Mixing
Differentiability of simple zero of a polynomial curve
$begingroup$
Suppose $f(x, t) = x^n + a_{n-1}(t) x^{n-1} + dots + a_0(t) in mathbb R[x]$ where each $a_j(t) in C^{infty}(mathbb R)$. If $x_0 in mathbb C$ is a simple zero of $f(x, t_0)$, I want to know if there is a $C^{infty}$ function $eta: I to mathbb C$ such that $eta(t_0) = x_0$ and $eta(t)$ is a zero for $f(x, t)$ for every $t in I$ and $I$ is some interval with $t_0 in I$.
Clearly, a zero should be a function over the coefficients and thus a function over $t$. By chain rule, we can formally differentiate with respect to $t$ and get $f(x(t))' = f'(x(t)) x'(t)$ here $x$ can be equivalently viewed as $eta$. At $t_0$, $f'(x(t_0)) = f'(x_0) neq 0$. I want to say by inverse function theorem, $eta$ is defined for some $I$. But I got suspicious over the first step. I am not convinced why I can do that?
abstract-algebra complex-analysis polynomials
asked Jan 19 at 0:59
MyCindy2012MyCindy2012
729
$endgroup$
add a comment |
$begingroup$
Suppose $f(x, t) = x^n + a_{n-1}(t) x^{n-1} + dots + a_0(t) in mathbb R[x]$ where each $a_j(t) in C^{infty}(mathbb R)$. If $x_0 in mathbb C$ is a simple zero of $f(x, t_0)$, I want to know if there is a $C^{infty}$ function $eta: I to mathbb C$ such that $eta(t_0) = x_0$ and $eta(t)$ is a zero for $f(x, t)$ for every $t in I$ and $I$ is some interval with $t_0 in I$.
Clearly, a zero should be a function over the coefficients and thus a function over $t$. By chain rule, we can formally differentiate with respect to $t$ and get $f(x(t))' = f'(x(t)) x'(t)$ here $x$ can be equivalently viewed as $eta$. At $t_0$, $f'(x(t_0)) = f'(x_0) neq 0$. I want to say by inverse function theorem, $eta$ is defined for some $I$. But I got suspicious over the first step. I am not convinced why I can do that?
abstract-algebra complex-analysis polynomials
asked Jan 19 at 0:59
MyCindy2012MyCindy2012
729
$endgroup$
1
$begingroup$
When there is a formula to work out the roots, is obvious. But what about other situation?
$endgroup$
– Hugo
Jan 19 at 1:17
$begingroup$
Very good question, endorsed, +1!!! But I'm having a little trouble deciphering what it is you refer to as "the first step". Care to explicate? Thanks!
$endgroup$
– Robert Lewis
Jan 19 at 2:18
1
$begingroup$
Why there is such a smooth function, I guess.
$endgroup$
– Hugo
Jan 19 at 2:49
1
$begingroup$
@RobertLewis: If I am not mistaken, there should be a continuous function $eta: I to mathbb C$ satisfying the prescribed condition. But I don't know whether I can just naively take derivatives.
$endgroup$
– MyCindy2012
Jan 19 at 3:25
add a comment |
$begingroup$
Suppose $f(x, t) = x^n + a_{n-1}(t) x^{n-1} + dots + a_0(t) in mathbb R[x]$ where each $a_j(t) in C^{infty}(mathbb R)$. If $x_0 in mathbb C$ is a simple zero of $f(x, t_0)$, I want to know if there is a $C^{infty}$ function $eta: I to mathbb C$ such that $eta(t_0) = x_0$ and $eta(t)$ is a zero for $f(x, t)$ for every $t in I$ and $I$ is some interval with $t_0 in I$.
Clearly, a zero should be a function over the coefficients and thus a function over $t$. By chain rule, we can formally differentiate with respect to $t$ and get $f(x(t))' = f'(x(t)) x'(t)$ here $x$ can be equivalently viewed as $eta$. At $t_0$, $f'(x(t_0)) = f'(x_0) neq 0$. I want to say by inverse function theorem, $eta$ is defined for some $I$. But I got suspicious over the first step. I am not convinced why I can do that?
abstract-algebra complex-analysis polynomials
asked Jan 19 at 0:59
MyCindy2012MyCindy2012
729
$endgroup$
Suppose $f(x, t) = x^n + a_{n-1}(t) x^{n-1} + dots + a_0(t) in mathbb R[x]$ where each $a_j(t) in C^{infty}(mathbb R)$. If $x_0 in mathbb C$ is a simple zero of $f(x, t_0)$, I want to know if there is a $C^{infty}$ function $eta: I to mathbb C$ such that $eta(t_0) = x_0$ and $eta(t)$ is a zero for $f(x, t)$ for every $t in I$ and $I$ is some interval with $t_0 in I$.
