text:
\section{Definite Integral from 0 to Half Pi of Logarithm of Sine x by Cosine of 2nx}
Tags: Definite Integrals involving Logarithm Function, Definite Integrals involving Sine Function

\begin{theorem}
For $n \in \N_{>0}$:
:$\displaystyle \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \ \d x = -\frac \pi {4 n}$
\end{theorem}

\begin{proof}
First we have:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to 0} \map \ln {\sin x} \sin 2 n x
      | r = \lim_{x \mathop \to 0} \frac {\map \ln {\sin x} } {\csc 2 n x}
      | c = {{Defof|Cosecant}}
}}
{{eqn | r = \lim_{x \mathop \to 0} \frac {\cot x} {- 2 n \cot 2 n x \csc 2 n x}
      | c = [[L'Hôpital's Rule/Corollary 2]]
}}
{{eqn | r = \lim_{x \mathop \to 0} \frac {\cos x} {- 2 n \cos 2 n x} \frac {\sin^2 2 n x} {\sin x}
      | c = {{Defof|Cosecant}}, {{Defof|Cotangent}}
}}
{{eqn | l = \lim_{x \mathop \to 0} \frac {\sin^2 2 n x} {\sin x}
      | r = \lim_{x \mathop \to 0} \frac {4 n \sin 2 n x \cos 2 n x} {\cos x}
      | c = [[L'Hôpital's Rule]]
}}
{{eqn | r = 0
}}
{{end-eqn}}
hence $\displaystyle \lim_{x \mathop \to 0} \map \ln {\sin x} \sin 2 n x = 0$.
Thus:
{{begin-eqn}}
{{eqn | l = \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \ \d x
      | r = \frac 1 {2 n} \int_0^{\pi/2} \map \ln {\sin x} \ \map \d {\sin 2 n x}
      | c = [[Primitive of Cosine Function/Corollary]]
}}
{{eqn | r = \frac 1 {2 n} \paren {\bigintlimits {\map \ln {\sin x} \sin 2 n x} 0 {\pi/2} - \int_0^{\pi/2} \sin n x \ \map \d {\map \ln {\sin x} } }
      | c = [[Integration by Parts]]
}}
{{eqn | r = -\frac 1 {2 n} \int_0^{\pi/2} \sin n x \ \map \d {\map \ln {\sin x} }
      | c = From above
}}
{{eqn | r = -\frac 1 {2 n} \int_0^{\pi/2} \sin 2 n x \frac {\cos x} {\sin x} \ \d x
      | c = [[Primitive of Cotangent Function]]
}}
{{eqn | r = -\frac 1 {2 n} \int_0^{\pi/2} \frac {\sin \paren {2 n + 1} x + \sin \paren {2 n - 1} x} {2 \sin x} \ \d x
      | c = [[Simpson's Formulas/Sine by Cosine]]
}}
{{eqn | r = -\frac 1 {4 n} \int_0^{\pi} \frac {\sin \paren {\paren {2 n + 1} u/2} + \sin \paren {\paren {2 n - 1} u/2} } {2 \sin \paren {u/2} } \ \d u
      | c = [[Integration by Substitution|Substituting]] $u = 2 x$
}}
{{eqn | r = -\frac 1 {4 n} \int_0^{\pi} \paren {\frac 1 2 + \sum_{k \mathop = 1}^n \map \cos {k u} + \frac 1 2 + \sum_{k \mathop = 1}^{n - 1} \map \cos {k u} } \ \d u
      | c = [[Sum of Cosines of Multiples of Angle]]
}}
{{eqn | r = -\frac 1 {4 n} \int_0^{\pi} 1 \ \d u
      | c = All integrals involving $\cos k u$ evaluates to $0$
}}
{{eqn | r = -\frac \pi {4 n}
      | c = [[Integral of Constant/Definite]]
}}
{{end-eqn}}
{{qed}}
[[Category:Definite Integrals involving Logarithm Function]]
[[Category:Definite Integrals involving Sine Function]]
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\end{proof}

config: wiki


TITLE: Give the usual estimator of \theta?
QUESTION [0 upvotes]: I am given a random sample $X_1, \ldots, X_n$ (of size n) from the exponential distribution whose pdf is $f_\theta(x) = \frac{1}{\theta} e^{-\frac{x}{\theta}}$ for $0 < x < \infty.$
I am asked to give the "usual estimator of $\theta.$"
The answer is: "Since $X \sim \mathsf{Exp}(\frac{1}{\theta})$, we know that $E(X) = \theta$ and we can estimate $\theta$ with the empirical sample mean."
Why isn't the answer to this question $\frac{1}{\theta}$ since the exponential distribution has an expectation of $\frac{1}{\theta}$?

