How to fit the data?












1












$begingroup$


Here is my attempt.



data = {{0., 2.61}, {0.1, 2.62}, {0.2, 2.62}, {0.3, 2.62}, {0.4, 

2.63}, {0.5, 2.63}, {0.6, 2.74}, {0.7, 2.98}, {0.8, 3.66}, {0.9, 

5.04}, {1., 7.52}, {1.1, 10.74}, {1.2, 12.62}, {1.3, 10.17}, {1.4,

 5}, {1.5, 2.64}, {1.6, 11.5}, {1.65, 35.4}};

NonlinearModelFit[data,a*Cosh[b*x^c*Sin[d*x^e]], {{a, 3}, {b, 0.2}, {c, 1}, {d, 1}, 

{e, 3}}, x, Method -> "LevenbergMarquardt"]



NonlinearModelFit::nrjnum: The Jacobian is not a matrix of real numbers at {a,b,c,d,e} = {3.,0.2,1.,1.,3.}.




The same issues appear with other initial points. As far as I know it, a good fit with the residuals of quantity $0.01$ exists and can be found with Mathcad (not MATLAB).






fitting







share|improve this question









$endgroup$








  • 2




    $begingroup$
    The Jacobian contains Log[x] and your data contains x == 0
    $endgroup$
    – Coolwater
    1 hour ago












  • $begingroup$
    @Coolwater: There are other methods of fitting.
    $endgroup$
    – user64494
    33 mins ago
















1












$begingroup$


Here is my attempt.



data = {{0., 2.61}, {0.1, 2.62}, {0.2, 2.62}, {0.3, 2.62}, {0.4, 

2.63}, {0.5, 2.63}, {0.6, 2.74}, {0.7, 2.98}, {0.8, 3.66}, {0.9, 

5.04}, {1., 7.52}, {1.1, 10.74}, {1.2, 12.62}, {1.3, 10.17}, {1.4,

 5}, {1.5, 2.64}, {1.6, 11.5}, {1.65, 35.4}};

NonlinearModelFit[data,a*Cosh[b*x^c*Sin[d*x^e]], {{a, 3}, {b, 0.2}, {c, 1}, {d, 1}, 

{e, 3}}, x, Method -> "LevenbergMarquardt"]



NonlinearModelFit::nrjnum: The Jacobian is not a matrix of real numbers at {a,b,c,d,e} = {3.,0.2,1.,1.,3.}.




The same issues appear with other initial points. As far as I know it, a good fit with the residuals of quantity $0.01$ exists and can be found with Mathcad (not MATLAB).






fitting







share|improve this question









$endgroup$








  • 2




    $begingroup$
    The Jacobian contains Log[x] and your data contains x == 0
    $endgroup$
    – Coolwater
    1 hour ago












  • $begingroup$
    @Coolwater: There are other methods of fitting.
    $endgroup$
    – user64494
    33 mins ago














1












1








1





$begingroup$


Here is my attempt.



data = {{0., 2.61}, {0.1, 2.62}, {0.2, 2.62}, {0.3, 2.62}, {0.4, 

2.63}, {0.5, 2.63}, {0.6, 2.74}, {0.7, 2.98}, {0.8, 3.66}, {0.9, 

5.04}, {1., 7.52}, {1.1, 10.74}, {1.2, 12.62}, {1.3, 10.17}, {1.4,

 5}, {1.5, 2.64}, {1.6, 11.5}, {1.65, 35.4}};

NonlinearModelFit[data,a*Cosh[b*x^c*Sin[d*x^e]], {{a, 3}, {b, 0.2}, {c, 1}, {d, 1}, 

{e, 3}}, x, Method -> "LevenbergMarquardt"]



NonlinearModelFit::nrjnum: The Jacobian is not a matrix of real numbers at {a,b,c,d,e} = {3.,0.2,1.,1.,3.}.




The same issues appear with other initial points. As far as I know it, a good fit with the residuals of quantity $0.01$ exists and can be found with Mathcad (not MATLAB).






fitting







share|improve this question









$endgroup$




Here is my attempt.



data = {{0., 2.61}, {0.1, 2.62}, {0.2, 2.62}, {0.3, 2.62}, {0.4, 

2.63}, {0.5, 2.63}, {0.6, 2.74}, {0.7, 2.98}, {0.8, 3.66}, {0.9, 

5.04}, {1., 7.52}, {1.1, 10.74}, {1.2, 12.62}, {1.3, 10.17}, {1.4,

 5}, {1.5, 2.64}, {1.6, 11.5}, {1.65, 35.4}};

NonlinearModelFit[data,a*Cosh[b*x^c*Sin[d*x^e]], {{a, 3}, {b, 0.2}, {c, 1}, {d, 1}, 

{e, 3}}, x, Method -> "LevenbergMarquardt"]



NonlinearModelFit::nrjnum: The Jacobian is not a matrix of real numbers at {a,b,c,d,e} = {3.,0.2,1.,1.,3.}.




