Return to Answer

I think you can get the Puiseux expansion by solving $t = x^{1/m}$ and setting $y = \sum a_r x^{r/m}$. Then I'd guess there are $m$ branches if $a_1 \neq 0$.

See How to Compute a Puiseux Expansion on the arXiv. These are connected with resolutions of singularities on curves.

---

In the paper, they try to solve $2x^4 + x^2y + 4 xy ^2 + 4 y^3 = 0$ as a function $y(x)$ in terms of $x$:

$$y(x) = c_1 x^{\gamma_1} + c_2 x^{\gamma_1 + \gamma_2} + c_3 x^{\gamma_1 + \gamma_2 + \gamma_3 } + \dots$$ The draw the newton polygon (in ASCII format)

.....
o....
x....
xo...
xxo..
xxxo.

The possible slopes are -2, -1. So they make a guess: $y = x^2(c_1 + y_1)$ where somehow $y_1(x)$ is "smaller" so we could ignore it. $$ 2x^4 + c_1 x^4 + 4c_1 x^5 + 4c_1 x^6 \approx (2 + c_1)x^4 = 0 $$ Let's try $c_1 = -2$. Then it's possible to write an equation in $x, y_1$ and repeat this procedure. \begin{eqnarray} x^4(2 + c_1 + y_1) + 4 x^5(c_1 + y_1)^2 + 4x^6(c_1 + y_1)^3 &=& 0 \\\\ y_1 + 16 x - 16 x y_1 + 4 x y_1^2 -32x^2+48 x^2 y_1 - 24x^2 y_1^2+ 4 x^2 y_1^3 &=& 0 \\\\ \end{eqnarray}

Looking at the Newton polygon in $x,y_1$, or equivalently by noticing $$ y_1 + 16 x \approx 0 $$ they conclude $y_1(x) = c_1 x + c_2 x^2 + \dots $ is a power series with only one branch.

So there are only two branches for $y(x)$. In the first step we could have ignored the last two terms, $2x^4 + x^2 y \approx 0$ or $y \approx -2 x^2$.

I think you can get the Puiseux expansion by solving $t = x^{1/m}$ and setting $y = \sum a_r x^{r/m}$. Then I'd guess there are $m$ branches if $a_1 \neq 0$.

See How to Compute a Puiseux Expansion on the arXiv. These are connected with resolutions of singularities on curves.

---

In the paper, they try to solve $2x^4 + x^2y + 4 xy ^2 + 4 y^3 = 0$ as a function $y(x)$ in terms of $x$:

$$y(x) = c_1 x^{\gamma_1} + c_2 x^{\gamma_1 + \gamma_2} + c_3 x^{\gamma_1 + \gamma_2 + \gamma_3 } + \dots$$ The draw the newton polygon (in ASCII format)

.....
o....
x....
xo...
xxo..
xxxo.

The possible slopes are -2, -1. So they make a guess: $y = x^2(c_1 + y_1)$ where somehow $y_1(x)$ is "smaller" so we could ignore it. $$ 2x^4 + c_1 x^4 + 4c_1 x^5 + 4c_1 x^6 \approx (2 + c_1)x^4 = 0 $$ Let's try $c_1 = -2$. Then it's possible to write an equation in $x, y_1$ and repeat this procedure. \begin{eqnarray} x^4(2 + c_1 + y_1) + 4 x^5(c_1 + y_1)^2 + 4x^6(c_1 + y_1)^3 &=& 0 \\\\ y_1 + 16 x - 16 x y_1 + 4 x y_1^2 -32x^2+48 x^2 y_1 - 24x^2 y_1^2+ 4 x^2 y_1^3 &=& 0 \\\\ \end{eqnarray}

Looking at the Newton polygon in $x,y_1$, or equivalently by noticing $$ y_1 + 16 x \approx 0 $$ they conclude $y_1(x) = c_1 x + c_2 x^2 + \dots $ is a power series with only one branch.

So there are only two branches for $y(x)$. In the first step we could have ignored the last two terms, $2x^4 + x^2 y \approx 0$ or $y \approx -2 x^2$. We don't feel guilty about doing these Taylor-like approximations since the curve has a singularity at $(x,y)= (0,0)$ anyway.

I think you can get the Puiseux expansion by solving $t = x^{1/m}$ and setting $y = \sum a_r x^{r/m}$. Then I'd guess there are $m$ branches if $a_1 \neq 0$.

See How to Compute a Puiseux Expansion on the arXiv. These are connected with resolutions of singularities on curves.

---

In the paper, they try to solve $2x^4 + x^2y + 4 xy ^2 + 4 y^3 = 0$ as a function $y(x)$ in terms of $x$:

$$y(x) = c_1 x^{\gamma_1} + c_2 x^{\gamma_1 + \gamma_2} + c_3 x^{\gamma_1 + \gamma_2 + \gamma_3 } + \dots$$ The draw the newton polygon (in ASCII format)

xxxxx
oxxxx
.xxxx
.oxxx
..oxx
...ox

The possible slopes are -2, -1. So they make a guess: $y = x^2(c_1 + y_1)$ where somehow $y_1(x)$ is "smaller" so we could ignore it. $$ 2x^4 + c_1 x^4 + 4c_1 x^5 + 4c_1 x^6 \approx (2 + c_1)x^4 = 0 $$ Let's try $c_1 = -2$. Then it's possible to write an equation in $x, y_1$ and repeat this procedure. \begin{eqnarray} x^4(2 + c_1 + y_1) + 4 x^5(c_1 + y_1)^2 + 4x^6(c_1 + y_1)^3 &=& 0 \\\\ y_1 + 16 x - 16 x y_1 + 4 x y_1^2 -32x^2+48 x^2 y_1 - 24x^2 y_1^2+ 4 x^2 y_1^3 &=& 0 \\\\ \end{eqnarray}

Looking at the Newton polygon in $x,y_1$, or equivalently by noticing $$ y_1 + 16 x \approx 0 $$ they conclude $y_1(x) = c_1 x + c_2 x^2 + \dots $ is a power series with only one branch.

