Show that $limlimits_ntoinftyleftlangle N,fracp_n-prightrangle =0$ on a regular surface Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.Why study “curves” instead of 1-manifolds?Techniques to check if a surface is regularA condition on parametrisations of regular surfacesParametrised vs Regular SurfacesIntuition behind transverse planes and regular curvesRegular surfaces in Differential GeometryDifferential Geometry Regular Surface problemShow that a surface of revolution can always be parametrized so that $E = E(v), F = 0$ and $G = 1$.Why does this property of reparametrizations of surfaces matter?If a regular surface lie on side of a plane, then the plane is tagent plane
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Show that $limlimits_ntoinftyleftlangle N,fracp_n-prightrangle =0$ on a regular surface
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.Why study “curves” instead of 1-manifolds?Techniques to check if a surface is regularA condition on parametrisations of regular surfacesParametrised vs Regular SurfacesIntuition behind transverse planes and regular curvesRegular surfaces in Differential GeometryDifferential Geometry Regular Surface problemShow that a surface of revolution can always be parametrized so that $E = E(v), F = 0$ and $G = 1$.Why does this property of reparametrizations of surfaces matter?If a regular surface lie on side of a plane, then the plane is tagent plane
$begingroup$
I am trying to do the following question, because I think it is very interesting as it formalises the idea that the tangent space is practically indistinguishable from the surface if one looks in a very small neighbourhood of a point. The question is from "differential Geometry of Curves and Surfaces" by Tapp.
My first idea was to show that $limlimits_ntoinftyfracp_n-p=:vin T_pS$ but I don't see why/if this limit exists.
I would be very greatful for some hints or even a solution.
Thanks a lot in advance!:)
Definition of regular surface:
I tried to use $p_n-p=sigma(x_n)-sigma(x))=d_psigma(x_n-x)+o(|x_n-x|)$ to rewrite $langle N,fracp_n-prangle$

differential-geometry
$endgroup$
add a comment |
$begingroup$
I am trying to do the following question, because I think it is very interesting as it formalises the idea that the tangent space is practically indistinguishable from the surface if one looks in a very small neighbourhood of a point. The question is from "differential Geometry of Curves and Surfaces" by Tapp.
My first idea was to show that $limlimits_ntoinftyfracp_n-p=:vin T_pS$ but I don't see why/if this limit exists.
I would be very greatful for some hints or even a solution.
Thanks a lot in advance!:)
Definition of regular surface:
I tried to use $p_n-p=sigma(x_n)-sigma(x))=d_psigma(x_n-x)+o(|x_n-x|)$ to rewrite $langle N,fracp_n-prangle$

differential-geometry
$endgroup$
add a comment |
$begingroup$
I am trying to do the following question, because I think it is very interesting as it formalises the idea that the tangent space is practically indistinguishable from the surface if one looks in a very small neighbourhood of a point. The question is from "differential Geometry of Curves and Surfaces" by Tapp.
My first idea was to show that $limlimits_ntoinftyfracp_n-p=:vin T_pS$ but I don't see why/if this limit exists.
I would be very greatful for some hints or even a solution.
Thanks a lot in advance!:)
Definition of regular surface:
I tried to use $p_n-p=sigma(x_n)-sigma(x))=d_psigma(x_n-x)+o(|x_n-x|)$ to rewrite $langle N,fracp_n-prangle$

differential-geometry
$endgroup$
I am trying to do the following question, because I think it is very interesting as it formalises the idea that the tangent space is practically indistinguishable from the surface if one looks in a very small neighbourhood of a point. The question is from "differential Geometry of Curves and Surfaces" by Tapp.
My first idea was to show that $limlimits_ntoinftyfracp_n-p=:vin T_pS$ but I don't see why/if this limit exists.
I would be very greatful for some hints or even a solution.
Thanks a lot in advance!:)
Definition of regular surface:
I tried to use $p_n-p=sigma(x_n)-sigma(x))=d_psigma(x_n-x)+o(|x_n-x|)$ to rewrite $langle N,fracp_n-prangle$