The same issues appear with other initial points. As far as I know it, a good fit with the residuals of quantity $0.01$ exists and can be found with Mathcad (not MATLAB).






fitting




fitting






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 1 hour ago









user64494user64494

3,28111021




3,28111021








  • 2




    $begingroup$
    The Jacobian contains Log[x] and your data contains x == 0
    $endgroup$
    – Coolwater
    1 hour ago












  • $begingroup$
    @Coolwater: There are other methods of fitting.
    $endgroup$
    – user64494
    33 mins ago














  • 2




    $begingroup$
    The Jacobian contains Log[x] and your data contains x == 0
    $endgroup$
    – Coolwater
    1 hour ago












  • $begingroup$
    @Coolwater: There are other methods of fitting.
    $endgroup$
    – user64494
    33 mins ago








2




2




$begingroup$
The Jacobian contains Log[x] and your data contains x == 0
$endgroup$
– Coolwater
1 hour ago






$begingroup$
The Jacobian contains Log[x] and your data contains x == 0
$endgroup$
– Coolwater
1 hour ago














$begingroup$
@Coolwater: There are other methods of fitting.
$endgroup$
– user64494
33 mins ago




$begingroup$
@Coolwater: There are other methods of fitting.
$endgroup$
– user64494
33 mins ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

Below err is defined as the residual sum of squares. Using ?NumericQ on one of the arguments prevents FindMinimum from exact differentiation. In that way no problem arises even though the point with x == 0 is included.



err[a_, b_, c_, d_, e_?NumericQ] = Total[(a*Cosh[b*#^c*Sin[d*#^e]] - #2)^2 & @@@ data];



res[x_] = a*Cosh[b*x^c*Sin[d*x^e]] /. Last[FindMinimum[err[a, b, c, d, e], {

                            {a, 3, 1/100, 25}, {b, 2, 1/100, 25}, {c, 1, 1/100, 25},

                            {d, 23/10, 1/100, 25}, {e, 2, 1/100, 25}}, Method -> "InteriorPoint"]]



Show[Plot[res[x], {x, 0, 1.65}, PlotRange -> All], ListPlot[data]]







share|improve this answer









$endgroup$













  • $begingroup$
    Thank you. Likely a somewhat better fit than your $2.75699 cosh left(1.66039 x^{2.92883} sin left(4.74005 x^{0.70619}right)right) $ is $[a= 2.61748239217892,b= 1.71949328471199,c= 2.30924398627606,d= 1.50333104215966,e= 1.84597270613482]$. Can you kindly compare the results?
    $endgroup$
    – user64494
    35 mins ago













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1 Answer
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1 Answer
1






active

oldest

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active

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active

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3












$begingroup$

Below err is defined as the residual sum of squares. Using ?NumericQ on one of the arguments prevents FindMinimum from exact differentiation. In that way no problem arises even though the point with x == 0 is included.



err[a_, b_, c_, d_, e_?NumericQ] = Total[(a*Cosh[b*#^c*Sin[d*#^e]] - #2)^2 & @@@ data];



res[x_] = a*Cosh[b*x^c*Sin[d*x^e]] /. Last[FindMinimum[err[a, b, c, d, e], {

                            {a, 3, 1/100, 25}, {b, 2, 1/100, 25}, {c, 1, 1/100, 25},

                            {d, 23/10, 1/100, 25}, {e, 2, 1/100, 25}}, Method -> "InteriorPoint"]]



Show[Plot[res[x], {x, 0, 1.65}, PlotRange -> All], ListPlot[data]]







share|improve this answer









$endgroup$













  • $begingroup$
    Thank you. Likely a somewhat better fit than your $2.75699 cosh left(1.66039 x^{2.92883} sin left(4.74005 x^{0.70619}right)right) $ is $[a= 2.61748239217892,b= 1.71949328471199,c= 2.30924398627606,d= 1.50333104215966,e= 1.84597270613482]$. Can you kindly compare the results?
    $endgroup$
    – user64494
    35 mins ago


















3












$begingroup$

Below err is defined as the residual sum of squares. Using ?NumericQ on one of the arguments prevents FindMinimum from exact differentiation. In that way no problem arises even though the point with x == 0 is included.



