Using Frobenius method to solve the Legendre differential equation
$begingroup$
I'm tasked with solving the Legendre differential equation, and
Using $c=0$, obtain a series of even powers of $x$ (with $a_1=0$).
I found this exercise to be good at highlighting what I found confusing about this method. Firstly, I find the following indicial equations:
$$a_0 (c-1)c = 0$$
$$a_1(c+1)c = 0$$
I am instructed to choose $c=0$. Doing so, I don't see how this implies $a_1=0.$ Setting $c=0$ instantly satisfies the second equation $forall a_1 in mathbb R$. If I chose $c=1$, that would indeed imply $a_1 = 0$, but I don't see how $c=0$ does. That aside, I end up with the canonical recurrence relation
$$a_{n+2} = frac{n(n+1) - l(l+1)}{(n+2)(n+1)}a_n$$
And I went to check that even and odd indices of $a_n$ were non-zero. They were, so I don't see how I need to find an odd and even part for $c=0$. By forcing $a_1=0$, I can express an even part just by observing the even terms of recurrence relation. However, further I am tasked with:
Using $c=1$, obtain a series of odd powers of $x$ (with $a_1=0$).
Doing this forces $a_1$ to be zero, and I would then think that I'd only have odd powers of $x$ since $a_1 = 0 implies a_{n + 2} = 0 implies a_3 = a_5 = ... = 0$. However, I am meant to prove the recurrence relation:
$$a_{n+2} = frac{(n+1)(n+2) - l(l+1)}{(n+2)(n+3)}a_n$$
Which I don't recognize. This here highlights my confusions with this method:
- Why is my first recurrence relation not going to encompass both the even and odd parts if $a_{n+2} ne 0 $ for starting at both $a_1$ and $a_0$?
- Is there some non-arbitrary need that if $c=0 implies a_1 = 0$?
- If my answer may only contain odd or even powers of $x$ but $a_{n+2} ne 0 $ for starting at both $a_1$ and $a_0$, then how do I know if I only have an odd or even power of $x$ for my recurrence relation?
- Generally, what are the parametrisations of $c$ such that the solution will only encompass odd, even, or both powers of $x$?
legendre-polynomials frobenius-method
asked Dec 13 '18 at 3:22
sangstarsangstar
853215
$endgroup$
add a comment |
$begingroup$
I'm tasked with solving the Legendre differential equation, and
Using $c=0$, obtain a series of even powers of $x$ (with $a_1=0$).
I found this exercise to be good at highlighting what I found confusing about this method. Firstly, I find the following indicial equations:
$$a_0 (c-1)c = 0$$
$$a_1(c+1)c = 0$$
I am instructed to choose $c=0$. Doing so, I don't see how this implies $a_1=0.$ Setting $c=0$ instantly satisfies the second equation $forall a_1 in mathbb R$. If I chose $c=1$, that would indeed imply $a_1 = 0$, but I don't see how $c=0$ does. That aside, I end up with the canonical recurrence relation
$$a_{n+2} = frac{n(n+1) - l(l+1)}{(n+2)(n+1)}a_n$$
And I went to check that even and odd indices of $a_n$ were non-zero. They were, so I don't see how I need to find an odd and even part for $c=0$. By forcing $a_1=0$, I can express an even part just by observing the even terms of recurrence relation. However, further I am tasked with:
Using $c=1$, obtain a series of odd powers of $x$ (with $a_1=0$).
Doing this forces $a_1$ to be zero, and I would then think that I'd only have odd powers of $x$ since $a_1 = 0 implies a_{n + 2} = 0 implies a_3 = a_5 = ... = 0$. However, I am meant to prove the recurrence relation:
$$a_{n+2} = frac{(n+1)(n+2) - l(l+1)}{(n+2)(n+3)}a_n$$
Which I don't recognize. This here highlights my confusions with this method:
- Why is my first recurrence relation not going to encompass both the even and odd parts if $a_{n+2} ne 0 $ for starting at both $a_1$ and $a_0$?
- Is there some non-arbitrary need that if $c=0 implies a_1 = 0$?
- If my answer may only contain odd or even powers of $x$ but $a_{n+2} ne 0 $ for starting at both $a_1$ and $a_0$, then how do I know if I only have an odd or even power of $x$ for my recurrence relation?
- Generally, what are the parametrisations of $c$ such that the solution will only encompass odd, even, or both powers of $x$?
legendre-polynomials frobenius-method
asked Dec 13 '18 at 3:22
sangstarsangstar
853215
$endgroup$
add a comment |
$begingroup$
I'm tasked with solving the Legendre differential equation, and
Using $c=0$, obtain a series of even powers of $x$ (with $a_1=0$).
I found this exercise to be good at highlighting what I found confusing about this method. Firstly, I find the following indicial equations:
$$a_0 (c-1)c = 0$$
$$a_1(c+1)c = 0$$
I am instructed to choose $c=0$. Doing so, I don't see how this implies $a_1=0.$ Setting $c=0$ instantly satisfies the second equation $forall a_1 in mathbb R$. If I chose $c=1$, that would indeed imply $a_1 = 0$, but I don't see how $c=0$ does. That aside, I end up with the canonical recurrence relation
$$a_{n+2} = frac{n(n+1) - l(l+1)}{(n+2)(n+1)}a_n$$
And I went to check that even and odd indices of $a_n$ were non-zero. They were, so I don't see how I need to find an odd and even part for $c=0$. By forcing $a_1=0$, I can express an even part just by observing the even terms of recurrence relation. However, further I am tasked with:
Using $c=1$, obtain a series of odd powers of $x$ (with $a_1=0$).
