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Results tagged with Search options answers only user 125
101 results

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

I am sure there will be some input that will break the following method, but here is a way to use Pick or Select: Let sols = {-(((0.25 + 0.5 I) x^2 y^3)/z), ((3 + I) x^3 y)/w^2, 3 - 4 I, (5.` x^5 …
answered May 22 '12 by kglr
apQ = Equal @@ Differences @ # &; lista /. Solve[a3 + a6 == 34 && a4 + a9 == 50 && apQ @ lista, lista][] {3, 7, 11, 15, 19, 23, 27, 31, 35} FindSequenceFunction @ % -1 + 4 #1 &
answered Mar 16 '18 by kglr
A combination of Reap/Sow and Check using a modified version of the OP's example: counter = 50; d = ConstantArray[0, {counter}]; sol = {0, 0}; add = {0.5, 0.5}; data = Reap[ NestWhileList[{#[] + …
answered Sep 30 '12 by kglr
coords = {n, k} /. Solve[{Rationalize[4.5 == (n*k + n - 1)*0.1], n >= 0, k >= 0}, {n, k}, Integers] {{1, 45}, {2, 22}, {23, 1}, {46, 0}} ContourPlot[4.5 == (n*k + n - 1)*0.1, {n, 0, 46}, {k, …
answered Jun 29 '18 by kglr
You can try ContourPlot3D: ContourPlot3D[a x^5 + b Sin[x] + 3 == 0, {a, -1, 1}, {b, -1, 1}, {x, -2 Pi, 2 Pi}, AxesLabel->{"a","b","x"}] Alternatively, use ContourPlot to show combinations o …
answered Jun 19 by kglr
Try Solve[{PAR == program + TimeValue[PAR, rate2, term]*reserve, rate2 > 0}, rate2] {{rate2 -> 0.0025}}
answered Aug 14 by kglr
Using MeshFunctions: ContourPlot[{2 Abs[x] + Abs[y] == 1, Abs[x] + 2 Abs[y] == 1}, {x, -1, 1}, {y, -1, 1}, MeshFunctions -> {ConditionalExpression[#2, Abs[#] < 1] &, ConditionalExpression[#, Ab …
answered Jul 30 '17 by kglr
ArrayReshape[Tuples /@ ({#, Solve[(w^2*c^2 /. #)-1 == 0, w]} & /@ {{c -> 1}, {c -> 2}}), {4, 2}] or Partition[Flatten[Tuples /@ ({#,Solve[(w^2*c^2 /. #)-1 == 0, w]} & /@ {{c -> 1}, {c -> 2}})], 2] …
answered Sep 22 '14 by kglr
Define post as ClearAll[post]; post[t_]: = First[p /. Solve[tpos[p] == t, p]] before using post[t].
answered May 30 by kglr
With a large value for PlotPoints we can also use purely graphical approach to get the roots: Normal[Plot[s4[100, T], {T, 120, 250}, PlotRange -> All, MeshFunctions -> {#2 &}, Mesh -> {{0}}, …
answered Jul 4 '18 by kglr
You can check if a member of the candidate list of factors appear in FactorList of the input polynomial; if it does, you can get its exponent from the second entry of the corresponding element in Fac …
answered Feb 19 by kglr
NestWhile and NestWhileList are the built-in functions that feature arguments that can be used to capture your requirements (looping until a condition is satisfied where the condition can depend on va …
answered Sep 14 '12 by kglr
Syntactically correct version of sols2 is sols2 = Table[FindRoot[fun'[z] + 0.5 == m[[i]], {z, sols[[j, 1, 2]]}], {i, Length[m]}, {j, Length@sols}] using fun[z] as in Mark's answer. However, t …
answered May 16 '12 by kglr
res = {A < -0.756026 || 0. < A < 1.94105, A < -0.757879 || 0. < A < 7.36113}; res /. {Or -> List, Less -> (Last[{##}] &)} // Flatten Sequence @@@ res[[All, All, -1]] ArgMax[{A, #}, A] & /@ # & /@ re …
answered Jul 18 '17 by kglr
Factor x^p out of the last term to get w == a p (a + b (y/x)^p)^(1/p-1) then use Eliminate[{{w == a p (a + b z^p)^(1/p - 1), z == y/x}}, z] to get ((-a + ((a p)/w)^(p/(-1 + p)))/b)^(1/p) == …
answered Sep 2 '14 by kglr

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