Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Problem

This is an exercise on page 150 of Cox/Little/O'Shea's Ideal, varieties and algorithms: an introduction to computational algebraic geometry and commutative algebra, 3rd ed.

I get lost in this problem...

For part (a), I find the equation of the envelope to be $$g_1=-27 x^3 + 16 x^5 - 72 x^3 y + 16 x^3 y^2 - 64 x y^3=0,$$ by computing a Groebner basis for $<f=(x - t)^2 + (y - t^2)^2-t^2,\frac{\partial f}{\partial t}>.$

The whole Groebner basis is $$\{g_1=-27 x^3 + 16 x^5 - 72 x^3 y + 16 x^3 y^2 - 64 x y^3, g_2=27 x^2 - 16 x^4 + 72 t x y + 24 x^2 y + 32 t x y^2 - 16 x^2 y^2, g_3=9 t x^2 - 6 x^3 + 4 t x^2 y - 8 x y^2, -9 x^2 + 8 t x^3 - 24 t x y - 4 x^2 y, g_4=3 t x - 2 x^2 + 2 t^2 y - 2 y^2, g_5=3 t^2 x - 2 t x^2 + x y, g_6=2 t^3 - x - 2 t y\}.$$

I don't know how to show that "the envelope is the union of two varieties" algebraically. I plotted the picture of of $(x - t)^2 + (y - t^2)^2 =t^2$, which looks like enter image description here

Note that the image consists of two separate parts, so one can "imagine" that "the envelope is the union of two varieties." But how can we prove it formally?

For part (b), first I notice that $g_6=0$ is quadratic in $t$, so that a given partial solution $(x,y)$ extends in at most two ways to a complete solution.

Then I tried to find the singular points of $V(g_1)$, which are given by $g_1=\frac{\partial g_1}{\partial x}=\frac{\partial g_2}{\partial x}=0.$ In order to find the solutions of these equations, I computed a Groebner basis for them, which is $$\{h_1=729 y^6 + 3888 y^7 + 4320 y^8 + 1792 y^9 + 256 y^{10}, h_2=729 x y^3 + 3888 x y^4 + 4320 x y^5 + 1792 x y^6 + 256 x y^7, h_3=531441 x^2 + 419904 y^3 - 1119744 y^4 + 3234816 y^5 - 694270080 y^6 - 931055616 y^7 - 414746624 y^8 - 61521920 y^9\}$$

Note that $h_1$ includes only $y$, so from $h_1=0$, we get $y=-\frac{9}{4},-\frac{1}{4},0.$ Since the leading coefficient of $h_3$ in x is a constant, these partial solutions of $y$ extends. Then I am lost. I don't know the outline of how to solve this problem.

Above is all of my work. Looking forward to your advice and help. Thanks in advance.

share|improve this question
    
What you should do as far as pictures is place a number of circles in the $xy$ plane, say $t = 0, \pm \frac{1}{4}, \; \pm \frac{1}{2}, \; \pm \frac{3}{4}, \; \pm 1, \; \pm \frac{5}{4}, \; \pm \frac{3}{2}, $ a few more. All with $t \neq 0$ are tangent to the positive $y$ axis where $x=0.$ Not the three dimensional image you show above. –  Will Jagy Apr 1 '12 at 5:28
    
Your formula for $g_1$ suggests, as $y$ grows, $$ x \approx \sqrt{4 y + 2 - \frac{1}{4y}}, $$ looks pretty accurate for $y > 2.$ So you may have this. –  Will Jagy Apr 1 '12 at 7:37
1  
What @Will said... –  J. M. Apr 3 '12 at 16:32

3 Answers 3

up vote 1 down vote accepted
+50

The envelope is that of a family of circles with centers on a plane curve ($y=x^2$) that divides the plane into two pieces. Thus the envelope consists of two components, an upper and a lower envelope, that intersect at the point of the curve $(0,0)$ where the circle has radius $0$.

Algebraically, the equation of the envelope that you found factorizes as $g_1 (x,y) = x h(x,y) = 0$ with $h(x,y)$ an irreducible polynomial of degree 4. The two factors are the equations of the Zariski closures of the upper and lower components of the envelope. The upper envelope is the non-negative $y$-axis, and $x=0$ is the equation for the smallest algebraic set containing those points. The lower envelope is the set of all real points on the curve $h(x,y)=0$.

share|improve this answer

Here is an answer for part $\mathsf{a}$ and part of part $\mathsf{b}$:

The envelope for the family of circles $$ (x-t)^2+(y-t^2)^2=t^2\tag{1} $$ also satisfies the partial of $(1)$ with respect to $t$ (times $-\frac12$): $$ (x-t)+2t(y-t^2)=-t\tag{2} $$ Thus, $$ 0=(x-t)^2+(y-t^2)^2-t^2=((4t^2+1)(y-t^2)+4t^2)(y-t^2)\tag{3} $$ Therefore, either $$ y=t^2\text{ and }x=0\tag{4} $$ or $$ y=\frac{4t^4-3t^2}{4t^2+1}\text{ and }x=\frac{8t^3}{4t^2+1}\tag{5} $$ In the diagram below, the variety for $(4)$ is in blue and the variety for $(5)$ is in red.

