Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$? A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral:
$$\int\limits_0^\infty \frac{\sin x} x \,\mathrm  dx = \frac \pi 2$$
Well, can anyone prove this without using Residue theory?  I actually thought of using the series representation of $\sin x$:
$$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm  dt$$
but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.

I believe this can also be solved using double integrals.

It is possible (if I remember correctly) to justify switching the order of integration to give the equality:

$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Notice that
$$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$

This leads us to

$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$
Now the right hand side can be found easily, using integration by parts.

$$\begin{align*}
I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\
&= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\
&= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}.
\end{align*}$$
Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$
Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  By "closing" a contour in the first quadrant ( a quarter circle of radius $\ds{R}$ ):

\begin{align}
\int_{0}^{R\ >\ 0}{\sin\pars{x} \over x}\,\dd x & =
\Im\int_{0}^{R}{\expo{\ic x} - 1\over x}\,\dd x
\\[5mm] & =
-\,\Im\
\overbrace{\int_{0}^{\pi/2}{\exp\pars{\ic R\expo{\ic\theta}} - 1 \over R\expo{\ic\theta}}\,R\expo{\ic\theta}\ic\,\dd\theta}
^{\ds{\mbox{along the arc}}}\ -\
\Im\
\overbrace{\int_{R}^{0}{\expo{-y} - 1 \over \ic y}\,\ic\,\dd y}
^{\ds{\mbox{along the}\ y\ \mbox{axis}}}
\\[5mm] & =
-\,\Re\int_{0}^{\pi/2}\bracks{\exp\pars{\ic R\cos\pars{\theta}}
\exp\pars{ -R\sin\pars{\theta}} - 1}\,\dd\theta
\\[5mm] & =
{\pi \over 2} -
\Re\int_{0}^{\pi/2}\exp\pars{\ic R\cos\pars{\theta}}
\exp\pars{ -R\sin\pars{\theta}}\,\dd\theta = \bbx{\pi \over 2}
\end{align}

  Note that

\begin{align}
0 & < \verts{\int_{0}^{\pi/2}\exp\pars{\ic R\cos\pars{\theta}}
\exp\pars{ -R\sin\pars{\theta}}\,\dd\theta} <
\int_{0}^{\pi/2}\exp\pars{-R\sin\pars{\theta}}\,\dd\theta
\\[5mm] & <
\int_{0}^{\pi/2}\exp\pars{-\,{2R \over \pi}\,\theta}\,\dd\theta =
{\expo{-R} - 1 \over -2R/\pi}\,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\Large\to}\,\,\,{\large 0}
\end{align}


Independence: Discrete Random Variables I am taking my first course in Measure Theory and I am struggling with the following problem:

Consider the following random variables $X_1,\dots,X_k$ for which countable subsets $A_1,\dots,A_k$ are defined such that:
$P(X_j \in A_j)=1\ \  \forall j$
Show that $X_1,\dots,X_k$ are independent iff for all $a_j \in A_j$
$P(X_j=a_j,\forall j)=\prod_{i=1}^{k}P(X_j=a_j)$

I know that the independence of random variables can be written as:
$P(X_j\leq x_j,\forall j)=\prod_{i=1}^{k}P(X_j \leq x_j)$ for $x_j \in \mathbb{R}$
However, I am having trouble taking this into a discrete universe. Thanks in advance for any advice.

Let $Y_j=1_{(X_j=a_j)},j=1(1)k$, then $E(Y_j)=P(X_j=a_j)$. Note that since $X_1, X_2, \dots ,X_k$ are independent then $Y_1, Y_2, \dots , Y_k$ are too.

Now $E(Y_1Y_2\dots Y_k)=E(Y_1)E(Y_2)\dots E(Y_k)$ gives the proof of "only if" part.

Let $B_{ij}=\{X_i=a_j\}$, $a_j\in A_i$. From the expression given, we can conclude that the $\sigma$-algebra generated by $A_j$ are also independent. Then $X_1, X_2, \dots ,X_k$ are independent.


Question on vector cross product. Show that $\big((\mathbf{u} \times (\mathbf{u}\times \mathbf{v})) \times \mathbf{v}\big) \times (\mathbf{u} \times \mathbf{v})=0$.

From wolfram, it gives zero but there's no details. How to prove this?

Using the identity
$$\textbf{A}\times (\textbf{B}\times \textbf{C})= (\textbf{A}.\textbf{C})\textbf{B}-(\textbf{A}.\textbf{B})\textbf{C}$$
we get,
$\begin{align}
((\textbf{u}\times(\textbf{u}\times \textbf{v}))\times \textbf{v})\times(\textbf{u}\times \textbf{v})\\&=(((\textbf{u}.\textbf{v})\textbf{u}-(\textbf{u}.\textbf{u})\textbf{v})\times\textbf{v})\times(\textbf{u}\times \textbf{v})\\&=((\textbf{u}.\textbf{v})(\textbf{u}\times\textbf{v})-(\textbf{u}.\textbf{u})(\textbf{v}\times\textbf{v}))\times(\textbf{u}\times \textbf{v})\\&=(\textbf{u}.\textbf{v})(\textbf{u}\times\textbf{v})\times(\textbf{u}\times \textbf{v})=0\\
\end{align}$

Hint:
$a\times b=0$ if and only if $a$ and $b$ are parallel.


How to solve set theory question given a problem solving question Sarah has decided to open an ice cream shop.She has taken a tally and finds that she has
10 flavors that contain chocolate
5 flavors that contain nuts
2 flavors that contain chocolate and nuts
11 flavors that do not contain chocolate or nuts
Determine the total number of ice cream  flavors Sarah has
my work:
So to determine the total number of flavors you simply add all the flavors to 10+5+2+11=28
so the answer is 28,Sarah has 28 total flavors in her ice cream shop

There are four types of flavors:  Those that have chocolate and nuts.  Call that $A$.  Those the contain chocolate and no nuts.  Call that $B$.  Those that have nuts but no chocolate.  Call that $C$. And those that have neither chocolate nor nuts. Call that $D$.
We know:
10 flavors that contain chocolate.  $A$ and $B$ contain chocolate but $C$ and $D$ do not.  So $A+B=10$.
5 flavors that contain nuts.  $A$ and $C$ contain nuts but $B$ and $D$ do not.  So $A+C = 5$.
2 flavors contain chocolate and nuts.  That's $A$.  $A$ are all the flavors that contain chocolate and nuts.  So $A = 2$.
11 flavors that do not contain chocolate or nuts.  That's $D$.  $D$ are all the flavors that contain neither chocolate nor nuts.  So $D = 11$.
So $A + B = 10$,  $A+C=5$, $A = 2$, And $D=11$.  So replace $A$ with $2$:
$2 + B=10$,  $2+C=5$ so $B=8$ and $C = 3$.
now there are $2$ flavors with chocolate and nuts, $8$ flavors with chocolate but no nuts, $3$ with nuts but no chocolate.  And $11$ with neither chocolate or nuts.
So there are $2 + 8 +3+11 = 24$ total.
Now read your notes about "Inclusion-Exclusion principle".
Total Number =  Number of flavors with Chocolate + Number of Flavors with nuts - number flavors with both + number of flavors with neither= $10 + 5 -2 + 11=24$
You dont add the flavors with both because you counted them when you counted the chocolate flavores.  and you counted them again when you counted the nut flavors.  Because you counted them twice when you should have only counted them once, you must subtract them.  So the total is $10 + 5 -2 + 11$.

