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Legendre Series Recurrence Relation Divergence at $x=pm1$, using Gauss test
$begingroup$
How to show that the Legendre Series solution $y_{even}$ and $y_{odd}$, diverges as $x = pm1 $.
$y_{even} = sum_{j=0,2,ldots}^infty a_jx^j$, where $a_{j+2}=frac{j(j+1)-n(n+1)}{(j+1)(j+2)}a_j$.
$y_{odd}=sum_{j=1,3,ldots}^infty a_jx^j$, where $a_{j+2}=frac{(j+1)(j+2)-n(n+1)}{(j+2)(j+3)}a_j$.
My Work:
I have shown that for both the $y_{even}$, and the $y_{odd}$, solutions, that $lim |frac{a_{j+1}}{a_j}|$, as $j rightarrow infty$ is $1$.
I cannot find the next part? How to proceed?
P.S. This problem is from Arfken. Problem #8.3.1 (7th Ed.)
Update:
I have been thinking about this for a while, and I have come up with this. I am using the Gauss test.
I am taking the $y_{even}$ solution, and defining $alpha = left|frac{a_{2j}}{a_{2j+2}}right| = frac{(2j+1)(2j+2)}{(2j-n)(2j+n-1)}$.
Now after some manipulation, I get my $alpha$ to be,
$$alpha=1+frac1j+frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$$
Noe according to the Gauss test, one has to check that my third term, viz., $frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$, which, I should be able to manipulate to $frac{C_j}{j^r}$; for $lim_{jtoinfty}C_j$, should be finite, then only I can say that, the series does diverge for $xrightarrowpm1$.
Okay, thought for some more time, and I have a solution to this problem. Now I can manipulate the third term in $alpha$, which would lead me to,
$$frac{n(n+1)(1+frac1j)}{j^2 left(4+frac2j-frac{n(n+1)}{j^2} right) }$$
which I can identify straight as my $frac{c_j}{j^r}$ term in my Gauss test. So here $r=2$, and $c_j=frac{n(n+1)(1+frac1j)}{left(4+frac2j-frac{n(n+1)}{j^2} right) }$. And when I do take the $lim_{jtoinfty}c_j=frac{n(n+1)}4$, so it does remain bounded.
So now I can draw a conclusion about the convergence/divergence at $xrightarrowpm1$ for the Legendre ODE. Now my $h=1$, and $r(equiv2)>1$, and also $lim_{jtoinfty}=frac{n(n+1)}4$, so I can say with confirmation, that at $x=pm1$, the series diverges!
This completes the proof, I guess.
ordinary-differential-equations recurrence-relations
asked Jan 13 '15 at 5:16
sbpsbp
22119
$endgroup$
|
show 1 more comment
$begingroup$
How to show that the Legendre Series solution $y_{even}$ and $y_{odd}$, diverges as $x = pm1 $.
$y_{even} = sum_{j=0,2,ldots}^infty a_jx^j$, where $a_{j+2}=frac{j(j+1)-n(n+1)}{(j+1)(j+2)}a_j$.
$y_{odd}=sum_{j=1,3,ldots}^infty a_jx^j$, where $a_{j+2}=frac{(j+1)(j+2)-n(n+1)}{(j+2)(j+3)}a_j$.
My Work:
I have shown that for both the $y_{even}$, and the $y_{odd}$, solutions, that $lim |frac{a_{j+1}}{a_j}|$, as $j rightarrow infty$ is $1$.
I cannot find the next part? How to proceed?
P.S. This problem is from Arfken. Problem #8.3.1 (7th Ed.)
Update:
I have been thinking about this for a while, and I have come up with this. I am using the Gauss test.
I am taking the $y_{even}$ solution, and defining $alpha = left|frac{a_{2j}}{a_{2j+2}}right| = frac{(2j+1)(2j+2)}{(2j-n)(2j+n-1)}$.
Now after some manipulation, I get my $alpha$ to be,
$$alpha=1+frac1j+frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$$
Noe according to the Gauss test, one has to check that my third term, viz., $frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$, which, I should be able to manipulate to $frac{C_j}{j^r}$; for $lim_{jtoinfty}C_j$, should be finite, then only I can say that, the series does diverge for $xrightarrowpm1$.
