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 Mar28 comment Proving that a sequence is increasing Cleaner than what I had wrote. +1 :) Also, I had a small hole in my writing now that I think about it... Mar20 comment Can I use Burnside's Theorem? Or should I take a different approach for this proof? Thank you for the response. Just to make sure I am understanding this: $3|[G : G_x]$ since $|G|= 3^3$ And we know that $|G|/|G_x|$ must be an integer, resulting in either $3^3$, $3^2$ or $3$, hence $|X'| \equiv 32 \mod 3$ meaning $|X'|$ has a strong lower bound of at least $2$ (since 32 mod 3 = 2)? Mar19 comment How do you apply an element on the left of a permutation? @ZachGershkoff: I was testing some small cases for a larger problem. I do not believe I am allowed to discuss the question that motivated this question, but I believe there was probably a flaw in the original question. Mar19 comment How do you apply an element on the left of a permutation? Hmm, I was testing out small cases for a larger problem. After reviewing your answer and the comments, I believe the question that motivated this question has a flaw in it. I would post the original but, I do not believe I am allowed to do so. Thanks for the help! Mar18 comment What about my proof is “nonsense”? @MikeMiller: I was thinking, that $x = ah$ for some $a \in G$ and $h \in H (\leq G)$, by closure of multiplication it has to be a member of $G$? Does this not make that much sense? Mar17 comment When does function composition commute? I think the example is correct, but the math is wrong: $f(g(x)) = (x^2)^3 = g(f(x)) = (x^3)^2 = x^6$ Mar16 comment Help with proving property of Rubik's cube. @vadim123: Now that you say it, I can see it. There always exists a side that can be rotated four times for $C_2$. But you can do this to to get two other sides that don't touch $C_1$ and touch all 7 other cubicles. Thank you for the help! Mar14 comment How should I continue my proof of this cycle property? (And did I make a mistake?) Also, I would edit it (except it doesn't meet the minimum 6 chars needed), but I am fairly certain you mean $x(a_i)$ not $x(a_1)$. Mar14 comment How should I continue my proof of this cycle property? (And did I make a mistake?) Thanks, I get it know! Sorry for the late reply some stuff came up. Mar1 comment Use $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$ Wait, I think I'm reading this wrong, but if $k$ is fixed, then it is either going to be $\frac{\pi^4}{90}$ or the negative of that. Do you mean $-1^n$? Feb27 comment Apply Cauchy-Schwartz to vector? Sorry this is a better link: en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Rn Feb27 comment Apply Cauchy-Schwartz to vector? They have both. Under special cases they have $\mathbb{R}^n$ Feb27 comment Apply Cauchy-Schwartz to vector? They have the case for $\mathbb{R}^n$ on wikipedia: en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality With the proof. Feb23 comment Put this word problem into math terms: A man goes to a stream… Specifically the part about 2. Die Hard Feb23 comment Put this word problem into math terms: A man goes to a stream… You may want to look at this: ocw.mit.edu/courses/electrical-engineering-and-computer-science/… Feb18 comment Where does the function $f(x) =\Big[\frac{1}{2}*x\Big]$ contain discontinuities, left or right continuous? @amWhy: Sorry, recently went of a released exam that used awkward notation. Used brackets as fractional part. Feb17 comment Proving the sines and cosines of the special angles for the unit circle Because adj and opp are both <1 in those cases Feb17 comment When will point with velocity hit line? Constant velocity? Feb13 comment Prove $\mathbb{Z} \times \mathbb{Z} / \left\langle (6,9)\right\rangle$ has an element of order 3 Oh, I was thinking the quotient of two sets, so we would remove them completely, I guess group notation is slightly different then. Thank you for also clarifying there is no multiplication. :) Feb9 comment Integrating absolute complex exponential function @user90593: I have not seen the term before either. Sorry.