Warning Mathematica FittedModel: The precision of the argument function (MachinePrecision) is less than...
0
Fellows,
I couldn't figure why I am having the warning message from the following code in Mathematica:
data = {{0, 1}, {1, 0.02307044673005989`}, {2,
0.00784879347316981`}, {3, 0.0061305265946403195`}, {4,
0.0008550610216054799`}, {5, 0.00010928133254420425`}, {6,
0.000011431049984759768`}, {7, 1.93788101788827`*^-6}, {8,
1.6278670621771263`*^-6}, {9, 2.6661469926370584`*^-7}, {10,
3.443821224260662`*^-8}, {11, 7.413060538191399`*^-9}, {12,
1.4031525687948224`*^-9}, {13, 5.973790450062338`*^-10}, {14,
1.7434844383850214`*^-10}, {15, 2.6053424128998922`*^-11}, {16,
9.887095524831592`*^-12}, {17, 1.2318024865446659`*^-12}, {18,
2.2125640342387203`*^-13}, {19, 1.3176590670511745`*^-13}, {20,
2.7354393146500743`*^-14}};
fit = NonlinearModelFit[data, a + b Exp[-x/c], {a, b, c}, x,
MaxIterations -> [Infinity], PrecisionGoal -> MachinePrecision,
WorkingPrecision -> MachinePrecision];
fit["BestFitParameters"] (* THE WARNING APPEARS AFTER CALLING THIS FUNCTION *)
The warning message is:
FittedModel: The precision of the argument function (MachinePrecision) is less than WorkingPrecision (MachinePrecision).
Thanks in advance.
wolfram-mathematica precision non-linear-regression
asked Nov 23 '18 at 8:48
T.O.PuelT.O.Puel
52
add a comment |
0
Fellows,
I couldn't figure why I am having the warning message from the following code in Mathematica:
data = {{0, 1}, {1, 0.02307044673005989`}, {2,
0.00784879347316981`}, {3, 0.0061305265946403195`}, {4,
0.0008550610216054799`}, {5, 0.00010928133254420425`}, {6,
0.000011431049984759768`}, {7, 1.93788101788827`*^-6}, {8,
1.6278670621771263`*^-6}, {9, 2.6661469926370584`*^-7}, {10,
3.443821224260662`*^-8}, {11, 7.413060538191399`*^-9}, {12,
1.4031525687948224`*^-9}, {13, 5.973790450062338`*^-10}, {14,
1.7434844383850214`*^-10}, {15, 2.6053424128998922`*^-11}, {16,
9.887095524831592`*^-12}, {17, 1.2318024865446659`*^-12}, {18,
2.2125640342387203`*^-13}, {19, 1.3176590670511745`*^-13}, {20,
2.7354393146500743`*^-14}};
fit = NonlinearModelFit[data, a + b Exp[-x/c], {a, b, c}, x,
MaxIterations -> [Infinity], PrecisionGoal -> MachinePrecision,
WorkingPrecision -> MachinePrecision];
fit["BestFitParameters"] (* THE WARNING APPEARS AFTER CALLING THIS FUNCTION *)
The warning message is:
FittedModel: The precision of the argument function (MachinePrecision) is less than WorkingPrecision (MachinePrecision).
Thanks in advance.
wolfram-mathematica precision non-linear-regression
asked Nov 23 '18 at 8:48
T.O.PuelT.O.Puel
52
add a comment |
0
0
0
Fellows,
I couldn't figure why I am having the warning message from the following code in Mathematica:
data = {{0, 1}, {1, 0.02307044673005989`}, {2,
0.00784879347316981`}, {3, 0.0061305265946403195`}, {4,
0.0008550610216054799`}, {5, 0.00010928133254420425`}, {6,
0.000011431049984759768`}, {7, 1.93788101788827`*^-6}, {8,
1.6278670621771263`*^-6}, {9, 2.6661469926370584`*^-7}, {10,
3.443821224260662`*^-8}, {11, 7.413060538191399`*^-9}, {12,
1.4031525687948224`*^-9}, {13, 5.973790450062338`*^-10}, {14,
1.7434844383850214`*^-10}, {15, 2.6053424128998922`*^-11}, {16,
9.887095524831592`*^-12}, {17, 1.2318024865446659`*^-12}, {18,
2.2125640342387203`*^-13}, {19, 1.3176590670511745`*^-13}, {20,
2.7354393146500743`*^-14}};
fit = NonlinearModelFit[data, a + b Exp[-x/c], {a, b, c}, x,
MaxIterations -> [Infinity], PrecisionGoal -> MachinePrecision,
WorkingPrecision -> MachinePrecision];
fit["BestFitParameters"] (* THE WARNING APPEARS AFTER CALLING THIS FUNCTION *)
The warning message is:
FittedModel: The precision of the argument function (MachinePrecision) is less than WorkingPrecision (MachinePrecision).
