Problem in using FindFit
I have the following set of data
data=0,0,0,0,2,1,0,4,2.247,0,6,3.627,0,8,5.031,1,0,3.346;
where the values are n, L,$varepsilon$ and satisfy the following equations
$E(n,L) = 2n+1 + sqrtL(L+1)-frac34(L)^2 + 1 + beta_0^4$
e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]
$varepsilon = fracE(n,L)-E(0,0)E(0,2)-E(0,0)$,
where $beta_0$ should be determined. I don't know how I can use FindFit command of Mathematica to find the best value of $beta_0$ to have the best fit for $varepsilon$.
fitting
add a comment |
I have the following set of data
data=0,0,0,0,2,1,0,4,2.247,0,6,3.627,0,8,5.031,1,0,3.346;
where the values are n, L,$varepsilon$ and satisfy the following equations
$E(n,L) = 2n+1 + sqrtL(L+1)-frac34(L)^2 + 1 + beta_0^4$
e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]
$varepsilon = fracE(n,L)-E(0,0)E(0,2)-E(0,0)$,
where $beta_0$ should be determined. I don't know how I can use FindFit command of Mathematica to find the best value of $beta_0$ to have the best fit for $varepsilon$.
fitting
add a comment |
I have the following set of data
data=0,0,0,0,2,1,0,4,2.247,0,6,3.627,0,8,5.031,1,0,3.346;
where the values are n, L,$varepsilon$ and satisfy the following equations
$E(n,L) = 2n+1 + sqrtL(L+1)-frac34(L)^2 + 1 + beta_0^4$
e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]
$varepsilon = fracE(n,L)-E(0,0)E(0,2)-E(0,0)$,
where $beta_0$ should be determined. I don't know how I can use FindFit command of Mathematica to find the best value of $beta_0$ to have the best fit for $varepsilon$.
fitting
I have the following set of data
data=0,0,0,0,2,1,0,4,2.247,0,6,3.627,0,8,5.031,1,0,3.346;
where the values are n, L,$varepsilon$ and satisfy the following equations
$E(n,L) = 2n+1 + sqrtL(L+1)-frac34(L)^2 + 1 + beta_0^4$
e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]
$varepsilon = fracE(n,L)-E(0,0)E(0,2)-E(0,0)$,
where $beta_0$ should be determined. I don't know how I can use FindFit command of Mathematica to find the best value of $beta_0$ to have the best fit for $varepsilon$.
fitting
fitting
edited Nov 12 at 7:19
Coolwater
14.6k32552
14.6k32552
asked Nov 12 at 6:13
Hadi Sobhani
32417
32417
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]
FindFit[data, (e[n, L] - e[0, 0])/(e[0, 2] - e[0, 0]), b0, n, L]
b0 -> 1.3514967
Which seems reasonable in view of the residuals:
Plot[Evaluate[(e[#, #2] - e[0, 0])/(e[0, 2] - e[0, 0]) - #3 & @@@ data], b0, 0, 3]
The brown and purple residual has bigger slope around the roots in the plots. Hence for Mathematica to minimize the sum of squares in the y-dimension, the mean of the 2 data points that correspond to the big slopes are cared more about than the others. It is purpose specific whether this is appropriate. If it isn't you can add the NormFunction
-option to FindFit
.
Thank you dear @Coolwater. How does Mathematica recognize n,L for each data?
– Hadi Sobhani
Nov 12 at 7:25
Also, how does it recognize that the expr is the 3rd value of every data element?
