What is the expected value of a probability density function (PDF) itself? The expected value of a function can be found by integrating the product of the function with the probability density function (PDF). What if I want to find the expected value of the PDF itself? This is probably stupidly simple but I am lacking an insight.

Let me explain why I am asking this. In Monte Carlo integration, the expected value of the following term, $F$, gives us the integral $\int_a^b f(x)dx$:

$F = \frac{1}{N}\sum_{i=1}^{N}\frac{f(x_i)}{p(x_i)}$, where $p(x)$ is a PDF from which are drawing samples. We use this to estimate the value of an otherwise difficult to compute integral by averaging samples drawn from a PDF.

In the proof of this, the book I am following goes as follows:

$E[F] = E\left[\frac{1}{N}\sum_{i=1}^{N}\frac{f(x_i)}{p(x_i)}\right] \quad(1)\\
      = \frac{1}{N}\sum_{i=1}^{N}E\left[\frac{f(x_i)}{p(x_i)}\right] \quad(2)\\
      = \frac{1}{N}\sum_{i=1}^{N}\dfrac{\int_a^b f(x)p(x)dx}{p(x)} \quad(3)\\
      = \frac{1}{N}\sum_{i=1}^{N}\int_a^b f(x)dx \quad(4)\\
      = \int_a^b f(x)dx \quad(5)$

The place where I am lost how we transition from step 2 to step 3. I am assuming that we are distributing the expected value to the numerator and the denominator of the fraction. In the numerator, we then have $E[f(x_i)]$, which can be equated to $E[f(x)]$ as $x_i$ is a random variable (we can do this, right?). And we know that $E[f(x)] = \int_a^b f(x)p(x)dx$, so I see where the numerator in step 3 comes from. But I cannot see how we obtain $E[p(x_i)] = p(x)$. That's why I asked how can we determine the expected value of a PDF.

At this point in my edit of the question, I realized that it may be incorrect to distribute the expected value over the ratio. In this case, we do not have the problem that is asked. Does the following make sense then?

$E\left[\frac{f(x_i)}{p(x_i)}\right] = E[g(x_i)]\\
 = \int_a^b g(x)p(x)dx\\
 = \int_a^b \frac{f(x)}{p(x)}p(x)dx\\
 = \int_a^b f(x)dx$

If this is correct, I guess I have answered my own question but it would be great to get confirmation.

A random variable $X$ has eventually an expectation which - if $X$ has a PDF $f$ - can be found as: $$\mathbb EX=\int xf(x)dx$$

Further for a suitable function $g:\mathbb R\to\mathbb R$ we can find the expectation of $g(X)$ by: $$\mathbb Eg(X)=\int g(x)f(x)dx\tag1$$
You want to find the "expectation of the PDF". However, for that the PDF must be a random variable, so what you ask is absurd.

It is a fact though that $f(X)$ is a random variable. Applying $(1)$ its expectation can be found as:$$\int f(x)f(x)dx=\int f(x)^2dx$$Is this what you want maybe? Btw, I never engaged any meaningful stuff on this.

Random events have two characteristics: a value and a probability. It's clear in the discrete case; a normal die has a 1/6 probability of rolling each of one through six. You can find the expected value of one roll, it's $\frac{1+2+3+4+5+6}{6}$. But you can't find the expected value of the probabilities, because it's just not a meaningful question.

The same is true for continuous random events. "the function" is the value of the event, and the PDF is the probability. So you can find the expected value of the event, with the understanding that its values all have probability given by the PDF.

To expand a little bit, you can think of the pdf as representing values instead, but then you would need to specify a probability function for those values to occur, so you would need a different PDF.


Can a metric be evaluated at infinity? I just stared a new math course, and we began by reviewing metric spaces. The professor gave us the set $X = (-\infty,\infty)   $ , so all real numbers, and asked us to come up with any metric for it. When we suggested $|x-y|$, the simplest we could think of, he said that we were wrong because this failed to work for infinity. His alternative was a metric defined by $arctan|x-y|$ as this would return a non-infinite value when $+\infty$ or $-\infty$ was used for $x$ or $y$. This really confused me since I've taken a course on Real Analysis and this notion was never once mentioned. Also, it doesn't make sense for me intuitively since you can't "plug in infinity" since infinity is a concept, not a number. Thoughts?

On $\mathbb{R} = (-\infty, \infty)$, the function defined by $d(x,y) = |x - y|$ is certainly a metric. It's called the Euclidean metric.

For the extended reals $\overline{\mathbb{R}} = [-\infty, \infty]$, the same definition does not yield a metric since a metric must evaluate to a finite number. In this case, you need to modify your definition to something like $d(x,y) = \arctan|x-y|$.

You must think that $\infty$ and $-\infty$ are indeed points and they are not just concepts. Your professor is asking about a metric on $X$ that preserve the usual topology on $\mathbb{R}$ and such that preserve our usual understanding about infinities, I mean that $-\infty<x<\infty$. The trick is to find some continuos subadditive (think about triangular inequality) increasing function $f$ on $[0,\infty]$ such that $\lim_{x\rightarrow\infty} f(x)$ exists and is finite. Then every infinite point has finite distance to each other point.


Rule for multiplying rational numbers The rule for multiplying rational numbers is this:

$\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$

Can the rule be proven or is it meant to be taken as a given?

Edit: Where $b\neq 0$ and $d\neq 0$.

One knows that $a/b$ may be defined as the number $x$ such that $bx=a$, and that $c/d$ may be defined as the number $y$ such that $dy=c$.

If one wants the multiplication on these objects to be associative and commutative as it is on the integers, one should ask that $ac=(bx)(dy)=(bd)(xy)$ hence that the object $xy$ fits the definition of $(ac)/(bd)$.

It is more or less taken as a given. It fits the intuition and the construction of the rational numbers (which contains the definition of this multiplication) was generalized to arbitrary commutative rings (+ the choice of a multiplicative subset). This construction is called the localization (see wikipedia).


Let $\theta=\frac{2 \pi}{67}$ consider the rotation matrix $A$. What is $A^{2010}$? Let $\theta=\frac{2 \pi}{67}$. Consider the matrix
  $$A = \begin{pmatrix}
\cos\theta  & \sin\theta\\
 -\sin \theta&  \cos \theta
\end{pmatrix} $$
  Then the matrix $A^{2010}$ is?

My approach
$$A^2 = \begin{pmatrix}
\cos2\theta  & \sin2\theta\\
 -\sin 2\theta&  \cos 2\theta
\end{pmatrix} $$

$2010$ is a multiple of $67$. So I'm trying to convert in such format by taking $A$ matrix as power of $2$ and multiplying. But $2048$ is the nearest $2$ power term. So help please. I dont know if you can understand what I did!! Sorry for being not explaining properly

Observe that when you multiply the matrices
$$\begin{pmatrix}
\cos\theta  & \sin\theta\\
 -\sin \theta&  \cos \theta
\end{pmatrix} *
\begin{pmatrix}
\cos\phi  & \sin\phi\\
 -\sin \phi&  \cos \phi
\end{pmatrix}
=
\begin{pmatrix}
\cos(\theta+\phi)  & \sin(\theta+\phi)\\
 -\sin (\theta+\phi)&  \cos (\theta+\phi)
\end{pmatrix}.$$

(You can show this with the multiangle formulae for $\sin$ and $\cos$.) So we have

$$A^{67} = \begin{pmatrix}
\cos67\theta  & \sin67\theta\\
 -\sin 67\theta&  \cos 67\theta
\end{pmatrix}
=
\begin{pmatrix}
\cos2\pi  & \sin2\pi\\
 -\sin 2\pi&  \cos 2\pi
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} = I.$$
$$\therefore A^{2010} = (A^{67})^{30} = I.$$
(I tried to put the end bit as a 'spoiler' which you have to mouse over to view, but I couldn't work out how to!)

As mentioned by other people, applying the matrix
$$M_\theta = \begin{pmatrix}
\cos\theta  & \sin\theta\\
 -\sin \theta&  \cos \theta
\end{pmatrix}$$
has the action of rotation a vector (anticlockwise) by $\theta$. Thus doing it repeatedly simply means that they sum: $M_\theta * M_\phi = M_{\theta + \phi}$.

Hopefully this helps! :)

Building on what was said, you can view $R(\theta)$ as a rotation matrix in the complex plane. Indeed, for a vector written as $v=x+iy$ , then the rotation matrix is simply the operator $e^{i\theta}$.
This is shown by Euler's formula.
From this, you know that ${(e^{i\theta})}^n=(e^{ni\theta})$ which links back to what was shown.


Is the intersection between a subgroup, and a normal subgroup, normal in the parent group? Given $H, N \subset G$ and $N \lhd G$ is there some underlying fact or theorem for why $H \cap N$ would or would not be normal in $G$? My reasoning would be that it would be normal to $G$ as; $\forall x\in H \cap N, x \in N$, and thus it would be normal. Is this sufficent to prove this true or would I need further reasoning?

I know $H \cap N \lhd H$ due to the second isomorphism theorem, could this be extended to G some how?

Thank you for your time.

This is generally false: Consider $G=S_{n}$ for $n\ge3$ and $H=\langle(ab)\rangle=\{e,(ab)\}$
where $(ab)\in S_{n}$ is a two cycle.

$H\leq G$ but is not a normal subgroup (why ?).

Let $A$ be the set of all two cycles in $S_{n}$ and $N=\langle A\rangle$then
$N\trianglelefteq G$ (why ?)

Now
$$
H\subseteq N\implies H\cap N=H
$$

thus the intersection is not normal in $G$ as $H$ is not.

Note: the fact that $H\cap N\trianglelefteq H$ is simple and we do
not need the second isomorphism theorem - if $h\in H$ then
$$
h^{-1}Hh=H
$$

since $H$ is a group.

$$
h^{-1}Nh=N
$$

since this is true for all $g\in G$ and in particular to elements
of $h$. thus if $x\in N\cap H$ and $h\in H$ we have it that
$$
h^{-1}xh\in H\cap N
$$

which proved the normality of $H\cap N$ in $H$.

In general, I don't know about sufficient and necessary conditions
for this to hold. There are, however, some simple sufficient conditions
such as that that the order of $H,N$ are co prime (why ?)

Although your solution is correct, your $N$ is just $S_n$ (although it is true that this is a normal subgroup and the rest of the proof works). To generate $S_n$, we don't even need all two cycles, only $(1 2)$, $(1 3)$, $\dots$, $(1 n)$ is enough (because if you want to move $i$ to $j$, where $i \neq 1$, you only need to apply $(1 i)$ and then $(1 j)$. The remaining element is for $1$).


Statements possibly equivalent to the $\epsilon{-}N$ definition of limit of a sequence Which of these statements are equivalent to the $\epsilon{-}N$ definition of limit of a sequence?

  (a) For every integer $m>0$, there in an integer $N>0$ such that if $n > N$ then
  $|a_n - \alpha| < \frac1m$

  (b) For each $0<\epsilon <1$, there in an integer $N>0$ such that if $n>N$ then $|a_n - \alpha| < 3\epsilon$

  (c) For each $0<\epsilon <1$, there in an integer $N>0$ such that if $n>N$ then $|a_n - \alpha| < \frac{1}{\epsilon}$

  (d) For each $\epsilon > 0$, there in an integer $N>0$ such that if $n>N$ then $|a_n - \alpha| < \frac{1}{\epsilon}$

  (e) For each $\epsilon > 0$, there in an integer $N>0$ such that if $n>N^2$ then $|a_n - \alpha| < \epsilon$

  (f) For each $\epsilon > 0$, there in an integer $N>0$ such that if $n=N+1$ then $|a_n - \alpha| < \epsilon$

Can anyone help me with this? To me all of these seem equivalent to the original
$\epsilon-N$ definition where $\alpha$ is the limit of the sequence. In all the cases we find an $N$ such that for all $n>N$ the statement $|a_n - \alpha| < something$ is true. In (e) and (f) $n>N^2$ and $n = N+1$, respectively, but in both cases it is still greater than $n>N$ so I don't see how this effects the statements.

The definition of limit of a sequence $(a_n)_{n\in \mathbb{N}}$ is:
$$\lim_{n \to +\infty} a_n = \alpha$$ means:

  for every $\varepsilon > 0$, "no matter how small", $(a_n)_{n\in \mathbb{N}}$ is, from some point on, within $\varepsilon$ of $\alpha$.

"$x$ is within $\varepsilon$ of $y$" means $|y -x| < e$.

A condition or property is true a sequence $(a_n)_{n\in \mathbb{N}}$ "from some point on" means that there's an $N \in \mathbb{N}$ such that the property or condition is true about $a_n$ for all $n \ge N$. The $a_i$ can fiddle around for some finite number of $i$, but eventually (starting at $N$) the $a_i$ have to settle down and from that point ($N$) on, for all $n \ge N$, $a_n$ fulfills the condition.

Loosely: to within whatever degree of accuracy or precision we require, the sequence eventually gets that close to $\alpha$ and stays that close forever after.

With this in mind, let's run through the statements you list. They're all attempts at expressing that the sequence converges to $\alpha$.

Statements (a) - (f)

(a) True (equivalent). For any real $\varepsilon$ there's an integer $m$ with $\frac 1 m \le \varepsilon$.

(b) True. Sufficiently small $\varepsilon$ will be smaller than $1$. We're interested in shrinking the neighborhood around $\alpha$ and ensuring that no matter how small it is, from some point the on the sequence gets into that neighborhood and stays in that neighborhood.

(c) False (not equivalent): This says that for any arbitrarily large real number $r$ (namely, $1/\varepsilon$), the sequence is eventually within distance $r$ of the limit. But... big deal. This doesn't tell us that the sequence gets as close as we require to $\alpha$ and stays that close forever, which is what convergence means.

(d) True. Unlike (c), $\varepsilon$ can be taken large, so $1/\varepsilon$ can be arbitrarily small.

