Fixed point iteration-finding $g(x)$












2












$begingroup$


I have an equation $f(x)= x^4 +2x^3 -5x^2 -7x-5=0 $ and I want to solve it (numerically) at [1.5,3] using the fixed point iteration-method, so I want to find a function g(x) such that $f(x)=0$ is equivalent to $g(x)=x$. However I've tried many different $g(x)$'s but can't seem to find one that converges to the solution (which is about 2.19).
First of all, anyone has any idea which $g(x)$ I could use?
Second is there a trick to finding it?

Thanks in advance :)






numerical-methods fixed-point-theorems fixed-point-iteration







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$endgroup$

















    2












    $begingroup$


    I have an equation $f(x)= x^4 +2x^3 -5x^2 -7x-5=0 $ and I want to solve it (numerically) at [1.5,3] using the fixed point iteration-method, so I want to find a function g(x) such that $f(x)=0$ is equivalent to $g(x)=x$. However I've tried many different $g(x)$'s but can't seem to find one that converges to the solution (which is about 2.19).
    First of all, anyone has any idea which $g(x)$ I could use?
    Second is there a trick to finding it?

    Thanks in advance :)






    numerical-methods fixed-point-theorems fixed-point-iteration







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I have an equation $f(x)= x^4 +2x^3 -5x^2 -7x-5=0 $ and I want to solve it (numerically) at [1.5,3] using the fixed point iteration-method, so I want to find a function g(x) such that $f(x)=0$ is equivalent to $g(x)=x$. However I've tried many different $g(x)$'s but can't seem to find one that converges to the solution (which is about 2.19).
      First of all, anyone has any idea which $g(x)$ I could use?
      Second is there a trick to finding it?

      Thanks in advance :)






      numerical-methods fixed-point-theorems fixed-point-iteration







      share|cite|improve this question











      $endgroup$




      I have an equation $f(x)= x^4 +2x^3 -5x^2 -7x-5=0 $ and I want to solve it (numerically) at [1.5,3] using the fixed point iteration-method, so I want to find a function g(x) such that $f(x)=0$ is equivalent to $g(x)=x$. However I've tried many different $g(x)$'s but can't seem to find one that converges to the solution (which is about 2.19).
      First of all, anyone has any idea which $g(x)$ I could use?
      Second is there a trick to finding it?

      Thanks in advance :)






      numerical-methods fixed-point-theorems fixed-point-iteration




      numerical-methods fixed-point-theorems fixed-point-iteration






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 11 '16 at 15:10









      InsideOut

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      4,94131033










      asked Dec 11 '16 at 15:08









      DimitrisDimitris

      6121719




      6121719






















          2 Answers
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          $begingroup$

          To get the solution to converge, besides $g(x)=x$ you need $|g'(x)| lt 1$ at the root. The distance from the root is multiplied by about $|g'(x)|$ every iteration, so if that is less than $1$ it will converge. Powers change quickly and roots slowly, so a natural try is to write
          $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
          x=sqrt[4]{-2x^3+5x^2+7x+5}$$
          It converges nicely starting at $0, 1$ and $3.5$ but fails starting from $4$ because the piece under the root goes negative. Another try, which I have not tested, would be to write $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
          x=(-2x^3+5x^2+7x+5)/x^3$$ It is a bit of an art. If you don't know the root, you can't evaluate the derivative of your proposed $g(x)$ at it.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            The scholars of old had a philosophy of avoiding differences if manually computing roots of equations. That way a treatment of quadratic or cubic equations could fill multiple pages considering the various cases, but avoided the discussion of roots of negative numbers.



            In your equation one would separate by this philosophy
            $$
            x^4+2x^3=5x^2+7x+5
            $$
            and solve the higher degree side, for instance as
            $$
            x=g(x)=sqrt[Large3,!]{frac{5x^2+7x+5}{x+2}}=sqrt[Large3,!]{5x-3+frac{11}{x+2}}
            $$
            This as fixed-point iteration maps the interval $[2,3]$ into itself an has thus a fixed point. The derivative of $g$ is smaller than $frac{5}{12}<frac12$ providing sufficient speed of convergence. A numerical series confirms this result.



