The parabolic equation \( y = x^2 + 1 \) has no real roots, the curve doesn't have any intersections with the x axis (it has no abscissae on y = 0, in old fashioned terminology). But \( y = x^2 - 1 \) does intersect the x axis in two places. This is intuitive because the y intersection is -1 and \( x^2 \) is positive for all x. The equation \( x^2 - 1 = 0 \) is algebraically equivalent to \( 1 - x^2 = 0 \), and both equations have the same solutions: {-1, 1}, so they intersect the x-axis at the same two points. However they describe two different curves when instead of 0, they are equal to (some value of) y, one of these is the reflection of \( y = x^2 + 1 \) through y = 1. The latter (or the initial) equation has solutions: {-i, i} where i lies on the positive imaginary axis in the complex plane, and the reflection through y = 1 in \( \mathbb R^2 \) is equivalent to a rotation of \( \frac {\pi} {2} \) in \( \mathbb C \), which takes {-1, 1} to {-i, i}. Plots of functions like \( y = x^2 - 1 \) or its reflection through y = 0, \( y = 1 - x^2 \) are not like the polynomials \( x^2 - 1 = 0, 1 - x^2 = 0 \), which are equivalent if you multiply by -1.
My point is there is a way to visualise complex roots. The reflection has two real roots, but you could imagine that they represent the complex roots, rotated into the real plane.
I don't know what the 9th grade is, and I started learning French when I was 9. And, so . . . So what about complex roots of polynomials of degree > 2? Did they teach you anything about whether there is a simple realization of roots if you can reflect (or invert) the curve it represents (i.e. when y is not zero)? Does the existence or non-existence of a useful reflection in two dimensions for degree 3 or 4 polynomials "as functions", say anything someone who isn't in the 9th grade any more would say is interesting at all?
The ninth grade is a reference to (for me) 15 year old students in public school. My school system was a little more advanced than many in the US. For instance we started learning set theory in the 4th grade (10 years old). It was not very effective due to the teaching staff's abilities, but I did learn the basics then. For me, 9th grade was a algebra year, after a year of geometry (8th grade). Spare me the stories of how the French school system works. I don't really care. And we were obviously schooled on the imaginary roots of second degree equations. You are correct that we did not discuss 3rd or 4th degree equations but re-reading your original post neither did you.
My my. But I thought it was about as relevant as you mentioning what you learned about in grade school. I had an English education too, thank you. French is a second language, not a great second mind you. All that I suspect is in the "so what" category as well. No I didn't. My bad. But rpenner has opened a recent thread on roots, so I didn't figure I would need to get too carried away.
One other observation: reflecting \( y =f(x) = x^2 + 1 \) through y = 1 is algebraically equivalent to translating the parabola by a factor of -2 along the y axis. A reflection or a translation changes the sign of the discriminant, by changing the sign of a or b (not both) in \( ax^2 + b \). Does this generalize?
I guess I did get ramble a bit, but in my defense this seemed like a very informal thread. As mathman implied, your original post seems to go nowhere. Just a collection of observations about basic algebra with no obvious goal. And as I was saying, everything you have written so far was more or less covered in algebra class when I was a child. When do you get to the punch line.
When you were a child studying algebra, did they teach you about the equivalence (an isomorphism) between \( \mathbb {R^2} \) and the complex plane? Or if that isn't good enough for a punch line, how about "there is no spoon"?