(e) True: Given $\varepsilon$, there is $M$ (changing variable names) such that from $N = M^2$ on, $a_n$ is within $\varepsilon$ of $\alpha$ for $n > N$.

(f) False: it says "$n = N+1$" which doesn't say anything about the "eventual" behavior of the sequence. (Exercise: find a counterexample.)


One may see that for the initial inequality to hold true it is sufficient to prove that
$$
1-\frac{(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right)}{16 x(1-x)^2 }<0,\quad x \in (0,1), \tag1
$$ then setting
$$
\begin{align}
&f(x)=16 x(1-x)^2-(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right)
\end{align}
$$ one gets$$
\begin{align}
&f'(x)=700-2044 x+2535 x^2-1668 x^3+575 x^4-78 x^5 \in [20,700] \tag2
\\\\&f'(x)>0\implies f \nearrow, \quad x \in (0,1), \quad f(0)=-216, \quad f(1)=-8,
\end{align}
$$ giving $$ f(x)<0, \quad  x \in (0,1), \tag3$$
which yields $(1)$ then yielding the initial inequality.

Remark. By setting $t=1-x$ in $f'(x)$ above, one gets a polynomial with positive coefficients:

  $$
20+68 t+201 t^2+148 t^3+185 t^4+78 t^5 >0,\quad t \in (0,1),
$$

proving $(2)$.

Hint #1:  Assume to the contrary that
$$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)\left(2x(1-x)\right)^2} \geq 1$$
and then derive a contradiction to $0 < x < 1$.

Hint #2:  (Note: This edit is in response to Joshua Lochner.)
So the proof by contradiction actually goes like this.  Assume to the contrary that there exists an $x \in \mathbb{R}$ such that
$$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)\left(2x(1-x)\right)^2} \geq 1.$$
By this WolframAlpha computation, this implies that
$$\left\{x \geq 3.9483\right\} \lor \left\{1 < x \leq 1.41066\right\} \lor \left\{x < 0\right\},$$
which contradicts the assumption $0 < x < 1$.  Hence, we conclude that
$$\forall x \in (0, 1), \hspace{0.5in} \frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)\left(2x(1-x)\right)^2} < 1$$


Integration of fractions and cancellation How come $$\int\frac{3}{6+3x}\mathrm{d}x=\mathrm{ln}(2+x)+C$$
and not $\ \mathrm{ln}(6+3x)+C$ ?

I understand you can cancel the fraction but why should this make a difference?

I've tried searching through all my notes on this and online but the terms I am using do not return anything useful.

$$\ln(6+3x)+C=\ln(3[2+x])+C=\ln(2+x)+(\ln 3+C)$$

In other words, both answers are correct, they just use different arbitrary constants.

It is divided by $3$ and thus follows $\log(2+x)$


Hint: $1$ has more than one possible square root, and you need to check them all when you do a calculation like you are doing.

Hint:

$$\sqrt{9}=3 \land -3$$

People tend to forget these simple things...


Basically disconnected space which is not extremally disconnected Space $X$ is basically disconnected if every cozero-set has an open closure.

Every extremally disconnected space is basically disconnected But i think the converse fails. The one-point compactification $\alpha D(\tau)$ of an uncountable discrete space is a counterexample?

That doesn’t quite work: if $C$ is a countably infinite subset of $D(\tau)$, $C$ is a cozero-set in $\alpha D(\tau)$ whose closure isn’t open.

Take $X$ to be the one-point Lindelöfization of an uncountable discrete space, with non-isolated point $p$: nbhds of $p$ are the sets $X\setminus C$ for countable $C\subseteq(X\setminus\{p\})$. $X$ is a $P$-space, meaning that $G_\delta$-sets in $X$ are open, so every zero-set in $X$ is open, and therefore every cozero-set in $X$ is closed, and $X$ is basically disconnected.

Now let $X\setminus\{p\}=U\cup V$, where $U$ and $V$ are uncountable and disjoint. $U$ and $V$ are open, but $\operatorname{cl}U=U\cup\{p\}$ and $\operatorname{cl}V=V\cup\{p\}$ are not disjoint, so $X$ is not extremally disconnected.

you can find examples in the Gillman and Jerison's book: Ring of continuous functions


Secant method with two ODE's of degree 2 - matlab $$\frac{d^2r}{dt^2}-r\left(\frac{d\phi}{dt}\right)^2=G\cos\alpha-g\frac{R^2}{r^2}$$
$$r\frac{d^2\phi}{dt^2}+2\frac{dr}{dt}\frac{d\phi}{dt}=G\sin\alpha$$

The two ODE's above are given. I have written them as 4 ODE's of degree one and solved the equation with ODE45.

I've plotted the graph and now I want to find where the equation intersects with the horizontal line $y=1$. I want to use the secant method becuase then i do not have to calculate the derivative of my equation. Normally I would just take: "my equation" = 1, but how do I do it in this case when I got two equations?

Thanks

I'm going to assume you have converted from your polar coordinates to x-y coordinates.

In such a case, let y be the vector of $y$ values at each time step $t_i$. To find where $y$ is close to $1$, simply do the following:

[mn,idx] = min(abs(y-1));

This gives you the minimum absolute distance as well as the index of the element, stored in idx. You can then interpolate from the neighboring indices in whichever way you'd like.

Try this. (You will need to subtract 1 from your vector, because the function finds the zero crossing.index is supposed to be t vector in your case)

function [ out ] = VectorZero( index, vector )

out=[];
for i=1:length(index)-1

if(vector(i)*vector(i+1)<0)
    out=[out (index(i+1)*abs(vector(i))+index(i)*abs(vector(i+1)))/(abs(vector(i))+abs(vector(i+1)))];
end

if(vector(i)==0)
    out=[out index(i)];
end
end

I have written a few functions that find the index of min/max in a vector, finds the index of a value in a vector etc. Matlab doesn't seem to have a default function for things. Or maybe I didn't look hard enough.


Big subalgebras of the free polynomial algebra Let $R$ be a Noetherian commutative unital ring. Let $n\geq 1$ be an integer. Suppose that $R[x_1, \dots, x_n]$ has a $R$-subalgebra $A$ such that $R[x_1, \dots, x_n]$ is a free finitely generated $A$-module. Is it true that $A$ is isomorphic to $R[x_1, \dots, x_n]$ as an ungraded $R$-algebra? Is this true for $n=2$ at least?

Examples I have in mind are $R[x_1^{i_1},\dots, x_n^{i_n}]\subset R[x_1,\dots, x_n]$ for $i_1, \dots, i_n\geq 1$ for which this is true.

One can note that $A$ has the same Krull dimension as $R[x_1,\dots, x_n]$.

EDIT: Under the assumptions $A$ is finitely presented. The answer should give an explicit presentation for $A$.

This is true when $n=2$ and the ground ring is an algebraically closed field of characteristic $0$ by https://projecteuclid.org/download/pdf_1/euclid.ojm/1200773129


Eigenvectors of a $2 \times 2$ matrix when the eigenvalues are not integers How can I calculate the eigenvectors of the following matrix?

$$\begin{bmatrix}1& 3\\3& 2\end{bmatrix}$$

I calculated the eigenvalues. I got

$$\lambda_1 =  4.541381265149109$$

$$\lambda_2 = -1.5413812651491097$$

But, now I don't know how to get the eigenvectors. When I create a new matrix after I subtracted Lambda value from all the members of the matrix on the main diagonal and tried to solve the homogeneous system of equations, I get only null vector for both $\lambda_1$ and $\lambda_2$....

When I used this website for calculating eigenvalues and eigenvectors. I got these eigenvectors

$(0.6463748961301958, 0.7630199824727257)$ for $\lambda_1$

$(-0.7630199824727257, 0.6463748961301957)$ for $\lambda_2$

.... but have no idea how to calculate them by myself...
Is it even possible? ....or it's possible to calculate it numerically?

Indeed, as Chris Eagle and Michael pointed out to you, calculators are not always your best friend.

Instead, if you do your maths with the characteristic equation, you'll find out that the eigenvalues look nicer this way:

$$
\lambda = \frac{3 \pm \sqrt{37}}{2}
$$

And it's not at all impossible to find the eigenvectors. For instance the one with the $+$ sign, you could start like this:

$$
\begin{pmatrix}
1 - \dfrac{3 + \sqrt{37}}{2}  &   3                            & \vert & 0  \\
3                             &   2 - \dfrac{3 + \sqrt{37}}{2} & \vert & 0
\end{pmatrix}
$$

Hint: After some easy simplifications, you'll find out that it's very useful to multiply one of the rows by $1 - \sqrt{37}$ and that you can write the corresponding eigenvector as simple as this: $(1 - \sqrt{37}, -6)$.

By definition an eigenvector for eigenvalue $\lambda$, is a member of nullspace of $A - \lambda I$, if you can compute nullspaces, you can compute eigenvectors.


Noetherian Schemes Property Let $X$ be locally Noetherian Scheme, therefore for every $a \in X$ there exist an affine open set $U =\mathrm{Spec}(R)$ such that $R$ noetherian.

Why is then every $S_V:= \Gamma(V, \mathcal{O}_X)$ for each affine open $V$ noetherian?

I will reduce this into a statement of commutative algebra for you.

It is enough to prove: If $X=\operatorname{Spec}(A)$ is locally noetherian, then $A$ is noetherian. We may cover $X$ by open affines $\operatorname{Spec}(B)$ with $B$ noetherian. Such an affine open may be covered by principal open subsets $D(f)$, with $f\in A$, of $X$. If $\varphi\colon A\to B$ denotes the morphism of rings corresponding to the inclusion $i\colon \operatorname{Spec}(B)\subset \operatorname{Spec}(A)$, then $D(f)=i^{-1}(D(f))=D(\varphi(f))$, i.e. $B_{\varphi(f)}\cong A_f$, so that $A_f$ is noetherian. This shows that we may cover $X$ by $D(f_1),\ldots,D(f_n)$ with $A_{f_i}$ noetherian, in other words: We are given elements $f_1,\ldots,f_n\in A$ generating the unit ideal such that $A_{f_i}$ is noetherian for $i=1,\ldots,n$. Deduce from this that $A$ is noetherian.

Let $a\in X$ and $V$ be an open affine n'hood of $a$. Since $X$ is locally Noetherian, $V=Spec(R)$ for some Noetherian ring $R$ and $\Gamma(V,\mathcal{O}_{X})\cong R$ (since $V$ is affine) hence it is Noetherian.


Confusion in understanding why empty set is a subset of every set statement 1 : If $x \in \emptyset$ then $x \in A$.

statement 2 : If $x \in \emptyset$ then $x \notin A$.

I know that both statements are true since the hypothesis is false.

But first statement says that $\emptyset \subset A$ while second statement says that $\emptyset$ is not a subset of $A$.

My question is why we prefer $\emptyset \subset A$ over the other implication that $\emptyset$ is not a subset of $A$? Thanks.

The statement

  (For all $x$,) if $x\in B$ then $x\notin A$.

does not say that $B$ is not a subset of $A$. In order to say that $B$ is not a subset of A, you would need to say

  There is an $x\in B$ such that $x\notin A$.

This is vacuously false then $B$ is $\varnothing$.

In elementary set theory, each set can be defined as $A=\{x\mid P(x)\}$ where $P(x)$ is a predicate with free variable $x$, and each subset as $B=\{x\mid x\in A\wedge Q(x)\}$ with predicate $Q$.

Define $\emptyset_A=\{x\mid x\in A\wedge x\not\in A\}$. This is a subset of $A$, the empty subset of $A$. For any other set $B$, define similarly $\emptyset_B=\{x\mid x\in B\wedge x\not\in B\}$ as the empty subset of $B$.

But $\emptyset_A=\emptyset_B$, since $x\in\emptyset_A\Leftrightarrow x\in A\wedge x\not\in A\Leftrightarrow x\in B\wedge x\not\in B\Leftrightarrow x\in\emptyset_B$. In the 2nd equivalence, a false assertion is replaced by another false assertion.

Hence, the empty subset of a set $A$ is independent of the set $A$ itself and so we can safely speak of the empty set by putting $\emptyset=\emptyset_A$.


Percentage of primes among the natural numbers How high is the percentage of primes in $\mathbb{N}$?

($\mathbb{N} := \lbrace { 1, 2, 3, \ldots \rbrace }$ ; a prime is only divisible by itself and 1 in $\mathbb{N}$)

The percentage has to be lower than 50% as all even numbers (except for 2) aren't primes.
So the percentage has to be lower than $1 - (\frac{1}{2} + (\frac{1}{3} - \frac{1}{6})) = 1 - (\frac{1}{2} + \frac{1}{6}) = 1 - (\frac{1}{2} + \frac{1}{2 \cdot 3}) = \frac{1}{3}$

I guess it will be something like that:

$$\frac{\text{primes in } \mathbb{N}}{\text{numbers in } \mathbb{N}} = 1 - \sum_{i=\text{first Prime}}^\text{primes} \frac{1}{\prod_{j=\text{first prime}}^{i\text{-th prime}} j}$$

But calculating this (exact) value goes definitely beyond my math skills. Can somebody help me?

People are quoting the Prime Number Theorem here, but that is pretty serious overkill.  Even Chebyshev's theorem is a level above, in my opinion.
In my undergraduate number theory course, I prove that the prime numbers have density zero in the positive integers (including a careful statement of what this means).  The proof is given as Theorem 6 in these notes (Wayback Machine).  In fact it builds on the OP's observation that the density must be at most $\frac{1}{2}$ because half of all numbers are divisible by $2$.  Similarly the density must be at most $\frac{1}{3}$ because every prime number $p > 5$ is relatively prime to $6$, and $\frac{\varphi(6)}{6} = \frac{1}{3}$.  One can get a complete proof by showing that for each $\epsilon > 0$, there exists a positive integer $d$ such that $\frac{\varphi(d)}{d} < \epsilon$.  This was proved earlier in the course: it is Proposition 6 in this set of notes (Wayback Machine).  The proof doesn't use any analytic fact deeper than the divergence of the harmonic series.

