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Science Forum Index » Mathematics Forum » Orthogonal Polynomials
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Message |
| Lukas Horosiewicz |
 Posted: Sat Dec 27, 2003 6:08 pm |
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I have a problem. I don't see how the author finds the closed form
expression for the polynomials P_n(\lambda) in the book by Akhiezer --
'The Classical Moment Problem' at page 3. I have scanned the first 5
pages (although one only needs 1,2,3 to get the background
information):
http://hem.bredband.net/lukhor/moment/
Thankful for any help.
Happy new year! |
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| Michael Albert |
| Posted: Sun Dec 28, 2003 12:29 pm |
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On Sat, 27 Dec 2003, Lukas Horosiewicz wrote:
Quote: I have a problem. I don't see how the author finds the closed form
expression for the polynomials P_n(\lambda) in the book by Akhiezer --
'The Classical Moment Problem' at page 3. I have scanned the first 5
pages (although one only needs 1,2,3 to get the background
information):
http://hem.bredband.net/lukhor/moment/
Thankful for any help.
Happy new year!
First, a quick summary for those who don't have the relevant pages
in front of them. Consider the space of polynomials
R(\lambda) = p_0 + p_1 \lamba + p_2 \lambda^2 + ... + p_n \lambda^n.
We consider a linear functional S defined by a sequence of numbers
s_0, s_1, ..., defined by
S[R] = s_0 p_0 + s_1 p_1 + ....
where the sum is finite because the polynomial has only a finite number
of non-zero coefficients. Now we can define a bilinear form, given
polynomials R_1 and R_2, as S[ R_1 R_2 ]; that is, we muliply the
polynomials in the usual sense, then apply the linear functional
S[]. The coefficients s_0,..., are choosen so that the form is
positive definite.
Now the author wishes to choose an orthogonal basis, and gives
the following explicit formulae:
P_0 (\lambda) = 1
P_n (\lambda) = 1/ \sqrt{ D_{n-1} D_{n} } \times
| s_0 s_1 ... s_n |
| s_1 s_2 ... s_{n+1} |
| ... |
| s_{n-1} s_{n} ... s_{2n-1} |
| 1 \lambda ... \lambda^n |
where D_{m} is defined as the determinant of the matrix
| s_0 s_1 ... s_m |
| s_1 s_2 ... s_{m+1} |
| ... |
| s_{m} s_{m+1} ... s_{2m} |.
Now the author notes that P_n is of degree 'n' (as is clear by
expanding the matrix using the bottom row), so that to
prove orthogonality (leaving aside normalization for the moment) it
is sufficient to prove that P_n is orthogonal to 1, \lambda, \lambda^2,
...., \lambda^{n-1}. Agreed?
Now what happens when we multiply P_n by \lambda^m? This is
given on page 4 [let us call this equation#1],
P_n (\lambda) \lambda^m = 1/ \sqrt{ D_{n-1} D_{n} } \times
| s_0 s_1 ... s_n |
| s_1 s_2 ... s_{n+1} |
| ... |
| s_{n-1} s_{n} ... s_{2n-1} |
| \lambda^m \lambda^{m+1} ... \lambda^{m+n} |
Note: if we expand the determinant along using the last row, we
would get P_n(\lambda) \lambda^n = a polynominal in terms
of \lambda^{m}...\lambda^{m+n} where the coefficients are
the minors of the above matix (with appropriate signs).
The author then asks the reader "to apply the functional S[] to
both sides of it", so that the left side becomes the inner product
of P_n(\lambda) and \lambda^m, so that we obtain [this will be
our equation#2]
S[ P_n (\lambda) \lambda^m ] = 1/ \sqrt{ D_{n-1} D_{n} } \times
| s_0 s_1 ... s_n |
| s_1 s_2 ... s_{n+1} |
| ... |
| s_{n-1} s_{n} ... s_{2n-1} |
| s_{m} s_{m+1} ... s_{m+n} |.
To see this equalility, consider the expansion of this matrix
and the previous matrix (from equation#1 above) using the last
row--the minors are the same in both instances (and contain only
s_i's, no \lambda's!) and we simply have replaced \lambda^m with
s_m, \lambda^{m+1} with s_{m+1}, etc. Now, if m<n, then the
last row of this matrix will be identical to one of the previous
rows, and the determinant will be zero! Thus P_n is orthogonal
to \lambda^m for m<n, and thus P_n is orthogonal to P_m for m<n.
Now, for the normalization, the author notes at the bottom of
page 3 that P_n can be written as
P_n(\lambda) = \sqrt{ D_{n-1}/D_{n} } \lambda^n + R_{n-1}(\lambda)
where R_{n-1} has degree of at most 'n-1'. This is again obtain
by expanding the definition of P_n(\lambda) along the bottom row
of the matrix, although we only need to specify the coefficient
of \lambda^n exactly. Thus,
S[ P_n(\lambda) P_n(\lambda) ]
= S[ P_n(\lambda) * ( \sqrt{ D_{n-1}/D_{n} } \lambda^n +
R_{n-1}(\lambda) ) ]
= S[ P_n(\lambda) * ( \sqrt{ D_{n-1}/D_{n} } \lambda^n ) ]
+ S[ P_n(\lambda) * R_{n-1}(\lambda) ) ].
Since P_n is othogonal to all polynomials of degree less than n,
the second term vanishes. The first term can be computed by
returning to our equation #2 and using the definitions of
D_{n} and D_{n-1}.
Hope that helps.
Best wishes and happy new year.
-Mike |
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