# Search Results

Results tagged with Search options answers only user 34893
13 results

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

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
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
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
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 …
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
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
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
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
NSolve[x UnitStep[x] == 5, x] (* {{x -> 5.}} *)