Not completely sure about this but I am working on proving it:

6.598827715%

That is, e raised to the negative e (expressed as a percentage)

The number of primes is the likelihood of zero for a Poisson Distribution with Lambda of e.


Prove that $(S \cap T = \varnothing) \land (S \cup T = T) \rightarrow S = \varnothing$. Logically, the following proposition makes sense:

$(S \cap T = \varnothing) \land (S \cup T = T) \rightarrow S = \varnothing$

Or, in english, if sets $S$ and $T$ share no elements, and the union of the sets is equal to one of the sets, then the other set must be empty.

However, what I'm struggling with is how to translate this english into mathematical symbols.

Would anyone know how to solve this with mathematical symbols?

$$ (S \cap T = \varnothing)\tag{1}$$
$$(S \cup T = T) \tag{2}$$

Now:
Starting with $(2)$
$$S \cup T = T \implies S \subseteq T \implies S \cap T = S$$

Since we have $(1): S\cap T = \varnothing$, then $$S\cap T = S = \varnothing$$

Informally, assume that S is non-empty. Then $S \cup T$ is not T, which contradicts the assumption.


Steps to solve this recurrence relation? I have the following question:

I am aware that I have to find the characteristic polynomial of this equation but I do not understand how to deal with $64 . 3^{n-4}$ so could anyone explain how to solve a recurrence relation of this type?

Also, since this problem expands no $a_{n-4}$ does this mean the characteristic polynomial will be something with $x^4$ ??

After getting the homogeneous solution $a_{h,n}$ from the characteristic equation ($a_{h,n}$ is the solution of the equation $a_{h,n}=6\ a_{h,n-2} + 8\ a_{h,n-3}+ 3\ a_{h,n-4}$), we are just assuming a particular solution is $a_{p,n}=c_n \cdot 3^n$. All we need to do is  apply to the equation and find the $c_n$.

$$ a_{p,n}=6 a_{p,n-2} + 8 a_{p,n-3} + 3 a_{p,n-4} + 64\cdot 3^{n-4} \\ \Longrightarrow c_n\ 3^n=6c_{n-2}3^{n-2}+8c_{n-3}3^{n-3}+3c_{n-4}3^{n-4}+64\cdot3^{n-4} \\ \Longrightarrow c_n= 6c_{n-2}/9+8c_{n-3}/27+3c_{n-4}/81+64/81 $$

If you will get $a_{p,n}$ properly, the original solution is follow

$$a_n=a_{h,n}+a_{p,n}$$

For the entire procedure,

Find a homogeneous solution from the characteristic polynomial. $a_{h,n}$
Do not apply initial conditions because it is not complete.
Find a particular solution from the above method. $a_{p,n}$
Create a general solution using $a_n=a_{h,n}+a_{p,n}$
Solve the general solution using the initial conditions.


Can we find all the irreducible polynomials of $F_2[x]$? Can we find all the irreducible polynomials of $F_2[x]$ of a degree $n$?

Is the number of irreducible polynomial of $F_2[x]$ Infinite?

I was to find if there is any degree $n$ such that there is no irreducible polynomial of $F_2[x]$ of that degree.

Can anyone help me by giving some hints?   I think my three doubts are related to each other.

The first question is hard.  To answer the second and third question:

The Necklace polynomial $M(2,n)$ simultaneously represents:

The number of aperiodic necklaces which can be made by arranging $n$ colored beads, from $2$ colors; and
The number of monic irreducible polynomials of degree $n$ over a finite field with $2$ elements.

It is clear that this is strictly positive for all $n>0$ (e.g. one white bead and the rest black beads), hence there is at least one monic irreducible polynomial of each positive degree (over $F_2$).  In particular, there are infinitely many monic irreducible polynomials (over $F_2$).


A question about series ratio test Could you please give me some hint how to deal with this question:

Suppose $\left|\frac {a_{n+1}}{a_n}\right|\le c_n$ for each n and $c_n<1$.

May we conclude that $\left|\frac {a_{n+1}}{a_n}\right|\le c_n<1$ and by ratio test series $\sum a_n$ converges absolutely ?
Or there is counter-example ?

Thanks.

Define a sequence $a_1 = 3$, and $a_{n+1} = a_n - \dfrac{a_n}{n}$ for $n \geq 1$ then $\dfrac{a_{n+1}}{a_n} = 1 -\dfrac{1}{n}$, and $\lim_{n \to \infty} |\dfrac{a_{n+1}}{a_n}| = 1$. This example shows that the above convergence test can not be used since the limit of the ratio could be $1$ while the ratio test requires that it be less than $1$ for the series to be convergent.

It doesn't matter that you have used the "intermediary variable" $c_n$. You have shown that $$|\frac{a_{n+1}}{a_n}| < 1$$ and if this is true also for $n \to \infty$ then, by the ratio test, the series converges absolutely.


Determine if $\sum_{n=2}^{\infty} \frac{(-1)^n}{n+(-1)^n}$ converges or diverges I'm having a lot of trouble figuring this one out.

Determine if  $\sum_{i=2}^{\infty} \frac{(-1)^n}{n+(-1)^n}$ converges or diverges

Both ratio and root test are inconclusive and I'm at a loss. Can anyone help me?

Let

$$S_n=\sum_{k=2}^n \frac{(-1)^n}{n+(-1)^n}$$

Take the sum of two consecutive terms:

$$\frac{(-1)^{2n}}{2n+(-1)^{2n}}+\frac{(-1)^{2n+1}}{2n+1+(-1)^{2n+1}}=\frac{1}{2n+1}-\frac{1}{2n}=\frac{-1}{2n(2n+1)}$$

Thus $S_{2n+1}$ converges. Since the terms tend to $0$, $S_{2n}$ is also converging, to the same limit.

Terms are alternatively positive and negative, decrease in absolute value and converge to $0$, so the sseries converges (this is the alternating series test), but it does not converge absolutely.


Finding area between polar curves using double integral The question asks to find the area inside $r = 1 + \sin\theta$ and outside $r = 2 \sin\theta$ using double integrals.

In my attempt, I found the intersection to be $\theta = \frac{\pi}{2}$.

I bounded $r$ as $[2 \sin \theta, 1 + \sin \theta]$, since I couldn't find another intersection for the curves, I bounded $\theta$ as $[-\frac{\pi}{2}, \frac{\pi}{2}]$ to solve for half of the area required.

Then multiplied everything by $2$ to get the full area. I ended up with $\frac{-\pi}{2}$ while the correct answer is $\frac{\pi}{2}$.

What am I doing incorrectly?

The polar equation $$r(\theta):=2\sin\theta\qquad(0\leq\theta\leq 2\pi)$$ describes a circle of radius $1$ centered at $(0,1)$, covered twice. On the other hand, the area inside the curve $$r(\theta):=1+\sin\theta\geq0\qquad(0\leq\theta\leq 2\pi)$$ is given by
$${1\over2}\int_0^{2\pi}(1+2\sin\theta+\sin^2\theta)\>d\theta={1\over2}(2\pi+0+\pi)={3\pi\over2}\ .$$
The area $A$ in question therefore has the value ${\displaystyle{\ 3\pi\over2}-\pi={\pi\over2}}$.

If you want to compute $A$ in terms of a double integral you should stay with polar coordinates. You then obtain
$$A=2\left(\int_{-\pi/2}^0\int_0^{1+\sin\theta} r\>dr\ d\theta+\int_0^{\pi/2}\int_{2\sin\theta}^{1+\sin\theta} r\>dr\ d\theta\right)\ .$$


Ring having no nilpotents Consider the ring $\mathbb{C}[x_1,\dots,x_n]/I$, where $I=(\sum_{i=1}^nx_i^2)$. I want to understand for what $n$ does this ring contain no nilpotents (i.e., for what $n$ is this algebra reduced).

I have no idea how to approach this problem. If $f+I$ is a nilpotent then $(f+I)^k=I$, but how to deduce the information about $n$ from here?..

the algebra $k[x_1,...,x_n]/I$ is reduced if and only if $I$ is a radical ideal. this from "Elementray algebraic geometry" Klaus Hulek, page 39


Reason for the diferentiability of function on open interval in Rolle's Theorem "Let $f(x)$ be a continuous function on $[a,b]$ and differentiable on $]a,b[$ ,further let $f(a)=f(b)$. Then there exist at least one number $c \in ]a,b[$ such that $f'(c)=0$". The above statement is the Rolle's Theorem. Why we take open interval for differentiability rather then close interval. What's the reason behind it??

See the proof of the theorem. If $f$ isn't constant, the point where $f$ reaches the max or the min $\in(a,b)$, so is a local extremum...

Simply put: there is no two-side derivative for the bounders of the closed interval. only one side derivative exists for $a$ and $b$. so we must exclude the two points.


Solve the equations set in complex numbers. 1) Solve the equation $x^2+4=0$ set in complex numbers.
My solution, pretty sure this is right
$$x^2+4=0$$
$$x^2=-4$$
$$x=2i$$

But I have issues with this one:

2) Solve the equation $x^2+x+1=0$ set in complex numbers.

Not quite sure what to do here, I tried calculating delta:

$$\Delta = 1-4$$
$$\Delta = -3$$
$$\sqrt{\Delta}=\sqrt{3}i$$

But delta is negative so theres no solutions.

What's the right method for this example?

1) Solve the equation $x^2+4=0$ set in complex numbers.
  My solution, pretty sure this is right
  $$x^2+4=0$$
  $$x^2=-4$$
  $$x=2i$$

Just as the equation $x^2-4=0$ has two solutions (namely $x=\pm 2$), you're missing the solution $x=-2i$.

  But I have issues with this one:

  2) Solve the equation $x^2+x+1=0$ set in complex numbers.

If the discriminant $\Delta = b^2-4ac$ of a quadratic equation $ax^2+bx+c=0$ with real coefficients ($a,b,c \in \mathbb{R}$) is negative, then the (two) complex (conjugate) solutions are given by:
$$\frac{-b\pm\sqrt{-\Delta}i}{2a}$$

So for 1), remember that:

$x^2 = -4 \Rightarrow x = \pm \sqrt{-4} = \pm 2i$

As for 2), you can use the quadratic equation to your advantage here, which you can look up how to derive if you need to do things from first principles.

Write our equation in the following form $ax^2 + bx + c = 0$, then:

$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a} \Rightarrow x = \frac{-1\pm \sqrt{1-4\cdot1\cdot1}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm\sqrt{3}i}{2}$.

However, on a note of pedantry, I would like to remind you that there is no "right method" so to speak. Any method that doesn't contain a false piece of logic is ultimately valid, though obviously some methods are quicker and easier than others.

I hope I helped. Feel free to ask any questions, i'm more than happy to clear anything up for you.


If $x'(t) \leq f(x(t))$, then $x(t)\leq y(t)$ for which $y'(t)=f(y(t))$? Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be some Lipschitz function, and assume $x(t)$ satisfies $$x'(t) \leq f(x(t))\ {\rm with }\ x(0) = a$$ Let $y(t)$ be the unique solution to $y'(t) = f(y(t))$ with $y(0)=a$, does it then hold that $$x(t) \leq y(t)$$ for all $t\geq 0$?

If you consider $h(t):=x(t)-y(t)$ you have that

$|h’(t)|=|x’(t)-y’(t)|=|x’(t)-f(y(t))|$

$\leq |f(x(t))-f(y(t))|\leq |x(t)-y(t)|=|h(t)|$

So

$\frac{|h’(t)|}{|h(t)|}\leq 1$

And for each $\epsilon>0$

$log(|h(t)|)-log(|h(\epsilon)|)\leq t$

so

$|h(t)|\leq |h(\epsilon)|e^t$ and

$|x(t)|\leq |y(t)| + |h(\epsilon)|e^t$

So for $\epsilon\to 0$, by continuity of $h$, you have that

$ |x(t)|\leq |y(t)|+ $|h(0)|e^t$ =|y(t)|$

So

$|x(t)|<|y(t)|$

Since $x'(0)\leq f(a)=y'(0)$, then $x(t)\leq y(t)$ for small $t>0$.

If $x(t_0)=y(t_0)$ for some $t_0>0$, then $x'(t_0)\leq f(x(t_0)) =
y'(t_0)$ so that $x(t_0+t)\leq y(t_0+t)$ for small $t>0$.

[add] If $0<x'(0)=y'(0)$ and $x(t) > y(t)$ for all $ 0< t\leq
\varepsilon$, then there is $t_1<t_2$ s.t. $x(t_1)=y(t_2)$.

Hence $$ x'(t_1) < f(x(t_1))=y'(t_2) $$ which is a contradiction.


Why is the infinite period Fourier series the Fourier transform? We can find the Fourier transform as the limit for infinite period of a Fourier series as explained in https://class.ece.uw.edu/235dl/EE235/Project/lesson15old/lesson15.html

I do the derivation they do:

$f_p$ is a $L$ periodic function. I define $$k_n = \frac{2 \pi n}{L}\qquad \text{and}\qquad c_n = \frac{1}{L} \int_{-L/2}^{L/2} f_p(x) e^{-i k_n x}~ dx$$

$$f_p(x)=\sum_{n=-\infty}^{+ \infty} c_n e^{i k_n x}$$

Now, we take the period going to infinity and define $$f(x)=\lim_{L \to + \infty} f_p(x)$$

$$\implies f(x)=\lim_{L \to + \infty} \sum_{n=-\infty}^{+ \infty} (k_{n+1}-k_n) \left( \frac{1}{2 \pi} \int_{-L/2}^{L/2}dx f_p(x) e^{-i k_n x} \right) e^{i k_n x}$$

Up to this point I totally agree and I see that $k_{n+1}-k_n=\delta k$ will play the role of the $dk$ in the integration. But this is more a "feeling" than a proof. Because they say that at this point we recognize a Riemann sum and we thus recognize the Fourier transform:

$$f(x)=\int_{-\infty}^{+ \infty} dk \left( \frac{1}{2 \pi} \int_{-\infty}^{+\infty}dx f(x) e^{-i k x} \right) e^{i k x}$$

This is what I don't understand.