– J42161217
Nov 12 at 7:34
1
FindFit
assumes its first argument has the formvar1, var2, ..., varN, expr, ... , var1, var2, ..., varN, expr
wherevar1, var2, ..., varN
is the 4th argument ofFindFit
– Coolwater
Nov 12 at 7:36
Ok! Given b -> 1.27225, 1.29505, 1.28573, 1.40411 having Mean=1.31428 and Medean=1.29039 do you think Mathematica did a good job? Anyway +1 from me
– J42161217
Nov 12 at 7:52
@J42161217 It uses least squares, see edit
– Coolwater
Nov 12 at 8:05
add a comment |
You can also use NMinimize
. First we need to write cost function, i.e. residual.
data = 0, 0, 0, 0, 2, 1, 0, 4, 2.247, 0, 6, 3.627, 0, 8,
5.031, 1, 0, 3.346;
e[n_, L_] := 2 n + 1 + Sqrt[L (L + 1) - 3/4 L^2 + 1 + b0^4]
cost[b0_] =Sum[(e @@data[[i, 1 ;; 2]] - (data[[i, 3]] (e[0, 2] - e[0, 0]) +
e[0, 0]))^2, i, 6];
(*or Total[(e[#1, #2] - (#3 (e[0, 2] - e[0, 0]) + e[0, 0]))^2 & @@@ data]*)
fit = NMinimize[cost[b0] , b0]
0.0196376, b0 -> 1.35462
Since your cost function has only one variable you can also use grid search.
Ordering[val,1]
gives position of min value.
b0Val = Range[0, 10, 0.0001];
val = cost[b0Val];
b0Val[[Ordering[val, 1]]]
1.3546
Note that there is another min at b0=-1.3546
b0Val = Range[-1000, 1000, 0.001];
val = cost[b0Val];
b0Val[[Ordering[val, 2]]]
-1.3546, 1.3546
We can plot cost function
$textcost(b0)=left(-5.031 left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+25right)^2\+left(-3.627
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+
sqrttextb0^4+16right)^2\+left(2-3.346
left(sqrttextb0^4+4-sqrttextb0^4+1right)right)^2+left(-2.247
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+9right)^2$
Plot[cost[b0], b0, -10, 10]
add a comment |
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2 Answers
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2 Answers
2
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e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]
FindFit[data, (e[n, L] - e[0, 0])/(e[0, 2] - e[0, 0]), b0, n, L]
b0 -> 1.3514967
Which seems reasonable in view of the residuals:
Plot[Evaluate[(e[#, #2] - e[0, 0])/(e[0, 2] - e[0, 0]) - #3 & @@@ data], b0, 0, 3]
The brown and purple residual has bigger slope around the roots in the plots. Hence for Mathematica to minimize the sum of squares in the y-dimension, the mean of the 2 data points that correspond to the big slopes are cared more about than the others. It is purpose specific whether this is appropriate. If it isn't you can add the NormFunction
-option to FindFit
.
Thank you dear @Coolwater. How does Mathematica recognize n,L for each data?
– Hadi Sobhani
Nov 12 at 7:25
Also, how does it recognize that the expr is the 3rd value of every data element?
– J42161217
Nov 12 at 7:34
1
FindFit
assumes its first argument has the formvar1, var2, ..., varN, expr, ... , var1, var2, ..., varN, expr
wherevar1, var2, ..., varN
is the 4th argument ofFindFit
– Coolwater
Nov 12 at 7:36
Ok! Given b -> 1.27225, 1.29505, 1.28573, 1.40411 having Mean=1.31428 and Medean=1.29039 do you think Mathematica did a good job? Anyway +1 from me
– J42161217
Nov 12 at 7:52
@J42161217 It uses least squares, see edit
– Coolwater
Nov 12 at 8:05
add a comment |
e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]
FindFit[data, (e[n, L] - e[0, 0])/(e[0, 2] - e[0, 0]), b0, n, L]
b0 -> 1.3514967
Which seems reasonable in view of the residuals:
Plot[Evaluate[(e[#, #2] - e[0, 0])/(e[0, 2] - e[0, 0]) - #3 & @@@ data], b0, 0, 3]
The brown and purple residual has bigger slope around the roots in the plots. Hence for Mathematica to minimize the sum of squares in the y-dimension, the mean of the 2 data points that correspond to the big slopes are cared more about than the others. It is purpose specific whether this is appropriate. If it isn't you can add the NormFunction
-option to FindFit
.
Thank you dear @Coolwater. How does Mathematica recognize n,L for each data?
– Hadi Sobhani
Nov 12 at 7:25
Also, how does it recognize that the expr is the 3rd value of every data element?
