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.

Related Posts Plugin for WordPress, Blogger...

Leave a Reply