User:Egm6322.s12.sungsik

Problem 8.6
Show that

$$ P_{2k}(x) is even$$

$$ P_{2k+1}(x) is odd$$

where

$$ P_n(x) = \sum^{[n/2]}_{i=0}(-1)^i\frac{(2n-2i)!x^(n-2i)}{2^ni!(n-i)!(n-2i)!} $$

Even Function $$ F(x) = -F(x) $$

Odd Function $$ F(x) = F(-x) $$

When n=2k

$$ P_{2k}(x) = \sum^{[2k/2]}_{i=0}(-1)^i\frac{(4k-2i)!x^(2k-2i)}{2^2ki!(2k-i)!(2k-2i)!} $$

above equation, x term is the factor to deciede even or odd function.

Therefore, extract x term and expand it

Let assume that k = 5

Then, equation is reduced below

$$\sum_{i=0}^{[5]}x^{10-2i}= x^{10} + x^8 + x^6 + x^4 + x^2 + x^0 $$

since 6 terms are even functions, therefore, $$ P_{2k} $$ is even function

when n=2k+1

$$ P_{2k+1}(x) = \sum^{[2k+1/2]}_{i=0}(-1)^i\frac{(2(2k+1)-2i)!x^{((2k+1)-2i)}}{2^(2k+1)i!(2k+1)-i)!((2k+1)-2i)!} $$

extract x term and expand it

Assuming k=5

$$\sum_{i=0}^{[5]}x^{11-2i}= x^{11} + x^9 + x^7 + x^5 + x^3 + x^1 $$

Since those terms are odd function, therefore, $$P_{2k+1} $$ is odd function