– J42161217
Nov 12 at 7:34
1
FindFit
assumes its first argument has the formvar1, var2, ..., varN, expr, ... , var1, var2, ..., varN, expr
wherevar1, var2, ..., varN
is the 4th argument ofFindFit
– Coolwater
Nov 12 at 7:36
Ok! Given b -> 1.27225, 1.29505, 1.28573, 1.40411 having Mean=1.31428 and Medean=1.29039 do you think Mathematica did a good job? Anyway +1 from me
– J42161217
Nov 12 at 7:52
@J42161217 It uses least squares, see edit
– Coolwater
Nov 12 at 8:05
add a comment |
e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]
FindFit[data, (e[n, L] - e[0, 0])/(e[0, 2] - e[0, 0]), b0, n, L]
b0 -> 1.3514967
Which seems reasonable in view of the residuals:
Plot[Evaluate[(e[#, #2] - e[0, 0])/(e[0, 2] - e[0, 0]) - #3 & @@@ data], b0, 0, 3]
The brown and purple residual has bigger slope around the roots in the plots. Hence for Mathematica to minimize the sum of squares in the y-dimension, the mean of the 2 data points that correspond to the big slopes are cared more about than the others. It is purpose specific whether this is appropriate. If it isn't you can add the NormFunction
-option to FindFit
.
e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]
FindFit[data, (e[n, L] - e[0, 0])/(e[0, 2] - e[0, 0]), b0, n, L]
b0 -> 1.3514967
Which seems reasonable in view of the residuals:
Plot[Evaluate[(e[#, #2] - e[0, 0])/(e[0, 2] - e[0, 0]) - #3 & @@@ data], b0, 0, 3]
The brown and purple residual has bigger slope around the roots in the plots. Hence for Mathematica to minimize the sum of squares in the y-dimension, the mean of the 2 data points that correspond to the big slopes are cared more about than the others. It is purpose specific whether this is appropriate. If it isn't you can add the NormFunction
-option to FindFit
.
edited Nov 12 at 8:02
answered Nov 12 at 7:22
Coolwater
14.6k32552
14.6k32552
Thank you dear @Coolwater. How does Mathematica recognize n,L for each data?
– Hadi Sobhani
Nov 12 at 7:25
Also, how does it recognize that the expr is the 3rd value of every data element?
– J42161217
Nov 12 at 7:34
1
FindFit
assumes its first argument has the formvar1, var2, ..., varN, expr, ... , var1, var2, ..., varN, expr
wherevar1, var2, ..., varN
is the 4th argument ofFindFit
– Coolwater
Nov 12 at 7:36
Ok! Given b -> 1.27225, 1.29505, 1.28573, 1.40411 having Mean=1.31428 and Medean=1.29039 do you think Mathematica did a good job? Anyway +1 from me
– J42161217
Nov 12 at 7:52
@J42161217 It uses least squares, see edit
– Coolwater
Nov 12 at 8:05
add a comment |
Thank you dear @Coolwater. How does Mathematica recognize n,L for each data?
– Hadi Sobhani
Nov 12 at 7:25
Also, how does it recognize that the expr is the 3rd value of every data element?
– J42161217
Nov 12 at 7:34
1
FindFit
assumes its first argument has the formvar1, var2, ..., varN, expr, ... , var1, var2, ..., varN, expr
wherevar1, var2, ..., varN
is the 4th argument ofFindFit
– Coolwater
Nov 12 at 7:36
Ok! Given b -> 1.27225, 1.29505, 1.28573, 1.40411 having Mean=1.31428 and Medean=1.29039 do you think Mathematica did a good job? Anyway +1 from me
– J42161217
Nov 12 at 7:52
@J42161217 It uses least squares, see edit
– Coolwater
Nov 12 at 8:05
Thank you dear @Coolwater. How does Mathematica recognize n,L for each data?
– Hadi Sobhani
Nov 12 at 7:25
Thank you dear @Coolwater. How does Mathematica recognize n,L for each data?