My answer is 13.
Sarah has 2 ice-creams that contain chocolate and nuts, 10 that contain chocolate (the previous 2 are included in here) and 5 that contain nuts (again the first 2 are included in here). So for now you have 2 + 8 + 3 = 13.
Now, if you count the ice-creams that contain only one flavour you have 11, which is exactly what she has, so you do not add more ice-creams to the 13 you already counted. (It is said the 11 do not contain chocolate or nuts, so it can be one of them).


$na_n=\frac{5}{3}a_{n-1}-na_{n-1}$

Multiply by $x^n$ both sides and sum from $n=1$ to $\infty$

$\sum\limits_{n=1}^\infty na_n x^n=\frac{5}{3}\sum\limits_{n=1}^\infty a_{n-1}x^n-\sum\limits_{n=1}^\infty na_{n-1}x^n$

Reindex of the RHS:

$\sum\limits_{n=1}^\infty na_nx^n=\frac{5}{3}x\sum\limits_{n=0}^\infty a_{n}x^n-x\sum\limits_{n=0}^\infty (n+1)a_{n}x^n$

After sorting the eqution:

$\sum\limits_{n=1}^\infty na_nx^n+x\sum\limits_{n=0}^\infty na_{n}x^n=\frac{2}{3}x\sum\limits_{n=0}^\infty a_{n}x^n$

$(1+x)\sum\limits_{n=0}^\infty na_nx^n=\frac{2}{3}x\sum\limits_{n=0}^\infty a_{n}x^n$

Use that $nx^{n-1}=\frac{dx^n}{dx}$ and let $f(x)=\sum\limits_{n=0}^\infty a_nx^n$

We have the following differential eqution:

$(x+1)\frac{df(x)}{dx}=\frac{2}{3}f(x)$

Finally $f(x)=(x+1)^\frac{2}{3}+c$

$\sum\limits_{n=1}^\infty a_n x^n$ is convergent.


How to tell if two matrices are equal up to a permutation Given two real rectangular matrices A, B how can I tell if they are equal up to a permutation of their rows/column without trying all possible permutations?

(This is closely related to the question I asked yesterday but which seems too specific/complicated to receive a response. Maybe I'll be luckier with this simpler one!)

This includes the graph isomorphism problem for bipartite graphs, so is likely to be hard. (To see this, lookat the adjacency matrix A which is $1$ at position $(i,j)$ if vertex number $i$ on the left side is connected to vertex number $j$ on the right side.)

Some heuristics from GIP carry over to this case. Of course, for most matrices just looking at column and row sums will help. In greater generality computing the spectrum of both matrices can help (although not in finding an actual rearrangement).

I think this problem is related to the problem of equivalence of two graphs (just as Darij Grinberg mentioned above), and I think I've found a pretty fast polynomial solution for the graphs. I wrote up the explanation starting with http://babkin-cep.blogspot.com/2018/06/graph-equivalence-1-overall-algorithm.html (the informal proof of why I think it works is in the parts 3 and 4, and in case if someone notices any errors in it, I would be interested to hear about them). It can probably be adapter to the matrices as well.


Proof of a generic limit with log and exponential How do I prove that

$\lim_{n \to \infty} \frac{\log_a n}{n^b}=0$

If $a>1$, you could substitute $n=a^t$ and the limit becomes

$$\lim_{t\to\infty} \frac{t}{a^{bt}}.$$

If $0<a<1$, use $n=a^{-t}.$  Now use L'hospital (or series) if that's allowed, or perhaps a concavity argument.

Just change basis to basis $\;e\;$ , say (and assume $\;b>0\;$, as commented):

$$\frac{\log_an}{n^b}=\frac{\frac{\log n}{\log b}}{e^{b\log n}}=\frac{\log n}{\log b\cdot\left(e^b\right)^{\log n}}\xrightarrow[n\to\infty]{}0$$


Need help proving that an intersection has a single element. Let $S=\{s : 0 < s < 1 \}$, and $A_s = \{x : s < x < 1/s \}$.

Claim I want to prove:

$$\bigcap_{s \in S} A_s = \{1\} \, . $$

I'm not sure how to demonstrate this rigorously. However, I do understand that if we pick an $s$ very close to $0$ we will get a very wide interval.  If  we pick numbers close to $1$ we get very narrow intervals.

First. show that $1 \in A_s$ for all $s \in S$.

Next, show that for any $y \ne 1$, you can find an $s \in S$ such that $y \notin A_s$.

For example, the following choice will work: $$s = \begin{cases}
y & \text{if } 0 < y < 1 \\
1/y & \text{if } 1 < y \\
1/2 & \text{otherwise.} \\
\end{cases}$$

Any number less than 1 take its inverse it is greater than 1 so everfy set contains 1 intersection of those contains 1
Also if s ->1 then 1 / s ->1 since 1 / s is continuous around 1
The result follows


Convergence of composition of sequence of functions on a compact subset of $\mathbb{R}$. Let $\lbrace f_n\rbrace$ and $\lbrace g_n\rbrace $ be two sequences of real-valued continuous functions on a compact subset of $\mathbb{R}$ such that they converge pointwise to $f$ and $g$ respectively, where $f$ and $g$ are continuous functions on $\mathbb{R}$. Then what can we say about the composite function $f_n \circ g_n$. Can we say that it will converge to $f\circ g$.

A similar question is asked here Convergence of composition of functions sequences. But here the continuity of functions $f$ and $g$ was not given. Please at least give me some hint to solve this.

Let $f_n(x)=nx$  for $0 \leq x \leq 1/n$, $2-nx$  for $ 1/n \leq x \leq 2/n$ and $0$ for $x >2/n$. $g_n(x)=\frac 1 n$. Let $f(x)=0$ for all $x$ and $g(x)=0$ for all $x$. Then $f_n(g_n(x))\to 1\neq f(g(x))$.