Thanks in advance.
wolfram-mathematica precision non-linear-regression
asked Nov 23 '18 at 8:48
T.O.PuelT.O.Puel
52
Fellows,
I couldn't figure why I am having the warning message from the following code in Mathematica:
data = {{0, 1}, {1, 0.02307044673005989`}, {2,
0.00784879347316981`}, {3, 0.0061305265946403195`}, {4,
0.0008550610216054799`}, {5, 0.00010928133254420425`}, {6,
0.000011431049984759768`}, {7, 1.93788101788827`*^-6}, {8,
1.6278670621771263`*^-6}, {9, 2.6661469926370584`*^-7}, {10,
3.443821224260662`*^-8}, {11, 7.413060538191399`*^-9}, {12,
1.4031525687948224`*^-9}, {13, 5.973790450062338`*^-10}, {14,
1.7434844383850214`*^-10}, {15, 2.6053424128998922`*^-11}, {16,
9.887095524831592`*^-12}, {17, 1.2318024865446659`*^-12}, {18,
2.2125640342387203`*^-13}, {19, 1.3176590670511745`*^-13}, {20,
2.7354393146500743`*^-14}};
fit = NonlinearModelFit[data, a + b Exp[-x/c], {a, b, c}, x,
MaxIterations -> [Infinity], PrecisionGoal -> MachinePrecision,
WorkingPrecision -> MachinePrecision];
fit["BestFitParameters"] (* THE WARNING APPEARS AFTER CALLING THIS FUNCTION *)
The warning message is:
FittedModel: The precision of the argument function (MachinePrecision) is less than WorkingPrecision (MachinePrecision).
Thanks in advance.
wolfram-mathematica precision non-linear-regression
wolfram-mathematica precision non-linear-regression
asked Nov 23 '18 at 8:48
T.O.PuelT.O.Puel
52
asked Nov 23 '18 at 8:48
T.O.PuelT.O.Puel
52
asked Nov 23 '18 at 8:48
T.O.PuelT.O.Puel
52
asked Nov 23 '18 at 8:48
T.O.PuelT.O.Puel
52
asked Nov 23 '18 at 8:48
T.O.PuelT.O.Puel
52
52
add a comment |
add a comment |
1 Answer
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The problem is that many of the data values are small and close to machine precision. You can try linear fitting to the Log of the data values.
data2 = {First[#], Log[Last[#]]} & /@ data;
lm = LinearModelFit[data2, x, x]
Show[ListPlot[data2], Plot[lm[x], {x, 0, 20}]]
The first data point {0, 1} does not look right. Are you sure it is correct?
answered Nov 24 '18 at 0:27
Rohit NamjoshiRohit Namjoshi
32119
Thanks @Rohit. Besides still not understanding the reason, I'll keep in mind that the small numbers + machine precision issue is not a good match. I believe that, in the present case, LinearModelFit works nice because the constantaina + b Exp[-x/c]is basically zero, otherwise NonLinearModelFit would be required. Ps.: the first number is correct by construction, thanks for asking.
– T.O.Puel
Nov 26 '18 at 8:41
add a comment |
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1 Answer
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1 Answer
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active
oldest
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The problem is that many of the data values are small and close to machine precision. You can try linear fitting to the Log of the data values.
data2 = {First[#], Log[Last[#]]} & /@ data;
lm = LinearModelFit[data2, x, x]
Show[ListPlot[data2], Plot[lm[x], {x, 0, 20}]]
The first data point {0, 1} does not look right. Are you sure it is correct?
answered Nov 24 '18 at 0:27
Rohit NamjoshiRohit Namjoshi
32119
Thanks @Rohit. Besides still not understanding the reason, I'll keep in mind that the small numbers + machine precision issue is not a good match. I believe that, in the present case, LinearModelFit works nice because the constantaina + b Exp[-x/c]is basically zero, otherwise NonLinearModelFit would be required. Ps.: the first number is correct by construction, thanks for asking.
