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14

The mesh seems to be fine and you can see that it is by doing: region = ImplicitRegion[! (Norm[{x, y, z}] < 1), {{x, -5, 5}, {y, -5, 5}, {z, 0, 5}}]; m = DiscretizeRegion[region, {{-2, 2}, {-2, 2}, {0, 1}}] To view as wireframe you can do: Needs["NDSolve`FEM`"] mesh = ToElementMesh[m] // Quiet; Then: Show[mesh["Wireframe"]] If you want to ...


10

Note that sin is not a function in Mathematica, instead use Sin with a capital S and functions use square brackets to hold their arguments. So Sin[x] instead of sin(x) or even sinx as you have it. You can use NestWhile or NestWhileList (for the list of values) also for this problem: f[x_] := x - 0.8 - 0.2 Sin[x] newtonsMethod[foo_, k_, s_: 0.0001] := ...


10

Here are a few additions to @RunnyKine suggestions. If you are ever in doubt about the quality of a mesh (an ElementMesh to be exact) you can query the mesh. Needs["NDSolve`FEM`"] region = ImplicitRegion[! (Norm[{x, y, z}] < 1), {{x, -5, 5}, {y, -5, 5}, {z, 0, 5}}]; mesh = ToElementMesh[region]; Min[mesh["Quality"]] 0.004439742441262357` So the ...


9

The interpolation step seems to be unnecessary because the integral into which it enters can be equally well approximated as a Riemann sum. So to get really fast results you could do the following: {r, h} = Transpose[hrdata]; d = Differences[r]; Clear[s]; s[q_] := (4 Pi 0.83 )/q Total[d Rest[r Sin[q r] (h - 1)]] ans2 = Table[{i, s[i]}, {i, 0.05, 11.4, ...


8

Mathematica should render the secondary structure in the usual way when you import pdb files. I dont know why this doesn't work with the example you provided. I thought there might be a size limit for the protein but I managed to import much bigger proteins such as 1YHU and they got rendered without problems... strange. ...


7

Here is the same approach as Jens, but using Association and assuming uniform grid data: st[hr_?MatrixQ, rho_, min_, max_, step_] := Module[{h = <|Rule @@@ hr|>, s, freq, dr = hr[[2, 1]] - hr[[1, 1]]}, s = Function[q, 4 Pi rho /q Tr[dr # Sin[q #] (h[#] - 1) & /@ Keys[h]]]; freq = Range[min, max, step]; Transpose[{freq, s /@ freq}] ] Now: ...


6

There is a difficulty with the statement of the problem. Generally the problem can be solved as shown below. In this case there is a stipulation that $1 < x < 5$ and $1 < y < 5$. Unfortunately the solution to the system does not satisfy these constraints (also shown below). If we agree to use only numerical techniques and pretend that Solve ...


6

Here is another way to look at the OP's plot, which came to me after reading Daniel Lichtblau's comment under the question. His comment is worth emphasizing, which I think the following will show. (Mr.Wizard alludes to the issue in his remark on validity near the end of his answer.) Let's say that machine precision is binary64, which has 53 bits of ...


6

The reason is that you're using machine-precision input, so the result you get will always be machine precision. N[LegendreP[5, 0.1], 20] 0.17882875 Precision /@ {0.1, 0.17882875`} {MachinePrecision, MachinePrecision} To get result in arbitrary precision you can use exact input or non-machine precision input (which is what happened when you ...


5

The sorting (ordering) done by Union is different for different forms of expressions, e.g., analytic versus numeric expressions for a number. Union[{2., (Sqrt[5] + 1)/2}] {2., 1/2 (1 + Sqrt[5])} % // N {2., 1.61803} Union[{2., (Sqrt[5] + 1.0)/2}] {1.61803, 2.} SortBy[{2., (Sqrt[5] + 1)/2}, N] {1/2 (1 + Sqrt[5]), 2.}


4

eldo already posed and deleted something like this, but I think it works well in many cases: poly = Expand @ FromDigits[RandomComplex[2 + 2 I, 6], x]; poly /. n_?NumberQ :> Round[n] (1 + I) + (1 + I) x + I x^2 + (1 + I) x^3 + x^4 + (1 + I) x^5 Note that this will round exponents as well. More robust is the method of rhermans, which might also be ...


