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 9362

Questions on the use of Mathematica to construct models for approximating empirical data.

1
vote
data = {{1, 1, 4}, {1, 3, 4}, {2, 1, 4}, {2, 3, 3}, {3, 7, 4}, {3, 3, 2}}; expr = a*x + b*y + c*x*y + d*y^2 + e; f[x_, y_] = expr /. FindFit[data, expr, {a, b, c, d, e}, {x, y}] // Rational …
answered Dec 15 '14 by Bob Hanlon
10
votes
dataskmc2 = {{9.65827, 0.551402}, {10.2803, 0.602804}, {11.4566, 0.953271}, {12.6648, 1.3972}, {13.8468, 1.13551}, {15.0618, 0.845794}, {16.1433, 0.817757}, {17.4852, 0.981308}, {18.6631, …
answered Oct 19 '17 by Bob Hanlon
2
votes
dataset1 = Table[{x1[n], y1[n]}, {n, 5}]; poly2[x_] = a*x^2 + b*x + c; rms1 = Norm[(poly2 /@ dataset1[[All, 1]]) - dataset1[[All, 2]]]/ Sqrt[Length[dataset1]]; rms2 = RootMeanSquare[(poly2 /@ da …
answered Jul 14 '14 by Bob Hanlon
2
votes
data = {{0, 0.201519}, {0.693147, 0.339104}, {1.09861, 0.390401}, {1.38629, 0.410394}, {1.60944, 0.412307}, {1.79176, 0.417754}, {1.94591, 0.435408}, {2.07944, 0.444448}, {2.19722, 0.44 …
answered May 17 '17 by Bob Hanlon
4
votes
Clear[a, b, c, f, x] data = {{16, 2508}, {18, 2518}, {20, 3000}, {22, 3423}, {24, 3507}, {26, 3400}, {28, 3500}, {30, 3883}, {32, 3823}, {34, 3646}, {36, 3708}, {38, 3333}, {40, 3517}, {42, …
answered Oct 11 '18 by Bob Hanlon
2
votes
data = Transpose[{x, y}]; Clear[a, b, c, m] nlm = NonlinearModelFit[ data, {a*Exp[-(t - m)^2/(2 b^2)] + c, a > 0, b > 0, 0 < c < 1, 1432 < m < 1456}, {a, b, c, m}, t]; nlm[t] (* 0.00166378 …
answered Oct 15 '18 by Bob Hanlon
1
vote
One option is to use FindFormula $Version (* "11.1.1 for Mac OS X x86 (64-bit) (April 18, 2017)" *) data = {{0.99823, 1.005}, {1.0221, 1.31}, {1.0469, 1.76}, {1.0727, 2.5}, {1.0993, 3.72}, { …
answered Aug 21 '17 by Bob Hanlon
5
votes
HSI = {{1., 4.502201319666759}, {2., 4.498737206541754}, {3., 4.4964025063938955}, {4., 4.491168609823489}, {5., 4.480225462058295}, {6., 4.472178038829604}, {7., 4.4743675246518775}, { …
answered Jan 5 by Bob Hanlon
2
votes
You have defined nlm as rules for the parameters rather than the model data = {{0, 1}, {1, 0}, {3, 2}, {5, 4}, {6, 4}, {7, 5}}; nlm = NonlinearModelFit[data, Log[a + b x^2], {a, b}, x]["BestFitParam …
answered May 16 '18 by Bob Hanlon
4
votes
d = 2.72973; a1 = 2.03251; b1 = 1.79216; c1 = 1.35974; Bernoullifast = 4.8311; Bernoullislow = 4.83111; expr = Simplify[ u^2 + d/(u*R^2)^(c - 1) - a/R + b (2 (1 - R^2)/(1 - R^2 u) + (R^2 - …
answered Jul 10 '18 by Bob Hanlon
6
votes
Options[FindFormula] (* {Method -> Automatic, TargetFunctions -> All, TimeConstraint -> Automatic, SpecificityGoal -> 0.8, RandomSeeding -> 1234, "Monitor" -> False, PerformanceGoal -> Automatic} …
answered May 16 by Bob Hanlon
1
vote
data = {{0, 0}, {1.1, 0.8}, {1.4, 1}, {1.7, 0.8}, {2.6, 0.2}, {3.6, 0.06}, {5, 0}}; Set the derivative to zero for first point, last point, and peak point. data2 = ({{#[[1]]}, #[[2]]} & /@ dat …
answered Nov 13 '18 by Bob Hanlon
1
vote
Use Interpolation Clear["Global`*"] dados = {{0, 0}, {1, 1000}, {2, -750}, {3, 250}, {4, -1000}, {5, 0}}; {xmin, xmax} = MinMax[dados[[All, 1]]] (* {0, 5} *) f = Interpolation[dados, Interpolation …
answered Nov 28 '18 by Bob Hanlon
5
votes
The function is ReplaceAll. Assuming that the encoding that you want is {"Yes"->1, "No"->0}: list = {{"No", 729.526, 44361.6, "No"}, {"Yes", 817.18, 12106.1, "No"}, {"No", 1073.55, 31767.1, "No …
answered Feb 12 '15 by Bob Hanlon
2
votes
Fit your data to a CDF using NonlinearModelFit data = {{406.833, 0.05}, {423.458, 0.1}, {436.375, 0.15}, {448.042, 0.2}, {459.583, 0.25}, {467.75, 0.3}, {479.083, 0.35}, {489.917, 0.4}, {50 …
answered Jun 24 '16 by Bob Hanlon

15 30 50 per page