Let $\lbrace f_n\rbrace$ and $\lbrace g_n\rbrace $ be two sequences of real-valued continuous functions on compact subset of $\mathbb{R}$ such that they converge pointwise to $f$ and $g$ respectively, where $f$ and $g$ are continuous functions on $\mathbb{R}$. Then what can we say about the composite function $f_n \circ g_n$. Can we say that it will converge to $f\circ g$?


lim sup and lim inf of sequence of sets. I was wondering if someone would be so kind to provide a very simple explanation of lim sup and lim inf of s sequence of sets. For a sequence of subsets $A_n$  of a set $X$, the $\limsup A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)$ and $\liminf A_n = \bigcup_{N=1}^\infty \left(\bigcap_{n \ge N} A_n\right)$. But I am having a hard time imagining what that really means unions of intersections and intersections of unions I think maybe causing the trouble. I read the version on Wikipedia but that didn't resolve this either.
Any help would be much appreciated.

A member of
$$
\bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n
$$
is a member of at least one of the sets
$$
\bigcap_{n\ge N} A_n,
$$
meaning it's a member of either $A_1\cap A_2 \cap A_3 \cap \cdots$ or $A_2\cap A_3 \cap A_4 \cap \cdots$ or $A_3\cap A_4 \cap A_5 \cap \cdots$ or $A_4\cap A_5 \cap A_6 \cap \cdots$ or $\ldots$ etc.  That means it's a member of all except finitely many of the $A$.

A member of
$$
\bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n
$$
is a member of all of the sets
$$
\bigcup_{n\ge N} A_n,
$$
so it's a member of $A_1\cup A_2 \cup A_3 \cup \cdots$ and of $A_2\cup A_3 \cup A_4 \cup \cdots$ and of $A_3\cup A_4 \cup A_5 \cup \cdots$ and of $A_4\cup A_5 \cup A_6 \cup \cdots$ and of $\ldots$ etc.  That means no matter how far down the sequence you go, it's a member of at least one of the sets that come later.  That means it's a member of infinitely many of them, but there might also be infinitely many that it does not belong to.

Another way to think about it is via indicator functions: let $f_n$ be the indicator function of the set $X_n$ (i.e. $f_n(x)=1$ if $x\in X_n$ and 0 otherwise). Then it is not difficult to check that $\lim \sup f_n$ is the indicator function of $\lim \sup X_n$ and similarly for limit inferior.


If a coin toss is observed to come up as heads many times, does that affect the probability of the next toss? A two-sided coin has just been minted with two different sides (heads and tails). It has never been flipped before. Basic understanding of probability suggests that the probability of flipping heads is .5 and tails is .5. Unexpectedly, you flip the coin a very large number of times and it always lands on heads. Is the probability of flipping heads/tails still .5 each? Or has it changed in favor of tails because the probability should tend to .5 heads and .5 tails as you approach an infinite number of trials?

I understand that flipping coins is generally a stochastic process, but does that change at all if you see a large number of trials bias to one side?

If you don't know whether it is a fair coin to start with, then it isn't a dumb question at all. (EDIT) You ask if the coin will be biased towards Tails to account for the all of the heads.  If the coin was fair, then the answer from tilper addresses this well, with the overall answer being "No".  Without that assumption of fairness, the overall answer becomes "No, and in fact we should believe the coin is biased towards heads.".

One way to think about this is by thinking of the probability $p$ of the coin landing heads to be a random variable. We can assume that we know absolutely nothing about the coin to start with, and take the distribution for $p$ to be a uniform random variable over $[0,1]$.  Then, after flipping the coin some number of times and collecting data, we change our distribution accordingly.

There is actually a distribution which does exactly this, called the Beta Distribution, which is a continuous distribution with probability density function
$$f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}$$
where $\alpha-1$ represents the number of heads we've recorded and $\beta-1$ the number of tails.  The number $B(\alpha,\beta)$ is just a constant to normalize $f$ (however, it is actually equal to $\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$).

The following graphic (from the wikipedia article) shows how $f$ changes with difference choices of $\alpha,\beta$:

As $\alpha \to \infty$ (i.e. you continue getting more heads) and $\beta$ stays constant (below I chose $\beta=5$), this will become extremely skewed in favor of $p$ being close to $1$.

Also, to play devil's advocate, remember there is also the Gambler's Fallacy theorem.

"I will wager a bet on either red or black. There is a 1 in 2 chance of either outcome. There has been 49 red, so the next outcome must be black!"

Each incidence is a specific, discrete event that is unrelated to the other (all other factors considered, in relation to the elimination of bias, undue influence and assumption of all variables equal across each occurence of the spin of the roulette wheel).

In situations were the outcome is truly random, the assumption of balancing is erroneous and fallacious logic.


is the image of path metrizable? Let $X$ be Hausdorff and $\alpha$ be a path in $X$.

How do I prove that $\alpha(I)$ is metrizable?

I know that $\alpha(I)$ is compact, connected, locally connected. I tried to prove this by applyin Urysohn's lemma.

Since $I$ is compact metriazble space, there exists a countable dense subset $D$ of $I$.

Fix $t_0\in I$ and $N$ be an open neighborhood of $\alpha(t_0)$. Since $D$ is dense, $\alpha^{-1}(N)\cap D \neq \emptyset$. Thus, $N\cap f(D) \neq \emptyset$. Hence, $f(D)$ is dense in $\alpha(I)$. Thus, $\alpha(I)$ is separable. However, $\alpha(I)$ need not be second countable in this case.

HOw do I prove it?

For $x,y\in \alpha(I)$ let $d(x,y)=\inf\{\,|s-t|:\alpha(s)=x,\alpha(t)=y\,\}$.


If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$

My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$

And,
$$x^2-5x+1=0$$
??


Integral on part of n-sphere Let $S^n$ be the $n$-sphere and $0<c<1$. Show that $$ \int_{\{x \in S^n |  c\le x^2_1+x^2_2\}} \ln \left (\frac{1}{\sqrt{1-x_1^2 -x^2_2}}-1\right ) dx < \infty$$

Since we are integrating over the n-sphere I'm not sure how to calculate this.


Is there a close form solution for this series? I want to find close form solution for this series
$$\sum_{n=1}^{\infty}\cos\left(\frac{x}{3^n}\right)\sin^2\left(\frac{x}{3^n}\right)$$
Thanks for your help.

Playing around with the series numerically (plotting it etc) it seems it's equal to $\frac12 \sin^2(\frac{x}{2})$. I don't have a proof right now (and I'm too lazy to work one out), but it seems to work. Please feel free to check it yourself ;-)

With this in mind, you might be able to come up with something.

Cheers!


Describe all subrings of the ring of integers I know that every subset $S \subseteq \mathbb{Z}$ is is a subring if it is a ring under the operations defined on the ring $(\mathbb{Z},+,*)$
that is, $( S ,+)$ is an abelian subgroup of $( \mathbb{Z,+} )$ and $\forall \  x,y,z \in S $ we have that

$$x*y \in S \  \text{closure} $$  $$ x*(y*z) = (x*y)*z \text{ associativity } $$ and $$ x*(y+z) = x*y+x*z \ , (y+z)*x = y*x+z*x  \ \text{distributivity}$$

I have read on this  website https://www.quora.com/How-to-describe-all-the-subrings-of-the-ring-of-integers that is enough to describe all subgroups of integers under addition, I think that is because if $x,y,z \in S \subseteq Z$
and $(\mathbb{Z},+,*)$ is a ring then we have that associativity and distributivity holds( Am I right? ). But what about closure?

or, why is enough to describe all additive subgroups of $(\mathbb{Z},+)$ ?