4

You don't give an example so I create my own: poly = Expand@FromDigits[RandomComplex[2 + 2 I, 6], x] (0.117797 + 0.674094 I) + (0.980296 + 1.90575 I) x + (0.190167 + 1.68039 I) x^2 + (1.65725 + 1.83193 I) x^3 + (1.07084 + 1.19757 I) x^4 + (0.473445 + 1.37764 I) x^5 Probably you can extract your coefficients first, and then ...


3

Amplifying on Chenminqi's answer g = 1/10; func = (p^2*Sqrt[1 - (2*p)/(-g + p*(1 + p))])/(-1 + (p/g)*(p - 1)) - p^2/((1 + p/g)*(1 + p))*Sqrt[1 - (2*p)/(g + p*(1 + p))] // Simplify; fd = FunctionDomain[func, p] This is equivalent to requiring that the arguments of Sqrt be positive fd == Reduce[ Thread[ Cases[func, Sqrt[x_] -> x, Infinity] ...


2

The finite element method can be used on this problem if we make a change of variables to convert the domain $[0, \infty)$ to a finite interval. I believe only MachinePrecision is available in FEM. Since AiryAi vanishes so rapidly, it will make a precise result for a large argument difficult to obtain. Another difficulty in obtaining a precise solution is ...


2

Second answer -- OK, the first answer was hogwash (the curious can inspect the edit history). It pays sometimes to write out the equation and think about it first. The code for this one looks so complicated, but the equations basically have the form (here d = 80000) rc'[t] == 2.05594*10^-10/rc[t] - 8189.14 rc[t] + 80000. rm[t], rm'[t] == -80000. rc[t] + ...


2

To get the list that you want f[x_] = x - .8 - .2* Sin[x] // Simplify; NewtonsMethodList[f_, x0_, n : _Integer : 20] := NestList[# - f[#]/f'[#] &, x0, n] (nml1 = NewtonsMethodList[f, Pi // N]) // InputForm {3.141592653589793, 1.1902654422649652, 0.9692779750858744, 0.9643361576782574, 0.9643338876957006, 0.9643338876952227, ...


2

If you want to see the convergents, I recommendFixedPointList. f[x_] := x - 0.8 - 0.2 Sin[x] newtonsMethodList[f_, x0_, n_] := With[{iter = # - f[#]/D[f[#], #]}, FixedPointList[iter &, x0, n]] newtonsMethodList[f, N @ Pi, 10] {3.14159, 1.19027, 0.969278, 0.964336, 0.964334, 0.964334, 0.964334} Notice that, although I set a limit of ten ...


2

fun = x - 0.8 - 0.2 Sin[x] newton1[fun_, n_] := With[{f = fun/D[fun, x]}, Nest[# - f /. x -> # &, 2., n]] newton1[fun, 10] 0.964334 newton2[fun_, n_] := With[{f = fun/D[fun, x]}, NestList[# - f /. x -> # &, 2., n]] ListLinePlot[newton2[fun, 10], AxesOrigin -> {0, 0}, Mesh -> All, MeshStyle -> ...


2

My first observation is that {{x -> 2.36147 - 1.11022*10^-16 I}, {x -> -2.52892 + 0. I}, {x -> 0.167449 + 0. I}} is not a set of solutions for x^3 - 5 x^2 - x + 1 == 0 This can be seen by plotting the polynomial Plot[x^3 - 5 x^2 - x + 1, {x, -1., 6.}] However, the problem of imaginary fuzz in the roots remains. Solve[x^3 - 5 x^2 - x ...


1

This is a limitation of Graphics itself. If the PlotRange problem is bypassed we still get an error: NN = 200; W[n1_] := NN!/(n1! (NN - n1!)) (1/2)^n1 (1/2)^(NN - n1) dat = Table[{n1, W[n1]}, {n1, 1, 10, 0.1}]; Graphics[Point @ dat] This at least indicates that the problem is with the coordinates, but again the error message is less than accurate such ...


1

Thanks for updating your Question. With the new, clearer example I believe I can see the issue. Analysis The first method uses evenper on Symbolic values that are in canonical order: r1 = evenper[{a, b, c, d}] {{a, b, c, d}, {a, c, d, b}, {a, d, b, c}, {b, a, d, c}, {b, c, a, d}, {b, d, c, a}, {c, a, b, d}, {c, b, d, a}, {c, d, a, b}, {d, a, c, b}, ...



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