Cramer Rao lower bound in Cauchy distribution I need to calculate the Cramer Rao lower bound of variance for the parameter $\theta$ of the distribution $$f(x)=\frac{1}{\pi(1+(x-\theta)^2)}$$

How do I proceed I have calculated $$4 E\frac{(X-\theta)^2}{1+X^2+\theta^2-2X\theta}$$

Can somebody help

Suppose $X\sim C(\theta,1)$, a Cauchy distribution with location $\theta$ and scale unity.

For $\theta\in\mathbb R$, the pdf of $X$ is

$$f_{\theta}(x)=\frac{1}{\pi(1+(x-\theta)^2)}\qquad,\,x\in\mathbb R$$

Clearly, $$\frac{\partial}{\partial\theta}\ln f_{\theta}(x)=-\frac{2(x-\theta)}{1+(x-\theta)^2}$$

Therefore,

$$E_{\theta}\left[\frac{\partial}{\partial\theta}\ln f_{\theta}(X)\right]^2=4E_{\theta}\left[\frac{X-\theta}{1+(X-\theta)^2}\right]^2\tag{1}$$

Now for every $\theta$,

\begin{align}
E_{\theta}\left[\frac{X-\theta}{1+(X-\theta)^2}\right]^2
&=\frac{1}{\pi}\int_{\mathbb R}\left[\frac{x-\theta}{1+(x-\theta)^2}\right]^2\frac{1}{1+(x-\theta)^2}\,\mathrm{d}x
\\&=\frac{1}{\pi}\int_{\mathbb R}\frac{(x-\theta)^2}{(1+(x-\theta)^2)^3}\,\mathrm{d}x\\&=\frac{2}{\pi}\int_0^\infty\frac{t^2}{(1+t^2)^3}\,\mathrm{d}t
\\&=\frac{1}{\pi}\int_0^\infty\frac{\sqrt u}{(1+u)^3}\,\mathrm{d}u
\\&=\frac{1}{\pi}B\left(\frac{3}{2},\frac{3}{2}\right)
\\&=\frac{1}{8}
\end{align}

So from $(1)$, we have the Fisher information

$$I(\theta)=E_{\theta}\left[\frac{\partial}{\partial\theta}\ln f_{\theta}(X)\right]^2=\frac{1}{2}\quad,\forall\,\theta$$

And Cramer-Rao lower bound for $\theta$ is $$\text{CRLB}(\theta)=\frac1{I(\theta)}=2 \quad,\forall\,\theta$$

In case $X_1,X_2,\ldots,X_n$ are i.i.d with pdf $f_{\theta}$, Fisher information in $\mathbf X=(X_1,\ldots,X_n)$ is

$$ I_{\mathbf X}(\theta)=n I(\theta)=\frac n2 \quad,\forall\,\theta$$

The Cramer-Rao bound for $\theta$ is then $\frac2n$ for every $\theta$.

First of all you should notice that there's no suficient estimator for the center of the bell $\theta$. Let's see this.
The likelihood for the Cauchy distribution is
$$L(x;\theta) = \prod _i^n \frac{1}{\pi\left [ 1+(x_i-\theta)^2 \right ]}, $$
and its logarithm is
$$\ln L(x;\theta) = -n \ln \pi -\sum_i^n\ln\left [ 1+(x_i-\theta)^2 \right ].$$
The estimator will maximize the likelihood and if there's a suficient estimator it's derivate can be factoriced i.e.:
$$\frac{\partial L(x;\theta)}{\partial \theta} = A(\theta)\left[t(x)-h(\theta)-b(\theta)\right], $$
where $A(\theta)$ is a function exclusively from the parameter, $t(x)$ is function exclusively of your data, $h(\theta$) is what you want to estimate and $b(\theta)$ is a possible bias.

If you derivate you should notice that the Cauchy distribution can not be factorized, but Cramer-Rao lets you find a bound for the variance, that is
$$ Var(t) \geq \frac{\left(\frac{\partial}{\partial \theta}(h+b)\right)^2}{E\left[(\frac{\partial}{\partial \theta}\ln L)^2\right]},$$
where the equality hold only if $\frac{\partial \ln L}{\partial \theta}$ can be factorized.

The most you can do is calculate the bound but it has no analytic closed solution for $\theta$


correcting a mistake in Spivak Spivak's Calculus on Manifolds asks the reader to prove this (problem 1-8, pp.4-5):

  If there is a basis $x_1, x_2, ..., x_n$ of $\mathbb{R}^n$ and numbers $\lambda_1, \lambda_2, ..., \lambda_n$ such that $T(x_i) = \lambda_i x_i$, $1 \leq i \leq n$, prove that $T$ is angle-preserving iff $\left| \lambda_i \right| = c, 1 \leq i \leq n$.

Here "angle-preserving" means that the linear map $T$ satisfies $$\frac{ \langle x, y \rangle}{\|x\| \|y\|} = \frac{ \langle T(x), T(y) \rangle}{\|T(x)\| \|T(y)\|},$$
and that $T$ is injective.

My first problem with this question is that the claim is false. Taking $n = 2$, $x_1 = (1, 0)$, $x_2 = (1,1)$, $T(x_1) = -x_1$, $T(x_2) = x_2$, and setting $x = x_1$, $y = x_1 + x_2$, the expression in the RHS above evaluates to $0$, while the expression in the LHS evaluates to $\frac{1}{\sqrt{2}}$.

My second, bigger problem is that I'm not really understanding what's going on. An earlier part of the problem had me show that norm-preserving matrices are angle-preserving; this I'm not sure I get. Thus, I'm not sure what true "version" of this statement the author had in mind (was he trying to get a converse?) and I don't know what to do.

Here's my guess:

Looking at some transformations in $\mathbb{R}^2$ (just drawing them), it looks like some of them could be "flip the sign of a basis vector" IF it's an orthogonal basis. However, I can't seem to recover the usual "rotation through an angle $\theta$" transformation this way, so I'm not sure that requiring the basis be orthogonal makes the statement true.

Also, I'm not sure how to take the inner product of vectors which aren't in standard coordinates. Or am I missing something here?

You are correct. This is one of the few errors in Spivak's Calculus on Manifolds. For this particular exercise, see the following questions:

Question about Angle-Preserving Operators
Action of angle-preserving linear transformation on basis vectors

copper.hat's answer to the first question cited in the above used the same counterexample as yours.

For your second question, I don't think there is any conflict. Spivak simply meant that if $T$ is a norm-preserving map, then it automatically enjoys the property of being angle-preserving. He did not say that the converse is true. In other words, norm preservation is a stronger condition than angle preservation. In fact, $T$ is norm-preserving map if and only if its matrix w.r.t. the standard basis is a real orthogonal matrix $Q$, and $T$ is angle-preserving if and only if its matrix w.r.t. the standard basis is $\lambda Q$ for some scalar $\lambda>0$ and some real orthogonal matrix $Q$ (for a proof, see the aforementioned answer by copper.hat).

I noticed this too. If you ignore the absolute value on lambda, it's a true statement which can be easily proven.


When does $(x^x)^x=x^{(x^x)}$ in Real numbers? I have tried to solve this equation:$(x^x)^x=x^{(x^x)}$ in real numbers I

got only $x=1,x=-1,x=2$ , are there others solutions ?

Note: $x$ is real number .

Thank you for your help .

Just a sketch for $x>0$:

Let $y=x^x$ and rewrite $x^y=y^x$ by taking $log$'s as $\log (x)/x = \log(y)/y$.

For each $x \in (1, e)$ there is a unique $y(x)$ such that $f(x) = f(y(x))$.

As a function of $y(x)$, you can show that it is increasing. So we are looking for solutions of $y(x) = x^x$. Because $g(x) = y(x)-x^x$ is increasing, it can have only one root. So there is only one positive $x$ that satisfies the equation with $x\neq 1$.

EDIT: For integer $x < 0$ non-integer, things get complicated because we will involve complex numbers and raising numbers to complex powers.

You have $(x^x)^x = x^{x^2}$. Assuming $x \not \in \{-1,1,1\}$,
$$x^{x^2} = x^{x^x}$$
happens if and only if the exponents agree, i.e. $x^2 = x^x$ which under our assumptions is equivalent to $x=2$.


Functional derivatives of Fourier transform I've tried to calculate this as well as been looking around online and I can't seem to find the calculation of this.   If $\mathfrak{F}$ is the Fourier transform:
$$
f\mapsto \int_{-\infty}^{\infty} f(x)e^{ix}dx,
$$
then what would the functional derivative
$$
\frac{\delta \mathfrak{F}}{\delta f}
$$
be?

The functional derivative is defined as the linear mapping
$$\left\langle \frac{\delta}{\delta f(x)} F[f](\xi), \phi(x) \right\rangle := \left. \frac{d}{d\lambda} F[f+\lambda\phi](\xi) \right|_{\lambda=0}$$

For $\mathfrak F[f](\xi) := \int_{-\infty}^{\infty} f(x)e^{ix\xi} \, dx$ we get
$$\begin{align}
\left. \frac{d}{d\lambda} \int_{-\infty}^{\infty} (f+\lambda\phi)(x)e^{ix\xi} \, dx \right|_{\lambda=0}
& = \left. \frac{d}{d\lambda} \int_{-\infty}^{\infty} \left(f(x)+\lambda\phi(x)\right)e^{ix\xi} \, dx \right|_{\lambda=0} \\
& = \left. \frac{d}{d\lambda} \left( \int_{-\infty}^{\infty} f(x) e^{ix\xi} \, dx + \lambda \int_{-\infty}^{\infty} \phi(x) e^{ix\xi} \, dx \right) \right|_{\lambda=0} \\
& = \left. \int_{-\infty}^{\infty} \phi(x) e^{ix\xi} \, dx \right|_{\lambda=0} \\
& = \int_{-\infty}^{\infty} \phi(x) e^{ix\xi} \, dx \\
& = \left\langle e^{ix\xi}, \phi(x) \right\rangle
\end{align}$$
Thus,
$$\frac{\delta}{\delta f(x)} \mathfrak F[f](\xi) = e^{ix\xi}$$

Are you looking for derivative of Fourier transform ?

considering $\omega$ insted of $f$ we can write accroding to the definition

  $$\begin{align}\frac{d \mathcal{F}(\omega)}{d\omega}&=\int_{-\infty}^{+\infty}f(t)\frac{d}{d\omega}\cdot e^{i\omega t}\,dt\\&=-i\int_{-\infty}^{+\infty}t.f(t)e^{i\omega t}\,dt\\&=-i \cdot\mathcal{IFT}\{t \,.f(t)\}\end{align}$$

And the reverse process can be

  $$\mathcal{FT}\left\{\frac{d f^n(t)}{d^nt}\right\}=(i\omega)^n.F(\omega)$$

Please let me know if somehow i can improve my answer , Thanks !


E-mails per day problem This is an exam problem from probability and statistics that I failed, and still cannot figure out. Thanks in advance.
The number of e-mails that a person receives in a work day (monday to friday) can be modeled with the poisson distribution with an average of 6 e-mails a day. The number of e-mails a person receives during the weekend can also be modeled with the poisson distribution with an average of 3 e-mails a day.
a) What is the probability that person does not receive any e-mails, if we do not know which day it is?
b) A day is chosen at random from the week. If that person didn't receive a an e-mail on that day, what is the probability that day is a work day?

Let $p_1=\exp(-6)$ be the probability of receiving $0$ emails on a week-day and $p_2=\exp(-3)$ be the probability of receiving $0$ emails on a weekend (from the definition of a Poisson random variable). If the day of the week is not known (but all days have the same probability), then the probability of receiving $0$ emails is $\tfrac{5}{7}p_1+\tfrac{2}{7}p_2$.
The probability that today is a workday AND no emails were received is $\tfrac{5}{7}p_1$ so the probability that today is a workday given that no emails were received is $\tfrac{\tfrac{5}{7}p_1}{\tfrac{5}{7}p_1+\tfrac{2}{7}p_2}$.


Deducing $((\neg a \to \neg b) \to ((\neg a \to b) \to b)))$ from axioms I have seen many questions here, using a different set of axioms than mine.

Here is mine :

$$1) (a \to (b \to a))$$
$$2) ((a \to (b \to c)) \to ((a \to b) \to (a \to c)))$$
$$3) ((\neg b \to \neg a) \to (a \to b))$$

Is it possible to deduce the following from these: $((a \to b) \to ((\neg a \to b) \to b))$.

I couldn't find a proof myself or online.

Thank you very much!!!

