $$KG_{2r-1,2}$$ has $$\textstyle n={2r-1\choose2}$$ vertices, and its chromatic number is $$2r-3$$ (see Kneser graph). The corresponding Turán graph $$T_{r-1}(n)$$ has chromatic number $$r-1$$. Therefore $$KG_{2r-1,2}$$ is not a Turán graph (for $$r\ge3$$), so it has strictly fewer edges than $$T_{r-1}(n)$$.
 * Of course, this construction doesn't produce examples for all values of n, but at least it gives an infinite family of maximal $$K_r$$-free graphs that are not Turán graphs, one for each value of r. —Bkell (talk) 06:40, 1 June 2011 (UTC)


 * Thank you Bkell, that's very helpful: although this is actually revision for an exam, and the question I'm doing asks specifically for a graph for each value of n and r with n > r: as such I feel like there might perhaps be a more obvious construction available which will work for all n and r. If anyone has any thoughts on this please do respond. Typeships17 (talk) 09:15, 1 June 2011 (UTC)


 * Never mind, got one :) Typeships17 (talk) 11:26, 2 June 2011 (UTC)