Mixing
integral of sum of function involving legendre polynomials
$begingroup$
Consider the integral $$int_{-1}^1left(sum_{j=1}^nsqrt{j(2j+1)}P_j(x)right)^2dx$$
How to evaluate the integral? Specifically, say, if $n=5$, then what would be the value? Which property of Legendre polynomials should I use here? Should I use completeness of Legendre polynomials here?
real-analysis calculus integration legendre-polynomials
edited Jan 31 at 10:38
YuiTo Cheng
2,3084937
asked Jan 31 at 10:23
vidyarthividyarthi
3,0801833
$endgroup$
add a comment |
$begingroup$
Consider the integral $$int_{-1}^1left(sum_{j=1}^nsqrt{j(2j+1)}P_j(x)right)^2dx$$
How to evaluate the integral? Specifically, say, if $n=5$, then what would be the value? Which property of Legendre polynomials should I use here? Should I use completeness of Legendre polynomials here?
real-analysis calculus integration legendre-polynomials
edited Jan 31 at 10:38
YuiTo Cheng
2,3084937
asked Jan 31 at 10:23
vidyarthividyarthi
3,0801833
$endgroup$
$begingroup$
Write the square of the sum over $j$ as a double sum over $j, k$, then use the orthonormality property of the Legendre polynomials to simplify the expression considerably.
$endgroup$
– Christoph
Jan 31 at 10:34
$begingroup$
@Christoph you mean write the subscript of $P$ as $k$?
$endgroup$
– vidyarthi
Jan 31 at 10:37
1
$begingroup$
Yes, using $left(sum_{j=1}^n a_jright)^2 = sum_{j,k=1}^n a_j a_k$.
$endgroup$
– Christoph
Jan 31 at 10:46
add a comment |
$begingroup$
Consider the integral $$int_{-1}^1left(sum_{j=1}^nsqrt{j(2j+1)}P_j(x)right)^2dx$$
How to evaluate the integral? Specifically, say, if $n=5$, then what would be the value? Which property of Legendre polynomials should I use here? Should I use completeness of Legendre polynomials here?
real-analysis calculus integration legendre-polynomials
edited Jan 31 at 10:38
YuiTo Cheng
2,3084937
asked Jan 31 at 10:23
vidyarthividyarthi
3,0801833
$endgroup$
Consider the integral $$int_{-1}^1left(sum_{j=1}^nsqrt{j(2j+1)}P_j(x)right)^2dx$$
How to evaluate the integral? Specifically, say, if $n=5$, then what would be the value? Which property of Legendre polynomials should I use here? Should I use completeness of Legendre polynomials here?
real-analysis calculus integration legendre-polynomials
real-analysis calculus integration legendre-polynomials
edited Jan 31 at 10:38
YuiTo Cheng
2,3084937
asked Jan 31 at 10:23
vidyarthividyarthi
3,0801833
edited Jan 31 at 10:38
YuiTo Cheng
2,3084937
asked Jan 31 at 10:23
vidyarthividyarthi
3,0801833
edited Jan 31 at 10:38
YuiTo Cheng
2,3084937
edited Jan 31 at 10:38
YuiTo Cheng
2,3084937
edited Jan 31 at 10:38
YuiTo Cheng
2,3084937
2,3084937
asked Jan 31 at 10:23
vidyarthividyarthi
3,0801833
asked Jan 31 at 10:23
vidyarthividyarthi
3,0801833
asked Jan 31 at 10:23
vidyarthividyarthi
3,0801833
3,0801833
$begingroup$
Write the square of the sum over $j$ as a double sum over $j, k$, then use the orthonormality property of the Legendre polynomials to simplify the expression considerably.
$endgroup$
– Christoph
Jan 31 at 10:34
$begingroup$
@Christoph you mean write the subscript of $P$ as $k$?
$endgroup$
– vidyarthi
Jan 31 at 10:37
1
$begingroup$
Yes, using $left(sum_{j=1}^n a_jright)^2 = sum_{j,k=1}^n a_j a_k$.
$endgroup$
– Christoph
Jan 31 at 10:46
add a comment |
$begingroup$
Write the square of the sum over $j$ as a double sum over $j, k$, then use the orthonormality property of the Legendre polynomials to simplify the expression considerably.
$endgroup$
– Christoph
Jan 31 at 10:34
$begingroup$
@Christoph you mean write the subscript of $P$ as $k$?
$endgroup$
– vidyarthi
Jan 31 at 10:37
1
$begingroup$
Yes, using $left(sum_{j=1}^n a_jright)^2 = sum_{j,k=1}^n a_j a_k$.
$endgroup$
– Christoph
Jan 31 at 10:46
$begingroup$
Write the square of the sum over $j$ as a double sum over $j, k$, then use the orthonormality property of the Legendre polynomials to simplify the expression considerably.
$endgroup$
– Christoph
Jan 31 at 10:34
$begingroup$
Write the square of the sum over $j$ as a double sum over $j, k$, then use the orthonormality property of the Legendre polynomials to simplify the expression considerably.