– T.O.Puel
Nov 26 '18 at 8:41
add a comment |
0
The problem is that many of the data values are small and close to machine precision. You can try linear fitting to the Log of the data values.
data2 = {First[#], Log[Last[#]]} & /@ data;
lm = LinearModelFit[data2, x, x]
Show[ListPlot[data2], Plot[lm[x], {x, 0, 20}]]
The first data point {0, 1} does not look right. Are you sure it is correct?
answered Nov 24 '18 at 0:27
Rohit NamjoshiRohit Namjoshi
32119
Thanks @Rohit. Besides still not understanding the reason, I'll keep in mind that the small numbers + machine precision issue is not a good match. I believe that, in the present case, LinearModelFit works nice because the constantaina + b Exp[-x/c]is basically zero, otherwise NonLinearModelFit would be required. Ps.: the first number is correct by construction, thanks for asking.
– T.O.Puel
Nov 26 '18 at 8:41
add a comment |
0
0
0
The problem is that many of the data values are small and close to machine precision. You can try linear fitting to the Log of the data values.
data2 = {First[#], Log[Last[#]]} & /@ data;
lm = LinearModelFit[data2, x, x]
Show[ListPlot[data2], Plot[lm[x], {x, 0, 20}]]
The first data point {0, 1} does not look right. Are you sure it is correct?
answered Nov 24 '18 at 0:27
Rohit NamjoshiRohit Namjoshi
32119
The problem is that many of the data values are small and close to machine precision. You can try linear fitting to the Log of the data values.
data2 = {First[#], Log[Last[#]]} & /@ data;
lm = LinearModelFit[data2, x, x]
Show[ListPlot[data2], Plot[lm[x], {x, 0, 20}]]
The first data point {0, 1} does not look right. Are you sure it is correct?
answered Nov 24 '18 at 0:27
Rohit NamjoshiRohit Namjoshi
32119
answered Nov 24 '18 at 0:27
Rohit NamjoshiRohit Namjoshi
32119
answered Nov 24 '18 at 0:27
Rohit NamjoshiRohit Namjoshi
32119
answered Nov 24 '18 at 0:27
Rohit NamjoshiRohit Namjoshi
32119
32119
Thanks @Rohit. Besides still not understanding the reason, I'll keep in mind that the small numbers + machine precision issue is not a good match. I believe that, in the present case, LinearModelFit works nice because the constantaina + b Exp[-x/c]is basically zero, otherwise NonLinearModelFit would be required. Ps.: the first number is correct by construction, thanks for asking.
– T.O.Puel
Nov 26 '18 at 8:41
add a comment |
Thanks @Rohit. Besides still not understanding the reason, I'll keep in mind that the small numbers + machine precision issue is not a good match. I believe that, in the present case, LinearModelFit works nice because the constantaina + b Exp[-x/c]is basically zero, otherwise NonLinearModelFit would be required. Ps.: the first number is correct by construction, thanks for asking.
– T.O.Puel
Nov 26 '18 at 8:41
Thanks @Rohit. Besides still not understanding the reason, I'll keep in mind that the small numbers + machine precision issue is not a good match. I believe that, in the present case, LinearModelFit works nice because the constant
a in a + b Exp[-x/c] is basically zero, otherwise NonLinearModelFit would be required. Ps.: the first number is correct by construction, thanks for asking.– T.O.Puel
Nov 26 '18 at 8:41
Thanks @Rohit. Besides still not understanding the reason, I'll keep in mind that the small numbers + machine precision issue is not a good match. I believe that, in the present case, LinearModelFit works nice because the constant
a in a + b Exp[-x/c] is basically zero, otherwise NonLinearModelFit would be required. Ps.: the first number is correct by construction, thanks for asking.– T.O.Puel
Nov 26 '18 at 8:41
add a comment |
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