## Problem Set 1

PROBLEMS

1.) The sum of two numbers is $18$ and there difference is $-4$. What are the two numbers?

2.) Find the values of $p, q$ and $r$ if:

$p + q + r = 3 \frac{1}{2}$
$pq + qr + rp = - 2 \frac{1}{2}$
$pqr = -2$

3.) Prove that $\displaystyle\frac{x + y}{2} \geq \sqrt{xy}$

4.) Define $cos(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$ and $sin(x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$

a.) Prove that $cos(-x) = cos(x)$
b.) Prove that $sin(x) = -sin(x)$

SOLUTIONS AND PROOFS
Posted October 13, 2009

1.) Solution: Let $x$ and $y$ be the two numbers. Then, $x + y = 18$ and $x - y = -4$. Adding the equations, we have $(x + y) + (x - y) = 18 + (-4) \Rightarrow 2x = 16 \Rightarrow x = 8$. Substituting it to the first equation gives us $y = 10$. Therefore, the two numbers are $6$ and $10$.

2.) Solution: From the given, $p, q$ and $r$ are roots of of the cubic equation $x^3 - \frac{7x^2}{2} - \frac{5x}{2} + 2.$ Factoring, we have $(x+1)(x-4)(2x + 1)=0.$ Therefore, $x = 1, 4$ or $-1/2$

3.) Proof: We know that the square of the difference of any two numbers is always positive or $0$. Let $x, y$ be any two numbers. Then, $(x - y)^2 \geq 0$. Expanding, we have $x^2 - 2xy + y^2 \geq 0$. Adding $4xy$ to both sides of the equation yields $x^2 + 2xy + y^2 \geq 4xy \Rightarrow (x + y)^2 \geq 4xy$. Getting the square root of both, we have, $x + y \geq 2 \sqrt{xy} \Rightarrow \frac{x + y}{2} \geq \sqrt{xy}. \blacksquare$

4.) Proof (a): We want $\cos (-x)$ so we will just replace $x$‘s with $-x$. Therefore, $cos(-x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-x)^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-1 \cdot x)^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (-1)^{2n} \cdot x^{2n}}{(2n)!} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}= cos(x). \blacksquare$

Proof of 4b is left as an exercise. It’s very similar to the proof of 4a.

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