fractional curve fitting of the function $y=a+bx^{alpha}$












0












$begingroup$


Assume I have a set of data $(x_{i},y_{i})$, $i=1,...,m$. How can we find the best values of the parameters $a$ and $b$ and $alpha$ such that the curve $y=a+bx^{alpha}$ best fits the data.



This question arises in many problems that use fractional calculus in modelling. I see it is almost always done numerically using a certain routine in Matlab or Maple. Can we do it analytically using the least squares minimization method ?



If $a=0$, we can linearize the model by taking the logarithm of both sides. The difficulty here is that, in practice, the parameter $a$ is unknown and needs to be determined.



Another question: Can Wolfram Alpha solve such a problem ??



Thanks a lot.






statistics regression least-squares weighted-least-squares







share|cite|improve this question









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    0












    $begingroup$


    Assume I have a set of data $(x_{i},y_{i})$, $i=1,...,m$. How can we find the best values of the parameters $a$ and $b$ and $alpha$ such that the curve $y=a+bx^{alpha}$ best fits the data.



    This question arises in many problems that use fractional calculus in modelling. I see it is almost always done numerically using a certain routine in Matlab or Maple. Can we do it analytically using the least squares minimization method ?



    If $a=0$, we can linearize the model by taking the logarithm of both sides. The difficulty here is that, in practice, the parameter $a$ is unknown and needs to be determined.



    Another question: Can Wolfram Alpha solve such a problem ??



    Thanks a lot.






    statistics regression least-squares weighted-least-squares







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Assume I have a set of data $(x_{i},y_{i})$, $i=1,...,m$. How can we find the best values of the parameters $a$ and $b$ and $alpha$ such that the curve $y=a+bx^{alpha}$ best fits the data.



      This question arises in many problems that use fractional calculus in modelling. I see it is almost always done numerically using a certain routine in Matlab or Maple. Can we do it analytically using the least squares minimization method ?



      If $a=0$, we can linearize the model by taking the logarithm of both sides. The difficulty here is that, in practice, the parameter $a$ is unknown and needs to be determined.



      Another question: Can Wolfram Alpha solve such a problem ??



      Thanks a lot.






      statistics regression least-squares weighted-least-squares







      share|cite|improve this question









      $endgroup$




      Assume I have a set of data $(x_{i},y_{i})$, $i=1,...,m$. How can we find the best values of the parameters $a$ and $b$ and $alpha$ such that the curve $y=a+bx^{alpha}$ best fits the data.



      This question arises in many problems that use fractional calculus in modelling. I see it is almost always done numerically using a certain routine in Matlab or Maple. Can we do it analytically using the least squares minimization method ?



      If $a=0$, we can linearize the model by taking the logarithm of both sides. The difficulty here is that, in practice, the parameter $a$ is unknown and needs to be determined.



      Another question: Can Wolfram Alpha solve such a problem ??



      Thanks a lot.






      statistics regression least-squares weighted-least-squares




      statistics regression least-squares weighted-least-squares






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 26 '18 at 3:43









      MedoMedo

      617214




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          $begingroup$

          Based on $n$ data points $(x_i,y_i)$, you want to fit the model $$y=a+bx^{alpha}$$ which is nonlinear which means that you need estimates to start the nonlinear regression and the question is then : how to get these estimates ?



          If you think about it, the model is nonlinear just because of $alpha$. So, suppose you give a value to $alpha$, define $t_i=x_i^{alpha}$ and you just face a linear model $y=a+b t$. So, for this arbitrary value of $alpha$, you van get $a(alpha$), $b(alpha$) and $SSQ(alpha)$.



          So, an idea is to try different values of $alpha$ and, plotting $SSQ(alpha)$, to look where it goes more or less through a minimum. Let us call $alpha_*$ this point. To it correspond $a_*$ and $b_*$. These are your estimates for the nonlinear regression.



          If you do not have such a tool, then contine the process zooming more and more around the minimum.



          It this clear for you ? If not, I could elaborate using an example you could add to the post.



