showing orthonormality of functions











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Let $ F$ be an unit sphere with $ F:= partial K_1(o) subset mathbb{R}^3 $



For all continous functions $f,g in C(F)$ define
$ <f,g>_n := int_F f(x)g(x) do(x) $



And let



$ F_1(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta cos phi $



$ F_2(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta sin phi $



$F_3(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta cos theta cos phi$



$x( theta, phi) $ is the presentation of $x$ in $F$ in polarcoordinates $ theta in (o, pi), phi in (0, 2pi) $



I need to show that those functions are pairwise orthnormal.



okay, let's put



$<F_1,F_2> = int_Fsqrt{ frac{3}{4 pi }} sin theta cos phisqrt{ frac{3}{4 pi }} sin theta sin phi do(x) $



Now I am confused, what to put for the bounds for the integral?










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    up vote
    1
    down vote

    favorite












    Let $ F$ be an unit sphere with $ F:= partial K_1(o) subset mathbb{R}^3 $



    For all continous functions $f,g in C(F)$ define
    $ <f,g>_n := int_F f(x)g(x) do(x) $



    And let



    $ F_1(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta cos phi $



    $ F_2(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta sin phi $



    $F_3(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta cos theta cos phi$



    $x( theta, phi) $ is the presentation of $x$ in $F$ in polarcoordinates $ theta in (o, pi), phi in (0, 2pi) $



    I need to show that those functions are pairwise orthnormal.



    okay, let's put



    $<F_1,F_2> = int_Fsqrt{ frac{3}{4 pi }} sin theta cos phisqrt{ frac{3}{4 pi }} sin theta sin phi do(x) $



    Now I am confused, what to put for the bounds for the integral?










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $ F$ be an unit sphere with $ F:= partial K_1(o) subset mathbb{R}^3 $



      For all continous functions $f,g in C(F)$ define
      $ <f,g>_n := int_F f(x)g(x) do(x) $



      And let



      $ F_1(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta cos phi $



      $ F_2(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta sin phi $



      $F_3(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta cos theta cos phi$



      $x( theta, phi) $ is the presentation of $x$ in $F$ in polarcoordinates $ theta in (o, pi), phi in (0, 2pi) $



      I need to show that those functions are pairwise orthnormal.



      okay, let's put



      $<F_1,F_2> = int_Fsqrt{ frac{3}{4 pi }} sin theta cos phisqrt{ frac{3}{4 pi }} sin theta sin phi do(x) $



      Now I am confused, what to put for the bounds for the integral?










      share|cite|improve this question













      Let $ F$ be an unit sphere with $ F:= partial K_1(o) subset mathbb{R}^3 $



      For all continous functions $f,g in C(F)$ define
      $ <f,g>_n := int_F f(x)g(x) do(x) $



      And let



      $ F_1(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta cos phi $



      $ F_2(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta sin phi $



      $F_3(x( theta, phi)):= sqrt{ frac{3}{4 pi }} sin theta cos theta cos phi$



      $x( theta, phi) $ is the presentation of $x$ in $F$ in polarcoordinates $ theta in (o, pi), phi in (0, 2pi) $



      I need to show that those functions are pairwise orthnormal.



      okay, let's put



      $<F_1,F_2> = int_Fsqrt{ frac{3}{4 pi }} sin theta cos phisqrt{ frac{3}{4 pi }} sin theta sin phi do(x) $



      Now I am confused, what to put for the bounds for the integral?







      real-analysis integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Dec 2 at 20:02









      constant94

      625




      625






















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          Note that You integrate over a manifold $F=partial(K_1(0))$ for which ( except for a set of measure zero) You have a chart:
          $$Psi:(0,pi)times(0,2pi)rightarrow F,(theta,varphi)mapsto (sinthetacosvarphi,sinthetasinvarphi,costheta).$$
          Now You compute
          $$frac{partialPsi}{partialtheta}=begin{pmatrix}costhetacosvarphi\ costhetasinvarphi\-sinthetaend{pmatrix},frac{partialPsi}{partialvarphi}=begin{pmatrix}-sinthetasinvarphi\sinthetacosvarphi\0end{pmatrix}$$
          and $|frac{partialPsi}{partialtheta}|^2=1,|frac{partialPsi}{partialvarphi}|^2=sin^2theta,langlefrac{partialPsi}{partialtheta},frac{partialPsi}{partialvarphi}rangle=0$ to get the metric tensor :
          $$G=begin{pmatrix}1&0\0&sin^2thetaend{pmatrix}.$$
          Now for example
          $$langle F_1,F_2rangle=int_{(0,pi)times(0,2pi)}F_1(theta,varphi)F_2(theta,varphi)sqrt(|det G|)d(theta,varphi)=$$
          $$int_0^{2pi}int_0^{pi}sqrt{frac{3}{4pi}}sinthetacosvarphisqrt{frac{3}{4pi}}sinthetasinvarphi|sintheta|dtheta dvarphi=$$
          $$=frac{3}{4pi}int_0^{pi}sin^3theta dthetaunderbrace{int_0^{2pi}cosvarphisinvarphi dvarphi}_{=0}=0$$
          and
          $$langle F_1,F_1rangle=int_0^{2pi}int_0^{pi}(sqrt{frac{3}{4pi}}sinthetacosvarphi)^2|sintheta|dtheta dvarphi=$$
          $$frac{3}{4pi}underbrace{int_0^{pi}sin^3theta dtheta}_{=frac{4}{3}}underbrace{int_0^{2pi}cos^2varphi dvarphi}_{=pi}=1$$.
          You proceed similar with the other products.