Thanks for your help!

I am going to admit non-unital subrings of $\Bbb Z$ to my discussion; if every subring contains $1$, then every subring is equal to $\Bbb Z$ and there is not much more to be written.

I claim that if

$S \subseteq \Bbb Z \tag 1$

is an additive subgroup, then it is of the form

$S = s \Bbb Z, \; \tag 2$

for some $s \in \Bbb Z$; for if

$S \ne \{ 0 \}, \tag 3$

there is some $0 \ne s \in S$; since $S$ is a subgroup,

$s \in S \Longleftrightarrow -s \in S, \tag 4$

so we may without loss of generality assume that

$s > 0; \tag 5$

since $S$ has positive elements, it has a smallest such; we may assume that $s$ is the same.  Then clearly

$ns \in S, \; \forall n \in \Bbb Z; \tag 6$

this may be seen by simply adding $s$ or $-s$ to itself $n$ times.  Thus

$s\Bbb Z \subset S; \tag 7$

now if there is some

$t \in S \setminus s \Bbb Z, \tag 8$

we may let $m \in \Bbb Z$ be the largest integer with

$ms < t; \tag 9$

then

$0 < t - ms < s; \tag{10}$

we cannot have $t - ms = 0$ or $t - ms = s$ since then $t \in s \Bbb Z$; but since $t \in S$,

$t - ms \in S, \tag{11}$

which contradicts the hypothesis that $s$ is the smallest positive element of $S$.  Therefore (2) binds.

Since every subring of $\Bbb Z$ is also an additive subgroup, we have shown that every subring of $\Bbb Z$ is of the form $s \Bbb Z$, which sets are clearly closed under addition and multiplication.

Also, every additive subgroup $S$, being of the form $S = s \Bbb Z$, is a subring as well since it is multiplicatively closed:  if $as, bs \in S$, then
$(as)(bs) = abs^2 = (abs)s \in s \Bbb Z$; the other ring axioms, associativity, commutativity, etc.,  are simply inherited from the ring $\Bbb Z$.

For the same reasons, that every subring is an additive subgroup etc., it suffices to find the additive subgroups of $\Bbb Z$.

I guess it is worth pointing out the the sets $S = s \Bbb Z$ are also the ideals of the principal ideal domain $\Bbb Z$.

All the subgroups of $\mathbb{Z}$ have the form $m\mathbb{Z}$ when $0\leq m\in\mathbb{Z}$.  It is pretty easy to see that every such subgroup is a subring. If $x,y\in m\mathbb{Z}$ then you can write $x=mp,y=mq$ when $p,q\in\mathbb{Z}$. And then:

$xy=mpmq=m^2pq=m(mpq)\in m\mathbb{Z}$

So $m\mathbb{Z}$ is closed under multiplication.


Axiom of Limitation of Size implies Axiom of Union? On The Wikipedia page for Axiom of limitation of size it states that this axiom can be used to prove the axiom of union: https://en.wikipedia.org/wiki/Axiom_of_limitation_of_size

  It implies the axiom schema of specification, axiom schema of replacement, axiom of global choice, and even, as noticed later by Azriel Levy, axiom of union[2] at one stroke.

I understand the first three axioms mentioned, but I haven't been able to find anything regarding the axiom of union. Does anyone know how this axiom implies the other? Thanks in advance.

Never mind, I did a little more searching and I finally found the information I was looking for. Sorry for the hassle. Here's where I found it: https://www.jstor.org/stable/2315201?seq=1#page_scan_tab_contents


Using the Existence and uniqueness theorem to find the interval of validity of a unique solution in a Linear ODE From the existness and uniqueness Theorem,the initial value problem

$$y'=3x(y-1)^{1/3}            ,          y(3)=-7$$

has a unique solution on some open interval that contains $x=3$. Find the solution and determine
the largest open interval on which it’s unique.

What i tried,
             First i tried to solve the equation by the seperable equation method to get,$$y=1+(x^2-5)^{1.5}$$. Then from here i used the existence and uniquness theorem to calculate $f(x,y)$ and $f_{y}$. From the calculations,when $y$ not equals to $1$, $f_{y}$ will be continuous, hence there will be a unique solution when $y$ not equals to $1$, according to the wxistness and uniquness theorem. However im stuck from here onwards as to finding the interval of validaty. Is my working correct. Could anyone explain. Thanks

once you have $y = 1 - (x^2 -5)^{3/2}$, you can see that $\sqrt 5 \le x < \infty$ is the maximal interval of existence.

The answer is correct and it gives the technique for finding the interval of validity. By the way it is the first answer to be posted solving the issue of the interval of validity.

I'll advance based on your solution. To find the interval of validity we need the solution of the ode

  $$ y = 1+ (x^2-5)^{3/2} $$

which implies imposing the condition

  $$ x^2-5 \geq 0 \implies x\in ( -\infty, -\sqrt{5} )\cup (\sqrt{5}, \infty ). $$

The above suggested that the interval of validity is $x\in (\sqrt{5}, \infty )$.


Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist? Does $1.0000000000\cdots 1$ (with  an infinite number of  $0$ in it) exist?

First let me tell you that the idea that an infinite sequence "ends with something" is a solid idea. It's a perfectly natural one. The point is that the sequence is not indexed by $\Bbb N$, anymore, but rather by $\Bbb N\cup\{\infty\}$, where $\infty$ is another point, which lies after all the natural numbers.

The point is that an "infinite sequence" is a very general notion. People just often like to think about sequences which are indexed only using the natural numbers (with their natural ordering, that is). But as you will progress in your studies you might meet other objects which are indexed using other infinite sets.

And the reason people often limit themselves to sequences indexed by the natural numbers is that for the real numbers (and similar concepts), these sequences are enough. In this case, of the real numbers, we have that each real number can be defined as a limit of decimal digits, as others have explained, and therefore $1.\underbrace{000\ldots}_{\text{infinite }0\text{'s}}1$ is not a definition of a real number.

Note that this is not the limit of $1+(\frac1{10})^n$, either, which colloquially might be written as $1+(\frac1{10})^\infty$. That limit would be the limit of $1.1,1.01,1.001,\ldots$ and you can see that at no point in this sequence there is a number with infinitely many $0$'s written after it. And indeed this limit would be equal to $1$.

