How to do an exponential fit for this data?Finding exponential model from dataFit a custom process to a data: inhomogeneous 2-state Markov chainFitting experimental data to ODEs by using NonlinearModelFitProblem with fitting NonlinearModelFit::nrlnum:Discover type of model for a set of dataTrouble fitting exponential with NonLinearModelFitNon linear fit/Find distribution parameters for large gradient data;Exponential fit for time seriesSelecting method and using Abs[] for FindFitNon Linear Model Fit - Fitting ODE to Data
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How to do an exponential fit for this data?
Finding exponential model from dataFit a custom process to a data: inhomogeneous 2-state Markov chainFitting experimental data to ODEs by using NonlinearModelFitProblem with fitting NonlinearModelFit::nrlnum:Discover type of model for a set of dataTrouble fitting exponential with NonLinearModelFitNon linear fit/Find distribution parameters for large gradient data;Exponential fit for time seriesSelecting method and using Abs[] for FindFitNon Linear Model Fit - Fitting ODE to Data
$begingroup$
Whenever I try to use FindFit or NonlinearModelFit for the data (shown below), I keep getting the following error thrown at me:
RecursionLimit::reclim: Recursion depth of 1024 exceeded.
I am using an exponential model to try and fit it.
data = 0.19676778483499557`, 0.20885602228466532`, 0.22210833221738427`,
0.23668665287868895`, 0.252783561788831`, 0.2706298263857738`,
0.2905043053747535`, 0.3127471056565888`, 0.3377773233068993`,
0.3661173611964135`, 0.39842687921844677`, 0.43555118981628316`,
0.4785919073537348`, 0.5290129571413393`, 0.5888048389535786`,
0.6607490640587161`;
The particular model that I'm using is FindFit[data, Exp[a*x] + b, a, b, x]
How can I stop getting this error thrown at me/get the model I'm trying to use to work?
fitting
asked May 17 at 12:47
Spencer KellerSpencer Keller
265
$endgroup$
add a comment |
$begingroup$
Whenever I try to use FindFit or NonlinearModelFit for the data (shown below), I keep getting the following error thrown at me:
RecursionLimit::reclim: Recursion depth of 1024 exceeded.
I am using an exponential model to try and fit it.
data = 0.19676778483499557`, 0.20885602228466532`, 0.22210833221738427`,
0.23668665287868895`, 0.252783561788831`, 0.2706298263857738`,
0.2905043053747535`, 0.3127471056565888`, 0.3377773233068993`,
0.3661173611964135`, 0.39842687921844677`, 0.43555118981628316`,
0.4785919073537348`, 0.5290129571413393`, 0.5888048389535786`,
0.6607490640587161`;
The particular model that I'm using is FindFit[data, Exp[a*x] + b, a, b, x]
How can I stop getting this error thrown at me/get the model I'm trying to use to work?
fitting
asked May 17 at 12:47
Spencer KellerSpencer Keller
265
$endgroup$
$begingroup$
Can you give the exact command you used to fit please? Otherwise it's difficult to tell what's going on. But as a first guess I'd suggest you try again with a fresh kernel.
$endgroup$
– Roman
May 17 at 12:51
$begingroup$
@Roman Updated the question.
$endgroup$
– Spencer Keller
May 17 at 12:55
2
$begingroup$
Runs fine on my computer. Please try your own code with a fresh kernel. Check?aand?bto see if there are any lingering definitions.
$endgroup$
– Roman
May 17 at 13:01
$begingroup$
@Roman I took your advice and used a new kernel. Worked just fine. Would you happen to know why using a new kernel worked?
$endgroup$
– Spencer Keller
May 17 at 13:21
$begingroup$
Lingering definitions ofaandb. Check them with?aand?b.
$endgroup$
– Roman
May 17 at 13:30
add a comment |
$begingroup$
Whenever I try to use FindFit or NonlinearModelFit for the data (shown below), I keep getting the following error thrown at me:
RecursionLimit::reclim: Recursion depth of 1024 exceeded.
I am using an exponential model to try and fit it.
data = 0.19676778483499557`, 0.20885602228466532`, 0.22210833221738427`,
0.23668665287868895`, 0.252783561788831`, 0.2706298263857738`,
0.2905043053747535`, 0.3127471056565888`, 0.3377773233068993`,
0.3661173611964135`, 0.39842687921844677`, 0.43555118981628316`,
0.4785919073537348`, 0.5290129571413393`, 0.5888048389535786`,
0.6607490640587161`;
The particular model that I'm using is FindFit[data, Exp[a*x] + b, a, b, x]
How can I stop getting this error thrown at me/get the model I'm trying to use to work?
fitting
asked May 17 at 12:47
Spencer KellerSpencer Keller
265
$endgroup$
Whenever I try to use FindFit or NonlinearModelFit for the data (shown below), I keep getting the following error thrown at me:
RecursionLimit::reclim: Recursion depth of 1024 exceeded.