It helps to know:

Two logical expressions are equivalent iff they have the same truth table
$(a \rightarrow b) \leftrightarrow (\neg a \vee b) $ (One can proof that using the truth table)
$(a \rightarrow b) \leftrightarrow (\neg b \rightarrow \neg a) $ (One can proof that using the truth table)
$(a \vee \neg a) =1$ (Tautology)
$\neg \neg a = a$ (Double negation)
$1$ is the neutral element of the logical conjuction: $1\wedge a = a$

Here is my approach to deduce the given expression from elementary propositions:

Let $a$ and $b$ be two propositions, with
$$a \rightarrow b$$
$$ \leftrightarrow \neg a \vee b$$
$$\leftrightarrow (\neg a \wedge b)\vee(\neg a \vee b)$$
$$\leftrightarrow (\neg (a \vee \neg b))\vee(\neg a \vee b)$$
$$\leftrightarrow (a \vee \neg b)\rightarrow (\neg a \vee b)$$
$$\leftrightarrow (a \vee \neg b)\rightarrow (b \vee \neg a)$$
$$\leftrightarrow (a \vee \neg b)\rightarrow (b \vee \neg a) \wedge 1 $$
$$\leftrightarrow (a \vee \neg b)\rightarrow (b \vee \neg a) \wedge (b \vee \neg b) $$
$$\leftrightarrow (a \vee \neg b)\rightarrow b \vee( \neg a \wedge \neg b) $$
$$\leftrightarrow (a \vee \neg b)\rightarrow b \vee \neg(a \vee b) $$
$$\leftrightarrow (a \vee \neg b)\rightarrow b \vee \neg(\neg \neg a \vee b) $$
$$\leftrightarrow (a \vee \neg b)\rightarrow b \vee \neg( \neg a \rightarrow b) $$
$$\leftrightarrow (a \vee \neg b)\rightarrow \neg( \neg a \rightarrow b) \vee b $$
$$\leftrightarrow (a \vee \neg b)\rightarrow (( \neg a \rightarrow b) \rightarrow b) $$
$$\leftrightarrow (\neg b \vee a)\rightarrow (( \neg a \rightarrow b) \rightarrow b) $$
$$\leftrightarrow (b \rightarrow a)\rightarrow (( \neg a \rightarrow b) \rightarrow b) $$
$$\leftrightarrow (\neg a \rightarrow \neg b)\rightarrow (( \neg a \rightarrow b) \rightarrow b)$$


Find nth derivative of $\frac{x^{n}}{(1-x)^{2}}$, please? I need to find the nth derivative of $\frac{x^{n}}{(1-x)^{2}}$ for $0<x<1$

So far, I tried the same method used for $\frac{x^{n}}{1-x}$ and here's what I got:
\begin{equation}
\frac{x^{n}}{(1-x)^{2}}=x^{n}(1+2x+3x^{2}+4x^{3}+....)=(x^{n}+2x^{n+1}+3x^{n+2}+...)
\end{equation}
Take nth derivative to get:
\begin{align}
\frac{\partial^n }{\partial x^n} \frac{x^{n}}{(1-x)^{2}} \notag\\ =
& (n!+2(n+1)!x+3\frac{(n+2)!}{2!}x^{2}+...) \notag\\ =
& n!(1+2(n+1)x+3\frac{(n+2)(n+1)}{2!}x^{2}+4\frac{(n+3)(n+2)(n+1)}{3!}x^{3}+..)
\end{align}
Next would be finding a a function whose taylor expansion is the series in the parenthesis but I couldn't think of one. Any ideas for a function or for another method to do this? Thanks!

Write $x = 1-t$, so you're looking at $$(-1)^n \dfrac{d^n}{dt^n} \dfrac{(1-t)^n}{t^2} = \dfrac{d^n}{dt^n}  \sum_{j=0}^n {n \choose j} (-1)^{n+j}   t^{j-2}$$
The terms with $j \ge 2$ all vanish, and you're left with
$$ \dfrac{d^n}{dt^n} \left((-1)^n (t^{-2} - n t^{-1})\right)$$

From there it's easy.  You can write the answer as
$$\dfrac{n!(nx+1)}{(1-x)^{n+2}}$$

You could write
$$\frac{\partial^n}{\partial x^n}x^n(1-x)^{-2}=\sum_{k=0}^n\binom{n}{k}\left(\frac{\partial^k}{\partial x^k}x^n\right)\left(\frac{\partial^{n-k}}{\partial x^{n-k}}(1-x)^{-2}\right)$$ and proceed from there. You'd have it as a finite sum.


Find countable set that is not in the additive, multiplicative closure of a finite set. I would like to find a sequence $a_0, a_1, a_2, \dots$ of elements of $[0,1]$, such that there is no finite set $B = \{b_0, \dots , b_n\}$ of elements of $[0,1]$ such that every $a_n$ can be written as a finite combination (using multiplication and summation) of elements of $B$.

For example, I think that $a_n = \frac{1}{n}$ should work. At least it is not possible to find a finite set of rationals from which you can make every number $\frac{1}{n}$ just by multiplication and addition. However, it becomes a bit more difficult if you also consider irrationals.

Although, I think it should be possible to solve it in an easier way, I also tried something with field extensions (and the degree). However, here it becomes difficult if $B$ consists of transcendental numbers.

We know that $\mathrm{tr\ deg}(\mathbb R/\mathbb Q)$ is infinite. Let $A=\{a_n: n\geq1\}\subset\mathbb R$ be an infinite algebraically independent subset over $\mathbb Q$. Changing $a_n$ by $a_n-\lfloor a_n\rfloor$ we can suppose that each $a_n\in[0,1)$. For any subset $B\subseteq [0,1]$, the ring closure $\mathbb Z[B]$ of $B$ is contained in $\mathbb Q(B)$, the subfield of $\mathbb R$ generated by $B$ over $\mathbb Q$, so if $\mathbb Z[B]$ contains $A$, then $\mathbb Q(B)$ also contains $A$. Consequently, by well-known properties of the transcendence degree we have the inequalities $|B|\geq\mathrm{tr\ deg}\bigl(\mathbb Q(B)/\mathbb Q\bigr)\geq |A|$, which shows that $B$ must be infinite.

Assuming you only multiply by naturals, any set that has $0$ as a limit point will work.  It will have an element smaller than any element of $B$.


How to plot an algebraic curve? Is there an algorithm to plot an algebraic curve $f(x, y)=0$ ?
If the degree of $f$ is 4 or less, then there is a formula for the solution, so I think I can plot the curve (I don't know there is a better algorithm or not). What about the general case?

If you use Wolfram Alpha, it is quite easy : you just as for the contour "line" where $f(x,y)=0$.
I added a small example here.

Yes there are many ways to do this. Here is one particular way:

Guess one root.
Apply some numerical optimization to find your way towards one true root from your guess.
When close enough to a root, iterate the following:

Take a small steps in the tangent direction.
Take steps in the gradient direction.

If your surface is in many dimensions you may want to somehow systematically generate some kind of a local ON gradient-tangent basis.
One good thing about this approach is that you get to choose your guess freely, you can pick for example a point close to the center of the screen or whichever area you want to visualize.


Find the region in the w-plane to which the line y = 1 is transformed by $\frac{1}{z}$ I tried to do the following:

$$w=\frac{1}{z}=\frac{x-iy}{x^2+y^2}$$

$\implies u = \frac{x}{x^2+y^2} and\space v = \frac{-y}{x^2+y^2}$

$\color{green}{need\space to\space transform\space the\space line\space y = 1\space into\space the\space w-plane}$

-I am not too sure about how to go about it, do i just assume that if we are dealing with $y$ then $x$ must be zero and so:

$\implies u = \frac{0}{0+1}=0 \space and\space v = -1$

-Thanks.

EDIT:

my notes says that lines should transform to circles, and circles should transform to lines, given what i have done, i have no idea how that can be possible.

We are not just dealing with $y$, and you don't only consider $x=0$.  That would only tell you where one point goes, namely $i$.   The equation $y=1$ describes an entire line, the horizontal line where the imaginary part is $1$.  It includes the point $52+i$, for example.  The problem is to determine the image under the inversion map of the set of all points of the form $x+i$.

The line $y=1$ is symmetric about the $y$-axis and $z\mapsto \frac1z$ preserves this symmetry, so $\frac1i=-i$ will lie on a vertical axis of symmetry for the image and will also have the greatest modulus of the points in the image (because $|1/z|=1/|z|$).  As $x\to\infty$ along the line $y=1$, $\frac1z\to0$, so the image will approach the point $0$.  The image will be in the lower half plane ($v<0$).  Even if you don't have yet a proof for why the image should be a circle, if you have heard of this fact, then the above gives you enough to guess what the image is, namely a circle with a diameter from $-i$ to $0$, but without $0$ included.

That circle would have equation $u^2+(v+\frac12)^2 =\frac14$, so let's check.  You have from your formulas for $u$ and $v$ that $u=\frac{x}{x^2+1}$ and $v=\frac{-1}{x^2+1}$.  Therefore
$$\begin{align*}u^2+\left(v+\frac12\right)^2&=\frac{x^2}{(x^2+1)^2}+\frac{(\frac12(x^2-1))^2}{(x^2+1)^2}\\
&=\frac{4x^2}{4(x^2+1)^2}+\frac{(x^2-1)^2}{4(x^2+1)^2}\\
&=\frac{4x^2+x^4-2x^2+1}{4(x^2+1)^2}\\
&=\frac{(x^2+1)^2}{4(x^2+1)^2}\\
&=\frac14.
\end{align*}$$

The term "region" is usually reserved for open subsets of $\mathbb C$, not for curves like circles and lines.  The line is mapped to the circle, but the half-plane regions on either side of the line get mapped to the regions inside and outside the circle.

The line $y=1$ is mapped into $$\frac {x-i}{x^2+1} $$  by substituting $y=1$ in your original expression.


Critical Numbers And Extreme Values Of A Piecewise Function? Please determine the critical numbers and extreme values of the :
$$
f(x) =
\begin{cases}
2-2x-x^2, & -2\le x \le 0 \\
|x-2|, & 0<x<3 \\
\frac {1}{3}(x-2)^3, & 3\le x \le 4
\end{cases}
$$
I even know that the derivative of $f(x)$ is$$ f(x) =
\begin{cases}
-2-2x, & -2\le x < 0 \\
\frac {x-2}{|x-2|}, & 0<x<3 \\
(x-2)^2, & 3< x \le 4
\end{cases}
$$
I think $x=0$ is the only critical number!

Thanks in advance!

Hint:

First notice that the function is continuous except at $x=3$.

Notice that the derivative vanishes at $x=-1$, that the left derivative does not equal the right derivative at $x=0$ and $x=2$, and since the function is not continuous at $x=3$, the derivative cannot exist there.

Further Hint:

Now, simply look at the function between these points:

a) at $x=-1$, there is a local maximum: $f(-2)=2$, $f(-1)=3$, and $f(0)=2$.

b) at $x=0$, the left derivative is $-2$ and the right derivative is $-1$.

c) at $x=2$, the left derivative is $-1$ and the right derivative is $+1$.

d) at $x=3$, the function jumps from $1$ to $\frac13$.

You should consider

$f(-2)=2\qquad$endpoint
$\color{#C00000}{f(-1)=3}\qquad$derivative is $0$
$\color{#C00000}{f(2)=0}\ \ \ \qquad$left derivative has different sign than right derivative
$f(3-)=1\qquad$left side of jump discontinuity
$f(3)=\frac13\ \,\qquad$right side of jump discontinuity
$f(4)=\frac83\ \,\qquad$endpoint

$\hspace{3cm}$

It seems like there is a relative minimum from the right of x=3.  From the left, the function isn't defined at x=3, so we can't have a relative maximum.


Evaluating indefinite integrals. If $$\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\dfrac{d}{b}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$ where $b,d$ are relatively prime find $b+d$.

My solution:

$$\displaystyle\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\int\dfrac{1}{x}\sqrt{\dfrac{x^3}{a^3-x^3}}dx$$
Then let $x^3=t\implies 3x^2\cdot dx=dt$

$$\implies\dfrac{1}{3}\int\dfrac{1}{x}\cdot\dfrac{3x^2}{x^2}\sqrt{\dfrac{x^3}{a^3-x^3}}dx\\=\dfrac{1}{3}\int\dfrac{1}{t}\sqrt{\dfrac{t}{a^3-t}}dt\\ =\dfrac{1}{3}\int{\dfrac{1}{\sqrt{ta^3-t^2}}}dt\\=\dfrac{1}{3}\int{\dfrac{1}{\sqrt{\frac{a^6}{4}-\bigg(t-\frac{a^3}{2}\bigg)^2}}}dt\\=\dfrac{1}{3}\sin^{-1}\bigg(\dfrac{2t-a^3}{a^3}\bigg)+C$$ $$=\dfrac{1}{3}\sin^{-1}\bigg[2\bigg(\dfrac{x^{3/2}}{a^{3/2}}\bigg)^2-1\bigg]+C\tag{1}$$

This is not of desired form but when I drew graph of: $\color{grey}{1.57+\sin^{-1}\left(2x^2-1\right)}$ and $\color{green}{2\sin^{-1}x}$, they coincides for $\color{red}{x>0}$. Green graph is for $\ 2\sin^{-1}x\ $.

So from $(1)$ $$\dfrac{1}{3}\sin^{-1}\bigg[2\bigg(\dfrac{x^{3/2}}{a^{3/2}}\bigg)^2-1\bigg]+C=\dfrac{2}{3}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$

If I didn't made any mistake then please help me getting $1.57+\sin^{-1}\left(2x^2-1\right)\equiv\ 2\sin^{-1}x\ $ for $x>0$ without any graphs.

Also, I found that if I substitute $x^3=a^3\sin^{2}x$ then we reach the desired form of the result directly. But I'm an enthusiast to continue my previous solution. Please help.

Thanks!

Since you already have the form you must get the answer into then why not get some hints from it.

$$\int \frac {\sqrt x}{\sqrt {a^3-x^3}} dx=\int \frac {\sqrt x}{a^{\frac 32}\sqrt {1-\left(\frac {x^{3/2}}{a^{3/2}}\right)^2 }} dx$$

Let $$\frac {x^{3/2}}{a^{3/2}}=t$$ hence $$dt=\frac 32\cdot \frac {\sqrt x}{a^{3/2}}dx$$

Hence the integral changes to $$\int \frac 23 \cdot \frac {1}{\sqrt {1-t^2}} dt $$

Hence the answer $b=3,d=2$ follows

Hint: As suggested in the comments, differentiate both sides of the equation.

