# Search Results

Results tagged with Search options answers only user 9362
30 results

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

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
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
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
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
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
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
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
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
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
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
Options[FindFormula] (* {Method -> Automatic, TargetFunctions -> All, TimeConstraint -> Automatic, SpecificityGoal -> 0.8, RandomSeeding -> 1234, "Monitor" -> False, PerformanceGoal -> Automatic} …
answered May 16 by Bob Hanlon
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
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
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
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

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