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Why doesn't Mathematica completely draw the fit?


How to Output Chi-Squared Statistics when using NonLinearModelFitFindfit doesn't give the good fit; Changing the starting values will not change the resultsFitting data to a complicated functionNonlinearModelFit errors and precisionHow to improve the nonlinear fit to a data setNon-linear-Model-Fit problem in mathematicaNon-linear model fit doesn't like my weightsErrors Using FindFit and FindRootNoLinearModelFit error messageWhy UnitCovert doesn't work with Fit?













1












$begingroup$


I have a question, and I really need your help, because I think I'm not understanding the reason...



I have a potential model..



F[x_] := k x^n


for the next graphic...



enter image description here



the fits shows a error and it doens't draw the curve completly...




NonlinearModelFit::sszero: The step size in the search has become less than the tolerance prescribed by the PrecisionGoal option, but the gradient is larger than the tolerance specified by the AccuracyGoal option. There is a possibility that the method has stalled at a point that is not a local minimum.




data2=0.000205, 81.41, 0.000422, 58.14, 0.00144, 30.51, 0.00198,25.93, 0.00396, 18.98, 0.00485, 16.83


and and I used this...



FindFit[data2, F[x], k, n, x]









share|improve this question











$endgroup$











  • $begingroup$
    Provide the code (copy and paste form) and the data that you used.
    $endgroup$
    – Bob Hanlon
    8 hours ago










  • $begingroup$
    try using explicit plot range: Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, PlotRange -> 0, .006, 0, 90, Epilog -> Red, PointSize[Large], Point@data2]?
    $endgroup$
    – kglr
    7 hours ago















1












$begingroup$


I have a question, and I really need your help, because I think I'm not understanding the reason...



I have a potential model..



F[x_] := k x^n


for the next graphic...



enter image description here



the fits shows a error and it doens't draw the curve completly...




NonlinearModelFit::sszero: The step size in the search has become less than the tolerance prescribed by the PrecisionGoal option, but the gradient is larger than the tolerance specified by the AccuracyGoal option. There is a possibility that the method has stalled at a point that is not a local minimum.




data2=0.000205, 81.41, 0.000422, 58.14, 0.00144, 30.51, 0.00198,25.93, 0.00396, 18.98, 0.00485, 16.83


and and I used this...



FindFit[data2, F[x], k, n, x]









share|improve this question











$endgroup$











  • $begingroup$
    Provide the code (copy and paste form) and the data that you used.
    $endgroup$
    – Bob Hanlon
    8 hours ago










  • $begingroup$
    try using explicit plot range: Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, PlotRange -> 0, .006, 0, 90, Epilog -> Red, PointSize[Large], Point@data2]?
    $endgroup$
    – kglr
    7 hours ago













1












1








1





$begingroup$


I have a question, and I really need your help, because I think I'm not understanding the reason...



I have a potential model..



F[x_] := k x^n


for the next graphic...



enter image description here



the fits shows a error and it doens't draw the curve completly...




NonlinearModelFit::sszero: The step size in the search has become less than the tolerance prescribed by the PrecisionGoal option, but the gradient is larger than the tolerance specified by the AccuracyGoal option. There is a possibility that the method has stalled at a point that is not a local minimum.




data2=0.000205, 81.41, 0.000422, 58.14, 0.00144, 30.51, 0.00198,25.93, 0.00396, 18.98, 0.00485, 16.83


and and I used this...



FindFit[data2, F[x], k, n, x]









share|improve this question











$endgroup$




I have a question, and I really need your help, because I think I'm not understanding the reason...



I have a potential model..



F[x_] := k x^n


for the next graphic...



enter image description here



the fits shows a error and it doens't draw the curve completly...




NonlinearModelFit::sszero: The step size in the search has become less than the tolerance prescribed by the PrecisionGoal option, but the gradient is larger than the tolerance specified by the AccuracyGoal option. There is a possibility that the method has stalled at a point that is not a local minimum.




data2=0.000205, 81.41, 0.000422, 58.14, 0.00144, 30.51, 0.00198,25.93, 0.00396, 18.98, 0.00485, 16.83


and and I used this...



FindFit[data2, F[x], k, n, x]






fitting data modeling






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 7 hours ago







Chris Schwenke

















asked 8 hours ago









Chris SchwenkeChris Schwenke

723




723











  • $begingroup$
    Provide the code (copy and paste form) and the data that you used.
    $endgroup$
    – Bob Hanlon
    8 hours ago










  • $begingroup$
    try using explicit plot range: Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, PlotRange -> 0, .006, 0, 90, Epilog -> Red, PointSize[Large], Point@data2]?
    $endgroup$
    – kglr
    7 hours ago
















  • $begingroup$
    Provide the code (copy and paste form) and the data that you used.
    $endgroup$
    – Bob Hanlon
    8 hours ago










  • $begingroup$
    try using explicit plot range: Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, PlotRange -> 0, .006, 0, 90, Epilog -> Red, PointSize[Large], Point@data2]?
    $endgroup$
    – kglr
    7 hours ago















