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7

It seems me that the answers of mathe and Yves Klett do not meet expectations of the author. The latter is as much as I have got it, to have a short analytical expression for the solution. Probably the author has an intention to use the result further in some analytical calculations, or to do something comparable. Am I right? If yes, one should first of ...


6

Solve[L == (3 W)/2 + 3/2 Sqrt[4 A^2 Pi^2 + W^2] - Sqrt[ 6 A^2 Pi^2 + 3 W^2 + 5 W Sqrt[4 A^2 Pi^2 + W^2]]/Sqrt[2], W, Quartics -> False] or Solve[L == (3 W)/2 + 3/2 Sqrt[4 A^2 Pi^2 + W^2] - Sqrt[ 6 A^2 Pi^2 + 3 W^2 + 5 W Sqrt[4 A^2 Pi^2 + W^2]]/Sqrt[2], W, Reals]


5

Many times I've found Mathematica to be way better solving equations, when trigonometric functions are written with exponential functions. This time is no exception: Solve[FullSimplify[TrigToExp[-2 (m^2/36 + n^2/4)^(1/4) Sin[1/2 ArcTan[n/2,-(m/6)]]==y]],n] {{n -> (m^2 - 9 y^4)/(18 y^2)}}


5

As @b.gatessucks said in the comments, there are two issues with your code. First, you'll need to define z as a function with SetDelayed, and also add in ComplexExpand: z[x_, y_] := ComplexExpand @ Abs[((1/3 x + I y) - 2 x)/(I y + 1/x)]; z[x,y] Sqrt[(25 x^2)/9 + y^2] / Sqrt[x^-2 + y^2] Additionally, ContourPlot holds its arguments (i.e. it doesn't ...


5

What have you tried so far? You can use Solve to solve for θ. Solve[a1 Sin[2θ] + a2 Sin[2ϕ] + a3 Cos[2θ] + a4 Cos[2ϕ] == a5, θ] /. _C -> 0 Since your equation have periods π, you can just let ϕ run between 0 and π, and add arbitrary multiples of π to the solutions. Another way You can plot it using ContourPlot. I used bounds 0 < θ < π and 0 ...


4

eqns = {(300 (1920 + 8 x - 21 y) (-80 + y))/(7 (30 + x)^3) + (2025 (208 x + 5 (-3446 + y)))/(52 (90 + y)^2) + (300 (4500 + 80 x - 21 z) (-100 + z))/(7 (30 + x)^3) - (1521 (15425 - 539 x + 50 z))/(539 (39 + z)^2) == 0, -((300 (1920 + 8 x - 21 y))/(7 (30 + x)^2)) - (2025 (-85 + x) (208 x + 5 (-3446 + ...


4

Mathematica 10 provides new functionality dealing with curves, (see e.g. the Vector Analysis tutorial) like ArcLength, ArcCurvature and especially FrenetSerretSystem: FrenetSerretSystem[{ x1, ..., xn}, t] gives the generalized curvatures and Frenet-Serret basis for the parametric curve x[t] i.e. it returns {{ k1, ..., k(n-1)}, { e1, ..., en}}, where ki ...


4

eqn1[w_, x_, y_, z_] := 3 x y'[x] - 2 z^2 + 3 w y[x] == 5 x eqn2 = eqn1[1, x, y, 2] sol = NDSolve[{eqn2, y[1] == 2}, y, {x, 1, 2}] Then, to get x when y[x] is 3: Solve[(y[x] /. sol) == 3, x] {{x -> 1.556466}} OR as suggested by Mr.Wizard, you can use FindRoot FindRoot[(y[x] /. sol) == 3, {x, 1}] {x -> 1.556466}


4

You may use the following: Reduce[(1 - I)/Sqrt[2] == ExpToTrig[Exp[I alpha]]*Tan[beta] && 0 < beta < Pi && 0 < alpha < 2 Pi]; {ToRules[%]} // FullSimplify {{alpha -> (7 π)/4, beta -> π/4}, {alpha -> (3 π)/4, beta -> (3 π)/4}} Although I'm not sure why it doesn't work without the ExpToTrig[] thing


3

Please check the result by doing the following eqn = FullSimplify[ 2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q - 4*a^2*b*n*p*Q^2 - 4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 + 6*a*b^2*n*p^2*Q^2 - 2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 + a^2*b*c*Q^2 + a*b^2*c*Q^2 - 2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2 == 0] // TraditionalForm This returns ...


