I'm sure that this problem is easier than I am making out to be, but I'm going over some review problems for an exit exam and I'm having a little trouble with this one.
Let the matrix $A$ be given by:
$$A = \begin{pmatrix} 1&4\\ 2&3 \end{pmatrix}$$
Find $A^n$ for general $n$.
I have the...
Yeah, now that I've seen it worked out it makes sense. Mine was certainly not leading to a simpler solution, I was just curious about how they were so sure that Stirling's formula and Abel summation were used to solve it. I think in any case the density of the prime numbers is required, as you...
This was actually along the lines of my first solution to the problem. I'm glad that you also share this idea.
It does seem like using Stirling's formula (and I'm not sure which form I'm supposed to use) is a bit unnecessary... Although I've heard there is a weaker form that is actually a...
I am in an independent study working through probabilistic graph theory and I am stuck on part of a theorem from chapter 4 of The Probabilistic Method by Joel Spencer and Noga Alon (specifically theorem 4.2.1).
In this context, $p$ is a prime number.
The part where I am confused comes from a...
I was asked to see if I could prove this was true, but I have been totally unable to. I can't find many results about extendability and so I have had a lot of trouble. After days of thinking about it without anything to show, I figured that I'd ask for some help here. Here's the problem:
A...
This makes some sense... Although I'm not sure what it means to "slide up". I am curious, though. Graph theory proofs are new to me so I'm trying to learn everything I can.
I was curious as to whether or not I could use contradiction. Showing that a cycle of length three occurring as...
I'm guessing that the girth of a cube will be 4 as well, since each face is a square. Beyond 3-dimensions, though, I can't really see what's going on. I know that the square is a subgraph, but not necessarily that it is a smallest cycle.
I'm not sure what you're asking, though. When $d=2$, I...
My problem is to show that, for $d\geq 2$, the girth of the d-dimensional hypercube graph, I'll call it $H_d$, is $4$.
I'm pretty sure I should use induction, since the base case is simply a cycle of length $4$. Then I suppose that the claim is true for some $d\geq 2$. So I need to show that...
My problem is this:
Find the integral curves of $\textbf{V} = (log(y+z),1,-1)$.
I first set up the system:
\frac{dx}{log(y+z)} = \frac{dy}{1} = \frac{dz}{-1}
I have two find two curves, $u_1$ and $u_2$ that work as integral curves.
The first, and most obvious, function is $u_1(x,y,z) = y +...
Well, I'm pretty sure I solved this a few days ago, but I forgot that posting here might help others or myself so here's the proof to the reverse implication. I used an earlier exercise that proved that Boolean lattices were join continuous. Turns out this problem also solved my next problem...
Well, as you said, you can technically split it up however you like, but as with most problems there are reasonably good choices you can make.
If you're splitting the interval, convenient choices are at the origin or at discontinuities, and you choose discontinuities because you can be more...
I'm pretty sure you can do the double limits if there are no discontinuities. I'm not positive though since I never did it that way.
As for the discontinuities, remember the rule for improper integrals:
If $\int_a^t f(x) ~dx$ exists for every $t > a$, then \int_a^{\infty} f(x) ~dx =...
My problem for this thread is:
Let $L$ be a Boolean lattice. Prove that $L$ is atomic if and only if the top element is the join of a set of atoms.
For the forward implication, I am already done. I used Zorn's lemma to show that the set, $\mathcal{F}$, of the elements in $L$ which are the...
Thank you for sticking with me! This is what I was trying to do. My professor gave me a very small hint and I can see it used in here... But I'm not sure how he expected me to get there. I'm still kind of new to this, so thanks also for the details at the end, the information is very helpful and...