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?
real-analysis integration
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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?
real-analysis integration
add a comment |
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?
real-analysis integration
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
real-analysis integration
asked Dec 2 at 20:02
constant94
625
625
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1 Answer
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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.
perfect answere ! thanks a lot
– constant94
Dec 3 at 17:49
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
perfect answere ! thanks a lot
– constant94
Dec 3 at 17:49
add a comment |
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.
perfect answere ! thanks a lot
– constant94
Dec 3 at 17:49
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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