          What you can also do, hoping that the errors are not too large is to take three points (say for example points $1$, $n$ and one somewhere in the middle (say point $m$) and write the three equations
          $$y_1=a+bx_1^{alpha}qquad
          y_m=a+bx_m^{alpha}qquad
          y_n=a+bx_n^{alpha}$$           So
          $$y_m-y_1=b (x_m^{alpha}-x_1^{alpha})qquad
          y_n-y_m=b (x_n^{alpha}-x_m^{alpha})$$
          $$frac {y_m-y_1 } {y_n-y_m }=frac {x_m^{alpha}-x_1^{alpha} } { x_n^{alpha}-x_m^{alpha}}$$ which is a single equation in $alpha$ that you can solve using a plot. This gives you $alpha_*$ and, going backward in he three equations, then $a_*$ and $b_*$.



          Now, you are ready !






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the insight.
            $endgroup$
            – Medo
            Nov 26 '18 at 23:52











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          1 Answer
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          active

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          1












          $begingroup$

          Based on $n$ data points $(x_i,y_i)$, you want to fit the model $$y=a+bx^{alpha}$$ which is nonlinear which means that you need estimates to start the nonlinear regression and the question is then : how to get these estimates ?



          If you think about it, the model is nonlinear just because of $alpha$. So, suppose you give a value to $alpha$, define $t_i=x_i^{alpha}$ and you just face a linear model $y=a+b t$. So, for this arbitrary value of $alpha$, you van get $a(alpha$), $b(alpha$) and $SSQ(alpha)$.



          So, an idea is to try different values of $alpha$ and, plotting $SSQ(alpha)$, to look where it goes more or less through a minimum. Let us call $alpha_*$ this point. To it correspond $a_*$ and $b_*$. These are your estimates for the nonlinear regression.



          If you do not have such a tool, then contine the process zooming more and more around the minimum.



          It this clear for you ? If not, I could elaborate using an example you could add to the post.



          What you can also do, hoping that the errors are not too large is to take three points (say for example points $1$, $n$ and one somewhere in the middle (say point $m$) and write the three equations
          $$y_1=a+bx_1^{alpha}qquad
          y_m=a+bx_m^{alpha}qquad
          y_n=a+bx_n^{alpha}$$           So
          $$y_m-y_1=b (x_m^{alpha}-x_1^{alpha})qquad
          y_n-y_m=b (x_n^{alpha}-x_m^{alpha})$$
          $$frac {y_m-y_1 } {y_n-y_m }=frac {x_m^{alpha}-x_1^{alpha} } { x_n^{alpha}-x_m^{alpha}}$$ which is a single equation in $alpha$ that you can solve using a plot. This gives you $alpha_*$ and, going backward in he three equations, then $a_*$ and $b_*$.



          Now, you are ready !






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the insight.
            $endgroup$
            – Medo
            Nov 26 '18 at 23:52
















          1












          $begingroup$

          Based on $n$ data points $(x_i,y_i)$, you want to fit the model $$y=a+bx^{alpha}$$ which is nonlinear which means that you need estimates to start the nonlinear regression and the question is then : how to get these estimates ?



          If you think about it, the model is nonlinear just because of $alpha$. So, suppose you give a value to $alpha$, define $t_i=x_i^{alpha}$ and you just face a linear model $y=a+b t$. So, for this arbitrary value of $alpha$, you van get $a(alpha$), $b(alpha$) and $SSQ(alpha)$.



          So, an idea is to try different values of $alpha$ and, plotting $SSQ(alpha)$, to look where it goes more or less through a minimum. Let us call $alpha_*$ this point. To it correspond $a_*$ and $b_*$. These are your estimates for the nonlinear regression.



          If you do not have such a tool, then contine the process zooming more and more around the minimum.



          It this clear for you ? If not, I could elaborate using an example you could add to the post.



          What you can also do, hoping that the errors are not too large is to take three points (say for example points $1$, $n$ and one somewhere in the middle (say point $m$) and write the three equations
          $$y_1=a+bx_1^{alpha}qquad
          y_m=a+bx_m^{alpha}qquad
          y_n=a+bx_n^{alpha}$$           So
          $$y_m-y_1=b (x_m^{alpha}-x_1^{alpha})qquad
          y_n-y_m=b (x_n^{alpha}-x_m^{alpha})$$
          $$frac {y_m-y_1 } {y_n-y_m }=frac {x_m^{alpha}-x_1^{alpha} } { x_n^{alpha}-x_m^{alpha}}$$ which is a single equation in $alpha$ that you can solve using a plot. This gives you $alpha_*$ and, going backward in he three equations, then $a_*$ and $b_*$.



          Now, you are ready !






