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










2












$begingroup$


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:
enter image description here



I tried to use $p_n-p=sigma(x_n)-sigma(x))=d_psigma(x_n-x)+o(|x_n-x|)$ to rewrite $langle N,fracp_n-prangle$



enter image description here










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    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:
    enter image description here



    I tried to use $p_n-p=sigma(x_n)-sigma(x))=d_psigma(x_n-x)+o(|x_n-x|)$ to rewrite $langle N,fracp_n-prangle$



    enter image description here










    share|cite|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$


      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:
      enter image description here



      I tried to use $p_n-p=sigma(x_n)-sigma(x))=d_psigma(x_n-x)+o(|x_n-x|)$ to rewrite $langle N,fracp_n-prangle$



      enter image description here










      share|cite|improve this question











      $endgroup$




      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:
      enter image description here



      I tried to use $p_n-p=sigma(x_n)-sigma(x))=d_psigma(x_n-x)+o(|x_n-x|)$ to rewrite $langle N,fracp_n-prangle$



      enter image description here







      differential-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 26 at 17:01







      Hans

















      asked Mar 26 at 15:48









      HansHans

      122110




      122110




















          1 Answer
          1






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          2












          $begingroup$

          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)q_n_i-qcdot fracq_n_i-q \
          &=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)q_n_i-q +dvarphileft(fracq_n_i-qq_n_i-qright)cdot fracq_n_i-q.
          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.)






          share|cite|improve this 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)q_n_i-qcdot fracq_n_i-q \
            &=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)q_n_i-q +dvarphileft(fracq_n_i-qq_n_i-qright)cdot fracq_n_i-q.
            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.)






            share|cite|improve this answer











            $endgroup$

















              2












              $begingroup$

              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)q_n_i-qcdot fracq_n_i-q \
              &=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)q_n_i-q +dvarphileft(fracq_n_i-qq_n_i-qright)cdot fracq_n_i-q.
              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.)






              share|cite|improve this answer











              $endgroup$















                2












                2








                2





                $begingroup$

                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)q_n_i-qcdot fracq_n_i-q \
                &=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)q_n_i-q +dvarphileft(fracq_n_i-qq_n_i-qright)cdot fracq_n_i-q.
                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.)






                share|cite|improve this answer











                $endgroup$



                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)q_n_i-qcdot fracq_n_i-q \
                &=fracvarphi(q_n_i)-varphi(q)-dvarphi(q_n_i-q)q_n_i-q +dvarphileft(fracq_n_i-qq_n_i-qright)cdot fracq_n_i-q.
                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.)







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 26 at 17:01

























                answered Mar 26 at 16:56









                Aloizio MacedoAloizio Macedo

                23.8k24088




                23.8k24088



























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