# Tag Info

## Hot answers tagged equation-solving

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: InternalInequalitySolverOptions[] InternalReduceOptions[] Internal`NSolveOptions[] (* {"ARSDecision" -> False, "BrownProjection" -> True, "CAD" -> True, "CADAlgebraicCoefficients" -> True, ...

5

The equation can be solved exactly by rationalizing the coefficients first r = Rationalize; n0 = r@9.54472*10^(-5); n1 = r@3.72396*10^(-4); n2 = r@7.431*10^(-4); n3 = r@1.4862*1^(-3); k0 = Exp[(Log[n1]*Log[P0] - Log[n0]*Log[P1])/(Log[n1/n0])]; gam0 = (Log[P0/P1])/(Log[n0/n1]); s0 = First@Solve[k0*gam0*n1^(gam0 - 1) - 1 == 0, P1, Reals]; s[p_] := P1 /. s0 ...

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] ...

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[ ...

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

The issue is that you're asking for a general solution for y only whereas your particular set of equations has a solution only for one specific value of x. You should have called Solve as: Solve[f[x, y] == g[x, y], {x, y}] During evaluation of Solve::ratnz: Solve was unable to solve the system with inexact coefficients. The answer was obtained by ...

2

FindInstance, as described by Mr.Wizard, gives solutions for a given perimeter, which may then be tested for uniqueness. However, Project Euler challenges you to consider and learn about other methods which solve the problem, usually faster and without a mysterious "black box". For example, this problem concerns right-angled triangles with integer sides, ...

2

It's works with the assumed n! Lets n = 5. n = 5; y = Sum[c[i]*x^i, {i, 0, n}] + O[x]^(n + 1); ODE = (D[y, {x, 2}])^2 + 1/2*D[y, x] + 1/2*y - p*x - q == 0; Y = FullSimplify@ Normal[y /. Solve[LogicalExpand[ODE], Table[c[i], {i, 1, n}]]] // Quiet Solution: {\frac{-10 c(2) x^5 (c(0)+2 p-2 q)^2+3 x^5 (c(0)+2 p-2 q)^3-32 c(2)^3 x^4 (3 x-5) (c(0)+2 ...

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

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

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

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

1

Though there may be an analytic approach to your problem I think for the generic case you can still make use of FindInstance; request two solutions and see how many it returns (zero, one, two). FindInstance[ {a^2 + b^2 == c^2, a + b + c == 1200, 0 < b <= a, c > 0}, {a, b, c}, Integers, 2 ] {{a -> 450, b -> 240, c -> 510}, {a -> ...

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