Mixing
Fixed point iteration-finding $g(x)$
$begingroup$
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
edited Dec 11 '16 at 15:10
InsideOut
4,94131033
asked Dec 11 '16 at 15:08
DimitrisDimitris
6121719
$endgroup$
add a comment |
$begingroup$
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
edited Dec 11 '16 at 15:10
InsideOut
4,94131033
asked Dec 11 '16 at 15:08
DimitrisDimitris
6121719
$endgroup$
add a comment |
$begingroup$
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
edited Dec 11 '16 at 15:10
InsideOut
4,94131033
asked Dec 11 '16 at 15:08
DimitrisDimitris
6121719
$endgroup$
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
edited Dec 11 '16 at 15:10
InsideOut
4,94131033
asked Dec 11 '16 at 15:08
DimitrisDimitris
6121719
edited Dec 11 '16 at 15:10
InsideOut
4,94131033
asked Dec 11 '16 at 15:08
DimitrisDimitris
6121719
edited Dec 11 '16 at 15:10
InsideOut
4,94131033
edited Dec 11 '16 at 15:10
InsideOut
4,94131033
edited Dec 11 '16 at 15:10
InsideOut
4,94131033
4,94131033
asked Dec 11 '16 at 15:08
DimitrisDimitris
6121719
asked Dec 11 '16 at 15:08
DimitrisDimitris
6121719
asked Dec 11 '16 at 15:08
DimitrisDimitris
6121719
6121719
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 11 '16 at 15:21
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
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
answered Jan 4 at 15:11
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
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.
answered Dec 11 '16 at 15:21
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 11 '16 at 15:21
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 11 '16 at 15:21
Ross MillikanRoss Millikan
293k23197371
$endgroup$
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.
answered Dec 11 '16 at 15:21
Ross MillikanRoss Millikan
293k23197371
edited Dec 11 '16 at 16:31
answered Dec 11 '16 at 15:21
Ross MillikanRoss Millikan
293k23197371
answered Dec 11 '16 at 15:21
Ross MillikanRoss Millikan
293k23197371
answered Dec 11 '16 at 15:21
Ross MillikanRoss Millikan
293k23197371
293k23197371
add a comment |
add a comment |
$begingroup$
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
answered Jan 4 at 15:11
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
$begingroup$
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
answered Jan 4 at 15:11
LutzLLutzL
57.2k42054
$endgroup$
add a comment |
$begingroup$
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
answered Jan 4 at 15:11
LutzLLutzL
57.2k42054
$endgroup$
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
answered Jan 4 at 15:11
LutzLLutzL
57.2k42054
answered Jan 4 at 15:11
LutzLLutzL
57.2k42054
answered Jan 4 at 15:11
LutzLLutzL
57.2k42054
answered Jan 4 at 15:11
LutzLLutzL
57.2k42054
57.2k42054
add a comment |
add a comment |
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