From what I understood from comment on this post, it is not true "in general" but only for some "nice" functions. Indeed, for example, the definition of the integral through Riemann sum is the following:

$$ \int_a^b f(x) dx = \lim_{N \to +\infty} \sum_{n=0}^N \frac{(b-a)}{N} f(a + (b-a)\frac{n}{N})$$

But putting both $a$ and $b$ going to $+ \infty$ and $- \infty$ respectively I don't find an analog expression as the infinite-period Fourier series.

My question:

I would like a proof based on the Riemann summations on which for a set of "nice" functions the equality is rigorously proved.

I tried to prove it for compact support functions but I miss something in my proof as you can see in my very last paragraph.

Thanks to @mathworker21 guidance, I managed to prove it for some class of functions.

I assume I work in an ensemble $\widetilde{\mathcal{D}}(\mathbb{R})$. In this ensemble the functions admit Fourier transform.

Definitions

I call $f$ a function in $\widetilde{\mathcal{D}}(\mathbb{R})$.

I have:

$$f(x)=\int_{-\infty}^{+\infty} \widehat{f}(k) e^{ikx}dk$$
$$\widehat{f}(k)=\frac{1}{2 \pi}\int_{-\infty}^{+\infty}  f(x) e^{-ikx}dx$$

I define $f_L$, the function that is build from $f$ periodized on $[-L/2,+L/2]$. I can thus write its Fourier series:

$$f_L(x)=\sum_{n=-\infty}^{+\infty} \frac{1}{L} c_{L}(k_n) e^{i k_n x} \text{ with } k_n=\frac{2 \pi n}{L} \text{ and } c_{L}(k_n) \equiv \int_{-L/2}^{+L/2} f(t)e^{-ik_n t}dt$$

I define a function $\widetilde{f}_L$:

$$\widetilde{f}_L=\sum_{n=-\infty}^{+\infty} \frac{1}{L} c_{\infty}(k_n) e^{i k_n x}$$

Derivation

Now, I want to prove that the infinite period Fourier serie match the Fourier transform, i.e: $\lim_{L \rightarrow + \infty} f_L(x) = f(x)$. It is equivalent to prove that $|f_L(x)-f(x)|$ goes to $0$ for $L$ infinite.

To prove it I do:

$$|f_L(x)-f(x)|=|\left(f_L(x)-\widetilde{f}_L\right) + \left(\widetilde{f}_L-f(x)\right)| \leq |f_L(x)-\widetilde{f}_L|+|\widetilde{f}_L-f(x)|$$

I will prove that for "good" functions $f$, the rhs converge to $0$ which will prove my statement. I start to deal with the second term:

$$\widetilde{f}_L(x) = \sum_{n=-\infty}^{+\infty} \frac{1}{L} c_{\infty}(k_n) e^{i k_n x}=\frac{1}{L} \sum_{n=-\infty}^{+\infty} g_x(\frac{n}{L})$$

Where:

$$g_x(\frac{n}{L})=c_{\infty}(2 \pi \frac{n}{L}) e^{i 2 \pi \frac{n}{L} x}$$

As explained here Riemann sum on infinite interval, if $u \rightarrow g_x(u)$ is such that $\int_{-\infty}^{+\infty} g_x'(u)du \leq + \infty$, then I have a Riemann summation, which means:

$$ \lim_{L \rightarrow +\infty} \frac{1}{L} \sum_{n=-\infty}^{+\infty} g_x(\frac{n}{L}) = \int_{-\infty}^{+\infty} g_x(u)du$$

Now, I have:

$$\lim_{L \rightarrow +\infty} \widetilde{f}_L(x)=\lim_{L \rightarrow +\infty} \frac{1}{L} \sum_{n=-\infty}^{+\infty} g_x(\frac{n}{L})=\int_{-\infty}^{+\infty} g_x(u)du\\=\int_{-\infty}^{+\infty} c_{\infty}(2 \pi u) e^{i 2 \pi u x}du=\frac{1}{2 \pi}\int_{-\infty}^{+\infty} c_{\infty}(k) e^{i k x}dk=f(x)$$

Thus, the second term in the triangular inequality vanishes.

It remains to prove that the first term vanishes.

We have:

$$|f_L(x)-\widetilde{f}_L|\leq \frac{1}{L} \sum_{n=-\infty}^{+\infty} |c_L(k_n)-c_{\infty}(k_n)|$$

We need to take "nice" function that allow this term to vanish for $L$ going to infinity.
One simple example is to take functions defined on a compact. For $L$ big enough, we have:

$$c_{L}(k_n)=\int_{-L/2}^{+L/2} f(t)e^{-ik_n t}dt=\int_{-\infty}^{+\infty} f(t)e^{-ik_n t}dt=c_{\infty}(k_n)$$

Thus, for $L$ big enough: $\sum_{n=-\infty}^{+\infty} |c_L(k_n)-c_{\infty}(k_n)|=0$.

And we would have proven that for $L$ going to infinity:
$$f_L(x) \rightarrow f(x)$$

In summary, sufficient condition we need

We need two different things to get to the result using this method:

First requirement:
$\lim_{L \rightarrow +\infty} \frac{1}{L} \sum_{n=-\infty}^{+\infty} g_x(\frac{n}{L})=\int_{-\infty}^{+\infty} g_x(u)du$, which can be true if the derivative of $g_x$ is integrable.

Second requirement:
$|f_L(x)-\widetilde{f}_L|$ , or more restrictively: $\sum_{n=-\infty}^{+\infty} |c_L(k_n)-c_{\infty}(k_n)|$ goes to $0$ for $L$ infinite. It can be done for $f$ having a compact support.

I am however not sure that the first requirement is true if $f$ is has a compact support ?

Let $f_p$ is a $L_0$ periodic function and
\begin{cases}
k_n = \dfrac{2 \pi n}{L_0}\\[4pt]
c_n = \dfrac{1}{L_0} \int\limits_{-\,^{L_0}/_2}^{^{L_0}/_2} \mathrm dx f_p(x) e^{-i k_n x}\\[4pt]
f_p(x)=\sum\limits_{n=-\infty}^{+ \infty} c_n e^{i k_n x}.
\end{cases}

Let the new period of $f_p(x)$ is $L=mL_0.$ Then
\begin{cases}
k_n = \dfrac{2 \pi n}{mL_0}\\[4pt]
c_n = \dfrac{1}{mL_0} \int\limits_{-\,^{mL_0}/_2}^{^{mL_0}/_2} \mathrm dx f_p(x) e^{-i k_n x}\\[4pt]
f_p(x)=\sum\limits_{n=-\infty}^{+ \infty} c_n e^{i k_n x},
\end{cases}

$$f_p(x)=\sum\limits_{n=-\infty}^{+ \infty}
\left(\dfrac{1}{mL_0} \int\limits_{-\,^{mL_0}/_2}^{^{mL_0}/_2} \mathrm dx f_p(x) e^{-i k_n x}\right) e^{i k_n x}$$
$$ = \sum\limits_{n=-\infty}^{+ \infty}
\dfrac{2\pi}{mL_0}\left(\dfrac{1}{2\pi} \int\limits_{-\,^{mL_0}/_2}^{^{mL_0}/_2} \mathrm dx f_p(x) e^{-i k_n x}\right) e^{i k_n x}.$$

Assume that
$$\lim\limits_{m\to\infty}\dfrac{1}{2\pi} \int\limits_{-\,^{mL_0}/_2}^{^{mL_0}/_2} \mathrm dx f_p(x) e^{-i k_n x} = F(k_n),$$

then the expression
$$f_p(x)=\sum\limits_{n=-\infty}^{+ \infty} F(k_n) e^{i k_n x} \Delta k_n$$

looks as the classic Riemann integral sum of the improper integral
$$f(x)=\int\limits_{-\infty}^{+ \infty}\mathrm dk F(k) e^{i k x},$$

or
$$f(x)=\int\limits_{-\infty}^{+ \infty}\mathrm dk \left( \frac{1}{2 \pi} \int\limits_{-\infty}^{+\infty}dx f(x) e^{-i k x} \right) e^{i k x}.$$

If all the nesessary conditions are provided, then the limiting transition $m\to\infty$ provides the limiting transition $L\to \infty.$

Therefore, considered approach looks correct.


Suppose $1>a_n>0$ for $n\in \mathbb{N}$. Prove that $\prod_{n=1}^\infty (1-a_n)$ converges if and only if $\sum_{n=1}^\infty a_n<\infty$. Suppose $1>a_n>0$ for $n\in \mathbb{N}$. Prove that $$\prod_{n=1}^\infty (1-a_n)$$ converges if and only if $\sum_{n=1}^\infty a_n<\infty$.

I know this question is similar to one I just posted but I am getting confused with the negatives.

Proof: $\Rightarrow$ Assume that $\prod_{n=1}^\infty (1-a_n)=0$ converges. Well,
$$\ln\left(\prod_{n=1}^\infty (1-a_n)\right)=\sum_{n=1}^\infty ln(1-a_n).$$ We know that for any $0<a<1$, $$-\frac{a}{1-a}<\ln(1-a)<-a.$$ So, $$-\sum_{n=1}^\infty \frac{a_n}{1-a_n}<\sum_{n=1}^\infty \ln(1-a_n)<-\sum_{n=1}^\infty a_n.$$ This implies that $$-\sum_{n=1}^\infty \frac{a_n}{1-a_n}<\sum_{n=1}^\infty \ln(1-a_n)<\infty$$ and so $$-\sum_{n=1}^\infty \frac{a_n}{1-a_n}<\infty$$ (i.e. $-\sum_{n=1}^\infty \frac{a_n}{1-a_n}$ converges).

We need to show that if $-\sum_{n=1}^\infty \frac{a_n}{1-a_n}$ converges, then $\sum_{n=1}^\infty \frac{-a_n}{1-a_n}$ converges. Well if $-\sum_{n=1}^\infty \frac{a_n}{1-a_n}$ converges then $$\lim_{n\to \infty}\frac{-a_n}{1-a_n}=0$$ and so is  $$\lim_{n\to \infty}\frac{a_n}{1-a_n}=0.$$ This implies that $\{a_n\}$ is bounded. So $a_n\leq M$ for all $n\geq 1$. Thus $$\frac{1}{1-M}a_n\leq \frac{a_n}{1-a_n}.$$ So $\lim_{n\to \infty} a_n=0$ which implies that $\sum_{n=1}^\infty a_n<\infty$.

$\Leftarrow$ Assume that $\sum_{n=1}^\infty a_n<\infty$.

I am not sure if I can say that that if $\sum_{n=1}^\infty a_n<\infty$
 then $-\sum_{n=1}^\infty a_n<\infty$.

Whenever $\sum_{i=1}^\infty a_i$ converges $\prod_{i=1}^\infty (1-a_i)$ converges as well. It follows from the fact that $s_n=\prod_{i=1}^n (1-a_i)$ satisfies $s_{n-1}-s_{n}<a_n$ followed by the fact that increasing bounded sequences converge. I believe this is the correct statement (and intended question).

Note that $\prod_{i=1}^\infty (1-a_i)$ converging doesn't imply the other way around. Pick $a_i=1/2$ for all $i$ as a counterexample.

This statement is incorrect. Pick $a_i=1/2$ for all $i$. $\sum_{i=1}^\infty a_i=\infty$. On the other hand $\prod_{i=1}^n(1-a_i)=0.5^n$ which converges to $0$ as $n\rightarrow\infty$.


Ways to get a sum of $59$ from $1$, $4$ and $6$ How many ways can you get a sum of $59$ using only the numbers $1$, $4$ and $6$?
Note: In this case, $1 + 4$ and $4 + 1$ are two different ways to sum $5$.
For example, there are $5$ ways to get $6$:
$1+1+1+1+1+1$
$1+1+4$
$4+1+1$
$1+4+1$
$6$

You can set up a recurrence: the last term of the sum is either $1$ or $4$ or $6$, so after the first few initial terms we have (using $c(n)$ for the count of the number of ways of getting the sum $n$) $$c(n)=c(n-1)+c(n-4)+c(n-6)$$This can be solved in various standard ways, or simply computed out.

You can find it from a spreadsheet.  Using fill down, list the numbers from -6 through 59 n a column.  The next column will have the number of ways to express the number in the first column.  Put $1$ opposite $0$ as there is one way to express $0$.  Then in the line below put =Up1+Up4+Up6 to get the number of ways to express $1$, which is $1$.  We left the blank lines so this works even for $1$.  Copy down and you have your answer of $2890628972$


Show that inverse image of a Lebesgue measurable function is Lebesgue-measurable I am struggling with this exercise. Can anyone please give me a hint?

  Suppose f is Lebesgue-Measurable. Show that $f^{-1}(B)$ is Lebesgue-
  measurable for any borel set B.

I do know that both the borel and lebesgue measurable sets are $\sigma$-algebra. And that the borel sigma algebra is generated by all the open sets. I also know that the collection of borel-sets is a proper subset of the collection of the lebesgue measurable sets.

I know that what it means to be lebesgue measurable is that $f^{-1}(O)$ is lebesgue measurable, for any open set $O$.

But the problem is that $B$ may not be open. And it may even not be a countable union of open sets, if it were I would just use that the lebesgue measurable sets for a $\sigma$-algebra.

Hint: Show that the collection $\mathcal C_f = \{U : f^{-1}(U) \text{ is measurable}\}$ forms a $\sigma$-algebra.  That is, show that this collection is closed under complements and countable unions.

Then, note that $\mathcal C_f$ is a $\sigma$-algebra containing the open sets.