Using generally applicable methods applicable to any quadratic equation reflected in any line in the 2-D plane, we can show that \( y = x^2 + 1 \) and \(y = 1 - x^2\) are mirror images via the line \(y = 1\). This should be no surprise because if the equations can be written in the form \(y - 1 = x^2, \; y-1 = -x^2\). On the other hand, \(y = x^2 -1 \) and \( y = 1 - x^2\) are related by a reflection in the line \(y=0\). Finally, \(y = x^2 + 1\) and \(y = x^2 -1\) are related both by an ordered pair of reflections in two parallel lines and by a single translation of 2 units in the direction perpendicular to those lines. How to reflect a point in R x + S y + T = 0. A point (x,y) and it's reflection in R x + S y + T = 0, (x', y'), are related by \(\begin{pmatrix} x' \\ y' \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{S^2 - R^2}{R^2 + S^2} & \frac{- 2 R S}{R^2 + S^2} & \frac{- 2 R T}{R^2 + S^2} \\ \frac{-2 R S}{R^2 + S^2} & \frac{R^2 - S^2}{R^2 + S^2} & \frac{-2 S T}{R^2 + S^2} \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix}\) where the 3×3 matrix is its own inverse. Demonstrating two parallel reflections is a translation of points. Given two parallel lines in the plane, we may write them as \(R x + S y + T = 0, R x + S y + U = 0\) giving: \( \begin{pmatrix} x' \\ y' \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{S^2 - R^2}{R^2 + S^2} & \frac{- 2 R S}{R^2 + S^2} & \frac{- 2 R U}{R^2 + S^2} \\ \frac{-2 R S}{R^2 + S^2} & \frac{R^2 - S^2}{R^2 + S^2} & \frac{-2 S U}{R^2 + S^2} \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{S^2 - R^2}{R^2 + S^2} & \frac{- 2 R S}{R^2 + S^2} & \frac{- 2 R T}{R^2 + S^2} \\ \frac{-2 R S}{R^2 + S^2} & \frac{R^2 - S^2}{R^2 + S^2} & \frac{-2 S T}{R^2 + S^2} \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix}\) But by the associative property of matrix multiplication we may multiply the reflections together to get a single transform matrix which is: \(\begin{pmatrix} 1 & 0 & \frac{2 R}{R^2 + S^2} (T - U) \\ 0 & 1 & \frac{2 S}{R^2 + S^2} (T - U) \\ 0 & 0 & 1 \end{pmatrix} \) which corresponds to a translation perpendicular to the parallel lines if \(T \neq U\). \( x' = x + \frac{2 R}{R^2 + S^2} (T - U), \quad y' = y + \frac{2 S}{R^2 + S^2} (T - U)\) How to reflect A x^2 + B xy + C y^2 + D x + E y + F = 0 in R x + S y + T = 0. Solutions to A x^2 + B xy + C y^2 + D x + E y + F = 0 are solutions to \( \begin{pmatrix} x \\ y \\ 1 \end{pmatrix}^{T} \begin{pmatrix} A & \frac{B}{2} & \frac{D}{2} \\ \frac{B}{2} & C & \frac{E}{2} \\ \frac{D}{2} & \frac{E}{2} & F \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0\) So it follows that we may look at the transformed quadratic form matrix \(\begin{pmatrix} \frac{S^2 - R^2}{R^2 + S^2} & \frac{- 2 R S}{R^2 + S^2} & \frac{- 2 R T}{R^2 + S^2} \\ \frac{-2 R S}{R^2 + S^2} & \frac{R^2 - S^2}{R^2 + S^2} & \frac{-2 S T}{R^2 + S^2} \\ 0 & 0 & 1 \end{pmatrix}^{T} \begin{pmatrix} A & \frac{B}{2} & \frac{D}{2} \\ \frac{B}{2} & C & \frac{E}{2} \\ \frac{D}{2} & \frac{E}{2} & F \end{pmatrix} \begin{pmatrix} \frac{S^2 - R^2}{R^2 + S^2} & \frac{- 2 R S}{R^2 + S^2} & \frac{- 2 R T}{R^2 + S^2} \\ \frac{-2 R S}{R^2 + S^2} & \frac{R^2 - S^2}{R^2 + S^2} & \frac{-2 S T}{R^2 + S^2} \\ 0 & 0 & 1 \end{pmatrix}\) or \(\frac{1}{\left( R^2 + S^2 \right)^2} \begin{pmatrix} A (R^2 - S^2)^2 + 2 B R S ( R^2 - S^2) + 4 C R^2 S^2 & \frac{ 4 (A - C)R S(R^2 - S^2) + B (4 R^2 S^2 - (R^2 - S^2)^2) ) }{2} & \frac{4 A R (R^2-S^2) T + 2 B S (3 R^2-S^2) T + 8 C R S^2 T -D (R^4-S^4) - 2 E R S(R^2+S^2)}{2} \\ \frac{ 4 (A - C)R S(R^2 - S^2) + B (4 R^2 S^2 - (R^2 - S^2)^2) ) }{2} & 4 A R^2 S^2 - 2 B R S ( R^2 - S^2) + C (R^2 - S^2)^2 & \frac{8 A R^2 S T - 2 B R (R^2-3 S^2) T - 4 C S (R^2-S^2) T - 2 D R S (R^2+S^2) + E (R^4 - S^4)}{2} \\ \frac{4 A R (R^2-S^2) T + 2 B S (3 R^2-S^2) T + 8 C R S^2 T -D (R^4-S^4) - 2 E R S(R^2+S^2)}{2} & \frac{8 A R^2 S T - 2 B R (R^2-3 S^2) T - 4 C S (R^2-S^2) T - 2 D R S (R^2+S^2) + E (R^4 - S^4)}{2} & 4 A R^2 T^2 + 4 B R S T^2 + 4 C S^2 T^2 - 2 D R (R^2 + S^2)T - 2 E S (R^2 + S^2) T + F (R^2 + S^2)^2 \end{pmatrix}\) and read off the reflection. From \(y = x^2 + 1\) we may rewrite this relation as \( 1 x^2 + 0 xy + 0 y^2 + 0 x + (-1) y + 1 = 0\) or \( \begin{pmatrix} x \\ y \\ 1 \end{pmatrix}^{T} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -\frac{1}{2} \\ 0 & -\frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0\). From \(y = 1\) we may rewrite this relation as \(0 x + 1 y - 1 = 0\). So we have \(A = 1, E = -1, F = 1, S = 1, T = -1\) and all other parameters equal to zero. Thus the transformed quadratic form is: \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - \frac{1}{2} \\ 0 & - \frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 2 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \\ 0 & \frac{1}{2} & -1 \end{pmatrix}\) which is equivalent to \( 1 x^2 + 0 xy + 0 y^2 + 0 x + (1) y + (-1) = 0\) or \(y = -x^2 + 1\) or \(y = 1 - x^2\).
Please Register or Log in to view the hidden image! This is properly called the graph of a polynomial function; the roots of the polynomial (not the function) lie along the x axis. This is something you learn fairly early about quadratic equations. Thinking about what the graph tells us about quadratics in general, this one looks like the parabola \( y = x^2 + c \) but translated so it isn't symmetric about the y axis. If the curve is translated by a positive factor (more than 2.25) along the y axis it will have a conjugate pair of roots in the complex plane; is there a simple way to visualise these roots? Suppose we have \( f(x) = x^2 - 2x +2 \). This looks like \( x^2 + 1 \) translated by 1 unit to the right. If we subtract this unit distance, the reflection symmetry says the imaginary part of the conjugate pair of roots is {-i, i}. Add the real unit back and we have {(1 - i), (1 + i)}. That's a bit fast and loose but sure enough: \( (x - (i + 1))(x - (i - 1)) = x^2 - 2x +2 \) With translation and reflection, quadratic polynomials and their roots do seem to transform generally (viz rpenner's post). What about cubics?
