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

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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); x_c = cos(C); y_c = 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(y_cy₂y₃ + y_cx₂x₃ + x_cy₂x₃ - x_cx₂y₃)

y₁ = Dy₂

supplemented by the trivial x₁²+y₁² = x₂²+y₂² = x₃²+y₃² = x_c²+y_c² = 1 giving 6 algebraic equations involving the 6 unknowns x₁, y₁, x₂, y₂, x₃, y₃ and the known constants A, B, x_c, y_c, D.

The equations are f_i(x₁,y₁,x₂,y₂,x₃,y₃) = 0 for i=1,..,6, where

f₁ = - x₁y₃A + y₁(x₃A - 1)

f₂ = x₂(x₃y_c - y₃x_c)B + y₂(x₃Bx_c + y₃By_c - 1)

f₃ = - y₁ + y₂D

f₄ = x₁² + y₁² - 1

f₅ = x₂² + y₂² - 1

f₆ = x₃² + y₃² - 1

f₇ = x_c² + y_c² - 1

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

To eliminate x₁ compute

f_1a = (x₁y₃A + y₁(x₃A-1))f₁ = - x₁²y₃²A² + y₁²(x₃A-1)² = - x₁²y₃²A² + y₁²(x₃²A²-x₃A2+1 )

f_4a = y₃²A²f₄ = y₃²A²(x₁² + y₁² - 1) = x₁²y₃²A² + y₁²y₃²A² - y₃²A²

f₈ = f_1a + f_4a = y₁²(x₃²A²-x₃A2 +1) + y₁²y₃²A² - y₃²A² = y₁²((x₃²+y₃²)A²-x₃A2+1) - y₃²A²

f_6a = y₁²A²f₆ = y₁²(x₃² + y₃² - 1)A²

f₉ = f₈ - f_6a = 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:

f_3a = (y₁+y₂D)f₃ = (y₁+y₂D)(-y₁+y₂D) = - y₁² + y₂²D²

f₁₀ = f₉ + (-x₃A2+A²+1)f_3a = 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:

f_2a = (-x₂(x₃y_c-y₃x_c)B + y₂(x₃Bx_c+y₃By_c-1))f₂ = - x₂²(x₃y_c-y₃x_c)²B² + y₂²(x₃Bx_c+y₃By_c-1)²

f_5a = (x₃y_c-y₃x_c)²B²f₅ = (x₃y_c-y₃x_c)²B²(x₂²+y₂²-1) = x₂²(x₃y_c-y₃x_c)²B² + y₂²(x₃y_c-y₃x_c)²B² - (x₃y_c-y₃x_c)²B²

f₁₁ = f_2a + f_5a = y₂²((x₃Bx_c+y₃By_c-1)² + (x₃y_c-y₃x_c)²B²) - (x₃y_c-y₃x_c)²B² = y₂²a₁₁ - (x₃y_c-y₃x_c)²B²

where, using f₆=f₇=0,

a₁₁ = (x₃Bx_c+y₃By_c-1)² + (x₃y_c-y₃x_c)²B² = - x₃Bx_c2 + y₃By_c2 + B²+1

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

Eliminate y₂:

f_10a = a₁₁f₁₀ = a₁₁(y₂²a₁₀ - y₃²A²) = y₂²a₁₀a₁₁ - y₃²a₁₁A²

f_11a = a₁₀f₁₁ = a₁₀(y₂²a₁₁ - (x₃y_c-y₃x_c)²B²) = y₂²a₁₀a₁₁ - a₁₀(x₃y_c-y₃x_c)²B²

f₁₂ = - f_10a + f_11a = y₃²a₁₁A² - a₁₀(x₃y_c-y₃x_c)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≤y₃≤1, then sin(θ₃)=y₃ define the angle θ₃. 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". [1] 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 $\textstyle n={2r-1 \choose 2}$ vertices, and its chromatic number is $2r-3$ (see Kneser graph#Properties). The corresponding Turán graph $T_{r-1}(n)$ has chromatic number $r-1$. Therefore $KG_{2r-1,2}$ is not a Turán graph (for $r\geq 3$), 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)