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Feb
14
revised An Exercise of Finite Groups
deleted 57 characters in body
Feb
13
answered An Exercise of Finite Groups
Feb
13
answered Obvious Group and subgroup questions.
Feb
13
comment Step by step for solving $\sum^n_{i=1}{(a+b)^i}$?
¿Have you tried to prove it by induction?
Feb
13
answered Why this “interpolation” is correct?
Feb
13
answered True or false - Function analytic in bounded domain is bounded.
Feb
13
revised non linear optimization
No need for caps in title.
Feb
13
suggested approved edit on non linear optimization
Feb
12
comment Identifying Proof Method and Implementing It
What you mention should work perfectly. As for the name of the method, I don't know, but I've seen that sometimes be called prove by inspection. Which is what you do when you analise all possible cases...
Feb
12
comment Group tables for a group of four elements.
@user61854 Yes, in my case 1, you have that: $a^2=b^2=c^2=1$, so: $ab=c, bc=a$, it's abelian, so the whole table is there. In my case 2, $a$ is the generator, so $a=a,a^2=b,a^3=c$, and you can compose to get the rest of the combinations.
Feb
12
revised Let $A$ be a $3 \times 3$ matrix with real entries such that…
deleted 1 characters in body
Feb
12
answered Let $A$ be a $3 \times 3$ matrix with real entries such that…
Feb
12
comment Finding the $g'$ of 2 functions
Except for the line that @GerryMyerson has said, I think they're correct. The difference between these and the expressions for a function that may be used to see, is that here you're getting to a differential equation for $g$, instead of the function itself.
Feb
11
comment Group tables for a group of four elements.
@amWhy I would have never imagined, Lagrange always comes really soon. I guess that then he will have to do every possible table to get to the same conclusion.
Feb
11
answered Group tables for a group of four elements.
Feb
11
comment Complex Analysis - confusion over conjugate
To get to that polar form you're supposing that $z\not =0$, if $z=0$ then the function is 0/0, so it's not analytical in that point.
Feb
11
answered Let $m \in \mathbb{Z^+} , n \in \mathbb{Z^+}$ and let $d=\gcd(m,n)$. Prove that $m\mathbb{Z}+n\mathbb{Z}=d\mathbb{Z}$
Feb
10
comment For each $n \ge 1$ compute $Z(S_n)$
@CrystalSeluini You're welcome, I updated my answer to explain what $Z(G)$ is, as you asked it in a comment. Good luck!
Feb
10
revised For each $n \ge 1$ compute $Z(S_n)$
added 469 characters in body
Feb
10
answered For each $n \ge 1$ compute $Z(S_n)$