The simplest type of singular point of a curve $\gamma$ is an ordinary cusp: a point $p$ of $\gamma$, corresponding to a parameter value $t_0$, say, is an ordinary cusp if $\gamma '(t_0) = 0$ and the vectors $\gamma ''(t_0)$ and $\gamma ''' (t_0)$ are linearly independent (in particular, these vectors must both be non-zero).
Show that the curve $\gamma (t) = (t^m, t^n)$, where $m$ and $n$ are positive integers, has an ordinary cusp at the origin if and only if $(m, n) = (2, 3)$ or $(3, 2)$.
We have that $\gamma '(t)=(mt^{m-1}, nt^{n-1})$. The origin correpsonds to $t=0$. So $\gamma '(0)=(0,0)$.
We have that $\gamma ''(t)=(m(m-1)t^{m-2}, n(n-1)t^{n-2})$ and $\gamma '''(t)=(m(m-1)(m-2)t^{m-3}, n(n-1)(n-2)t^{n-3})$.
For $t=0$ we have that $\gamma ''(0)=(0,0)$ and $\gamma '''(0)=(0,0)$. But these two vectors should be both non-zero and linearly independent. How can that be?
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EDIT1:
If $\gamma$ has an ordinary cusp at a point $p$, so does any reparametrization of $\gamma$.
Since $\gamma$ has an ordinary cusp at $p=\gamma (t_0)$ the following conditions hold:
- $\gamma '(t_0) = 0$
- $\gamma '' (t_0) \neq 0$, $\gamma ''' (t_0)\neq 0$
- $\gamma ''(t_0)$ and $\gamma '''(t_0)$ are linearly independent
Since $\tilde{\gamma}$ is a reparametrization of $\gamma$, then there exists a smooth bijective map $\phi$ such that $\tilde{\gamma}(t)=\gamma (\phi (t))$.
Let $t=\phi (\tilde{t})$ and $\psi=\phi^{-1}$ such that $\tilde{t}=\psi (t)$. Deriving the equation $\phi (\psi (t))=t$ in respect to $t$ and using the chain rule we get $$\frac{d\phi}{d\tilde{t}} \frac{d\psi}{dt}=1$$ So $\frac{d\phi}{d\tilde{t}}$ is never zero.
Let $\tilde{\gamma}(\tilde{t})=\gamma (\phi (\tilde{t}))$. Using the chain rule we get $$\frac{d\tilde{\gamma}}{d\tilde{t}}=\frac{d\gamma}{dt}\frac{d\phi}{d\tilde{t}}$$
Then the second derivative is the following:
$$\frac{d^2\tilde{\gamma}}{d\tilde{t}^2}=\frac{d}{d\tilde{t}}\left (\frac{d\gamma}{dt}\frac{d\phi}{d\tilde{t}}\right )=\frac{d}{d\tilde{t}}\left (\frac{d\gamma}{dt}\right )\frac{d\phi}{d\tilde{t}}+\frac{d\gamma}{dt}\frac{d}{d\tilde{t}}\left (\frac{d\phi}{d\tilde{t}}\right )=\frac{d^2\gamma}{d\tilde{t}^2}\left (\frac{d\phi}{dt}\right )^2+\frac{d\gamma}{dt}\frac{d^2\phi}{d\tilde{t}^2}$$
Then $$\frac{d^2\tilde{\gamma}}{d\tilde{t}^2} \large |_{\tilde{t}=\psi (t_0)} =\frac{d^2\gamma}{d\tilde{t}^2}\large |_{t=t_0} \left (\frac{d\phi}{dt}\right )^2\large |_{\tilde{t}=\psi (t_0)} +\frac{d\gamma}{dt}\large |_{t=t_0}\frac{d^2\phi}{d\tilde{t}^2}\large |_{\tilde{t}=\psi (t_0)} \\ \Rightarrow \frac{d^2\tilde{\gamma}}{d\tilde{t}^2} \large |_{\tilde{t}=\psi (t_0)} =\frac{d^2\gamma}{d\tilde{t}^2}\large |_{t=t_0} \left (\frac{d\phi}{dt}\right )^2\large |_{\tilde{t}=\psi (t_0)}$$
