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10

One quick way for rational functions is to leverage built-in control system functions: TransferFunctionPoles[TransferFunctionModel[1/(1 + s^2), s]][[1, 1]] {-I, I}


8

Yo can go by solving the inverse of the function and then Count the solution. f[x_] := x/(a - x)^3/(b - x); div = x /. Solve[1/f[x] == 0, x]; poles = Union[div] Count[div, #] & /@ poles {a,b} {3,1} Depending on how complicated your function is, you may have to go for numerical treatment, like NSolve or NRoots. Tally As Guess who it is ...


7

The equation has infinite number of nontrivial roots. You can query a subset of them, though: Solve[1 + x^2 Tan[x] == 0 && -10 < x < 10, x, Reals] (* {{x -> Root[{1 + #1^2 Tan[#1] &, -9.4360086156910331017}]}, {x -> Root[{1 + #1^2 Tan[#1] &, -6.3083089552381513776}]}, {x -> Root[{1 + #1^2 Tan[#1] &, ...


6

Here is a partial answer. I believe for Method -> {opts}, opts may be any of the following, and they will have whatever effect they have: Internal`InequalitySolverOptions[] Internal`ReduceOptions[] Internal`NSolveOptions[] (* {"ARSDecision" -> False, "BrownProjection" -> True, "CAD" -> True, "CADAlgebraicCoefficients" -> True, ...


6

This is best I can do so far. The system is linear so LinearSolve is a natural thing to try. arrays = CoefficientArrays[eqs, vars] (* {SparseArray[< 2 >, {9}], SparseArray[< 35 >, {9, 9}]} *) solv1 = Thread[ vars -> LinearSolve[arrays[[2]], -arrays[[1]], Method -> "CofactorExpansion"]]; // AbsoluteTiming (* {0.28347, Null} ...


4

I know nothing about the range of the data for your problem. Assuming that the values for Rho, Mu and z are real numbers you can gain insight into your problem by combining bbgodfrey's comment with a plot using Manipulate. For example, if Rho and Mu are known parameters you can see how the solution varies as you change the value of z. Manipulate[ ...


4

With Mathematica version 10 you could do it as follows: y[x_] := 1/5 E^(2/5 x) sol = Maximize[ArcCurvature[{x, y[x]}, x], x] (* {8/375 Sqrt[2/3] E^(1/2 (-3 Log[2] + 4 Log[5])), {x -> -(5/4) (3 Log[2] - 4 Log[5])}} *) Plot[y[x], {x, -5, 10}, Epilog -> {Red, PointSize[.02], Point[{sol[[2, 1, 2]], y[sol[[2, 1, 2]]]}]}, AspectRatio -> Automatic] ...


3

How are you calculating the velocity around the cylinder I am using Plot[Norm[f[20 + 5 Cos[\[Theta]], 20 + 5 Sin[ \[Theta]]]], {\[Theta], 0, 2 \[Pi]}] to give Which I think is correct. For the pressure we need the correct form for Bernoulli. Where you take the values of pressure at infinity as 0 but ignore the velocity at infinity. I am also ...


3

b m - a m - p n (p (b - a) + p b - p b m - p a + p a m) == 5 b FullSimplify@Solve[% /. a -> b - x, x][[1]] /. Rule -> Equal /. x -> b - a {-a + b == (5 b)/(m + (-2 + m) n p^2)}


3

This appears to give you one approximate solution in a few seconds sol=NMinimize[ Norm[Abs[eta1]^2 + Abs[gamma1]^2 - 1] + Norm[Abs[eta2]^2 + Abs[gamma2]^2 - 1] + Norm[Abs[eta3]^2 + Abs[gamma3]^2 - 1] + Norm[Abs[tau1]^2 + Abs[kappa1]^2 - 1] + Norm[Abs[tau2]^2 + Abs[kappa2]^2 - 1] + Norm[Abs[tau3]^2 + Abs[kappa3]^2 -1 ] + Norm[Dot[q1, Conjugate[q1]] - ...


2

Using @Guess comment answer, but spelled out in more detail: a = {{1, -3, 2, -2}, {3, -2, 0, -1}, {2, 36, -28, 27}, {1, -3, 22, 5}}; L = {{1, 0, 0, 0}, {b, 1, 0, 0}, {c, d, 1, 0}, {e, f, g, 1}}; U = {{1, -3, 2, -2}, {0, h, i, j}, {0, 0, k, l}, {0, 0, 0, m}}; result = {L, U} /. First @ Solve[L.U == a]; MatrixForm /@ result Note that matrix multiplication ...


1

There is a LUDecomposition function in Mathematica: a = {{1, -3, 2, -2}, {3, -2, 0, -1}, {2, 36, -28, 27}, {1, -3, 22, 5}}; n = Dimensions[a][[1]]; {LU, MP, MC} = LUDecomposition[a] LU = First[LUDecomposition[a]] L = LU * SparseArray[{i_, j_}/; j<i -> 1, {n, n}] + IdentityMatrix[n] U = LU * SparseArray[{i_, j_}/; j>=i -> 1, {n, n}] L.U


1

To put Daniel's comment into code: x = Array[C, Length@First@m]; constr = First@Solve[m.x == b] (* {C[6] -> -(2/7) - C[1] - C[2] - C[3] - C[4] - C[5], <<26>>, C[56] -> -(11/21)-C[1]-C[2]-C[3]-C[7]-C[8]-C[12]-C[22]-C[23]-C[27]-C[37]} *) Reduce[Thread[x >= 0] /. constr, Variables[x /. constr]] (* False *) Easy, fast, and ...


1

It seems FindInstance uses different methods to find 1 instance and to find more than one instance. constraint = {a > 0, b >= 0, c >= a + b, a >= c/4}; Trace[ FindInstance[constraint, {a, b, c}, Integers, 1], _Reduce`LinearDiophantineInstance | _Reduce`ReduceInstance, TraceInternal -> True ] Trace[ FindInstance[constraint, {a, b, ...


1

The title of your question states that you will be satisfied with an instance of {p, n} that satisfies Sum[p^k/k!, {k, 0, n}] > 0. Here is a function to find any number of such instances: instances[m_Integer?Positive] := Block[{p, n, k}, FindInstance[Sum[p^k/k!, {k, 0, n}] > 0 && p ∈ Reals && n ∈ Integers, {n, p}, m]] And ...


1

I would use {x1, x2} = x /. solutions The key is ReplaceAll.



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