$endgroup$
– Christoph
Jan 31 at 10:34
$begingroup$
@Christoph you mean write the subscript of $P$ as $k$?
$endgroup$
– vidyarthi
Jan 31 at 10:37
$begingroup$
@Christoph you mean write the subscript of $P$ as $k$?
$endgroup$
– vidyarthi
Jan 31 at 10:37
1
1
$begingroup$
Yes, using $left(sum_{j=1}^n a_jright)^2 = sum_{j,k=1}^n a_j a_k$.
$endgroup$
– Christoph
Jan 31 at 10:46
$begingroup$
Yes, using $left(sum_{j=1}^n a_jright)^2 = sum_{j,k=1}^n a_j a_k$.
$endgroup$
– Christoph
Jan 31 at 10:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As pointed out in the comments by @Christoph, the answer is simple owing to the orthogonality of Legendre polynomials. We have:
$$left(sum_{j=1}^nsqrt{j(2j+1)}P_j(x)right)^2=sum_{j,k=1}^nsqrt{j(2j+1)}P_j(x)sum_{k=1}^nsqrt{k(2k+1)}P_k(x)$$
Now, On integration, owing to orthogonality and the result $int_{-1}^1P_n^2dx=frac{2}{2n+1}$, we obtain,
the required integral equal to $$2sum_{j=1}^nj=n(n+1)$$
answered Jan 31 at 11:47
vidyarthividyarthi
3,0801833
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
As pointed out in the comments by @Christoph, the answer is simple owing to the orthogonality of Legendre polynomials. We have:
$$left(sum_{j=1}^nsqrt{j(2j+1)}P_j(x)right)^2=sum_{j,k=1}^nsqrt{j(2j+1)}P_j(x)sum_{k=1}^nsqrt{k(2k+1)}P_k(x)$$
Now, On integration, owing to orthogonality and the result $int_{-1}^1P_n^2dx=frac{2}{2n+1}$, we obtain,
the required integral equal to $$2sum_{j=1}^nj=n(n+1)$$
answered Jan 31 at 11:47
vidyarthividyarthi
3,0801833
$endgroup$
add a comment |
$begingroup$
As pointed out in the comments by @Christoph, the answer is simple owing to the orthogonality of Legendre polynomials. We have:
$$left(sum_{j=1}^nsqrt{j(2j+1)}P_j(x)right)^2=sum_{j,k=1}^nsqrt{j(2j+1)}P_j(x)sum_{k=1}^nsqrt{k(2k+1)}P_k(x)$$
Now, On integration, owing to orthogonality and the result $int_{-1}^1P_n^2dx=frac{2}{2n+1}$, we obtain,
the required integral equal to $$2sum_{j=1}^nj=n(n+1)$$
answered Jan 31 at 11:47
vidyarthividyarthi
3,0801833
$endgroup$
add a comment |
$begingroup$
As pointed out in the comments by @Christoph, the answer is simple owing to the orthogonality of Legendre polynomials. We have:
$$left(sum_{j=1}^nsqrt{j(2j+1)}P_j(x)right)^2=sum_{j,k=1}^nsqrt{j(2j+1)}P_j(x)sum_{k=1}^nsqrt{k(2k+1)}P_k(x)$$
Now, On integration, owing to orthogonality and the result $int_{-1}^1P_n^2dx=frac{2}{2n+1}$, we obtain,
the required integral equal to $$2sum_{j=1}^nj=n(n+1)$$
answered Jan 31 at 11:47
vidyarthividyarthi
3,0801833
$endgroup$
As pointed out in the comments by @Christoph, the answer is simple owing to the orthogonality of Legendre polynomials. We have:
$$left(sum_{j=1}^nsqrt{j(2j+1)}P_j(x)right)^2=sum_{j,k=1}^nsqrt{j(2j+1)}P_j(x)sum_{k=1}^nsqrt{k(2k+1)}P_k(x)$$
Now, On integration, owing to orthogonality and the result $int_{-1}^1P_n^2dx=frac{2}{2n+1}$, we obtain,
the required integral equal to $$2sum_{j=1}^nj=n(n+1)$$
answered Jan 31 at 11:47
vidyarthividyarthi
3,0801833
answered Jan 31 at 11:47
vidyarthividyarthi
3,0801833
answered Jan 31 at 11:47
vidyarthividyarthi
3,0801833
answered Jan 31 at 11:47
vidyarthividyarthi
3,0801833
3,0801833
add a comment |
add a comment |
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$begingroup$
Write the square of the sum over $j$ as a double sum over $j, k$, then use the orthonormality property of the Legendre polynomials to simplify the expression considerably.
$endgroup$
– Christoph
Jan 31 at 10:34
$begingroup$
@Christoph you mean write the subscript of $P$ as $k$?
$endgroup$
– vidyarthi
Jan 31 at 10:37
1
$begingroup$
Yes, using $left(sum_{j=1}^n a_jright)^2 = sum_{j,k=1}^n a_j a_k$.
$endgroup$
– Christoph
Jan 31 at 10:46