Okay, thought for some more time, and I have a solution to this problem. Now I can manipulate the third term in $alpha$, which would lead me to,
$$frac{n(n+1)(1+frac1j)}{j^2 left(4+frac2j-frac{n(n+1)}{j^2} right) }$$
which I can identify straight as my $frac{c_j}{j^r}$ term in my Gauss test. So here $r=2$, and $c_j=frac{n(n+1)(1+frac1j)}{left(4+frac2j-frac{n(n+1)}{j^2} right) }$. And when I do take the $lim_{jtoinfty}c_j=frac{n(n+1)}4$, so it does remain bounded.
So now I can draw a conclusion about the convergence/divergence at $xrightarrowpm1$ for the Legendre ODE. Now my $h=1$, and $r(equiv2)>1$, and also $lim_{jtoinfty}=frac{n(n+1)}4$, so I can say with confirmation, that at $x=pm1$, the series diverges!
This completes the proof, I guess.
ordinary-differential-equations recurrence-relations
asked Jan 13 '15 at 5:16
sbpsbp
22119
$endgroup$
$begingroup$
How are those $;n;$ in the recursive formula for the coefficients related to $;j;$ ?
$endgroup$
– Timbuc
Jan 13 '15 at 11:36
$begingroup$
@Timbuc: Its nothing but the $n(n+1)$ term in the Legendre Differential Equation.
$endgroup$
– sbp
Jan 13 '15 at 11:46
$begingroup$
still I can't see how $;n;$ is related to $;j;$ : is it the same $;n;$ for all $;j$'s orwhat?
$endgroup$
– Timbuc
Jan 13 '15 at 11:52
$begingroup$
Check the Legendre DE; $n$, is some parameter, say $1$. You might remember $P_n(x)$, as the Legendre polynomials!
$endgroup$
– sbp
Jan 13 '15 at 12:02
$begingroup$
So $;n;$ is the same for each and every $;j;$ , right? I'm not acquainted with Legendre's DE.
$endgroup$
– Timbuc
Jan 13 '15 at 12:29
|
show 1 more comment
$begingroup$
How to show that the Legendre Series solution $y_{even}$ and $y_{odd}$, diverges as $x = pm1 $.
$y_{even} = sum_{j=0,2,ldots}^infty a_jx^j$, where $a_{j+2}=frac{j(j+1)-n(n+1)}{(j+1)(j+2)}a_j$.
$y_{odd}=sum_{j=1,3,ldots}^infty a_jx^j$, where $a_{j+2}=frac{(j+1)(j+2)-n(n+1)}{(j+2)(j+3)}a_j$.
My Work:
I have shown that for both the $y_{even}$, and the $y_{odd}$, solutions, that $lim |frac{a_{j+1}}{a_j}|$, as $j rightarrow infty$ is $1$.
I cannot find the next part? How to proceed?
P.S. This problem is from Arfken. Problem #8.3.1 (7th Ed.)
Update:
I have been thinking about this for a while, and I have come up with this. I am using the Gauss test.
I am taking the $y_{even}$ solution, and defining $alpha = left|frac{a_{2j}}{a_{2j+2}}right| = frac{(2j+1)(2j+2)}{(2j-n)(2j+n-1)}$.
Now after some manipulation, I get my $alpha$ to be,
$$alpha=1+frac1j+frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$$
Noe according to the Gauss test, one has to check that my third term, viz., $frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$, which, I should be able to manipulate to $frac{C_j}{j^r}$; for $lim_{jtoinfty}C_j$, should be finite, then only I can say that, the series does diverge for $xrightarrowpm1$.
Okay, thought for some more time, and I have a solution to this problem. Now I can manipulate the third term in $alpha$, which would lead me to,
$$frac{n(n+1)(1+frac1j)}{j^2 left(4+frac2j-frac{n(n+1)}{j^2} right) }$$
which I can identify straight as my $frac{c_j}{j^r}$ term in my Gauss test. So here $r=2$, and $c_j=frac{n(n+1)(1+frac1j)}{left(4+frac2j-frac{n(n+1)}{j^2} right) }$. And when I do take the $lim_{jtoinfty}c_j=frac{n(n+1)}4$, so it does remain bounded.