– Hadi Sobhani
Nov 12 at 7:25
Also, how does it recognize that the expr is the 3rd value of every data element?
– J42161217
Nov 12 at 7:34
Also, how does it recognize that the expr is the 3rd value of every data element?
– J42161217
Nov 12 at 7:34
1
1
FindFit
assumes its first argument has the form var1, var2, ..., varN, expr, ... , var1, var2, ..., varN, expr
where var1, var2, ..., varN
is the 4th argument of FindFit
– Coolwater
Nov 12 at 7:36
FindFit
assumes its first argument has the form var1, var2, ..., varN, expr, ... , var1, var2, ..., varN, expr
where var1, var2, ..., varN
is the 4th argument of FindFit
– Coolwater
Nov 12 at 7:36
Ok! Given b -> 1.27225, 1.29505, 1.28573, 1.40411 having Mean=1.31428 and Medean=1.29039 do you think Mathematica did a good job? Anyway +1 from me
– J42161217
Nov 12 at 7:52
Ok! Given b -> 1.27225, 1.29505, 1.28573, 1.40411 having Mean=1.31428 and Medean=1.29039 do you think Mathematica did a good job? Anyway +1 from me
– J42161217
Nov 12 at 7:52
@J42161217 It uses least squares, see edit
– Coolwater
Nov 12 at 8:05
@J42161217 It uses least squares, see edit
– Coolwater
Nov 12 at 8:05
add a comment |
You can also use NMinimize
. First we need to write cost function, i.e. residual.
data = 0, 0, 0, 0, 2, 1, 0, 4, 2.247, 0, 6, 3.627, 0, 8,
5.031, 1, 0, 3.346;
e[n_, L_] := 2 n + 1 + Sqrt[L (L + 1) - 3/4 L^2 + 1 + b0^4]
cost[b0_] =Sum[(e @@data[[i, 1 ;; 2]] - (data[[i, 3]] (e[0, 2] - e[0, 0]) +
e[0, 0]))^2, i, 6];
(*or Total[(e[#1, #2] - (#3 (e[0, 2] - e[0, 0]) + e[0, 0]))^2 & @@@ data]*)
fit = NMinimize[cost[b0] , b0]
0.0196376, b0 -> 1.35462
Since your cost function has only one variable you can also use grid search.
Ordering[val,1]
gives position of min value.
b0Val = Range[0, 10, 0.0001];
val = cost[b0Val];
b0Val[[Ordering[val, 1]]]
1.3546
Note that there is another min at b0=-1.3546
b0Val = Range[-1000, 1000, 0.001];
val = cost[b0Val];
b0Val[[Ordering[val, 2]]]
-1.3546, 1.3546
We can plot cost function
$textcost(b0)=left(-5.031 left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+25right)^2\+left(-3.627
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+
sqrttextb0^4+16right)^2\+left(2-3.346
left(sqrttextb0^4+4-sqrttextb0^4+1right)right)^2+left(-2.247
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+9right)^2$
Plot[cost[b0], b0, -10, 10]
add a comment |
You can also use NMinimize
. First we need to write cost function, i.e. residual.
data = 0, 0, 0, 0, 2, 1, 0, 4, 2.247, 0, 6, 3.627, 0, 8,
5.031, 1, 0, 3.346;
e[n_, L_] := 2 n + 1 + Sqrt[L (L + 1) - 3/4 L^2 + 1 + b0^4]
cost[b0_] =Sum[(e @@data[[i, 1 ;; 2]] - (data[[i, 3]] (e[0, 2] - e[0, 0]) +
e[0, 0]))^2, i, 6];
(*or Total[(e[#1, #2] - (#3 (e[0, 2] - e[0, 0]) + e[0, 0]))^2 & @@@ data]*)
fit = NMinimize[cost[b0] , b0]
0.0196376, b0 -> 1.35462
Since your cost function has only one variable you can also use grid search.
