Prove that this function is a contraction
$begingroup$
Let $f in C^2([a,b])$ be a function with the following properties:
(i) $f(a)f(b) <0 $
(ii) $f'(x) neq 0$ $forall x$
(iii) $frac{|f(a)|} {|f'(a)|} <(b-a)$ and $frac{|f(b)|} {|f'(b)|} <(b-a)$
(iv) $ f''(x) leq 0$ $forall x$
Then $h(x) := x-frac{f(x)} {f'(x)} $ is a contraction.
What I did is:
Suppose wlog $f(a)<0<f(b)$ and so $f'(x)>0$. We have
$|h'(x)| leq... leq L<1$
I don't know how to do the middle part, maybe we have $frac{|f(x)|} {|f'(x)|} <(b-a)$ $forall x$? (not sure)
Curiosity: thie statement will prove a theorem of converge for the Newton method
calculus functions contraction-operator
asked Dec 28 '18 at 4:09
Naj KampNaj Kamp
1178
$endgroup$
add a comment |
$begingroup$
Let $f in C^2([a,b])$ be a function with the following properties:
(i) $f(a)f(b) <0 $
(ii) $f'(x) neq 0$ $forall x$
(iii) $frac{|f(a)|} {|f'(a)|} <(b-a)$ and $frac{|f(b)|} {|f'(b)|} <(b-a)$
(iv) $ f''(x) leq 0$ $forall x$
Then $h(x) := x-frac{f(x)} {f'(x)} $ is a contraction.
What I did is:
Suppose wlog $f(a)<0<f(b)$ and so $f'(x)>0$. We have
$|h'(x)| leq... leq L<1$
I don't know how to do the middle part, maybe we have $frac{|f(x)|} {|f'(x)|} <(b-a)$ $forall x$? (not sure)
Curiosity: thie statement will prove a theorem of converge for the Newton method
calculus functions contraction-operator
asked Dec 28 '18 at 4:09
Naj KampNaj Kamp
1178
$endgroup$
$begingroup$
We can prove that $frac{|f(x)|}{|f'(x)|} < b-a$ for all $x$. Indeed, $f$ is increasing since $f' >0$ while $f'$ is decreasing by assumption, so that $frac{f}{f'}$ is increasing. Since the inequality is satisfied at both endpoints, it must be true for all $x in [a,b]$.
$endgroup$
– Michh
Dec 29 '18 at 5:43
add a comment |
$begingroup$
Let $f in C^2([a,b])$ be a function with the following properties:
(i) $f(a)f(b) <0 $
(ii) $f'(x) neq 0$ $forall x$
(iii) $frac{|f(a)|} {|f'(a)|} <(b-a)$ and $frac{|f(b)|} {|f'(b)|} <(b-a)$
(iv) $ f''(x) leq 0$ $forall x$
Then $h(x) := x-frac{f(x)} {f'(x)} $ is a contraction.
What I did is:
Suppose wlog $f(a)<0<f(b)$ and so $f'(x)>0$. We have
$|h'(x)| leq... leq L<1$
I don't know how to do the middle part, maybe we have $frac{|f(x)|} {|f'(x)|} <(b-a)$ $forall x$? (not sure)
Curiosity: thie statement will prove a theorem of converge for the Newton method
calculus functions contraction-operator
asked Dec 28 '18 at 4:09
Naj KampNaj Kamp
1178
$endgroup$
Let $f in C^2([a,b])$ be a function with the following properties:
(i) $f(a)f(b) <0 $
(ii) $f'(x) neq 0$ $forall x$
(iii) $frac{|f(a)|} {|f'(a)|} <(b-a)$ and $frac{|f(b)|} {|f'(b)|} <(b-a)$
(iv) $ f''(x) leq 0$ $forall x$
Then $h(x) := x-frac{f(x)} {f'(x)} $ is a contraction.
What I did is:
Suppose wlog $f(a)<0<f(b)$ and so $f'(x)>0$. We have
$|h'(x)| leq... leq L<1$
I don't know how to do the middle part, maybe we have $frac{|f(x)|} {|f'(x)|} <(b-a)$ $forall x$? (not sure)
Curiosity: thie statement will prove a theorem of converge for the Newton method
calculus functions contraction-operator
calculus functions contraction-operator
asked Dec 28 '18 at 4:09
Naj KampNaj Kamp
1178
asked Dec 28 '18 at 4:09
Naj KampNaj Kamp
1178
edited Dec 28 '18 at 4:24
Naj Kamp
asked Dec 28 '18 at 4:09
Naj KampNaj Kamp
1178
asked Dec 28 '18 at 4:09
Naj KampNaj Kamp
1178
asked Dec 28 '18 at 4:09
Naj KampNaj Kamp
1178
1178
$begingroup$
We can prove that $frac{|f(x)|}{|f'(x)|} < b-a$ for all $x$. Indeed, $f$ is increasing since $f' >0$ while $f'$ is decreasing by assumption, so that $frac{f}{f'}$ is increasing. Since the inequality is satisfied at both endpoints, it must be true for all $x in [a,b]$.
$endgroup$
– Michh
Dec 29 '18 at 5:43
add a comment |
$begingroup$
We can prove that $frac{|f(x)|}{|f'(x)|} < b-a$ for all $x$. Indeed, $f$ is increasing since $f' >0$ while $f'$ is decreasing by assumption, so that $frac{f}{f'}$ is increasing. Since the inequality is satisfied at both endpoints, it must be true for all $x in [a,b]$.
$endgroup$
– Michh
Dec 29 '18 at 5:43
$begingroup$
We can prove that $frac{|f(x)|}{|f'(x)|} < b-a$ for all $x$. Indeed, $f$ is increasing since $f' >0$ while $f'$ is decreasing by assumption, so that $frac{f}{f'}$ is increasing. Since the inequality is satisfied at both endpoints, it must be true for all $x in [a,b]$.
$endgroup$
– Michh
Dec 29 '18 at 5:43
$begingroup$
We can prove that $frac{|f(x)|}{|f'(x)|} < b-a$ for all $x$. Indeed, $f$ is increasing since $f' >0$ while $f'$ is decreasing by assumption, so that $frac{f}{f'}$ is increasing. Since the inequality is satisfied at both endpoints, it must be true for all $x in [a,b]$.
$endgroup$
– Michh
Dec 29 '18 at 5:43
add a comment |
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$begingroup$
We can prove that $frac{|f(x)|}{|f'(x)|} < b-a$ for all $x$. Indeed, $f$ is increasing since $f' >0$ while $f'$ is decreasing by assumption, so that $frac{f}{f'}$ is increasing. Since the inequality is satisfied at both endpoints, it must be true for all $x in [a,b]$.
$endgroup$
– Michh
Dec 29 '18 at 5:43