Recent content by Elucidus

I think you're in for a lot of swearing and bloody knuckles. Since x^x is positive for all x > 0 then x^{(x^x)} is continuous for x > 0 and therefore Riemann integrable. But trying to find any sort of friendly resolution to it is a fool's errand. Numeric approximation is your best hope since...

Cantor proved that the real algebraiic numbers are denumerable in his 1874 paper "On a Property of the Totality of All Real Algebraic Numbers" the proof of which is nicely summarized in "The Calculus Gallery" by William Dunham. The Cartesian product of denumerable sets is also denumerable by...

The function f(x) = \left\{ \begin{array}{rl} 1, & x \text{ rational} \\ 0, & x \text{ irrational} \end{array} (aka the rational comb function) fails since it is not continuous anywhere. The function sought needs to be continuous everywhere except at 1, 1/2, 1/3, 1/4, etc. The...

I will point out the functions that are being cooked up are discontinuous at the natural numbers, but the original challenge was for the function to be discontinuous at x = 1, 1/2, 1/3, 1/4,... , but continuous at x = 0 and at all other numbers (I assume for all other real numbers). I have...

Your tables are not clear enough for me to make out exactly what you are doing. I'm also not convinved there is a precedence rule between \Rightarrow and \Leftrightarrow. Part (c) may end up being ambiguous. --Elucidus

Assuming |r| < 1 then \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} Differentiation both sides with respect to r gives: \sum_{n=1}^{\infty} n \cdot r^{n-1} = \frac{1}{(1-r)^2} This should give you a push in the right direction. (Warning: Be careful of your initial index.) --Elucidus

I believe HallsofIvy means question 3 here. --Elucidus

When an expression evaluates to something of the form 0/0, it is called an indeterminate quotient. There are more advanced methods (usually seen midway through Calculus I) that can handle these better. It is unfortunately the case that nothing about the limit can be deduced from the knowledge...

x = 16 is a solution of \log_4(x^2) = (\log_4 x)^2. Both sides equal 4. HallsofIvy's comment though gives the hint for turning this into a (factorable) quadratic equation in log4x which has 2 solutions, namely 1 and 16. --Elucidus

Hints: (a) This is a combination, (8+9) choose 7. (b) (committee of 7 with 6 women) + (committee of 7 with 7 women) (c) (all committees) - (those with both Bob and Alice). --Elucidus

Two of your answers are incorrect. Question 4 is correct, but question 6 is not. Use the hint in number 6 and let x = 0. Do you see why it must be false? I leave you to find the other incorrect answer. --Elucidus

This is false. -ab means -(ab). If you meant raising -a to the bth power, you have to write (-a)b. --Elucidus

I assume you are asked to show p \rightarrow (q \rightarrow r) \Rightarrow (p \rightarrow q) \rightarrow r. Proofs involving conclusions of the form "if A then B" are usually best proven by assuming the premises of the claim and A and then showing B is a consequence. Basically: Given p...

Yeah. I knew I needed (5) but blew it on (4). Monotonicity was staring at me, taunting me. I went and looked back at my scratch work and right there on page 2 is u \leq v \Rightarrow f(u) \leq f(v) I will mention though that (1) and (2) should be 1) Prove that f(1) = 0 or 1. 2)...

You can use the following theorems: For any real number a and positive integer m: \lim_{x \rightarrow a^+} \frac{1}{(x-a)^m} = \infty \lim_{x \rightarrow a^-} \frac{1}{(x-a)^m} = \left\{ \begin{array}{rl} \infty, & \text{if }m \text{ is even} \\ -\infty, & \text{if }m \text{ is odd}...