Clearly, a zero should be a function over the coefficients and thus a function over $t$. By chain rule, we can formally differentiate with respect to $t$ and get $f(x(t))' = f'(x(t)) x'(t)$ here $x$ can be equivalently viewed as $eta$. At $t_0$, $f'(x(t_0)) = f'(x_0) neq 0$. I want to say by inverse function theorem, $eta$ is defined for some $I$. But I got suspicious over the first step. I am not convinced why I can do that?
abstract-algebra complex-analysis polynomials
abstract-algebra complex-analysis polynomials
asked Jan 19 at 0:59
MyCindy2012MyCindy2012
729
asked Jan 19 at 0:59
MyCindy2012MyCindy2012
729
edited Jan 19 at 4:13
MyCindy2012
asked Jan 19 at 0:59
MyCindy2012MyCindy2012
729
asked Jan 19 at 0:59
MyCindy2012MyCindy2012
729
asked Jan 19 at 0:59
MyCindy2012MyCindy2012
729
729
1
$begingroup$
When there is a formula to work out the roots, is obvious. But what about other situation?
$endgroup$
– Hugo
Jan 19 at 1:17
$begingroup$
Very good question, endorsed, +1!!! But I'm having a little trouble deciphering what it is you refer to as "the first step". Care to explicate? Thanks!
$endgroup$
– Robert Lewis
Jan 19 at 2:18
1
$begingroup$
Why there is such a smooth function, I guess.
$endgroup$
– Hugo
Jan 19 at 2:49
1
$begingroup$
@RobertLewis: If I am not mistaken, there should be a continuous function $eta: I to mathbb C$ satisfying the prescribed condition. But I don't know whether I can just naively take derivatives.
$endgroup$
– MyCindy2012
Jan 19 at 3:25
add a comment |
1
$begingroup$
When there is a formula to work out the roots, is obvious. But what about other situation?
$endgroup$
– Hugo
Jan 19 at 1:17
$begingroup$
Very good question, endorsed, +1!!! But I'm having a little trouble deciphering what it is you refer to as "the first step". Care to explicate? Thanks!
$endgroup$
– Robert Lewis
Jan 19 at 2:18
1
$begingroup$
Why there is such a smooth function, I guess.
$endgroup$
– Hugo
Jan 19 at 2:49
1
$begingroup$
@RobertLewis: If I am not mistaken, there should be a continuous function $eta: I to mathbb C$ satisfying the prescribed condition. But I don't know whether I can just naively take derivatives.
$endgroup$
– MyCindy2012
Jan 19 at 3:25
1
1
$begingroup$
When there is a formula to work out the roots, is obvious. But what about other situation?
$endgroup$
– Hugo
Jan 19 at 1:17
$begingroup$
When there is a formula to work out the roots, is obvious. But what about other situation?
$endgroup$
– Hugo
Jan 19 at 1:17
$begingroup$
Very good question, endorsed, +1!!! But I'm having a little trouble deciphering what it is you refer to as "the first step". Care to explicate? Thanks!
$endgroup$
– Robert Lewis
Jan 19 at 2:18
$begingroup$
Very good question, endorsed, +1!!! But I'm having a little trouble deciphering what it is you refer to as "the first step". Care to explicate? Thanks!
$endgroup$
– Robert Lewis
Jan 19 at 2:18
1
1
$begingroup$
Why there is such a smooth function, I guess.
$endgroup$
– Hugo
Jan 19 at 2:49
$begingroup$
Why there is such a smooth function, I guess.
$endgroup$
– Hugo
Jan 19 at 2:49
1
1
$begingroup$
@RobertLewis: If I am not mistaken, there should be a continuous function $eta: I to mathbb C$ satisfying the prescribed condition. But I don't know whether I can just naively take derivatives.
$endgroup$
– MyCindy2012
Jan 19 at 3:25
$begingroup$
@RobertLewis: If I am not mistaken, there should be a continuous function $eta: I to mathbb C$ satisfying the prescribed condition. But I don't know whether I can just naively take derivatives.
$endgroup$
– MyCindy2012
Jan 19 at 3:25
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
This is essentially the implicit function theorem in this context.