Edit: If you do this, you obtain $$\left(\frac{x}{a^3-x^3}\right)^{1/2}=\frac db \frac {1}{\sqrt{1-[\left(\frac xa\right)^{3/2}]^2}}\cdot \left(\frac1 a\right)^{3/2}\frac32 x^{1/2}.$$ Hopefully, you can rearrange this to obtain $$\frac{x^{1/2}}{(a^3-x^3)^{1/2}}=\frac{3d}{2b}\cdot \frac{x^{1/2}}{(a^3-x^3)^{1/2}},$$ which implies $$\frac db=\frac23,$$ so that $d=2$ and $b=3$ since $d$ and $b$ are coprime.


Generalized way of solving this types of equations $x^3 +y^4 =z^5$ $$x^3 +y^4 =z^5$$
How can I solve this equation.I only know trial and error method, but it's not a generalized way.
Please tell me a generic way to solve this type of equation.

The equation $x^3 + y^4 = z^5$ has infinitely many solutions in positive integers. An infinite family of solutions is generated by
$$x = a(a^3 + b^4)^{8k},\qquad y=b(a^3 + b^4)^{6k},\qquad z=(a^3 + b^4)^{5k}.$$
There are probably other solutions; I doubt that an exhaustive list of solutions is known.

Beal's conjecture would imply that the equation has no relatively prime integer solutions. But this conjecture remains unproved, and there is a $1 million prize for a proof or counterexample.


Related Trigonometric Rates A port and a radar station are $3 \text{ mi}$ apart on a straight shore running east and west. A ship leaves the port at noon traveling at a rate of $13 \frac{\text{mi}}{\text{hr}}$. Find the rate of change of the tracking angle $\theta$ between the shore and the line between the radar station and the ship at $12:30 \text{ PM}$, assuming the ship maintains its speed and course.

The following graphic represents this scenario:

$\hspace{1.95in}$

Let $\dfrac{ds}{dt} = 13 \frac{\text{mi}}{\text{hr}}$.

The ship is given to move from $A \to C$ at $\dfrac{ds}{dt}$ departing at noon and arriving at $12:30 \text{ PM}$ such that:

$$s = \dfrac{13}{2}$$

Let the segment $\overline{CB} = x$.

By the law of cosines:
$$x = \sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}$$

Then apply the law of sines to find that:
$$\dfrac{\sin{\left( \frac{\pi}{4} \right)}}{x} = \dfrac{\sin{\theta}}{s}$$

Through substitution and manipulation of the equality its found that:
$$\dfrac{s \cdot \sin{\left( \frac{\pi}{4} \right)}}{\sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}} = \sin{\theta}$$

Take the derivative with respect to time of both sides to find that:
$$
\\ \begin{align*}
\\ \frac{d}{dt}\left[ \arcsin \left( \dfrac{s \cdot \sin{\left( \frac{\pi}{4} \right)}}{\sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}} \right) \right] &= \frac{d}{dt}\left[ \theta \right]
\\
\\ \dfrac{\dfrac{1}{\sqrt{2}\sqrt{s^2-3\sqrt{2}s+9}} - \dfrac{s^2}{\sqrt{2}{(s^2-3\sqrt{2}s+9)}^{3/2}}}{\sqrt{1-\dfrac{s^2}{2(s^2-3\sqrt{2}s+9)}}}\cdot\frac{ds}{dt} &= \frac{d\theta}{dt}
\\ \end{align*}
$$

Which Wolfram evaluates to be $-4.5193 \frac{\text{rad}}{\text{hr}}$ as seen here.

This doesn't appear to be correct, however.

I'm uncertain of the error I've made, so if you could explicitly state at what point I've made an error in my approach, that would be very helpful. All that I'm certain of is that this is not the expected answer, but my approach appears proper to me (though it is clearly not).

let the distance between the ship and the radar station be $y.$ so we have $$y^2 = s^2 + 3^2 - 2 \times 3 \times s\cos 45^\circ  = s^2 + 9 -3\sqrt 2s$$ differencing gives $$2y\frac{dy}{dt} = 2s\frac{ds}{st} - 3\sqrt 2 \frac{ds}{dt} \tag 1$$ now we put the data $$\frac{ds}{dt} = 13, s = 6.5, y = \sqrt{6.5^2 + 9 - 3\sqrt 2 \times 6.5}= 4.865$$

putting all these in $(1),$ we get $$2\times 4.865\frac{dy}{dt}=13\times 13-3\sqrt 2 \times 13=113.845\to \frac{dy}{dt} = 11.700\, \text{ mile/hour} $$

$\bf edit:$ if you need $\frac{d\theta}{dt},$  we can use $$\frac{s}{\sin \theta} = \frac{y}{\sin 45^\circ}\to\sin 45^\circ \frac{ds}{dt} = y\cos \theta \frac{d\theta}{dt} + \sin \theta \frac{dy}{dt} \tag 2$$  we will first compute $\theta.$
we have $$\cos \theta = \frac{-\frac{6.5}{\sqrt 2} + 3}{y} = -0.328, \sin \theta = \frac{6.5}{\sqrt 2 y}=0.944$$
now subbing $$13\frac{\sqrt 2}2=4.865\times (-0.328)\frac{d\theta}{dt}+0.944\times 11.700 $$

i get $$\frac{d\theta}{dt} = 1.165\, radian /hour = 66.787^\circ/hour$$


Is $\sin \infty$ an indeterminate form? I had a long discussion in chat over what seems like a simple question:

  Is $\sin \infty$ an indeterminate form?

What do you think? :)

I'm labelling this as (soft-question) to be safe, but it should have a clear answer depending on the definition of indeterminate form used.
The term "indeterminate form" is seldom used in post-calculus mathematics, but I believe it has one or more accepted definitions, which are either informal or formal. So any answer which takes a standard definition and argues the case would be interesting to me.

How to make a good answer:

State the definition of indeterminate form, either from an online or textbook source, or a definition you came up with on your own.
Determine, using your stated definition, whether $\sin \infty$ is indeterminate or not.

EDIT: What do I mean by $\sin (\infty)$?

It's not a well-defined expression, but neither are any of the other indeterminate forms: $\frac{0}{0}$ doesn't exist, $1^\infty$, doesn't exist, and so on. So the question is whether this expression -- which is not well-defined, just like any other indeterminate form -- is an indeterminate form.

Yes.  An indeterminate form is an expression such that if you replace the constants appearing in the expression with sequences which approach those values, then the limit of the expression is not uniquely determined.  In the case of $\sin(\infty)$, if we replace $\infty$ by the sequence $a_n=\pi n$ we get a limit $$\lim_{n\to\infty}\sin(\pi n)=0.$$  If we instead take the sequence $b_n=2\pi n+\pi/2$ we get $$\lim_{n\to\infty}\sin(2\pi n+\pi/2)=1.$$  If we instead take the sequence $c_n=n$ then we get a limit $$\lim_{n\to\infty}\sin(n)$$ which does not exist.

The point here is that if you are trying to evaluate a limit of the form $$\lim \sin(\text{something})$$ where you know that the "something" is approaching $\infty$, you cannot tell what the answer is from just this information--the limit could be any number between $-1$ and $1$, or it could not exist.  This is just like the more familiar indeterminate forms like $\frac{0}{0}$ which are taught in calculus: if you have a limit $$\lim\frac{\text{something}}{\text{something else}}$$ where both "something" and "something else" approach $0$, that is not enough information to determine the limit.

It is perhaps worth remarking that with this definition, an indeterminate form is equivalently an expression such that if you replace the constants with variables ranging over the real numbers, then the limit as these variables jointly approach their values does not exist (where "exist" includes the possibility of being $\pm\infty$).  In other words, to say that $\sin(\infty)$ is an indeterminate form really just means that $\lim\limits_{x\to\infty}\sin(x)$ does not exist (where here $x$ approaches $\infty$ in the real numbers, in contrast to the sequential limits we had before).  Indeed, if you can have multiple different sequential limits, then the limit over the real numbers cannot exist.  Conversely, if the limit over the real numbers does not exist, then by compactness of $[-\infty,\infty]$ the limit must still accumulate somewhere in $[-\infty,\infty]$ and so must accumulate at two different values, and then you can choose sequences for which the limit approaches two different values in $[-\infty,\infty]$.

since oo is not a real number sin(oo) is meaningless. If you meane the limit for x->oo , it is indeterminate, since the value can not be determined.


Reference: In every free homotopy class is a unique minimizing closed geodesic Does anyone know a reference for the following result:

Let $M$ be a compact hyperbolic manifold/manifold with strict negative curvature . Then in every non-trivial free homotopy class of $M$ there exists a unique minimizing closed geodesic.

The existence is proofed for example in Do Carmo's Riemannian Geometry, for which the negative curvature is not needed.
But for the uniquness I've not yet found a reference.
Can anyone help?

This question has already been answered, but I just wanted to point out to others who may arrive at this question while reading do Carmo that this result in fact follows from the results in Chapter 12.

Theorem 2.2 tells you that if you that every compact manifold has some closed geodesic in a given free homotopy class.

Proposition 2.6 tells you that given a covering transformation, $\alpha$, of the universal cover $\tilde{M}$ of $M$, with $M$ compact, there is a geodesic which is fixed by $\alpha$. We can associate covering transformations to free homotopy classes in $M$ in the usual way. If you go through the proof of this proposition, you'll see that without changing anything, the proof tells you that in fact for every closed geodesic in this free homotopy class, the lift to $\tilde{M}$ is preserved by $\alpha$ (this is not what the proposition says, but it follows from the proof).

Lemma 3.3 tells you that if $K < 0$ on $M$, then given any covering transformation $\alpha$ of $\tilde{M}$, there is at most one geodesic fixed by $\alpha$.

Put this altogether and you get that if $K < 0$ and $M$ is compact, then there is exactly one closed geodesic in a given free homotopy class (since free homotopy classes correspond to covering transformations and invariant geodesics in $\tilde{M}$ correspond to closed geodesics in $M$ in the corresponding free homotopy class).

I think the following works:

Nonpositive curvature implies that the distance function is convex. So if you have two minimizing geodesics then you actually have a 1-parameter family of minimizing geodesics. Then look at the Jacobi field of such a 1-parameter family, say with one endpoint fixed. In the case of nonpositive curvature the Jacobi equation will yield a contradiction.


What is the length of a point on the real number line? Since an interval is made up of an infinite number of points, I am considering the relation of the length of an interval and the length of a point, this lead me to ask what is the length of a point on the real number line ?

The nested interval theorem made me feel the length of a point should be $0$ because of $\lim_{{n\to\infty}}(b_{n}\!-\!a_{n})=0$

While if the length of a point  on the real number line is $0$, then I get a contradiction : Supposing we remove the point $0$ on the real number line, then there is a gap there, and the width of the gap is $0$ since the length of a point  on the real number line is $0$, however I think the width of the gap being $0$ is equivalent to there being no such gap on the real number line, so this leads to a contradiction.

What's wrong here? Does a point on the real number line have a width? If so, what is the length of a point on the real number line? Infinitesimal?

Much like many other 'paradoxes' you'll find that carefully formalizing the problem makes it go away. Whatever counter intuitive phenomena persist is not paradoxical, but simply counter intuitive.

So, what do you mean by length of a point? Well, whatever it is it seems reasonable to define it to have the following properties: 1) the length of an interval $[a,b]$ is $b-a$; and 2) if the point $p$ is contained in an interval, than the length of the point is $\le $ the length of the interval. We further assume all lengths are measured by non-negative real numbers, so there are no infinitesimals around at all.

Now, from these it follows that the length of any point is $0$. So, any notion of length of a point conforming to the above must assign length $0$ to each point. That is a theorem.

Consequently, the situation you describe, where removing a point having length $0$ results in a broken line is a correct description. It is not a paradox but rather a counter intuitive situation. Well, what shall we do with it? You can examine our assumptions and change them, or hone your intuition. In this case, I suggest the latter.

Yes your are absolutely correct.I argue that length of a point is infinitesimal,so is breadth of a line segment.An interval(a,b) is a disjoint union of points.If point has a length 0, 0+0+.......infinity =b-a-finite is impossible.So it should be infinitesimal.Same argument holds for breadth of a line segment.A rectangle of sides a and b whose area is ab is formed by placing a line segment of length a arbitrarily close to one another.If the line segment has breadth 0 then 0+0+......=ab-finite is impossible. similarly we can prove that thickness of a plane rectangular area is infinitesimal. My final conclusion is that length,breadth, thickness of a point is infinitesimal.point has no length is wrong.it is arbitrarily small  . I think this proof clarifies your doubt


Determine if $y = x^2$ is injective I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).

However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.

Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.

This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($\sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.

However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.

Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?

The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x \neq y$.  The contrapositive fails as well because you have $x \neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.

I think that the syntax of the definition from your class is the point of confusion. The class definition depends on functions being defined as
$$
f(x)=(\text{expression of }x)
$$ rather than as
$$
y=(\text{expression of }x).
$$ Consequently, the "$y$" in "$f(y)$" is just some dummy variable that gets input into $f$ and is unrelated to $x$. So, for your example, $f(x)=x^2$, $f(x)= 4$ for $x=2$ and $f(y)=f(4)=16$ for $y=4$. Again, note that $y$ is unrelated to $x$.