$\hspace{5cm}$varieties

Inspecting $(4)$ and $(5)$, it can be seen that each point on the variety for $(4)$ is tangent to two circles (except at $y=0$): one for each solution to $y=t^2$. Each point on the variety for $(5)$ is tangent only to one circle since we can solve $x=\frac{8t^3}{4t^2+1}$ uniquely for each $x$. This is supported by the diagram above.


Solve $\mathbf{(4)}$ for $\mathbf{x}$ in terms of $\mathbf{y}$:

The curve in $(4)$ is part of the variety $$ x=0\tag{6} $$ Solve $\mathbf{(5)}$ for $\mathbf{x}$ in terms of $\mathbf{y}$:

From $(5)$, we get $$ \frac{4y+1}{4y+9}=\left(1-\frac{4}{4t^2+3}\right)^2\tag{7} $$ and $$ \frac{4y^2}{4y^2+3x^2}=\left(1-\frac{6}{4t^2+3}\right)^2\tag{8} $$ Combining $(7)$ and $(8)$ yields $$ 3\sqrt{\frac{4y+1}{4y+9}}-2\sqrt{\frac{4y^2}{4y^2+3x^2}}=1\tag{9} $$ from which we get $$ \frac{40y+18\mp6\sqrt{16y^2+40y+9}}{4y+9}=\frac{16y^2}{4y^2+3x^2} $$ $$ \frac{4y+9}{64}(40y+18\pm6\sqrt{16y^2+40y+9})=4y^2+3x^2 $$ $$ 32x^2=108-(4y-9)^2\pm(4y+9)\sqrt{(4y+1)(4y+9)}\tag{10} $$ The branch for '$+$' is in red and the branch for '$-$' is in green.

$\hspace{3cm}$solutions

The curve in $(10)$ is the variety $$ (32x^2+(4y-9)^2-108)^2=(4y+1)(4y+9)^3 $$ $$ 16x^4-27x^2-72x^2y+16x^2y^2-64y^3=0\tag{11} $$

share|improve this answer
    
In what sense is the lower envelope only a part of the degree 4 curve? At least from the pictures it seems that all real points of that curve lie on the envelope. –  zyx Apr 3 '12 at 21:43
    
@zyx: I had thought that $y\le-\frac94$, which makes $\sqrt{(4y+1)(4y+9)}$ real, might add some more to the variety. However, I see that upon closer inspection, for $y\le-\frac94$, the RHS of $(10)$ is always negative. Thus, the "part" is the "whole". :-) –  robjohn Apr 3 '12 at 22:03
    
Thanks. My puzzlement was due to the observation that the lower envelope (like the upper envelope) has only one connected component in the Euclidean topology. If any points or segments were missing from the lower envelope it would be split into several pieces, which "should" not happen. –  zyx Apr 3 '12 at 22:36
    
$y \geq -1/4,$ achieved at $x = \pm 1/2$ –  Will Jagy Apr 3 '12 at 22:45
    
@Will: I seem to be missing the point of your comment. Would you care to elaborate? –  robjohn Apr 3 '12 at 22:51

I don't know about Groebner bases. In calculus we have determined the envelope as follows: Given a family of curves $$F(x,y,c):=(x-c)^2 +(y-c^2)^2-c^2 =0$$ (in this case a family of circles $\gamma_c\,$), parametrized by $c\ne0$ we should look for the points $(x,y)\in{\mathbb R}^2$ where the function $c:=\phi(x,y)$ giving the "number" $c$ of the curve passing through $(x,y)$ is somewhat fishy. By the implicit function theorem, for the corresponding triples $(x,y,c)$ one would necessarily have $$F_c(x,y,c)=0\ , \quad{\rm i.e.,}\quad 2x+2c(y-c^2)=0\ .$$ Points of this kind on the circle $\gamma_c$ are obtained by intersecting $\gamma_c$ with the line $2x+2c (y-c^2)=0$. One obtains the points $(0,c^2)$ (doubly covering the positive $y$-axis) and $$P_c:=\Bigl({2c^3\over c^2+1},{c^2(c^2-1)\over c^2+1}\Bigr)\qquad(c\ne0)\ .$$ Here is a picture of the curve $c\mapsto P_c\,$:

enter image description here

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.