REPLY [1 votes]: Other comments or observations.
Assume the following. You have a independent and identically distributed random variables, say $X_1, \ldots, X_n.$ Suppose that these random variables follow some distribution $F$ on $\mathbf{R},$ so that $\mathbf{P}(X_1 \in \mathrm{A}) = \int\limits_\mathrm{A} F(dx).$ Suppose $F$ has a finite first moment, that is $\int\limits_{\mathbf{R}} |x| F(dx) < \infty$ and call $\mu = \int\limits_\mathbf{R} x F(dx)$ the first moment of $F.$ The "method of moments" estimator of $\mu$ is, by definition, $\bar X = \bar X(n):= \frac{1}{n} \sum\limits_{i = 1}^n X_i.$ Simple linearity of the expectation shows at once that $\mathbf{E}(\bar X) = \mu$ and the law of large numbers show that $\lim\limits_{n \to\infty} \bar X = \mu.$ These two properties are deisrable, the first one is called "unbiasdness" and the second one is called "consistency." Therefore, for any distribution whatsoever, the sample mean define a sequence of estimators that are unbiased and the sequence is consistent.
Now, if is often the case that statisticians assume much more than the previous paragraph, explicitly, they assume that $F(dx) = f_\theta(x) dx,$ where $\theta \in \Theta \subset \mathbf{R}^q,$ and $(x, \theta) \to f_\theta(x)$ is smooth in some sense. Under these circumstances, they construct the "likelihood function" which is $L(X_1, \ldots, X_n; \theta) = \prod\limits_{i = 1}^n f_\theta(X_i)$ and maximise this function in $\theta$ for observed $X_1, \ldots, X_n.$ This procedure constructs $\hat \theta = \hat \theta(X_1, \ldots, X_n) \in \Theta$ which is known as the "maximum likelihood estimate" of the parameter $\theta.$ Under general conditions on $(x, \theta) \mapsto f_\theta(x),$ the maximum likelihood estimate is the best estimator, and it is often the case (but not always!) that when $\theta = \int\limits_\mathbf{R} x f_\theta(x) dx$ ($\theta$ is the population mean when the "true model" is $f_\theta$) then $\hat \theta = \bar X.$ An important result regarding estimators of a "parameter" is that estimators should not depend on the parameter, but solely on the sample (they are functions of $(X_1,\ldots, X_n)$ but not of $(X_1, \ldots, X_n, \theta)$).


TITLE: Can a charged particle be trapped by a standing EM wave?
QUESTION [1 upvotes]: Can a charged partìcle be trapped between two extreme points (peaks) of a standing wave so because of ite inertia it goes like alternate current stright up and down and so possibly generates its own field?

REPLY [1 votes]: Though not exactly what you asked, I hope that the case of static electric and magnetic fields mutually perpendicular to each other (the easiest case) provides some useful insight for you.
This case has been discussed in detail by (for example) Griffiths Introduction to Electrodynamics (4th Edition): If you consider placing a charge at rest in an static electric and magnetic fields that are mutually perpendicular to each other what happens? The charge is at rest so there is no magnetic force ($q \bf{v} \times {\bf B}$ with the speed initially zero) and the electric force ($q \bf{E}$) causes the charge to move in the direction of the field. In the attached figure (taken from here) the electric field is in the y direction and the magnetic field is in the z direction.

Note that the charge is pushed along the direction that is perperdicular to the electric force. In fact the motion of the particle is in a direction perpendicular to both the electric and magnetic fields.
As the charge begins to move in the y direction the magnetic force causes it move in a direction perpendicular to its motion, a bit in the x direction and a bit in the negative y direction. Eventually the charge is moving in the opposite direction to the electric field and the electric field acts in a direction to cause the charge to slow down to zero. Then you're back at the beginning: charge in the presence of the electric and magnetic field, though now moving in the x direction. This repeats itself and the resulting motion is called cycloidal motion.
If you know some electromagnetism this won't be too much of a surprise since the Poynting vector, ${\bf S}$, is proportional to ${\bf E} \times {\bf B}$ which is in the x direction.
According to here, the Poynting vector "represents the directional energy flux (the energy transfer per unit area per unit time) of an electromagnetic field".
Your case of interest is a bit more complicated since the electric and magnetic fields, though still mutually perpendicular, now both depend upon both time and space but the general principle holds: the energy flux of the electromagnetic field acts in the direction ${\bf E} \times {\bf B}$.
For the case of a standing wave the motion of the charge will be up and down the resonantor that supports the wave: at times ${\bf E} \times {\bf B}$ will be along the (say) x axis and at other times along the negative x direction.


TITLE: Is $f(x)=\sum_{n\geq 1}\frac{(-x)^n}{n^2+1}$ convex at $x=0$?
QUESTION [2 upvotes]: Let $\sum_{n=1}^{\infty}\frac{(−1)^n}{ n^2+1} x^n$ be the Taylor series of $f(x)$ about $0$. Then, is it that, $f(x)$ is concave up at $x = 0$?

REPLY [0 votes]: $$ f(x)=\sum_{n\geq 1}\frac{(-1)^n x^n}{n^2+1} = \sum_{n\geq 1}(-1)^n x^n \int_{0}^{+\infty}\sin(t)e^{-nt}\,dt = -\int_{0}^{+\infty}\frac{x\sin(t)}{x+e^t}\,dt$$
hence:
$$ f'(x) = -\int_{0}^{+\infty}\frac{e^t \sin(t)}{(x+e^t)^2}\,dt,\qquad f''(x)=2\int_{0}^{+\infty}\frac{e^t\sin(t)}{(x+e^t)^3}\,dt $$
and as soon as $x$ grants that
$$ \forall t\geq 0, \qquad \frac{e^{t+\pi}}{(x+e^{t+\pi})^3}<\frac{e^t}{(x+e^t)^3} $$
we have that the integral defining $f''(x)$ is positive, hence $f(x)$ is convex.
That surely happens if $x\in(-1,1)$, that is the interior of the interval of convergence of the given power series. We may also notice that $g_x(t)=\frac{e^t}{(x+e^t)^3}$ is a decreasing function on $\mathbb{R}^+$ for every $x\geq 0$, hence not only $f$, but also its analytic extension on $[-1,+\infty)$ given by $-\int_{0}^{+\infty}\frac{x\sin(t)}{x+e^t}$ are convex functions on the whole interior of the domain.