I am using an exponential model to try and fit it.
data = 0.19676778483499557`, 0.20885602228466532`, 0.22210833221738427`,
0.23668665287868895`, 0.252783561788831`, 0.2706298263857738`,
0.2905043053747535`, 0.3127471056565888`, 0.3377773233068993`,
0.3661173611964135`, 0.39842687921844677`, 0.43555118981628316`,
0.4785919073537348`, 0.5290129571413393`, 0.5888048389535786`,
0.6607490640587161`;
The particular model that I'm using is FindFit[data, Exp[a*x] + b, a, b, x]
How can I stop getting this error thrown at me/get the model I'm trying to use to work?
fitting
fitting
asked May 17 at 12:47
Spencer KellerSpencer Keller
265
asked May 17 at 12:47
Spencer KellerSpencer Keller
265
edited May 17 at 12:54
Spencer Keller
asked May 17 at 12:47
Spencer KellerSpencer Keller
265
asked May 17 at 12:47
Spencer KellerSpencer Keller
265
asked May 17 at 12:47
Spencer KellerSpencer Keller
265
265
$begingroup$
Can you give the exact command you used to fit please? Otherwise it's difficult to tell what's going on. But as a first guess I'd suggest you try again with a fresh kernel.
$endgroup$
– Roman
May 17 at 12:51
$begingroup$
@Roman Updated the question.
$endgroup$
– Spencer Keller
May 17 at 12:55
2
$begingroup$
Runs fine on my computer. Please try your own code with a fresh kernel. Check?aand?bto see if there are any lingering definitions.
$endgroup$
– Roman
May 17 at 13:01
$begingroup$
@Roman I took your advice and used a new kernel. Worked just fine. Would you happen to know why using a new kernel worked?
$endgroup$
– Spencer Keller
May 17 at 13:21
$begingroup$
Lingering definitions ofaandb. Check them with?aand?b.
$endgroup$
– Roman
May 17 at 13:30
add a comment |
$begingroup$
Can you give the exact command you used to fit please? Otherwise it's difficult to tell what's going on. But as a first guess I'd suggest you try again with a fresh kernel.
$endgroup$
– Roman
May 17 at 12:51
$begingroup$
@Roman Updated the question.
$endgroup$
– Spencer Keller
May 17 at 12:55
2
$begingroup$
Runs fine on my computer. Please try your own code with a fresh kernel. Check?aand?bto see if there are any lingering definitions.
$endgroup$
– Roman
May 17 at 13:01
$begingroup$
@Roman I took your advice and used a new kernel. Worked just fine. Would you happen to know why using a new kernel worked?
$endgroup$
– Spencer Keller
May 17 at 13:21
$begingroup$
Lingering definitions ofaandb. Check them with?aand?b.
$endgroup$
– Roman
May 17 at 13:30
$begingroup$
Can you give the exact command you used to fit please? Otherwise it's difficult to tell what's going on. But as a first guess I'd suggest you try again with a fresh kernel.
$endgroup$
– Roman
May 17 at 12:51
$begingroup$
Can you give the exact command you used to fit please? Otherwise it's difficult to tell what's going on. But as a first guess I'd suggest you try again with a fresh kernel.
$endgroup$
– Roman
May 17 at 12:51
$begingroup$
@Roman Updated the question.
$endgroup$
– Spencer Keller
May 17 at 12:55
$begingroup$
@Roman Updated the question.
$endgroup$
– Spencer Keller
May 17 at 12:55
2
2
$begingroup$
Runs fine on my computer. Please try your own code with a fresh kernel. Check
?a and ?b to see if there are any lingering definitions.$endgroup$
– Roman
May 17 at 13:01
$begingroup$
Runs fine on my computer. Please try your own code with a fresh kernel. Check
?a and ?b to see if there are any lingering definitions.$endgroup$
– Roman
May 17 at 13:01
$begingroup$
@Roman I took your advice and used a new kernel. Worked just fine. Would you happen to know why using a new kernel worked?
$endgroup$
– Spencer Keller
May 17 at 13:21
$begingroup$
@Roman I took your advice and used a new kernel. Worked just fine. Would you happen to know why using a new kernel worked?
$endgroup$
– Spencer Keller
May 17 at 13:21
$begingroup$
Lingering definitions of
a and b. Check them with ?a and ?b.$endgroup$
– Roman
May 17 at 13:30
$begingroup$
Lingering definitions of
a and b. Check them with ?a and ?b.$endgroup$
– Roman
May 17 at 13:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Try
fit = FindFit[data, a + b Exp[c x], a, b, c, x]
Show[ListPlot[data],Plot[a + b Exp[c x] /. fit, x, 1, Length[data]]]
answered May 17 at 12:56
Ulrich NeumannUlrich Neumann
10.1k617
$endgroup$
$begingroup$
Thank you. That worked beautifully!
