How to interpolate/extrapolate a complex function?












0












$begingroup$


Let us assume that we have $f:mathbb{R} to mathbb{R}$. Also let us assume that $x_1in mathbb{R}$ and $x_2in mathbb{R}$ are given too. With this we can calculate $y_1 = f(x_1)$ and $y_2 = f(x_2)$.



Now if we want to interpolate/extrapolate an in-between value we could take a value like e.g. $alpha in mathbb{R}$ (but for interpolation $alpha$ would be in range $[0,1]$).



With this we could calculate the interpolated values $x_{irp},y_{irp} in mathbb{R}$ like:



$$x_{irp}=alpha cdot x_1 + (1- alpha) cdot x_2$$
$$y_{irp}=alpha cdot y_1 + (1- alpha) cdot y_2$$



Now my question:



Does the same approach also works for complex numbers too? In this case instead of $mathbb{R}$ we would have $mathbb{C}$.



(I know indeed that for multidimensional input and output this approach works, but for complex numbers too?)






complex-numbers interpolation







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let us assume that we have $f:mathbb{R} to mathbb{R}$. Also let us assume that $x_1in mathbb{R}$ and $x_2in mathbb{R}$ are given too. With this we can calculate $y_1 = f(x_1)$ and $y_2 = f(x_2)$.



    Now if we want to interpolate/extrapolate an in-between value we could take a value like e.g. $alpha in mathbb{R}$ (but for interpolation $alpha$ would be in range $[0,1]$).



    With this we could calculate the interpolated values $x_{irp},y_{irp} in mathbb{R}$ like:



    $$x_{irp}=alpha cdot x_1 + (1- alpha) cdot x_2$$
    $$y_{irp}=alpha cdot y_1 + (1- alpha) cdot y_2$$



    Now my question:



    Does the same approach also works for complex numbers too? In this case instead of $mathbb{R}$ we would have $mathbb{C}$.



    (I know indeed that for multidimensional input and output this approach works, but for complex numbers too?)






    complex-numbers interpolation







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let us assume that we have $f:mathbb{R} to mathbb{R}$. Also let us assume that $x_1in mathbb{R}$ and $x_2in mathbb{R}$ are given too. With this we can calculate $y_1 = f(x_1)$ and $y_2 = f(x_2)$.



      Now if we want to interpolate/extrapolate an in-between value we could take a value like e.g. $alpha in mathbb{R}$ (but for interpolation $alpha$ would be in range $[0,1]$).



      With this we could calculate the interpolated values $x_{irp},y_{irp} in mathbb{R}$ like:



      $$x_{irp}=alpha cdot x_1 + (1- alpha) cdot x_2$$
      $$y_{irp}=alpha cdot y_1 + (1- alpha) cdot y_2$$



      Now my question:



      Does the same approach also works for complex numbers too? In this case instead of $mathbb{R}$ we would have $mathbb{C}$.



      (I know indeed that for multidimensional input and output this approach works, but for complex numbers too?)






      complex-numbers interpolation







      share|cite|improve this question









      $endgroup$




      Let us assume that we have $f:mathbb{R} to mathbb{R}$. Also let us assume that $x_1in mathbb{R}$ and $x_2in mathbb{R}$ are given too. With this we can calculate $y_1 = f(x_1)$ and $y_2 = f(x_2)$.



      Now if we want to interpolate/extrapolate an in-between value we could take a value like e.g. $alpha in mathbb{R}$ (but for interpolation $alpha$ would be in range $[0,1]$).



      With this we could calculate the interpolated values $x_{irp},y_{irp} in mathbb{R}$ like:



      $$x_{irp}=alpha cdot x_1 + (1- alpha) cdot x_2$$
      $$y_{irp}=alpha cdot y_1 + (1- alpha) cdot y_2$$



      Now my question:



      Does the same approach also works for complex numbers too? In this case instead of $mathbb{R}$ we would have $mathbb{C}$.



      (I know indeed that for multidimensional input and output this approach works, but for complex numbers too?)






      complex-numbers interpolation




      complex-numbers interpolation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 '18 at 13:47









      PiMathCLanguagePiMathCLanguage

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          $begingroup$

          Yes, exactly the same method will work (with exactly the same limitations that this is a really bad estimate in general).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is there a better approach too? I mean like for the $mathbb{R}$ numbers with a square line approximation? How would this one work for $mathbb{C}$?
            $endgroup$
            – PiMathCLanguage
            Nov 26 '18 at 13:51













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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          Yes, exactly the same method will work (with exactly the same limitations that this is a really bad estimate in general).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is there a better approach too? I mean like for the $mathbb{R}$ numbers with a square line approximation? How would this one work for $mathbb{C}$?
            $endgroup$
            – PiMathCLanguage
            Nov 26 '18 at 13:51


















          1












          $begingroup$

          Yes, exactly the same method will work (with exactly the same limitations that this is a really bad estimate in general).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is there a better approach too? I mean like for the $mathbb{R}$ numbers with a square line approximation? How would this one work for $mathbb{C}$?
            $endgroup$
            – PiMathCLanguage
            Nov 26 '18 at 13:51
















          1












          1








          1





          $begingroup$

          Yes, exactly the same method will work (with exactly the same limitations that this is a really bad estimate in general).






          share|cite|improve this answer









          $endgroup$



          Yes, exactly the same method will work (with exactly the same limitations that this is a really bad estimate in general).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 13:49









          user3482749user3482749

          4,047818




          4,047818












          • $begingroup$
            Is there a better approach too? I mean like for the $mathbb{R}$ numbers with a square line approximation? How would this one work for $mathbb{C}$?
            $endgroup$
            – PiMathCLanguage
            Nov 26 '18 at 13:51




















          • $begingroup$
            Is there a better approach too? I mean like for the $mathbb{R}$ numbers with a square line approximation? How would this one work for $mathbb{C}$?
            $endgroup$
            – PiMathCLanguage
            Nov 26 '18 at 13:51


















          $begingroup$
          Is there a better approach too? I mean like for the $mathbb{R}$ numbers with a square line approximation? How would this one work for $mathbb{C}$?
          $endgroup$
          – PiMathCLanguage
          Nov 26 '18 at 13:51






          $begingroup$
          Is there a better approach too? I mean like for the $mathbb{R}$ numbers with a square line approximation? How would this one work for $mathbb{C}$?
          $endgroup$
          – PiMathCLanguage
          Nov 26 '18 at 13:51




















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