Finding the modulus of a complex number $1-e^{itheta }$

I have an answer for this question, but I'm not very confident with it:

$\left | 1-e^{i\theta } \right |^{2}$ i.e. find the square of the modulus of 1-$e^{i\theta }$

$e^{i\theta } = R(cos\theta +isin\theta )$, R=1

= $cos\theta +isin\theta$

In cartesian form:

$a = Rcos\theta \Rightarrow cos\theta$

$b = Rsin\theta \Rightarrow sin\theta$

$\therefore 1-e^{i\Theta } = 1-cos\theta-isin\theta$

$\Rightarrow \left | 1-e^i\theta \right | = \left | 1-cos\theta-isin\theta \right |$

Where the real part, a, = $1-cos\theta$
and the imaginary, b, = $-sin\theta$

The modulus of a complex number being $\sqrt{a^{2} +b^{2}}$

$\therefore \left | 1-cos\theta-isin\theta \right | = \sqrt{(1-cos\theta)^{2} + (-sin\theta)^{2}}$

$\Rightarrow \left | 1-cos\theta-isin\theta \right |^{2} = (1-cos\theta)^{2} + (-sin\theta)^{2} \Rightarrow 1-2cos\theta +cos^{2}\theta +sin^{2}\theta \Rightarrow (\because cos^{2}\theta +sin^{2}\theta = 1) = 2-2cos\theta$

Is this correct? Or is my approach incorrect? Thanks in advance for any feedback/help :)

Answer

Your argument is lengthy, but sound.

If just the square of the modulus is needed, you can consider that $|z|^2=z\bar{z}$, so
$$
|1-e^{i\theta}|^2=(1-e^{i\theta})(1-e^{-i\theta})=
1-(e^{i\theta}+e^{-i\theta})+1=2-2\cos\theta
$$

If also the argument is needed, the trick with $1-e^{i\theta}$ (but also $1+e^{i\theta}$) is to set

$$
\theta=2\alpha
$$
and rewrite
$$
1-e^{i\theta}=1-e^{2i\alpha}=
-e^{i\alpha}(e^{i\alpha}-e^{-i\alpha})=-2i(\sin\alpha)e^{i\alpha}
$$
Since $0\le\theta<2\pi$, we have $0\le\alpha<\pi$, so $\sin\alpha\ge0$. Hence the modulus is $2\sin\alpha$ and, from $-i=e^{3i\pi/2}$, the argument is
$$

\alpha+\frac{3\pi}{2}
$$
(up to an integer multiple of $2\pi$).

Thus the square of the modulus is
$$
4\sin^2\alpha=4\sin^2\frac{\theta}{2}=4\frac{1-\cos\theta}{2}=
2(1-\cos\theta)
$$

integration - Calculating a Complicated Integral of Two Variables

I have encountered the following integral
$$I=\int_{0}^{2\pi} \int_{0}^{2\pi} \frac{c_1 + c_2 \cos\theta_2}{-a + \cos\theta_1 + \cos\theta_2} \cos(m\theta_1) \cos(n\theta_2) d\theta_1 d\theta_2$$

where $a>2$ and I am having a hard time moving forward. I would be happy with an analytic solution, converting this to some non-elementary but known special function, or finding the asymptotic behavior of the integral as $a \rightarrow 2$ from the positive reals.

I was able to perform the $\theta_1$ integration by looking in Gradshteyn and Ryzhik and this yields something proportional to
$$\int_{0}^{2\pi} \frac{c_1 + c_2\cos\theta_{2}}{\sqrt{a^2 - 2 a\cos(\theta_{2})-\sin(\theta_{2})^2}} \cos\left[n\theta_{q'}\right]\\
\times \left(\sqrt{a^2 - 2 a\cos(\theta_{2})-\sin(\theta_{2})^2} -a + \cos(\theta_{2}) \right)^{m}d\theta_{2}.$$
Trying to find this form in a table was hopeless, so I changed variables to $x=\cos\theta_2$. Having to expand $\cos(n\theta_2)$ introduced a sum, and the resulting integral is proportional to
$$\int_{-1}^{1} \frac{dx}{\sqrt{1-x^2}} \frac{c_1 + c_1 x}{\sqrt{\left(-a + x \right)^2 -1 }} \\
\times \left(\sqrt{\left(-a + x\right)^2 -1} -a + x) \right)^{m}\sum\limits_{\substack{r=0 \\ 2r \leq n}} (-1)^r \binom{n}{2r} x^{n-2r} (1-x^2)^r.$$
I am pretty stumped at this point.