differential-geometry
differential-geometry
edited Mar 26 at 17:01
Hans
asked Mar 26 at 15:48
HansHans
122110
122110
add a comment |
add a comment |
1 Answer
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$begingroup$
It may very well be that $lim_n to infty fracp_n-p$ does not exist. (To see this, consider an oscillation near a point.)
But by taking subsequences, we can prove the result, even if the limit above does not exist by itself. Fix a subsequence $p_n_i$ of $p_n$.
First, taking a parametrization $varphi$ near $p$, we have that $p_n_i$ is in that chart for $i$ sufficiently big. Letting $q_n_i=varphi^-1(p_n_i)$, we know that
beginalign
fracp_n_i-p&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)+dvarphi(q_n_i-q)cdot frac \
&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q) +dvarphileft(fracq_n_i-qright)cdot frac.
endalign
By passing to subsequences yet again (twice), this converges to
$Kcdot dvarphi(h)$ for some $h in mathbbR^2$, $K in mathbbR$. Therefore, the limit of this subsequence $fracp_n_i_j-p$ of $p_n_i$ belongs to $T_pS$ and it follows that
$$lim_j to infty leftlangle N, fracp_n_i_j-prightrangle=0.$$
Letting $a_n=leftlangle N, fracp_n-prightrangle$, we have thus proved that every subsequence of $a_n$ has a further subsequence converging to $0$. This proves that $a_n$ converges to zero. (This is a general fact about sequences - see here, for example.)
$endgroup$
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$begingroup$
It may very well be that $lim_n to infty fracp_n-p$ does not exist. (To see this, consider an oscillation near a point.)
But by taking subsequences, we can prove the result, even if the limit above does not exist by itself. Fix a subsequence $p_n_i$ of $p_n$.
First, taking a parametrization $varphi$ near $p$, we have that $p_n_i$ is in that chart for $i$ sufficiently big. Letting $q_n_i=varphi^-1(p_n_i)$, we know that
beginalign
fracp_n_i-p&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)+dvarphi(q_n_i-q)cdot frac \
&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q) +dvarphileft(fracq_n_i-qright)cdot frac.
endalign
By passing to subsequences yet again (twice), this converges to
$Kcdot dvarphi(h)$ for some $h in mathbbR^2$, $K in mathbbR$. Therefore, the limit of this subsequence $fracp_n_i_j-p$ of $p_n_i$ belongs to $T_pS$ and it follows that
$$lim_j to infty leftlangle N, fracp_n_i_j-prightrangle=0.$$
Letting $a_n=leftlangle N, fracp_n-prightrangle$, we have thus proved that every subsequence of $a_n$ has a further subsequence converging to $0$. This proves that $a_n$ converges to zero. (This is a general fact about sequences - see here, for example.)
$endgroup$
add a comment |
$begingroup$
It may very well be that $lim_n to infty fracp_n-p$ does not exist. (To see this, consider an oscillation near a point.)
But by taking subsequences, we can prove the result, even if the limit above does not exist by itself. Fix a subsequence $p_n_i$ of $p_n$.
First, taking a parametrization $varphi$ near $p$, we have that $p_n_i$ is in that chart for $i$ sufficiently big. Letting $q_n_i=varphi^-1(p_n_i)$, we know that
beginalign
fracp_n_i-p&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)+dvarphi(q_n_i-q)cdot frac \
&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q) +dvarphileft(fracq_n_i-qright)cdot frac.
endalign
By passing to subsequences yet again (twice), this converges to
$Kcdot dvarphi(h)$ for some $h in mathbbR^2$, $K in mathbbR$. Therefore, the limit of this subsequence $fracp_n_i_j-p$ of $p_n_i$ belongs to $T_pS$ and it follows that
$$lim_j to infty leftlangle N, fracp_n_i_j-prightrangle=0.$$
Letting $a_n=leftlangle N, fracp_n-prightrangle$, we have thus proved that every subsequence of $a_n$ has a further subsequence converging to $0$. This proves that $a_n$ converges to zero. (This is a general fact about sequences - see here, for example.)
$endgroup$
add a comment |
$begingroup$
It may very well be that $lim_n to infty fracp_n-p$ does not exist. (To see this, consider an oscillation near a point.)
But by taking subsequences, we can prove the result, even if the limit above does not exist by itself. Fix a subsequence $p_n_i$ of $p_n$.
First, taking a parametrization $varphi$ near $p$, we have that $p_n_i$ is in that chart for $i$ sufficiently big. Letting $q_n_i=varphi^-1(p_n_i)$, we know that
beginalign
fracp_n_i-p&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)+dvarphi(q_n_i-q)cdot frac \
&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q) +dvarphileft(fracq_n_i-qright)cdot frac.
endalign
By passing to subsequences yet again (twice), this converges to
$Kcdot dvarphi(h)$ for some $h in mathbbR^2$, $K in mathbbR$. Therefore, the limit of this subsequence $fracp_n_i_j-p$ of $p_n_i$ belongs to $T_pS$ and it follows that
$$lim_j to infty leftlangle N, fracp_n_i_j-prightrangle=0.$$
Letting $a_n=leftlangle N, fracp_n-prightrangle$, we have thus proved that every subsequence of $a_n$ has a further subsequence converging to $0$. This proves that $a_n$ converges to zero. (This is a general fact about sequences - see here, for example.)
$endgroup$
It may very well be that $lim_n to infty fracp_n-p$ does not exist. (To see this, consider an oscillation near a point.)
But by taking subsequences, we can prove the result, even if the limit above does not exist by itself. Fix a subsequence $p_n_i$ of $p_n$.
First, taking a parametrization $varphi$ near $p$, we have that $p_n_i$ is in that chart for $i$ sufficiently big. Letting $q_n_i=varphi^-1(p_n_i)$, we know that
beginalign
fracp_n_i-p&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)+dvarphi(q_n_i-q)cdot frac \
&=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q) +dvarphileft(fracq_n_i-qright)cdot frac.
endalign
By passing to subsequences yet again (twice), this converges to
$Kcdot dvarphi(h)$ for some $h in mathbbR^2$, $K in mathbbR$. Therefore, the limit of this subsequence $fracp_n_i_j-p$ of $p_n_i$ belongs to $T_pS$ and it follows that
$$lim_j to infty leftlangle N, fracp_n_i_j-prightrangle=0.$$
Letting $a_n=leftlangle N, fracp_n-prightrangle$, we have thus proved that every subsequence of $a_n$ has a further subsequence converging to $0$. This proves that $a_n$ converges to zero. (This is a general fact about sequences - see here, for example.)
edited Mar 26 at 17:01
answered Mar 26 at 16:56
Aloizio Macedo♦Aloizio Macedo
23.8k24088
23.8k24088
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