Question

Let $P_n(x)$ be a function satisfying $\sum_{n=0}^{\infty} h^nP_n(x) = (1-2hx+h^2)^{-1/2}, |x| \leq 1, |h| < 1/3$. Then the unit digit of, $P_3(10)$, is

Answer 1

5

Answer 2

The given generating function is: $\sum_{n=0}^{\infty} h^nP_n(x) = (1-2hx+h^2)^{-1/2}$ This is the generating function for Legendre polynomials, $P_n(x)$. We need to find the unit digit of $P_3(10)$.

We can find $P_3(x)$ by expanding the given generating function using the binomial theorem. Recall the binomial expansion: $(1-u)^{-1/2} = 1 + \frac{1}{2}u + \frac{1}{2}\frac{3}{2}\frac{1}{2!}u^2 + \frac{1}{2}\frac{3}{2}\frac{5}{2}\frac{1}{3!}u^3 + \dots$ $(1-u)^{-1/2} = 1 + \frac{1}{2}u + \frac{3}{8}u^2 + \frac{5}{16}u^3 + \dots$ Let $u = 2hx - h^2$. Substitute this into the expansion: $(1-2hx+h^2)^{-1/2} = 1 + \frac{1}{2}(2hx-h^2) + \frac{3}{8}(2hx-h^2)^2 + \frac{5}{16}(2hx-h^2)^3 + \dots$ We are looking for $P_3(x)$, which is the coefficient of $h^3$. Let's expand the terms and collect the coefficients of $h^3$:

1. From the first term: $1$ (coefficient of $h^0$)

2. From the second term: $\frac{1}{2}(2hx-h^2) = hx - \frac{1}{2}h^2$. No $h^3$ term here.

3. From the third term: $\frac{3}{8}(2hx-h^2)^2$ $\frac{3}{8}(2hx-h^2)^2 = \frac{3}{8}((2hx)^2 - 2(2hx)(h^2) + (h^2)^2)$ $= \frac{3}{8}(4h^2x^2 - 4h^3x + h^4)$ The term with $h^3$ is $\frac{3}{8}(-4h^3x) = -\frac{3}{2}h^3x$. So, the coefficient of $h^3$ from this term is $-\frac{3}{2}x$.

4. From the fourth term: $\frac{5}{16}(2hx-h^2)^3$ $\frac{5}{16}(2hx-h^2)^3 = \frac{5}{16}((2hx)^3 - 3(2hx)^2(h^2) + 3(2hx)(h^2)^2 - (h^2)^3)$ We only need the term with $h^3$. This comes from the first part of the expansion $(2hx)^3$: $\frac{5}{16}(2hx)^3 = \frac{5}{16}(8h^3x^3) = \frac{5}{2}h^3x^3$ So, the coefficient of $h^3$ from this term is $\frac{5}{2}x^3$. Higher terms in the expansion like $h^4, h^5, h^6$ will not contribute to $P_3(x)$.

Combining the coefficients of $h^3$ from all terms, we get $P_3(x)$: $P_3(x) = \frac{5}{2}x^3 - \frac{3}{2}x$ $P_3(x) = \frac{1}{2}(5x^3 - 3x)$

Now, we need to find $P_3(10)$: $P_3(10) = \frac{1}{2}(5(10)^3 - 3(10))$ $P_3(10) = \frac{1}{2}(5(1000) - 30)$ $P_3(10) = \frac{1}{2}(5000 - 30)$ $P_3(10) = \frac{1}{2}(4970)$ $P_3(10) = 2485$ The unit digit of $P_3(10)$ is 5.