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In algebraic geometry, this construction is known as the tangent cone to the graph. More generally, suppose we have the zero set of any polynomial $f(x,y) = 0$, and assume $f(0,0)=0$. Then we can write

$f(x,y) = a_m (x,y) + a_{m+1}(x,y) +a_{m+2}(x,y) +\cdots$

where $a_i(x,y)$ is a homogeneous polynomial of degree $i$ and $a_m$ is nonzero. The zero set of $a_m$ is called the tangent cone to the curve at the origin. It is a product of $m$ linear forms (over $\mathbb{C}$), and $m=1$ exactly when the zero set is smooth at the origin. In this case, the tangent cone coincides with the tangent space.

From your point of view, when we substitute $x\mapsto x/c$ and $y\mapsto y/c$ it is clear that the term left in the limit is $a_m$.

We can of course find tangent cones at other points of the zero set by changing coordinates.

In general, for a smooth function $f$ you should be able to take a multivariate Taylor expansion and read off the tangent cone from the lowest degree part. This is where the difficulty comes in for actually defining the tangent line in terms of the tangent cone in a calculus class, as finding computing the Taylor expansion demands we already have a notion of derivative. This difficulty is obviously not seen in the case of polynomials, although recentering the Taylor expansion of a polynomial at a different point is perhaps easiest done with the aid of derivatives.

Higher dimensional analogues are also available without any real work, although in the singular case the tangent cone is much more interesting than just a union of hyperplanes: it will be a cone over some variety. The homogeneous polynomial $a_m(x_1,\ldots,x_n)$ typically doesn't factor into a product of linear forms when $n>2$.

Tangent cones are treated in any reasonable introduction to algebraic geometry, such as Harris' "First course" book or Shafarevich.

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In algebraic geometry, this construction is known as the tangent cone to the graph. More generally, suppose we have the zero set of any polynomial $f(x,y) = 0$, and assume $f(0,0)=0$. Then we can write

$f(x,y) = a_m (x,y) + a_{m+1}(x,y) +a_{m+2}(x,y) +\cdots$

where $a_i(x,y)$ is a homogeneous polynomial of degree $i$ and $a_m$ is nonzero. The zero set of $a_m$ is called the tangent cone to the curve at the origin. It is a product of $m$ linear forms (over $\mathbb{C}$), and $m=1$ exactly when the zero set is smooth at the origin. In this case, the tangent cone coincides with the tangent space.

From your point of view, when we substitute $x\mapsto x/c$ and $y\mapsto y/c$ it is clear that the term left in the limit is $a_m$.

We can of course find tangent cones at other points of the zero set by changing coordinates.

In general, for a smooth function $f$ you should be able to take a multivariate Taylor expansion and read off the tangent cone from the lowest degree part. This is where the difficulty comes in for actually defining the tangent line in terms of the tangent cone, as finding the Taylor expansion demands we already have a notion of derivative. This difficulty is obviously not seen in the case of polynomials, although recentering the Taylor expansion of a polynomial at a different point is perhaps easiest done with the aid of derivatives.

Higher dimensional analogues are also available without any real work, although in the singular case the tangent cone is much more interesting than just a union of hyperplanes: it will be a cone over some variety. The homogeneous polynomial $a_m(x_1,\ldots,x_n)$ typically doesn't factor into a product of linear forms when $n>2$.

Tangent cones are treated in any reasonable introduction to algebraic geometry, such as Harris' "First course" book or Shafarevich.

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In algebraic geometry, this construction is known as the tangent cone to the graph. More generally, suppose we have the zero set of any polynomial $f(x,y) = 0$, and assume $f(0,0)=0$. Then we can write

$f(x,y) = a_m (x,y) + a_{m+1}(x,y) +a_{m+2}(x,y) +\cdots$

where $a_i(x,y)$ is a homogeneous polynomial of degree $i$. i$and$a_m$is nonzero. The zero set of$a_m$is called the tangent cone to the curve at the origin. It is a product of$m$linear forms (over$\mathbb{C}$), and$m=1$exactly when the zero set is smooth at the origin. In this case, the tangent cone coincides with the tangent space. From your point of view, when we substitute$x\mapsto x/c$and$y\mapsto y/c$it is clear that the term left in the limit is$a_m$. We can of course find tangent cones at other points of the zero set by changing coordinates. In general, for a smooth function$f\$ you should be able to take a multivariate Taylor expansion and read off the tangent cone from the lowest degree part.

Higher dimensional analogues are also available without any real work, although in the singular case the tangent cone is much more interesting than just a union of hyperplanes: it will be a cone over some variety.

Tangent cones are treated in any reasonable introduction to algebraic geometry, such as Harris' "First course" book or Shafarevich.

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