So there are only two branches for $y(x)$. In the first step we could have ignored the last two terms, $2x^4 + x^2 y \approx 0$ or $y \approx -2 x^2$.

I think you can get the Puiseux expansion by solving $t = x^{1/m}$ and setting $y = \sum a_r x^{r/m}$. Then I'd guess there are $m$ branches if $a_1 \neq 0$.

See How to Compute a Puiseux Expansion on the arXiv. These are connected with resolutions of singularities on curves.

---

In the paper, they try to solve $2x^4 + x^2y + 4 xy ^2 + 4 y^3 = 0$ as a function $y(x)$ in terms of $x$:

$$y(x) = c_1 x^{\gamma_1} + c_2 x^{\gamma_1 + \gamma_2} + c_3 x^{\gamma_1 + \gamma_2 + \gamma_3 } + \dots$$ The draw the newton polygon (in ASCII format)

.....
o....
x....
xo...
xxo..
xxxo.

The possible slopes are -2, -1. So they make a guess: $y = x^2(c_1 + y_1)$ where somehow $y_1(x)$ is "smaller" so we could ignore it. $$ 2x^4 + c_1 x^4 + 4c_1 x^5 + 4c_1 x^6 \approx (2 + c_1)x^4 = 0 $$ Let's try $c_1 = -2$. Then it's possible to write an equation in $x, y_1$ and repeat this procedure. \begin{eqnarray} x^4(2 + c_1 + y_1) + 4 x^5(c_1 + y_1)^2 + 4x^6(c_1 + y_1)^3 &=& 0 \\\\ y_1 + 16 x - 16 x y_1 + 4 x y_1^2 -32x^2+48 x^2 y_1 - 24x^2 y_1^2+ 4 x^2 y_1^3 &=& 0 \\\\ \end{eqnarray}

Looking at the Newton polygon in $x,y_1$, or equivalently by noticing $$ y_1 + 16 x \approx 0 $$ they conclude $y_1(x) = c_1 x + c_2 x^2 + \dots $ is a power series with only one branch.

So there are only two branches for $y(x)$. In the first step we could have ignored the last two terms, $2x^4 + x^2 y \approx 0$ or $y \approx -2 x^2$.

I think you can get the Puiseux expansion by solving $t = x^{1/m}$ and setting $y = \sum a_r x^{r/m}$. Then I'd guess there are $m$ branches if $a_1 \neq 0$.

See How to Compute a Puiseux Expansion on the arXiv. These are connected with resolutions of singularities on curves.

I think you can get the Puiseux expansion by solving $t = x^{1/m}$ and setting $y = \sum a_r x^{r/m}$. Then I'd guess there are $m$ branches if $a_1 \neq 0$.

See How to Compute a Puiseux Expansion on the arXiv. These are connected with resolutions of singularities on curves.

---

In the paper, they try to solve $2x^4 + x^2y + 4 xy ^2 + 4 y^3 = 0$ as a function $y(x)$ in terms of $x$:

$$y(x) = c_1 x^{\gamma_1} + c_2 x^{\gamma_1 + \gamma_2} + c_3 x^{\gamma_1 + \gamma_2 + \gamma_3 } + \dots$$ The draw the newton polygon (in ASCII format)

xxxxx
oxxxx
.xxxx
.oxxx
..oxx
...ox

The possible slopes are -2, -1. So they make a guess: $y = x^2(c_1 + y_1)$ where somehow $y_1(x)$ is "smaller" so we could ignore it. $$ 2x^4 + c_1 x^4 + 4c_1 x^5 + 4c_1 x^6 \approx (2 + c_1)x^4 = 0 $$ Let's try $c_1 = -2$. Then it's possible to write an equation in $x, y_1$ and repeat this procedure. \begin{eqnarray} x^4(2 + c_1 + y_1) + 4 x^5(c_1 + y_1)^2 + 4x^6(c_1 + y_1)^3 &=& 0 \\\\ y_1 + 16 x - 16 x y_1 + 4 x y_1^2 -32x^2+48 x^2 y_1 - 24x^2 y_1^2+ 4 x^2 y_1^3 &=& 0 \\\\ \end{eqnarray}

Looking at the Newton polygon in $x,y_1$, or equivalently by noticing $$ y_1 + 16 x \approx 0 $$ they conclude $y_1(x) = c_1 x + c_2 x^2 + \dots $ is a power series with only one branch.

So there are only two branches for $y(x)$. In the first step we could have ignored the last two terms, $2x^4 + x^2 y \approx 0$ or $y \approx -2 x^2$.