Mixing
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Iterative method with squared function (edited)
$begingroup$
For solving a certain equation I've come up with this iterative method $$x^{n+1}=g^2(x^{n}),$$ where $g$ is given by
$$g(z)=frac{1}{2}left[sqrt{left(Acdottext{erf}(sqrt{z}/2)right)^2+Bcdotexpleft(-z/4right)}-Acdottext{erf}(sqrt{z}/2)right],$$ for certain coefficients $A,B>0$. The numerical method seems to work but I would like to prove that it is actually expected to work, using the fix point theorem I guess. I want to prove that $g^2$ is a contraction and I find
begin{align}
|x^{n+2}-x^{n+1}|&=|g^2(x^{n+1})-g^2(x^{n})|\
&<frac{A^2}{2}left|text{erf}left(sqrt{x^{n+1}}/2)right)^2-text{erf}left(sqrt{x^{n}}/2)right)^2right|\
&quad +frac{B}{4}left|expleft(-x^{n+1}/4right)-expleft(-x^{n}/4right)right|\
&quad +frac{A}{4}left|F(x^{n+1})-F(x^{n})right|,
end{align}
where $$F(z)=sqrt{left(Acdottext{erf}(sqrt{z}/2)right)^2+Bcdotexpleft(-z/4right)}cdottext{erf}(sqrt{z}/2).$$ The second term I can use the mean value theorem and I find, using $x^{n}>0$ $forall n$,
$$frac{B}{4}left|expleft(-x^{n+1}/4right)-expleft(-x^{n}/4right)right|<frac{B}{8}|x^{n+1}-x^{n}|.$$ Do I have to proceed the same way with the other two? Or is there a results which can help me out and tell me that the method will converge for sure?
Edit: I have tried another way. It turns out that $g$ satisfies:
begin{align}
&1.qquad g(0)=sqrt{B}/2,\
&2.qquad g(x)>0text{ for }x>0,\
&3.qquad g'(x)<0text{ for }x>0,
end{align}
the equation $x=g^2(x)$ has exactly one solution. Does this mean that the iterative method will converge?
real-analysis calculus fixed-point-theorems
asked Jan 31 at 11:00
MarcMarc
44139
$endgroup$
add a comment |
$begingroup$
For solving a certain equation I've come up with this iterative method $$x^{n+1}=g^2(x^{n}),$$ where $g$ is given by
$$g(z)=frac{1}{2}left[sqrt{left(Acdottext{erf}(sqrt{z}/2)right)^2+Bcdotexpleft(-z/4right)}-Acdottext{erf}(sqrt{z}/2)right],$$ for certain coefficients $A,B>0$. The numerical method seems to work but I would like to prove that it is actually expected to work, using the fix point theorem I guess. I want to prove that $g^2$ is a contraction and I find
begin{align}
|x^{n+2}-x^{n+1}|&=|g^2(x^{n+1})-g^2(x^{n})|\
&<frac{A^2}{2}left|text{erf}left(sqrt{x^{n+1}}/2)right)^2-text{erf}left(sqrt{x^{n}}/2)right)^2right|\
&quad +frac{B}{4}left|expleft(-x^{n+1}/4right)-expleft(-x^{n}/4right)right|\
&quad +frac{A}{4}left|F(x^{n+1})-F(x^{n})right|,
end{align}
where $$F(z)=sqrt{left(Acdottext{erf}(sqrt{z}/2)right)^2+Bcdotexpleft(-z/4right)}cdottext{erf}(sqrt{z}/2).$$ The second term I can use the mean value theorem and I find, using $x^{n}>0$ $forall n$,
$$frac{B}{4}left|expleft(-x^{n+1}/4right)-expleft(-x^{n}/4right)right|<frac{B}{8}|x^{n+1}-x^{n}|.$$ Do I have to proceed the same way with the other two? Or is there a results which can help me out and tell me that the method will converge for sure?