This is also different from the $0.999\ldots$ situation, since it is a sequence indexed by $\Bbb N$, which can be seen as the limit of its initial segments. Whereas a sequence indexed by $\Bbb N\cup\{\infty\}$ is not the limit of its initial segments, since none of them include information about the last digit.

So does it exist? Yes. It's just not a real number. It's a sequence of digits indexed by something other than $\Bbb N$.

Finally, Let me point out that as far as the concept of infinity goes in calculus, it's not quite unique. There is one infinity which signifies arbitrarily large values, another which signifies arbitrarily large negative values, there are infinities which ignore the sign at all, when you talk about a smooth function that can be differentiated infinitely many times, the infinity here is in fact "infinite sequence" rather than the infinities mentioned before, and it's a completely different type of infinity.

And there are other infinities which you might encounter, even in a calculus class.

Yes, if the infinite quantity of zeros are placed both before the decimal point and before the leading $1$.  Those six adjacent dots appear to me to just be small zeros filled in.  I'd like to use the notation that ... placed before a sequence means that it repeats indefinitely in the left direction. So then we would have:

...0$1.00000000000000001$ with the bold stuff being fixed in position.


$\lim_{x\to6^-}\big[\frac{\sqrt{36-x^2}}{x-6}\big]$ using L'Hopital's rule? Disclaimer: this is a homework problem, I at least need a bit of direction, I can't seem to get anywhere, I must be overlooking something.
$\lim_{x\to6^-}\biggl[\frac{\sqrt{36-x^2}}{x-6}\biggr]$
I've tried applying L'Hopital's rule since we end up with an indeterminate form if we evaluate at $x=6$
I took the first, second, and third derivative but the denominator will always force division by 0. I'm not sure what to do. I'd love to put in the first and second derivatives into this question but, I'm a bit sloppy with the 'Ol Mathjax!
EDIT: Found the big wet mistake: For L'Hopital's rule, I have been taking the derivative of the ENTIRE function, instead of the derivative of the numerator, divided by the derivative of the denominator! Don't do that folks!

I think you can do it without L'Hospital Rule. Try to multiply with $\frac{\sqrt{36-x^2}}{\sqrt{36-x^2}}$. You will then get $\frac{36-x^2}{(x-6)(\sqrt{36-x^2})}$. Then you can decompose $36-x^2$ to $(6-x)(6+x)$ and eliminate the $x-6$. You can then apply the limit to get $-\infty$.

The answer is -infinity. It's not in 0/0form neither in infinity/infinite form.
To get that form divide both numerator and denominator by sqrt(6-x) and u will get the desirable answer


Exponential distribution with rocket simulation I've come across the following question and I'm not sure how to go about solving it:

Rockets are launched, and the particles descend with acceleration due to gravity. Height/altitude can be found using $h(t) = 200 - (1/2)9.8t^2 + 8t$. Particle burn time has an exponential distribution: $B\sim \operatorname{Exp}(\lambda)$. $B$ is measured in seconds, and on average a particle burns for $2$ seconds.

Simulate the launching of many rockets with the objective of determining how often a particle reaches ground before burning out. Assume the result for one particle for each rocket simulates the result for many.

So far I have: $2 = \operatorname{E}(B) = 1/\lambda$, so $\lambda = 1/2$ and particles reach the ground after $7.25703$ seconds. I get that the $P(B = 7.25703) = 0.01327795\ldots$

Are the above calculations correct? I'm unsure how to continue. I'm trying to simulate this using R (using only stuff like dexp(), pexp(), qexp(), rexp()).

I think the probability required is actually $P(B>7.25703)=\operatorname{exp}(-\frac{1}{2}(7.25703))=0.0266338\approx0.0266$.

Exponential distribution is continuous so $P(B=t)=0$ for all $t>0$.

(Just curious, if you want a higher precision for your answer maybe your gravitational constant should be of higher precision too like $g\approx9.807$.)


Compute the Fourier transform of $f(t) = \sin t$? Compute the Fourier transform of $f(t) = \sin t$. Does that converge?

If you consider $f:=\sin$ as a function on $\mathbb R$ then it has a Fourier transform only in the distribution sense, as remarked in H.L.'s comment. If, however, you consider $\sin$ as a function on $\mathbb R/(2\pi)$ then its Fourier transform $\hat f$ is a function on $\mathbb Z$, namely the doubly infinite sequence $(c_k)_{k\in\mathbb Z}$ obtained by integration of $\sin$ against the functions $t\mapsto e^{-ikt}$: $$c_k={1\over 2\pi}\int_{-\pi}^\pi \sin(t)e^{-ikt}dt\qquad (k\in\mathbb Z).\qquad(*)$$ As $\sin$ is a nice function Fourier theory guarantees that we can get $\sin (t)$ back using the $c_k$ as follows:  $$\sin (t)=\sum_{k=-\infty}^\infty  c_k e^{ikt}.$$ Now, by Euler's formula, $\sin(t) =(e^{it}-e^{-it})/(2i)$, so we can immediately read off $\hat f(1)=c_1=-i/2$, $\hat f(-1)=c_{-1}= i/2$ and $\hat f(k)=c_k=0$ otherwise, without doing the integration (*).

$\sin(t)=\frac{e^{it}-e^{-it}}{2i}$. Since, $e^{it}$ and $e^{-it}$ are basis functions of the Fourier transform. From this easily can be computed using duality property and there will be no convergence issue.


Derivative of a function with bounded bilinear functions Let $X, X_1, X_2, Y $ be normed Vector spaces , $B : X_1 \times X_2 $  a bounded bilinear function and $f_i : X \longrightarrow X_i$ differentiable.

How do you show that the function $H:X \longrightarrow Y , H(x) := B(f_1(x),f_2(x)) $ is differentiable and how does the derivative look like?

It's the composition of differentiable functions, which implies it is differentiable. It's derivative in direction $v$ is

$$DH (x) v= B(f_1(x),Df_2(x)v) + B(Df_1(x)v, f_2(x))$$

which follows from the fact that the derivative of the bilinear map
$$ (v,w) \mapsto B(v,w) $$
is given by
$$ DB(v,w) (e,f) = B(v,f)+ B(e,w)$$

This latter statement follows from the definition of the derivative from the following equality:

$$B(v+e, w+f) = B(v,w) + B(v,f) + B(e,w) + B(e,f) \\
= B(v,w) + DB(v,w)(e,f) + o(e,f)$$
(see, e.g., bilinear map and differentiability, if in doubt)


Locally exact differential in a disk is exact I'm reading through Ahlfors' Complex Analysis text for self study, and I found difficulty with a proof.

In chapter 4 he defines a locally exact differential as a differential who is exact in some neighborhood of every point of its domain.

In the proof of theorem 16 however, he claims that such differential is exact in every open disk contained in the region. I fail to see why is it so: Each point should have its own radius, and these can get very small (?).

I can use the fact that a locally exact differential of class $C^1$ is closed and apply Poincare's lemma, but no differentiability assumptions were made in the statement of the theorem.