Finally, in order to show that $f(x)=x^2$ isn't injective, you can start with the definition or with its contrapositive, as you stated:

Starting with the definition,
$$
f(-2)=f(2)\:\text{ but }-2\:\text{ isn't equal to }2.$$
Starting with the contrapositive, let's consider $x=2$ and $y=-2$: then
$$
x\neq y\:\text{ but }\:f(x)=f(y)=4.
$$


Formula relating Euler characteristics $\chi(A)$, $\chi(X)$, $\chi(Y)$, $\chi(Y \cup_f X)$ when $X$ and $Y$ are finite. This is a followup to my question here.

Let $A$ be the subcomplex of a CW complex $X$, let $Y$ be a CW complex, let $f: A \to Y$ be a cellular map, and let $Y \cup_f X$ be the pushout of $f$ and the inclusion $A \to X$. Is there a formula relating the Euler characteristics $\chi(A)$, $\chi(X)$, $\chi(Y)$, and $\chi(Y \cup_f X)$ when $X$ and $Y$ are finite?

Indeed, there is. Let $Z=Y \cup_f X$ be the pushout of the diagram $X\leftarrow A\rightarrow Y$. I claim that $$\chi(Z)=\chi(X)+\chi(Y)-\chi(A).$$

To see this, let us first recall the following general fact

  For any chain complex $C_*$ it holds that $$\sum_n (-1)^n \text{rank }C_n = \sum_n (-1)^n \text{rank }H_n(C_*).$$

This is rather simple to prove and you can read about it, for example, in Hatcher's proof of Theorem 2.44.

Now consider the Mayer-Vietoris sequence corresponding to the above pushout:

$$\cdots \underbrace{H_1(Z)}_{C_3}\to \underbrace{H_0(A)}_{C_2}\to \underbrace{H_0(X)\oplus H_0(Y)}_{C_1}\to \underbrace{H_0(Z)}_{C_0}\to 0 $$

Any long exact sequence may trivially be considered a chain complex, so let's define $C_i$ as indicated above. Then since the homology of an exact chain complex vanishes, we have $\text{rank }H_n(C_*)=0$ for all $n$. Hence, we get
$$\begin{eqnarray}0&=&\sum_n (-1)^n \text{rank }H_n(C_*)
\\&\stackrel{\text{General fact}}{=}& \sum_n(-1)^n\text{rank }C_n
\\&\stackrel{\text{Def. of }C_i}{=}&\text{rank }H_0(Z) - (\text{rank }H_0(X) +\text{rank }H_0(Y)) + \text{rank }H_0(A) -\text{rank }H_1(Z)+\cdots
\\&\stackrel{\text{Euler-Poincare formula}}{=}&\chi(Z) - (\chi(X)+\chi(Y)) + \chi(A)
\end{eqnarray}$$

which implies the claim.

You don't need homology or Maier-Vietoris for this. Simply observe that the relative cells of $(Y\cup_F X, Y)$ are in bijection with those of $(X,A)$.


Does $\exists$ a differentiable function $f:\mathbb{R}\to \mathbb{R}, f(x) \neq x+c$ s.t. every interval $(a,b)$ contains a point $p$ with gradient 1? Does there exist a differentiable function $f:\mathbb{R} \to \mathbb{R}, f(x) \neq x+c$ such that every interval $(a,b)$ contains a point $p$ with gradient $1$?
I would guess no, but I've no idea how to prove it.
Now that I think about it, isn't this question similar to: Does there exist a non-constant differentiable function $f:\mathbb{R} \to \mathbb{R}$, such that every interval $(a,b)$ contains a point $p$ with gradient $0$? Which I may or may not have seen on the site somewhere - can't remember. I guess the disproof would be something like: $f(x)$ has gradient $0$ almost everywhere $\implies f(x)$ has unbounded variation somewhere (e.g. in some interval) $\implies$ f(x) is not everywhere differentiable. I'm not that familiar with bounded variation other than having skim-read this thread once, but bounded variation may in fact not be necessary to answer this question, I've no idea.
Maybe the mean value theorem is more relevant here.

The answer is that yes, such a function exists.
The starting point is the Pompeiu Derivative..  To summarize Wikipedia, Pompeiu constructed an everywhere differentiable strictly increasing function $g:[0,1]\rightarrow \mathbb{R}$ whose derivative $g'(x)$ is $0$ on a dense subset of $[0,1]$.  Call this dense set $D$.
Now, let $h:\mathbb{R}\rightarrow (0,1)$ be your favorite diffeomorphism (e.g., you could pick $h(x) = \frac{1}{\pi}\arctan(x) + \frac{1}{2}$.)
Set $f(x) = g(h(x)) + x$.   I claim that $f$ fulfills your criterion.  By the chain rule, $f'(x) = g'(h(x))h'(x) + 1$.  For $x\in h^{-1}(D)$, $g'(h(x)) = 0$, so $f'(x) = 1$ for $x\in h^{-1}(D)$.  Note that since $h$ is a diffeomorphism, it is a homeomorphism, so $h^{-1}(D)\subseteq \mathbb{R}$ is dense.  Lastly, since $g$ is strictly increasing, $f(x) -x$ is strictly increasing, so is not constant.

Is c a constant in (a,b) interval? If yes, you can try to prove it by contradiction, like using the function f(x) = x^2 in interval ]-1, 1[. I guess that's what you want.(Sorry about formatting, I'm ne here)


An equation of a tangent to the parabola $y^2=8x$ is $y=x+2$. the point on this line from which the other tangent Problem :

An equation of a tangent to the parabola $y^2=8x$ is $y=x+2$. the point on this line from which the other tangent to the parabola is perpendicular to the  given tangent is given by ...

Solution :

Slope at any point on the parabola from where tangent can be drawn can be taken by differentiating equation of parabola $y^2=8x$

Which gives $2y \frac{dy}{dx}=8 \Rightarrow \frac{dy}{dx}=4/y$

Slope of the given line is equal to the slope of the line at this point therefore, $y =4$ and $x=2$.

Please suggest whether this is the correct method of doing this and how can I proceed this problem further thanks..

Hints:

You know the gradient of the tangent, so you can find the slope of a line perpendicular to it.
You can use the same method you have already used to find the point where this second line is tangent to the parabola and so the equation of the second tangent.
You now have the equation for two straight lines, and find where they intersect.


Critical values of $f(x, y) = x^3 − 147xy + 343y^3$ The only critical values I can find are $(0,0$) and $(7,1$). Apparently there is another one, and I can't find it which is very concerning. Just so we are clear, critical values are where $f_x = 0$ and $f_y = 0$ or either one DNE.

$$f(x, y) = x^3 − 147xy + 343y^3$$

Partial for $x$:

$$\frac{\partial f}{\partial x}=3x^2-147y$$

Partial for $y$:

$$\frac{\partial f}{\partial y}=-147x+1029y^2$$

Critical points are when $\frac{\partial f}{\partial x}$ AND $\frac{\partial f}{\partial y}$ are equal to $0$. Hence:

$$3x^2-147y=0$$

$$1029y^2-147x=0$$

We can solve for one variable, let's just solve for $x$:

$$1029y^2=147x$$
$$x=7y^2$$

Now use it:

$$3(7y^2)^2-147y=0$$
$$147y^4-147y=0$$
$$147y(y^3-1)=0$$

Difference of cubes:

$$147y(y^3-1)=147y(y-1)(y^2+y+1)=0$$

Now we can find the three zeroes:

$$147y=0$$
$$y-1=0$$
$$y^2+y+1=0$$

We get that $y=0, y=1, y=\frac{-1\pm 3i}{2}$. From those, you can solve for $x$ by plugging back in.

I am down to my last try. I don't know how to solve for the last point. I got two points which where $x=0$ and $x=7$.


Inverse of Mellin transform with lower bound at $1$ I've seen two definitions of the Mellin transform: more commonly, $g$ is the Mellin transform of $f$ if $$g(s)=\int_0^\infty x^{s-1}f(x)\; dx,$$ or secondly, and more rarely, defined by the same integral with the lower bound at $1$ in place of $0$. The second type seems to be easier to work with in some cases because it avoids a possible singularity at $0$. Now, the inverse Mellin transform is well defined for the first definition. However, for the second, what is the formula for the inverse?

If $$g(s)=\int_1^\infty x^{s-1} f(x)\; dx=\int_0^\infty e^{st}f(e^t)\;dt,$$ then applying the Laplace inversion formula we have $$f(s)=\frac 1{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}s^{-x}g(x)\; dx$$ (for an appropriate constant $\gamma$) which is exactly the same as for the first definition of the Mellin transform. Clearly this is wrong; the two transforms are not identical. What is wrong here and what is the actual formula?

g(s) in the first formula and g(s) in second one are not the same.
For instance if f(x)=Exp(-x) the first gives g(s)=Gamma(s) - Euler Gamma and the second gives Incomplete Gamma g(s)=Gamma(s,1). Inversion formula is identical for both, because they are both Mellin transforms. In the first case it gives Exp(-x) from zero to infinity as a result and in the second case we obtain 0 from zero to one and Exp(-x) from one to infinity. For x = 1 we obtain Exp(-1)/2.


Is there a way to write $\dots 2(2(2(2+1)-1)+1)-1 \dots$ in closed form? It's something like a sequence I'm working with and a part of it has coefficients, which go $3$, $5$, $11$, $21$ and so on. I was wondering if it's possible to find a (closed) formula dependent on say $n$.

The only thing I can contribute is that $(-1)^n$ for the plus and minus ones doesn't seem to lead anywhere (since one has to include the value of the preceding addition/subtraction).

With closed formula I mean a formula where you can plug in your $n$ and obtain the corresponding coefficient immediately, without having to know any of the previous ones.

So observe that $u_{n+1}=2u_n-(-1)^{n}$ and $u_n=2u_{n-1}+(-1)^n$

Add these to obtain $u_{n+1}+u_n=2u_n+2u_{n-1}$ or $u_{n+1}-u_n-2u_{n-1}=0$

Unsurprisingly standard methods give $u_n=A2^n+B(-1)^n$ and if $u_1=3, u_2=5$ you can compute $A=\frac 43, B=-\frac 13$


can a parabolic curve have two turning points? the curve $v= 0.00005(t-200)^2 - 1$ seems to have only $1$ minimum point to me at $(200,-1)$ as a minimum point.
(i completed the squares to find it)
but in my book it says it also has a maximum point at $(0,1)$
$0\leq t\leq800$ (inclusive)
my question is how would you have found the maximum point? i learned the trigonometric graphs of sine and cosine and know that they have multiple turning points but how can a quadratic graph have multiple turning points?
could anyone please  explain why?

When looking for a curve's maxima or minima, you need to examine two things - inflection points and discontinuities. Inflection points are the turning points that you seem to be familiar with, which can be found through the typical use of derivatives. Derivatives don't exist at discontinuous points, however, so you need to check them individually. This parabolic function is discontinuous at the endpoints of t=0 and t=800, so you need to check values there. Basically, you can have maxima/minima at points where the slope is not equal to zero - these aren't "turning points", but rather endpoints or some other kind of discontinuity. You can see, for example, that the straight line x=y defined for [0,10] has no inflection points at all, but still has a maximum and minimum.


Linearisation of product of two continuous variable. I have a quadratic constraint which I want to linearise.

$x_1x_2=0$
where $0\leqslant x_1,x_2\leqslant1$.

What is the best way to do this?

The region described by $0\le x_1,x_2\le 1$ is a square and the condition $x_1x_2=0$ is equivalent to "$x_1=0$ or $x_2=0$" which restricts the solution space to the left and lower edges of the square. The smallest convex set containing these two edges is the triangle described by $x_1+x_2\le 1$.

Depending on your optimization problem (which you didn't tell us) it may be a good or a horrible idea to linearize the condition by $x_1+x_2\le 1$.

If this doesn't answer your question, please edit your post to clarify what you are asking for and include the full description of the optimization problem you are trying to solve or linearize.


Finding a confidence interval for shifted exponential distribution Let  $X_1,\ldots, X_n$ are i.i.d. random variables such that: $$f(x;\sigma ,\theta)=\frac{1}{\sigma}e^{\frac{-(x-\theta)}{\sigma}}, x\gt \theta$$ where $\sigma \gt 0 $ and $\theta \in R$ .
a) if $\theta $ is known, find a $100(1-\alpha)% $% confidence interval for $\sigma$. (Hint: use $\sum_{i=0}^n (X_i -\theta)$ or a modification)
b) if $\theta $ is unknown, find a $100(1-\alpha)% $% confidence interval for $\sigma$. (Hint: use $\sum_{i=0}^n (X_i -X_{(n)})$ or a modification)

a) Since $\theta$ is a location parameter then $X_i - \theta \sim Exp(1/\sigma)$ then $Y=\sum_{i=0}^n (X_i - \theta)  \sim Gamma(n,1/ \sigma)$, so $\frac{\sigma}{2}Y\sim Gamma (n,1/2) \sim {\chi^2}_{2n}$ and solving $P(a\le\frac{\sigma}{2}Y\le b)=1-\alpha $ I can find the condidence interval. Is it right?
b)I found that $X_{(1)}-\theta\sim Exp (n/\sigma)$ I was thinking sum $X_i - \theta$ and $X_{(1)}-\theta$ but due to $X_{(1)}$ and $X_i$ are not independent I couldn't compute the distribution. Any ideas?