If $f$ is Lebesgue-measurable, by the definition, it means that for each Lebesgue measurable set $X \subseteq R$, $f^{-1}(X)$ also is Lebesgue measurable. Since each Borel subset $B \subseteq R$ is Lebesgue measurable, we automatically get that $f^{−1}(B)$ is Lebesgue- measurable.


Use Fermat's Little Theorem to solve the equation $x^{13} \equiv 2\;(\mod\ 23)$ I have to use Fermat's Little Theorem to solve the equation
$$x^{13} \equiv 2\;(\mod\ 23)$$
But I have no idea how to do so.
I understand that $2^{23} \equiv 2\;(\mod\ 23)$.
But I don't know how to solve what the question is asking.

Fermat's little theorem says that if $x$ is relatively prime to $23$, then $x^{22}\equiv1\pmod{23}$.

Moreover, $$13\times17=221=22\times10+1.$$
So if
$$x^{13}\equiv2\pmod{23},$$
then
$$(x^{13})^{17}\equiv x^{22\times10+1}\equiv x^{1}\equiv2^{17}\pmod{23}.$$
Now $$\begin{align}2^2&\equiv4\\2^4&\equiv16\\2^8&\equiv256\equiv3\\
2^{16}&\equiv9\end{align}.$$
So $2^{17}\equiv18\pmod{23}$. As a consequence, we find that $18$ is a solution to $x^{13}\equiv2\pmod{23}$.

Hope this helps.

Let us find $a,b$ such that $13a+22b=1,$

By Observation $b=3, a=-5$

Using Fermat's Little Theorem,

$$x=x^{22(3)-5(13)}=(x^{22})^3(x^{13})^{-5}\equiv1^3\cdot2^{-5}\pmod{23}$$

Let us find $c,d$ such that $-32c+23d=1\implies c\equiv32^{-1}\pmod{23}$

Use my method from  Solving a Linear Congruence

Or using Fermat's Little Theorem, $2^{-5}\equiv2^{17}\pmod{23}$


The L deductive System in Propositional Logic I am trying to proof a few statements in the deductive system L, in propositional logic. The system contains 3 axioms (I, II, III below) and a few proven statements (1,2,3,4). In addition, the only inference rule is the modus ponens. In a book I found, the system differs from mine in the 3rd axiom. In the book, the 3rd axiom is (~b->~a)->((~b->a)->b) while my 3rd axiom is  (~b->~a)->(a->b). I wanted to ask, is it possible, given my axioms, statements the MP rule and the deduction theorem, to prove the 3rd axiom from the book? Thank you.

You will need Law of Clavius. For how to derive it using your System L, see help with some Hilbert style proofs in a propositional logic axiom system. Here's an answer to your question.

Theorem $\vdash(\neg\beta\rightarrow\neg\alpha)\rightarrow((\neg\beta\rightarrow\alpha)\rightarrow\beta)$

Proof:

$(\neg\beta\rightarrow\neg\alpha)$ (premise)
$(\neg\beta\rightarrow\alpha)$ (premise)
$(\neg\beta\rightarrow\neg\alpha)\rightarrow(\alpha\rightarrow\beta)$ (Axiom 3 in your System L)
$(\alpha\rightarrow\beta)$ (m.p. 1,3)
$(\neg\beta\rightarrow\beta)$ (Theorem/Statement 4 in your System L, combining 2,4)
$(\neg\beta\rightarrow\beta)\rightarrow\beta$ (Law of Clavius)
$\beta$ (m.p. 5,6) (conclusion)
Therefore $(\neg\beta\rightarrow\neg\alpha),(\neg\beta\rightarrow\alpha)\vdash\beta$ (1,2,7,)
$(\neg\beta\rightarrow\neg\alpha)\vdash(\neg\beta\rightarrow\alpha)\rightarrow\beta$ (deduction theorem)
$\vdash(\neg\beta\rightarrow\neg\alpha)\rightarrow((\neg\beta\rightarrow\alpha)\rightarrow\beta)$ (deduction theorem)

You would need some sort of rule of inference for you to change deductions from I., II., or III. to a form like p -> (q -> (r -> p).  Let '{' and '}' stand in place of the third parenthesis form types, considered by their size. Let '[' and ']' stand in place of the second form types, also considered by their size.

That might require a rule of inference to change

{p -> [q -> (r -> p)]}

into

p -> [q -> (r -> p)].

That last step does not accord with the rules of formation.

That all said, something similar does appear to seem possible:


show that $F= \underset{n\geqslant1}{\bigcap} \left\{ x \in X, d(x,F) < \frac{1}{n} \right\}.$ Let $F$ be a closed set from a metric space $(X,d)$ , show that $$F= \underset{n\geqslant1}{\bigcap} \left\{ x \in X, d(x,F) < \frac{1}{n} \right\}.$$
My attempt :  $\Rightarrow)$
for all $x \in X$,  $ x\in F\iff d(x,F)=0$, then  $x \in \underset{n\geqslant1}{\bigcap} \left\{ x \in X, d(x,F) < \frac{1}{n} \right\} $
$\Leftarrow)$ for all $n\geqslant 1$, $0\leqslant d(x,F) < \frac{1}{n} \xrightarrow[n\to \infty]{}0,$ then $0\leqslant d(x,F) <0 $ !! ( can a strict inequality became broad ? this is my question ? )

Generally, limits always transform strict into large hence the result.
(Strict always implies large actually)
Think of $a_n=\frac{1}{n}$ :
$$0<a_n$$ implies $$0\leq a_n$$ which is convenient with taking the limit


Find the number of subsets of {1, 2, . . . , n} with an odd number of elements. Find the number of subsets of {1, 2, . . . , n} with an odd number of elements.

I am looking at an example where they are asking for an even number of sets. Do the number of even and odd sets have to be equal? I am confused :/

If n=0, the answer is 0, since there is only one subset of the empty set (itself), and that has even size.

If n>0, then there are $2^{n-1}$ ways to choose a subset $S$ of $\{1,2,...,n-1\}$. For each of them, exactly one of $S, S \cup\{n\}$ has odd size.

Thus, there are $2^{n-1}$ such sets.

Say $S_n=\{1,2,\dots,n\}$. First assume $n$ is odd. Then $E\subset S_n$ has an odd number of elements if and only if $S_n\setminus E$ has an even number of elements. So the number of odd subsets is the same as the number of even subsets, hence $2^n/2=2^{n-1}$.

Now suppose $n$ is even and $E\subset S_n$. Let $$F=E\cap S_{n-1}.$$ For $E$ to be odd there are two possibilities: $F$ is even and $n\in E$ or $F$ is odd and $n\notin E$. So the number of odd subsets of $S_n$ is the same as the number of subsets of $S_{n-1}$, again $2^{n-1}$.


Do power series converge uniformly and absolutely on radius of convergence? I was reading this text on power series. On pages $182-183$, the author gives a proof of the following statement:

Let
$$
\sum_{n=0}^{\infty} a_n (x-c)^n
$$
be a power series. There is a non-negative, extended real number $0 \le R \le \infty$ such that the series converges absolutely for  for $0 ≤ |x − c| < R$ and diverges for
$|x − c| > R$. Furthermore, if $0 ≤ \rho < R$, then the power series converges uniformly
on the interval $|x − c| ≤ \rho$.

During the proof, the author first proves that the series converges absolutely by bounding $|a_nx^n|$ using inequalities leading to a geometric series, and then using that the common ratio is $<1$ to conclude the series is $< \infty$. After this, the author then proves that the series converges uniformly on $|x − c| \le \rho$ by again using inequalities that lead to a geometric series, but this time he uses the  Weierstrass M-test to conclude that the series converges uniformly.
This left me a bit confused because it is my understanding that from the M-test you can conclude that the series is both uniformly and absolutely convergent. And since the statement of the theorem allows for $|x − c| = \rho$, I don't understand why the author separates the statement (and the proof) into two parts when I believe you could simplify it to

Let
$$
\sum_{n=0}^{\infty} a_n (x-c)^n
$$
be a power series. There is a non-negative, extended real number $0 \le R \le \infty$ such that the series converges uniformly and absolutely for $0 ≤ |x − c| < R$ and diverges for
$|x − c| > R$.

Is there a reason to introduce the number $\rho$ which I'm missing? Or can the text actually be simplified like this? Thank you!

This final statement of yours is false. Take, for instance, the geometric series$$\sum_{n=0}^\infty x^n.\tag1$$The only $R$ for which it is both true that the series $(1)$ converges absolutely when $|x|<R$ and that it diverges when $|x|>R$ is $R=1$. However, the series $(1)$ does not converge uniformly when $|x|<1$. On the other hand, for each $\rho\in(0,1)$ the series $(1)$ does converge uniformly when $|x|\leqslant\rho$.
However, it is true that if you prove that a power series$$\sum_{n=0}^\infty a_n(x-c)^n\tag2$$converges uniformly when $|x-c|\leqslant\rho$, for each $\rho<R$ and if that proof uses the Weierstrass $M$-test, then you can deduce from that proof that the series $(2)$ converges absolutely when $|x-c|<R$.

Fixing $\rho < R$ allows you to use
$(\rho/R)^n \to 0$.


Why is it important for a matrix to be square? I am currently trying to self-study linear algebra. I've noticed that a lot of the definitions for terms (like eigenvectors, characteristic polynomials, determinants, and so on) require a square matrix instead of just any real-valued matrix. For example, Wolfram has this in its definition of the characteristic polynomial:

  The characteristic polynomial is the polynomial left-hand side of the characteristic equation $\det(A - I\lambda) = 0$, where $A$ is a square matrix.

Why must the matrix be square? What happens if the matrix is not square? And why do square matrices come up so frequently in these definitions? Sorry if this is a really simple question, but I feel like I'm missing something fundamental.

Remember that an $n$-by-$m$ matrix with real-number entries represents a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$ (or more generally, an $n$-by-$m$ matrix with entries from some field $k$ represents a linear map from $k^m$ to $k^n$). When $m=n$ - that is, when the matrix is square - we're talking about a map from a space to itself.

So really your question amounts to:

  Why are maps from a space to itself - as opposed to maps from a space to something else - particularly interesting?

Well, the point is that when I'm looking at a map from a space to itself inputs to and outputs from that map are the same "type" of thing, and so I can meaningfully compare them. So, for example, if $f:\mathbb{R}^4\rightarrow\mathbb{R}^4$ it makes sense to ask when $f(v)$ is parallel to $v$, since $f(v)$ and $v$ lie in the same space; but asking when $g(v)$ is parallel to $v$ for $g:\mathbb{R}^4\rightarrow\mathbb{R}^3$ doesn't make any sense, since $g(v)$ and $v$ are just different types of objects. (This example, by the way, is just saying that eigenvectors/values make sense when the matrix is square, but not when it's not square.)

As another example, let's consider the determinant. The geometric meaning of the determinant is that it measures how much a linear map "expands/shrinks" a unit of (signed) volume - e.g. the map $(x,y,z)\mapsto(-2x,2y,2z)$ takes a unit of volume to $-8$ units of volume, so has determinant $-8$. What's interesting is that this applies to every blob of volume: it doesn't matter whether we look at how the map distorts the usual 1-1-1 cube, or some other random cube.

But what if we try to go from $3$D to $2$D (so we're considering a $2$-by-$3$ matrix) or vice versa? Well, we can try to use the same idea: (proportionally) how much area does a given volume wind up producing? However, we now run into problems:

If we go from $3$ to $2$, the "stretching factor" is no longer invariant. Consider the projection map $(x,y,z)\mapsto (x,y)$, and think about what happens when I stretch a bit of volume vertically ...
If we go from $2$ to $3$, we're never going to get any volume at all - the starting dimension is just too small! So regardless of what map we're looking at, our "stretching factor" seems to be $0$.

The point is, in the non-square case the "determinant" as naively construed either is ill-defined or is $0$ for stupid reasons.

For a matrix $\mathbf{A}$ to be invertible it has to be an $n\times n$ square matrix with there existing an $n\times n$ square matrix $\mathbf{B}$ such that, by the operations of matrix multiplication, we have

$$\mathbf {AB} =\mathbf {BA} =\mathbf {I}_{n}\tag{1}$$
where $\mathbf {I}_{n}$ denotes the $n\times n$ identity matrix. This matrix $\mathbf{B}$ is then unique and is termed the inverse of $\mathbf{A}$, denoted by $\mathbf{A}^{-1}$. We see by (1) the matrices have to be square, else they would not commute. A square matrix has no inverse if and only if its determinant is $0$ and is then termed singular.

An important outcome of having invertible square matrices is that a group structure may be imposed on them. The set of all  $n\times n$ invertible matrices, over some field $F$, forms a group under matrix multiplication termed the general linear group of degree $n$, or $\text{GL}(n,F)$, these being the set of nonsingular $n\times n$ matrices. It can be seen the product of two invertible matrices is invertible, and that for an invertible matrix $(\mathbf{A}^{-1})^{-1}=\mathbf{A}$.

The special linear group, $\text{SL}(n,F)$, is the subset of $\text{GL}(n,F)$ whose matrices have determinant $1$, and is a normal subgroup of it.

Consider $\text{SL}(n,\mathbb{R})$: due to its matrices having determinant $1$, the linear maps from $\mathbb{R}^n\rightarrow\mathbb{R}^n$ under it are transformed with volume and orientation preserved.


Conjectures which can't be right or wrong Recently I was talking with some of my non-mathematician friends. On some very unrelated subject in order to make my point I said:

  "There are some conjectures in mathematics which are proven to be
  neither right or wrong."

They were really impressed, I was happy, but I don't know if I am right or wrong (I may conject that). Are there such statements and if there are can you tell me please?

What do you mean by neither right nor wrong?

Certainly for many important specific axiom systems $T$, such as first-order Peano arithmetic, or ZFC (by far the most widely used axiomatic set theory) there are sentences that are neither provable nor refutable from $T$.

