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A question about a claim from “Riemannian Geometry” by Petersen
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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.
riemannian-geometry smooth-functions
asked Jan 31 at 6:34
Anju GeorgeAnju George
1088
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add a comment |
$begingroup$
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.
riemannian-geometry smooth-functions
asked Jan 31 at 6:34
Anju GeorgeAnju George
1088
$endgroup$
1
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It looks unlikely ... is there anything else that Petersen assumes about $psi$?
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– Lord Shark the Unknown
Jan 31 at 7:39
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@LordSharktheUnknown- No. He just assumes $psi$ is smooth analytic
$endgroup$
– Anju George
Jan 31 at 13:51
add a comment |
$begingroup$
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.
riemannian-geometry smooth-functions
asked Jan 31 at 6:34
Anju GeorgeAnju George
1088
$endgroup$
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.
riemannian-geometry smooth-functions
riemannian-geometry smooth-functions
asked Jan 31 at 6:34
Anju GeorgeAnju George
1088
asked Jan 31 at 6:34
Anju GeorgeAnju George
1088
asked Jan 31 at 6:34
Anju GeorgeAnju George
1088
asked Jan 31 at 6:34
Anju GeorgeAnju George
1088
asked Jan 31 at 6:34
Anju GeorgeAnju George
1088
1088
1
$begingroup$
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
add a comment |
1
$begingroup$
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
1
1
$begingroup$
It looks unlikely ... is there anything else that Petersen assumes about $psi$?
$endgroup$
– Lord Shark the Unknown
Jan 31 at 7:39
$begingroup$
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
$begingroup$
@LordSharktheUnknown- No. He just assumes $psi$ is smooth analytic
$endgroup$
– Anju George
Jan 31 at 13:51
add a comment |
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1
$begingroup$
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