For part (a), note that $X_i-\theta$ are i.i.d $\mathsf{Exp}$ with mean $\sigma$. In other words, $\frac{2}{\sigma}(X_i-\theta)$ are i.i.d $\mathsf{Exp}$ with mean $2$, i.e. a $\chi^2_2$ variable. Therefore the correct pivot is $$T_1(\mathbf X,\theta,\sigma)=\frac{2}{\sigma}\sum_{i=1}^n (X_i-\theta)\sim \chi^2_{2n}$$

You can get a $100(1-\alpha)\%$ confidence interval for $\sigma$ starting from $$P_{\sigma}\left[\chi^2_{1-\alpha/2,2n}<T_1<\chi^2_{\alpha/2,2n}\right]=1-\alpha\quad\forall\,\sigma\,,$$

where $\chi^2_{\alpha,2n}$ is such that $P(\chi^2_{2n}>\chi^2_{\alpha,2n})=\alpha$.

For part (b), use the  pivot (see this answer for proof) $$T_2(\mathbf X,\theta,\sigma)=\frac{2}{\sigma}\sum_{i=1}^n (X_{(i)}-X_{(1)})=\frac{2}{\sigma}\sum_{i=1}^n(X_i-X_{(1)})\sim \chi^2_{2n-2}$$


5 digits numbers such that when the sum of digits divided by 4 leaves remainder 2. How many 5 digits numbers such that when the sum of digit divided by 4 leaves remainder 2.

Example:-
Consider a 5 digit number-
$(x1,x2,x3,x4,x5)$
Then $(x1+x2+x3+x4+x5)$ must be of form $(4n+2)$

I tried this
(x+x²+x³...+x^9)(1+x+x²+x³....+x^9)⁴

In this sum of coefficient of x^(2,6,10,14....42)

But this involve lot of calculation.!

Please some one provide me something different and smarter solution.

Call digits $0$ and $1$ small, and digits $2-9$ large.

Given $n\ge 1$, we divide the $n$-digit numbers into two sets: $S$, which contains all numbers consisting entirely of the small digits $0$ and $1$; and $L$, which contains all other numbers. Note that:

$|S|=2^{n-1}$, because each digit apart from the initial $1$ is either $0$ or $1$;
$|L|=9\cdot 10^{n-1}-|S|$, because there are $9\cdot 10^{n-1}$ $n$-digit numbers in total.

Now, if $c$ and $d$ are large digits, then the number of numbers whose first large digit is $c$ is equal to the number of numbers whose first large digit is $d$. And because the large digits are evenly distributed modulo $4$, this means that the digit sums of the numbers in $L$ are also evenly distributed modulo $4$. So the number of numbers in $L$ with a given digit sum modulo $4$ is $|L|/4$.

This just leaves $S$. But this is easy: the number of numbers in $S$ with digit sum $k$ is the number of ways of choosing $k-1$ positions for the $1$'s (given that the first digit has to be $1$). This is the binomial coefficient $\binom{n-1}{k-1}$.

Thus we see that the number of $n$-digit numbers with a digit sum equal to $m$ mod $4$ is equal to $|L|/4+\sum_k\binom{n-1}{k-1}$, where the sum is taken over all $k$ with $1\le k\le n$ and $k\equiv m$ mod $4$.

Here is a table of the number of $n$-digit numbers with a given digit sum modulo $4$, for $n=1$ to $6$:
$$
\begin{array}{c|lcr}
n & |L|/4 & 0\bmod 4 & 1\bmod 4 & 2\bmod 4 & 3\bmod 4 \\
\hline
1 & 2 & 2 & 3 & 2 & 2\\
2 & 22 & 22 & 23 & 23 & 22\\
3 & 224 & 224 & 225 &226 &225\\
4 & 2248 & 2249 & 2249 & 2251 & 2251\\
5 & 22496 & 22500 & 22498 & \color{red}{22500} & 22502\\
6 & 224992 & 225002 & 224998 & 224998 & 225002\\
\end{array}
$$

Let us first see the required numbers in

$10000-10010$

One or two digit number in front of five digit number represent the sum of digits of the five digit number

$10000-1

10001-2

10002-3

10003-4

10004-5

10005-6

10006-7

10007-8

10008-9

10009-10$

Number of required number in the above series are 3.
If we see the next series(2) of next 10 numbers. The required numbers are 3. And same for the series(3) (From 10020-10029) number of requires numbers are 2. And for fourth series no. Of required numbers are 2.

And the sequence of required numbers per ten first ten digits is-

$3 3 2 2 3 3 2 2 3 3 2 2......3 3 2 2$

Total number between $10000-99999$ are $90000$.

No.s of 2's and 3's are
=$\frac{90000}{10}=9000$

No.s of 2's  or 3's are
$\frac{9000}2=4500$

Therefore the required no.s are
$$4500*(2+3)=22500$$


Existence of matrices with non-zero principal minors The problem sounds very simple but I have yet to come to an answer.
Prove or disprove:

  For all $n$ there exists a matrix $A \in \mathbb{R}^{n \times n}$ with $\det(A) = 0$ such that all first minors $M_{i,j} \neq 0$.

Trivial example for $n=2$:
\begin{pmatrix}
1 & 1\\
1 & 1
\end{pmatrix}

Hint: assume the $n$ variable matrix looks the same (all ones), and compute it's determinant by the cofactor of minors method. We know what the determinant of any possible 2 by 2 minor is. What does this tell us?


100th derivative of $\frac{x^2+x}{2^x}$ at point 0 Problem:
$$\frac{\mathrm d^{100}}{\mathrm dx^{100}}\frac{x^2+x}{2^x}$$
Can anyone help me with this because I tried to use General Leibniz rule and I didn't get much better information and I also tried some proof by induction but also no success
Without Taylor


Which of the following fields are isomorphic? Let $\mathbb{Q},\mathbb{R}$ denote the fields of rational numbers and real numbers respectively. Which of the following fields are not isomorphic.
  (a) $\mathbb{Q}[x]/( x^2 + 1 )$ and $\mathbb{Q}[x]/( x^2 + x + 1 )$.
  (b) $\mathbb{R}[x]/( x^2 + 1 )$ and $\mathbb{R}[x]/( x^2 + x + 1 )$.

  Justify your answer.

My thoughts:

(a) $\mathbb{Q}[x]/( x^2 + 1 )$ is isomorphic to $\mathbb{Q}(i)$ and $\mathbb{Q}[x]/( x^2 + x + 1 )$ is isomorphic to $\mathbb{Q}(\sqrt{3},i)$. So they are not isomorphic.

(b) $\mathbb{R}[x]/( x^2 + 1 )$ is isomorphic to $\mathbb{R}(i)$ that is $\mathbb{C}$ and $\mathbb{R}[x]/ ( x^2 + x + 1 )$ is isomorphic to $\mathbb{R}(\sqrt{3} , i)$ that is $\mathbb{C}$.
So they are isomorphic.

Are my arguments correct?

One idea: In $\mathbb{Q}[x]/<x^2 + 1>$, all elements are of the form $a + bx$, where $a$ and $b$ are rational numbers, and $x^2 = -1$.

Show that the polynomial $p(t) = t^2 + t + 1$ does not have any solutions in this field. Of course, $p(t)$ does have solutions in the field $\mathbb{Q}[x]/<x^2 + x + 1>$, so if it didn't have any solutions in the first field, you would show the fields are not isomorphic.

To show that $p(t)$ doesn't have solutions in the first field, suppose $t = a + bx$ were a solution. Expand $t^2 + t + 1$, and show that this places impossible restrictions on the rational numbers $a$ and $b$.

The isomorphism between R[x]/(x^2+1) and R[x]/(x^2+x+1)
Both are isomorphic to C. First isomorphism is f(i) and for the other one it is f((i*sqrt(3)-1)/2)


How to solve this absolute value equation and summation question?? $$|2x − 3| − |x + 2| = 5$$

I have no idea. I didn't see anything like this in class. It is a practice question and something like it will come up on the exam can someone please show me the full solution and working (that is how I learn) Thanks in advance.

Hint: using the definition of the absolute value we get $$|2x-3|=\begin{cases} 2x-3, & 2x-3\geq 0 \\ -(2x-3), & 2x-3<0 \end{cases}=\begin{cases} 2x-3, & x\geq \frac{3}{2} \\ -2x+3, & x<\frac {3}{2}\end{cases}$$
$$ |x+2|=\begin{cases} x+2, & x+2\geq 0 \\ -(x+2) & x+2<0 \end{cases} = \begin{cases} x+2, & x\geq -2 \\ -x-2, & x<-2\end{cases}$$

This is all we need to simplify the equation. "Something happens" at $x=-2$ and $x=\frac{3}{2}$, one of the terms $|2x-3|$ or $|x+2|$ will change there. Using this we get 3 cases we have to consider:

$x<-2$
$-2\leq x < \frac{3}{2}$
$\frac{3}{2}\leq x$

Hint

if $x \geq 0$ then you would have $$2x-3 -(x+2) = 5$$

if $x < 0$ then you would have $$-(2x-3) +(x+2) =5$$


Formalizing converging norm Let $x \in \ell^2$. Define $(u_N)$ as the sequence of vectors $u_N \in U$ such that $u_n = x_n$ if $n \le N$, $0$ otherwise.

Then $\lVert u_N \rVert$ is an increasing function which converges to finite $\lVert x \rVert$.

The statement seems obvious, but how can I make this claim more rigorous? Saying the norm of $u_N$ increases doesn't seem convincing.

I suppose it depends on how norm is defined. If defined as an infinite sum which is defined as the limit of a finite summation, then the result follows by definition ...

$\sqrt {\sum\limits_{k=1}^{n} |x_k|^{2}}$ increases to $\sqrt {\sum\limits_{k=1}^{\infty} |x_k|^{2}}$. This is just deinition of the infinite sum $\sum\limits_{k=1}^{\infty} |x_k|^{2}$.


Show by Induction, but stuck on Induction Step (Don't know how to implement N = K+1) Prove by Induction that
  $$
1\cdot 2\cdot3  +  2\cdot3\cdot4 + \cdots + n \cdot(n+1)\cdot(n+2) = \frac{n  (n+1)(n+2)  (n+3)}{4}
$$

For my basis step, I set $$n=0$$
This leaves me with zero on both LHS and RHS, so I proceed.
Induction Hypothesis consists of plugging in $N = K$ assuming the basis step applies for $$ k \leq n $$

After this, I am stuck on the Induction Step. I try to solve for $n = k+1$, but I don't know what to do.

Due to requests asking me to show what I have tried:
$$
(n+1)(n+2)(n+3) = \frac{(n+1)(n+2)(n+3)(n+4)}{4}
$$
I don't know where to go from here. I don't think it is correct anyways.


Prove that Im(TS) = Im(T) if $S$ is surjective and $T$ is any linear transformation Given that $S: U \to V$ and $T: V \to W$ prove the above...

Not sure where to start with this. I know that since $S$ is surjective, that $Im(S) = V$, but image isn't distributive right? I.e. $\mbox{Im}(TS) = \mbox{Im}(T).\mbox{Im}(S)$.

Suggestions appreciated.

Let $TS(x) \in \mbox{Im}(TS)$. Then, $TS(x) = T(S(x))$, so that $TS(x) \in \mbox{Im}(T)$ also.

The other way,let $T(y) \in \mbox{Im}(T)$. For this $y$, there exists $z$ such that $S(z) = y$ by surjectivity of $S$. Now, $TS(z) = T(y) \in \mbox{Im}(TS)$.

Hence, if $S$ is surjective, $\mbox{Im } T = \mbox{Im } TS$. Of course, if $T$ were surjective but $S$ is not, then this is obviously not true, for example $S$ could be the operator sending all elements to zero.

Let Im=f . f(T)= {y such that y=f(T(x)) for some x}={y such that y=f(T(S(x)) for some x}(because S is surjective)=f(TS).


Determine if the sequence $\{a_n\}:a_n=(-3)^n$ is an increasing or a decreasing sequence Determine if the sequence $\{a_n\}:a_n=(-3)^n$ is an increasing or a decreasing sequence.
The $(n+1)^{st}$ term is $a_{n+1}=(-3)^{n+1}.$ So $\dfrac{a_{n+1}}{a_n}=\dfrac{(-3)^{n+1}}{(-3)^n}=\dfrac{(-3)^n(-3)^1}{(-3)^n}=-3<1.$ Doesn't this mean that $a_{n+1}<a_n$ for every $n$ which obviously isn't true? The sequence looks like $-3,9,-27,81...,$ so it isn't monotonic, but why the usual method tells us else? We can use it if $a_n\ne0$ for every $n$ which holds here.

It is because $\frac{a_{n+1}}{a_n} < 1$ does not tell you the sequence is decreasing.
What we would all like to do is to multiply both side by $a_n$, but WE CAN'T, since $a_n$ is not necessarily positive !
So $\frac{a_{n+1}}{a_n} < 1$ does not imply $a_{n+1} < a_n$.

As others have remarked we cannot use your method to conclude $a_{n+1} < a_{n}$. Writing out the first three terms $a_1=-3, a_2 = 9, a_3 = -27$ we see that $a_1 < a_2$ which means the sequence isn't decreasing. However, we also have $a_2 > a_3$ which means the sequence isn't increasing. In conclusion the sequence is neither increasing nor decreasing.


Getting a feel for the transformation A on vector x which lies outside of any eigenspace In one of his videos, after 13:25 Sal starts to talk about the interpretation of the eigenvectors and how they relate to a vector $x$ being transformed by the matrix $A$.

He then goes through showing what happens when $x$ would be in one of the eigenspaces.

My questions are:

what is the interpretation of the transformation being applied to a vector outside of any eigenspace, like any regular 3-dimensional $x$?
How could I "visualize" what happens to such a vector?
Is it related somehow to the characteristic polynomial? What is the meaning of the characteristic polynomial, beside giving us the eigenvalues at the intersections with the $x$-axis?

