Fixed-point method for $x=x+wf(x)$












0












$begingroup$


The equation $f(x)=x^3+x-1=0$ admites a unique solution $alphain[0;1]$. I want to approximate the solution $alpha$ using fixed-point method, for that, sitting
$x=g(x)=x+wf(x)$ an equivalent of $f(x)=0$. So we get for $g(0), g(1)in[0;1]$, $win[-1;0]$.



My question: $w$ belongs to which interval in order to apply the fixed-point method.






numerical-methods fixed-point-theorems







share|cite|improve this question











$endgroup$












  • $begingroup$
    And your take on this would be, what?
    $endgroup$
    – Did
    Jan 4 at 10:59










  • $begingroup$
    The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
    $endgroup$
    – HB khaled
    Jan 4 at 11:04










  • $begingroup$
    Yes. And? $ $ $ $
    $endgroup$
    – Did
    Jan 4 at 11:05










  • $begingroup$
    I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
    $endgroup$
    – HB khaled
    Jan 4 at 11:19


















0












$begingroup$


The equation $f(x)=x^3+x-1=0$ admites a unique solution $alphain[0;1]$. I want to approximate the solution $alpha$ using fixed-point method, for that, sitting
$x=g(x)=x+wf(x)$ an equivalent of $f(x)=0$. So we get for $g(0), g(1)in[0;1]$, $win[-1;0]$.



My question: $w$ belongs to which interval in order to apply the fixed-point method.






numerical-methods fixed-point-theorems







share|cite|improve this question











$endgroup$












  • $begingroup$
    And your take on this would be, what?
    $endgroup$
    – Did
    Jan 4 at 10:59










  • $begingroup$
    The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
    $endgroup$
    – HB khaled
    Jan 4 at 11:04










  • $begingroup$
    Yes. And? $ $ $ $
    $endgroup$
    – Did
    Jan 4 at 11:05










  • $begingroup$
    I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
    $endgroup$
    – HB khaled
    Jan 4 at 11:19
















0












0








0





$begingroup$


The equation $f(x)=x^3+x-1=0$ admites a unique solution $alphain[0;1]$. I want to approximate the solution $alpha$ using fixed-point method, for that, sitting
$x=g(x)=x+wf(x)$ an equivalent of $f(x)=0$. So we get for $g(0), g(1)in[0;1]$, $win[-1;0]$.



My question: $w$ belongs to which interval in order to apply the fixed-point method.






numerical-methods fixed-point-theorems







share|cite|improve this question











$endgroup$




The equation $f(x)=x^3+x-1=0$ admites a unique solution $alphain[0;1]$. I want to approximate the solution $alpha$ using fixed-point method, for that, sitting
$x=g(x)=x+wf(x)$ an equivalent of $f(x)=0$. So we get for $g(0), g(1)in[0;1]$, $win[-1;0]$.



My question: $w$ belongs to which interval in order to apply the fixed-point method.






numerical-methods fixed-point-theorems




numerical-methods fixed-point-theorems






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 11:29







HB khaled

















asked Jan 4 at 10:56









HB khaledHB khaled

12410




12410












  • $begingroup$
    And your take on this would be, what?
    $endgroup$
    – Did
    Jan 4 at 10:59










  • $begingroup$
    The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
    $endgroup$
    – HB khaled
    Jan 4 at 11:04










  • $begingroup$
    Yes. And? $ $ $ $
    $endgroup$
    – Did
    Jan 4 at 11:05










  • $begingroup$
    I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
    $endgroup$
    – HB khaled
    Jan 4 at 11:19




















  • $begingroup$
    And your take on this would be, what?
    $endgroup$
    – Did
    Jan 4 at 10:59










  • $begingroup$
    The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
    $endgroup$
    – HB khaled
    Jan 4 at 11:04










  • $begingroup$
    Yes. And? $ $ $ $
    $endgroup$
    – Did
    Jan 4 at 11:05










  • $begingroup$
    I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
    $endgroup$
    – HB khaled
    Jan 4 at 11:19


















$begingroup$
And your take on this would be, what?
$endgroup$
– Did
Jan 4 at 10:59




$begingroup$
And your take on this would be, what?
$endgroup$
– Did
Jan 4 at 10:59












$begingroup$
The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
$endgroup$
– HB khaled
Jan 4 at 11:04




$begingroup$
The fixed-point method cannot be applied for all $win[-1;0]$ see for example $w=-1$ and $w=-0.5$ do not work.
$endgroup$
– HB khaled
Jan 4 at 11:04












$begingroup$
Yes. And? $ $ $ $
$endgroup$
– Did
Jan 4 at 11:05




$begingroup$
Yes. And? $ $ $ $
$endgroup$
– Did
Jan 4 at 11:05












$begingroup$
I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
$endgroup$
– HB khaled
Jan 4 at 11:19






$begingroup$
I think that all $win [-1;-0.5]$ the method does not work, so $win [-0.5;0]$. And the other values of $w$?
$endgroup$
– HB khaled
Jan 4 at 11:19












1 Answer
1






active

oldest

votes


















2












$begingroup$

You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02


















2












$begingroup$

You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02
















2












2








2





$begingroup$

You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.






share|cite|improve this answer











$endgroup$



You can visualize this iteration using phase plots for the iteration $x_{k+1}=g(x_k)$.



phase portraits



The gallery shows behavior from periodic cycles and then convergence to the fixed point, the root of $f$, with fastest convergence around $w=-0.6$. You can quantify this by considering the contraction factor $$q=max_{xin[0,1]}|g'(x)|=max|1-wf'(x)|.$$ Then if $|2g(0.5)-1|=|2wf(0.5)|<1-q$ one has also $$g([0,1])subset[g(0.5)-0.5q,, g(0.5)+0.5q]subset[0,1]$$ by the mean value theorem, and thus a fixed point.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 14:08

























answered Jan 4 at 13:13









LutzLLutzL

59.6k42057




59.6k42057












  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02




















  • $begingroup$
    Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
    $endgroup$
    – HB khaled
    Jan 4 at 13:44










  • $begingroup$
    For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
    $endgroup$
    – LutzL
    Jan 4 at 14:02


















$begingroup$
Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
$endgroup$
– HB khaled
Jan 4 at 13:44




$begingroup$
Calculating $g'(x)=1+w(3x^2+1)$, this gives $max_{xin[0;1]}|g'(x)|=max|1+w(3x^2+1) |=|1+4w|$.
$endgroup$
– HB khaled
Jan 4 at 13:44












$begingroup$
For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
$endgroup$
– LutzL
Jan 4 at 14:02






$begingroup$
For a monotonous function crossing the zero axis you need to consider the absolute value at both interval ends, $q=max(|1+w|, |1+4w|)$. Then combine with $g(frac12)=frac12-frac38w$ to the condition $frac34|w|+max(|1+w|,|1+4w|)<1$. The left side is minimal for $w=-0.4$ with value $q=0.6$.
$endgroup$
– LutzL
Jan 4 at 14:02




















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