# A question about a claim from “Riemannian Geometry” by Petersen

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.

• It looks unlikely ... is there anything else that Petersen assumes about $\psi$? – Lord Shark the Unknown Jan 31 at 7:39
• @LordSharktheUnknown- No. He just assumes $\psi$ is smooth analytic – Anju George Jan 31 at 13:51