General term of $(1-2xt +x^2)^{- \frac{1}{2}} = \sum_{n=0}^{\infty} P_n (t) x^n$ and constructing the generating function. 
(a) Given that, with $|x|<1$, $$(1-2xt +x^2)^{- \frac{1}{2}} = \sum_{n=0}^{\infty} P_n (t) x^n, |x|<1 $$
If $-1<t<1$, show that $$P_0 (t) = 1,$$ $$ P_1(t) =t, $$ 
  $$P_{n+1} (t) = \frac{2n+1}{n+1} t \cdot P_n(t) - \frac{n}{n+1} \cdot P_{n-1}(t). \tag{4}$$ $n \geq 1$ .

Given that $\sum_{n=0}^{\infty} P_n (t) x^n$ is a power series of the form $\sum_{n=0}^{\infty} a_n (x-0)^n$. Where $a_n$, as it is assumed to be converging, is $$a_n = P_n(t) = \frac{f^{(n)}(0)}{n!}$$. 
Given that $f(x) = (1-2xt +x^2)^{- \frac{1}{2}}$, and $f^{(1)} (x)= (t-x) (1-2xt+x^2)^{-\frac{3}{2}}$
$$\implies a_0 = P_0(t) = \frac{f^{(0)}(0)}{0!} = 1$$
$$ \implies a_1= P_1(t) = \frac{f^{(1)}(0)}{1!} = t$$
.
Proving $(3)$, when $n=0$, 


Quoting Wikipedia:
     To obtain further terms without resorting to direct expansion of the Taylor series, [$\frac{1}{\sqrt{1-2tx+x^2}} = \sum_{n=0}^\infty P_n(t) x^n \tag{1}$] is differentiated with respect to $x$ on both sides and rearranged to obtain
    $$\frac{t-x}{\sqrt{1-2tx+x^2}} = (1-2tx+x^2) \sum_{n=1}^\infty n P_n(t) x^{n-1} \tag{2}$$
    Replacing the quotient of the square root with its definition in (1), and equating the coefficients of powers of $x$ in the resulting expansion gives 
    $$ (n+1) P_{n+1}(t) = (2n+1) t P_n(t) - n P_{n-1}(t) \tag{3}$$ 


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My understanding from Wikipedia, From (2) we have :
$$(t-x) \sum_{n=0}^{\infty} P_n(t)x^n = (1-2tx+x^2) \sum_{n=1}^\infty n P_n(t) x^{n-1} $$
I am trying to apply wikipedia approach that I placed above, I am not sure how to proceed after this?
Much appreciated for your input or help. 
 A: For part (a) I'd skip using induction and manipulate the series instead.
Define $\phi (x,t)$ as the generating function:
$$ \phi (x,t) = \dfrac{1}{\sqrt{1 - 2xt + x^2}}$$
Differentiate this with respect to $x$ to get
$$ \dfrac{\partial \phi}{\partial x} = \dfrac{t - x}{(1 - 2xt + x^2)^{3/2}}$$
which can be rearranged to 
$$ (1 - 2xt + x^2) \dfrac{\partial \phi}{\partial x} + (x - t) \phi = 0$$
Insert the series representation for $\phi$ into the above:
$$ (1 - 2xt + x^2) \sum_{n=0}^{\infty}n P_n(t) x^{n-1} + (x - t) \sum_{n=0}^{\infty} P_n(t)x^n = 0 $$
After this expand to five separate series and collect powers of $x$; there's some index shifting involved. The end result is:
$$P_1(t) - t P_0(t) + \sum_{n=1}^{\infty}[ (n+1) P_{n+1}(t) - (2n+1)tP_n(t) + n P_{n-1}(t)] x^n = 0$$
From this you can conclude that 
$$ P_1(t) - t P_0(t) = 0$$
and 
$$ (n+1) P_{n+1}(t) - (2n+1)tP_n(t) + n P_{n-1}(t) = 0$$
which you can rearrange.
For reference a lot of texts on mathematical physics cover special functions/orthogonal polynomials in detail. Special Functions: An Introduction to the Classical Functions of Mathematical Physics by Nico Temme is particularly thorough.
