Your personal quadratic



(10 pts by February 13)

Using the strategy for finding numbers a, b, and c that we used on the Your personal monic cubic problem, construct your own personal quadratic f(x) = ax^2+bx+c. Then find a numbers \alpha, \beta, and \gamma so that the function \varphi(x)=\alpha x + \beta conjugates f to g(x)=x^2 + \gamma.

Finally, compute the first five terms of the critical orbit of your f, the first five terms of the critical orbit of g, and show that \varphi maps one to the other.


My Initials are WSF. So from last time I have

a = 3

b = 2

c = 0

So my function is f(x) = 3x^2 + 2x.

Let g(x) = x^2 + \gamma and \varphi(x) = \alpha x+\beta.


f(\varphi(x)) = 3(\alpha x +\beta)^2+ 2(\alpha x + \beta) = (3\alpha^2)x^2 +(6\alpha\beta + 2\alpha)x + (3\beta^2 + 2\beta)

\varphi(g(x)) = \alpha(x^2 + \gamma) + \beta = \alpha x^2 + (\alpha\gamma + \beta)

Setting f(\varphi(x)) = \varphi(g(x)), we can construct the following system of equations:

3\alpha^2 = \alpha

6\alpha\beta + 2\alpha = 0

3\beta^2 + 2\beta = \alpha\gamma + \beta

Solving these equations gives us \alpha = 1/3, \beta = -1/3, and \gamma = 0.

So then g(x) = x^2 and \varphi(x) = (1/3)x - 1/3.

Lastly, we’ll check to see if we got the right \alpha, \beta, and \gamma.

f(\varphi(x)) = 3((1/3)x-1/3)^2+2((1/3)x-1/3) = (1/3)x^2 -1/3

\varphi(g(x)) = (1/3)x^2 - 1/3

We got it!

Now we will compute the first five terms of the critical orbits for f and g.

First, let’s find the critical points for f and g.

Note that f'(x) = 6x + 2 = 0 implies that x = -1/3 and g'(x) = 2x = 0 implies that x = 0.

Using Mathematica and our critical points, we generate the first five terms of the critical orbits for f and g.

For f we have:

NestList[3*#^2 + 2*# &, -1/3, 4]
{-(1/3), -(1/3), -(1/3), -(1/3), -(1/3)}

For g we have:

NestList[#^2 &, 0, 4]
{0, 0, 0, 0, 0}

Interestingly, our critical points for f and g were also fixed points.

Now, we will look at the image of the critical orbit of g, \{0,0,0,0,0\}, under \varphi.

Note that \varphi(0) = (1/3)\cdot 0 -1/3 = -1/3.

So then the image of \{0,0,0,0,0\} under \varphi is \{-(1/3),-(1/3),-(1/3),-(1/3),-(1/3)\}, which is the critical orbit of f.


Using the same (a, b, c) = (1,1,2) as last time:
f(x) = x^2+x+2,
g(x) = x^2+\gamma, and
\phi (x) = \alpha x + \beta.

Thus, f(\phi(x)) = (\alpha x + \beta)^2 + (\alpha x + \beta) + 2 = (\alpha^2x^2) + (2\alpha\beta x+\alpha x) +(\beta^2 + \beta +2) .
Also, \phi (g(x)) = \alpha (x^2+\gamma) + \beta = (\alpha x^2) + (\alpha \gamma + \beta).

Collecting similar terms, we have a system of equations:
\alpha ^2 = \alpha
2 \alpha\beta + \alpha =0
\beta ^2 + \beta + 2 = \alpha \gamma + \beta.
Solving this returns the real solutions of \alpha=1, \beta = -\frac{1}{2}, and \gamma=\frac{9}{4}.
So g(x) = x^2+\frac{9}{4} and \phi (x) = x -\frac{1}{2}.

As for the critical points of f and g, we check the zeroes of their derivatives.
f'(x) = 2x+1=0 \implies x =-\frac{1}{2}, which just like Ed_Boy’s, is our \beta value.
Likewise, g'(x) = 2x = 0 \implies x=0.

To find the critical orbits, we plug-and-chug.

Our f-orbit is \{-\frac{1}{2}, \frac{7}{4}, \frac{109}{16}, \frac{14,137}{256}, \frac{203,604,913}{65,536} \}.

Our g-orbit is \{0, \frac{9}{4}, \frac{117}{16}, \frac{14,265}{256}, \frac{203,637,681}{65,536} \}.

Looking at our critical orbit of g under \phi, we plug our g-values into \phi.

Our \phi ( g(x))-orbit is \{-\frac{1}{2}, \frac{7}{4}, \frac{109}{16}, \frac{14,137}{256}, \frac{203,604,913}{65,536} \}, exactly our f-orbit.

