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Questions on the analytic and numerical equation solving functions of Mathematica (Solve, Reduce, NSolve, FindRoot, DSolve, RSolve, etc.).

0
votes
Is this what you want? Using r[c_] := x /. NSolve[f[x] + c, x, Reals] Show[ Plot[r[c], {c, 0, 1/405 (-442 + 79 Sqrt[79])}], Plot[r[c], {c, 1/405 (-442 + 79 Sqrt[79]), 1}] , PlotRange -> All] I …
answered Jun 12 '18 by AccidentalFourierTransform
7
votes
Your integral doesn't exist: the integrand has a non-integrable singularity in the integration region. This singularity is given by the solution of g == n ϕ[x]: zero[n_] := FindRoot[g - n ϕ[x] == 0, …
answered Oct 13 '18 by AccidentalFourierTransform
3
votes
Use fKN[R_] := NSolve[(1/Sqrt[f] == (2*Log[R*Sqrt[f]] - 0.8)), f][[1, 1, 2]] // Quiet With this, LogLinearPlot[fKN[R], {R, 10^5, 10^6}] Note: in this case, FindRoot is actually faster than NSo …
answered Feb 7 '18 by AccidentalFourierTransform
1
vote
The equation is a typical example of a transcendental equation and, as such, it has no analytic solution for symbolic $a,b$. That being said, it does admit an asymptotic expansion for small $a$ (or la …
answered Jul 4 by AccidentalFourierTransform
3
votes
In addition to Henrik Schumacher's and Alexei Boulbitch's answers, we note the following. Due to the extreme scale set by $q/k_BT\sim 10^{-18}$, the root of your equation varies extremely slowly with …
answered Apr 30 '18 by AccidentalFourierTransform
1
vote
Well, you can define your own simplifying/expanding functions. For example, the following does the trick: simplify[exp_] := FullSimplify[FunctionExpand[exp], ComplexityFunction -> ((LeafCount[#] + 10 …
answered Sep 10 '18 by AccidentalFourierTransform
7
votes
Using the example in the OP, A = {{1, 0}, {1, 1}}; we compare the different options mentioned in the comments (plus one that is mine): Table[If[Mod[MatrixPower[A, n], 2] == IdentityMatrix[Length@A …
answered May 8 '18 by AccidentalFourierTransform
4
votes
Quick-and-dirty solution: y^2 - (x^2 + a x^2 y^2 + b y^2 x^3 + c y^3 x^2) /. y -> x + c1 x^2 + c2 x^3 + c3 x^4 + O[x]^5 // FullSimplify Solve[% == 0, {c1, c2, c3}] y -> x + c1 x^2 + c2 x^3 + …
answered Nov 4 '18 by AccidentalFourierTransform
4
votes
NSolve[x UnitStep[x] == 5, x] (* {{x -> 5.}} *)
answered Jul 1 by AccidentalFourierTransform
3
votes
Unless I misunderstood the question, this should do: Let ineq = ((β - x - 19/2) Piecewise[{{0.3333333333333333 (1. - 1. y), -1. < y < 1.}}, 0.] > (β - z - 23/2) Piecewise[{{1., -1. < w < 0.}, {0.3 …
answered Apr 29 '18 by AccidentalFourierTransform
2
votes
Most compact method: SetAttributes[Greater, Listable]; {1, 4, 5, -2} > 0 (* {True, True, True, False} *) SetAttributes[GreaterEqual, Listable] {-(r/(-d + k r)), -(r/(-d + k r)), 0} >= 0 // Reduce (* …
answered May 1 '18 by AccidentalFourierTransform
1
vote
Quick-and-dirty solution: Manipulate[ ListPlot[ Table[{#, (1 - #)^(1 - k + n) #^(-1 + k) Binomial[n, -1 + k] Hypergeometric2F1[1, -1 + k - n, k, #/(# - 1)]} &@ FindRoot[1/(p ( …
answered Apr 9 '18 by AccidentalFourierTransform
3
votes
Split the variables into {x, y, z}, {t} instead of {c, y, z, t}: Solve[{p == x + y + z + t && x > 0 && y >= x && z >= y && t > 0}, {x, y, z}, {t}, Integers] (* {{x -> 1, y -> 1, z -> 1}, {x -> 1, y …
answered Aug 17 by AccidentalFourierTransform