So now I can draw a conclusion about the convergence/divergence at $xrightarrowpm1$ for the Legendre ODE. Now my $h=1$, and $r(equiv2)>1$, and also $lim_{jtoinfty}=frac{n(n+1)}4$, so I can say with confirmation, that at $x=pm1$, the series diverges!
This completes the proof, I guess.
ordinary-differential-equations recurrence-relations
asked Jan 13 '15 at 5:16
sbpsbp
22119
$endgroup$
How to show that the Legendre Series solution $y_{even}$ and $y_{odd}$, diverges as $x = pm1 $.
$y_{even} = sum_{j=0,2,ldots}^infty a_jx^j$, where $a_{j+2}=frac{j(j+1)-n(n+1)}{(j+1)(j+2)}a_j$.
$y_{odd}=sum_{j=1,3,ldots}^infty a_jx^j$, where $a_{j+2}=frac{(j+1)(j+2)-n(n+1)}{(j+2)(j+3)}a_j$.
My Work:
I have shown that for both the $y_{even}$, and the $y_{odd}$, solutions, that $lim |frac{a_{j+1}}{a_j}|$, as $j rightarrow infty$ is $1$.
I cannot find the next part? How to proceed?
P.S. This problem is from Arfken. Problem #8.3.1 (7th Ed.)
Update:
I have been thinking about this for a while, and I have come up with this. I am using the Gauss test.
I am taking the $y_{even}$ solution, and defining $alpha = left|frac{a_{2j}}{a_{2j+2}}right| = frac{(2j+1)(2j+2)}{(2j-n)(2j+n-1)}$.
Now after some manipulation, I get my $alpha$ to be,
$$alpha=1+frac1j+frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$$
Noe according to the Gauss test, one has to check that my third term, viz., $frac{n(n+1)(j+1)}{(4j^2+2j-n(n+1))j}$, which, I should be able to manipulate to $frac{C_j}{j^r}$; for $lim_{jtoinfty}C_j$, should be finite, then only I can say that, the series does diverge for $xrightarrowpm1$.
Okay, thought for some more time, and I have a solution to this problem. Now I can manipulate the third term in $alpha$, which would lead me to,
$$frac{n(n+1)(1+frac1j)}{j^2 left(4+frac2j-frac{n(n+1)}{j^2} right) }$$
which I can identify straight as my $frac{c_j}{j^r}$ term in my Gauss test. So here $r=2$, and $c_j=frac{n(n+1)(1+frac1j)}{left(4+frac2j-frac{n(n+1)}{j^2} right) }$. And when I do take the $lim_{jtoinfty}c_j=frac{n(n+1)}4$, so it does remain bounded.
So now I can draw a conclusion about the convergence/divergence at $xrightarrowpm1$ for the Legendre ODE. Now my $h=1$, and $r(equiv2)>1$, and also $lim_{jtoinfty}=frac{n(n+1)}4$, so I can say with confirmation, that at $x=pm1$, the series diverges!
This completes the proof, I guess.
ordinary-differential-equations recurrence-relations
ordinary-differential-equations recurrence-relations
asked Jan 13 '15 at 5:16
sbpsbp
22119
asked Jan 13 '15 at 5:16
sbpsbp
22119
edited Jan 14 '15 at 19:51
sbp
asked Jan 13 '15 at 5:16
sbpsbp
22119
asked Jan 13 '15 at 5:16
sbpsbp
22119
asked Jan 13 '15 at 5:16
sbpsbp
22119
22119
$begingroup$
How are those $;n;$ in the recursive formula for the coefficients related to $;j;$ ?
$endgroup$
– Timbuc
Jan 13 '15 at 11:36
$begingroup$
@Timbuc: Its nothing but the $n(n+1)$ term in the Legendre Differential Equation.
