We know that the Legendre Polynomial is $P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n}[(x^2-1)^n]$. The Shifted Legendre Polynomial $\tilde P_n(x)$ is defined as $P_n(2x-1)$. Can you please tell me how to get $\tilde P_n(x) = \frac{1}{n!}\frac{d^n}{dx^n}[(x^2-x)^n]$? Thanks.
Shifted Legendre Polynomial
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legendre-polynomials
1 Answers
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$$P_n(X) = \frac{1}{2^nn!}\frac{d^n}{dX^n}[(X^2-1)^n]$$ With $X=(2x-1)$
$$\tilde P_n(x)=P_n(2x-1)=P_n(X)=\frac{1}{2^nn!}\frac{d^n}{d(2x-1)^n}[\left((2x-1)^2-1\right)^n]$$ $d(2x-1)=2dx \quad\to\quad d(2x-1)^n=2^n dx^n$
$\left((2x-1)^2-1\right)^n=(4x^2-4x)^n=4^n(x^2-x)^n$
$$\tilde P_n(x)=\frac{1}{2^n n!}\frac{d^n}{2^n dx^n}[4^n(x^2-x)^n]=\frac{4^n}{2^n 2^n n!}\frac{d^n}{dx^n}[(x^2-x)^n]$$
$$\tilde P_n(x) = \frac{1}{n!}\frac{d^n}{dx^n}[(x^2-x)^n]$$