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The Legendre function of the second kind, $Q_n(z)$, along with the usual Legendre polynomial $P_n(z)$, are the two linearly independent solutions of the Legendre differential equation.

$Q_n(z)$ can be expressed in the following form:

$$Q_n(z)=P_n(z)\mathrm{artanh}\,z-W_{n-1}(z)$$

where $W_{n-1}(z)$ can be expressed either as

$$W_{n-1}(z)=\sum_{k=1}^n \frac{P_{k-1}(z) P_{n-k}(z)}{k}$$

or as

$$W_{n-1}(z)=\sum_{k=0}^{n-1} \frac{(H_n-H_k)(n+k)!}{2^k (n-k)! (k!)^2} (z-1)^k$$

where $H_k$ is the $k$-th harmonic number, $H_k=\sum\limits_{j=1}^k \frac1{j}$.

My questions:

  1. Mathematica returns a rather complicated expression for $W_{n-1}(z)$ involving the unknown solution of a certain recurrence (i.e. DifferenceRoot[]). Is there possibly a simpler form for this polynomial?

  2. Might there be a (hopefully simple) $n$-term recurrence that generates these polynomials?


Addendum:

After staring long and hard at Pietro's answer, I feel now that my second question was sorta kinda dumb; I already knew that both Legendre functions satisfied the same difference equation, so it stands to reason that a linear combination of them should also be a solution to that recurrence.

I now would like to expand my first question a bit: is it possible to express $W_n(z)$ as a single hypergeometric function (e.g. ${}_p F_q$ or some of the fancy multivariate ones), perhaps with one of the parameters being a negative integer? For instance, $P_n(z)$ is expressible as a Gaussian hypergeometric function ${}_2 F_1$ with one of the numerator parameters being a negative integer. Might there be something similar for the $W_n$?

I would also like to consider an additional question: are the $W_n$ orthogonal polynomials with respect to some weight function $\omega(x)$ and an associated support interval $(a,b)$? That is, if

$$\int_a^b\omega(t)W_j(t)W_k(t)\mathrm dt=0,\qquad j\neq k$$

for some $\omega(x)$ and some interval $(a,b)$, what is this weight function and its support interval?

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Note that the $P_n(z)$ have generating function $f:=\frac{1}{\sqrt{1-2zt+t^2}}$. This leads to a GF for the $W_n$: by your second formula they have a generating function which is a product of $f$ and its antiderivative (wrto $t$). –  Pietro Majer Sep 4 '11 at 8:06

1 Answer 1

The generating function of the sequence $W_n(z)$ (shifted) is $$w(t,z):=(1-2zt+t^2)^{-\frac{1}{2}}\log\bigg(\frac{-z+t+(1-2zt+t^2)^\frac{1}{2}}{1-z}\bigg) =\sum_{n=1}^\infty\, W_{n-1} (z)\, t^n\, .$$ It verifies a simple linear first order differential equation: $$(1-2zt+t^2)w_t + (t-z)w=1$$ that translates into a three-term linear recurrence for the $W_n\, $:

$$(n+1)W_n=(2n+1)zW_{n-1}-nW_{n-2}\qquad $$

with the initial contitions $W_0=1$ and $W_1:=\frac{3}{2}z\, .$

edit. Note that one may start the above recurrence with $W_{-1}:=0$ and $W_0:=1$. Also note that, up to a shift, that recurrence is the same as the Legendre polynomials. This means that the polynomials $R_n:=W_{n-1}$ are the other linear independent solution to the recurrence of the Legendre polynomials, $$(n+1)y_{n+1}=(2n+1)zy_{n}-ny_{n-1}\, ,$$ that corresponds to the initial conditions $R_0=0$, $R_1=1$ (while $P_0=1$ and $P_1=z$). According to the notations of the general theory of orthogonal polynomials, these $R_n$ should be named "Legendre polynomials of the second kind" (not to be confused with the Legendre functions of the second kind).

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1  
Actually, since the $Q_n$ satisfy the same recurrence relation of the $P_n$ (mathworld.wolfram.com/LegendreFunctionoftheSecondKind.html) it is not surprising that so do $W_{n-1}=P_n\arctan(z) -Q_n$. –  Pietro Majer Sep 4 '11 at 22:31
    
Thanks a lot, Pietro! –  J. M. Sep 5 '11 at 2:16

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