$begingroup$
Provide the code (copy and paste form) and the data that you used.
$endgroup$
– Bob Hanlon
8 hours ago




$begingroup$
Provide the code (copy and paste form) and the data that you used.
$endgroup$
– Bob Hanlon
8 hours ago












$begingroup$
try using explicit plot range: Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, PlotRange -> 0, .006, 0, 90, Epilog -> Red, PointSize[Large], Point@data2]?
$endgroup$
– kglr
7 hours ago




$begingroup$
try using explicit plot range: Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, PlotRange -> 0, .006, 0, 90, Epilog -> Red, PointSize[Large], Point@data2]?
$endgroup$
– kglr
7 hours ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

Add the option PlotRange (say PlotRange -> 0, .006, 0, 90), and



use data2 as Epilog:



Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90,
Epilog -> Red, PointSize[Large], Point@data2]


enter image description here



Alternatively, use Plot first in Show:



Show[Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90],
ListPlot[data2, PlotStyle -> Directive[Red, PointSize[Large]]]]



same picture







share|improve this answer









$endgroup$












  • $begingroup$
    thanks a lot @kglr, it was very helpfull :)
    $endgroup$
    – Chris Schwenke
    6 hours ago










  • $begingroup$
    @ChrisSchwenke, my pleasure. Thank you for the accept.
    $endgroup$
    – kglr
    6 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Add the option PlotRange (say PlotRange -> 0, .006, 0, 90), and



use data2 as Epilog:



Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90,
Epilog -> Red, PointSize[Large], Point@data2]


enter image description here



Alternatively, use Plot first in Show:



Show[Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90],
ListPlot[data2, PlotStyle -> Directive[Red, PointSize[Large]]]]



same picture







share|improve this answer









$endgroup$












  • $begingroup$
    thanks a lot @kglr, it was very helpfull :)
    $endgroup$
    – Chris Schwenke
    6 hours ago










  • $begingroup$
    @ChrisSchwenke, my pleasure. Thank you for the accept.
    $endgroup$
    – kglr
    6 hours ago















6












$begingroup$

Add the option PlotRange (say PlotRange -> 0, .006, 0, 90), and



use data2 as Epilog:



Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90,
Epilog -> Red, PointSize[Large], Point@data2]


enter image description here



Alternatively, use Plot first in Show:



Show[Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90],
ListPlot[data2, PlotStyle -> Directive[Red, PointSize[Large]]]]



same picture







share|improve this answer









$endgroup$












  • $begingroup$
    thanks a lot @kglr, it was very helpfull :)
    $endgroup$
    – Chris Schwenke
    6 hours ago










  • $begingroup$
    @ChrisSchwenke, my pleasure. Thank you for the accept.
    $endgroup$
    – kglr
    6 hours ago













6












6








6





$begingroup$

Add the option PlotRange (say PlotRange -> 0, .006, 0, 90), and



use data2 as Epilog:



Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90,
Epilog -> Red, PointSize[Large], Point@data2]


enter image description here



Alternatively, use Plot first in Show:



Show[Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90],
ListPlot[data2, PlotStyle -> Directive[Red, PointSize[Large]]]]



same picture







share|improve this answer









$endgroup$



Add the option PlotRange (say PlotRange -> 0, .006, 0, 90), and



use data2 as Epilog:



Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90,
Epilog -> Red, PointSize[Large], Point@data2]


enter image description here



Alternatively, use Plot first in Show:



Show[Plot[Evaluate[F[x] /. FindFit[data2, F[x], k, n, x]], x, 0, .006, 
PlotRange -> 0, .006, 0, 90],
ListPlot[data2, PlotStyle -> Directive[Red, PointSize[Large]]]]



same picture








share|improve this answer












share|improve this answer



share|improve this answer










answered 7 hours ago









kglrkglr

198k10223449




198k10223449











  • $begingroup$
    thanks a lot @kglr, it was very helpfull :)
    $endgroup$
    – Chris Schwenke
    6 hours ago










  • $begingroup$
    @ChrisSchwenke, my pleasure. Thank you for the accept.
    $endgroup$
    – kglr
    6 hours ago
















  • $begingroup$
    thanks a lot @kglr, it was very helpfull :)
    $endgroup$
    – Chris Schwenke
    6 hours ago










  • $begingroup$
    @ChrisSchwenke, my pleasure. Thank you for the accept.
    $endgroup$
    – kglr
    6 hours ago















$begingroup$
thanks a lot @kglr, it was very helpfull :)
$endgroup$
– Chris Schwenke
6 hours ago




$begingroup$
thanks a lot @kglr, it was very helpfull :)
$endgroup$
– Chris Schwenke
6 hours ago












$begingroup$
@ChrisSchwenke, my pleasure. Thank you for the accept.
$endgroup$
– kglr
6 hours ago




$begingroup$
@ChrisSchwenke, my pleasure. Thank you for the accept.
$endgroup$
– kglr
6 hours ago

















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