Question

prove that Pn'(-1)=
$( - 1)^(n)$
n(n+1)/2​

Answer 1

Answer:

-1/2

Step-by-step explanation:

-1(-1+1)/2

-1/2

=-1/2

Answer 2

Answer:

we proved that $P'_{n} (-1) = \frac{n(n+1)}{2}$ and $P'_{n} (-1) = (-1)^{n}$

Step-by-step explanation:

Explanation :

We know that the generating function $P_{n} (x)$ = $(1 -2hx +h^{2} )^{\frac{-1}{2} }$

So, it is define as $(1 -2hx +h^{2} )^{\frac{-1}{2} }$ = ∑$P_{n} (x)h^{n}$ ......(i)

Where $P_{n} (x)$is the  cofficient of $h^{n}$.

Step 1:

put x = -1 in (i)

∴ $P_{n} (-1)$ = $(1 -2h(-1) +h^{2} )^{\frac{-1}{2} }$

                =$(1 +2h+h^{2} )^{\frac{-1}{2} }$

                = $[(1+h)^{2} ]^{\frac{-1}{2} }$

               =  $(1+h)^{-1}$

       ∑$P_{n} (-1)h^{n}$       = ∑ $(1^{-1}) h^{n}$

Hence , $P'_{n} (-1)$ = $(-1)^{n}$ proved .

Step2:

Since , $P_{n} (x)$ is a solution of legendre equation

∴ $(1-x^{2} )P_{n} ^{''} (x)- 2xP'_{n} (x) +n(n+1)P_{n} (x) = 0$

put x = -1

then ,

$0 -2 P'_{n}(1) +n(n+1) = 0$

⇒$P'_{n} (-1) = \frac{n(n+1)}{2}$  proved.

Final answer :

Hence, here we prove that $P'_{n} (-1) = \frac{n(n+1)}{2}$ and $P'_{n} (-1) = (-1)^{n}$.