$endgroup$
– sbp
Jan 13 '15 at 11:46
$begingroup$
still I can't see how $;n;$ is related to $;j;$ : is it the same $;n;$ for all $;j$'s orwhat?
$endgroup$
– Timbuc
Jan 13 '15 at 11:52
$begingroup$
Check the Legendre DE; $n$, is some parameter, say $1$. You might remember $P_n(x)$, as the Legendre polynomials!
$endgroup$
– sbp
Jan 13 '15 at 12:02
$begingroup$
So $;n;$ is the same for each and every $;j;$ , right? I'm not acquainted with Legendre's DE.
$endgroup$
– Timbuc
Jan 13 '15 at 12:29
|
show 1 more comment
$begingroup$
How are those $;n;$ in the recursive formula for the coefficients related to $;j;$ ?
$endgroup$
– Timbuc
Jan 13 '15 at 11:36
$begingroup$
@Timbuc: Its nothing but the $n(n+1)$ term in the Legendre Differential Equation.
$endgroup$
– sbp
Jan 13 '15 at 11:46
$begingroup$
still I can't see how $;n;$ is related to $;j;$ : is it the same $;n;$ for all $;j$'s orwhat?
$endgroup$
– Timbuc
Jan 13 '15 at 11:52
$begingroup$
Check the Legendre DE; $n$, is some parameter, say $1$. You might remember $P_n(x)$, as the Legendre polynomials!
$endgroup$
– sbp
Jan 13 '15 at 12:02
$begingroup$
So $;n;$ is the same for each and every $;j;$ , right? I'm not acquainted with Legendre's DE.
$endgroup$
– Timbuc
Jan 13 '15 at 12:29
$begingroup$
How are those $;n;$ in the recursive formula for the coefficients related to $;j;$ ?
$endgroup$
– Timbuc
Jan 13 '15 at 11:36
$begingroup$
How are those $;n;$ in the recursive formula for the coefficients related to $;j;$ ?
$endgroup$
– Timbuc
Jan 13 '15 at 11:36
$begingroup$
@Timbuc: Its nothing but the $n(n+1)$ term in the Legendre Differential Equation.
$endgroup$
– sbp
Jan 13 '15 at 11:46
$begingroup$
@Timbuc: Its nothing but the $n(n+1)$ term in the Legendre Differential Equation.
$endgroup$
– sbp
Jan 13 '15 at 11:46
$begingroup$
still I can't see how $;n;$ is related to $;j;$ : is it the same $;n;$ for all $;j$'s orwhat?
$endgroup$
– Timbuc
Jan 13 '15 at 11:52
$begingroup$
still I can't see how $;n;$ is related to $;j;$ : is it the same $;n;$ for all $;j$'s orwhat?
$endgroup$
– Timbuc
Jan 13 '15 at 11:52
$begingroup$
Check the Legendre DE; $n$, is some parameter, say $1$. You might remember $P_n(x)$, as the Legendre polynomials!
$endgroup$
– sbp
Jan 13 '15 at 12:02
$begingroup$
Check the Legendre DE; $n$, is some parameter, say $1$. You might remember $P_n(x)$, as the Legendre polynomials!
$endgroup$
– sbp
Jan 13 '15 at 12:02
$begingroup$
So $;n;$ is the same for each and every $;j;$ , right? I'm not acquainted with Legendre's DE.
$endgroup$
– Timbuc
Jan 13 '15 at 12:29
$begingroup$
So $;n;$ is the same for each and every $;j;$ , right? I'm not acquainted with Legendre's DE.
$endgroup$
– Timbuc
Jan 13 '15 at 12:29
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
See About the Legendre differential equation where the recurrence relations above are derived from entering $y=sum_{j=0}^infty a_jx^j$ into the Legendre differential equation.
Solutions to the equation are $y=P_n(x)$ which is a polynomial of degree $n$, and $y=Q_n(x)$, which is constructed of polynomials and logarithms.
$P_n(1)=1$ and $P_n(-1)=pm 1$ so in these cases the series does not diverge.
But indeed the $Q_n(x)$ series diverges for $x=pm 1$.