I tried doing some further expansions, but I got infinite sums of infinite sums, so this didn't seem like progress (the hypergeometric ${}_{3}F_{2}$ and the regularized ${}_{2}\tilde{F}_{1}$ appeared in this approach).

There are reduction formulas for integrals, but many of them don't seem to apply, as I have two distinct square roots ($\sqrt{1-x^2}$ and $\sqrt{\left(-a + x\right)^2 -1}$). However, my integrand is a rational function of these square roots and $x$.

There is a small parameter in the problem, namely $\epsilon>0$ in $a = 2 +\epsilon$. The integral diverges for $a=2$, so this is a singular perturbation problem (if this approach is even helpful).

I would appreciate any suggestions or advice on this problem!

Answer

Extracting the leading order is not hard. Indeed, let $f : \mathbb{R}^2 \to \mathbb{C}$ be a continuous function such that $f(k_1, k_2) = f(k_1 + 2\pi, k_2) = f(k_1, k_2 + 2\pi)$ for all $k_1, k_2 \in \mathbb{R}$. Then with $\|f\| = \sup |f|$,

$$ \int_{0}^{2\pi}\int_{0}^{2\pi} \frac{f(k)}{a-\cos k_1 - \cos k_2} \, dk_1 dk_2
= \int_{B(0,\pi)} \frac{f(k)}{a - 2 + \frac{1}{2}|k|^2} \, dk + \mathcal{O}(\|f\|). $$

Now applying the polar coordinates change followed by the substitution $r=\sqrt{2(a-2)u}$,

\begin{align*}
\int_{B(0,\pi)} \frac{f(k)}{a - 2 + \frac{1}{2}|k|^2} \, dk
& = \int_{0}^{\pi} \frac{r}{a-2+\frac{1}{2}r^2} \left( \int_{S^1} f(r\omega) \, d\omega \right) \, dr \\
&= \int_{0}^{\frac{\pi^2}{2(a-2)}} \left( \int_{S^1} f(\omega\sqrt{2(a-2)u}) \, d\omega \right) \, \frac{du}{u+1} \\
&\sim 2\pi f(0) \log \left( \frac{\pi^2}{2(a-2)} \right) \\

&\sim -2\pi f(0) \log(a-2).
\end{align*}

As an alternative direction, I guess that a probabilistic interpretation may perhaps help analyze the behavior of $I$. Indeed,

$$ G^p(x,y) = \frac{1}{2\pi^2} \int_{0}^{2\pi}\int_{0}^{2\pi} \frac{e^{-ik\cdot y}}{\frac{2}{1-p} - \cos(k_1) - \cos(k_2)} \, dk $$

is the Green function of the 2-D simple random walk on $\mathbb{Z}^2$ killed at rate $p$.

calculus - continuity of $f ' $

if $f$ is differentiable on $(a,b)$ can we say that $f'$ is continuous on $(a,b)$?

I tried some functions and it seems that we can say so but I'm not sure.

calculus - Meaning of differentials when treated separately from the Leibniz notation dy/dx

I’ve heard people say $dy/dx$ is not a fraction with $dy$ as the numerator and $dx$ as the denominator; that it is just notation representing the derivative of the function $y$ with respect to the variable $x$. The calculus book I am reading (Calculus: A Complete Course - Adams & Essex) says that even though $dy/dx$, defined as the limit of $\Delta y / \Delta x$, appears to be “meaningless” if we treat it as a fraction; it can still be “useful to be able to define the quantities $dy$ and $dx$ in such a way that their quotient is the derivative $dy/dx$”. It then goes on to define the differential $dy$ as “a function of $x$ and $dx$”, as follows: $$dy = \frac{dy}{dx}\,dx = f’(x)\,dx$$ What is the meaning of $dx$ here? It is now an independent variable, yet it seemingly is not one supplied by most functions I would work with. In a later chapter, focused on using differentials to approximate change, the book gives the following: $$\Delta y = \frac{\Delta y}{\Delta x}\,\Delta x \approx \frac{dy}{dx}\,\Delta x = f’(x)\,\Delta x$$ This makes sense to me. The change in $y$, $\Delta y$, at/near a point can be approximated by multiplying the derivative at that point by some change in
$x$, $\Delta x$, at/near the point. Here $\Delta y$ and $\Delta x$ are real numbers, so there is no leap in understanding that is necessary. The problem with the definition of $dy$ is that the multiplication is not between a derivative and a real number, such as in the approximation of $\Delta y$, but a product of a derivative and an object that is not explicitly defined. Because I do not understand what $dx$ is, I can’t use it to build an understanding of what $dy$ is. I also have no real understanding of what $dy$ is meant to be, so I cannot work backwards to attach some meaning to $dx$. Is $dy$ meant to be some infinitesimal quantity? If so, how can we be justified in using it when most of the book is restricted to the real numbers? Are $dy$ and $dx$ still related to the limits of functions, or are they detached from that meaning? Later in the chapter on using differentials to approximate change, the book says it is convenient to “denote the change in $x$ by $dx$ instead of $\Delta x$”. We can just switch out $\Delta x$ for $dx$? Why is it convenient to do this? More importantly, how are we justified in doing this?

