How to find g(x) and aux function h(x) when doing fixed point interation?












0












$begingroup$


I'm learning fixed point iteration (first and second form).



My teacher said there are two forms:



g(x) = x - f(x)

g(x) = x - h(x)f(x)


where h(x) is an aux function.



suppose f(x) = x^2 + 2x -3.



Let f(x) = 0.



x^2 + 2x -3 = 0

x^2  = - 2x + 3

x = 3/x -2


So we let g(x) = 3/x -2, for a fixed point, and root, right?



To solve, we iterate and see convergence:



xk+1 = 3/xk -2


and pick an initial xk.



My teacher said that in second form:



from f(x) = x^2 + 2x -3:



g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)


with the aux function h(x) = -1 / (x^2 - 5)



Where did he get h(x) = -1 / (x^2 - 5) from? And where is 



g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)


from?



thanks!






fixed-point-theorems fixedpoints







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    0












    $begingroup$


    I'm learning fixed point iteration (first and second form).



    My teacher said there are two forms:



    g(x) = x - f(x)
    
g(x) = x - h(x)f(x)


    where h(x) is an aux function.



    suppose f(x) = x^2 + 2x -3.



    Let f(x) = 0.



    x^2 + 2x -3 = 0
    
x^2  = - 2x + 3
    
x = 3/x -2


    So we let g(x) = 3/x -2, for a fixed point, and root, right?



    To solve, we iterate and see convergence:



    xk+1 = 3/xk -2


    and pick an initial xk.



    My teacher said that in second form:



    from f(x) = x^2 + 2x -3:



    g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)


    with the aux function h(x) = -1 / (x^2 - 5)



    Where did he get h(x) = -1 / (x^2 - 5) from? And where is 



    g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)


    from?



    thanks!






    fixed-point-theorems fixedpoints







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm learning fixed point iteration (first and second form).



      My teacher said there are two forms:



      g(x) = x - f(x)
      
g(x) = x - h(x)f(x)


      where h(x) is an aux function.



      suppose f(x) = x^2 + 2x -3.



      Let f(x) = 0.



      x^2 + 2x -3 = 0
      
x^2  = - 2x + 3
      
x = 3/x -2


      So we let g(x) = 3/x -2, for a fixed point, and root, right?



      To solve, we iterate and see convergence:



      xk+1 = 3/xk -2


      and pick an initial xk.



      My teacher said that in second form:



      from f(x) = x^2 + 2x -3:



      g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)


      with the aux function h(x) = -1 / (x^2 - 5)



      Where did he get h(x) = -1 / (x^2 - 5) from? And where is 



      g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)


      from?



      thanks!






      fixed-point-theorems fixedpoints







      share|cite|improve this question









      $endgroup$




      I'm learning fixed point iteration (first and second form).



      My teacher said there are two forms:



      g(x) = x - f(x)
      
g(x) = x - h(x)f(x)


      where h(x) is an aux function.



      suppose f(x) = x^2 + 2x -3.



      Let f(x) = 0.



      x^2 + 2x -3 = 0
      
x^2  = - 2x + 3
      
x = 3/x -2


      So we let g(x) = 3/x -2, for a fixed point, and root, right?



      To solve, we iterate and see convergence:



      xk+1 = 3/xk -2


      and pick an initial xk.



      My teacher said that in second form:



      from f(x) = x^2 + 2x -3:



      g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)


      with the aux function h(x) = -1 / (x^2 - 5)



      Where did he get h(x) = -1 / (x^2 - 5) from? And where is 



      g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)


      from?



      thanks!






      fixed-point-theorems fixedpoints




      fixed-point-theorems fixedpoints






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      asked Dec 13 '18 at 17:20









      Jay PatelJay Patel

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          $begingroup$


          My teacher said that in second form: from f(x) = x^2 + 2x -3:



          [one could consider] g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



          with the aux[iliary] function h(x) = -1 / (x^2 - 5)



          Where did he get h(x) = -1 / (x^2 - 5) from? And where is



          g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



          [coming] from?




          This is most probably not what they said, and it seems that typos due to carelessness are killing you here.



