6
You can divide both sides by T. This reduces the number of variables to 2:
$$ j=\frac{J}{T},\quad b=\frac{B}{T}.$$
To visualize inequality, one can use ContourPlot. The inequality is fulfilled in the green area.
ContourPlot[Sqrt[b^2 + j^2] Cosh[2 j] - j Sinh[2 Sqrt[b^2 + j^2]],{j,0,1},{b,0,1},
PlotRange->{-3,1},
Contours->20,
ColorFunctionScaling->...
4
LeastSquare seems to use MKL lib internally and is already parallelized. Run the following code and observe cpu load:
SeedRandom[1] ;
size = 3000 ;
m = RandomReal[{-1,1},{size,size}] ;
b = RandomReal[{-1,1},size] ;
LeastSquares[m,b] ; // AbsoluteTiming
Check your parallel options and compare with observed cpu load:
SystemOptions["ParallelOptions"]
...
4
Mathematica is correct. In general, there will only be the trivial solution.
You are trying so solve an equation $ M x = b $ with $ b= 0$. This will have a nontrivial solution if and only if $\mathrm{det} M = 0$, because otherwise the matrix can be inverted, i.e. there exists a matrix $M^{-1}$ such that $M M^{-1} = M^{-1} M = I$, where $I$ is the identity ...
2
d = 3;
Clear[A]; Clear[theta]
Array[A, {d, d}]; Array[theta, {d, d}];
For[m = 0, m < d, m++,
For[n = 0, n < d, n++,
A[m, n] = Exp[Sum[theta[i, j], {i, 0, m}, {j, 0, n}]]
]
];
(mat = Table[A[i, j], {i, 0, d - 1}, {j, 0, d - 1}]) // MatrixForm
Dimensions[mat]
(* {3, 3} *)
Det[mat] // Simplify
2
The equations are consistent, as can be shown as follows. First, note that the first six expressions, {g1, g2, g3, g4, g5, g6}, are linear in {px, py, pz, ptx, pty, ptz}. The coefficient matrix of the first six expressions with respect to these variables is
m = Last@CoefficientArrays[{g1, g2, g3, g4, g5, g6}, {px, py, pz, ptx, pty, ptz}] // Normal;
The ...
2
Combining equations we obtain a conventional linear equation in fully unique way without any constraints on the matrices:
$$\left\{
\begin{array}{l}XA^\dagger C^\dagger=ABJD^\dagger-BJB^\dagger C^\dagger,\\
X C^\dagger + B J D^\dagger = 0;\end{array}\right.$$
or transposing
$$\left\{
\begin{array}{l}CAX^\dagger=D J B^\dagger A^\dagger-C B JB^\dagger,\\
C X^\...
1
Instead of Labeled, use Text
Clear["Global`*"];
eqns = {(1 - x)/2, -2 - 3 x};
pt = {x, eqns[[1]]} /.
Solve[Equal @@ eqns, x, Reals][[1]]
(* {-1, 1} *)
plot = Plot[eqns, {x, -3, 1}, PlotRange -> {-1, 3}, PlotStyle -> Thick,
PlotLabels -> {y == (1 - x)/2, y == -2 - 3 x}, AxesLabel -> {x, y},
ImageSize -> {Automatic, 200}];...
1
With Method->Reduce you get the result as ConditionalExpressions.
Solve[Simplify[
a.X + X.a\[ConjugateTranspose] + b.ji.b\[ConjugateTranspose]] ==
ConstantArray[0, {n, n}] &&
Simplify[X.c\[ConjugateTranspose] + b.ji.d\[ConjugateTranspose]] ==
ConstantArray[0, {n, no}], Flatten[X], Method -> Reduce]
Alternatively, you can ...
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