abstract algebra - Order of element of multiplicative group of finite field mod polynomial

If $K$ is a finite field of size $q$ and $f$ is a degree $n$ polynomial in $K[x]$, then we can form the quotient field by modding out this polynomial. Elements of this quotient field are of the form $\phi + (f)$ where $f$ is the ideal generated by $f$. I've been trying to figure out what the order of $x^2 + (f)$ within the multiplicative group of this field, but I am struggling with how to incorporate the coefficients of this arbitrary polynomial.

I would like to understand the case where $f$ is any polynomial, but I'm struggling to even do the simpler case of when $f$ is monic and irreducible.

calculus - Tedious undefined limit without L'Hospital $mathop {lim }limits_{x to frac{pi }{2}} ,,frac{{tan ,(x)}}{{ln ,(2x - pi )}}$

When I try to calculate this limit:
$$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \,\,\frac{{\tan \,(x)}}{{\ln \,(2x - \pi )}}$$

I find this:
$$\begin{array}{l}
L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \,\,\frac{{\tan \,(x)}}{{\ln \,(2x - \pi )}}\\
\text{variable changing}\\
y = 2x - \pi \\
x \to \frac{\pi }{2}\,\,\,\, \Rightarrow \,\,\,y \to 0\\
\text{so:}\\
L = \mathop {\lim }\limits_{y \to 0} \,\,\frac{{\tan \,\left( {\frac{{y + \pi }}{2}} \right)}}{{\ln \,(y)}} = \mathop {\lim }\limits_{y \to 0} \,\,\frac{{\tan \,\left( {\frac{y}{2} + \frac{\pi }{2}} \right)}}{{\ln \,(y)}}\\
= \mathop {\lim }\limits_{y \to 0} \,\,\frac{{ - \cot\,\left( {\frac{y}{2}} \right)}}{{\ln \,(y)}} = - \mathop {\lim }\limits_{y \to 0} \,\,\frac{{\csc (y) + \cot (y)}}{{\ln \,(y)}}\\

= \frac{{ \pm \infty \pm \infty }}{{ - \infty }} = ??
\end{array}$$
and in the latter part I get stuck,

should be obtained using mathematical software $L= \pm \infty$

how I justify without L'Hospital?

Answer

The change of variables is a good start! Write
$$

-\frac{\cot \frac y2}{\ln y} = -2 \cos\frac y2 \cdot \frac{\frac y2}{\sin\frac y2} \cdot \frac1{y\ln y}.
$$
The first factor has limit $-2$ as $y\to0$, by continuity; the second factor has limit $1$ as $y\to0$, due to the fundamental limit result $\lim_{x\to0} \frac{\sin x}x = 1$; and the denominator of the last factor tends to $0$ as $y\to0+$ (and is undefined as $y\to0-$). Therefore the whole thing tends to $-\infty$.

This depends upon two fundamental limits, namely $\lim_{x\to0} \frac{\sin x}x = 1$ and $\lim_{x\to0+} x\ln x = 0$. The first can be established by geometrical arguments, for sure. I'd have to think about the second one, but presumably it has a l'Hopital-free proof as well.

contest math - Inequality with sums and products

Let a,b,c,d positive real numbers such that

$a+b and
$(a+b)(c+d).
Prove that
$(a+b)cd>(c+d)ab$

Source: book on olympiads
I tried manipulating the given statements and i obtained
$3ab
To use this i would need to show that $cd<3(a+b)$ but i am stuck.

sequences and series - Find sum of $1 + cos theta + frac{1}{2!}cos 2theta + cdots$

Find the sum of following series:

$$1 + \cos \theta + \frac{1}{2!}\cos 2\theta + \cdots$$

where $\theta \in \mathbb R$.

My attempt: I need hint to start.