          Actually, to solve $f(x)=0$ with $f(x)=x^2 + 2x -3$, they most probably suggested to solve $g(x)=x$ with $$g(x) = x + frac{x^2 + 2x -3}{x^2 - 5}$$ that is, a function $g$ exactly of the general form $g(x)=x-f(x)h(x)$, for the auxiliary function $$h(x) = -frac1{x^2 - 5}$$ They chose $h$ arbitrarily.




          Supplementary material: This choice of auxiliary function $h$ is not especially suited to this precise function $f$, if you ask me. A better choice would be to get rid of the $x^2$ term in $f(x)$, using $$g_c(x)=x-frac{x^2 + 2x -3}{x+c}$$ for some fixed positive $c$, that is, $g_c(x)=x-f(x)h_c(x)$ with $$h_c(x)=frac1{x+c}$$ Now, a simple computation shows that $$g_c(x)=frac{(c-2)x+3}{x+c}$$ hence the value $c=2$ is interesting, leading to the iteration of the function $$g_2(x)=frac3{x+2}$$
          Note finally that for every initial condition $x_0geqslant0$, the sequence defined recursively by $x_{n+1}=g_2(x_n)$ converges to the positive root $x^*$ of $f$ at a geometric rate $frac13$ (and that $x^*=1$, but this is irrelevant to the construction and to the remark).






          share|cite|improve this answer









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            $begingroup$


            My teacher said that in second form: from f(x) = x^2 + 2x -3:



            [one could consider] g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



            with the aux[iliary] function h(x) = -1 / (x^2 - 5)



            Where did he get h(x) = -1 / (x^2 - 5) from? And where is



            g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



            [coming] from?




            This is most probably not what they said, and it seems that typos due to carelessness are killing you here.



            Actually, to solve $f(x)=0$ with $f(x)=x^2 + 2x -3$, they most probably suggested to solve $g(x)=x$ with $$g(x) = x + frac{x^2 + 2x -3}{x^2 - 5}$$ that is, a function $g$ exactly of the general form $g(x)=x-f(x)h(x)$, for the auxiliary function $$h(x) = -frac1{x^2 - 5}$$ They chose $h$ arbitrarily.




            Supplementary material: This choice of auxiliary function $h$ is not especially suited to this precise function $f$, if you ask me. A better choice would be to get rid of the $x^2$ term in $f(x)$, using $$g_c(x)=x-frac{x^2 + 2x -3}{x+c}$$ for some fixed positive $c$, that is, $g_c(x)=x-f(x)h_c(x)$ with $$h_c(x)=frac1{x+c}$$ Now, a simple computation shows that $$g_c(x)=frac{(c-2)x+3}{x+c}$$ hence the value $c=2$ is interesting, leading to the iteration of the function $$g_2(x)=frac3{x+2}$$
            Note finally that for every initial condition $x_0geqslant0$, the sequence defined recursively by $x_{n+1}=g_2(x_n)$ converges to the positive root $x^*$ of $f$ at a geometric rate $frac13$ (and that $x^*=1$, but this is irrelevant to the construction and to the remark).






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$


              My teacher said that in second form: from f(x) = x^2 + 2x -3:



              [one could consider] g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



              with the aux[iliary] function h(x) = -1 / (x^2 - 5)



              Where did he get h(x) = -1 / (x^2 - 5) from? And where is



              g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



              [coming] from?




              This is most probably not what they said, and it seems that typos due to carelessness are killing you here.



              Actually, to solve $f(x)=0$ with $f(x)=x^2 + 2x -3$, they most probably suggested to solve $g(x)=x$ with $$g(x) = x + frac{x^2 + 2x -3}{x^2 - 5}$$ that is, a function $g$ exactly of the general form $g(x)=x-f(x)h(x)$, for the auxiliary function $$h(x) = -frac1{x^2 - 5}$$ They chose $h$ arbitrarily.