Any help will be appriciated.

All you need is a continuous $1$-form $\omega$. Since a disk is convex, you know that $\int_{\partial R}\omega=0$ for any rectangle $R$ contained in the disk. But this is all you need to define a potential function. See Theorem 1 in the same chapter.

A differential $P\mathrm dx+Q\mathrm dy$ is said to be locally exact in the given region if it is exact in some neighbourhood of each point in that region.


Given $A$ and $B$, how many positive integers $N$ such that $N\times B$ has at least one divisior $D$ that lies in $N \lt D \le A$? For two integers $A$ and $B$, how can we  find the number of positive integers $N$ such that $N\times B$ has at least one divisior $D$ that lies in $N \lt D \le A$?

For example, if $A = 100$ and $B = 11$ then the answer is $41$.

Are you expecting some simple math formula for the following value?
$$\left|\left\{n\in\mathbb{N}\ :\ \exists n_1,n_2,d\in\mathbb{N}\ :\ n=n_1 n_2,\ \ d\mid B,\ \ n_1<d,\ \ n_2\le\frac Ad\right\}\right|$$
Hard to believe such formula exists.
But it can be computed :-)


What is the name for a shape that is like a capsule, but with two different radii? I'm looking for the name of a shape that is like a capsule, but where each circle can have different radii. The shape could be described using two circles (two centers and two radii). Something like this (this is a 2D shape, not a lame attempt at 3D):

You can just say it is the convex hull of two circles.

I have 3 ideas from which you may be able use to settle upon a suitable name:

a conrod or tapered conrod (as in the big-end small-end connecting rod);

it resembles a pull tab of a zipper; and,

it also resembles the tab pull of a soda can.


Cover $(0, +\infty )$ by open sets Cover $(0, +\infty)$ by open sets $U_\alpha$ such that for any $\epsilon > 0$ there are points $x, y \in (0, +\infty)$ with $|x-y|<\epsilon$, not both belonging to the same $U_\alpha$

The distance beteen $x$ and $y$ is $\epsilon$ so would it work if we took our open setss $U_\alpha=B_{\frac{\epsilon}{2}}(\alpha)=\{x \in \mathbb{R}| d(x, \alpha) <\frac{\epsilon}{2} \}$

So $|x-y|<\epsilon$ but we can take our open ball as having a radius$\frac{\epsilon}{2} $

I feel like I may have missed something - is my answer correct?

$$\mathscr U:=\{B_{1/n}(n):n\in\mathbb N\}\cup \{(n,n+1):n\in\omega\}$$ is an open cover of $(0,\infty)$.

Now let $\epsilon>0$. There exists $n\in\omega$ such that $1/n<\epsilon/3$. Then $n-\epsilon/3$ and $n+\epsilon/3$ are distance less than $\epsilon$ apart but are not in any single member of $\mathscr U$.

I approached the problem by thinking about simple open covers - open covers consisting of intervals. I noted that such open cover cannot have a finite subcover if we want your property. So then I thought about my favorite such open cover of $\mathbb R$, which is all the sets $(n,n+1)$ and $(n−1/2,n+1/2)$ (pictured in my head). Then I just thought about shrinking the $(n−1/2,n+1/2)$  intervals so that I can always get close-together $x$ and $y$ at the "ends" of one of them.

Your definition of $U_{\alpha}$ is not clear, is $\alpha$ a point or is it an index?

Besides that, you want to be able to find these two points for every $\epsilon$.

If I understand correctly, what you need is a family of sets such that you can always find two points very close to each other such that they correspond to different $U_{\alpha}$'s. You can go overkill and try $U_{x,\epsilon}:=B_{\epsilon}(x)\cap (0,+\infty)=\{y\in (0,+\infty)\mid d(x,y)<\epsilon\}$, for every point $x\in (0,+\infty)$ and every $\epsilon >0$.

So now let $\epsilon >0 $, take any $x\in (0,+\infty)$ and the neighborhood $U_{x,\frac{\epsilon}{4}}$ and $y:=x+\frac{\epsilon}{2}$ and the neighborhood $U_{y,\frac{\epsilon}{4}}$, you know that $d(x,y)=\frac{\epsilon}{2} < \epsilon $ and that $U_{x,\frac{\epsilon}{2}}\cap U_{y,\frac{\epsilon}{4}}=\emptyset$


Solving $x^2 - 16 x+55> 0$ for $x$ Solving $x^2 - 16 x+55> 0$ for $x$

my work

$$(x-11)(x-5) > 0$$

then  x >11 and x > 5

is that correct ???

Draw out the parabola $y=x^2-16x+55$ , using a graphics calculator

We can clearly see that when $x>11$ and $x<5$ the inequality is satisfied.

Yes, the first part of your argument is right

But Carefully introspect the case that 2 negative numbers when multiplied also gives a positive number number

I recommend using the wavy curve method to check your solution to  the above inequality

$\quad x\quad \in \quad (-\infty ,5)\cup (11,\infty )$


Mean value theorem with integrals? Here is the question:

  Let $f:[0, 1]\rightarrow \mathbb{R}$ be a continuous function satisfying
  $$\int_0^1 (1-x)f(x) \,dx = 0$$
  Show that there exists $c\in (0, 1)$ such that
  $$\int_0^c xf(x)\,dx = cf(c)$$

I'm pretty sure that the problem wants me to use the mean value theorem of some type. So I tried to consider a function $F(t)$ that would give the form
$$F'(t) = \int_0^t xf(x)\,dx -tf(t)$$
so that I would be able to say $F'(c) = 0$ for some $c\in (0, 1)$, using Rolle's theorem. But this gave me
$$F(t) = \int_0^t ((t-1)x-x^2)f(x)\,dx$$
which didn't really help me proceed any further.

I have also tried setting
$$F(t) = \int_0^t (t-x)f(x)\,dx$$
in hopes of using Rolle's theorem, since $F(0)=F(1)=0$. But $F'(t)$ wasn't really the required form. I've also tried other different forms so that I could apply mean value theorem for integrals, or Cauchy's mean value theorem. But I couldn't derive the correct form to solve the problem.

Maybe I have missed something? Or can someone provide me a different approach to this problem? Thanks in advance.