Answer

Hint:
$$
1+\cos x + \frac{1}{2!}\cos 2x + \ldots = \Re(e^{0ix} + e^{1ix} + \frac{1}{2!}e^{2ix} + \ldots)=\Re e^{e^{ix}}
$$

$$
e^{ix}=\cos x+i\sin x\\\Longrightarrow e^{e^{ix}} = e^{\cos x}e^{i\sin x}=e^{\cos x}(\cos(\sin x)+i\sin(\sin x))
$$
Your sum is

$$
e^{\cos x}\cos(\sin x)
$$

sequences and series - Simpler derivation of $sum_{n=1}^infty frac{1}{n^2} = frac{pi^2}{6}$

I know that the equality $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ can be proved in numerous ways by using the Fourier series. However, is there a way to derive it using more fundamental tools? I've tried:
$$ \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \int_0^1 x^{n-1} dx \int_0^1 y^{n-1} dy = \int_0^1 dx\int_0^1 dy \frac{1}{1-xy}$$
and by changing variables I was able to write it in several other forms:$$ = -\int_0^1\frac{\ln(1-x)}{x} dx = \int_0^\infty \frac{\ln (1+t)}{t(1+t)} dt = \int_0^\infty \frac{u }{e^u -1}du $$
but that's as far as I could get.

I specifically don't want to use Fourier series. More fundamental complex analysis, like contour integration, is fine.

Answer

Yes, this can be done. Actually, that's what Tom Apostol did in an article he published in 1983; you can read the proof here (it's the first proof). He used a change of variable ($(x,y)\mapsto(x+y,x-y)$) in order to compute the integral$$\iint_{[0,1]\times[0,1]}\frac1{1-xy}\,\mathrm dx\,\mathrm dy.$$

elementary number theory - Euclid Algorithm to Find Muliplicative Inverse

Here I am trying to find the multiplicative inverse of 19 respect to 29.

$$19x \equiv 1 \pmod{29} $$

What I tried

\begin{align*}
29 &= 1(19) + 10\\\
19 &= 1(10) + 9\\\
10 &= 1(9) + 1.
\end{align*}

From backtracking, I came up with the

\begin{align*}
1 &= 2(29) - 3(19)\\\
\end{align*}

However, 3 is not a multiplicative inverse of the 29. Where am I making a mistake?

I looked many answers including this answer; however, couldn't figure out my mistake.

Answer

What you have found indeed is that $-3\equiv 26$ is the multiplicative inverse of $19$ $\mod 29$.

Prove the convergence of $prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} $ and Find Its Limit

Suppose $p> 1$ and the sequence $\{x_n\}_{n=1}^{\infty}$ has a general term of
$$x_n=\prod\limits^{n}_{k=1}{\left(1+\frac{k}{n^p}\right)} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space n=1,2,3, \cdots$$
Show that the sequence $\{x_n\}_{n=1}^{\infty}$ converges, and hence find
$$\lim_{n\rightarrow\infty}{x_n}$$
which is related to $p$ itself.

I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $p$-series. And also the question seems difficult to find its limit because the denominator is of $p^{th}$ power. How do I deal it?

Answer

We have that

$$\prod\limits^{n}_{k=1}{\left(1+\frac{k}{n^p}\right)}=e^{\sum^{n}_{k=1}{\log \left(1+\frac{k}{n^p}\right)}}$$

and

$$\sum^{n}_{k=1}{\log \left(1+\frac{k}{n^p}\right)}=\sum^{n}_{k=1} \left(\frac{k}{n^p}+O\left(\frac{k^2}{n^{2p}}\right)\right)=$$$$=\frac{n(n-1)}{2n^{p}}+O\left(\frac{n^3}{n^{2p}}\right)=\frac{1}{2n^{p-2}}+O\left(\frac{1}{n^{2p-3}}\right)$$

therefore the sequence converges for $p\ge 2$

• for $p=2 \implies x_n \to \sqrt e$


• for $p>2 \implies x_n \to 1$

and diverges for $1.

Refer also to the related