Problem in using FindFit












4














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



$E(n,L) = 2n+1 + sqrt{L(L+1)-frac{3}{4}(L)^2 + 1 + beta_0^4}$



e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]


$varepsilon = frac{E(n,L)-E(0,0)}{E(0,2)-E(0,0)}$,



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







share|improve this question





























    4














    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



    $E(n,L) = 2n+1 + sqrt{L(L+1)-frac{3}{4}(L)^2 + 1 + beta_0^4}$



    e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]


    $varepsilon = frac{E(n,L)-E(0,0)}{E(0,2)-E(0,0)}$,



    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







    share|improve this question



























      4












      4








      4







      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



      $E(n,L) = 2n+1 + sqrt{L(L+1)-frac{3}{4}(L)^2 + 1 + beta_0^4}$



      e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]


      $varepsilon = frac{E(n,L)-E(0,0)}{E(0,2)-E(0,0)}$,



      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







      share|improve this question















      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



      $E(n,L) = 2n+1 + sqrt{L(L+1)-frac{3}{4}(L)^2 + 1 + beta_0^4}$



      e[n_, L_] = 2n + 1 + Sqrt[L(L + 1) - 3/4 L^2 + 1 + b0^4]


      $varepsilon = frac{E(n,L)-E(0,0)}{E(0,2)-E(0,0)}$,



      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 12 '18 at 7:19









      Coolwater

      14.7k32553




      14.7k32553










      asked Nov 12 '18 at 6:13









      Hadi Sobhani

      37327




      37327






















          2 Answers
          2






          active

          oldest

          votes


















          6














          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.






          share|improve this answer























          • Thank you dear @Coolwater. How does Mathematica recognize {n,L} for each data?
            – Hadi Sobhani
            Nov 12 '18 at 7:25










          • Also, how does it recognize that the expr is the 3rd value of every data element?
            – J42161217
            Nov 12 '18 at 7:34






          • 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 '18 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 '18 at 7:52










          • @J42161217 It uses least squares, see edit
            – Coolwater
            Nov 12 '18 at 8:05



















          2














          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



          $text{cost}(b0)=left(-5.031 left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+25}right)^2\+left(-3.627
          left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+
          sqrt{text{b0}^4+16}right)^2\+left(2-3.346
          left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)right)^2+left(-2.247
          left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+9}right)^2$



          Plot[cost[b0], {b0, -10, 10}]


          enter image description here






          share|improve this answer























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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            6














            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.






            share|improve this answer























            • Thank you dear @Coolwater. How does Mathematica recognize {n,L} for each data?
              – Hadi Sobhani
              Nov 12 '18 at 7:25










            • Also, how does it recognize that the expr is the 3rd value of every data element?
              – J42161217
              Nov 12 '18 at 7:34






            • 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 '18 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 '18 at 7:52










            • @J42161217 It uses least squares, see edit
              – Coolwater
              Nov 12 '18 at 8:05
















            6














            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.






            share|improve this answer























            • Thank you dear @Coolwater. How does Mathematica recognize {n,L} for each data?
              – Hadi Sobhani
              Nov 12 '18 at 7:25










            • Also, how does it recognize that the expr is the 3rd value of every data element?
              – J42161217
              Nov 12 '18 at 7:34






            • 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 '18 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 '18 at 7:52










            • @J42161217 It uses least squares, see edit
              – Coolwater
              Nov 12 '18 at 8:05














            6












            6








            6






            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.






            share|improve this answer














            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.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 12 '18 at 8:02

























            answered Nov 12 '18 at 7:22









            Coolwater

            14.7k32553




            14.7k32553












            • Thank you dear @Coolwater. How does Mathematica recognize {n,L} for each data?
              – Hadi Sobhani
              Nov 12 '18 at 7:25










            • Also, how does it recognize that the expr is the 3rd value of every data element?
              – J42161217
              Nov 12 '18 at 7:34






            • 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 '18 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 '18 at 7:52










            • @J42161217 It uses least squares, see edit
              – Coolwater
              Nov 12 '18 at 8:05


















            • Thank you dear @Coolwater. How does Mathematica recognize {n,L} for each data?
              – Hadi Sobhani
              Nov 12 '18 at 7:25










            • Also, how does it recognize that the expr is the 3rd value of every data element?
              – J42161217
              Nov 12 '18 at 7:34






            • 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 '18 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 '18 at 7:52










            • @J42161217 It uses least squares, see edit
              – Coolwater
              Nov 12 '18 at 8:05
















            Thank you dear @Coolwater. How does Mathematica recognize {n,L} for each data?
            – Hadi Sobhani
            Nov 12 '18 at 7:25




            Thank you dear @Coolwater. How does Mathematica recognize {n,L} for each data?
            – Hadi Sobhani
            Nov 12 '18 at 7:25












            Also, how does it recognize that the expr is the 3rd value of every data element?
            – J42161217
            Nov 12 '18 at 7:34




            Also, how does it recognize that the expr is the 3rd value of every data element?
            – J42161217
            Nov 12 '18 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 '18 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 '18 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 '18 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 '18 at 7:52












            @J42161217 It uses least squares, see edit
            – Coolwater
            Nov 12 '18 at 8:05




            @J42161217 It uses least squares, see edit
            – Coolwater
            Nov 12 '18 at 8:05











            2














            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



            $text{cost}(b0)=left(-5.031 left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+25}right)^2\+left(-3.627
            left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+
            sqrt{text{b0}^4+16}right)^2\+left(2-3.346
            left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)right)^2+left(-2.247
            left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+9}right)^2$



            Plot[cost[b0], {b0, -10, 10}]


            enter image description here






            share|improve this answer




























              2














              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



              $text{cost}(b0)=left(-5.031 left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+25}right)^2\+left(-3.627
              left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+
              sqrt{text{b0}^4+16}right)^2\+left(2-3.346
              left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)right)^2+left(-2.247
              left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+9}right)^2$



              Plot[cost[b0], {b0, -10, 10}]


              enter image description here






              share|improve this answer


























                2












                2








                2






                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



                $text{cost}(b0)=left(-5.031 left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+25}right)^2\+left(-3.627
                left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+
                sqrt{text{b0}^4+16}right)^2\+left(2-3.346
                left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)right)^2+left(-2.247
                left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+9}right)^2$



                Plot[cost[b0], {b0, -10, 10}]


                enter image description here






                share|improve this answer














                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



                $text{cost}(b0)=left(-5.031 left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+25}right)^2\+left(-3.627
                left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+
                sqrt{text{b0}^4+16}right)^2\+left(2-3.346
                left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)right)^2+left(-2.247
                left(sqrt{text{b0}^4+4}-sqrt{text{b0}^4+1}right)-sqrt{text{b0}^4+1}+sqrt{text{b0}^4+9}right)^2$



                Plot[cost[b0], {b0, -10, 10}]


                enter image description here







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 12 '18 at 15:38

























                answered Nov 12 '18 at 12:05









                Okkes Dulgerci

                4,1851816




                4,1851816






























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