Mixing
Why does $int_{-1}^1 ((1-x^2)(P_m'P_n-P_n'P_m))',dx = 0$ for Legendre polynomials?
$begingroup$
I was looking at a proof of the orthogonality of the Legendre polynomials in Lebedev's Special Functions and their Applications:
I can't understand why the integral of the first term vanishes. I appreciate any help.
real-analysis functional-analysis legendre-polynomials
edited Jan 19 at 22:15
J.G.
28.4k22845
asked Jan 19 at 20:50
Luke CollinsLuke Collins
749419
$endgroup$
|
show 1 more comment
$begingroup$
I was looking at a proof of the orthogonality of the Legendre polynomials in Lebedev's Special Functions and their Applications:
I can't understand why the integral of the first term vanishes. I appreciate any help.
real-analysis functional-analysis legendre-polynomials
edited Jan 19 at 22:15
J.G.
28.4k22845
asked Jan 19 at 20:50
Luke CollinsLuke Collins
749419
$endgroup$
1
$begingroup$
Start by finding the antiderivative of the first term (this should easy since the entire term is given as a derivative). Then evaluate the result at -1 and 1 (evaluating the Legendre polynomials exactly will turnout to be less important for the problem).
$endgroup$
– Testcase
Jan 19 at 21:14
$begingroup$
It follows from $int_{-1}^{1} F'(x) dx = F(1) - F(-1)$ for any $C^{1}$ function $F$.
$endgroup$
– Seewoo Lee
Jan 19 at 21:17
$begingroup$
There's a typo in your title that would make this trivial.
$endgroup$
– J.G.
Jan 19 at 21:22
$begingroup$
I was ignoring the outer prime for some reason. Must be tired!
$endgroup$
– Luke Collins
Jan 19 at 22:13
$begingroup$
@J.G. It is trival!
$endgroup$
– Luke Collins
Jan 19 at 22:14
|
show 1 more comment
$begingroup$
I was looking at a proof of the orthogonality of the Legendre polynomials in Lebedev's Special Functions and their Applications:
I can't understand why the integral of the first term vanishes. I appreciate any help.
real-analysis functional-analysis legendre-polynomials
edited Jan 19 at 22:15
J.G.
28.4k22845
asked Jan 19 at 20:50
Luke CollinsLuke Collins
749419
$endgroup$
I was looking at a proof of the orthogonality of the Legendre polynomials in Lebedev's Special Functions and their Applications:
I can't understand why the integral of the first term vanishes. I appreciate any help.
real-analysis functional-analysis legendre-polynomials
real-analysis functional-analysis legendre-polynomials
edited Jan 19 at 22:15
J.G.
28.4k22845
asked Jan 19 at 20:50
Luke CollinsLuke Collins
749419
edited Jan 19 at 22:15
J.G.
28.4k22845
asked Jan 19 at 20:50
Luke CollinsLuke Collins
749419
edited Jan 19 at 22:15
J.G.
28.4k22845
edited Jan 19 at 22:15
J.G.
28.4k22845
edited Jan 19 at 22:15
J.G.
28.4k22845
28.4k22845
asked Jan 19 at 20:50
Luke CollinsLuke Collins
749419
asked Jan 19 at 20:50
Luke CollinsLuke Collins
749419
asked Jan 19 at 20:50
Luke CollinsLuke Collins
749419
749419
1
$begingroup$
Start by finding the antiderivative of the first term (this should easy since the entire term is given as a derivative). Then evaluate the result at -1 and 1 (evaluating the Legendre polynomials exactly will turnout to be less important for the problem).
$endgroup$
– Testcase
Jan 19 at 21:14
$begingroup$
It follows from $int_{-1}^{1} F'(x) dx = F(1) - F(-1)$ for any $C^{1}$ function $F$.
$endgroup$
– Seewoo Lee
Jan 19 at 21:17
$begingroup$
There's a typo in your title that would make this trivial.
$endgroup$
– J.G.
Jan 19 at 21:22
$begingroup$
I was ignoring the outer prime for some reason. Must be tired!
$endgroup$
– Luke Collins
Jan 19 at 22:13
$begingroup$
@J.G. It is trival!
$endgroup$
– Luke Collins
Jan 19 at 22:14
|
show 1 more comment
1
$begingroup$
Start by finding the antiderivative of the first term (this should easy since the entire term is given as a derivative). Then evaluate the result at -1 and 1 (evaluating the Legendre polynomials exactly will turnout to be less important for the problem).
$endgroup$
– Testcase
Jan 19 at 21:14
$begingroup$
It follows from $int_{-1}^{1} F'(x) dx = F(1) - F(-1)$ for any $C^{1}$ function $F$.
$endgroup$
– Seewoo Lee
Jan 19 at 21:17
$begingroup$
There's a typo in your title that would make this trivial.
$endgroup$
– J.G.
Jan 19 at 21:22
$begingroup$
I was ignoring the outer prime for some reason. Must be tired!