             k         x[k]        x[k]-x[k-1]     1.4*0.3^k
            

            
 0      2.0
            
 1      2.1363293408   0.1363293408     0.138
            
 2      2.1786508930   0.04232155218    0.0414
            
 3      2.1915435027   0.0128926097     0.01242
            
 4      2.1954485170   0.003905014239   0.003726
            
 5      2.1966292396   0.001180722584   0.0011178
            
 6      2.1969860556   0.0003568160362  0.00033534
            
 7      2.1970938687   0.0001078131484  0.000100602
            
 8      2.1971264433   3.25745335e-05   3.01806e-05
            
 9      2.1971362852   9.841886902e-06  9.05418e-06
            
10      2.1971392587   2.973559468e-06  2.716254e-06
            
11      2.1971401571   8.984094482e-07  8.148762e-07
            
12      2.1971404286   2.714387324e-07  2.4446286e-07





            share|cite|improve this answer









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              $begingroup$

              To get the solution to converge, besides $g(x)=x$ you need $|g'(x)| lt 1$ at the root. The distance from the root is multiplied by about $|g'(x)|$ every iteration, so if that is less than $1$ it will converge. Powers change quickly and roots slowly, so a natural try is to write
              $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
              x=sqrt[4]{-2x^3+5x^2+7x+5}$$
              It converges nicely starting at $0, 1$ and $3.5$ but fails starting from $4$ because the piece under the root goes negative. Another try, which I have not tested, would be to write $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
              x=(-2x^3+5x^2+7x+5)/x^3$$ It is a bit of an art. If you don't know the root, you can't evaluate the derivative of your proposed $g(x)$ at it.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                To get the solution to converge, besides $g(x)=x$ you need $|g'(x)| lt 1$ at the root. The distance from the root is multiplied by about $|g'(x)|$ every iteration, so if that is less than $1$ it will converge. Powers change quickly and roots slowly, so a natural try is to write
                $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
                x=sqrt[4]{-2x^3+5x^2+7x+5}$$
                It converges nicely starting at $0, 1$ and $3.5$ but fails starting from $4$ because the piece under the root goes negative. Another try, which I have not tested, would be to write $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
                x=(-2x^3+5x^2+7x+5)/x^3$$ It is a bit of an art. If you don't know the root, you can't evaluate the derivative of your proposed $g(x)$ at it.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  To get the solution to converge, besides $g(x)=x$ you need $|g'(x)| lt 1$ at the root. The distance from the root is multiplied by about $|g'(x)|$ every iteration, so if that is less than $1$ it will converge. Powers change quickly and roots slowly, so a natural try is to write
                  $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
                  x=sqrt[4]{-2x^3+5x^2+7x+5}$$
                  It converges nicely starting at $0, 1$ and $3.5$ but fails starting from $4$ because the piece under the root goes negative. Another try, which I have not tested, would be to write $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
                  x=(-2x^3+5x^2+7x+5)/x^3$$ It is a bit of an art. If you don't know the root, you can't evaluate the derivative of your proposed $g(x)$ at it.






                  share|cite|improve this answer











                  $endgroup$



                  To get the solution to converge, besides $g(x)=x$ you need $|g'(x)| lt 1$ at the root. The distance from the root is multiplied by about $|g'(x)|$ every iteration, so if that is less than $1$ it will converge. Powers change quickly and roots slowly, so a natural try is to write
                  $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
                  x=sqrt[4]{-2x^3+5x^2+7x+5}$$
                  It converges nicely starting at $0, 1$ and $3.5$ but fails starting from $4$ because the piece under the root goes negative. Another try, which I have not tested, would be to write $$x^4 +2x^3 -5x^2 -7x-5=0\x^4=-2x^3+5x^2+7x+5\
                  x=(-2x^3+5x^2+7x+5)/x^3$$ It is a bit of an art. If you don't know the root, you can't evaluate the derivative of your proposed $g(x)$ at it.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 11 '16 at 16:31

























                  answered Dec 11 '16 at 15:21









                  Ross MillikanRoss Millikan

                  293k23197371




                  293k23197371























                      1












                      $begingroup$

                      The scholars of old had a philosophy of avoiding differences if manually computing roots of equations. That way a treatment of quadratic or cubic equations could fill multiple pages considering the various cases, but avoided the discussion of roots of negative numbers.



                      In your equation one would separate by this philosophy
                      $$
                      x^4+2x^3=5x^2+7x+5
                      $$
                      and solve the higher degree side, for instance as
                      $$
                      x=g(x)=sqrt[Large3,!]{frac{5x^2+7x+5}{x+2}}=sqrt[Large3,!]{5x-3+frac{11}{x+2}}
                      $$
                      This as fixed-point iteration maps the interval $[2,3]$ into itself an has thus a fixed point. The derivative of $g$ is smaller than $frac{5}{12}<frac12$ providing sufficient speed of convergence. A numerical series confirms this result.



                       k         x[k]        x[k]-x[k-1]     1.4*0.3^k
                      

                      
 0      2.0
                      
 1      2.1363293408   0.1363293408     0.138
                      
 2      2.1786508930   0.04232155218    0.0414
                      
 3      2.1915435027   0.0128926097     0.01242
                      
 4      2.1954485170   0.003905014239   0.003726
                      
 5      2.1966292396   0.001180722584   0.0011178
                      
 6      2.1969860556   0.0003568160362  0.00033534
                      
 7      2.1970938687   0.0001078131484  0.000100602
                      
 8      2.1971264433   3.25745335e-05   3.01806e-05
                      
 9      2.1971362852   9.841886902e-06  9.05418e-06
                      
10      2.1971392587   2.973559468e-06  2.716254e-06
                      
11      2.1971401571   8.984094482e-07  8.148762e-07
                      
12      2.1971404286   2.714387324e-07  2.4446286e-07





                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        The scholars of old had a philosophy of avoiding differences if manually computing roots of equations. That way a treatment of quadratic or cubic equations could fill multiple pages considering the various cases, but avoided the discussion of roots of negative numbers.



