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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.

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

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