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












0














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
























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










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










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










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


















0














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
























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










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










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










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
















0












0








0







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















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












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










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










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










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




















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










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










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










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


















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




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












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




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












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




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












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






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












1 Answer
1






active

oldest

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2














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























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










  • 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 $q=0.6$.
    – LutzL
    Jan 4 at 14:02













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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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























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










  • 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 $q=0.6$.
    – LutzL
    Jan 4 at 14:02


















2














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























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










  • 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 $q=0.6$.
    – LutzL
    Jan 4 at 14:02
















2












2








2






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














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

56.5k42054




56.5k42054












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










  • 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 $q=0.6$.
    – LutzL
    Jan 4 at 14:02




















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










  • 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 $q=0.6$.
    – LutzL
    Jan 4 at 14:02


















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




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












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 $q=0.6$.
– LutzL
Jan 4 at 14:02






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 $q=0.6$.
– LutzL
Jan 4 at 14:02




















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