The eigenspaces for a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ are the simplest building blocks for that transformation.  In fact sometimes they are too simple, in the sense that you may not be able to write $T$ as a direct sum of eigenspaces.  You can do so iff $T$ is diagonalizable.  This really is the simplest case so should be well understood first.

If your transformation is diagonalizable but not scalar -- i.e., if it is not just a single eigenspace -- then most vectors in your space will not be eigenvectors.  In general the best description I can think of for such a vector simply comes from the fact that it is a sum of eigenvectors -- indeed you can chose a basis for $\mathbb{R}^n$ of eigenvectors in this case -- so the transformation will act on it by scaling each component eigenvector by a different amount.  I would try thinking about the effect of the diagonal matrix $\left[ \begin{array}{cc} 2 & 0 \\ 0 & \frac{-1}{3} \end{array} \right]$ on a random vector $[x,y]^T$ and drawing some pictures to try to get comfortable.

When there is no basis of eigenvectors things get more complicated.  There are really two different things that could stop you from having a basis of eigenvectors.  First, over a field like $\mathbb{R}$ you simply may not have any eigenvalues at all.  The basic example of this is rotation in the plane (by any angle which is not an integer multiple of $\pi$), and this is worth thinking about.  By taking direct sums you can build linear transformations on $\mathbb{R}^{n}$ for all even $n$ without eigenvalues.  However, when $n$ is odd you must have at least one eigenvalue, because the eigenvalues are the roots of the characteristic polynomial and an odd degree polynomial over $\mathbb{R}$ must have at least one root.

However, it is possible that your characteristic polynomial factors completely into linear factors (for instance this always happens if you take $\mathbb{C}$ as your scalar field, by the Fundamental Theorem of Algebra) but that you still don't have a basis of eigenvectors.  This is about where a standard first linear algebra course tends to end, so you don't normally get a clear explanation of what is going on here but rather some rather unenlightening talk about "algebraic and geometric multiplicities".  If you go on to take a second course -- or better yet, a course in abstract algebra covering the structure theory of modules over a PID -- then you will hear the rest of the story.

It happens that I had the occasion to write a reasonably middlebrow exposition of the theory of invariant subspaces about a month ago: you can find it here.  In these notes I was influenced by Sheldon Axler's article "Down with determinants!" and the book that followed it, although in my interest in the case of a general ground field I go a bit beyond his ambitions...and at one point I explain why determinants really help a lot, as far as I can see.  But I definitely follow Axler's approach of not defining the characteristic polynomial as the determinant of $t I - T$: I agree with him that this loses all contact with geometric explanation.  In fact, the polynomial referred to in the previous sentence I call the Cayley-Hamilton Polynomial, and I give a completely different definition of the characteristic polynomial $\chi(t)$ in terms of things that are too elaborate to get into here, like primary decompositions and complete flags of invariant subspaces).  From this "geometric" definition of the characteristic polynomial it is rather straightforward to show that it is divisible by the minimal polynomial, or equivalently that $\chi(T) = 0$.  By the Cayley-Hamilton Theorem I mean the result which says that the characteristic polynomial is always equal to the Cayley-Hamilton polynomial.

If you find this exposition too brisk or abstract -- there are admittedly few examples, and if I recall I even neglected to type out some companion matrices -- then you might try Axler's book Linear Algebra Done Right instead.  He spends quite a long time telling the story, but he tells it with admirable care and insight.

I didn't have a correct definition of eigenspaces. I'm not an expert on the last part, but I'll attempt to explain anyway.

If A has a limiting matrix $A^\infty$ (powers can't flip back and forth, for example), then for every (column) vector v, $A^\infty v$ is an eigenvector of A.
So the vectors that are not in any eigenspace tend towards the eigenspaces asymptotically.
"Meaning is not in things, but in between them." - Norman O. Brown. Matrices and polynomials both form rings. Rings are often studied geometrically by analogy with the polynomial ring, and how the 0-sets of polynomials behave best as curves, like x^2+y^2-1=0. Sending every matrix to its characteristic polynomial is a morphism between curves, called a regular map (as is sending it to that polynomial's discriminant, see this question), and the fact that it vanishes at every point of the n-by-n matrix algebra treated as the affine plane of dimension $n^2$ is proof of the Cayley-Hamilton theorem from a different perspective.


Local martingale but not martingale On wikipedia there is an example of a local martingale which is not a martingale, but I do not understand why it is a local martingale. We have the process

$
 X_t = \begin{cases}
  W_{\min(t/(1-t),T)} &\text{for } 0 \le t < 1,\\
  -1 &\text{for } 1 \le t < \infty.
 \end{cases}$

where $(W_t)$ is a standard Brownian motion and $T = \inf\{ t : W_t = −1 \}$.

The expectation is

$\mathbb{E} X_t = \begin{cases}
  0 &\text{for } 0 \le t < 1,\\
  -1 &\text{for } 1 \le t < \infty.
 \end{cases}$

This expectation is clearly discontinuous. So we have that it is not a martingale.

Now we will conclude that it is a local martingale with localizing sequence $ \tau_k = \min \{ t : X_t = k \}$ if there is such $t$, otherwise $τ_k = k$.
However, I can not figure out why this is true.

I would appreciate some help.

I also think the wikipedia example a little bit confusing. It seems the localizing sequence can not give a martingale results since when $t\in(0,1), X_t^{\tau_n}$ is again a martingale with mean 0 and when $t\geq 1, $ it has mean -1. I have checked the reference book in that wikipedia page, and the book gives a different localizing sequence.

Details have been discussed in this thread, and I think it is quite clear.

How to show the following process is a local martingale but not a martingale?

wrt your general question: A local martingale is not necessarily integrable, whereas a martingale is (by definition).
This all boils down to the statement that a limit of a sequence of integrable random variables need not be integrable!


Find the Distribution of the following; Give that $Z_1,Z_2,...,Z_n$ are independent identically distributed standard Gaussian random variables with mean 0 and variance 1.
  Find the Distribution of $$X=\dfrac{(Z_1+Z_2)^2}{(Z_1-Z_2)^2}$$

Now first thing I did was open it up, but no common distributions I know are looking familiar. Any hints would be appreciated.

$Z_1 + Z_2, Z_1 - Z_2$ are jointly Gaussian and luckily their convariance is $0$, so they are also independent, then it's easy to see the quotient of their squares follows a F-distribution with parameter $(1,1)$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
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\begin{align}
\color{#66f}{\large\pp\pars{X}}&\equiv
\int_{-\infty}^{\infty}\dd Z_{1}\,{\exp\pars{-Z_{1}^{2}/2}\over \root{2\pi}}
\int_{-\infty}^{\infty}\dd Z_{2}\,{\exp\pars{-Z_{2}^{2}/2}\over \root{2\pi}}\,
\delta\pars{X - {\pars{Z_{1} + Z_{2}}^{2} \over \pars{Z_{1} - Z_{2}}^{2}}}
\\[5mm]&={1 \over 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\exp\pars{-\,{Z_{1}^{2} + Z_{2}^{2} \over 2}}
\delta\pars{X - {\bracks{Z_{1} + Z_{2}}^{2} \over \bracks{Z_{1} - Z_{2}}^{2}}}
\,\dd Z_{1}\,\dd Z_{2}
\\[5mm]&={1 \over 2\pi}\int_{0}^{2\pi}\int_{0}^{\infty}
\exp\pars{-\,{\rho^{2} \over 2}}
\delta\pars{X - {\bracks{\cos\pars{\theta} + \sin\pars{\theta}}^{2} \over \bracks{\cos\pars{\theta} - \sin\pars{\theta}}^{2}}}\rho\,\dd\rho\,\dd\theta
\\[5mm]&=\color{#66f}{\large{\Theta\pars{X} \over 2\pi}\int_{-\pi}^{\pi}
\delta\pars{%
X - {\bracks{\cos\pars{\theta} + \sin\pars{\theta}}^{2} \over \bracks{\cos\pars{\theta} - \sin\pars{\theta}}^{2}}}
\,\dd\theta}
\\[5mm]&\mbox{because}\quad
\int_{0}^{\infty}\exp\pars{-\,{\rho^{2} \over 2}}\rho\,\dd\rho = 1
\end{align}

  Also,
  \begin{align}
&\color{#66f}{\large\pp\pars{X}}={\Theta\pars{x} \over 2\pi}\int_{-\pi}^{\pi}
\delta\pars{%
X - {\cos^{2}\pars{\theta - \pi/4} \over \cos^{2}\pars{\theta + \pi/4}}}\,\dd\theta
\\[5mm]&={\Theta\pars{x} \over 2\pi}\bracks{\int_{0}^{\pi}\delta\pars{%
X - {\cos^{2}\pars{\theta - \pi/4}\over\cos^{2}\pars{\theta + \pi/4}}}\,\dd\theta
+\int_{0}^{\pi}\delta\pars{%
X - {\cos^{2}\pars{\theta + \pi/4}\over\cos^{2}\pars{\theta - \pi/4}}}\,\dd\theta}
\\[5mm]&={\Theta\pars{x} \over 2\pi}\bracks{\int_{-\pi/2}^{\pi/2}\!\delta\pars{%
X - {\sin^{2}\pars{\theta - \pi/4}\over\sin^{2}\pars{\theta + \pi/4}}}\,\dd\theta
+\int_{-\pi/2}^{\pi/2}\!\delta\pars{%
X - {\sin^{2}\pars{\theta + \pi/4}\over\sin^{2}\pars{\theta - \pi/4}}}\,\dd\theta}
\\[5mm]&={\Theta\pars{x} \over \pi}\bracks{\int_{0}^{\pi/2}\!\delta\pars{%
X - {\sin^{2}\pars{\theta - \pi/4}\over\sin^{2}\pars{\theta + \pi/4}}}\,\dd\theta
+\int_{0}^{\pi/2}\!\delta\pars{%
X - {\sin^{2}\pars{\theta + \pi/4}\over\sin^{2}\pars{\theta - \pi/4}}}\,\dd\theta}
\\[5mm]&={\Theta\pars{x} \over \pi}\bracks{\int_{-\pi/4}^{\pi/4}\!\delta\pars{%
X - {\sin^{2}\pars{\theta}\over\cos^{2}\pars{\theta}}}\,\dd\theta
+\int_{-\pi/4}^{\pi/4}\!\delta\pars{%
X - {\cos^{2}\pars{\theta}\over\sin^{2}\pars{\theta}}}\,\dd\theta}
\\[5mm]&={2\Theta\pars{x} \over \pi}\bracks{\int_{0}^{\pi/4}\!\delta\pars{%
X - \tan^{2}\pars{\theta}}\,\dd\theta
+\int_{0}^{\pi/4}\!\delta\pars{X - \cot^{2}\pars{\theta}}\,\dd\theta}
\\[5mm]&={2\Theta\pars{x} \over \pi}\left\{\Theta\pars{1 - X}\int_{0}^{\pi/4}{\delta\pars{\theta - \arctan\pars{\root{X}}} \over
\verts{2\tan\pars{\theta}\sec^{2}\pars{\theta}}}\,\dd\theta\right.
\\&\left.\phantom{{2\Theta\pars{x} \over \pi}\left[\right.}
+\Theta\pars{X - 1}\int_{0}^{\pi/4}
{\delta\pars{\theta - {\rm arccot}\pars{\root{X}}} \over
\verts{2\cot\pars{\theta}\bracks{-\csc^{2}\pars{\theta}}}}\,\dd\theta\right\}
\\[5mm]&={2\Theta\pars{x} \over \pi}\bracks{%
{\Theta\pars{1 - X} \over 2\root{X}\pars{X + 1}}
+{\Theta\pars{X - 1} \over 2\root{X}\pars{X + 1}}}
=\color{#66f}{\large{\Theta\pars{X} \over \pi}\,{1 \over \root{X}\pars{X + 1}}}
\end{align}


if G is generated by {a,b} and ab=ba, then prove G is Abelian if G is generated by {a,b} and ab=ba, then prove G is Abelian

All elements of G will be of the form $a^kb^j,\,\, k,j \in \mathbb{Z}^+$

so I need to get $a^kb^j = b^ja^k$

First notice that $G=<a,b>$ does not imply directly that all elements of $G$ are of the form $a^kb^j$.

Here we have to different assertion the first one is $G=<a,b> \,\,\,(1)$ and the second one is $ab=ba\,\,\, (2)$. If this two assertions are true than the $G$ is Abelian. To understand the consequences of $(2)$ over $G$ we need first to understand what $G$ look like without $(2)$ and then assuming $(2)$ we simplify $(1)$.