                        In your equation one would separate by this philosophy
                        $$
                        x^4+2x^3=5x^2+7x+5
                        $$
                        and solve the higher degree side, for instance as
                        $$
                        x=g(x)=sqrt[Large3,!]{frac{5x^2+7x+5}{x+2}}=sqrt[Large3,!]{5x-3+frac{11}{x+2}}
                        $$
                        This as fixed-point iteration maps the interval $[2,3]$ into itself an has thus a fixed point. The derivative of $g$ is smaller than $frac{5}{12}<frac12$ providing sufficient speed of convergence. A numerical series confirms this result.



                         k         x[k]        x[k]-x[k-1]     1.4*0.3^k
                        

                        
 0      2.0
                        
 1      2.1363293408   0.1363293408     0.138
                        
 2      2.1786508930   0.04232155218    0.0414
                        
 3      2.1915435027   0.0128926097     0.01242
                        
 4      2.1954485170   0.003905014239   0.003726
                        
 5      2.1966292396   0.001180722584   0.0011178
                        
 6      2.1969860556   0.0003568160362  0.00033534
                        
 7      2.1970938687   0.0001078131484  0.000100602
                        
 8      2.1971264433   3.25745335e-05   3.01806e-05
                        
 9      2.1971362852   9.841886902e-06  9.05418e-06
                        
10      2.1971392587   2.973559468e-06  2.716254e-06
                        
11      2.1971401571   8.984094482e-07  8.148762e-07
                        
12      2.1971404286   2.714387324e-07  2.4446286e-07





                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The scholars of old had a philosophy of avoiding differences if manually computing roots of equations. That way a treatment of quadratic or cubic equations could fill multiple pages considering the various cases, but avoided the discussion of roots of negative numbers.



                          In your equation one would separate by this philosophy
                          $$
                          x^4+2x^3=5x^2+7x+5
                          $$
                          and solve the higher degree side, for instance as
                          $$
                          x=g(x)=sqrt[Large3,!]{frac{5x^2+7x+5}{x+2}}=sqrt[Large3,!]{5x-3+frac{11}{x+2}}
                          $$
                          This as fixed-point iteration maps the interval $[2,3]$ into itself an has thus a fixed point. The derivative of $g$ is smaller than $frac{5}{12}<frac12$ providing sufficient speed of convergence. A numerical series confirms this result.



                           k         x[k]        x[k]-x[k-1]     1.4*0.3^k
                          

                          
 0      2.0
                          
 1      2.1363293408   0.1363293408     0.138
                          
 2      2.1786508930   0.04232155218    0.0414
                          
 3      2.1915435027   0.0128926097     0.01242
                          
 4      2.1954485170   0.003905014239   0.003726
                          
 5      2.1966292396   0.001180722584   0.0011178
                          
 6      2.1969860556   0.0003568160362  0.00033534
                          
 7      2.1970938687   0.0001078131484  0.000100602
                          
 8      2.1971264433   3.25745335e-05   3.01806e-05
                          
 9      2.1971362852   9.841886902e-06  9.05418e-06
                          
10      2.1971392587   2.973559468e-06  2.716254e-06
                          
11      2.1971401571   8.984094482e-07  8.148762e-07
                          
12      2.1971404286   2.714387324e-07  2.4446286e-07





                          share|cite|improve this answer









                          $endgroup$



                          The scholars of old had a philosophy of avoiding differences if manually computing roots of equations. That way a treatment of quadratic or cubic equations could fill multiple pages considering the various cases, but avoided the discussion of roots of negative numbers.



                          In your equation one would separate by this philosophy
                          $$
                          x^4+2x^3=5x^2+7x+5
                          $$
                          and solve the higher degree side, for instance as
                          $$
                          x=g(x)=sqrt[Large3,!]{frac{5x^2+7x+5}{x+2}}=sqrt[Large3,!]{5x-3+frac{11}{x+2}}
                          $$
                          This as fixed-point iteration maps the interval $[2,3]$ into itself an has thus a fixed point. The derivative of $g$ is smaller than $frac{5}{12}<frac12$ providing sufficient speed of convergence. A numerical series confirms this result.



                           k         x[k]        x[k]-x[k-1]     1.4*0.3^k
                          

                          
 0      2.0
                          
 1      2.1363293408   0.1363293408     0.138
                          
 2      2.1786508930   0.04232155218    0.0414
                          
 3      2.1915435027   0.0128926097     0.01242
                          
 4      2.1954485170   0.003905014239   0.003726
                          
 5      2.1966292396   0.001180722584   0.0011178
                          
 6      2.1969860556   0.0003568160362  0.00033534
                          
 7      2.1970938687   0.0001078131484  0.000100602
                          
 8      2.1971264433   3.25745335e-05   3.01806e-05
                          
 9      2.1971362852   9.841886902e-06  9.05418e-06
                          
10      2.1971392587   2.973559468e-06  2.716254e-06
                          
11      2.1971401571   8.984094482e-07  8.148762e-07
                          
12      2.1971404286   2.714387324e-07  2.4446286e-07






                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 4 at 15:11









                          LutzLLutzL

                          57.2k42054




                          57.2k42054






























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