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the insight.
            $endgroup$
            – Medo
            Nov 26 '18 at 23:52














          1












          1








          1





          $begingroup$

          Based on $n$ data points $(x_i,y_i)$, you want to fit the model $$y=a+bx^{alpha}$$ which is nonlinear which means that you need estimates to start the nonlinear regression and the question is then : how to get these estimates ?



          If you think about it, the model is nonlinear just because of $alpha$. So, suppose you give a value to $alpha$, define $t_i=x_i^{alpha}$ and you just face a linear model $y=a+b t$. So, for this arbitrary value of $alpha$, you van get $a(alpha$), $b(alpha$) and $SSQ(alpha)$.



          So, an idea is to try different values of $alpha$ and, plotting $SSQ(alpha)$, to look where it goes more or less through a minimum. Let us call $alpha_*$ this point. To it correspond $a_*$ and $b_*$. These are your estimates for the nonlinear regression.



          If you do not have such a tool, then contine the process zooming more and more around the minimum.



          It this clear for you ? If not, I could elaborate using an example you could add to the post.



          What you can also do, hoping that the errors are not too large is to take three points (say for example points $1$, $n$ and one somewhere in the middle (say point $m$) and write the three equations
          $$y_1=a+bx_1^{alpha}qquad
          y_m=a+bx_m^{alpha}qquad
          y_n=a+bx_n^{alpha}$$           So
          $$y_m-y_1=b (x_m^{alpha}-x_1^{alpha})qquad
          y_n-y_m=b (x_n^{alpha}-x_m^{alpha})$$
          $$frac {y_m-y_1 } {y_n-y_m }=frac {x_m^{alpha}-x_1^{alpha} } { x_n^{alpha}-x_m^{alpha}}$$ which is a single equation in $alpha$ that you can solve using a plot. This gives you $alpha_*$ and, going backward in he three equations, then $a_*$ and $b_*$.



          Now, you are ready !






          share|cite|improve this answer









          $endgroup$



          Based on $n$ data points $(x_i,y_i)$, you want to fit the model $$y=a+bx^{alpha}$$ which is nonlinear which means that you need estimates to start the nonlinear regression and the question is then : how to get these estimates ?



          If you think about it, the model is nonlinear just because of $alpha$. So, suppose you give a value to $alpha$, define $t_i=x_i^{alpha}$ and you just face a linear model $y=a+b t$. So, for this arbitrary value of $alpha$, you van get $a(alpha$), $b(alpha$) and $SSQ(alpha)$.



          So, an idea is to try different values of $alpha$ and, plotting $SSQ(alpha)$, to look where it goes more or less through a minimum. Let us call $alpha_*$ this point. To it correspond $a_*$ and $b_*$. These are your estimates for the nonlinear regression.



          If you do not have such a tool, then contine the process zooming more and more around the minimum.



          It this clear for you ? If not, I could elaborate using an example you could add to the post.



          What you can also do, hoping that the errors are not too large is to take three points (say for example points $1$, $n$ and one somewhere in the middle (say point $m$) and write the three equations
          $$y_1=a+bx_1^{alpha}qquad
          y_m=a+bx_m^{alpha}qquad
          y_n=a+bx_n^{alpha}$$           So
          $$y_m-y_1=b (x_m^{alpha}-x_1^{alpha})qquad
          y_n-y_m=b (x_n^{alpha}-x_m^{alpha})$$
          $$frac {y_m-y_1 } {y_n-y_m }=frac {x_m^{alpha}-x_1^{alpha} } { x_n^{alpha}-x_m^{alpha}}$$ which is a single equation in $alpha$ that you can solve using a plot. This gives you $alpha_*$ and, going backward in he three equations, then $a_*$ and $b_*$.



          Now, you are ready !







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 5:05









          Claude LeiboviciClaude Leibovici

          120k1157132




          120k1157132












          • $begingroup$
            Thank you for the insight.
            $endgroup$
            – Medo
            Nov 26 '18 at 23:52


















          • $begingroup$
            Thank you for the insight.
            $endgroup$
            – Medo
            Nov 26 '18 at 23:52
















          $begingroup$
          Thank you for the insight.
          $endgroup$
          – Medo
          Nov 26 '18 at 23:52




          $begingroup$
          Thank you for the insight.
          $endgroup$
          – Medo
          Nov 26 '18 at 23:52


















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