
Mathematics Stack Exchange
Q&A for people studying math at any level and professionals in related fields
discrete mathematics - Show $|u^n| = n|u|$ for all strings $u$ and …
Can anyone please help me with this homework question on automata from Peter Linz? Use induction on $n$ to show that $|u^n| = n|u|$ for all strings $u$ and all $n$.
modular arithmetic - Prove that that $U (n)$ is an abelian group ...
Prove that that $U(n)$, which is the set of all numbers relatively prime to $n$ that are greater than or equal to one or less than or equal to $n-1$ is an Abelian ...
Prove that the sequence (1+1/n)^n is convergent [duplicate]
Mar 27, 2019 · I know the proof using binomial expansion and then by monotone convergence theorem. But i want to collect some other proofs without using the binomial expansion. *if you …
Prove that the order of $U (n)$ is even when $n>2$.
Right! I like that: $ (n-1)^2=n^2-2n+1=n (n-2)+1 \equiv 1 (\bmod {n})$. I was skeptical of the line "However, we know (I forgot the theorem's name) that the number of elements of order 2 is …
Proof that $U (n)$ is connected - Mathematics Stack Exchange
Thanks for the link @muzzlator. I've just had a look at it and it's very interesting (and seems a lot simpler), however it uses methods a little different to those that I have been using for the …
How do we calculate factorials for numbers with decimal places?
I was playing with my calculator when I tried $1.5!$. It came out to be $1.32934038817$. Now my question is that isn't factorial for natural numbers only? Like $2!$ is $2\\times1$, but how do …
Find convolution of u [n]-u [n-2] and u [n]-u [n-2]
Start asking to get answers Find the answer to your question by asking. Ask question
$U (n)$ is not isomorphic to $SU (n) \times S^1$ [duplicate]
Yes, it is more natural to use the determinant, see the duplicate. There the same doubt was mentioned, that the proof with centers is not so "insightful". On the other hand, the definition of …
If a series converges, then the sequence of terms converges to $0$.
@NeilsonsMilk, ah, it did not even occur to me that this involves a step. See, where I learned mathematics, it is not unusual to first define when a sequence converges to zero (and we have …