Mathematica非线性拟合,data = Table[{{1.081533825,47},{1.079100598,48},{1.071849984,52},{1.056458082,64},{1.045335019,76},{1.028929244,100},{1.014458888,127},{1.006932388,143},{1.003074162,147},{1.000767214,150},{0.997194416,151},{0.995034543,150

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Mathematica非线性拟合,data = Table[{{1.081533825,47},{1.079100598,48},{1.071849984,52},{1.056458082,64},{1.045335019,76},{1.028929244,100},{1.014458888,127},{1.006932388,143},{1.003074162,147},{1.000767214,150},{0.997194416,151},{0.995034543,150
Mathematica非线性拟合,
data = Table[{{1.081533825,47},{1.079100598,48},{1.071849984,
52},{1.056458082,64},{1.045335019,76},{1.028929244,
100},{1.014458888,127},{1.006932388,143},{1.003074162,
147},{1.000767214,150},{0.997194416,151},{0.995034543,
150},{0.991240084,147},{0.990238086,144},{0.984973538,
136},{0.976430116,118},{0.969102296,102},{0.952910248,
77},{0.944898179,68},{0.946304714,69},{0.945015271,
68},{0.943611609,67},{0.941977953,65},{0.939651508,
63},{0.938028121,62}}];
拟合的曲线方程(实际是受迫振动的幅频特性曲线):a/Sqrt[(1 - x)^2 + bx]
一楼的是软件的结果,可是不符合实际散点啊,按这个结果很明显函数值应该是负的,而实际散点都是正的。你把散点和拟合曲线画在一张图上看看

Mathematica非线性拟合,data = Table[{{1.081533825,47},{1.079100598,48},{1.071849984,52},{1.056458082,64},{1.045335019,76},{1.028929244,100},{1.014458888,127},{1.006932388,143},{1.003074162,147},{1.000767214,150},{0.997194416,151},{0.995034543,150
data = {{1.081533825,47},{1.079100598,48},{1.071849984,
52},{1.056458082,64},{1.045335019,76},{1.028929244,
100},{1.014458888,127},{1.006932388,143},{1.003074162,
147},{1.000767214,150},{0.997194416,151},{0.995034543,
150},{0.991240084,147},{0.990238086,144},{0.984973538,
136},{0.976430116,118},{0.969102296,102},{0.952910248,
77},{0.944898179,68},{0.946304714,69},{0.945015271,
68},{0.943611609,67},{0.941977953,65},{0.939651508,
63},{0.938028121,62}};
FindFit[data,a/Sqrt[(1 - x)^2 + b x],{a,b},x]
{a -> -14.8065,b -> -223.813}
这样就能求出参数a、b的值
常用的Mathematica拟合函数有两个,一个是Fit[],再一个就是FindFit[],自己到帮助文件里具体再看看吧

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