Define $F, G : [0,1] \to \mathbb{R}$ as $F(x) = \displaystyle\int_{0}^{x}f(t)dt$ and $G(x) = \displaystyle \int_{0}^{x}tf(t)dt.$ Integrating the second integral by parts, allows us to write:

$G(x) = xF(x) - \displaystyle \int_{0}^{x}F(t)dt.$

Note that $G(0) =0.$ Our aim is to find another zero of $G$, say $b$, and then apply Rolle's theorem to the function $e^{-x}G(x)$ on the interval $[0, b].$

Claim: There exists $b \in (0,1)$ such that $G(b) =0.$

Proof of claim: Suppose not. Then since $G$ is continuous, it cannot change sign on $(0,1]$ so WLOG, assume $G(x) >0 \quad \forall \, x >0.$

Since $F$ is continuous on a closed and bounded interval $[0, 1]$, it attains its bounds. Let $d \in [0,1]$ be a point of minimisation of $F.$ Two cases arise:

1) $d=0$

Then $\forall \, t \in [0,1] \quad F(t) \geq F(0) =0.$ By the mean value theorem for integrals, $\, \exists \, 0< x_{0} < 1$ such that $\displaystyle \int_{0}^{1}F(t)dt = F(x_{0}).$ The given condition can be stated as $\displaystyle \int_{0}^{1}F(t)dt =0$, hence $F(x_{0}) =0.$

By assumption, $G(x_{0})>0$ which implies $\displaystyle \int_{0}^{x_{0}}F(t)dt< 0$ which by the mean value theorem again implies that $x_{0}F(x_{1})< 0$ for some $x_{1} \in (0, x_{0})$ and thus $F(x_{1}) <0, $ a contradiction. Thus $d=0$ is an impossibility and hence $d>0$ which brings us to case 2.

2) $d>0$

For all $t \in [0,1]$ we have $F(t) \geq F(d).$ Integrating over $[0,x]$ we get $\displaystyle \int_{0}^{x}F(t)dt \geq xF(d)$ and rearranging this yields $G(x) \leq x(F(x) - F(d))$ from which we get that $G(d) \leq 0$ for $d>0,$ a contradiction. Hence the claim holds $\blacksquare$.

With the claim in hand, we apply Rolle's theorem to the function $h(x) = e^{-x}G(x)$ on the interval $[0, b]$ to get $c \in (0, b)$ such that $h'(c) =0$ which implies $e^{-c}(G'(c) - G(c))=0$ from which it follows that $\displaystyle \int_{0}^{c}tf(t)dt = cf(c).$

You can use Taylor Lagrange equality between expansion :

I set $$ F(t)\triangleq \int_{0}^t(1-x)f(x)dx$$

So as $F(t)$ is differentiable , by Lagrange there exists $c \in ]0;1[$

So that $$ F(1)=F(0)+F'(c)$$

Which respectively gives :

$$ cf(c)=f(c) $$ so $f(c)=0$ or $c=1$. Yet $c$ isn't equal to $1$ according to the theorem,

So know, you have to proove

$$  \int_0^cxf(x)dx=0$$


consider the mobius transformation $f(z) =\frac{1}{z} $ , $z \in \mathbb{C}$, $z \neq 0$.choose the correct options consider the  mobius transformation  $f(z) =\frac{1}{z} $  , $z \in \mathbb{C}$, $z \neq 0$. If C  denotes  a circle  with positive radius  passing through the  origin, then  $f$ map  $C \setminus \{0\}$ to

choose the correct options

$1.$Circle

$2.$ a line

$3.$ a  line passing  through the  origin

$4.$ a  line   not passing  through the  origin

My attempt :  i take  $z = e^{i\theta}$,$f(z) = 1/z  = 1/e^{i\theta}=e^{-i\theta}$

that $f(z)  =  \cos\theta  - i\sin\theta$

After that im  not  able  to proceed  further pliz help me

any hints/solution will be appreciated thanks u

What's the point of taking $z=e^{i\theta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.

It is easy to show that $A(x^2+y^2)+Bx+Cy+D=0$ under $\frac1z$ goes to $D(u^2+v^2)+Bu-Cv+A=0$

So as the circle does not pass through the origin, $A$ and either $B$ or $C$ are non-zero and $D=0$

Hence, options 2 and 4 true.


Why is E(Y|X) a RANDOM VARIABLE? Is E(Y|X) a RANDOM VARIABLE ?

X is a random variable so I suppose E(Y|X) is also a random variable since its value depends on X ?

Am I right ? If yes, Then what distribution does it follow and what is its mean and variance ?

Thanks!

Yes, $E(Y|X)$ is a random variable, because its value is a function of the value that $X$ takes. Specifics of its distribution will depend on what $X$ and $Y$ are, and particularly the relationship between them.

A random variable is a function from the sample space to the real numbers, i.e. $X:\Omega\to\mathbb{R}$.

For a random variable $Y$ and an event $\{X=x\}\subseteq\Omega$ we can compute the expected value
$$
\mathbb{E}(Y|X=x)=\sum_{y}y\mathbb{P}(Y=y|X=x).
$$
Then for every $\omega\in\Omega$,
$$
\mathbb{E}(Y|X)(\omega)=u(X(ω)),
$$
where $u(x)=E(Y∣X=x)$ for every $x\in\mathbb{R}$. Therefore $\mathbb{E}(Y|X)$ is also a function from $\Omega$ to $\mathbb{R}$ and is a random variable.

In other words, the random variable $\mathbb{E}(Y|X)$ assignes probability $\mathbb{P}(X=x)$ to the value $\mathbb{E}(Y|X=x)$ for every $x\in\mathbb{R}$.

Edit: This explanation is for discrete and one-dimensional random variables, but the same idea can be applied for continuous or general settings.


Limit as t approaches 0 of $-\sin(\pi t)/(t)$ Sorry I am a highschool student who isn't too good at math.

I know that the limit $\sin(t)/(t)$ as $t$ approaches 0 is 1. But how in the world would I evaluate

$$-\sin(πt)/(t)$$

as $t$ approches 0. I have not yet L'Hospital's rule yet so would there be an approach? Is it possible to factor out π from sin(πx)?

Note that

${\sin{\pi t}\over t}=\pi{\sin{\pi t}\over \pi t}$

If we nos set $u=\pi t$ we get

$${\sin{\pi t}\over t}=\pi{\sin{u}\over u}$$

And $u\to 0$ when $t\to 0$ so ${\sin{u}\over u}\to 1$. We therefore have

$$\lim_{t\to 0}{\sin{\pi t}\over t}=\pi$$

If you have learned about Taylor expansions, you could also Taylor expand about t = 0 to find the limit. Like so :

$\sin {\pi t} \approx (\pi t) - \frac{(\pi t)^3}{6} + ...$

Substituting this in to your original expression and only taking the first term in the above approximation,

$\lim_{t\rightarrow0}-\frac{\sin {\pi t}}{t} \approx \lim_{t\rightarrow0}-\frac{\left(\pi t\right)}{t} = -\pi$.

Where the last step just involves cancelling out the $t$s.


Proof with 3D vectors Let ${a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, ${b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and ${c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if
${a} \times {b} + {b} \times {c} + {c} \times {a} ={0}.$

Hello,

Is there any other way to do this problem without bashing?  I can't seem to find a nice and slick solution to this.

Thanks in advance!