The condition that $x_0$ is a simple root of $f(x,t_0)$ is the statement that $(partial f/partial x)neq 0$ at $(x_0,t_0)$.
As in the proof of the implicit function theorem, you can start with $eta_0(t)=x_0$ and use successive iterates of the Newton-Raphson method to provide better and better approximations of the desired $eta(t)$ as
$$
eta_{k+1}(t) = eta_k(t) - frac{f(eta_k(t),t)}{(partial f/partial t)(x_0,t_0)}
$$
Obviously, this will only work sufficiently close to $t_0$.
answered Jan 19 at 4:59
KapilKapil
37114
$endgroup$
add a comment |
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$begingroup$
This is essentially the implicit function theorem in this context.
The condition that $x_0$ is a simple root of $f(x,t_0)$ is the statement that $(partial f/partial x)neq 0$ at $(x_0,t_0)$.
As in the proof of the implicit function theorem, you can start with $eta_0(t)=x_0$ and use successive iterates of the Newton-Raphson method to provide better and better approximations of the desired $eta(t)$ as
$$
eta_{k+1}(t) = eta_k(t) - frac{f(eta_k(t),t)}{(partial f/partial t)(x_0,t_0)}
$$
Obviously, this will only work sufficiently close to $t_0$.
answered Jan 19 at 4:59
KapilKapil
37114
$endgroup$
add a comment |
$begingroup$
This is essentially the implicit function theorem in this context.
The condition that $x_0$ is a simple root of $f(x,t_0)$ is the statement that $(partial f/partial x)neq 0$ at $(x_0,t_0)$.
As in the proof of the implicit function theorem, you can start with $eta_0(t)=x_0$ and use successive iterates of the Newton-Raphson method to provide better and better approximations of the desired $eta(t)$ as
$$
eta_{k+1}(t) = eta_k(t) - frac{f(eta_k(t),t)}{(partial f/partial t)(x_0,t_0)}
$$
Obviously, this will only work sufficiently close to $t_0$.
answered Jan 19 at 4:59
KapilKapil
37114
$endgroup$
add a comment |
$begingroup$
This is essentially the implicit function theorem in this context.
The condition that $x_0$ is a simple root of $f(x,t_0)$ is the statement that $(partial f/partial x)neq 0$ at $(x_0,t_0)$.
As in the proof of the implicit function theorem, you can start with $eta_0(t)=x_0$ and use successive iterates of the Newton-Raphson method to provide better and better approximations of the desired $eta(t)$ as
$$
eta_{k+1}(t) = eta_k(t) - frac{f(eta_k(t),t)}{(partial f/partial t)(x_0,t_0)}
$$
Obviously, this will only work sufficiently close to $t_0$.
answered Jan 19 at 4:59
KapilKapil
37114
$endgroup$
This is essentially the implicit function theorem in this context.
The condition that $x_0$ is a simple root of $f(x,t_0)$ is the statement that $(partial f/partial x)neq 0$ at $(x_0,t_0)$.
As in the proof of the implicit function theorem, you can start with $eta_0(t)=x_0$ and use successive iterates of the Newton-Raphson method to provide better and better approximations of the desired $eta(t)$ as
$$
eta_{k+1}(t) = eta_k(t) - frac{f(eta_k(t),t)}{(partial f/partial t)(x_0,t_0)}
$$
Obviously, this will only work sufficiently close to $t_0$.
answered Jan 19 at 4:59
KapilKapil
37114
answered Jan 19 at 4:59
KapilKapil
37114
answered Jan 19 at 4:59
KapilKapil
37114
answered Jan 19 at 4:59
KapilKapil
37114
37114
add a comment |
add a comment |
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1
$begingroup$
When there is a formula to work out the roots, is obvious. But what about other situation?
$endgroup$
– Hugo
Jan 19 at 1:17
$begingroup$
Very good question, endorsed, +1!!! But I'm having a little trouble deciphering what it is you refer to as "the first step". Care to explicate? Thanks!
$endgroup$
– Robert Lewis
Jan 19 at 2:18
1
$begingroup$
Why there is such a smooth function, I guess.
$endgroup$
– Hugo
Jan 19 at 2:49
1
$begingroup$
@RobertLewis: If I am not mistaken, there should be a continuous function $eta: I to mathbb C$ satisfying the prescribed condition. But I don't know whether I can just naively take derivatives.
$endgroup$
– MyCindy2012
Jan 19 at 3:25