$$\frac{d^3\tilde{\gamma}}{d\tilde{t}^3}= \cdots =\frac{d^3\gamma}{dt^3}\left (\frac{d\phi}{d\tilde{t}}\right )^3+3\frac{d^2\gamma}{dt^2}\frac{d\phi}{d\tilde{t}}\frac{d^2\phi}{d\tilde{t}^2}+\frac{d\gamma}{dt}\frac{d^3\phi}{d\tilde{t}^3}$$
Then $$\frac{d^3\tilde{\gamma}}{d\tilde{t}^3}\large |_{\tilde{t}=\psi (t_0)}=\frac{d^3\gamma}{dt^3}\large |_{t=t_0} \left (\frac{d\phi}{d\tilde{t}}\right )^3\large |_{\tilde{t}=\psi (t_0)} +3\frac{d^2\gamma}{dt^2}\large |_{t=t_0}\frac{d\phi}{d\tilde{t}}\large |_{\tilde{t}=\psi (t_0)} \frac{d^2\phi}{d\tilde{t}^2}\large |_{\tilde{t}=\psi (t_0)}+\frac{d\gamma}{dt}\large |_{t=t_0}\frac{d^3\phi}{d\tilde{t}^3}\large |_{\tilde{t}=\psi (t_0)} \\ \Rightarrow \frac{d^3\tilde{\gamma}}{d\tilde{t}^3}\large |_{\tilde{t}=\psi (t_0)}=\frac{d^3\gamma}{dt^3}\large |_{t=t_0} \left (\frac{d\phi}{d\tilde{t}}\right )^3\large |_{\tilde{t}=\psi (t_0)} +3\frac{d^2\gamma}{dt^2}\large |_{t=t_0}\frac{d\phi}{d\tilde{t}}\large |_{\tilde{t}=\psi (t_0)} \frac{d^2\phi}{d\tilde{t}^2}\large |_{\tilde{t}=\psi (t_0)}$$
Are all the derivatives correct?
How could we continue to show that $\frac{d^2\tilde{\gamma}}{d\tilde{t}^2} \large |_{\tilde{t}=\psi (t_0)}$ and $\frac{d^3\tilde{\gamma}}{d\tilde{t}^3}\large |_{\tilde{t}=\psi (t_0)}$ are non-zero and linearly independent?
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EDIT2:
I am facing some difficulties showing that $\frac{d^3\tilde{\gamma}}{d\tilde{t}^3}\large |_{\tilde{t}=\psi (t_0)}$ is non-zero. We have that $$\frac{d^3\tilde{\gamma}}{d\tilde{t}^3}\large |_{\tilde{t}=\psi (t_0)}=\frac{d^3\gamma}{dt^3}\large |_{t=t_0} \left (\frac{d\phi}{d\tilde{t}}\right )^3\large |_{\tilde{t}=\psi (t_0)} +3\frac{d^2\gamma}{dt^2}\large |_{t=t_0}\frac{d\phi}{d\tilde{t}}\large |_{\tilde{t}=\psi (t_0)} \frac{d^2\phi}{d\tilde{t}^2}\large |_{\tilde{t}=\psi (t_0)}$$
We know that $\frac{d^3\gamma}{dt^3}\large |_{t=t_0}$ and $\left (\frac{d\phi}{d\tilde{t}}\right )^3\large |_{\tilde{t}=\psi (t_0)}$ are non-zero. Is the term $\frac{d^2\phi}{d\tilde{t}^2}\large |_{\tilde{t}=\psi (t_0)}$ non-zero, so that the product $\frac{d^2\gamma}{dt^2}\large |_{t=t_0}\frac{d\phi}{d\tilde{t}}\large |_{\tilde{t}=\psi (t_0)} \frac{d^2\phi}{d\tilde{t}^2}\large |_{\tilde{t}=\psi (t_0)}$ is also non-zero?
The first, second, and third derivatives of $\gamma$ with respect to $s$ are $$ \left(\frac{\gamma'}{s'},\frac{\gamma''s'-\gamma's''}{s'^3},\frac{\left(\gamma'''s'-\gamma's'''\right)s'^3-3s'^2s''\left(\gamma''s'-\gamma's''\right)}{s'^7}\right)\tag{1} $$ If $\gamma'=0$, $(1)$ simplifies to $$ \left(0,\frac{\gamma''}{s'^2},\frac{\gamma'''s'-3s''\gamma''}{s'^4}\right)\tag{2} $$ Suppose that $$ a\,\frac{\gamma''}{s'^2}+b\,\frac{\gamma'''s'-3s''\gamma''}{s'^4}=0\tag{3} $$ Then $$ \left(\frac{a}{s'^2}-\frac{3bs''}{s'^4}\right)\gamma''+\frac{b}{s'^3}\,\gamma'''=0\tag{4} $$ Since $\gamma''$ and $\gamma'''$ are independent, $a=b=0$, which means that the second and third derivatives of $\gamma$ with respect to $s$ must also be linearly independent.