For example: $Q_0(x)=frac{1}{2} ln(frac{1+x}{1-x})$
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 17 '15 at 20:27
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
See About the Legendre differential equation where the recurrence relations above are derived from entering $y=sum_{j=0}^infty a_jx^j$ into the Legendre differential equation.
Solutions to the equation are $y=P_n(x)$ which is a polynomial of degree $n$, and $y=Q_n(x)$, which is constructed of polynomials and logarithms.
$P_n(1)=1$ and $P_n(-1)=pm 1$ so in these cases the series does not diverge.
But indeed the $Q_n(x)$ series diverges for $x=pm 1$.
For example: $Q_0(x)=frac{1}{2} ln(frac{1+x}{1-x})$
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 17 '15 at 20:27
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
$begingroup$
See About the Legendre differential equation where the recurrence relations above are derived from entering $y=sum_{j=0}^infty a_jx^j$ into the Legendre differential equation.
Solutions to the equation are $y=P_n(x)$ which is a polynomial of degree $n$, and $y=Q_n(x)$, which is constructed of polynomials and logarithms.
$P_n(1)=1$ and $P_n(-1)=pm 1$ so in these cases the series does not diverge.
But indeed the $Q_n(x)$ series diverges for $x=pm 1$.
For example: $Q_0(x)=frac{1}{2} ln(frac{1+x}{1-x})$
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 17 '15 at 20:27
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
$begingroup$
See About the Legendre differential equation where the recurrence relations above are derived from entering $y=sum_{j=0}^infty a_jx^j$ into the Legendre differential equation.
Solutions to the equation are $y=P_n(x)$ which is a polynomial of degree $n$, and $y=Q_n(x)$, which is constructed of polynomials and logarithms.
$P_n(1)=1$ and $P_n(-1)=pm 1$ so in these cases the series does not diverge.
But indeed the $Q_n(x)$ series diverges for $x=pm 1$.
For example: $Q_0(x)=frac{1}{2} ln(frac{1+x}{1-x})$
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 17 '15 at 20:27
Maestro13Maestro13
1,081724
$endgroup$
See About the Legendre differential equation where the recurrence relations above are derived from entering $y=sum_{j=0}^infty a_jx^j$ into the Legendre differential equation.
Solutions to the equation are $y=P_n(x)$ which is a polynomial of degree $n$, and $y=Q_n(x)$, which is constructed of polynomials and logarithms.
$P_n(1)=1$ and $P_n(-1)=pm 1$ so in these cases the series does not diverge.
But indeed the $Q_n(x)$ series diverges for $x=pm 1$.
For example: $Q_0(x)=frac{1}{2} ln(frac{1+x}{1-x})$
edited Apr 13 '17 at 12:21
Community♦
1
answered Jan 17 '15 at 20:27
Maestro13Maestro13
1,081724
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Jan 17 '15 at 20:27
Maestro13Maestro13
1,081724
answered Jan 17 '15 at 20:27
Maestro13Maestro13
1,081724
answered Jan 17 '15 at 20:27
Maestro13Maestro13
1,081724
1,081724
add a comment |
add a comment |
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$begingroup$
How are those $;n;$ in the recursive formula for the coefficients related to $;j;$ ?
$endgroup$
– Timbuc
Jan 13 '15 at 11:36
$begingroup$
@Timbuc: Its nothing but the $n(n+1)$ term in the Legendre Differential Equation.
$endgroup$
– sbp
Jan 13 '15 at 11:46
$begingroup$
still I can't see how $;n;$ is related to $;j;$ : is it the same $;n;$ for all $;j$'s orwhat?
$endgroup$
– Timbuc
Jan 13 '15 at 11:52
$begingroup$
Check the Legendre DE; $n$, is some parameter, say $1$. You might remember $P_n(x)$, as the Legendre polynomials!
$endgroup$
– sbp
Jan 13 '15 at 12:02
$begingroup$
So $;n;$ is the same for each and every $;j;$ , right? I'm not acquainted with Legendre's DE.
$endgroup$
– Timbuc
Jan 13 '15 at 12:29