Why is this partial derivative zero? (Algebraic functions)
Why is $F'(a,z_i) ne 0$?
An algebraic function $y=f(x)$ is defined by the algebraic equation
$$ F(x,y) := g_n(x)y^n + g_{n-1}y^{n-1} + cdots + g_0(x) = 0 $$
where $g_j$ are polynomials. In what follows for simplicity we will
assume that $g_n(x)equiv 1$ (then the roots will not escape to infinity).
Let a point $nin mathbb{C}$ be such that the equation $F(x,y)=0$ has $n$
different roots $y=z_1,dots,z_n$. then $F'_y(a,z_i)ne 0$ and Implicit
Function Theorem asserts that for any $x$ from some ...
It looks to me like it should be something obvious but I’ve been trying to figure it out a while now. Have I missed something or is this from some deeper result a second year undergraduate such as myself may not have seen?
Original image
algebraic-curves
edited Dec 9 at 20:08
Somos
12.9k11034
asked Dec 9 at 18:37
oliver maclean
31
add a comment |
Why is $F'(a,z_i) ne 0$?
An algebraic function $y=f(x)$ is defined by the algebraic equation
$$ F(x,y) := g_n(x)y^n + g_{n-1}y^{n-1} + cdots + g_0(x) = 0 $$
where $g_j$ are polynomials. In what follows for simplicity we will
assume that $g_n(x)equiv 1$ (then the roots will not escape to infinity).
Let a point $nin mathbb{C}$ be such that the equation $F(x,y)=0$ has $n$
different roots $y=z_1,dots,z_n$. then $F'_y(a,z_i)ne 0$ and Implicit
Function Theorem asserts that for any $x$ from some ...
It looks to me like it should be something obvious but I’ve been trying to figure it out a while now. Have I missed something or is this from some deeper result a second year undergraduate such as myself may not have seen?
Original image
algebraic-curves
edited Dec 9 at 20:08
Somos
12.9k11034
asked Dec 9 at 18:37
oliver maclean
31
add a comment |
Why is $F'(a,z_i) ne 0$?
An algebraic function $y=f(x)$ is defined by the algebraic equation
$$ F(x,y) := g_n(x)y^n + g_{n-1}y^{n-1} + cdots + g_0(x) = 0 $$
where $g_j$ are polynomials. In what follows for simplicity we will
assume that $g_n(x)equiv 1$ (then the roots will not escape to infinity).
Let a point $nin mathbb{C}$ be such that the equation $F(x,y)=0$ has $n$
different roots $y=z_1,dots,z_n$. then $F'_y(a,z_i)ne 0$ and Implicit
Function Theorem asserts that for any $x$ from some ...
It looks to me like it should be something obvious but I’ve been trying to figure it out a while now. Have I missed something or is this from some deeper result a second year undergraduate such as myself may not have seen?
Original image
algebraic-curves
edited Dec 9 at 20:08
Somos
12.9k11034
asked Dec 9 at 18:37
oliver maclean
31
Why is $F'(a,z_i) ne 0$?
An algebraic function $y=f(x)$ is defined by the algebraic equation
$$ F(x,y) := g_n(x)y^n + g_{n-1}y^{n-1} + cdots + g_0(x) = 0 $$
where $g_j$ are polynomials. In what follows for simplicity we will
assume that $g_n(x)equiv 1$ (then the roots will not escape to infinity).
Let a point $nin mathbb{C}$ be such that the equation $F(x,y)=0$ has $n$
different roots $y=z_1,dots,z_n$. then $F'_y(a,z_i)ne 0$ and Implicit
Function Theorem asserts that for any $x$ from some ...
It looks to me like it should be something obvious but I’ve been trying to figure it out a while now. Have I missed something or is this from some deeper result a second year undergraduate such as myself may not have seen?
