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visits member for 1 year, 6 months
seen Apr 11 at 5:38

UCR Fourth Year student.


Apr
11
comment Showing that $\lambda(n)|\phi(n)$ where $n$ is a positive integer.
In the course I'm in we don't have the machinery of Groups. That would have made the proof nicer.
Apr
11
asked Showing that $\lambda(n)|\phi(n)$ where $n$ is a positive integer.
Mar
28
accepted $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
Mar
28
comment $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
Because $t=Ord_n(ab)$.
Mar
28
comment $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
So the definition of order, for relatively prime $b$ and $n$, is the least positive integer $x$ such that $b^{x} \equiv 1 \mod n$. Multiples of $x$ will still make it one, but they won't be the least such.
Mar
28
comment $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
My reasoning is that $b^t \equiv 1$. Hence $t$ must be some multiple of $s$.
Mar
28
asked $Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$
Mar
18
accepted Showing that $\sum\limits_{n=1}^{\infty}f_n-f_{n+1}$ converges normally if $f_n$ converges normally
Mar
18
asked Showing that the sequence $z^n$ is normally but not uniformly convergent
Mar
18
accepted Property of a normally convergence squence of functions
Mar
18
comment Showing that $\sum\limits_{n=1}^{\infty}f_n-f_{n+1}$ converges normally if $f_n$ converges normally
Would it be correct to say that because $\lim\limits_{m\to \infty}s_m=f_1-f$. where $f_{m+1} \to f$ normally we know that $\sum\limits_{n=1}^{\infty}(f_n-f_{n+1})$ converges normally?
Mar
18
comment Showing that $\sum\limits_{n=1}^{\infty}f_n-f_{n+1}$ converges normally if $f_n$ converges normally
So this leaves us with $f_1-f$. How does this show normal convergence of the series?
Mar
18
accepted Properties of periodic functions
Mar
18
comment Showing that $\sum\limits_{n=1}^{\infty}f_n-f_{n+1}$ converges normally if $f_n$ converges normally
@Henry what do you mean?
Mar
18
comment Showing that $\sum\limits_{n=1}^{\infty}f_n-f_{n+1}$ converges normally if $f_n$ converges normally
@Marc If I subtract the partial sum I'm left with the $n+1$ term. Is this what you mean?
Mar
18
comment Showing that $\sum\limits_{n=1}^{\infty}f_n-f_{n+1}$ converges normally if $f_n$ converges normally
@RobertIsrael Am I thinking about this incorrectly by looking at the partial sums?
Mar
18
asked Showing that $\sum\limits_{n=1}^{\infty}f_n-f_{n+1}$ converges normally if $f_n$ converges normally
Mar
18
comment Property of a normally convergence squence of functions
I want to see if I can fill in all the details. Let $z_0 \in K \subseteq U$. Since $f_n$ converges normally in $U$, it converges uniformly on $K$. Let $\epsilon >0$, then there exists an $N$ such that for $n \geq N$, we have $|f_n(z_n)-f(z_n)|+|f(z_n)+f(z_0)|$. The left portion of the sum is less than $\epsilon/2$ for sufficiently large $n$ since $f_n$ is uniformly convergent on $K$. The right hand side is bounded by $\epsilon/2$ since $f$ is continuous since $f_n$ is uniformly convergent on $K$. Hence $|f_n(z_n)-f(z_0)|<\epsilon$.
Mar
17
comment Property of a normally convergence squence of functions
I don't understand the notation on the double modulus at the end of the inequality.
Mar
17
asked Property of a normally convergence squence of functions