Calculate the order of the point

[evinda]

Hello! (Wave)

I want to calculate the order of the point $M$ at the curve $f(x,y) \in \mathbb{R}[x,y]$, when:

$$M(0,1), f=(x^2-1)^2-y^2(3-2y)$$

That's what I have tried:

$$f(x,y)=x^4-2x^2+1-3y^2+2y^3$$

$$f_4(x,y)=x^4$$

$$f_3(x,y)=2y^3$$

$$f_2(x,y)=-2x^2-3y^2$$

$$f_0(x,y)=1$$We check if $M$ is singular.

$$f(0,1)=0$$
$$f_x(x,y)=4x^3-4x$$
$$f_x(0,1)=0$$
$$f_y(x,y)=-6y+6y^2$$
$$f_y(0,1)=0$$

Taylor expansion of $f(x,y)$ at the point $M(0,1)$

$$f(x,y)=f(0,1)+[f_x(0,1)(x-0)+f_y(0,1)(y-1)]+ \dots$$

$$f(x,y)=-2x^2+3(y-1)^2+2(y-1)^3+x^4$$

So, the order of $M$ is $2$.

Could you tell me if it is right or if I have done something wrong? (Thinking)

 

[I like Serena]

$$f(x,y)=-2x^2+3(y-1)^2+2(y-1)^3+x^4$$

So, the order of $M$ is $2$.

Could you tell me if it is right or if I have done something wrong? (Thinking)

Hii (Smile)

I'm not really aware of anything called "the order of a point", but since your polynomial has $x^4$ as its highest order term, shouldn't the order be $4$? (Wondering)

 

[evinda]

Hii (Smile)

I'm not really aware of anything called "the order of a point", but since your polynomial has $x^4$ as its highest order term, shouldn't the order be $4$? (Wondering)

According to my notes:

Let $f$ be an algebraic curve, $f(x,y) \in \mathbb{C}[x,y]$.
$M \in \mathbb{C}^2$ singular point of the curve
$\Leftrightarrow \left\{\begin{matrix}
f(x,y)=0\\ \frac{\partial f}{\partial x}(M)=0\\ \frac{\partial f}{\partial y}(M)=0

\end{matrix}\right.$

Each polynomial $f(x,y) \in K[x,y]$ can be written in the form:

$$f(x,y)=f_m(x,y)+f_{m+1}(x,y)+ \dots + f_n(x,y)$$

where $f_i(x,y)$ is a homogeneous polynomial of degree $i$ and $f_m(x,y) \not\equiv 0, f_n(x,y) \not\equiv 0$.

The order of the singular point $M=(a,b)$ of the curve $f$ is defined as the natural number $m$. (Nerd)

 

[I like Serena]

According to my notes:

Let $f$ be an algebraic curve, $f(x,y) \in \mathbb{C}[x,y]$.
$M \in \mathbb{C}^2$ singular point of the curve
$\Leftrightarrow \left\{\begin{matrix}
f(x,y)=0\\ \frac{\partial f}{\partial x}(M)=0\\ \frac{\partial f}{\partial y}(M)=0

\end{matrix}\right.$

Each polynomial $f(x,y) \in K[x,y]$ can be written in the form:

$$f(x,y)=f_m(x,y)+f_{m+1}(x,y)+ \dots + f_n(x,y)$$

where $f_i(x,y)$ is a homogeneous polynomial of degree $i$ and $f_m(x,y) \not\equiv 0, f_n(x,y) \not\equiv 0$.

The order of the singular point $M=(a,b)$ of the curve $f$ is defined as the natural number $m$. (Nerd)

Aha!
That makes sense. (Smile)

In that case I believe it is correct that the order of $M$ is $2$. (Wasntme)

 

[evinda]

Aha!
That makes sense. (Smile)

In that case I believe it is correct that the order of $M$ is $2$. (Wasntme)

Nice! Could we also use this formula:

$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$

for the taylor expansion? (Thinking)

 

[I like Serena]

Nice! Could we also use this formula:

$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$

for the taylor expansion? (Thinking)

That formula only has $x$ in it, and no $y$. (Worried)We can do it, but we'll need the generalized version of that formula for several variables:
[box=]Multi dimensional form of Taylor expansion
$$f(\mathbf x)=\sum_{n=0}^{\infty} \frac 1 {n!}\mathbf D^nf(\mathbf a)(\mathbf x-\mathbf a)^n$$
where $\mathbf x = (x,y)$, $\mathbf a = (a,b)$, and $\mathbf D^nf$ is the generalized derivative for multiple dimensions.[/box] (Dull)In this case it expands to:
\begin{aligned}f(x,y) = f(a,b) &+ \Big(f_x(a,b)(x-a) + f_y(a,b)(y-b)\Big) \\
&+ \frac 1{2!}\Big(f_{xx}\cdot (x-a)^2 + 2f_{xy}\cdot (x-a)(y-b) + f_{yy}\cdot (y-b)^2\Big) \\
&+ ...
\end{aligned}
(Wasntme)

 

[evinda]

That formula only has $x$ in it, and no $y$. (Worried)We can do it, but we'll need the generalized version of that formula for several variables:
$$f(\mathbf x)=\sum_{n=0}^{\infty} \frac 1 {n!}\mathbf D^nf(\mathbf a)(\mathbf x-\mathbf a)^n$$
where $\mathbf x = (x,y)$, $\mathbf a = (a,b)$, and $\mathbf D^nf$ is the generalized derivative for multiple dimensions. (Dull)In this case it expands to:
\begin{aligned}f(x,y) = f(a,b) &+ \Big(f_x(a,b)(x-a) + f_y(a,b)(y-b)\Big) \\
&+ \frac 1{2!}\Big(f_{xx}\cdot (x-a)^2 + 2f_{xy}\cdot (x-a)(y-b) + f_{yy}\cdot (y-b)^2\Big) \\
&+ ...
\end{aligned}
(Wasntme)

A ok! And which is the formula of $\mathbf D^nf$? (Thinking)

 

[I like Serena]

A ok! And which is the formula of $\mathbf D^nf$? (Thinking)

$Df$ is the vector of particle derivatives:
$$Df = (f_x, f_y)$$

$D^2f$ is the matrix of all possible second particle derivatives:
$$D^2f = \begin{bmatrix}f_{xx} & f_{xy} \\ f_{xy} & f_{yy}\end{bmatrix} = \left(\frac{\partial^2}{\partial x_i \partial x_j}\right)$$

$D^3f$ is a 2x2x2 matrix (aka tensor) with all possible combinations of the partial derivatives:
$$D^3f = \left(\frac{\partial^3}{\partial x_i \partial x_j \partial x_k}\right)$$

And so on. (Wasntme)

So for instance:
$$D^3f(\mathbf a) (\mathbf x - \mathbf a)^3 = \sum_i \sum_j \sum_k \frac{\partial^3f(\mathbf a)}{\partial x_i \partial x_j \partial x_k}(x_i-a_i)(x_j-a_j)(x_k-a_k)$$
(Sweating)