              Supplementary material: This choice of auxiliary function $h$ is not especially suited to this precise function $f$, if you ask me. A better choice would be to get rid of the $x^2$ term in $f(x)$, using $$g_c(x)=x-frac{x^2 + 2x -3}{x+c}$$ for some fixed positive $c$, that is, $g_c(x)=x-f(x)h_c(x)$ with $$h_c(x)=frac1{x+c}$$ Now, a simple computation shows that $$g_c(x)=frac{(c-2)x+3}{x+c}$$ hence the value $c=2$ is interesting, leading to the iteration of the function $$g_2(x)=frac3{x+2}$$
              Note finally that for every initial condition $x_0geqslant0$, the sequence defined recursively by $x_{n+1}=g_2(x_n)$ converges to the positive root $x^*$ of $f$ at a geometric rate $frac13$ (and that $x^*=1$, but this is irrelevant to the construction and to the remark).






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$


                My teacher said that in second form: from f(x) = x^2 + 2x -3:



                [one could consider] g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



                with the aux[iliary] function h(x) = -1 / (x^2 - 5)



                Where did he get h(x) = -1 / (x^2 - 5) from? And where is



                g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



                [coming] from?




                This is most probably not what they said, and it seems that typos due to carelessness are killing you here.



                Actually, to solve $f(x)=0$ with $f(x)=x^2 + 2x -3$, they most probably suggested to solve $g(x)=x$ with $$g(x) = x + frac{x^2 + 2x -3}{x^2 - 5}$$ that is, a function $g$ exactly of the general form $g(x)=x-f(x)h(x)$, for the auxiliary function $$h(x) = -frac1{x^2 - 5}$$ They chose $h$ arbitrarily.




                Supplementary material: This choice of auxiliary function $h$ is not especially suited to this precise function $f$, if you ask me. A better choice would be to get rid of the $x^2$ term in $f(x)$, using $$g_c(x)=x-frac{x^2 + 2x -3}{x+c}$$ for some fixed positive $c$, that is, $g_c(x)=x-f(x)h_c(x)$ with $$h_c(x)=frac1{x+c}$$ Now, a simple computation shows that $$g_c(x)=frac{(c-2)x+3}{x+c}$$ hence the value $c=2$ is interesting, leading to the iteration of the function $$g_2(x)=frac3{x+2}$$
                Note finally that for every initial condition $x_0geqslant0$, the sequence defined recursively by $x_{n+1}=g_2(x_n)$ converges to the positive root $x^*$ of $f$ at a geometric rate $frac13$ (and that $x^*=1$, but this is irrelevant to the construction and to the remark).






                share|cite|improve this answer









                $endgroup$




                My teacher said that in second form: from f(x) = x^2 + 2x -3:



                [one could consider] g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



                with the aux[iliary] function h(x) = -1 / (x^2 - 5)



                Where did he get h(x) = -1 / (x^2 - 5) from? And where is



                g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)



                [coming] from?




                This is most probably not what they said, and it seems that typos due to carelessness are killing you here.



                Actually, to solve $f(x)=0$ with $f(x)=x^2 + 2x -3$, they most probably suggested to solve $g(x)=x$ with $$g(x) = x + frac{x^2 + 2x -3}{x^2 - 5}$$ that is, a function $g$ exactly of the general form $g(x)=x-f(x)h(x)$, for the auxiliary function $$h(x) = -frac1{x^2 - 5}$$ They chose $h$ arbitrarily.




                Supplementary material: This choice of auxiliary function $h$ is not especially suited to this precise function $f$, if you ask me. A better choice would be to get rid of the $x^2$ term in $f(x)$, using $$g_c(x)=x-frac{x^2 + 2x -3}{x+c}$$ for some fixed positive $c$, that is, $g_c(x)=x-f(x)h_c(x)$ with $$h_c(x)=frac1{x+c}$$ Now, a simple computation shows that $$g_c(x)=frac{(c-2)x+3}{x+c}$$ hence the value $c=2$ is interesting, leading to the iteration of the function $$g_2(x)=frac3{x+2}$$
                Note finally that for every initial condition $x_0geqslant0$, the sequence defined recursively by $x_{n+1}=g_2(x_n)$ converges to the positive root $x^*$ of $f$ at a geometric rate $frac13$ (and that $x^*=1$, but this is irrelevant to the construction and to the remark).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 21 '18 at 10:49









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