I should really show that the relation between \( x^2 - 2x + 2 \) and \( x^2 + 1 \) is the same graph shifted left or right (i.e. parallel to the x axis), one unit. Just substitute \( x \pm 1 \) appropriately into either polynomial: \( (x-1)^2 + 1 = x^2 - 2x + 2\) \( (x+1)^2 - 2(x+1) + 2 = x^2 + 2x + 1 - 2x - 2 + 2 = x^2 + 1 \) Substituting x+1 into the translated polynomial gives the form which is symmetric about y; it corresponds to subtracting 1 from the x coordinates. This is the real part you add back to the pair of imaginary numbers given by the reflection about y = 1 (or generally y = c). The real part of the pair of complex roots is given by the translation distance of the lowest part of the quadratic curve from x = 0.
rpenner discusses quadratic, cubic and quartic polynomials in terms of sums of compositions of roots; roots don't determine unique polynomials since, given any pair of points on y = 0, there is a possibly infinite set of quadratic curves passing through that pair. Given any polynomial, scaling it by some value doesn't change the roots. Consider compositions of linear or degree one polynomials, in a quadratic such as the one in the above diagram: \( y = x^2 - x - 2 \). This factors "easily" into \( y = (x + 1)(x - 2) \). Each term separately describes a line with positive slope (or the points such a line intersects on x = 0, y = 0), so the line y = x + 1 is parallel to y = x - 2. Change the sign of one of these terms, the lines are no longer parallel but intersect at x + 1 = 2 - x => x = 1/2, and their composition is a reflection (through the x axis) of \( y = (x + 1)(x - 2) \). The quadratic curve now fits on a cone. Does a composition of quadratics fit quartics onto "compositions of cones"? Suppose you deform the quadratic (in the diagram) smoothly by making a dent near the lowest point and adding two more points of inflection. That's like deforming the surface of the cone so it has a "groove" along it. Each groove means two more inflection points (or critical points) in the curve fitted to the surface, so it works for polynomials of even degree.
About the isomorphism between \( \mathbb R^2 \) and \( \mathbb C \). There is a way to show they are algebraically equivalent using the polynomial \( x^2 +1 \) as an ideal of the ring of polynomials over \( \mathbb R \), notated as: \( \mathbb R\)\([x] \). Why is \( \langle x^2 +1 \rangle \) an ideal (i asks myself)? There are two ways to see this, one is by using long division (the division algorithm), the other is by the congruence (an equivalence) \( x^2 \equiv -1 \). Surprisingly, replacing occurrences of \( x^2 \), in some polynomial, with -1 leaves the same remainder as long division. For instance, \( x^3 \) becomes \( -x \), and so on. This makes it a shortcut, since the quotient is a 'multiple' of \( x^2 +1 \). So division by this ideal makes any polynomial in \( \mathbb R \)\([x] \). into some multiple of \( x^2 +1 \) (algebraically zero), plus a linear remainder like ax + b. So that, quotienting \( \mathbb R \)\([x] \) by \( \langle x^2 +1 \rangle \) is now a ring whose elements have the form \( \langle x^2 +1 \rangle + ax + b \).
Yeah, so what the equivalence \( x^2 \equiv -1 \) does is "force" x to be equivalent to i, the imaginary root of -1. Then \( \langle x^2 + 1 \rangle + ax + b \equiv 0 + ai + b \). Any polynomial in \( \mathbb R\)\([x] \) is mapped to some complex number, or formally: \( \mathbb R\)\([x] / \langle x^2 + 1 \rangle \equiv \mathbb C \). We map any \( f(x) \) to an equivalent \( f(i) \). So \( f(x) = x^2 + 1 \) is an ordinary parabolic curve in \( \mathbb R^2 \), \( x^2 + 1 = 0 \) is an irreducible polynomial with coefficients in \( \mathbb R^2 \), its reflection through a horizontal line (a line tangent to f(x) at the critical point) has roots {-1, 1}, and the quotient \( \frac {1 - x^2} { \langle x^2 + 1 \rangle } \) has a remainder of 2, the distance along y between \( 1 - x^2 \) and \( x^2 - 1 \).
Damn, I hate copy paste sometimes. The last para should say: "\( x^2 + 1 = 0 \) is an irreducible polynomial with coefficients in \( \mathbb R \)".
