Why does $int_{-1}^1 ((1-x^2)(P_m'P_n-P_n'P_m))',dx = 0$ for Legendre polynomials?

1

$begingroup$

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

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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

1

$begingroup$

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

share|cite|improve this question

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

1

1

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Jan 19 at 22:15

J.G.

28.4k22845

asked Jan 19 at 20:50

Luke CollinsLuke Collins

749419

share|cite|improve this question

edited Jan 19 at 22:15

J.G.

28.4k22845

asked Jan 19 at 20:50

Luke CollinsLuke Collins

749419

share|cite|improve this question

share|cite|improve this question

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

active

oldest

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3

$begingroup$

$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$

share|cite|improve this answer

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 | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

3

$begingroup$

$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$

share|cite|improve this answer

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 | 

3

$begingroup$

$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$

share|cite|improve this answer

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 | 

3

3

3

$begingroup$

$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$

share|cite|improve this answer

answered Jan 19 at 21:18

Marvin CohenMarvin Cohen

146117

$endgroup$

$$ int_{-1}^1 Big((1 - x^2)(P_m'P_n - P_n'P_m) Big)' mathrm{d}x = Big[(1 - x^2)(P_m'P_n - P_n'P_m)Big]_{-1}^1 $$
$$= Big[(1 - 1^2)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(1 - (-1)^2)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big]$$
$$ = Big[(0)(P_m'(1)P_n(1) - P_n'(1)P_m(1))Big] - Big[(0)(P_m'(-1)P_n(-1) - P_n'(-1)P_m(-1))Big] $$
$$= 0 - 0 $$
$$= 0 $$

share|cite|improve this answer

answered Jan 19 at 21:18

Marvin CohenMarvin Cohen

146117

share|cite|improve this answer

share|cite|improve this answer

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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