Fixed-point iteration: $sqrt{varepsilon} = sqrt{c - varepsilon} tan (a sqrt{c - varepsilon})$
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Let $a, c > 0$. Use the Banach fixed-point theorem to show that $$sqrt{varepsilon} = sqrt{c - varepsilon} tan (a sqrt{c - varepsilon})$$ has at least one solution $varepsilon in (0, c)$ for $c$ small enough and find an a-priori bound. Hint: Set $x = sqrt{varepsilon}$ and transform the equation into a fixed-point equation for $x$.
I tried to show that $|f (x^2) - f (y^2)| < |x - y|$ for $x, y$ small, where $f (varepsilon) := sqrt{c - varepsilon} tan (a sqrt{c - varepsilon})$, with the mean value theorem. However, $f'$ is complicated and I fail to see how $|f' (z^2)| < 1$ for some $z$, $c$ small enough (let alone the property $f : [0, c] rightarrow [0, c]$).
Alternatively, I tried to show the Banach properties for $x tan (a x)$ (which gives an easier derivative) and substitute subsequently $x = sqrt{c - varepsilon}$. Assuming that this is admissible, I fail to show, though, that $|sqrt{c - varepsilon} - sqrt{c - varepsilon'}| le |sqrt{varepsilon} - sqrt{varepsilon'}|$.
fixed-point-theorems fixedpoints
asked Nov 18 at 11:17
Andy
606
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Let $a, c > 0$. Use the Banach fixed-point theorem to show that $$sqrt{varepsilon} = sqrt{c - varepsilon} tan (a sqrt{c - varepsilon})$$ has at least one solution $varepsilon in (0, c)$ for $c$ small enough and find an a-priori bound. Hint: Set $x = sqrt{varepsilon}$ and transform the equation into a fixed-point equation for $x$.
I tried to show that $|f (x^2) - f (y^2)| < |x - y|$ for $x, y$ small, where $f (varepsilon) := sqrt{c - varepsilon} tan (a sqrt{c - varepsilon})$, with the mean value theorem. However, $f'$ is complicated and I fail to see how $|f' (z^2)| < 1$ for some $z$, $c$ small enough (let alone the property $f : [0, c] rightarrow [0, c]$).
Alternatively, I tried to show the Banach properties for $x tan (a x)$ (which gives an easier derivative) and substitute subsequently $x = sqrt{c - varepsilon}$. Assuming that this is admissible, I fail to show, though, that $|sqrt{c - varepsilon} - sqrt{c - varepsilon'}| le |sqrt{varepsilon} - sqrt{varepsilon'}|$.
fixed-point-theorems fixedpoints
asked Nov 18 at 11:17
Andy
606
This is probably a very naive approach and not the one you're looking for, but would it be sufficient to prove that there exists an number $x_1$ absolutely less than $pi/2a$ such that $x_1 tan(ax_1)<sqrt{c-{x_1}^2}$ and that there also exists another number $x_2$ also absolutely less than $pi/2a$ such that $x_2 tan(ax_2)>sqrt{c-{x_2}^2}$ ? The proof would then follow from the continuity of $xtan(ax)$ on the interval $(-pi/2a,pi/2a)$, if my thinking is correct.
– Aaron Quitta
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $a, c > 0$. Use the Banach fixed-point theorem to show that $$sqrt{varepsilon} = sqrt{c - varepsilon} tan (a sqrt{c - varepsilon})$$ has at least one solution $varepsilon in (0, c)$ for $c$ small enough and find an a-priori bound. Hint: Set $x = sqrt{varepsilon}$ and transform the equation into a fixed-point equation for $x$.
I tried to show that $|f (x^2) - f (y^2)| < |x - y|$ for $x, y$ small, where $f (varepsilon) := sqrt{c - varepsilon} tan (a sqrt{c - varepsilon})$, with the mean value theorem. However, $f'$ is complicated and I fail to see how $|f' (z^2)| < 1$ for some $z$, $c$ small enough (let alone the property $f : [0, c] rightarrow [0, c]$).
