Petersen says the following on pg 13 of "Riemannian Geometry" by Petersen:
If $\frac{\psi^2-1}{t^2}$ is a smooth function of $t$ at $t=0$, then $\psi^{(1)}=\psi^{(3)}=\dots=\psi^{(odd)}=0$
I don't understand this claim. What if $\psi=(1+t^2+t^3)$? The function $\frac{\psi^2-1}{t^2}$ still seems to be smooth, and some odd derivatives of $\psi$ are non-zero.