The definition of the group $<a,b>$ is the following :

  $$<a,b>=\big\{a^{i_1}b^{j_1}\cdots a^{i_n}b^{j_n}\Big / n\in \mathbb{N}, (i_1,j_1,\cdots,i_n,j_n)\in \mathbb{Z}\big\} $$

Let's suppose that $ab=ba$, we pick two elements $x,y$ of the group $<a,b>$ we can write :
$$x=a^{i_1}b^{j_1}\cdots a^{i_n}b^{j_n}\\y=a^{l_1}b^{k_1}\cdots a^{l_m}b^{k_m} $$

and we want to prove that $xy=yx$ , the idea is to prove it by induction on $$n(x)=\sum_{i=1,\cdots,n}|i_k|+|j_k|$$

$n(x)=0$ implies that $x$ is the neutral element.
$n(x)=1$ then $x=a$ or $x=b$, because of the symmetric role of $a$ and $b$ we can assume WLOG that  $x=a$, we have $ab=ba$ so $ab^k=ab.b^{k-1}=b (b.a^{k-1})$using this, we can prove for all $k\in \mathbb{Z}$ that $ab^k=b^ka$ hence :
$$ay=a.a^{l_1}b^{k_1}\cdots a^{l_m}b^{k_m}=a^{l_1}.a.b^{k_1}\cdots a^{l_m}b^{k_m}=a^{l_1}b^{k_1}.a.\cdots a^{l_m}b^{k_m}\\ =a^{l_1}b^{k_1}\cdots a^{l_m}b^{k_m}.a=y.a $$
because $a$ commutes with the powers of both $a$ and $b$, so $xy=yx$
Now suppose that the result is true for all $t$ such that $n(t)=k$, Let $x$ be an element of $<a,b>$ such that $n(x)=k+1$ so we can write either $x=tb$, $x=tb^{-1}$, $x=ta$ or $x=ta^{-1}$ for some $t$ such that $n(t)=k$,and we apply the induction hypothesis in all four cases to conclude that $xy=yx$

Other than induction you can try this. The center of $G$ name it $Z(G)$ contains $\{a,b\}$ the generators of the grup $G$ and also it is generated by them. Also $Z(G)\leq G$ ,thus $Z(G)=G$. This means that $G$ is abelian.


Calculating Overtake Time for Two Cars Yes I've seen several similar questions but they all seem to not have the same provided information as what I'm looking for. Given my very rusty math skills I've been unable to convert the answers I've found there to my use.
Car A is traveling at 20mph. Car B is traveling at 40mph and is 1 mile behind car A. How long will it take for Car B to overtake Car A?
I'm really looking for the correct formula here. So I could reword this to look more like:
Car A is traveling at a mph. Car B is traveling at b mph and is d miles behind car A. How much time (t) in hours will it take Car B to overtake Car A?
Thanks ahead of time. I've been trying to turn this into a formula and have been humbled by my inability to wrap my ahead around what seems to me to be a simple question.

Start by snowing B is approaching A at b - a mph.


What is the cardinality of an infinite product of sets? We have $(A_i)_{i\in I}$ a family of sets. We can suppose that the axiom of choice is true. I am trying to prove that for every $\kappa$ infinite cardinal, we have $card(\prod_{i \in I} A_i)<\kappa'$ for a certain $\kappa '$ cardinal (to be determined) if and only if $\forall i\in I,card(A_i)<\kappa$ and $card(I)<\kappa$. How can one prove this ? Is this even true ?

The cardinality of the product of a family of sets is easily proved to be the $\sup$ of the cardinalities of the sets in the family (provided they are infinite of course) or the cardinality of $I$, whatever is bigger.

If $\kappa$ is a regular cardinal, then you can set $\kappa' = \kappa$, because by definition no sequence of less than $\kappa$ cardinals all smaller that $\kappa$ can sum $\kappa$.

If on the other side $\kappa$ is singular (that is, its cofinality is strictly less than $\kappa$) then you can have that $\sup(card(A_i)) = \kappa$ if $I$ is the cofinality of $\kappa$. so you can set $\kappa'$ to be the successor of $\kappa$.

I'll leave you to fill out the details.

Edit:

Thinking a bit more, it seems that I am completely wrong!

One of the reasons is that (provided enough $A_i$ are at least $2$) then

$$
\left| \prod_{i\in I} A_i\right| ≥ \left| \mathscr{P}(I) \right|
$$

and that number may be anything...


Why 5.8 (decimal) is 101.11001100... (binary)? I would like to know why 5.8, in decimal 101.11001100..., in binary, a recurring decimal?

interesting question. so cut out inessentials first, as with food preparation. the integer part $5$ is irrelevant. you have $0.8 = \frac45$ cut out the $4$ as irrelevant because powers of two map into left or right shifts in base $2$ notation.

so the essence of your question concerns $\frac15$

first you must find $inf\{n \in \mathbb{N}\ : 5 \mid 2^n -1\})$. by inspection this is $4$. so your binary fraction must have period $4$. it is easy to see the relevant expansion is:
$$
\frac15= 3\sum_{k=1}^{\infty} \frac1{2^{4k}}
$$

i.e.
$$
0.0011001100110011...
$$
hence by shifing two places leftwards, you have
$$
\frac45=0.11001100110011...
$$
and if you are really that bothered:
$$5.8 = 101.110011001100...
$$

you can see that the obvious possibility of expressing the reciprocal of any odd integer in binary implies that
$$
 \forall m \in \mathbb{N}, inf\{n \in \mathbb{N}\ : 2m+1 \mid 2^n -1\})
$$
always exists.

can you prove this algebraically? there is some interesting stuff going on here which concerns the fact that $2$ is never a zero divisor in the finite ring:
$$
\frac{\mathbb{Z}}{(2m+1)\mathbb{Z}} = \mathbb{Z}_{2m+1}
$$

As to the left of the decimal point, each position represents ones, tens, hundreds, etc., each position to the right represents tenths, hundredths, and so on.  5.8 is another way of saying five and eight tenths, or simplified, five and four fifths.

In binary, similarly, positions to the right of the decimal point (binary point?) represent halfs, quarters, eighths and so on.  So how do you convert four fifths to binary?  Pretty much the same way you convert integers into binary, except by using multiplication instead of division:

Take the number, multiply it by two.  If the result is greater than or equal to one, the next digit in the binary number is one, otherwise it's zero.

If that numbers exactly one, you're done, otherwise take the remainder and repeat.

No matter how many times you iterate this process on 4/5, you'll never get rid of that remainder:

4/5 * 2 = 8/5 -> 1 + 3/5
3/5 * 2 = 6/5 -> 1 + 1/5
1/5 * 2 = 2/5 -> 0 + 2/5
2/5 * 2 = 4/5 -> 0 + 4/5  -> at this point everything repeats.

Thus you get the recurring decimal.


Semiprime numbers which, along with their prime factors, generate many semiprimes by concatenation There's something quite interesting about the number $1191$:

this number is a semiprime ($1191= 3 \cdot 397$),
the concatenation of its prime factors in any order are semiprimes ($3397$ and $3973$ are semiprimes), but not just that,
if you concatenate $1191$ and its prime factors $3$ and $397$ in any order, the result is always a semiprime (i.e. $11913397$, $11913973$, $31191397$, $39711913$, $33971191$, and $39731191$ are all semiprimes).

I call such numbers semiprime-to-the-bones, or STB.
Definitions: Let $s = p \cdot q$ be a semiprime.

I call $s$ an STB number if these following 8 concatenations are all semiprimes: $p \mathop{^\smallfrown} q$, $q \mathop{^\smallfrown} p$, $s \mathop{^\smallfrown} p \mathop{^\smallfrown} q$, $s \mathop{^\smallfrown} q \mathop{^\smallfrown} p$, $p \mathop{^\smallfrown} s \mathop{^\smallfrown} q$, $q \mathop{^\smallfrown} s \mathop{^\smallfrown} p$, $p \mathop{^\smallfrown} q \mathop{^\smallfrown} s$, $q \mathop{^\smallfrown} p \mathop{^\smallfrown} s$, where $a \mathop{^\smallfrown} b$ means the number whose decimal representation is the decimal representation of $a$ followed by that of $b$ (i.e. concatenation).

I call $s$ a nice STB number if it is an STB number whose reversal is also semiprime.

I call $s$ a super STB number if it is an STB number and the following are all also semiprimes:
$s \mathop{^\smallfrown} p$,
$s \mathop{^\smallfrown} q$,
$p \mathop{^\smallfrown} s$,
$q \mathop{^\smallfrown} s$.

I've searched numbers up to $3000$, but the only STB number I've found is $1191$.

Questions:

Can you find a larger example of STB numbers (or nice STB numbers).
Do super STB numbers exist?

I want give you a valuable information, I did some observations on super STB numbers, and I conclude that there are only two types of super STB numbers:

(Let S=M*N be a super STB number),the first type is super STB numbers  that is divisible by 3 (i.e. S=3*N), which is Joriki or Bernard has found one of them (9=3*3, very trivial one and weak ?).

The second type is super STB numbers that is not divisible by 3 but with S+M+N divisible by 3, do you know what it means? It means that we need all these six concatenations S.M.N, S.N.M, M.S.N, N.S.M, M.N.S, and N.M.S to be all semiprimes of the form 3c, yes we need six semiprimes of the form 3c, The reason is this:

S= M*N, we MUST avoid S+M or S+N to be divisible by 3, because if one of them (S+M or S+N) divisible by 3, then either S.M or S.N will have more than two prime factors ( remember that 9= 3*3 is type 1 super STB number). Dots here means concatenation.

Okay folks, so just check those and you will find the first non trivial super STB number. :D


Is this probability theorem correct? Let me preface my question by saying that I am not a formally trained mathematician, so please forgive my informal statement of my problem.
I have the following 'theorem' of an idea that I believe to be pretty evident in probability.

Given an event E
Given the Set of all real positive integers R and
Given the set of all possible outcomes of a random Event N
Given the number of favorable outcomes n where E occurs.
p(E) = n/N
for 0 < p(E) <= 1
the probability that E will occur in R trials is 1.

I have three questions:

Is there something logically incorrect about my theorem?
Is this a well known theorem, and if so what is it called?
How can I write this theorem, more succinctly using symbolic or mathematical notation?

This is called the Infinite monkey theorem. The statement is that a monkey hitting keys at random on a typewriter keyboard for an infinite amount of time will almost surely type a given text, such as the complete works of William Shakespeare. One sometimes states the theorem with an infinite amount of monkeys but this is a mistake since one monkey (living forever) is enough. What is infinite in the theorem is the amount of time during which the keys are hit at random.

Like it's stated above, the theorem resembles that of the infinite monkey theorem.

The theorem is purely a proposition than an unlimited number of monkeys can “almost surely” produce a specific text such as Macbeth or even the entire works of Shakespeare, provided that they are given sufficient typewriters and time.

The probability of a complete universe full of monkeys typing an entire complete work of Shakespeare such as Macbeth/Hamlet is so tiny that the chance of it happening is extremely low (time taken to perform the feat should be longer than the age of the universe). But do remember that the mention is “extremely low” which technically, isn’t a zero.

There is a direct and straightforward proof of this theorem. Remember that if two events are independent, the probability of occurrence of both the events is equal to the product of the probabilities of each event happening independently. A simple example:

Chance of rain in place A on a specific day= 0.5.

Chance of rain in place B on a specific day= 0.4

Chances of rain happening on the same day in BOTH places= 0.5 x 0.4= 0.20.

Hope you get the point. Now let’s get back to the theorem.

Imagine that a typewriter has 50 keys. The word that you have to type is “SLEEPY.” According to this theorem, we’ll have to press the typewriter keys randomly and at the same time, generate the required word. We have ample time and resources on our hand. So the only limitation is patience, but that’s not the point. Our point is to prove that it CAN be done. If the typewriter keys are pressed independently and randomly, each key has an equal chance of being pressed.

So, the chance of the first letter “S” getting typed= 1/50.

Chance of the 2nd letter “L” getting typed= 1/50.

The same is true for other letters in a word SLEEPY. Therefore, the chance of getting the word right is-

(1/50) × (1/50) × (1/50) × (1/50) × (1/50) × (1/50) = (1/50)6 = 1/15 625 000 000.

Very very very small but it is NOT zero. Therefore it is very possible.

This is basically just a theorem IMO. It's almost impossible to implement test this concept in the real world and expect a correct result.

Source- The Infinite Monkey Theorem


Calculating $w=\sqrt {-i}$ I am looking at the following:
$w=\sqrt {-i}$.

Rewriting
$w=\sqrt {-i}=\sqrt{\exp(-i\frac{\pi}{2}+ik2\pi )}$

$w=\exp(-i\frac{\pi}{4}) \exp(i k  \pi)$

How do I know which is the correct answer for $w$?

Let's suppose we take the principal branch.

They are both correct. The numbers you have found are $\pm( \frac{\sqrt2}{2} -i\frac{\sqrt2}{2})$. There is not a natural single value.


Taylor series of $\sqrt{\ln\left(\frac{1}{x}\right)}$ I am trying to compute the Taylor series of:

$\sqrt{\ln\left(\frac{1}{x}\right)}$

I have computed the derivatives and evaluated them in $x=1/e$ but I cannot find the formula for the sequence of the coefficients.

Any other ideas?

Thanks!

PS: The approximation should be valid over the interval $x \in [0,1]$

This is a note on how you might compute the derivatives using implicit differentiation. It isn't really an answer, but was too long for a comment, and someone may see how to use this in an answer.

$$y=\left(-\ln(x)\right)^{\frac 12}$$

$$y'=\frac 12\left(-\ln(x)\right)^{-\frac 12}\cdot-\frac 1x=-\frac 1{2xy}$$

Whence $$2xyy'=-1$$ and $$2yy'+2xy'^2+2xyy''=0$$ From which you can cancel a factor two and keep going to compute higher derivatives.

Using $2xyy'=-1$ you can multiply through by $y'$ and obtain:

$$2yy'^2+2xy'^3-y''=0$$ or $$y''=2xy'^3+2yy'^2$$ Which you can differentiate as often as you like.

Hint: the general Taylor series expansion is

$$
\begin{align}
f(x) &= f(x_0) + {f'(x-x_0)(x-x_0)} + {f''(x_0) (x-x_0)^2\over 2!} + \cdots\\
&= \sum_{n=0}^\infty {{d^n\over dx^n}f(x)|_{x=x_0} \times (x-x_0)^n \over n!}
\end{align}
$$

This works wherever all the derivatives exist at $x_0$ and the function is continuous between $x$ and $x_0$.

