Question

(a) Use binomial series to formally show that

${\left(1-2xt+t^2\right)}^{-1/2}=\sum _{n=0}^{\infty }{P}_n\left(x\right)t^n$

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(b) Use the result obtained in part (a) to show that$P_n\left(1\right)=1$androle="math" localid="1664256224564" $P_n(-1)={(-1)}^n$. See Properties (ii) and (iii) on page 272.

Short answer

$P_n(-1)={(-1)}^n$

Step-by-step solution

Definition of Power series solutions

If $x=x_0$is an ordinary point of the differential equation , we can always find two linearly independent solutions in the form of a power series centered at $x_0$, that is,

$y=\sum _{n=0}^x{C}_n{(x-x_0)}^n.$

A power series solution converges at least on some interval defined by$\left|\begin{array}{c}x-x_0\end{array}\right|<R$, where R is the distance from$x_0$to the closest singular point.

Using Binomial theorem to solve the problem

Using binomial theorem ${(x=y)}^2=\sum _{k=0}^{\infty }\left(\begin{array}{c}r\\ k\end{array}\right)x^{rk}{y}^k$ where $\left(\begin{array}{c}r\\ k\end{array}\right)=\frac{r(r-1)...(r-k+1)}{k!}$we find

$-1/2=(t^2-2xt)+\frac{(-1/2)(-3/2)}{2!}{(t^2-2xt)}^2+\frac{(-1/2)(-3/2)(-5/2)}{3!}{(t^2-2xt)}^2+..$

$$\begin{gathered} \left[1+(t^2-2xt)\right]=1-\frac{1}{2}=1-\frac{1}{2}=1+xt=\sum _{n=0}^{\infty }{P}_n(t^2-2xt)+\frac{3}{8}{(t^2-2xt)}^2-\frac{5}{16}-\frac{5}{16}{(t^2-2xt)}^3+..+\frac{1}{2}(3x^2-1)t^2+\frac{1}{2}(5x^2-3x)t^3+... \\ \left(x\right)t^n \end{gathered}$$

(b)

Substituting 1 for x in role="math" localid="1664259643703" ${\left(1-2xt+t^2\right)}^{-1/2}$, we obtain

role="math" localid="1664260261907" ${\left(1-2t+t^2\right)}^{-1/2}={(1-t)}^{-1}=\frac{1}{1-t}=1+t+t^2+t^3+...\underset{n=0}{\overset{\left|t\right|<\infty }{=\sum t^n}}$

Now, we can see that

role="math" localid="1664260516052" $\sum _{n=0}^{\infty }{P}_n\left(1\right)t^n\stackrel{\left(a\right)}{=}{\left(1-2t+t^2\right)}^{-1/2}\stackrel{2}{=}\sum _{n=0}^{\infty }{t}^n$

From this, we have that

$P_n\left(1\right)=1$

In the same way, substituting $-1$for $x$in${\left(1-2xt+t^2\right)}^{-1/2}$, we obtain

${\left(1-2t+t^2\right)}^{-1/2}={(1+t)}^{-1}=\frac{1}{1-t}=1-t+t^2-t^3+...\underset{n=0}{\overset{\left|t\right|<\infty }{=\sum {(-1)}^n{t}^n}}$

Now, we can see that

$\sum _{n=0}^{\infty }{P}_n(-1)t^n\stackrel{\left(a\right)}{=}{\left(1+2t+t^2\right)}^{-1/2}\stackrel{\left(3\right)}{=}\underset{n=0}{\overset{\infty }{\sum {(-1)}^n}}{t}^n$

From this, we have that

$P_n(-1)={(-1)}^n$