          share|cite|improve this answer























          • perfect answere ! thanks a lot
            – constant94
            Dec 3 at 17:49











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          1 Answer
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          1 Answer
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          active

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          active

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          up vote
          1
          down vote



          accepted










          Note that You integrate over a manifold $F=partial(K_1(0))$ for which ( except for a set of measure zero) You have a chart:
          $$Psi:(0,pi)times(0,2pi)rightarrow F,(theta,varphi)mapsto (sinthetacosvarphi,sinthetasinvarphi,costheta).$$
          Now You compute
          $$frac{partialPsi}{partialtheta}=begin{pmatrix}costhetacosvarphi\ costhetasinvarphi\-sinthetaend{pmatrix},frac{partialPsi}{partialvarphi}=begin{pmatrix}-sinthetasinvarphi\sinthetacosvarphi\0end{pmatrix}$$
          and $|frac{partialPsi}{partialtheta}|^2=1,|frac{partialPsi}{partialvarphi}|^2=sin^2theta,langlefrac{partialPsi}{partialtheta},frac{partialPsi}{partialvarphi}rangle=0$ to get the metric tensor :
          $$G=begin{pmatrix}1&0\0&sin^2thetaend{pmatrix}.$$
          Now for example
          $$langle F_1,F_2rangle=int_{(0,pi)times(0,2pi)}F_1(theta,varphi)F_2(theta,varphi)sqrt(|det G|)d(theta,varphi)=$$
          $$int_0^{2pi}int_0^{pi}sqrt{frac{3}{4pi}}sinthetacosvarphisqrt{frac{3}{4pi}}sinthetasinvarphi|sintheta|dtheta dvarphi=$$
          $$=frac{3}{4pi}int_0^{pi}sin^3theta dthetaunderbrace{int_0^{2pi}cosvarphisinvarphi dvarphi}_{=0}=0$$
          and
          $$langle F_1,F_1rangle=int_0^{2pi}int_0^{pi}(sqrt{frac{3}{4pi}}sinthetacosvarphi)^2|sintheta|dtheta dvarphi=$$
          $$frac{3}{4pi}underbrace{int_0^{pi}sin^3theta dtheta}_{=frac{4}{3}}underbrace{int_0^{2pi}cos^2varphi dvarphi}_{=pi}=1$$.
          You proceed similar with the other products.






          share|cite|improve this answer























          • perfect answere ! thanks a lot
            – constant94
            Dec 3 at 17:49















          up vote
          1
          down vote



          accepted










          Note that You integrate over a manifold $F=partial(K_1(0))$ for which ( except for a set of measure zero) You have a chart:
          $$Psi:(0,pi)times(0,2pi)rightarrow F,(theta,varphi)mapsto (sinthetacosvarphi,sinthetasinvarphi,costheta).$$
          Now You compute
          $$frac{partialPsi}{partialtheta}=begin{pmatrix}costhetacosvarphi\ costhetasinvarphi\-sinthetaend{pmatrix},frac{partialPsi}{partialvarphi}=begin{pmatrix}-sinthetasinvarphi\sinthetacosvarphi\0end{pmatrix}$$
          and $|frac{partialPsi}{partialtheta}|^2=1,|frac{partialPsi}{partialvarphi}|^2=sin^2theta,langlefrac{partialPsi}{partialtheta},frac{partialPsi}{partialvarphi}rangle=0$ to get the metric tensor :
          $$G=begin{pmatrix}1&0\0&sin^2thetaend{pmatrix}.$$
          Now for example
          $$langle F_1,F_2rangle=int_{(0,pi)times(0,2pi)}F_1(theta,varphi)F_2(theta,varphi)sqrt(|det G|)d(theta,varphi)=$$
          $$int_0^{2pi}int_0^{pi}sqrt{frac{3}{4pi}}sinthetacosvarphisqrt{frac{3}{4pi}}sinthetasinvarphi|sintheta|dtheta dvarphi=$$
          $$=frac{3}{4pi}int_0^{pi}sin^3theta dthetaunderbrace{int_0^{2pi}cosvarphisinvarphi dvarphi}_{=0}=0$$
          and
          $$langle F_1,F_1rangle=int_0^{2pi}int_0^{pi}(sqrt{frac{3}{4pi}}sinthetacosvarphi)^2|sintheta|dtheta dvarphi=$$
          $$frac{3}{4pi}underbrace{int_0^{pi}sin^3theta dtheta}_{=frac{4}{3}}underbrace{int_0^{2pi}cos^2varphi dvarphi}_{=pi}=1$$.
          You proceed similar with the other products.






