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8
votes
Accepted
How to obtain the asymptotics of Legendre polynomials directly from their generating function
You may write $2x=a+1/a$ for certain $a$, $|a|>1$ (I guess you mean $|x|>1$), then
$$
\frac1{\sqrt{1-2tx+t^2}}=
\frac1{\sqrt{(1-at)(1-a^{-1}t)}}\\= \sum (-1)^n{-1/2\choose n}a^nt^n\cdot \sum (-1)^n{-1 …
6
votes
Accepted
Looking for bound in integral involving Legendre polynomial
We have
$$
\int_{0}^1 \int_{0}^1 \frac{p_n(x) p_n(y)}{1-xy} dx dy=
\int_{0}^1 \int_{0}^1 p_n(x) p_n(y)\sum_{k=0}^\infty(xy)^k dx dy=
\sum_{k=0}^\infty \left(\int_0^1 p_n(x)x^kdx\right)^2.
$$
Next, in …
1
vote
Accepted
Integral Transform with associated Legendre Function of second kind as kernel
[Disclaimer: the below text is not mine, but by Vladimir Petrov, who does not have MO account]
Consider generalized Mehler-Fock transform:
$$
\begin{cases}
F(\xi,\,\mu)&=\intop_1^\infty f(y)P^{-\mu}_ …
1
vote
Accepted
Equality cannot hold unless $x \in \{-1,1\}$ and/or Wronskian is not zero
Let $y_1< z_1< y_2< z_2< \dots< y_{l-m}\leq z_{l-m}< y_{l-m+1}$ be roots of polynomials $f=(1-x^2)^{-m/2} P_l^m$ and $g=(1-x^2)^{-m/2} P_{l+1}^m$ ($y$'s are roots of $g$, $z$'s are roots of $f$). They …