Doing this forces $a_1$ to be zero, and I would then think that I'd only have odd powers of $x$ since $a_1 = 0 implies a_{n + 2} = 0 implies a_3 = a_5 = ... = 0$. However, I am meant to prove the recurrence relation:
$$a_{n+2} = frac{(n+1)(n+2) - l(l+1)}{(n+2)(n+3)}a_n$$
Which I don't recognize. This here highlights my confusions with this method:
- Why is my first recurrence relation not going to encompass both the even and odd parts if $a_{n+2} ne 0 $ for starting at both $a_1$ and $a_0$?
- Is there some non-arbitrary need that if $c=0 implies a_1 = 0$?
- If my answer may only contain odd or even powers of $x$ but $a_{n+2} ne 0 $ for starting at both $a_1$ and $a_0$, then how do I know if I only have an odd or even power of $x$ for my recurrence relation?
- Generally, what are the parametrisations of $c$ such that the solution will only encompass odd, even, or both powers of $x$?
legendre-polynomials frobenius-method
asked Dec 13 '18 at 3:22
sangstarsangstar
853215
$endgroup$
I'm tasked with solving the Legendre differential equation, and
Using $c=0$, obtain a series of even powers of $x$ (with $a_1=0$).
I found this exercise to be good at highlighting what I found confusing about this method. Firstly, I find the following indicial equations:
$$a_0 (c-1)c = 0$$
$$a_1(c+1)c = 0$$
I am instructed to choose $c=0$. Doing so, I don't see how this implies $a_1=0.$ Setting $c=0$ instantly satisfies the second equation $forall a_1 in mathbb R$. If I chose $c=1$, that would indeed imply $a_1 = 0$, but I don't see how $c=0$ does. That aside, I end up with the canonical recurrence relation
$$a_{n+2} = frac{n(n+1) - l(l+1)}{(n+2)(n+1)}a_n$$
And I went to check that even and odd indices of $a_n$ were non-zero. They were, so I don't see how I need to find an odd and even part for $c=0$. By forcing $a_1=0$, I can express an even part just by observing the even terms of recurrence relation. However, further I am tasked with:
Using $c=1$, obtain a series of odd powers of $x$ (with $a_1=0$).
Doing this forces $a_1$ to be zero, and I would then think that I'd only have odd powers of $x$ since $a_1 = 0 implies a_{n + 2} = 0 implies a_3 = a_5 = ... = 0$. However, I am meant to prove the recurrence relation:
$$a_{n+2} = frac{(n+1)(n+2) - l(l+1)}{(n+2)(n+3)}a_n$$
Which I don't recognize. This here highlights my confusions with this method:
- Why is my first recurrence relation not going to encompass both the even and odd parts if $a_{n+2} ne 0 $ for starting at both $a_1$ and $a_0$?
- Is there some non-arbitrary need that if $c=0 implies a_1 = 0$?
- If my answer may only contain odd or even powers of $x$ but $a_{n+2} ne 0 $ for starting at both $a_1$ and $a_0$, then how do I know if I only have an odd or even power of $x$ for my recurrence relation?
- Generally, what are the parametrisations of $c$ such that the solution will only encompass odd, even, or both powers of $x$?
legendre-polynomials frobenius-method
legendre-polynomials frobenius-method
asked Dec 13 '18 at 3:22
sangstarsangstar
853215
asked Dec 13 '18 at 3:22
sangstarsangstar
853215
edited Dec 13 '18 at 15:16
sangstar
asked Dec 13 '18 at 3:22
sangstarsangstar
853215
asked Dec 13 '18 at 3:22
sangstarsangstar
853215
asked Dec 13 '18 at 3:22
sangstarsangstar
853215
853215
add a comment |
add a comment |
1 Answer
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Details can be found in my answer to another question related to Legendre equation:
about-the-legendre-differential-equation
answered Dec 24 '18 at 14:58
Maestro13Maestro13
1,081724
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Details can be found in my answer to another question related to Legendre equation:
about-the-legendre-differential-equation
answered Dec 24 '18 at 14:58
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
$begingroup$
Details can be found in my answer to another question related to Legendre equation:
about-the-legendre-differential-equation
answered Dec 24 '18 at 14:58
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
$begingroup$
Details can be found in my answer to another question related to Legendre equation:
about-the-legendre-differential-equation
answered Dec 24 '18 at 14:58
Maestro13Maestro13
1,081724
$endgroup$
Details can be found in my answer to another question related to Legendre equation:
about-the-legendre-differential-equation
answered Dec 24 '18 at 14:58
Maestro13Maestro13
1,081724
answered Dec 24 '18 at 14:58
Maestro13Maestro13
1,081724
answered Dec 24 '18 at 14:58
Maestro13Maestro13
1,081724
answered Dec 24 '18 at 14:58
Maestro13Maestro13
1,081724
1,081724
add a comment |
add a comment |
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