Edit: I have tried another way. It turns out that $g$ satisfies:
begin{align}
&1.qquad g(0)=sqrt{B}/2,\
&2.qquad g(x)>0text{ for }x>0,\
&3.qquad g'(x)<0text{ for }x>0,
end{align}
the equation $x=g^2(x)$ has exactly one solution. Does this mean that the iterative method will converge?
real-analysis calculus fixed-point-theorems
asked Jan 31 at 11:00
MarcMarc
44139
$endgroup$
add a comment |
$begingroup$
For solving a certain equation I've come up with this iterative method $$x^{n+1}=g^2(x^{n}),$$ where $g$ is given by
$$g(z)=frac{1}{2}left[sqrt{left(Acdottext{erf}(sqrt{z}/2)right)^2+Bcdotexpleft(-z/4right)}-Acdottext{erf}(sqrt{z}/2)right],$$ for certain coefficients $A,B>0$. The numerical method seems to work but I would like to prove that it is actually expected to work, using the fix point theorem I guess. I want to prove that $g^2$ is a contraction and I find
begin{align}
|x^{n+2}-x^{n+1}|&=|g^2(x^{n+1})-g^2(x^{n})|\
&<frac{A^2}{2}left|text{erf}left(sqrt{x^{n+1}}/2)right)^2-text{erf}left(sqrt{x^{n}}/2)right)^2right|\
&quad +frac{B}{4}left|expleft(-x^{n+1}/4right)-expleft(-x^{n}/4right)right|\
&quad +frac{A}{4}left|F(x^{n+1})-F(x^{n})right|,
end{align}
where $$F(z)=sqrt{left(Acdottext{erf}(sqrt{z}/2)right)^2+Bcdotexpleft(-z/4right)}cdottext{erf}(sqrt{z}/2).$$ The second term I can use the mean value theorem and I find, using $x^{n}>0$ $forall n$,
$$frac{B}{4}left|expleft(-x^{n+1}/4right)-expleft(-x^{n}/4right)right|<frac{B}{8}|x^{n+1}-x^{n}|.$$ Do I have to proceed the same way with the other two? Or is there a results which can help me out and tell me that the method will converge for sure?
Edit: I have tried another way. It turns out that $g$ satisfies:
begin{align}
&1.qquad g(0)=sqrt{B}/2,\
&2.qquad g(x)>0text{ for }x>0,\
&3.qquad g'(x)<0text{ for }x>0,
end{align}
the equation $x=g^2(x)$ has exactly one solution. Does this mean that the iterative method will converge?
real-analysis calculus fixed-point-theorems
asked Jan 31 at 11:00
MarcMarc
44139
$endgroup$
For solving a certain equation I've come up with this iterative method $$x^{n+1}=g^2(x^{n}),$$ where $g$ is given by
$$g(z)=frac{1}{2}left[sqrt{left(Acdottext{erf}(sqrt{z}/2)right)^2+Bcdotexpleft(-z/4right)}-Acdottext{erf}(sqrt{z}/2)right],$$ for certain coefficients $A,B>0$. The numerical method seems to work but I would like to prove that it is actually expected to work, using the fix point theorem I guess. I want to prove that $g^2$ is a contraction and I find
begin{align}
|x^{n+2}-x^{n+1}|&=|g^2(x^{n+1})-g^2(x^{n})|\
&<frac{A^2}{2}left|text{erf}left(sqrt{x^{n+1}}/2)right)^2-text{erf}left(sqrt{x^{n}}/2)right)^2right|\
&quad +frac{B}{4}left|expleft(-x^{n+1}/4right)-expleft(-x^{n}/4right)right|\
&quad +frac{A}{4}left|F(x^{n+1})-F(x^{n})right|,
end{align}
where $$F(z)=sqrt{left(Acdottext{erf}(sqrt{z}/2)right)^2+Bcdotexpleft(-z/4right)}cdottext{erf}(sqrt{z}/2).$$ The second term I can use the mean value theorem and I find, using $x^{n}>0$ $forall n$,
$$frac{B}{4}left|expleft(-x^{n+1}/4right)-expleft(-x^{n}/4right)right|<frac{B}{8}|x^{n+1}-x^{n}|.$$ Do I have to proceed the same way with the other two? Or is there a results which can help me out and tell me that the method will converge for sure?
Edit: I have tried another way. It turns out that $g$ satisfies:
begin{align}
&1.qquad g(0)=sqrt{B}/2,\
&2.qquad g(x)>0text{ for }x>0,\
&3.qquad g'(x)<0text{ for }x>0,
end{align}
the equation $x=g^2(x)$ has exactly one solution. Does this mean that the iterative method will converge?
real-analysis calculus fixed-point-theorems
real-analysis calculus fixed-point-theorems
asked Jan 31 at 11:00
MarcMarc
44139
asked Jan 31 at 11:00
MarcMarc
44139
edited Jan 31 at 13:43
Marc
asked Jan 31 at 11:00
MarcMarc
44139
asked Jan 31 at 11:00
MarcMarc
44139
asked Jan 31 at 11:00
MarcMarc
44139
44139
add a comment |
add a comment |
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