3

With the slightly modified input pde1 = 3*D[z[w, x, y], y] == 2 (z[w, x, y] - 1) + (1 - y^2 w) x soln1 = z[w, x, y] /. First@DSolve[pde1, z[w, x, y], {w, x, y}] soln5 = soln1 /. {x -> 0} eqn1 = soln5 == 2.7*soln1 and a replacement of the integration constant ContourPlot[ x /. Solve[eqn1 /. {C[1][w, 0] -> 1, C[1][w, x] -> 1}, x, Method -> ...


3

Reduce has its own syntax for specifying constrains and domains. Please try this: Reduce[2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q - 4*a^2*b*n*p*Q^2 - 4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 + 6*a*b^2*n*p^2*Q^2 - 2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 + a^2*b*c*Q^2 + a*b^2*c*Q^2 - 2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2 == 0 && 0 < a ...


3

Maybe something like this? I don't know if this is what you meant? sol=Reduce[x1 >= 0 &&-4*x1 <= 16 &&4*x1 >= 16 ||x1 <= 0 &&4*x1 <= 16 &&-4*x1 >= 16, {x1}]; sol[[1, 2]] (*-4*) sol[[2, 2]] (*4*)


3

The current answer by Chip Hurst deals well with the case of solving for the roots of a (possibly) complex-valued function of a real variable, which NSolve struggles with. However, the more general case of finding roots of a complex-valued equation of a complex variable is a good bit more complicated, and it requires a stronger approach. One thing to note ...


2

Setting expr equal to the OP's expression, the equations are given by Flatten@expr == 0. expr = {{(300 (1920+8 x-21 y) (-80+y))/(7 (30+x)^3)+(2025 (208 x+5 (-3446+y)))/(52 (90+y)^2)+(300 (4500+80 x-21 z) (-100+z))/(7 (30+x)^3)-(1521 (15425-539 x+50 z))/(539 (39+z)^2)},{-((300 (1920+8 x-21 y))/(7 (30+x)^2))-(2025 (-85+x) (208 x+5 (-3446+y)))/(52 ...


2

Version 9. Specifying a domain may speed things up or may greatly slow things down. Cases[N[Solve[{...}, {x,y,z}], 30], {x->_Real, y->_Real, z->_Real}] Result in 28 seconds. N[Solve[{...}, {x,y,z}, Reals], 30] Stopped it after 6 hours with no result.


2

We can adapt Sjoerd's solution to the question, Table - find index of the maximum element. Other methods may be found here: List manipulation: position & max value combination. tt1 = Flatten[ Table[Thread@{x, y, z /. Solve[z^2 == x^2 y - z, z, Method -> Reduce]}, {x, 0, 5, 1}, {y, 0, 5, 1}], 2]; Then this yields {x, y, max}: tt1 ~Part~ ...


2

You are getting no answers at all. Lets get you some answers quickly, by giving up going to Infinity first. In[1]:= j = 8; NMinimize[ Norm[3/2 Sum[k^(5/2) E^-((B+H/2)k) Integrate[x^2 E^(3/2 H x^2 k), {x, 0, 1}], {k, 1, j}]/ Sum[k^(5/2) E^-((B+H/2)k) Integrate[E^(3/2 H x^2 k), {x, 0, 1}], {k, 1, j}]-1/2-H]+ Norm[ Sum[k^(3/2) E^-((B + H/2) k) ...


2

I have to post a separate answer because my approach in this one is very different from my other answer's. It is based largely on the method that Frank Kampus used here to solve a sudoku, using Backtrack from the Combinatorica package in one method, and Outer/Select in the other method. Warning: it's a long answer. I managed to solve the puzzle but only ...


2

Below is my (unsuccessful) attempt at solving the problem. I think this is an interesting problem that the MMA experts on here could help solve, and in the process will hopefully reveal the capabilities of MMA. 1) I borrowed from the answer in the linked Q&A the fact that the total of all elements in the matrix will be 468. This means that the row ...