Ordering[val,1]
gives position of min value.
b0Val = Range[0, 10, 0.0001];
val = cost[b0Val];
b0Val[[Ordering[val, 1]]]
1.3546
Note that there is another min at b0=-1.3546
b0Val = Range[-1000, 1000, 0.001];
val = cost[b0Val];
b0Val[[Ordering[val, 2]]]
-1.3546, 1.3546
We can plot cost function
$textcost(b0)=left(-5.031 left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+25right)^2\+left(-3.627
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+
sqrttextb0^4+16right)^2\+left(2-3.346
left(sqrttextb0^4+4-sqrttextb0^4+1right)right)^2+left(-2.247
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+9right)^2$
Plot[cost[b0], b0, -10, 10]
add a comment |
You can also use NMinimize
. First we need to write cost function, i.e. residual.
data = 0, 0, 0, 0, 2, 1, 0, 4, 2.247, 0, 6, 3.627, 0, 8,
5.031, 1, 0, 3.346;
e[n_, L_] := 2 n + 1 + Sqrt[L (L + 1) - 3/4 L^2 + 1 + b0^4]
cost[b0_] =Sum[(e @@data[[i, 1 ;; 2]] - (data[[i, 3]] (e[0, 2] - e[0, 0]) +
e[0, 0]))^2, i, 6];
(*or Total[(e[#1, #2] - (#3 (e[0, 2] - e[0, 0]) + e[0, 0]))^2 & @@@ data]*)
fit = NMinimize[cost[b0] , b0]
0.0196376, b0 -> 1.35462
Since your cost function has only one variable you can also use grid search.
Ordering[val,1]
gives position of min value.
b0Val = Range[0, 10, 0.0001];
val = cost[b0Val];
b0Val[[Ordering[val, 1]]]
1.3546
Note that there is another min at b0=-1.3546
b0Val = Range[-1000, 1000, 0.001];
val = cost[b0Val];
b0Val[[Ordering[val, 2]]]
-1.3546, 1.3546
We can plot cost function
$textcost(b0)=left(-5.031 left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+25right)^2\+left(-3.627
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+
sqrttextb0^4+16right)^2\+left(2-3.346
left(sqrttextb0^4+4-sqrttextb0^4+1right)right)^2+left(-2.247
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+9right)^2$
Plot[cost[b0], b0, -10, 10]
You can also use NMinimize
. First we need to write cost function, i.e. residual.
data = 0, 0, 0, 0, 2, 1, 0, 4, 2.247, 0, 6, 3.627, 0, 8,
5.031, 1, 0, 3.346;
e[n_, L_] := 2 n + 1 + Sqrt[L (L + 1) - 3/4 L^2 + 1 + b0^4]
cost[b0_] =Sum[(e @@data[[i, 1 ;; 2]] - (data[[i, 3]] (e[0, 2] - e[0, 0]) +
e[0, 0]))^2, i, 6];
(*or Total[(e[#1, #2] - (#3 (e[0, 2] - e[0, 0]) + e[0, 0]))^2 & @@@ data]*)
fit = NMinimize[cost[b0] , b0]
0.0196376, b0 -> 1.35462
Since your cost function has only one variable you can also use grid search.
Ordering[val,1]
gives position of min value.
b0Val = Range[0, 10, 0.0001];
val = cost[b0Val];
b0Val[[Ordering[val, 1]]]
1.3546
Note that there is another min at b0=-1.3546
b0Val = Range[-1000, 1000, 0.001];
val = cost[b0Val];
b0Val[[Ordering[val, 2]]]
-1.3546, 1.3546
We can plot cost function
$textcost(b0)=left(-5.031 left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+25right)^2\+left(-3.627
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+
sqrttextb0^4+16right)^2\+left(2-3.346
left(sqrttextb0^4+4-sqrttextb0^4+1right)right)^2+left(-2.247
left(sqrttextb0^4+4-sqrttextb0^4+1right)-sqrttextb0^4+1+sqrttextb0^4+9right)^2$
Plot[cost[b0], b0, -10, 10]
edited Nov 12 at 15:38
answered Nov 12 at 12:05
Okkes Dulgerci
3,7951816
3,7951816
add a comment |
add a comment |
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