But for example any sentence $\varphi$ in the first-order language of number theory is either true or false in the natural numbers. We just don't happen to know, for many such $\varphi$, whether $\varphi$ is true or false in $\mathbb{N}$. And first-order Peano arithmetic, though very powerful, is provably not strong enough to decide the question for all $\varphi$. Indeed no recursively axiomatized theory is strong enough to determine the answer for all $\varphi$. However, that has no direct connection with truth in $\mathbb{N}$.

The proposition (p$\rightarrow$q), where it is not known whether "p" is true or false, and it is not known where "q" is true or false, is neither right nor wrong, nor could it get proven either way in classical (and plenty of other types of) logic.


Proving Inequality using Induction $a^n-b^n \leq na^{n-1}(a-b)$ I was trying to prove this inequality using induction, but couldn't do.

Question: Suppose $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then:

$$a^n-b^n \leq na^{n-1}(a-b)$$

You will want to use that $$a^n-b^n=(a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}$$

What can you say about the powers of $a,b$ given $0<b<a$?*

SPOILER

   Since $0<b<a$, we have $0<b^k<a^k$

thus

  $$\begin{align} a^n-b^n&=(a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}\\&<(a-b)\sum_{k=0}^{n-1}a^{n-k-1}a^{k}\\&=(a-b)\sum_{k=0}^{n-1}a^{n-1}\\&=(a-b)na^{n-1}\end{align}$$

The proof proceeds by induction. Here we will need to make use of two base
cases, since it turns out that $n = 1$ needs to be treated separately.

(1) First base case $(n = 1)$: here $a^1 −b^1 \le 1 \times a^0(a−b) = a − b$,

so the statement holds.

(2) Second base case $(n = 2)$:

$a^2 − b^2 = (a + b)(a − b)$

$< 2a(a − b)$    since $a > b$

$= na^{n−1}(a − b)$, for $n = 2$

(3) Inductive step: Now suppose that $a^n − b^n < na^{n−1}(a − b)$, where $0<  b < a$. Then

$a^{n−1} − b^(n+1)= a^{n−1}- a^n \times b + a^n \times b − b^(n+1) = a^n(a − b) + b(a^n − b^n)$

$< a^n(a − b) + b^n \times a^{n−1}(a − b)$ by the inductive hypothesis

$< a^n(a − b) + ana^{n−1}(a − b)$ since $b < a$

$= (n + 1)a^n(a − b)$

so that $a^{n−1} − b^{n−1} < (n + 1)a^n(a − b)$ for $n > 2$.


Find possible areas of triangle given radius of circumscribed circle Question: In triangle ABC, AB=4, AC=5, and the radius R of the circumscribed circle is equal to √7. Find all possible values of the area of triangle ABC.

Through using the sine rule, i found the angle of B to be approximately $70.89339$, the angle of C to be approximately $49.10661$ and the angle of A to be $60$. Using the cosine law, I found the missing side length BC to be $√21$. Using Heron's law, I then found the area to be $5√3$.

However, this question requires more than one area. I am confused as to how to obtain the second area?

$AB$ and $AC$ are two chords of a circle of radius $\sqrt7$. Picture this circle together with its diameter through $A$. The two chords can either be on the same side or opposite sides of this radius. You’ve computed the area of the resulting triangle for only one of these cases.

You're right, there is only one possible answer.
Since $M$ is the circumcenter of $\triangle ABC $, $MD $ is the perpendicular bisector of $\overline{AB} $. Applying the sine law in $\triangle ABC $ then gives $$|\measuredangle AMD|=\arcsin (\frac {2}{\sqrt {7}}). $$
Analogously, we obtain $$|\measuredangle EMD|=\arcsin (\frac {2.5}{\sqrt {7}}). $$
Overall this leads to $|\measuredangle EMD|= \frac {2\pi}{3} $ and therefore we get $|\measuredangle BAC|=|\measuredangle DAE|=\frac {\pi}{3 } $.
Hence, the area of our triangle comes out to be $\frac {1}{2} \cdot 4 \cdot 5 \cdot \sin (\frac {\pi}{3 }) =5\sqrt {3}$, which is exactly your result.


Expected Value dice rolling You roll a die until you get the number five. What is the expected value of the maximum roll you see(don't include 5, we only consider rolls upto and before the 5). I created a solution to this problem, although I am still confused because of the dependency on the number of rolls. Any commentary would be appreciated

Suppose that it takes $n$ rolls to see a 5. Then
$E(n) = \frac{1}{6}^n \cdot 1 + \frac{1}{6} \cdot n \cdot \frac{2}{6}^{n-1} \cdot 2 + \frac{1}{6} \cdot n \cdot \frac{3}{6}^{n-1} \cdot 3 + \frac{1}{6} \cdot n \cdot \frac{4}{6}^{n-1} \cdot 4 + (1-\frac{5}{6}^n) \cdot 6$.
The probabilities in the expected value function are computed as the probability that number will be the highest out of n rolls. For instance for 2 being the highest roll, we have to have two appear in the sequence of $n$ rolls, and then have all the other rolls be two or less, which corresponds to probability $\frac{1}{6} \cdot n \cdot \frac{2}{6}^{n-1}$. Now my question is how this depends on the number of rolls. For the overall expected value do we simply use the expected number of flips to roll a five, which is six, and then plug 5 into the equation to get the overall expected value? (when I do I get 4.42)

(as another thought, should i integrate the expected value function from $n=0$ to $\infty$?)

Let's assume you have a five-sided die, with faces $\{1,2,3,4,6\}$.  You roll it $k$ times.  What is the expected value of the max roll? Well, it's going to depend on $k$.
\begin{equation}
P(max=1) = \frac{1}{5^k}
\end{equation}
\begin{equation}
P(max=2) = \left(\frac{2}{5}\right)^k - \frac{1}{5^k}
\end{equation}
\begin{equation}
P(max=3) = \left(\frac{3}{5}\right)^k - \left(\frac{2}{5}\right)^k
\end{equation}
\begin{equation}
P(max=4) = \left(\frac{4}{5}\right)^k - \left(\frac{3}{5}\right)^k
\end{equation}
\begin{equation}
P(max=6) = 1 - \left(\frac{4}{5}\right)^k
\end{equation}
\begin{equation}
E[max] = 6 - 2 \left(\frac{4}{5}\right)^k - \left(\frac{3}{5}\right)^k - \left(\frac{2}{5}\right)^k - \frac{1}{5^k}
\end{equation}
Basically, the probability that the $m$th highest value is the maximum is the probability that all the rolls were at most that large, minus the probability that all the rolls were smaller than that value.

Now, we want to sum over $k$, using the probability that the $k+1$st roll is a 5 and none of the others are.
\begin{equation}
E = \sum_{k=0}^\infty \left( 6 - 2 \left(\frac{4}{5}\right)^k - \left(\frac{3}{5}\right)^k - \left(\frac{2}{5}\right)^k - \frac{1}{5^k} \right) * \left( \frac{5}{6} \right)^k \frac{1}{6}
= \frac{1}{6} * \left( 6 * 6 - 2 * 3 - 2 - \frac{3}{2} - \frac{6}{5} \right) = \frac{1}{6} * ( 25.3 ) \approx 4.2167
\end{equation}

Let $P_k$ be the probability that the first instnace to see a 5 was the kth roll, then $E=\sum_{k=1}^\infty kP_k\ where\ P_k=\frac{1}{6} (\frac{5}{6})^{k-1}$


A Pell equation inside a Pell equation While working on another problem (see https://mathoverflow.net/questions/143599/solving-the-quartic-equation-r4-4r3s-6r2s2-4rs3-s4-1), I found the following equation to be solved:
$$ \bigl((r+s)^2-2s^2\bigr)^2 - 2(2rs)^2 = 1.  \tag{*}$$

Conjecture: There are no solutions with $r > 3$; put another way the solution $(r,s)=(3,2)$ is the only solution with $r>s>0$.

Clearly, $(*)$ is the Pell equation $u^2-2v^2=1$, the fundamental solution of which is [perhaps coincidentally?] $(u,v)=(3,2)$. However, in this case, every other solution $u$ is of the same form, i.e., $u=(r+s)^2-2s^2$. This leads me to believe there is some sort of elementary descent argument that can prove the conjecture. Can anyone point me in the right direction?

Thanks, Kieren.

EDITED to incorporate Will Jagy's observation about "every [other] solution".

EDIT: In addition to (*), the following similar Pell equation is simultaneously true:
$$
    \bigl((r-s)^2-2s^2\bigr)^2 - 2(r^2-s^2)^2 = -1.
$$

EDIT (2014/06/05): Is there a way to prove that these two simultaneous Pell equations force $(r,s)$ to be a solution to $X^2-2Y^2=\pm1$?

[Edited mostly to add the uniqueness of the $(119,120,169)$ triangle
and the elliptic-curve connection]

The conjecture is correct, and equivalent to a famous theorem of Ljunggren
(1942, see references below).  No really simple proof is known,
and it might be that such a proof is not possible; the existence of
several solutions such as $(3,2)$ suggests it's hard to get a proof
with elementary techniques $-$ for instance an attempt to use just
congruences is bound to fail because any congruence condition must
allow the two orbits $\{ (0,\pm1), (1, \pm0) \}$ and
$\{(\pm3,\pm2), (\pm2,\mp3)$ of small solutions.

The left-hand side $\left( (r+s)^2 - 2s^2 \right)^2 - 2(2rs)^2$
is a homogeneous quartic polynomial
$$
Q(r,s) = r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4.
$$
Hence we are dealing with the
Thue equation
$Q(r,s) = 1$.  Thue proved that such an equation has only finitely many
solutions.  His proof is ineffective, i.e. does not yield a method for
provably finding all solutions; but later results do make this possible.
One technique is implemented in several computer algebra packages,
including gp in which the command

thue(thueinit(r^4+4*r^3-6*r^2-4*r+1),1)

yields the expected list

[[-2, 3], [2, -3], [0, 1], [0, -1], [3, 2], [-3, -2], [1, 0], [-1, 0]]

and reports only 14 milliseconds' computing time on my laptop.
But this relies on some quite difficult theorems to bound the search.

The quartic $Q$ is quite nice: it has fourfold symmetry generated by
$(r,s) \mapsto (s,-r)$, and the roots of $Q(r,1)=0$ generate the
cyclic quartic field ${\bf Q}(\cos \pi/8) = \frac12 \sqrt{2+\sqrt 2}$.
But ironically this symmetry makes it hard to use a more elementary
technique, Skolem's $p$-adic method, which often succeeds in showing
that a moderately large list $L$ of solutions is complete by showing
that any other solution must be congruent to one of the solutions in $L$
modulo an arbitrarily large power of $p$.  Ljunggren thus resorted to
applying the method not over ${\bf Q}(\cos \pi/8)$ but over some quadratic
extension, which made his proof even harder to carry out and to follow
(though still not as hard as the bounds exploited by the thue function).

The equation in the title of Ljunggren's paper is not $Q(r,s)=1$ but
$X^2 + 1 = 2Y^4$, which has large integer solutions $(X,Y) = (\pm 239, \pm 13)$
[NB $(1+\sqrt2)^7 = 239 + 13^2 \sqrt 2$] in addition to the clear
$(X,Y) = (\pm 1, \pm 1)$.  Ljungreen showed that there are no other solutions.
Thanks to the identity
$$
Q(r,s)^2 + Q(r,-s)^2 = 2(r^2+s^2)^4
$$
it follows that any solution of $Q(r,s) = 1$ has $r^2+s^2 = 1$ or $13$,
which soon yields the desired result.

Conversely any integer solution of $2X^2 + 1 = Y^4$
yields an integer solution of $Q(r,s) = 1$, as
Steiner and Tzanakis show in Section II of their paper; they use
arithmetic in the Gaussian ring ${\bf Z}[i]$, but one can instead
write the Ljunggren's equation as a primitive Pythagorean triple
$((X-1)/2, (X+1)/2, Y^2)$ and apply the parametrization $(m^2-n^2,2mn,m^2+n^2)$
of Pythagorean triples twice, first to obtain
$$
m^2-n^2-2mn = \pm 1, \quad m^2+n^2 = Y^2
$$
and then to the resulting triple $(m,n,Y)$.  Section III then gives a proof
of the solution of $Q(r,s) = 1$ which the authors claim is
simpler than Ljunggren's, but it requires "a deep result
of Mignotte and Waldschmidt on linear forms in logarithms" (presumably
the same result that the gp program exploits), so the
Steiner-Tzanakis proof ends up even more complicated than Ljunggren's.

  Ljunggren, W.:, Zur Theorie der Gleichung $x^2 + 1 = Dy^4$,
  Avh. Norske, Vid. Akad. Oslo 1 #5 (1942)

  Steiner, R., and Tzanakis, N.:
  Simplifying the Solution of Ljunggren's Equation $X^2 + 1 = 2Y^4$,
  Journal of Number Theory 37 (1991), 123-132

P.S. 1) The Ljunggren/Pythagorean-triple connection also gives
the following equivalent formulation: There is a unique nearly-isosceles
Pythagorean triangle, namely $(119, 120, 169)$, whose hypotenuse is a square.
(I use "nearly isosceles" to mean $(a,b,c)$ with $a-b = \pm 1$.)

2) If $Q(r,s) = 1$ then $(r,s,1)$ is a integer solution of $Q(r,s) = u^2$.
If we seek rational solutions,
we get an elliptic curve $E$ isomorphic with $y^2 = x^3 - 2x$
[explicitly $u/s^2 = t^2+2t-1 - 2x$ where $t=r/s=(y-x)/(x-2)$].
By standard descent techniques one shows that $E$ has infinitely many points,
each of which is either $kP$ or $kP+T$ where $P$ is the point $(x,y)=(-1,1)$
and $T$ is the 2-torsion point $(x,y)=(0,0)$.  The symmetries of the problem
let us assume without loss of generality that it's $kP$ with $k$ odd.
At this point we can use $p$-adic techniques to show that $k = \pm 1$,
which recovers the known solutions.  For example, if we use $p=3$
then it seems that the numerator of $u/s^2$ is congruent to $1 \bmod 3^f$
iff $k \equiv \pm 1 \bmod 3^{f-1}$, whence $u=1$ happens only
at the known $k = \pm 1$.  A proof along these lines, though still
far from elementary, might satisfy Mordell's desire for a proof
simpler than Ljunggren's, as well as Kieren MacMillan's hope for
a proof that uses descent (though descent is not the only ingredient).

