Mixing
Gram-Schimdt get the Legendre polynomial
$begingroup$
When I apply the Gram-Schmidt algorithm I don't understand why I don't get the Legendre polynomials. When I apply this algorithm I always get monic polynomials whereas the Legendre polynomials aren't all monics.
So we want to get an orthogonal basis for the inner product : $langle f,grangle:=int fg$. To so we are going to apply G-S algorithm on the basis: $(1,X,X^2,...)$
We have : $P_0 = 1$, $P_1 = X$ and
$$P_2 = X^2 - frac{langle X^2, 1rangle}{|1 |}1 - frac{langle X^2, Xrangle}{|X |^2}X$$
Hence $P_2$ is a monic polynomial here while the second Legendre polynomial isn't monic.
And more generally we have :
$$P_n = X^n - sum_{k = 0}^{n-1} frac{langle X^n, P_irangle }{| P_i |^2} P_i$$
So all the polynomials I get thanks to G-S are monic.
So where am I going wrong here?
Thank you !
linear-algebra polynomials vector-spaces orthogonality
edited Jan 23 at 15:55
Adrian Keister
5,27371933
asked Jan 23 at 15:47
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
|
show 7 more comments
$begingroup$
When I apply the Gram-Schmidt algorithm I don't understand why I don't get the Legendre polynomials. When I apply this algorithm I always get monic polynomials whereas the Legendre polynomials aren't all monics.
So we want to get an orthogonal basis for the inner product : $langle f,grangle:=int fg$. To so we are going to apply G-S algorithm on the basis: $(1,X,X^2,...)$
We have : $P_0 = 1$, $P_1 = X$ and
$$P_2 = X^2 - frac{langle X^2, 1rangle}{|1 |}1 - frac{langle X^2, Xrangle}{|X |^2}X$$
Hence $P_2$ is a monic polynomial here while the second Legendre polynomial isn't monic.
And more generally we have :
$$P_n = X^n - sum_{k = 0}^{n-1} frac{langle X^n, P_irangle }{| P_i |^2} P_i$$
So all the polynomials I get thanks to G-S are monic.
So where am I going wrong here?
Thank you !
linear-algebra polynomials vector-spaces orthogonality
edited Jan 23 at 15:55
Adrian Keister
5,27371933
asked Jan 23 at 15:47
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
1
$begingroup$
After G-S, you should normalize these $P$.
$endgroup$
– xbh
Jan 23 at 15:54
$begingroup$
@xbh Thank you for your answe, but what do you mean by normalize ? What calculation I need to do in order to normalize theses polynmials ?. Thank you !
$endgroup$
– dghkgfzyukz
Jan 23 at 16:02
$begingroup$
A vector with length/module $1$ is called a normalized vector. So for $boldsymbol x$, its normalization is $boldsymbol x / Vert boldsymbol x Vert$.
$endgroup$
– xbh
Jan 23 at 16:07
$begingroup$
Note that Legendre polynomials are generally normalised such that $P_n(1) = 1$. However, I don't know why such an unusual normalisation is chosen
$endgroup$
– Damien
Jan 23 at 16:38
$begingroup$
@Damien $X^2-1/3$ is a monic polynomial.
$endgroup$
– dghkgfzyukz
Jan 23 at 16:45
|
show 7 more comments
$begingroup$
When I apply the Gram-Schmidt algorithm I don't understand why I don't get the Legendre polynomials. When I apply this algorithm I always get monic polynomials whereas the Legendre polynomials aren't all monics.
So we want to get an orthogonal basis for the inner product : $langle f,grangle:=int fg$. To so we are going to apply G-S algorithm on the basis: $(1,X,X^2,...)$
We have : $P_0 = 1$, $P_1 = X$ and
$$P_2 = X^2 - frac{langle X^2, 1rangle}{|1 |}1 - frac{langle X^2, Xrangle}{|X |^2}X$$
Hence $P_2$ is a monic polynomial here while the second Legendre polynomial isn't monic.
And more generally we have :
$$P_n = X^n - sum_{k = 0}^{n-1} frac{langle X^n, P_irangle }{| P_i |^2} P_i$$
So all the polynomials I get thanks to G-S are monic.
So where am I going wrong here?
Thank you !
linear-algebra polynomials vector-spaces orthogonality
edited Jan 23 at 15:55
Adrian Keister
5,27371933
asked Jan 23 at 15:47
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
When I apply the Gram-Schmidt algorithm I don't understand why I don't get the Legendre polynomials. When I apply this algorithm I always get monic polynomials whereas the Legendre polynomials aren't all monics.
So we want to get an orthogonal basis for the inner product : $langle f,grangle:=int fg$. To so we are going to apply G-S algorithm on the basis: $(1,X,X^2,...)$
We have : $P_0 = 1$, $P_1 = X$ and
$$P_2 = X^2 - frac{langle X^2, 1rangle}{|1 |}1 - frac{langle X^2, Xrangle}{|X |^2}X$$
Hence $P_2$ is a monic polynomial here while the second Legendre polynomial isn't monic.
And more generally we have :
$$P_n = X^n - sum_{k = 0}^{n-1} frac{langle X^n, P_irangle }{| P_i |^2} P_i$$
So all the polynomials I get thanks to G-S are monic.
So where am I going wrong here?
