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Aug
22
awarded  calculus
Aug
13
answered Simple way to generate a sequence where the 1st number is $x_1$, the tenth number is $x_{10}>x_1$ and the 20th number is above some order of magnitude
Aug
5
answered Rewrite $\sin^4 x$ so that it involves only the first power of cosine?
Jul
13
comment Find all postive integers $n$ such $(2n+7)\mid (n!-1)$
Yes, you are right about that mistake.
Jul
13
revised Find all postive integers $n$ such $(2n+7)\mid (n!-1)$
deleted 41 characters in body
Jul
13
revised Find all postive integers $n$ such $(2n+7)\mid (n!-1)$
added 327 characters in body
Jul
13
answered Find all postive integers $n$ such $(2n+7)\mid (n!-1)$
Jul
13
comment Find all postive integers $n$ such $(2n+7)\mid (n!-1)$
I gave a reason in my comment. What is your question?
Jul
13
comment Find all postive integers $n$ such $(2n+7)\mid (n!-1)$
You cannot have $m=2n+7$ factor into any factors less than $n$. Otherwise, if $d>1$ divides $m$, $d|n!-1$ implies $d|1$ which is a contradiction.
Jul
13
answered Suppose $A$ is an $n \times n$ matrix. Prove that similarity transformation is reflexive. ie $A=P^{-1}AP $
Jul
13
revised Suppose $A$ is an $n \times n$ matrix. Prove that similarity transformation is reflexive. ie $A=P^{-1}AP $
edited tags; edited title
Jul
12
comment for each $\epsilon >0$ there is a $\delta >0$ such that whenever $m(A)<\delta$, $\int_A f(x)dx <\epsilon$
Yes, you are right, though I am a tremendous fan of continuous functions. Just trying to bridge to an answer using the OP's thoughts!
Jul
12
answered for each $\epsilon >0$ there is a $\delta >0$ such that whenever $m(A)<\delta$, $\int_A f(x)dx <\epsilon$
Jul
11
answered Why does the “separation of variables” method for DEs work?
Jul
11
comment Why does the “separation of variables” method for DEs work?
Congress passed a law in '79 making it legal.
Jul
10
comment How to prove $\lim_{n \to \infty} \big(\text{sin}(\frac{1}{n+1})+…+\text{sin}(\frac{1}{2n}) \big)$ exists?
You might want to note the limit evaluates to ln 2 by using Riemann Sums.
Jul
9
comment There is no injective morphism from $\mathbb S_7$ to $\mathbb A_8$
@AndreasBlass Makes sooo much more sense! Thanks!
Jul
9
comment There is no injective morphism from $\mathbb S_7$ to $\mathbb A_8$
I realize now that since the kernel of a group action fixes all elements of the set it is acting on, that is, $a \in \ker f$ means $a(gH)=gH$ for all $g \in A_8$, I now don't understand why $|\ker f| \geq 4$. I thought transitivity meant that would work, but it seems to me that's only if you are talking about $\ker f'$.
Jul
9
comment There is no injective morphism from $\mathbb S_7$ to $\mathbb A_8$
Here's where I'm confused. You can either have the group action $f:A_8 \times A_8/H \to A_8/H$ or you can have the group homomorphism $f':A_8 \to A_8/H$ with $f(a,gH)=(ag)H$ and $f'(a) = aH$. Wouldn't looking at $\ker f'$ be enough for our purposes? It isn't all of $A_8$ since $a \notin H$ means $f'(a) \neq H$ and it isn't the identity group since $h \in H$ means $f'(h)=H$ and $H$ is not trivial. So this should do just as much, that is getting a normal subgroup of $A_8$, without group actions.
Jul
9
comment There is no injective morphism from $\mathbb S_7$ to $\mathbb A_8$
So I get why the kernel of the group action is non trivial, which is enough to prove the original claim, but why is there the upper bound of the size of the kernel of 24?