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 $\varphi(x) = \alpha x+\beta$.

Then

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

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

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.

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