Thank you !
linear-algebra polynomials vector-spaces orthogonality
linear-algebra polynomials vector-spaces orthogonality
edited Jan 23 at 15:55
Adrian Keister
5,27371933
asked Jan 23 at 15:47
dghkgfzyukzdghkgfzyukz
16112
edited Jan 23 at 15:55
Adrian Keister
5,27371933
asked Jan 23 at 15:47
dghkgfzyukzdghkgfzyukz
16112
edited Jan 23 at 15:55
Adrian Keister
5,27371933
edited Jan 23 at 15:55
Adrian Keister
5,27371933
edited Jan 23 at 15:55
Adrian Keister
5,27371933
5,27371933
asked Jan 23 at 15:47
dghkgfzyukzdghkgfzyukz
16112
asked Jan 23 at 15:47
dghkgfzyukzdghkgfzyukz
16112
asked Jan 23 at 15:47
dghkgfzyukzdghkgfzyukz
16112
16112
1
$begingroup$
After G-S, you should normalize these $P$.
$endgroup$
– xbh
Jan 23 at 15:54
$begingroup$
@xbh Thank you for your answe, but what do you mean by normalize ? What calculation I need to do in order to normalize theses polynmials ?. Thank you !
$endgroup$
– dghkgfzyukz
Jan 23 at 16:02
$begingroup$
A vector with length/module $1$ is called a normalized vector. So for $boldsymbol x$, its normalization is $boldsymbol x / Vert boldsymbol x Vert$.
$endgroup$
– xbh
Jan 23 at 16:07
$begingroup$
Note that Legendre polynomials are generally normalised such that $P_n(1) = 1$. However, I don't know why such an unusual normalisation is chosen
$endgroup$
– Damien
Jan 23 at 16:38
$begingroup$
@Damien $X^2-1/3$ is a monic polynomial.
$endgroup$
– dghkgfzyukz
Jan 23 at 16:45
|
show 7 more comments
1
$begingroup$
After G-S, you should normalize these $P$.
$endgroup$
– xbh
Jan 23 at 15:54
$begingroup$
@xbh Thank you for your answe, but what do you mean by normalize ? What calculation I need to do in order to normalize theses polynmials ?. Thank you !
$endgroup$
– dghkgfzyukz
Jan 23 at 16:02
$begingroup$
A vector with length/module $1$ is called a normalized vector. So for $boldsymbol x$, its normalization is $boldsymbol x / Vert boldsymbol x Vert$.
$endgroup$
– xbh
Jan 23 at 16:07
$begingroup$
Note that Legendre polynomials are generally normalised such that $P_n(1) = 1$. However, I don't know why such an unusual normalisation is chosen
$endgroup$
– Damien
Jan 23 at 16:38
$begingroup$
@Damien $X^2-1/3$ is a monic polynomial.
$endgroup$
– dghkgfzyukz
Jan 23 at 16:45
1
1
$begingroup$
After G-S, you should normalize these $P$.
$endgroup$
– xbh
Jan 23 at 15:54
$begingroup$
After G-S, you should normalize these $P$.
$endgroup$
– xbh
Jan 23 at 15:54
$begingroup$
@xbh Thank you for your answe, but what do you mean by normalize ? What calculation I need to do in order to normalize theses polynmials ?. Thank you !
$endgroup$
– dghkgfzyukz
Jan 23 at 16:02
$begingroup$
@xbh Thank you for your answe, but what do you mean by normalize ? What calculation I need to do in order to normalize theses polynmials ?. Thank you !
$endgroup$
– dghkgfzyukz
Jan 23 at 16:02
$begingroup$
A vector with length/module $1$ is called a normalized vector. So for $boldsymbol x$, its normalization is $boldsymbol x / Vert boldsymbol x Vert$.
$endgroup$
– xbh
Jan 23 at 16:07
$begingroup$
A vector with length/module $1$ is called a normalized vector. So for $boldsymbol x$, its normalization is $boldsymbol x / Vert boldsymbol x Vert$.
$endgroup$
– xbh
Jan 23 at 16:07
$begingroup$
Note that Legendre polynomials are generally normalised such that $P_n(1) = 1$. However, I don't know why such an unusual normalisation is chosen
$endgroup$
– Damien
Jan 23 at 16:38
$begingroup$
Note that Legendre polynomials are generally normalised such that $P_n(1) = 1$. However, I don't know why such an unusual normalisation is chosen
$endgroup$
– Damien
Jan 23 at 16:38
$begingroup$
@Damien $X^2-1/3$ is a monic polynomial.
$endgroup$
– dghkgfzyukz
Jan 23 at 16:45
$begingroup$
@Damien $X^2-1/3$ is a monic polynomial.
$endgroup$
– dghkgfzyukz
Jan 23 at 16:45
|
show 7 more comments
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1
$begingroup$
After G-S, you should normalize these $P$.
$endgroup$
– xbh
Jan 23 at 15:54
$begingroup$
@xbh Thank you for your answe, but what do you mean by normalize ? What calculation I need to do in order to normalize theses polynmials ?. Thank you !
$endgroup$
– dghkgfzyukz
Jan 23 at 16:02
$begingroup$
A vector with length/module $1$ is called a normalized vector. So for $boldsymbol x$, its normalization is $boldsymbol x / Vert boldsymbol x Vert$.
$endgroup$
– xbh
Jan 23 at 16:07
$begingroup$
Note that Legendre polynomials are generally normalised such that $P_n(1) = 1$. However, I don't know why such an unusual normalisation is chosen
$endgroup$
– Damien
Jan 23 at 16:38
$begingroup$
@Damien $X^2-1/3$ is a monic polynomial.
$endgroup$
– dghkgfzyukz
Jan 23 at 16:45