2

Funny problem. I like it. But there is must be something else in the condition of the problem? Or I misunderstand you. You say that I am guaranteed that w1, w2 exist Ok. Let's set v1 = {0, 0, 0, 1}; v2 = {1, 0, 0, 0}; it's independent vectors. Then find w1 and w2 in general case for this example: w1 = v1*a1 + v2*b1 w2 = v1*a2 + v2*b2 ...


2

a[0] = 1; a[1] = a[2] = a[3] = a[5] = 0; a[4] = g2/20; funcs = Table[a[n] == Sum[(m - 1) a[m] b[n - m], {m, 0, n}], {n, 0, 5}] var = b /@ Range[0, 5] Solve[funcs, var]


2

Something like this: ComplexExpand[ Solve[ { Re[(1 - Sqrt[x - I y])/(1 + Sqrt[x - I y])] == A, Im[(1 - Sqrt[x - I y])/(1 + Sqrt[x - I y])] == B }, {A, B} ], TargetFunctions -> {Re, Im} ] (* {{A -> 1/((1 + (x^2 + y^2)^(1/4)*Cos[(1/2)*ArcTan[x, -y]])^2 + Sqrt[x^2 + y^2]*Sin[(1/2)*ArcTan[x, -y]]^2) - ...


2

There is a third argument for Solve but you can use NSolve to get a numerical solution in the Reals: NSolve[x^3 - 1 == x, x, Reals] {{x -> 1.32471796}} OR Solve for exact Root solutions Solve[x^3 - 1 == x, x, Reals] {{x -> Root[-1 - #1 + #1^3 &, 1]}} OR as Michael E2 noted in the comments, using ToRadicals will give you something ...


2

Wolfram support has confirmed to me that there is a known problem with NSolve in version 10. Hopefully it will be fixed soon.


2

added some numeric interval limits for y Clear@"`*" eqn2[w_, x_, y_, z_] := 3*z'[y] == 2 (z[y] - 1) + (1 - y^2 w) x eqn3 = eqn2[1, 1, y, z] sol = NDSolve[{eqn3, z[0] == 1}, z[y], {y, 1, 5} (*added interval*)] Plot[Evaluate[z[y] /. sol], {y, 1, 5}, PlotRange -> All] z3 = z[y] /. sol /. y -> 3


2

There is a problem with boundary conditions. Changing them and fixing a few typos you can get: eq = {-f[x]^2 + f[x]^4 + x^4 Derivative[1][f][x]^2 + (x^3) /(-1 + x) f[x] ((1 - 2 x) f'[x] - (-1 + x) x f''[x]) == 0, f[1/10] == 1, f'[1/10] == -1/10}; eq // Column // TraditionalForm s = NDSolve[eq, f, {x, 1/10, 9/10}]; Plot[Evaluate[f[x] /. ...


2

I don't want to compete with Algohi's nice answer, but - as to my experience - Reduce can be almost always replaced with Simplify or FullSimplify: res = Simplify[x1 >= 0 && -4*x1 <= 16 && 4*x1 >= 16 || x1 <= 0 && 4*x1 <= 16 && -4*x1 >= 16] Cases[res, _?NumberQ, -1] {-4, 4}


2

The reasons for the change in the behavior of ParallelTable are subtle. The main source of the problem is that in funcB, the argument k_ is not protected with ?NumericQ like this: funcB[t_?NumericQ, k_?NumericQ] := (* a solution *) funcB[t, k] = Exp[NIntegrate[funcA[et, k], {et, tini, t}]] But more on that later. The problem does not appear in the ...


2

One shotgun approach is to sic Simplify or FullSimplify onto your solution: sol1 = Solve[ L == (3 W)/2 + (3 Sqrt[4 A^2 Pi^2 + W^2])/2 - Sqrt[6 A^2 Pi^2 + 3 W^2 + 5 W Sqrt[4 A^2 Pi^2 + W^2]]/Sqrt[2], W]; sol2 = Simplify[sol1]; LeafCount /@ {sol1, sol2} ByteCount /@ {sol1, sol2} {3849, 3077} {111720, 92840} (Note: FullSimplify is still ...



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