How to determine double points on a curve?

Question

I want to determine all the double points of the curve $f : \mathbb{R} \to \mathbb{R}^2$ that is given by $f(t) = (t^3 - t , t^2 - 1)$. Graphically, the output $(0,0)$ should occur twice, which gives $t = \pm 1$. However, this does not serve as a full proof I would assume. Unfortunately, I can't find any adequate information about this topic on the internet.

Hopefully some of you can help me. Thank you in advance.

Answer 1

Graphing helps one discover true statements, such as the true statement "the output $(0,0)$ occurs twice, which gives $t=\pm 1$". Presumably you can verify the truth of that statement by explicit calculation; that calculation constitutes full proof of that statement.

However, you're not done. You have asked to determine all double points of that curve. Presumably your graph also tells you the true statement here: $t = \pm 1$ gives the only double point of the curve. This must still be proved.

So, you must take two unknown input values $s,t \in \mathbb R$, then form the equation $$f(s)=f(t) $$ $$(s^3-s,s^2-1) = (t^3-t,t^2-1) $$ which breaks into two numerical equations \begin{align*} s^3 - s &= t^3 - t \\ s^2 - 1 &= t^2 - 1 \end{align*} Thinking of this as a system of 2 equations in the 2 unknowns $s,t$, you must find all possible pairs of numbers $(s,t)$ which are simultaneous solutions of this pair of equations, and such that $s \ne t$ (these equations are of course always true when $s=t$, but that does not give you a double point). If your graph is not lying, the only solutions should be $s=1$, $t=-1$ and $s=-1,t=+1$ which gives just one double point.

Solving this system with algebra is not so hard, perhaps you can carry it through now?

Answer 2

You might note that $x = t y$, so if $y \ne 0$, $t = x/y$ is uniquely determined.

Answer 3

I think I got it, just wanna check if this is correct. The second equations is $s^2 - 1 = t^2 - 1$. We can cancel the 1's and obtain $s^2 = t^2$. The only way we can get a double point from this is by concluding $s = -t$, which satisfies the equation and ensures that $s \neq t$, as long as $t \neq 0$. Plugging this into the first equation, we get $(-t)^3 - (-t) = -t^3 + t = t^3 - t$. This can be rearranged to $t^3 - t = 0$, which yields the solutions $t = 0,1$. We have noted above that $t$ cannot be $0$, thus $t = 1$ and $s = -1$ or the other way around.