Recent content by Chris L T521

The power on $x$ is $\frac{1}{2n-1}$ which implies these are odd radicals for $n\geq 2$ (when $n=1$, the expression is just $x$). So if $x>0$, $x^{\frac{1}{2n-1}}>0$ and since $\frac{1}{2n-1}\rightarrow 0$ as $n\rightarrow\infty$, we find that $x^{\frac{1}{2n-1}}\rightarrow 1$ (since $a^0=1$...

Keep in mind that $x\in[-1,1]$. Note that \[\lim_{n\to\infty} x^{\frac{1}{2n-1}}=\begin{cases}-1, & x\in[-1,0),\\ 0, & x=0,\\ 1, & x\in(0,1].\end{cases}\] So in the end, we find that \[x\lim_{n\to\infty}x^{\frac{1}{2n-1}} = \begin{cases}x, & x\geq 0,\\ -x,& x<0\end{cases} = |x|.\] I hope this...

By definition of the derivative, \[\begin{aligned}\mathbf{F}^{\prime}(t) &= \lim_{h\to 0}\frac{\mathbf{F}(t+h)-\mathbf{F}(t)}{h}\\ &= \lim_{h\to 0}\frac{\mathbf{f}(\mathbf{c}+(t+h)\mathbf{u})-\mathbf{f}(\mathbf{c}+t\mathbf{u})}{h}\\ &= \lim_{h\to...

If I'm not mistaken, the order of $U(n)$ is $\varphi(n)$, which is the Euler's totient function. If $n$ is prime, then $\varphi(n) = n-1$. If $p$ is prime and $n=p^k$, then $\varphi(n) = p^{k-1}(p-1)$. Also, if $\gcd(m,n)=1$, $\varphi(mn) = \varphi(m)\varphi(n)$. Using these properties of...

2 doesn't work; if $\displaystyle\lim_{x\to 0} g(x)=2$, then $\displaystyle\lim_{x\to 0}\frac{4-g(x)}{x} \rightarrow \frac{2}{0}$, which is undefined. For $\displaystyle\lim_{x\to 0}\frac{4-g(x)}{x}=1$, $x$ must be a factor of $4-g(x)$. Suppose $4-g(x) = x f(x)$, where $g(x)$ and $f(x)$ both...

The types of probabilities that must be computed here are empirical probabilities. The first values you have listed would be the more appropriate way of stating the answers. Since these are mutually exclusive events, you can use this form of the addition rule: $P(I\cup B \cup S\cup G) =...

Very good. This is the correct answer. Again, this is the correct answer. Keep up the good work!

There's nothing wrong with this part. I would suggest something different for this answer. Note that John's age is x and his mother's age is x+25. What do you get when you add them together? I see nothing wrong with this part either, given the information you've provided in the problem. I...

Hello everyone, These are some formula/reference sheets I've made as an adjunct math instructor for courses I've taught over the past 4 years. Feel free to use them as you'd like. Statistics -- Dice & Card Table; Reference Sheet Finite Mathematics -- Reference Sheet Discrete Mathematics --...

Hi karush, For 15., consider the following definitions: \(A\cap B = \{x\mid x\in A\wedge x\in B\}\) \(A\setminus B = \{x\mid x\in A \wedge x\notin B\}\) It follows that \(\begin{aligned}A\cap(B\setminus C) &= \{x\mid x\in A\wedge(x\in B \wedge x\notin C)\}\\ &= \{x\mid (x\in A\wedge x\in...

Hello everyone, I have a question that seems simple at first but is "hard" for me to answer. I'm teaching a general education math course at a community college next semester, and one of the key components to that course is symbolic logic. I'm pretty confident in my knowledge and ability to...

Hm, it's not good. :-/ I would start off like this: \[\begin{aligned}\|(x,y) + (x^{\prime},y^{\prime})\|^2 &= \langle (x,y) + (x^{\prime},y^{\prime}),(x,y) + (x^{\prime},y^{\prime})\rangle \\ &= \langle (x,y),(x,y)\rangle + \langle (x,y),(x^{\prime},y^{\prime})\rangle + \langle...

If you can show instead that $\|(x,y)+(x^{\prime},y^{\prime})\|^{\color{red}2} \leq (\|(x,y)\|+\|(x^{\prime},y^{\prime})\|)^{\color{red}2}$, then it immediately follows that $\|(x,y)+(x^{\prime},y^{\prime})\| \leq \|(x,y)\|+\|(x^{\prime},y^{\prime})\|$. Do you know how to proceed from here?

This week's problem was correctly answered by Euge. You can find his solution below. It suffices to show that $\{0,1\}^{\Bbb N}$ is homeomorphic to the Cantor set, $C$. Define a map $f : \{0,1\}^{\Bbb N} \to C$ by setting $$f((a_n)) = \sum_{n = 1}^\infty \frac{2a_n}{3^n}.$$ It has an...

This will be my last Graduate POTW submission here on MHB; I'm running out of ideas (after doing this for 124 weeks), and I think it's time to get someone fresh in here to do things from now on. It's been a pleasure doing this for roughly 2.5 years now, and I hope you guys give the person who...