For Pell's equation: $x^2-2y^2=1$ following decision $(x_1;y_1)$ You can find previous knowledge $(x;y)$ formula:

$x_1=3x\pm4y$

$y_1=2x\pm3y$

We get a system. In this case we can set any shape these decisions - as we want. Well, it will be necessary to solve this system.Although you can not solve. You can see that at each step on the right is a polynomial of n-th degree, and left the second degree. In integers no solutions. Rather if there is one.


Can Rolle's Theorem be true for the critical point where derivative doesnt exist? there is the problem that I met

At 0 the derivative of f(x) doesn't exist so 0 is the critical number but the conclusion of Rolle's theorem is the f'(c) (here c=0) must be 0. Are there any explaination in this case

A critical point is a point where the derivative exists and equals zero. So $c=0$ is not a critical point of $f$. The point here is that $f$ fails to satisfy all of the assumptions of Rolle's theorem, and indeed the conclusion fails, too.

If f is differentiable, it also be continuous and have critical point. According to Fermat theorem, if f is differentiable at this point it's derivative must be 0. So your example does not meet the prerequisite.


Prove that $\sqrt{5521}-1$ is an irrational number I'm stuck on a textbook problem where I'm asked to prove that $\sqrt{5521} − 1$ is an irrational number by contradiction and prime factorization.

I tried following steps the teacher did in her class $10 + $$\sqrt{5}$ , and I made it to the end of the first steps but I couldn't get past that as I didn't know what to do after ending up with $5521b^2 = a^2$. For her example, she did four cases for when the remainder is $1$, $2$, $3$, and $4$ but I don't understand where she is getting these values from. My guess is she did every number up to but not including $5$, but the number I'm given is $5521$ and that seems like an excessive number of cases, which makes me believe that I'm not doing this right. I don't even know if the steps she did on her slides would also apply to any irrational number, specifically the one I'm asked to solve.

Any help or push in the right direction would be appreciated.

You're right on track! You have $5521b^2= a^2$. Now let's factor $5521$...and we're done! It's prime. Now you know $5521$ divides the left side (it's sitting right there), so $5521$ must divide the right side. But as I'm sure you know, a prime number divides a number squared if and only if it divides the number itself, so $5521$ divides $a$ (because it divides $a^2$).

But then $5521$ times some number must give you $a$. Let's say $5521M=a$ for some integer $M$. Plug in this for $a$, do some cancelling, and see if you can get $5521$ to divide $b$, because then you have your contradiction. [After all, you assumed $a/b$ was in lowest form, so no number should divide them both, except maybe $\pm 1$.]

You're right on track! You have $5521b^2= a^2$. Now let's factor $5521$...and we're done! It's prime. Now you know $5521$ divides the left side (it's sitting right there), so $5521$ must divide the right side. But as I'm sure you know, a prime number divides a number squared if and only if it divides the number itself, so $5521$ divides $a$ (because it divides $a^2$).

But then $5521$ times some number must give you $a$. Let's say $5521M=a$ for some integer $M$. Plug in this for $a$, do some cancelling, and see if you can get $5521$ to divide $b$, because then you have your contradiction. [After all, you assumed $a/b$ was in lowest form, so no number should divide them both, except maybe $\pm 1$.]

EDIT Thanks you the comment for pointing something out. Given the proof slides, they prove a prime divides a square if and only if it divides the number is given as a specific case example. But this is more general, and let's prove that fact now... [In fact, we will prove something better! If $p$ divides a product of two numbers $a,b$, i.e. if $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$.]

Lemma. If $p$ divides the product of two integers, $ab$, then $p$ divides $a$ or $p$ divides $b$.

Proof. Well, either $p$ divides $a$ or it doesn't. If $p$ divides $a$, we're done! Because then $p$ divides at least one of them. So we're left in the case that $p$ does not divide $a$.

Well, then we can use the Euclidean Algorithm (which you probably talked about when learning about division), to find integers $x,y$ so that $xp+ya=1$. Now let's multiple by $b$. Then we have
$$
xbp + yab=b
$$
But $p$ divides $xbp$ (there's a $p$ right there!) and it divides $yab$ (because $p$ divides $ab$ by assumption). Then $p$ must divide the sum $xbp + yab$ because it divides each of the numbers $xbp,yab$ individually. But then $p$ divides the left side, so it must divide the right side, which is $b$. But then again, $p$ divides at least one of them! QED

But what does this have to do with your problem? Well, you have $p$ divides $a^2$. But $a^2$ is a product of two numbers, namely $a$ and itself. Because $p$ divides $a^2= a\cdot a$, it must divide one of them. But both of them are $a$! Therefore, $p$ divides $a$.

PS. This works the other way too. If $p$ divides $a$, then $a= Mp$ for some integer $M$. But then $a^2=(Mp)^2=M^2p^2=(M^2p)p$ so that $p$ divides $a^2$ also. Hence, $p$ divides $a^2$ if and only if $p$ divides $a$.

Let, $\sqrt{5521}–1$= $\frac{p}{q}$, where $p$ and $q$ are co-prime integers.
 So,$\sqrt{5521}$=$\frac{p}{q}$+$1$

Squaring both side......

So, 5521=$\frac{p^2}{q^2}$+$1$+2$\frac{p}{q}$.

So, 5520=$\frac{p^2}{q^2}$+2$\frac{p}{q}$

Since, 5520 is integer...
So, $\frac{p^2}{q^2}$+2$\frac{p}{q}$must be an integer, which is not possible .

Hence, Contradiction!!


Homomorphisms between $\mathbb{Z} \star \mathbb{Z} $ and $\mathbb{Z} _2 $ What does a homomorphism between the free product $\mathbb{Z} \star \mathbb{Z} $ and $\mathbb{Z}_2 $ look like? I'm having trouble trying to do anything with this.

Think about the epimorphism $\Bbb Z\rightarrow \Bbb Z_2: x\mapsto x\mod 2$.
For $\Bbb Z^2\rightarrow\Bbb Z_2$ there are two obvious possibilities
$(x,y)\mapsto x+y \mod 2$ and $(x,y)\mapsto x\cdot y \mod 2$
since $\mod n$, $n\geq 2$, is a congruence relation, i.e. compatible with the operations of addition and multiplication.


If $X,Y\sim B$ so $X+Y\sim B$? Let $X,Y$ be random independent variables, such that $X\sim B(n,0.5),Y\sim B(n,0.2)$

  $X+Y$ is also binomial distributed?

My start:

$$X+Y\in \{0,1,2,3,\dots\}$$

$$P(X=k)=\binom{n}{k} 0.5^k(1-0.5)^{n-k}=\binom{n}{k} 0.5^n$$

$$P(X=t)=\binom{n}{t} 0.5^t(1-0.5)^{n-t}=\binom{n}{t}0.5^n$$

Let as denote $S:=X+Y\,\,\,\,\,\,\,\,\,\,\,\,\,\,$
$P(S=\mathscr{L})$

I'm stuck here

It is not binomial distributed. In general though there is a very reliable method so establish if a linear combination of independent variables follows some distribution: compute their charachteristic function!

So in this case
$$E[e^{iu(X+Y)}] = E[e^{iuX}]E[e^{iuY}]  = ((1-p_1) + p_1e^{iu})^n((1 - p_2) + p_2e^{iu})^n$$ and with some algebra you can readily see that this is not in the form $((1-p) + pe^{iu})^n$

I think you are stuck because the result you are trying to show isn't quite right. To the best of my knowledge, there is no named distribution that $X + Y$ follows, and it most definitely is not binomial.

As a simple illustration why, simply consider the $n=1$ case, which has PMF $P(X+Y=0) = 0.4$, $P(X + Y = 1) = 0.35$, $P(X + Y= 2) = 0.25$. If this were binomial, $n$ would have to equal 2 because the support of this distribution is $\{0,1,2\}$ and $p$ would have to equal $0.5$ because $P(X + Y=2) = p^2 \implies p = 0.5$. However, if we looked at $P(X+Y=0) = 1-p^2 \implies p = \sqrt{0.6}$ which draws a contradiction.

The only case where summing binomials gives you another binomial is when the probabilities are equal, thus if $X_1 \sim B(n_1, p)$ and $X_2 \sim B(n_2, p)$ then $X_1 + X_2 \sim B(n_1 + n_2, p)$.


Show the locus of all points such that $x^2+y^2+z^2=1$ and $x^2-y^2-z=0$ is smooth (Implicit Function Theorem) Let $\mathcal C$ be the locus of all point $(x,y,z) \in \mathbb R^3$ with
  $$x^2+y^2+z^2=1 \text{ and } x^2-y^2-z=0.$$
  Show that $\mathcal C$ is a smooth curve in the following sense: For each point $p \in \mathcal C$, there exist open sets $U \subset \mathbb R$, $W \subset \mathbb R^3$ and a $\mathcal C^1$ map $f:U \to W$ such that $p \in W$ and $f(U)= \mathcal C \cap W$.

My idea was to define a function $\Phi:\mathbb R^3 \to \mathbb R^2$ by $\Phi(x,y,z)=(x^2+y^2+z^2-1, x^2-y^2-z)$. Then $\Phi^{-1}(0)=\mathcal C$. Now I want to apply the Implicit Function Theorem but I run into a problem:

Let $\phi_1$ and $\phi_2$ denote the coordinate functions of $\Phi$. Then

$$\det \begin{bmatrix} \frac{\partial \phi_1}{\partial y} & \frac{\partial \phi_1}{\partial z} \\ \frac{\partial \phi_2}{\partial y} & \frac{\partial \phi_2}{\partial z} \end{bmatrix} = \det \begin{bmatrix} 2y & 2z \\ -2y & -1 \end{bmatrix} = 4yz-2y = 2y(2z-1). $$

So my determinant is zero when $y=0$ and when $z=\frac{1}{2}$ and I can't apply the Implicit Function Theorem at those points.

I think there might be a way to use the Implicit Function Theorem twice to get two functions $g,h: \mathbb R \to \mathbb R$ such that $\Phi(x,g(x),h(x)) = 0$. Then I would be able to define $f:\mathbb R \to \mathbb R^3$ by $f(x) = (x, g(x), h(x))$. Is this possible? Or do I need to change my function $\Phi$?

The two surfaces are a sphere and a hyperboloid. Their intersection is a perfectly smooth curve, by any reasonable definition of "smooth".

To see what I mean, go to Wolfram Alpha and type in

Plot3d[{Sqrt[1-x^2-y^2], x^2-y^2}, {x,-1.1,1.1}, {y,-1.1,1.1}]

It seems clear that the curve can be parameterized by polar angle $\theta$ in the $xy$ plane, though finding the parametric equations involves a bit of brute-force algebra. In fact, the parametric equations are:

\begin{align}
x &= \sqrt2 w \cos\theta \\
y &= \sqrt2 w \sin\theta \\
z &= 2w^2 \cos 2\theta
\end{align}

where
$$
w = \frac{1}{\sqrt{1+\sqrt{3+2\cos{4\theta}}}}
$$

These functions are clearly $C_\infty$. It is easy to check that $x^2+y^2+z^2=1$ and that $z=x^2-y^2$.


Existence of a denumerble partition. Is the following Proof Correct?

NOTE: In the text prior to the following exercise we have already proved that $\mathbf{Z^+}\times\mathbf{Z^+}$ is equinumerous to $\mathbf{Z^+}$.

Theorem. Given that $A$ is denumerable. There exists a partition $P$ of $A$ such that $P$ is denumerable and for all $X\in P$, $X$ is denumerable.

Proof. Since $A$ is denumerable there exists a bijection $h:\mathbf{Z^+}\to A$ furthermore from our initial discussion in $7.1$ we know that $\mathbf{Z^+}\times\mathbf{Z^+}\thicksim\mathbf{Z^+}$ consequently there exists a bijection $f:\mathbf{Z^+}\times\mathbf{Z^+}\to\mathbf{Z^+}$ in summary we have a bijection $h\circ f:\mathbf{Z^+}\times\mathbf{Z^+}\to A$. Now consider the following definition.
$$\forall k\in\mathbf{Z^+}\left(\mathcal{E}_k = \{(h\circ f)(k,j)\in\mathbf{Z^+}|j\in\mathbf{Z^+}\}\right)$$

We now prove that $P = \cup_{k\in\mathbf{Z^+}}\mathcal{E}_k = A$. Evidently $P \subseteq A$. Let $a\in A$ be arbitrary, the surjectivity of $A$ implies that for some $(i,j)\in\mathbf{Z^+}\times\mathbf{Z^+}$, $(h\circ f)(i,j) = a$ equivalently $a\in \mathcal{E}_i$ implying that $A\subseteq P$ and by extension $P = A$.

We now prove that $P$ is pairwise-disjoint. Assume for some $i\in\mathbf{Z^+}$ and $j\in\mathbf{Z^+}$ that $\mathcal{E}_i\cap\mathcal{E}_j\neq\varnothing$ and $i\neq j$ equivalently there exists some $q\in\mathcal{E}_i\cap\mathcal{E}_j$ thus for some $m,n\in\mathbf{Z^+}$, $(h\circ f)(i,m)=q=(h\circ f)(j,n)$ but the injectivity of $h\circ f$ implies that $(i,m) = (j,n)$ and by extension $i = j$ resulting in a contradiction.