Alternatively, I tried to show the Banach properties for $x tan (a x)$ (which gives an easier derivative) and substitute subsequently $x = sqrt{c - varepsilon}$. Assuming that this is admissible, I fail to show, though, that $|sqrt{c - varepsilon} - sqrt{c - varepsilon'}| le |sqrt{varepsilon} - sqrt{varepsilon'}|$.
fixed-point-theorems fixedpoints
asked Nov 18 at 11:17
Andy
606
Let $a, c > 0$. Use the Banach fixed-point theorem to show that $$sqrt{varepsilon} = sqrt{c - varepsilon} tan (a sqrt{c - varepsilon})$$ has at least one solution $varepsilon in (0, c)$ for $c$ small enough and find an a-priori bound. Hint: Set $x = sqrt{varepsilon}$ and transform the equation into a fixed-point equation for $x$.
I tried to show that $|f (x^2) - f (y^2)| < |x - y|$ for $x, y$ small, where $f (varepsilon) := sqrt{c - varepsilon} tan (a sqrt{c - varepsilon})$, with the mean value theorem. However, $f'$ is complicated and I fail to see how $|f' (z^2)| < 1$ for some $z$, $c$ small enough (let alone the property $f : [0, c] rightarrow [0, c]$).
Alternatively, I tried to show the Banach properties for $x tan (a x)$ (which gives an easier derivative) and substitute subsequently $x = sqrt{c - varepsilon}$. Assuming that this is admissible, I fail to show, though, that $|sqrt{c - varepsilon} - sqrt{c - varepsilon'}| le |sqrt{varepsilon} - sqrt{varepsilon'}|$.
fixed-point-theorems fixedpoints
fixed-point-theorems fixedpoints
asked Nov 18 at 11:17
Andy
606
asked Nov 18 at 11:17
Andy
606
asked Nov 18 at 11:17
Andy
606
asked Nov 18 at 11:17
Andy
606
asked Nov 18 at 11:17
Andy
606
606
This is probably a very naive approach and not the one you're looking for, but would it be sufficient to prove that there exists an number $x_1$ absolutely less than $pi/2a$ such that $x_1 tan(ax_1)<sqrt{c-{x_1}^2}$ and that there also exists another number $x_2$ also absolutely less than $pi/2a$ such that $x_2 tan(ax_2)>sqrt{c-{x_2}^2}$ ? The proof would then follow from the continuity of $xtan(ax)$ on the interval $(-pi/2a,pi/2a)$, if my thinking is correct.
– Aaron Quitta
yesterday
add a comment |
This is probably a very naive approach and not the one you're looking for, but would it be sufficient to prove that there exists an number $x_1$ absolutely less than $pi/2a$ such that $x_1 tan(ax_1)<sqrt{c-{x_1}^2}$ and that there also exists another number $x_2$ also absolutely less than $pi/2a$ such that $x_2 tan(ax_2)>sqrt{c-{x_2}^2}$ ? The proof would then follow from the continuity of $xtan(ax)$ on the interval $(-pi/2a,pi/2a)$, if my thinking is correct.
– Aaron Quitta
yesterday
This is probably a very naive approach and not the one you're looking for, but would it be sufficient to prove that there exists an number $x_1$ absolutely less than $pi/2a$ such that $x_1 tan(ax_1)<sqrt{c-{x_1}^2}$ and that there also exists another number $x_2$ also absolutely less than $pi/2a$ such that $x_2 tan(ax_2)>sqrt{c-{x_2}^2}$ ? The proof would then follow from the continuity of $xtan(ax)$ on the interval $(-pi/2a,pi/2a)$, if my thinking is correct.
– Aaron Quitta
yesterday
This is probably a very naive approach and not the one you're looking for, but would it be sufficient to prove that there exists an number $x_1$ absolutely less than $pi/2a$ such that $x_1 tan(ax_1)<sqrt{c-{x_1}^2}$ and that there also exists another number $x_2$ also absolutely less than $pi/2a$ such that $x_2 tan(ax_2)>sqrt{c-{x_2}^2}$ ? The proof would then follow from the continuity of $xtan(ax)$ on the interval $(-pi/2a,pi/2a)$, if my thinking is correct.
– Aaron Quitta
yesterday
add a comment |
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This is probably a very naive approach and not the one you're looking for, but would it be sufficient to prove that there exists an number $x_1$ absolutely less than $pi/2a$ such that $x_1 tan(ax_1)<sqrt{c-{x_1}^2}$ and that there also exists another number $x_2$ also absolutely less than $pi/2a$ such that $x_2 tan(ax_2)>sqrt{c-{x_2}^2}$ ? The proof would then follow from the continuity of $xtan(ax)$ on the interval $(-pi/2a,pi/2a)$, if my thinking is correct.
– Aaron Quitta
yesterday