What exactly is a differential?
And https://math.blogoverflow.com/2014/11/03/more-than-infinitesimal-what-is-dx/ both contain discussions that are beyond my understanding. My problem arose in an introductory textbook, I find it strange that we can just swap out different symbols when we go to great lengths to say they are different entities.

In What is the practical difference between a differential and a derivative? Arturo Magidin’s answer says that it is not “literally true” that $dy = \frac{dy}{dx}\,dx$ and that replacing $\Delta x$ with $dx$ is an “abuse of notation”. If that is the case, then the quotient of $dy$ and $dx$ would not be $\frac{dy}{dx}$, but $\frac{dy}{dx}\frac{dx}{\Delta x}$, right?

Proof series decreases by induction

I have a question regarding a proof by induction.
We have to see whether or not the following series converges.
$$U_n = \frac{1 \cdot 4 \cdot 7 \cdots (3n - 2)}{2 \cdot 5 \cdot 8 \cdots (3n-1)}$$
I was trying to do this by proving that this series has a lower limit of $0$ and is decreasing. It's easy enough to see that it has a lower limit of $0$ but proving (by induction) that this series is decreasing has proven to be difficult. I understand that I need to prove that $u_n > u_{n+1}$ but I have no idea how to go about doing this.

abstract algebra - Two number fields with trivial intersection, not linearly disjoint but....

Do there exist two finite extensions of $\mathbb{Q}$, $L=\mathbb{Q}(\alpha)$ and $K=\mathbb{Q}(\beta)$ such that $L\cap K=\mathbb{Q}$, the minimal polynomial $\mu_{\alpha}(X)\in \mathbb{Q}[X]$ of $\alpha$ over $\mathbb{Q}$ is not irreducible over $K$, but $\mu_\alpha(\beta)\neq 0$ ?

Answer

Sure. Let $\alpha=\root3\of2$, let $\gamma$ be a nonreal cube root of 2, and let $\beta=\gamma+1$.

Real Analysis Monotone Convergence Theorem Question

I am working on the following exercise for an introductory Real Analysis course.

Suppose that $x_0 \geq 2$ and $x_n = 2 + \sqrt{x_{n-1}-2}$ for $n \in \mathbb{N}$. Use the Monotone Convergence Theorem to prove that either $\lim_{n \to \infty}x_n=2$ or $3$ or $\infty$.

I need some help to move in the right direction. I'll start with what I know so far. My first approach was to break the question into different cases.

I've found that if $x_0=2$, then $x_n =2$ for all $n \in \mathbb{N}$. So, if this is the case, the series is neither decreasing nor increasing? Furthermore, I've found that if $x_0 = 3$ then $x_n =3$ for all $n \in \mathbb{N}$. For $2 < x_0 < 3$, the series is monotone increasing, and converges to $3$. For $x_0 >3$, the series is monotone decreasing and converges to $3$. You will notice none of the cases I've given result in $\lim_{n \to \infty}x_n= \infty$. I have not proven any of the above.

Now, the Monotone Convergence Theorem states, according to my textbook:

A monotone increasing sequence that is bounded above converges.

A monotone decreasing sequence that is bounded below converges.

Now, if I can show that the series is monotone increasing/decreasing and bounded above/below for the different cases, I believe I can use the following.

$$L=\lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} 2+\sqrt{x_n -2}=2 + \sqrt{\lim_{n \to \infty}x_n-2}=2+ \sqrt{L-2}$$

So,

$$L=2+\sqrt{L-2}$$

Which, solving for $L$, yields the solutions $L=2$ and $L=3$. This makes sense according to what I claimed earlier. Once again, I have not found under what conditions the series is divergent. Any help would be appreciated.

Also, any advice as to proving that the series is either monotone increasing and bounded above, or monotone decreasing and bounded below, under the different conditions for $x_0$, would be appreciated.