In addition given that $h\circ f$ is a bijection it is not difficult to see that for any $k\in\mathbf{Z^+}$, $g_k:\mathbf{Z^+}\to\mathcal{E}_k$ defined as follows
$$g_k(i) = (h\circ f)(k,i)$$
is a valid bijection from $\mathbf{Z^+}$ to $\mathcal{E}_k$ implying that $\mathcal{E_k}$ is denumerable.

Yes it is a proof .Why do you doubt it ?


Find $\lim_{x \to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)}$ using L'Hospital's Rule. I am having a hard time figuring out how to find $\lim_{x \to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)}$ using L'Hospital's Rule. I differentiate the numerator and denominator, but I'm not sure what to do with the result. This is my first time working with a limit where $x \to a^+$ instead of a known number like zero. If I just plug in $a^+$ the answer seems to become $\cos(a)$ but I need to use L'Hospital's Rule.

$$\lim_{x \to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)} = \lim_{x \to a^+} \frac{\frac{d}{dx}\cos(x)\ln(x-a)}{\frac{d}{dx}\ln(e^x-e^a)}$$

$$\frac{d}{dx} \ln(x - a)=\frac{x}{x - a} \to \frac{d}{dx}\cos(x)\ln(x-a) = -\sin(x)\ln(x-a)+\frac{\cos(x)}{x-a}$$

$$\frac{d}{dx}\ln(e^x-e^a) = \frac{e^x}{e^x-e^a}$$

$$\lim_{x \to a^+} \frac{\frac{d}{dx}\cos(x)\ln(x-a)}{\frac{d}{dx}\ln(e^x-e^a)} = \lim_{x \to a^+}\frac{-\sin(x)\ln(x-a)+\frac{\cos(x)}{x-a}}{\frac{e^x}{e^x-e^a}}$$

This is where I get stuck. How can I handle this?

Set $x-a=h$ in $$\dfrac{\ln(x-a)}{a+\ln(e^{x-a}-1)}$$

Then apply L'hospital rule

The cosine can be set to cos(a), while the remaining
can be calculated with the rule.
Applied twice -first we get
exp (-x) (exp x - exp a)/(x - a)
Then second time - we find exp(a-x).
Which has limit 1
Hence the final answer is:
cos(a)


$\bigcap_{n\in\mathbb{N}} I^n = (0)$ if and only if no zero divisor of $R$ is of the form $1-z$ with $z\in I$. Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that
  $$\bigcap_{n=1}^{\infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $z\in I$.

First I'll suppose that the intersection is $(0)$. Let $z\in I$ and let $0\neq r\in R$ such that $r(1-z)=0$. Then $r=rz$ and so $r\in I$.  Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.

Any help would be much appreciated! I'm studying for a qual and need all the help I can get.

Yes, the observation that $r=rz$ $\implies$ $r\in I$ is very useful, since we can use the equation again to get that $r\in I^2$, and thus $r\in I^3$, and so on. Hence $r\in \bigcap_{n=1}^\infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)

Edit

Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).

Let $$I^\infty = \bigcap_{n=1}^\infty I^n.$$
Observe that clearly $I(I^\infty) = I^\infty$. Then, since $R$ is Noetherian, $I^\infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.

Thus there exists $r\in R$ with $r-1\in I$ such that $rI^\infty =0$. But then $r-1=i$ for some $i\in I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^\infty = 0$ implies that $I^\infty=0$ as desired.

If there exists some y $\in$ I such that (1-y) is a zero divisor, then there exists an x $\not=$ 0 such that x(1-y) = 0. Then x = xy $\in$ I. Thus x $\in$ I $\cap$ I$^2$. It follows that x $\in$ $\displaystyle\bigcap_n$ I$^n$.


Proving associativity of matrix multiplication I'm trying to prove that matrix multiplication is associative, but seem to be making mistakes in each of my past write-ups, so hopefully someone can check over my work.

Theorem. Let $A$ be $\alpha \times \beta$, $B$ be $\beta \times \gamma$, and $C$ be $\gamma \times \delta$. Prove that $(AB)C = A(BC)$.

Proof. Define general entries of the matrices $A$, $B$, and $C$ by $a_{g,h}$, $b_{i,j}$, and $c_{k,m}$, respectively. Then, for the LHS:
\begin{align*}
& (AB)_{\alpha, \gamma} = \sum\limits_{p=1}^{\beta} a_{\alpha,p} b_{p,\gamma} \\
& \left((AB)C\right)_{\alpha, \delta} = \sum\limits_{n=1}^{\gamma} \left(AB\right)_{\alpha, n} c_{n, \delta} = \sum\limits_{n=1}^{\gamma} \left(\sum\limits_{p=1}^{\beta} a_{\alpha,p} b_{p,n} \right) c_{n, \delta} = \sum\limits_{n=1}^{\gamma} \sum\limits_{p=1}^{\beta} \left(a_{\alpha,p} b_{p,n}\right) c_{n, \delta}.
\end{align*}
For the RHS:
\begin{align*}
& \left(BC\right)_{\beta, \delta} = \sum\limits_{n=1}^{\gamma} b_{\beta, n} c_{n, \delta} \\
& \left(A\left(BC\right)\right)_{\alpha,\delta} = \sum\limits_{p=1}^{\beta} a_{\alpha,p} (BC)_{p, \delta} = \sum\limits_{p=1}^{\beta} a_{\alpha,p} \left(\sum\limits_{n=1}^{\gamma} b_{p, n} c_{n, \delta} \right) = \sum\limits_{p=1}^{\beta} \sum\limits_{n=1}^{\gamma} a_{\alpha,p} \left(b_{p, n} c_{n, \delta} \right).
\end{align*}
Assuming I have written these correctly, we can make two observations: first, the summands are equivalent, as multiplication is associative. Second, the order of the summations doesn't matter when we're summing a finite number of entries. Thus, $(AB)C = A(BC)$.

How does this look?

Your proof is fine.

We can change the order of summation as the sum is finite. When we mention multiplication is associative, we might want to mention multiplicative of which object, such as multiplicative of real numbers or complex number.

A matrix represents a linear transformation. The product of two matrices represents the composition of the operation the first matrix in the product represents and the operation the second matrix in the product represents in that order but composition is always associative.


Euclidean Algorithm with modular powers I have these two questions which go as:

1) Find $x \in \{0,1...16\}$ such that $x^{13}\equiv2\pmod{17}$

2) Find $x \in \{0,1...22\}$ such that $x^{13}\equiv3\pmod{23}$

I was wondering what the best way to approach these were, I was thinking using the Euclidean Algorithm?

This is what I have so far for 1):

We know hcf$((17-1),13)=1$, so $1=13s-16t$ for some s,t. But not sure how to go further? Maybe Fermat's Little Theroem but again not sure. Thank you!

Using Fermat's little theorem the first equation is equivalent to
$$x^{-3}\equiv2\mod{17}$$
$$2x^3\equiv1\mod{17}$$
$$x^3\equiv9\mod{17}$$
$$\therefore x\equiv-2\equiv15 \mod{17}$$
The second example seems like one can only solve with trial and error.

For the second example,
$$
16^{13}\equiv 3 \bmod 23.
$$
Also here we can compute $16^{11}\equiv 1\bmod 23$ as descibed in the comments, so that $16^{13}\equiv 16^2\equiv 3\bmod 23$.


Differential Equations, first order I am taking an Engineering Math class. When it comes of differential equations with $y^2$ included in the question I am lost on the procedure used to solve it. Bernoulli method does work because its not in form required and homogeneous equations doesn't work because of different degrees.

These are a few that had me stumped (Solving any one will be appreciated.)

$y' = (y^2 +9)/(xy+3y)$
$y \cdot \ln(x)y' = (y^2 +36) /x$
$y'=(x^2 y^4 - x^2) / y^3$

Thanks to anyone out there who can help.

For number one, here's a basic concept to help you get started:

$$y' = \frac{y^2 + 9}{xy + 3y} $$

$$y' = \frac{y^2 + 9}{y(x + 3)} $$

NOTE THE SEPARATION

$$\frac{y}{y^2+9}*\frac{dy}{dx} = \frac{y^2 + 9}{y(x + 3)}*\frac{y}{y^2+9}$$

$$\int \frac{y}{y^2+9}dy = \int \frac{1}{x+3} dx$$

$$\frac{\ln\left|{y^2+9}\right|}{2} = \ln\left|x+3\right| + C$$

$$\ln\left|{y^2+9}\right| = 2\ln\left|x+3\right| + C$$

$$y^2 + 9 =C(x+3)^2$$

Can you use this technique to solve the other problems? Comment if you have questions.

HINT 1

Start with the first one. Multiply across to get $(xy+3y)y'=y^2+9$. Now notice that the lhs has a factor $y$ and that $y y'$ is what you get if you differentiate $y^2$. What does that suggest?

Now look at the second one. Again we have a $y y'$. What happens if we try the same trick again?


Is it possible to obtain the sample min/max given the sample mean? Suppose we have iid $X_1, X_2, \dots, X_n$ samples of a standard normal distribution ($N(0, 1)$).

I am trying to get the sample minimum and the sample maximum, given that I know the value of n and the sample mean ($\bar{X}$). In summary, my goal is to obtain:

$E[\min(X_1, \dots, X_n)| \bar{X} = c ]$ for a known value of $c$
$E[\max(X_1, \dots, X_n)| \bar{X} = c ]$ for a known value of $c$

My first approach was using the expectation of the $k$-th order statistic $E[X_{(k)}]$, but I can't relate this expectation with the one I want to obtain.

Thanks

No.  Not even if you know the standard deviation.
For example:

-1.323, -.5, 0, .5, 1.323

Has a mean of zero and a sample standard deviation of close to 1

And:

-1.199,-.75,0,.75,1.199

Has a mean of zero and a sample standard deviation of close to 1

So starting with just the mean, sample size, and normal shape is not enough to restrict the data to just one possible max and min.


How much information is lost in this binary sequence? Given a sequence of $X= y_1, y_2 , ...,y_m$ binary data points. That means $y_i = 0 \text{ or } y_i=1$ for $i\in \{1,...,m\}$
Say that we select $n<m$ datapoints, 'damage' our file such that we are not certain about the outcomes of those $n$ points. How much information is now lost?

For example, say we have a sequence 000000. Then we damage this file
such that we are not certain about the first 2 values any more (the
file now could be 000000, 010000, 100000 or 110000
Does this mean that we lose $\frac{2^2}{2^6}=4/64=$ 6,25% information?

My question is: In general, how much information is lost when we are not certain about $n$ chosen outcomes of the sequence. Alternatively asked: how much information do the $m-n$ datapoints contain when compared to the original $m$ datapoints?
And what about if the $m$ datapoints are not binary, but for example can take $3$, $4$ or $r\in \mathbb{N}$ values?
Im looking for a clear, comprehensive answer, in a general formula. Thank you :-)

Each entry $Y_k\in\{0,1\}$ of your vector $X=(Y_1,\ldots ,Y_m)$ incorporates $1$ bit of information; therefore a "measurement" of $X$ is worth $m$ bits. This statement implicitly assumes that the $Y_k$ are independent and assume the value $0$ or $1$ with probability ${1\over2}$ each, in other words: that all $2^m$ possible words $X$ are equiprobable.

If $n$ of the outcomes $y_k$ are corrupted, and you know the affected set $J\subset[m]$, then the information loss is $n$ bits, since such a measurement $x$ does not specify a point in $\{0,1\}^m$ exactly, but admits $2^n$ points compatible with $x$.

When it is only known that $|J|=n$, but $J$ is unspecified otherwise, then things are much more complicated.

When $2$ is replaced by some $r>1$ in these arguments the information gained or lost per "correct digit" amounts to $\log_2 r$ bits.

First, notice that if a point is damaged such that it retains its original value, then there is no data loss. Now if $n$ out of $m$ point are perturbed we get an upper bound of $n/m$ data loss. But this supposes that the binary points were flipped (ie. $1$ became $0$ and vice-versa).

If damaged points are values taken at random in the set of possible values (here $\{0,1\}$) we get $P(dataloss)=1/2$. Hence the expected data loss would be $n/2m$.

Now, if the set of possible values is of cardinality $\mid X\mid$ the expected dataloss when $n$ elements out of $m$ are perturbed would be
$$E(\%\;of\;dataloss)=\frac{\mid X\mid-1}{\mid X\mid}\cdot\frac{n}{m}$$

This is, of course, assuming values in $X$ are equiprobable.

If the number of perturbed values is a random variable, then the formula can be rewritten as
$$E(\%\;of\;dataloss)=\frac{\mid X\mid-1}{\mid X\mid}P(digit\;is\;perturbed)\cdot M$$

But, the expected number of points you could not trust would be : $P(digit\;is\;perturbed)\cdot M$.


How many numbers can be formed using digits 1 2 3 4 5 6 7 8 9 such that they are in increasing order(e.g. 12345, 578)? Help me in how to solve this problem please. I will be grateful to you :)

Consider the string $123456789$. Observe that each of the $9$ digits in the string can safely be either included or not included, since failing to include a digit will not violate the increasing order of the digits. Hence, there are $2^9$ possible strings that contain ($0$ or $1$ or $2$ or ... or $9$) digits. However, we likely want to omit the case where there are $0$ digits, which leaves us with a final answer of:
$$
2^9 - 1 = 511
$$

numbers with digits in increasing order where they are in a.p. With $r=1 ...$ If it's a $5$ digit no and no starts with $5$ , then the number would be $56789$ . .. As I explained the question properly , now m pretty sure , u urself would be able to solve d problem , one digit numbers would not be considered .. $2$ digit no.- $8$ $(12,23,34,45,56,67,78,89)$
$3$ digit no.- $7(123,234,345,456,567,678,789)$,
$4$ digit no.- $6$, $5$ digit no -$5$, $6$ digit no -$4$, $7$ digit no-$3$, $8$ digit no-$2$, $9$ digit no -$1(123456789)$ ,... So total $=1+2+3+•••+8=36 $