err[a_, b_, c_, d_, e_?NumericQ] = Total[(a*Cosh[b*#^c*Sin[d*#^e]] - #2)^2 & @@@ data];



res[x_] = a*Cosh[b*x^c*Sin[d*x^e]] /. Last[FindMinimum[err[a, b, c, d, e], {

                            {a, 3, 1/100, 25}, {b, 2, 1/100, 25}, {c, 1, 1/100, 25},

                            {d, 23/10, 1/100, 25}, {e, 2, 1/100, 25}}, Method -> "InteriorPoint"]]



Show[Plot[res[x], {x, 0, 1.65}, PlotRange -> All], ListPlot[data]]







share|improve this answer









$endgroup$













  • $begingroup$
    Thank you. Likely a somewhat better fit than your $2.75699 cosh left(1.66039 x^{2.92883} sin left(4.74005 x^{0.70619}right)right) $ is $[a= 2.61748239217892,b= 1.71949328471199,c= 2.30924398627606,d= 1.50333104215966,e= 1.84597270613482]$. Can you kindly compare the results?
    $endgroup$
    – user64494
    35 mins ago
















3












3








3





$begingroup$

Below err is defined as the residual sum of squares. Using ?NumericQ on one of the arguments prevents FindMinimum from exact differentiation. In that way no problem arises even though the point with x == 0 is included.



err[a_, b_, c_, d_, e_?NumericQ] = Total[(a*Cosh[b*#^c*Sin[d*#^e]] - #2)^2 & @@@ data];



res[x_] = a*Cosh[b*x^c*Sin[d*x^e]] /. Last[FindMinimum[err[a, b, c, d, e], {

                            {a, 3, 1/100, 25}, {b, 2, 1/100, 25}, {c, 1, 1/100, 25},

                            {d, 23/10, 1/100, 25}, {e, 2, 1/100, 25}}, Method -> "InteriorPoint"]]



Show[Plot[res[x], {x, 0, 1.65}, PlotRange -> All], ListPlot[data]]







share|improve this answer









$endgroup$



Below err is defined as the residual sum of squares. Using ?NumericQ on one of the arguments prevents FindMinimum from exact differentiation. In that way no problem arises even though the point with x == 0 is included.



err[a_, b_, c_, d_, e_?NumericQ] = Total[(a*Cosh[b*#^c*Sin[d*#^e]] - #2)^2 & @@@ data];



res[x_] = a*Cosh[b*x^c*Sin[d*x^e]] /. Last[FindMinimum[err[a, b, c, d, e], {

                            {a, 3, 1/100, 25}, {b, 2, 1/100, 25}, {c, 1, 1/100, 25},

                            {d, 23/10, 1/100, 25}, {e, 2, 1/100, 25}}, Method -> "InteriorPoint"]]



Show[Plot[res[x], {x, 0, 1.65}, PlotRange -> All], ListPlot[data]]








share|improve this answer












share|improve this answer



share|improve this answer










answered 56 mins ago









CoolwaterCoolwater

14.8k32553




14.8k32553












  • $begingroup$
    Thank you. Likely a somewhat better fit than your $2.75699 cosh left(1.66039 x^{2.92883} sin left(4.74005 x^{0.70619}right)right) $ is $[a= 2.61748239217892,b= 1.71949328471199,c= 2.30924398627606,d= 1.50333104215966,e= 1.84597270613482]$. Can you kindly compare the results?
    $endgroup$
    – user64494
    35 mins ago




















  • $begingroup$
    Thank you. Likely a somewhat better fit than your $2.75699 cosh left(1.66039 x^{2.92883} sin left(4.74005 x^{0.70619}right)right) $ is $[a= 2.61748239217892,b= 1.71949328471199,c= 2.30924398627606,d= 1.50333104215966,e= 1.84597270613482]$. Can you kindly compare the results?
    $endgroup$
    – user64494
    35 mins ago


















$begingroup$
Thank you. Likely a somewhat better fit than your $2.75699 cosh left(1.66039 x^{2.92883} sin left(4.74005 x^{0.70619}right)right) $ is $[a= 2.61748239217892,b= 1.71949328471199,c= 2.30924398627606,d= 1.50333104215966,e= 1.84597270613482]$. Can you kindly compare the results?
$endgroup$
– user64494
35 mins ago






$begingroup$
Thank you. Likely a somewhat better fit than your $2.75699 cosh left(1.66039 x^{2.92883} sin left(4.74005 x^{0.70619}right)right) $ is $[a= 2.61748239217892,b= 1.71949328471199,c= 2.30924398627606,d= 1.50333104215966,e= 1.84597270613482]$. Can you kindly compare the results?
$endgroup$
– user64494
35 mins ago




















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