The problem<\/strong><\/p>\n and you’d like to fit the function<\/p>\n using nonlinear least squares. \u00a0You’re starting guesses for the parameters are p1=1 and P2=0.2<\/p>\n For now, we are primarily interested in the following results:<\/p>\n Future updates of these posts will show how to get other results such as confidence intervals. Let me know what you are most interested in.<\/p>\n Solution in Maple<\/strong><\/p>\n Maple’s user interface is quite user friendly and it uses non-linear optimization routines from The Numerical Algorithms Group under the hood. Here’s the code to get the parameter values and sum of squares of residuals<\/p>\n which gives the result<\/p>\n Various other outputs can be specified in the output vector such as:<\/p>\n The meanings of the above are mostly obvious.\u00a0 In those cases where they aren’t, you can look them up in the documentation link below.<\/p>\nxdata = -2,-1.64,-1.33,-0.7,0,0.45,1.2,1.64,2.32,2.9\r\nydata = 0.699369,0.700462,0.695354,1.03905,1.97389,2.41143,1.91091,0.919576,-0.730975,-1.42001<\/pre>\n
![\" F(p_1,p_2,x) = p_1 \cos(p_2 x)+p_2 \sin(p_1 x) \"](\"https:\/\/walkingrandomly.com\/wp-content\/ql-cache\/quicklatex.com-17f46b025dd91e024cf2dc04211e19ba_l3.png\" "\\"Rendered")
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with(Statistics):\r\n\r\nxdata := Vector([-2, -1.64, -1.33, -.7, 0, .45, 1.2, 1.64, 2.32, 2.9], datatype = float):\r\nydata := Vector([.699369, .700462, .695354, 1.03905, 1.97389, 2.41143, 1.91091, .919576, \r\n -.730975, -1.42001], datatype = float):\r\nNonlinearFit(p1*cos(p2*x)+p2*sin(p1*x), xdata, ydata, x, initialvalues = [p1 = 1, p2 = .2], \r\n output = [parametervalues, residualsumofsquares])<\/pre>\n
[[p1 = 1.88185090465902, p2 = 0.700229557992540], 0.05381269642]<\/pre>\n
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