$endgroup$
– Luke Collins
Jan 19 at 22:13
$begingroup$
@J.G. It is trival!
$endgroup$
– Luke Collins
Jan 19 at 22:14
1
1
$begingroup$
Start by finding the antiderivative of the first term (this should easy since the entire term is given as a derivative). Then evaluate the result at -1 and 1 (evaluating the Legendre polynomials exactly will turnout to be less important for the problem).
$endgroup$
– Testcase
Jan 19 at 21:14
$begingroup$
Start by finding the antiderivative of the first term (this should easy since the entire term is given as a derivative). Then evaluate the result at -1 and 1 (evaluating the Legendre polynomials exactly will turnout to be less important for the problem).
$endgroup$
– Testcase
Jan 19 at 21:14
$begingroup$
It follows from $int_{-1}^{1} F'(x) dx = F(1) - F(-1)$ for any $C^{1}$ function $F$.
$endgroup$
– Seewoo Lee
Jan 19 at 21:17
$begingroup$
It follows from $int_{-1}^{1} F'(x) dx = F(1) - F(-1)$ for any $C^{1}$ function $F$.
$endgroup$
– Seewoo Lee
Jan 19 at 21:17
$begingroup$
There's a typo in your title that would make this trivial.
$endgroup$
– J.G.
Jan 19 at 21:22
$begingroup$
There's a typo in your title that would make this trivial.
$endgroup$
– J.G.
Jan 19 at 21:22
$begingroup$
I was ignoring the outer prime for some reason. Must be tired!
$endgroup$
– Luke Collins
Jan 19 at 22:13
$begingroup$
I was ignoring the outer prime for some reason. Must be tired!
$endgroup$
– Luke Collins
Jan 19 at 22:13
$begingroup$
@J.G. It is trival!
$endgroup$
– Luke Collins
Jan 19 at 22:14
$begingroup$
@J.G. It is trival!
$endgroup$
– Luke Collins
Jan 19 at 22:14
|
show 1 more comment
1 Answer
1
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$begingroup$
$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$
answered Jan 19 at 21:18
Marvin CohenMarvin Cohen
146117
$endgroup$
$begingroup$
I can't believe I missed that. I was ignoring the outer prime. Oh well, time for bed.
$endgroup$
– Luke Collins
Jan 19 at 22:13
add a comment |
Your Answer
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$begingroup$
$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$
answered Jan 19 at 21:18
Marvin CohenMarvin Cohen
146117
$endgroup$
$begingroup$
I can't believe I missed that. I was ignoring the outer prime. Oh well, time for bed.
$endgroup$
– Luke Collins
Jan 19 at 22:13
add a comment |
$begingroup$
$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$
answered Jan 19 at 21:18
Marvin CohenMarvin Cohen
146117
$endgroup$
$begingroup$
I can't believe I missed that. I was ignoring the outer prime. Oh well, time for bed.
$endgroup$
– Luke Collins
Jan 19 at 22:13
add a comment |
$begingroup$
$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$
answered Jan 19 at 21:18
Marvin CohenMarvin Cohen
146117
$endgroup$
$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$
answered Jan 19 at 21:18
Marvin CohenMarvin Cohen
146117
answered Jan 19 at 21:18
Marvin CohenMarvin Cohen
146117
answered Jan 19 at 21:18
Marvin CohenMarvin Cohen
146117
answered Jan 19 at 21:18
Marvin CohenMarvin Cohen
146117
146117
$begingroup$
I can't believe I missed that. I was ignoring the outer prime. Oh well, time for bed.
$endgroup$
– Luke Collins
Jan 19 at 22:13
add a comment |
$begingroup$
I can't believe I missed that. I was ignoring the outer prime. Oh well, time for bed.
$endgroup$
– Luke Collins
Jan 19 at 22:13
$begingroup$
I can't believe I missed that. I was ignoring the outer prime. Oh well, time for bed.
$endgroup$
– Luke Collins
Jan 19 at 22:13
$begingroup$
I can't believe I missed that. I was ignoring the outer prime. Oh well, time for bed.
$endgroup$
– Luke Collins
Jan 19 at 22:13
add a comment |
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1
$begingroup$
Start by finding the antiderivative of the first term (this should easy since the entire term is given as a derivative). Then evaluate the result at -1 and 1 (evaluating the Legendre polynomials exactly will turnout to be less important for the problem).
$endgroup$
– Testcase
Jan 19 at 21:14
$begingroup$
It follows from $int_{-1}^{1} F'(x) dx = F(1) - F(-1)$ for any $C^{1}$ function $F$.
$endgroup$
– Seewoo Lee
Jan 19 at 21:17
$begingroup$
There's a typo in your title that would make this trivial.
$endgroup$
– J.G.
Jan 19 at 21:22
$begingroup$
I was ignoring the outer prime for some reason. Must be tired!
$endgroup$
– Luke Collins
Jan 19 at 22:13
$begingroup$
@J.G. It is trival!
$endgroup$
– Luke Collins
Jan 19 at 22:14