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alex.lupas@gmail.com science forum beginner
Joined: 23 Feb 2006
Posts: 47

Posted: Sat Jun 17, 2006 4:45 pm Post subject:
Legendre Polynomials..ineq.



If P_n(x):=c_n*((x^21)^n)^{(n)} , P_n(1)=1 ,
I need to prove that for all x in [0,1] the inequalities
P_{n+2}(x)*( P_{n+1}(x)+P_{n1}(x) ) =<
=< P_n(x)*( P_{n+3}(x)+P_{n+1}(x) ) , n in {1,2,...},
are verified ( ? ) . Thanking in advance ! 

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jared@math.uiuc.edu science forum beginner
Joined: 19 Jun 2006
Posts: 2

Posted: Mon Jun 19, 2006 6:45 pm Post subject:
Re: Legendre Polynomials..ineq.



alex.lupas@gmail.com wrote:
Quote:  If P_n(x):=c_n*((x^21)^n)^{(n)} , P_n(1)=1 ,
I need to prove that for all x in [0,1] the inequalities
P_{n+2}(x)*( P_{n+1}(x)+P_{n1}(x) ) =
=< P_n(x)*( P_{n+3}(x)+P_{n+1}(x) ) , n in {1,2,...},
are verified ( ? ) . Thanking in advance !

This is going to depend on the normalization, which you haven't
specified: What is c_n? Legendre polynomials are often normalized to
have L_2 norm 2(n+1)^(1). I believe this is c_n = (2^n n!)^(1) but
this is from memory. Other common normalizations are to make them all
monic or to make them have L_2 norm 1.
It is hard to say much without knowing exactly what is to be shown, but
I would try playing around with the following formula, which holds for
any orthogonal polynomial (correctly normalized):
(xy) \sum P_k(x) P_k(y) = P_(N+1)(x) P_N(y)  P_N(x) P_(N+1)(y)
Check Szego's book for the right normalization. Taking the limit as
y>x gives
\sum P_k(x) P_k(x) = Q_(N+1)(x) P_N(x)  Q_N(x) P_(N+1)(x)
where Q = dP/dx. This is at least reminiscent of the inequality you are
trying to prove, since it shows that the lefthand side, a difference of
products of two polynomials, is positive. There is a recurrence for the
Lpolynomials involving derivatives. Perhaps this together with the
above identity might prove useful... 

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alex.lupas@gmail.com science forum beginner
Joined: 23 Feb 2006
Posts: 47

Posted: Tue Jun 20, 2006 11:31 am Post subject:
Re: Legendre Polynomials..ineq.



jared@math.uiuc.edu wrote:
Quote:  alex.lupas@gmail.com wrote:
If P_n(x):=c_n*((x^21)^n)^{(n)} , P_n(1)=1 ,

Observe the normalization condition P_n(1)=1 .
This implies c_n= 1/(n!2^n) [easy !] 

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jared@math.uiuc.edu science forum beginner
Joined: 19 Jun 2006
Posts: 2

Posted: Fri Jun 23, 2006 12:30 am Post subject:
Re: Legendre Polynomials..ineq.



Quote:  Observe the normalization condition P_n(1)=1 .
This implies c_n= 1/(n!2^n) [easy !]

You're right  my bad. I tried the line of argument I suggested, but
couldn't get it to work.
Can I ask how this came up? 

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alex.lupas@gmail.com science forum beginner
Joined: 23 Feb 2006
Posts: 47

Posted: Sun Jun 25, 2006 3:20 pm Post subject:
Re: Legendre Polynomials..ineq.



jared@math.uiuc.edu wrote:
Quote:  Observe the normalization condition P_n(1)=1 .
This implies c_n= 1/(n!2^n) [easy !]
You're right  my bad. I tried the line of argument I suggested, but
couldn't get it to work. Can I ask how this came up?
====================== 
Let h(x):=(x^21)^n=(x1)^n(x+1)^n . Further denote
f(x):=(x1)^n , g(x):=(x+1)^n
C(n,k)=n(n1)...(nk+1)/k! , k,n in {0,1,...} .
Observe that for j in {0,1,...,n} we have
(#) ( (x+b)^n )^{(j)} = j!C(n,j)(x+b)^{nj} .
According to Leibniz formula [see also (#)]
h^{(n)}(x)=
=SUM_{k=0 to k=n}C(n,k)f^{(k)}(x)g^{(nk)}(x)=
=SUM k!(nk)!C(n,k)C(n,k)C(n,nk)(x1)^{nk}(x+1)^{k}.
For x=1 we find [see the last term from the sum]
h^{(n)}(1)=n!2^n .
But P_n(x)=c_n*[h(x)]^{(n)} and P_n(1)=1 .
Therefore 1=P_n(1)=c_n*n!2^n or c_n=1/(n!2^n) .Regards,Alex 

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