Nicolas Villanueva
• Member for 10 years, 9 months
• Last seen more than 10 years ago
• Atlanta, GA

Notice that if $x>1$, then the summation $$1+x+x^2+x^3+\ldots$$ diverges to $\infty$. And thus cannot be equal to a finite number.

We can multiply the numerator and denominator by $\cos(x)$ and get, $$\frac{\cos^4(x) - 2 \cdot \cos^2(x) + 1}{\cos^2(x) \cdot \sin^2(x)}$$ $$= \frac{\left(\cos^2(x) - 1 \right)^2}{\cos^2(x) \cdot \... View answer 5 votes Notice that if we have boys occupying every other seat, there are 2 total arrangements for boys and girls. If I take the arrangement where the first spot is a boy, we will have the following:$${\_\_}...

You think of each subset as a binary string of 0's and 1's, where the $i^{th}$ character in the string is 0 if the $i^{th}$ element in the set is not in the subset. So for your Inductive Hypothesis, ...

By simple arithmetic, $\frac{3 log_y(5) \cdot 2 log_y(5)}{6 log_y(5)} = \frac{6 log^2_y(5)}{6 log_y(5)} = log_y(5)$

This is an incomplete solution, but it's a starting point and I would be open to any further suggestions to get the final solution I hope my assumptions are correct, but I'm going to assume that you ...

Your first balanced flow equation is incorrect, it should be: $$x_1 - x_2 = 400$$

You should get a complete bipartite graph $G$ where $V(G) = V_1 \cup V_2$ where $V_1 = \left\{ 2,3,\ldots,10,J,Q,K,A \right\}$ and $V_2 = \left\{ Club, Heart, Diamond, Spade \right\}$. Thus a hand of ...

If we look at $N(t)$ to be the number of people in the room, then $S = \left\{ 0,1,2 \right\}$ with transitions of $\lambda$ from $0 \rightarrow 1$ and $1 \rightarrow 2$, and transitions of $\mu$ from ...

One possibility is to just have a $K_{3000}$ with self-loops on each vertex in $V$. Give the transition probability for each edge, $e \in E(G)$, to be $\frac{1}{3000}$.

Even if that first equation is a typo, that is the correct general solution. If a proof is necessary, I suggest using Induction on $n$.

25 total possibilities of picking 2 numbers from $\left[1,\ldots,5\right]$: Out of those 25 possibilities, which we'll represent as pairs $(x,y)$, we see that there are 9 ways of having either $(1,x)$...

By the Law of Sines and since $b$ is a right angle, $$len(A) = \frac{len(B)}{sin(\frac{\pi}{2}-a)}$$ where $0 \leq a <\pi$.

I believe that, $\left[ x \right]_{m(x)} = b(x) + \left(m(x)\right)$, where $b(x) \in F[\alpha]$ s.t. $deg(b(x)) < deg(m(x))$ Check here for more notes on Modular Arithmetic and Congruence ...

It means significantly smaller than, if I'm not mistaken.

This problem might be NP-Hard, since it's similar to finding the longest even Hamiltonian path, $p = (v \rightarrow w)$ , in $G$ s.t. $e = (v,w) \in E(G)$.

In the Euclidean Algorithm finding the solutions to $$3f+280y=1$$ is equivalent to finding the solutions to $$3a-280b=1$$ And it's not abnormal to have solutions that are negative since solutions to ...