          share|cite|improve this answer























          • perfect answere ! thanks a lot
            – constant94
            Dec 3 at 17:49













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Note that You integrate over a manifold $F=partial(K_1(0))$ for which ( except for a set of measure zero) You have a chart:
          $$Psi:(0,pi)times(0,2pi)rightarrow F,(theta,varphi)mapsto (sinthetacosvarphi,sinthetasinvarphi,costheta).$$
          Now You compute
          $$frac{partialPsi}{partialtheta}=begin{pmatrix}costhetacosvarphi\ costhetasinvarphi\-sinthetaend{pmatrix},frac{partialPsi}{partialvarphi}=begin{pmatrix}-sinthetasinvarphi\sinthetacosvarphi\0end{pmatrix}$$
          and $|frac{partialPsi}{partialtheta}|^2=1,|frac{partialPsi}{partialvarphi}|^2=sin^2theta,langlefrac{partialPsi}{partialtheta},frac{partialPsi}{partialvarphi}rangle=0$ to get the metric tensor :
          $$G=begin{pmatrix}1&0\0&sin^2thetaend{pmatrix}.$$
          Now for example
          $$langle F_1,F_2rangle=int_{(0,pi)times(0,2pi)}F_1(theta,varphi)F_2(theta,varphi)sqrt(|det G|)d(theta,varphi)=$$
          $$int_0^{2pi}int_0^{pi}sqrt{frac{3}{4pi}}sinthetacosvarphisqrt{frac{3}{4pi}}sinthetasinvarphi|sintheta|dtheta dvarphi=$$
          $$=frac{3}{4pi}int_0^{pi}sin^3theta dthetaunderbrace{int_0^{2pi}cosvarphisinvarphi dvarphi}_{=0}=0$$
          and
          $$langle F_1,F_1rangle=int_0^{2pi}int_0^{pi}(sqrt{frac{3}{4pi}}sinthetacosvarphi)^2|sintheta|dtheta dvarphi=$$
          $$frac{3}{4pi}underbrace{int_0^{pi}sin^3theta dtheta}_{=frac{4}{3}}underbrace{int_0^{2pi}cos^2varphi dvarphi}_{=pi}=1$$.
          You proceed similar with the other products.






          share|cite|improve this answer














          Note that You integrate over a manifold $F=partial(K_1(0))$ for which ( except for a set of measure zero) You have a chart:
          $$Psi:(0,pi)times(0,2pi)rightarrow F,(theta,varphi)mapsto (sinthetacosvarphi,sinthetasinvarphi,costheta).$$
          Now You compute
          $$frac{partialPsi}{partialtheta}=begin{pmatrix}costhetacosvarphi\ costhetasinvarphi\-sinthetaend{pmatrix},frac{partialPsi}{partialvarphi}=begin{pmatrix}-sinthetasinvarphi\sinthetacosvarphi\0end{pmatrix}$$
          and $|frac{partialPsi}{partialtheta}|^2=1,|frac{partialPsi}{partialvarphi}|^2=sin^2theta,langlefrac{partialPsi}{partialtheta},frac{partialPsi}{partialvarphi}rangle=0$ to get the metric tensor :
          $$G=begin{pmatrix}1&0\0&sin^2thetaend{pmatrix}.$$
          Now for example
          $$langle F_1,F_2rangle=int_{(0,pi)times(0,2pi)}F_1(theta,varphi)F_2(theta,varphi)sqrt(|det G|)d(theta,varphi)=$$
          $$int_0^{2pi}int_0^{pi}sqrt{frac{3}{4pi}}sinthetacosvarphisqrt{frac{3}{4pi}}sinthetasinvarphi|sintheta|dtheta dvarphi=$$
          $$=frac{3}{4pi}int_0^{pi}sin^3theta dthetaunderbrace{int_0^{2pi}cosvarphisinvarphi dvarphi}_{=0}=0$$
          and
          $$langle F_1,F_1rangle=int_0^{2pi}int_0^{pi}(sqrt{frac{3}{4pi}}sinthetacosvarphi)^2|sintheta|dtheta dvarphi=$$
          $$frac{3}{4pi}underbrace{int_0^{pi}sin^3theta dtheta}_{=frac{4}{3}}underbrace{int_0^{2pi}cos^2varphi dvarphi}_{=pi}=1$$.
          You proceed similar with the other products.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 at 2:27

























          answered Dec 2 at 20:50









          Peter Melech

          2,519813




          2,519813












          • perfect answere ! thanks a lot
            – constant94
            Dec 3 at 17:49


















          • perfect answere ! thanks a lot
            – constant94
            Dec 3 at 17:49
















          perfect answere ! thanks a lot
          – constant94
          Dec 3 at 17:49




          perfect answere ! thanks a lot
          – constant94
          Dec 3 at 17:49


















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