Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Favorites infavorites:mine
infavorites:1234
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with Search options answers only user 125

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

4
votes
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
1
vote
apQ = Equal @@ Differences @ # &; lista /. Solve[a3 + a6 == 34 && a4 + a9 == 50 && apQ @ lista, lista][[1]] {3, 7, 11, 15, 19, 23, 27, 31, 35} FindSequenceFunction @ % -1 + 4 #1 &
answered Mar 16 '18 by kglr
3
votes
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[{#[[1]] + …
answered Sep 30 '12 by kglr
4
votes
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
2
votes
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
1
vote
Try Solve[{PAR == program + TimeValue[PAR, rate2, term]*reserve, rate2 > 0}, rate2] {{rate2 -> 0.0025}}
answered Aug 14 by kglr
2
votes
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
3
votes
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
1
vote
Define post as ClearAll[post]; post[t_]: = First[p /. Solve[tpos[p] == t, p]] before using post[t].
answered May 30 by kglr
0
votes
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
1
vote
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
5
votes
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
4
votes
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
3
votes
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
3
votes
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

15 30 50 per page