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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^2-1)^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_{n-1}(x) ) =<

=< P_n(x)*( P_{n+3}(x)+P_{n+1}(x) ) , n in {1,2,...},

are verified ( ? ) . Thanking in advance !
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^2-1)^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_{n-1}(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):

(x-y) \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
L-polynomials involving derivatives. Perhaps this together with the
above identity might prove useful...
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^2-1)^n)^{(n)} , P_n(1)=1 ,

Observe the normalization condition P_n(1)=1 .
This implies c_n= 1/(n!2^n) [easy !]
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?
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^2-1)^n=(x-1)^n(x+1)^n . Further denote

f(x):=(x-1)^n , g(x):=(x+1)^n

C(n,k)=n(n-1)...(n-k+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)^{n-j} .

According to Leibniz formula [see also (#)]

h^{(n)}(x)=

=SUM_{k=0 to k=n}C(n,k)f^{(k)}(x)g^{(n-k)}(x)=

=SUM k!(n-k)!C(n,k)C(n,k)C(n,n-k)(x-1)^{n-k}(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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