$endgroup$
– Spencer Keller
May 17 at 13:01
$begingroup$
You're welcome.
$endgroup$
– Ulrich Neumann
May 17 at 13:02
$begingroup$
Or, equivalently, useExp[a*x + c] + bfor the model.
$endgroup$
– Bob Hanlon
May 17 at 13:20
add a comment |
$begingroup$
This is just an extended comment. While the fit might "look" good, an examination of the residuals vs the fit shows that there is still a lot more structure that maybe needs explaining:
ListPlot[Transpose[fit["PredictedResponse"], fit["FitResiduals"]],
Frame -> True, FrameLabel -> "Predicted response", "Fit residual"]
Three (of many) possibilities: (1) underlying curve is more complicated than first thought and (2) the measurement system has a floating bias, (3) the observations are correlated across the predictor variable (maybe that's "time"?).
The point is that to make inferences from regressions certain assumptions need to hold at least approximately. Such residual plots are just one kind of check on those assumptions to see if there are any gross deviations from those assumptions. A residual plot should be mandatory.
answered May 17 at 15:01
JimBJimB
19k12863
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try
fit = FindFit[data, a + b Exp[c x], a, b, c, x]
Show[ListPlot[data],Plot[a + b Exp[c x] /. fit, x, 1, Length[data]]]
answered May 17 at 12:56
Ulrich NeumannUlrich Neumann
10.1k617
$endgroup$
$begingroup$
Thank you. That worked beautifully!
$endgroup$
– Spencer Keller
May 17 at 13:01
$begingroup$
You're welcome.
$endgroup$
– Ulrich Neumann
May 17 at 13:02
$begingroup$
Or, equivalently, useExp[a*x + c] + bfor the model.
$endgroup$
– Bob Hanlon
May 17 at 13:20
add a comment |
$begingroup$
Try
fit = FindFit[data, a + b Exp[c x], a, b, c, x]
Show[ListPlot[data],Plot[a + b Exp[c x] /. fit, x, 1, Length[data]]]
answered May 17 at 12:56
Ulrich NeumannUlrich Neumann
10.1k617
$endgroup$
$begingroup$
Thank you. That worked beautifully!
$endgroup$
– Spencer Keller
May 17 at 13:01
$begingroup$
You're welcome.
$endgroup$
– Ulrich Neumann
May 17 at 13:02
$begingroup$
Or, equivalently, useExp[a*x + c] + bfor the model.
$endgroup$
– Bob Hanlon
May 17 at 13:20
add a comment |
$begingroup$
Try
fit = FindFit[data, a + b Exp[c x], a, b, c, x]
Show[ListPlot[data],Plot[a + b Exp[c x] /. fit, x, 1, Length[data]]]
answered May 17 at 12:56
Ulrich NeumannUlrich Neumann
10.1k617
$endgroup$
Try
fit = FindFit[data, a + b Exp[c x], a, b, c, x]
Show[ListPlot[data],Plot[a + b Exp[c x] /. fit, x, 1, Length[data]]]
answered May 17 at 12:56
Ulrich NeumannUlrich Neumann
10.1k617
answered May 17 at 12:56
Ulrich NeumannUlrich Neumann
10.1k617
answered May 17 at 12:56
Ulrich NeumannUlrich Neumann
10.1k617
answered May 17 at 12:56
Ulrich NeumannUlrich Neumann
10.1k617
10.1k617
$begingroup$
Thank you. That worked beautifully!
$endgroup$
– Spencer Keller
May 17 at 13:01
$begingroup$
You're welcome.
$endgroup$
– Ulrich Neumann
May 17 at 13:02
$begingroup$
Or, equivalently, useExp[a*x + c] + bfor the model.
$endgroup$
– Bob Hanlon
May 17 at 13:20
add a comment |
$begingroup$
Thank you. That worked beautifully!
$endgroup$
– Spencer Keller
May 17 at 13:01
$begingroup$
You're welcome.
$endgroup$
– Ulrich Neumann
May 17 at 13:02
$begingroup$
Or, equivalently, useExp[a*x + c] + bfor the model.
$endgroup$
– Bob Hanlon
May 17 at 13:20
$begingroup$
Thank you. That worked beautifully!
$endgroup$
– Spencer Keller
May 17 at 13:01
$begingroup$
Thank you. That worked beautifully!
$endgroup$
– Spencer Keller
May 17 at 13:01
$begingroup$
You're welcome.
