Prove that $(ϕ^n )' (α) = prod_{ i=0}^{n-1}ϕ'( ϕ^i (α)) $
$begingroup$
Prove that $displaystyle (phi^n )' (alpha) = prod_{ i=0}^{n-1}phi'( phi^i (alpha)) $ .
In particular, $alpha$ is a critical point of $phi^n$ if and only if one of the points $alpha, phi(alpha), ldots , phi^{n−1} (alpha)$ is a critical point of $phi$.
$phi(z) in Bbb C(z)$ is a rational function and $phi'(z)$ is the derivative of $phi(z)$
Here $phi^n=phicirc ldots circ phi text{($n$ times)}$
I was thinking that whether I have to use Taylor's theorem or not to prove the highlighted line or not.
As if we take the highlighted line to be true then:
$α$ is a critical point of $phi^n$ iff $displaystyle (phi^n )' (alpha)=0=prod_{ i=0}^{n-1}phi'( phi^i (alpha))$ iff $ phi'( phi^i (alpha))=0$ for some $i$ iff one of $alpha, phi(alpha), ldots , phi^{n−1} (alpha)$ is a critical point of $phi$.
Can you help me in proving the highlighted line?
calculus functions taylor-expansion dynamical-systems arithmetic-dynamics
edited Jan 10 at 7:29
El borito
668216
asked Dec 22 '18 at 3:09
GimgimGimgim
27113
$endgroup$
add a comment |
$begingroup$
Prove that $displaystyle (phi^n )' (alpha) = prod_{ i=0}^{n-1}phi'( phi^i (alpha)) $ .
In particular, $alpha$ is a critical point of $phi^n$ if and only if one of the points $alpha, phi(alpha), ldots , phi^{n−1} (alpha)$ is a critical point of $phi$.
$phi(z) in Bbb C(z)$ is a rational function and $phi'(z)$ is the derivative of $phi(z)$
Here $phi^n=phicirc ldots circ phi text{($n$ times)}$
I was thinking that whether I have to use Taylor's theorem or not to prove the highlighted line or not.
As if we take the highlighted line to be true then:
$α$ is a critical point of $phi^n$ iff $displaystyle (phi^n )' (alpha)=0=prod_{ i=0}^{n-1}phi'( phi^i (alpha))$ iff $ phi'( phi^i (alpha))=0$ for some $i$ iff one of $alpha, phi(alpha), ldots , phi^{n−1} (alpha)$ is a critical point of $phi$.
Can you help me in proving the highlighted line?
calculus functions taylor-expansion dynamical-systems arithmetic-dynamics
edited Jan 10 at 7:29
El borito
668216
asked Dec 22 '18 at 3:09
GimgimGimgim
27113
$endgroup$
4
$begingroup$
This is just the chain rule.
$endgroup$
– KReiser
Dec 22 '18 at 3:30
add a comment |
$begingroup$
Prove that $displaystyle (phi^n )' (alpha) = prod_{ i=0}^{n-1}phi'( phi^i (alpha)) $ .
In particular, $alpha$ is a critical point of $phi^n$ if and only if one of the points $alpha, phi(alpha), ldots , phi^{n−1} (alpha)$ is a critical point of $phi$.
$phi(z) in Bbb C(z)$ is a rational function and $phi'(z)$ is the derivative of $phi(z)$
Here $phi^n=phicirc ldots circ phi text{($n$ times)}$
I was thinking that whether I have to use Taylor's theorem or not to prove the highlighted line or not.
As if we take the highlighted line to be true then:
$α$ is a critical point of $phi^n$ iff $displaystyle (phi^n )' (alpha)=0=prod_{ i=0}^{n-1}phi'( phi^i (alpha))$ iff $ phi'( phi^i (alpha))=0$ for some $i$ iff one of $alpha, phi(alpha), ldots , phi^{n−1} (alpha)$ is a critical point of $phi$.
Can you help me in proving the highlighted line?
calculus functions taylor-expansion dynamical-systems arithmetic-dynamics
edited Jan 10 at 7:29
El borito
668216
asked Dec 22 '18 at 3:09
GimgimGimgim
27113
$endgroup$
Prove that $displaystyle (phi^n )' (alpha) = prod_{ i=0}^{n-1}phi'( phi^i (alpha)) $ .
In particular, $alpha$ is a critical point of $phi^n$ if and only if one of the points $alpha, phi(alpha), ldots , phi^{n−1} (alpha)$ is a critical point of $phi$.
$phi(z) in Bbb C(z)$ is a rational function and $phi'(z)$ is the derivative of $phi(z)$
Here $phi^n=phicirc ldots circ phi text{($n$ times)}$
I was thinking that whether I have to use Taylor's theorem or not to prove the highlighted line or not.
As if we take the highlighted line to be true then:
$α$ is a critical point of $phi^n$ iff $displaystyle (phi^n )' (alpha)=0=prod_{ i=0}^{n-1}phi'( phi^i (alpha))$ iff $ phi'( phi^i (alpha))=0$ for some $i$ iff one of $alpha, phi(alpha), ldots , phi^{n−1} (alpha)$ is a critical point of $phi$.
Can you help me in proving the highlighted line?
calculus functions taylor-expansion dynamical-systems arithmetic-dynamics
calculus functions taylor-expansion dynamical-systems arithmetic-dynamics
edited Jan 10 at 7:29
El borito
668216
asked Dec 22 '18 at 3:09
GimgimGimgim
27113
edited Jan 10 at 7:29
El borito
668216
asked Dec 22 '18 at 3:09
GimgimGimgim
27113
edited Jan 10 at 7:29
El borito
668216
edited Jan 10 at 7:29
El borito
668216
edited Jan 10 at 7:29
El borito
668216
668216
asked Dec 22 '18 at 3:09
GimgimGimgim
27113
asked Dec 22 '18 at 3:09
GimgimGimgim
27113
asked Dec 22 '18 at 3:09
GimgimGimgim
27113
27113
4
$begingroup$
This is just the chain rule.
$endgroup$
– KReiser
Dec 22 '18 at 3:30
add a comment |
4
$begingroup$
This is just the chain rule.
$endgroup$
– KReiser
Dec 22 '18 at 3:30
4
4
$begingroup$
This is just the chain rule.
$endgroup$
– KReiser
Dec 22 '18 at 3:30
$begingroup$
This is just the chain rule.
$endgroup$
– KReiser
Dec 22 '18 at 3:30
add a comment |
0
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4
$begingroup$
This is just the chain rule.
$endgroup$
– KReiser
Dec 22 '18 at 3:30