For this specific problem:

$$
f(x) = f({1\over e}) + f'({1\over e})(x-{1\over e}) + {f''({1\over e})(x-{1\over e})^2\over 2} + {f'''({1\over e})(x-{1\over e})^3\over 3!} + \cdots
$$


Some limits of a function implicitly given by the Mean Value Theorem Let $f: \mathbb{R} \to \mathbb{R}$ be (at least) differentiable and let $x > 0$. For any $\epsilon > 0$, the Mean Value Theorem furnishes a point $c_\epsilon \in (\epsilon, x)$ such that
$$f(x) - f(\epsilon) = f'(c_\epsilon)(x - \epsilon)$$
For the same reason, there is also a point $c \in (0,x)$ such that
$$f(x) - f(0) = f'(c)(x - 0) = f'(c)x$$

I have the following questions:

i) Under what conditions is it true that
$$\lim_{\epsilon \to 0} \, f'(c) - f'(c_\epsilon) = 0$$

ii) Can anything at all be said about the limit
$$\lim_{\epsilon \to 0} \frac{f'(c) - f'(c_\epsilon)}{\epsilon}$$

iii) In particular, is there a non-trivial class of functions such that
$$\lim_{\epsilon \to 0} \frac{f'(c) - f'(c_\epsilon)}{\epsilon} = 0$$

Feel free to impose any additional conditions on the function $f$. Of course, I realize that question ii) is very vague; I am not really expecting an answer to that one. However if you happen to know of any results related to this, I would be very happy if you shared them.

Thanks in advance.

EDIT: The original question doesn't quite make sense. There could of course be several points $t \in (\epsilon, x)$ such that $f(x) - f(\epsilon) = f'(t)(x - \epsilon) := L(x - \epsilon)$. However, if $f \in C^1(\mathbb{R})$ then $f'^{-1}(L)$ is closed. Since it is also contained in $(\epsilon, x)$ it is compact. So define $c_\epsilon$ to be the minimum of all such $t$ and do similarly for $c$. Of course it would also be interesting if anything can be said in the case where $c_\epsilon \in f'^{-1}(L)$ is an arbitrary choice.

i) Under all conditions. Consider:
$$\begin{array}{lcl} f'(c) & = & \frac{f(x)-f(0)}{x} & \text{by definition} \\
 & = & \frac{\lim_{\epsilon \to 0^+}(f(x))-f(\lim_{\epsilon \to 0^+}\epsilon)}{\lim_{\epsilon \to 0^+}(x)-\lim_{\epsilon \to 0^+}(\epsilon)} \\
 & = & \frac{\lim_{\epsilon \to 0^+}(f(x))-\lim_{\epsilon \to 0^+}(f(\epsilon))}{\lim_{\epsilon \to 0^+}(x)-\lim_{\epsilon \to 0^+}(\epsilon)} & \text{because} \ f\ \text{is differentiable, therefore continuous, at} \ 0^+ \\
 & = & \frac{\lim_{\epsilon \to 0^+}(f(x)-f(\epsilon))}{\lim_{\epsilon \to 0^+}(x-\epsilon)} \\
 & = & \lim_{\epsilon \to 0^+}\frac{f(x)-f(\epsilon)}{x-\epsilon} \\
 & = & \lim_{\epsilon \to 0^+}f'(c_\epsilon) & \text{by definition}
\end{array}$$

Therefore $0 = f'(c) - \lim_{\epsilon \to 0^+}f'(c_\epsilon) = \lim_{\epsilon \to 0^+}f'(c) - \lim_{\epsilon \to 0^+}f'(c_\epsilon) = \lim_{\epsilon \to 0^+}(f'(c) - f'(c_\epsilon))$.

ii)
$$\begin{array}{lcl} \lim_{\epsilon \to 0} \frac{f'(c) - f'(c_\epsilon)}{\epsilon} & = & \lim_{\epsilon \to 0} \frac{\frac{f(x)-f(0)}{x} - \frac{f(x)-f(\epsilon)}{x-\epsilon}}{\epsilon} & \text{by definition} \\
 & = & \lim_{\epsilon \to 0} \left( \frac{f(0)-f(x)}{x (x - \epsilon)} + \frac{f(\epsilon)-f(0)}{\epsilon (x - \epsilon)} \right ) & \text{by algebra} \\
 & = & \frac{f(0)-f(x)}{x^2} + \lim_{\epsilon \to 0} \frac{f(\epsilon)-f(0)}{\epsilon x} \\
 & = & -\frac{f'(c)}{x} + \frac{f'(0)}{x} \\
 & = & \frac{f'(0) - f'(c)}{x} \\
\end{array}$$

I'm reasonably confident that the above holds for any function meeting the assumptions. I tested it numerically for a number of "normal" functions (quadratics, sine, cosine, etc.), so at least there's nothing blatantly wrong with the algebra.

iii) It's clear from the above that this limit will equal zero if and only if $(x, f(x))$ is on the line that's tangent to $f$ at zero. The only way this can be a general property of $f$ is if $f$ is its tangent line at zero, that is, if $f$ is a line. But for other functions it can be true for specific values of $x$; for example, if $f(x) = \cos(x)$, then this limit equals zero whenever $x \in \{ 2n\pi \; | \; n \in \mathbb{N} \}$.

Edited to add: I mentioned above that I tested my answer for (ii) numerically for a few different choices of $f$ and $x$. Here is the C code that I used to do that (slightly modified, in an attempt to make it more readable by someone who doesn't know C):

#include <math.h>
#include <stdio.h>

#define PI (4 * atan(1))

double f(double x)
  { return cos(x); }

double f_prime(double x)
  { return -sin(x); }

double f_prime_c(double x)
  { return (f(x) - f(0)) / x; }

double f_prime_c_epsilon(double x, double epsilon)
  { return (f(x) - f(epsilon)) / (x - epsilon); }

int main()
{
  double x = PI / 3;

  // first, print five successive approximations to the limit:
  double epsilon;
  if(x > 1.0)
    epsilon = 1.0;
  else
    epsilon = x;
  for(int i = 0; i < 5; ++i)
  {
    epsilon /= 10;

    double approx =
      (f_prime_c(x) - f_prime_c_epsilon(x, epsilon)) / epsilon;

    printf("%f\t(epsilon=%f)\n", approx, epsilon);
  }

  printf("\n");

  // next, print what we *expect* the limit to be:

  printf("%f\t(epsilon=0; expected)\n", (f_prime(0) - f_prime_c(x)) / x);

  return 0;
}

To compile and run that, assuming you have a Linux-y system available, you would save it as a file named, say, fun_with_mvt.c, and run this command:

gcc -std=c99 -Wall fun_with_mvt.c -o fun_with_mvt && fun_with_mvt

And it should give output like this:

0.451338        (epsilon=0.100000)
0.455521        (epsilon=0.010000)
0.455903        (epsilon=0.001000)
0.455941        (epsilon=0.000100)
0.455945        (epsilon=0.000010)

0.455945        (epsilon=0; expected)

To choose a different $f$ (and $f'$), change the lines that read { return cos(x); } and { return -sin(x); }. To choose a different $x$, change the line that reads double x = PI / 3;.


Is my solution correct? [Compound Monthly Problem] "Find a formula for the present value of $n$ dollars $k$ years from now, assuming an interest rate of $r$% compounded monthly. You can check your answer by using , $n = 10000$, $k = 1$ , and $r = 8$, comparing the result to part (a). "

The result in part (a) was: $(1+0.0066666667)^12*x=10000. $ giving $x = 9233.61$

Answer:

The present value formula is $\frac{n}{(1+(\frac{r}{100}))^k}$ for $n$ dollars $k$ years from now, assuming an interest rate of $r$% compounded anually. However we must divide $r$ by $12$ since the interest is compounded monthly. Also we must multiply $k$ by $12$ since the interest is compounded monthly. Thus our present value formula of $n$ dollars $k$ years from now, assuming an interest rate of r% compounded monthly is $\frac{n}{(1+((\frac{r}{1200}))^{12k}}$

Does anyone have advice on how to clean up my answer or if it is wrong, why?


Maximizing Theta in a Summation Formula I need to take the first derivative of

$$\sum Y_i (\log(\Theta )) +(n-\sum Y_i)(\log(1-\Theta )) $$

with respect to theta, and then solve for theta. I believe this is my derivative...

$$\frac d{d\Theta} =\frac{ \sum Y_i}{ \Theta} +\frac{(n-\sum Y_i)}{1-\Theta} $$

but now (if that is right) I dont know how to solve for theta...

You have forgotten to derive $(1-\Theta)$, which is $-1$.

Therefore the derivative is $$d/d\Theta =\frac{ \sum_{i=1}^n Y_i}{ \Theta} \color{blue}{-}\frac{(n-\sum_{i=1}^n Y_i)}{1-\Theta}=0 $$

$$\frac{1-\Theta}{ \Theta} =\frac{(n-\sum_{i=1}^n Y_i)}{ \sum_{i=1}^n Y_i}$$

$\frac{1}{ \Theta}-1=\frac{n}{ \sum_{i=1}^n Y_i}-1 \quad |+1$

$\frac{1}{ \Theta}=\frac{n}{ \sum_{i=1}^n Y_i}$

$\Theta=\frac{\sum_{i=1}^n Y_i}{n}$

$\boxed{\Theta=\overline Y}$

You want to solve
$0
=d/d\Theta
=\frac{ \sum Yi}{ \Theta} +\frac{(n-\sum Yi)}{1-\Theta}
$.

Letting
$a = \sum Yi$
and
$b = (n-\sum Yi)$,
this is
$0
=\frac{a}{ \Theta} +\frac{b}{1-\Theta}
$.

Multiplying by
$\Theta(1-\Theta)$,
this becomes
$0
=a(1- \Theta) +b\Theta
=a +(b-a)\Theta
$,
so
$\Theta
=\frac{a}{a-b}
=\frac{\sum Yi}{\sum Yi-((n-\sum Yi))}
=\frac{\sum Yi}{2\sum Yi-n}
$.


Some basic book to start with modules? I am very interested in studying modules, which I have studied algebra, is basic theory of groups and rings, of Hungerford and Dummit. I was reading the Dummit the module, but I still struggled a bit in the section and the tensor product of exact sequences, as in the book of Atiyah. They know a simple book to begin with?

The best book I've ever seen on the topic is Module Theory: An Approach To Linear Algebra by T.S. Blyth. The book is extremely rigorous yet gentle and can be read after an undergraduate abstract algebra course. That's the book I'd start with.

Another book you might want to take a look at is Algebra: An Approach Via Modules by Steven Weintraub and W. Adkins. This is a very unusual and nearly forgotten graduate algebra text that presents a standard first year graduate course in algebra with total emphasis on module theory and how linear algebra and its generalizations unify algebra. I found the book very nicely written with a ton of concrete examples-it also has an outstanding final chapter on group representations. But to me, it's too classical and easy for a graduate course.There's no discussion of category theory and very little homological algebra. And the authors deliberately leave out field theory for another book (Weintraub's Galois Theory). I DO think it's a nice book to have and I'd consider using for an honors algebra course instead of Herstein or Artin. For your purposes, I think it's worth a very careful look since the book's overriding theme is module theory and how it relates all of algebra.

Those are the 2 books I'd use.

Sorry at first I didn't notice the rest of your text. you can read (Robert B. Ash, Abstract Algebra) and (herstein).


Examples for Hilbert's Quote Hilbert once said, “The art of doing mathematics consists in finding that special case which contains all the germs of generality.”

What would be (relatively) simple examples?

$$x_{n-1} =\frac{\alpha+\beta x_n +\gamma x_{n-1} +\delta x_{n-2}}{A+Bx_{n}+Cx_{n-1}+Dx_{n-2}}, n= 0, 1, ...,$$

where the parameters $\alpha, \beta, \gamma, \delta, A, B,C, D$ are non-negative real numbers and the initial conditions $m$ are arbitrary non-negative real numbers such that the denominator is always positive.

We are primarily concerned with the boundedness nature of solutions, the stability of the equilibrium points, the periodic character of the equation, and with convergence to periodic solutions including periodic trichotomies.

If we allow one or more of the parameters in the equation to be $0$, then we can see that the equation contains

$$(2^4 -1)(2^4 -1)= 225 $$

special cases, each with positive parameters and positive or non-negative initial conditions.

According to David Hilbert “The art of doing mathematics consists in finding that special case which contains all the germs of generality" and according to Paul Halmos 'The source of all good mathematics is the special case, the concrete example"

The special case of this equation contains a lot of the germs of generality of the theory of difference equations of order greater than one about which, at the beginning of the third millennium, we know no surprisingly little. The mathematics behind the special cases of this equation is also beautiful, surprising, and interesting.

The methods and techniques we develop to understand the dynamics of various special cases of rational difference equations and the theory that we obtain will also be useful in analysing the equation in any mathematical model that involves difference equation.

Take a positive definite quadratic form with integer matrix.  For example, $$w^2+x^2+y^2+z^2$$ or $$w^2+2x^2+5y^2+5z^2$$
It turns out that if this form represents each positive integer from 1 to 15, then it represents all positive integers.  This 15 theorem was first proved by Conway and Schneeberger in 1993, then again (much more simply and beautifully) by Manjul Bhargava in 2000.

Note: For the first example above, the answer is "yes"; for the second example, the answer is "no" -- it takes on every positive integer except 15.