$endgroup$
– Ulrich Neumann
May 17 at 13:02
$begingroup$
You're welcome.
$endgroup$
– Ulrich Neumann
May 17 at 13:02
$begingroup$
Or, equivalently, use
Exp[a*x + c] + b for the model.$endgroup$
– Bob Hanlon
May 17 at 13:20
$begingroup$
Or, equivalently, use
Exp[a*x + c] + b for the model.$endgroup$
– Bob Hanlon
May 17 at 13:20
add a comment |
$begingroup$
This is just an extended comment. While the fit might "look" good, an examination of the residuals vs the fit shows that there is still a lot more structure that maybe needs explaining:
ListPlot[Transpose[fit["PredictedResponse"], fit["FitResiduals"]],
Frame -> True, FrameLabel -> "Predicted response", "Fit residual"]
Three (of many) possibilities: (1) underlying curve is more complicated than first thought and (2) the measurement system has a floating bias, (3) the observations are correlated across the predictor variable (maybe that's "time"?).
The point is that to make inferences from regressions certain assumptions need to hold at least approximately. Such residual plots are just one kind of check on those assumptions to see if there are any gross deviations from those assumptions. A residual plot should be mandatory.
answered May 17 at 15:01
JimBJimB
19k12863
$endgroup$
add a comment |
$begingroup$
This is just an extended comment. While the fit might "look" good, an examination of the residuals vs the fit shows that there is still a lot more structure that maybe needs explaining:
ListPlot[Transpose[fit["PredictedResponse"], fit["FitResiduals"]],
Frame -> True, FrameLabel -> "Predicted response", "Fit residual"]
Three (of many) possibilities: (1) underlying curve is more complicated than first thought and (2) the measurement system has a floating bias, (3) the observations are correlated across the predictor variable (maybe that's "time"?).
The point is that to make inferences from regressions certain assumptions need to hold at least approximately. Such residual plots are just one kind of check on those assumptions to see if there are any gross deviations from those assumptions. A residual plot should be mandatory.
answered May 17 at 15:01
JimBJimB
19k12863
$endgroup$
add a comment |
$begingroup$
This is just an extended comment. While the fit might "look" good, an examination of the residuals vs the fit shows that there is still a lot more structure that maybe needs explaining:
ListPlot[Transpose[fit["PredictedResponse"], fit["FitResiduals"]],
Frame -> True, FrameLabel -> "Predicted response", "Fit residual"]
Three (of many) possibilities: (1) underlying curve is more complicated than first thought and (2) the measurement system has a floating bias, (3) the observations are correlated across the predictor variable (maybe that's "time"?).
The point is that to make inferences from regressions certain assumptions need to hold at least approximately. Such residual plots are just one kind of check on those assumptions to see if there are any gross deviations from those assumptions. A residual plot should be mandatory.
answered May 17 at 15:01
JimBJimB
19k12863
$endgroup$
This is just an extended comment. While the fit might "look" good, an examination of the residuals vs the fit shows that there is still a lot more structure that maybe needs explaining:
ListPlot[Transpose[fit["PredictedResponse"], fit["FitResiduals"]],
Frame -> True, FrameLabel -> "Predicted response", "Fit residual"]
Three (of many) possibilities: (1) underlying curve is more complicated than first thought and (2) the measurement system has a floating bias, (3) the observations are correlated across the predictor variable (maybe that's "time"?).
The point is that to make inferences from regressions certain assumptions need to hold at least approximately. Such residual plots are just one kind of check on those assumptions to see if there are any gross deviations from those assumptions. A residual plot should be mandatory.
answered May 17 at 15:01
JimBJimB
19k12863
answered May 17 at 15:01
JimBJimB
19k12863
answered May 17 at 15:01
JimBJimB
19k12863
answered May 17 at 15:01
JimBJimB
19k12863
19k12863
add a comment |
add a comment |
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$begingroup$
Can you give the exact command you used to fit please? Otherwise it's difficult to tell what's going on. But as a first guess I'd suggest you try again with a fresh kernel.
$endgroup$
– Roman
May 17 at 12:51
$begingroup$
@Roman Updated the question.
$endgroup$
– Spencer Keller
May 17 at 12:55
2
$begingroup$
Runs fine on my computer. Please try your own code with a fresh kernel. Check
?aand?bto see if there are any lingering definitions.$endgroup$
– Roman
May 17 at 13:01
$begingroup$
@Roman I took your advice and used a new kernel. Worked just fine. Would you happen to know why using a new kernel worked?
$endgroup$
– Spencer Keller
May 17 at 13:21
$begingroup$
Lingering definitions of
aandb. Check them with?aand?b.$endgroup$
– Roman
May 17 at 13:30