User:Egm6322.s12.TEAM1.KIM/HW1

Problem 5: Proof problems
 Solved without assistance 

Given
$$\displaystyle f = \sum_{i}g_i$$

Find
1. Show that if g is odd, then f is odd. 2. Show that if g is even, then f is even.

Solution
If g is odd, g(x) is following that $$\displaystyle g_i(-x) = -g_i(x)$$ $$\displaystyle f(-x) = \sum g_i(-x) = -\sum g_i(x) = -f(x) $$ This shows that if g is odd, f is also odd. If g is even, g(x) is following that $$\displaystyle g_i(-x) = g_i(x)$$ $$\displaystyle f(-x) = \sum g_i(-x) = \sum g_i(x) = f(x)$$ This result shows that if g is even, f is also even.

Given
$${{p}_{n\left( x \right)}}=\sum\limits_{i=0}^{\left[ {}^{n}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 2n-2i \right)!{{x}^{n-2i}}}{{{2}^{n}}i!\left( n-i \right)!\left( n-2i \right)!}}$$

Find
Show $$\displaystyle P_2k(x) $$ is even $$\displaystyle P_{2k+1}(x)$$ is odd When k = 0,1,2,3,....,

Solution
$${{p}_{n\left( x \right)}}=\sum\limits_{i=0}^{\left[ {}^{n}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 2n-2i \right)!{{x}^{n-2i}}}{{{2}^{n}}i!\left( n-i \right)!\left( n-2i \right)!}}$$ When n= 2k,the above equation is following that $${{p}_{2k\left( x \right)}}=\sum\limits_{i=0}^{\left[ {}^{2k}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i \right)!{{x}^{2k-2i}}}{{{2}^{2k}}i!\left( 2k-i \right)!\left( 2k-2i \right)!}} = \sum\limits_{i=0}^{\left[ {}^{2k}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i) \right)!{{x}^{2(k-i)}}}{{{2}^{2k}}i!\left( 2k-i \right)!\left( 2k-2i \right)!}}= \sum\limits_{i=0}^{\left[ {}^{2k}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i) \right)!{{(-x)}^2{(k-i)}}}{{{2}^{2k}}i!\left( 2k-i \right)!\left( 2k-2i \right)!}}$$ The above results show that if n is even, f(x) = f(-x). $$\displaystyle P_{2k}(x) = P_{2k}(-x)$$ When n = 2k+1, the equation is following that $${{p}_{2k+1\left( x \right)}}=\sum\limits_{i=0}^{\left[ {}^{2k+1}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i+2 \right)!{{x}^{2k-2i+1}}}{{{2}^{2k+1}}i!\left( 2k-i+1 \right)!\left( 2k-2i+1 \right)!}}=\sum\limits_{i=0}^{\left[ {}^{2k+1}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i+2 \right)!{{(x)x}^{2(k-i)}}}{{{2}^{2k+1}}i!\left( 2k-i+1 \right)!\left( 2k-2i+1 \right)!}}$$ When x= -x, $${{p}_{2k+1\left( x \right)}}= \sum\limits_{i=0}^{\left[ {}^{2k+1}\!\!\diagup\!\!{}_{2}\; \right]}{{{\left( -1 \right)}^{i}}\frac{\left( 4k-2i+2 \right)!{(-x){(-x)}^{2(k-i)}}}{{{2}^{2k+1}}i!\left( 2k-i+1 \right)!\left( 2k-2i+1 \right)!}}$$ This shows that $$\displaystyle P_{2k+1}(x) = -P_{2k+1}(-x)$$