Original image
algebraic-curves
algebraic-curves
edited Dec 9 at 20:08
Somos
12.9k11034
asked Dec 9 at 18:37
oliver maclean
31
edited Dec 9 at 20:08
Somos
12.9k11034
asked Dec 9 at 18:37
oliver maclean
31
edited Dec 9 at 20:08
Somos
12.9k11034
edited Dec 9 at 20:08
Somos
12.9k11034
edited Dec 9 at 20:08
Somos
12.9k11034
12.9k11034
asked Dec 9 at 18:37
oliver maclean
31
asked Dec 9 at 18:37
oliver maclean
31
asked Dec 9 at 18:37
oliver maclean
31
31
add a comment |
add a comment |
1 Answer
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oldest
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I think it means that fixing $x=a$, the polynomial (thought of as a variable in $y$) $F(a,y)=c(y-z_1(a))...(y-z_n(a))$ where $z_1(a),...,z_n(a)$ are distinct complex numbers. Then you can check that $F'(a,z_i(a)) neq 0$. (It's because the roots are distinct.)
By allowing $x$ to vary, the roots will depend on $x$, so you have in general functions $z_1(x),...,z_n(x)$.
answered Dec 9 at 20:22
maridia
1,06912
Ah thank you! Writing the polynomial in terms of its factors definitely should have occurred to me!
– oliver maclean
Dec 10 at 6:28
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think it means that fixing $x=a$, the polynomial (thought of as a variable in $y$) $F(a,y)=c(y-z_1(a))...(y-z_n(a))$ where $z_1(a),...,z_n(a)$ are distinct complex numbers. Then you can check that $F'(a,z_i(a)) neq 0$. (It's because the roots are distinct.)
By allowing $x$ to vary, the roots will depend on $x$, so you have in general functions $z_1(x),...,z_n(x)$.
answered Dec 9 at 20:22
maridia
1,06912
Ah thank you! Writing the polynomial in terms of its factors definitely should have occurred to me!
– oliver maclean
Dec 10 at 6:28
add a comment |
I think it means that fixing $x=a$, the polynomial (thought of as a variable in $y$) $F(a,y)=c(y-z_1(a))...(y-z_n(a))$ where $z_1(a),...,z_n(a)$ are distinct complex numbers. Then you can check that $F'(a,z_i(a)) neq 0$. (It's because the roots are distinct.)
By allowing $x$ to vary, the roots will depend on $x$, so you have in general functions $z_1(x),...,z_n(x)$.
answered Dec 9 at 20:22
maridia
1,06912
Ah thank you! Writing the polynomial in terms of its factors definitely should have occurred to me!
– oliver maclean
Dec 10 at 6:28
add a comment |
I think it means that fixing $x=a$, the polynomial (thought of as a variable in $y$) $F(a,y)=c(y-z_1(a))...(y-z_n(a))$ where $z_1(a),...,z_n(a)$ are distinct complex numbers. Then you can check that $F'(a,z_i(a)) neq 0$. (It's because the roots are distinct.)
By allowing $x$ to vary, the roots will depend on $x$, so you have in general functions $z_1(x),...,z_n(x)$.
answered Dec 9 at 20:22
maridia
1,06912
I think it means that fixing $x=a$, the polynomial (thought of as a variable in $y$) $F(a,y)=c(y-z_1(a))...(y-z_n(a))$ where $z_1(a),...,z_n(a)$ are distinct complex numbers. Then you can check that $F'(a,z_i(a)) neq 0$. (It's because the roots are distinct.)
By allowing $x$ to vary, the roots will depend on $x$, so you have in general functions $z_1(x),...,z_n(x)$.
answered Dec 9 at 20:22
maridia
1,06912
edited Dec 9 at 20:28
answered Dec 9 at 20:22
maridia
1,06912
answered Dec 9 at 20:22
maridia
1,06912
answered Dec 9 at 20:22
maridia
1,06912
1,06912
Ah thank you! Writing the polynomial in terms of its factors definitely should have occurred to me!
– oliver maclean
Dec 10 at 6:28
add a comment |
Ah thank you! Writing the polynomial in terms of its factors definitely should have occurred to me!
– oliver maclean
Dec 10 at 6:28
Ah thank you! Writing the polynomial in terms of its factors definitely should have occurred to me!
– oliver maclean
Dec 10 at 6:28
Ah thank you! Writing the polynomial in terms of its factors definitely should have occurred to me!
– oliver maclean
Dec 10 at 6:28
add a comment |
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