Thus, we’ve shown that the first five terms of the critical orbit of g and f are connected by the mapping \phi.


My a, b, and c values are 1, -1, and 1. That means my quadratic is f(x) = x^2-x+1.

With g(x) = x^2 + \gamma and an affine conjugacy function \phi(x)= \alpha x + \beta, then

f(\phi(x))=(\alpha x + \beta)^2+(\alpha x + \beta)+1= \alpha ^2x^2 +2 \alpha \beta x -\alpha x + \beta ^2 - \beta +1 and
\phi(g(x))= \alpha x^2 + \alpha \gamma + \beta.

The conjugacy relationship f(\phi(x)=\phi(g(x)) yeilds the real values 1, \frac{1}{2}, and \frac{1}{4} for \alpha, \beta, and \gamma.

So g(x) = x^2 + \frac{1}{4} and \phi (x) = x + \frac{1}{2}.

To verify these claims check if f(\phi(x)=\phi(g(x)) is valid.

f(\phi(x)) = (x+\frac{1}{2})^2 - (x+\frac{1}{2})+1 = x^2 +\frac{3}{4}

\phi(g(x)) = x^2 + \frac{1}{4}+\frac{1}{2} = x^2 + \frac{3}{4}. Looks like f(x) and g(x) are conjugate via the conjugacy \phi(x).

Since these are conjugate, there should be a relationship between their critical orbits.

Note that for f(x), the critical point is where f'(x)=0. f'(x)=0 \implies 2x -1 = 0 \implies x=1/2 is the critical point of f.

For g(x), the critical point is where g'(x)=0. g'(x)=0 \implies 2x =0 \implies x= 0 is the critical point of g.

The orbit of f from x = 1/2 is \{\frac{3}{4}, \frac{13}{16}, \frac{217}{256}, \frac{57073}{65536}, \frac{3811958497}{4294967296}\}, and the orbit of g from x=0 is \{\frac{1}{4}, \frac{5}{16}, \frac{89}{256}, \frac{24305}{65536}, \frac{1664474849}{4294967296} \}. Notice that when the orbit of g is plugged into \phi, the result is exactly the critical orbit of f under \phi.

It is clear that f and g are conjugate.


My initials (JAG) give values of a=0, b=-1, and c=0, which produces the non-quadratic quadratic f(x)=-x.

This is not very exciting for our purposes, and solving for the function’s conjugacy function \phi(x)=\alpha x+\beta quickly results in a trivial solution. So instead, since the set of all possible values for a, b, and c is [-1,0,1,2,3,4], I chose to substitute my previous values for a=3, b=-1, and c=4.

This gives us the quadratic f(x)=3x^2-x+4.

To say that \phi(x) conjugates f(x) to g(x)=x^2+\gamma, we must show that f(\phi(x)) = \phi(g(x)).

f(\phi(x)) = 3(\alpha x+\beta)^2 - (\alpha x+\beta) + 4 = (3\alpha^2)x^2 + (6\alpha\beta-\alpha)x + (3\beta^2-\beta+4)

\phi(g(x)) = \alpha(x^2+\gamma) + \beta = \alpha x^2 + \beta+\alpha

We can find a particular solution for f(\phi(x)) = \phi(g(x)) by setting their corresponding coefficients equal.

3\alpha^2 = \alpha \implies \alpha = 1/3
6\alpha\beta-\alpha = 0 \implies \beta = 1/6
3\beta^2-\beta+4 = \beta+\gamma \implies \gamma = 45/12

We then plug these values back into f(\phi(x)) and \phi(g(x)), which gives us the following:

f(\phi(x)) = 3(1/3)^2x + (6(1/3)(1/6)-1/3)x + 3(1/6)^2-1/6+4 = 1/3x^2 + 47/12
\phi(g(x)) = 1/3x^2 + 1/6+45/12 = 1/3x^2 + 47/12

This confirms that \phi(x) conjugates f(x) to g(x).

We find the critical points for f(x) and g(y) by solving each function’s derivative for 0:

f'(x) = 6x - 1
6x - 1 = 0 \implies x = 1/6
g'(x) = 2x
2x = 0 \implies x = 0

Computing the first 5 terms of each function’s critical orbit about these points gives us the following:

f(x): [1/6,47/12,2213/48,4865033/768,23667301429073/196608]
g(x): [0,15/4,285/16,82185/256,6754619985/65536]

Unfortunately, the deadline for this post looms 24 minutes away, and I’m having troubling with the last step of proving via these orbital computations that \phi maps f(x) to g(x), so it remains as mystery as of writing.


This topic was automatically closed after 5 days. New replies are no longer allowed.