The points are collinear iff the area of the triangle bounded by the points is zero.

$$\text{ Area } = ||(\vec a - \vec b ) \times (\vec a - \vec c ) ||$$

$$ = ||\vec a \times \vec a  - \vec a \times \vec c -  \vec b \times \vec a + \vec b \times \vec c ) ||   $$

$$ = ||0 + \vec c \times \vec a +  \vec a \times \vec b + \vec b \times \vec c ) ||   $$

In order to complete an "if and only if" proof, what you need to do is prove the statement in both directions. What you must prove is,

1) If $(x_a, y_a, z_a)$, $(x_b, y_b, z_b)$, and $(x_c, y_c, z_c)$ are collinear, then $a\times b + b\times c + c \times a = 0$.

2) If $a\times b + b\times c + c \times a = 0$, then $(x_a, y_a, z_a)$, $(x_b, y_b, z_b)$, and $(x_c, y_c, z_c)$ are collinear.

The way you start is by automatically assuming the "If" part is true. So for part 1), you're going to start by assuming that $(x_a, y_a, z_a)$, $(x_b, y_b, z_b)$, and $(x_c, y_c, z_c)$ are collinear. For part 2), you're going to assume that $a\times b + b\times c + c \times a = 0$. You don't need to prove $those$ parts because you're going to assume they're already true. What you must prove is the part after "then".


Gödel's paradox: Why is "a proof that some universal statement is unprovable" not a valid proof that this statement is true? Here is a paradox I have some difficulty resolving:

As far as I understand, by one of Gödel's incompleteness theorems, in a first order logic theory with Peano arithmetic, one can find some non-trivial universal closed sentences (starting with a "for all" quantifier, all variables being bound) that can be proven to be unprovable.

Consider such an unprovable universal statement of the form "For all x, P(x)". We proved that there can be no counter example of this statement, exactly because finding such a counter example would disprove the statement hence contradicting Gödel's theorem which said that this statement can not be proven nor disproven. Therefore the given statement must be true.

As one can observe, my previous paragraph is a valid sequence of arguments that explain why my considered universal statement must be true. This previous paragraph is, by the very definition of proof, a proof of the given statement. My conclusion is that either Gödel was wrong, or mathematics are inconsistent :)

What is wrong with my reasoning ? Can you explain why the second paragraph would not be a valid proof? Does it have something to do with metalanguage? Even if metalanguage is mixed with regular language, cannot all the metalanguage used here be encoded in a first order logic with Peano arithmetic, and seen as not part of a stronger theory ?

The key step in your argument is when you say that if there is a counterexample to $\forall x.P(x)$, then the theory $T$ we speak about can prove that it is in fact a counterexample.

The particular form of $P(x)$ that comes out of Gödel's construction has the property that whenever we have a particular numeral $\bar n$, then the formula $P(\bar n)$ will always be true, and $T$ proves it to be true.

However, that does not constitute a proof of $\forall x.P(x)$. To think it does invokes a hidden assumption that everything our $\forall x$ ranges over can be expressed as a numeral. For sure this is true about our intended interpretation of $T$ as the "actual natural numbers", but in order for $T$ to prove something it needs not only to be true in the intended interpretation, but also in every other interpretation that satisfies the axioms of $T$.

Here the reasoning runs into trouble, because we know that $T$ must have some models where some of the elements don't correspond to numerals. (This is so independently of the incompleteness theorem; the "usual compactness argument" shows that such a model must exist for any theory that can speak about the naturals). Therefore, the known fact that $P$ is true and provable about all numerals doesn't imply that $\forall x.P(x)$ is true in all models, and therefore we can't conclude that it ought to be provable. And so the entire argument falls apart.

Gödel's notion of an "$\omega$-consistent" theory requires that there is no formula $\varphi(x)$ such that $T$ both proves $\varphi(\bar n)$ about every numeral, and proves $\exists x.\neg\varphi(x)$ which is the same as $\neg \forall x.\varphi(x)$. One way to show that a $T$ is $\omega$-consistent is to show that it has some model where all elements are numerals; interpreted in such a model the Gödel sentence will be true. The original incompleteness proof explicitly assumed that $T$ is $\omega$-consistent and used that to argue that $T$ cannot prove $\neg\forall x.P(x)$. Later, Rosser found a way to avoid this assumption and instead just assume that $T$ is consistent.

Your argument breaks down when you reason, that there is no counter example. There may be very well counter examples, but all we can reason is, that we are unable to construct one of them.


Laplace transform of $\int_0^t f(s,t) ds$ I know how to find the Laplace transform of an integral of the form $\Gamma(t) := \int_0^t f(s) ds$, this is simply applying a formula, but I was wondering if a formula for the more general $\Gamma(t) := \int_0^t f(s,t) ds$ exists as well.

The particular case I am interested in is $\int_0^t f(s-t) g(s) ds$ for arbitrary functions $f$ and $g$.

Assuming that your complex argument is $s$ :

$$\mathcal{L}\bigg\{\int_0^t f(s-t) g(s) ds\bigg\}(ω) = \frac{\mathcal{L}\{f(0)g(t)\}(ω)}{ω}$$


propositional logic , don't know the answer Is the assertion "This statement is false" a proposition?

I think that it is a proposition because this("This Statement is false") may have truth values. the statement may be true or maybe false. therefore its proposition. but i don't know if i am correct. please correct me , if i'm wrong.

please explain the answer.

In modern logic, the predicate $True(\quad)$ is of "difficult usage", because in using it in a formal language we will encouter the problems connected with Self-Reference.
You can see in SEP the entry about the Liar Paradox ; it gives a review of most of philosophical and logical debates about this paradox.
Regarding "standard" logical approach to the Liar, see :

4.2.1 Tarski's hierarchy of languages
Traditionally, the main avenue for resolving the paradox within classical logic is Tarski's hierarchy of languages and metalanguages. Tarski concluded from the paradox that no language could contain its own truth predicate (in his terminology, no language can be ‘semantically closed’).
Instead, Tarski proposed that the truth predicate for a language is to be found only in an expanded metalanguage. For instance, one starts with an interpreted language $L_0$ that contains no truth predicate. One then ‘steps up’ to an expanded language $L_1$, which contains a truth predicate, but one that only applies to sentences of $L_0$. With this restriction, it is easy enough to define a truth predicate which completely accurately states the truth values of every sentence in $L_0$ and yields no paradox. Of course, this process does not stop. If we want to describe truth in $L_1$, we need to step up to $L_2$ to get a truth predicate for $L_1$. And so on. The process goes on indefinitely. At each stage, a new classical interpreted language is produced, which expresses truth for languages below it.
Why is there no Liar paradox in this sort of hierarchy of languages? Because the restriction that no truth predicate can apply to sentences of its own language is enforced as a syntactic one. Any sentence $\varphi$ equivalent to $\lnot Tr(\ulcorner \varphi \urcorner)$ is not syntactically well-formed. There is no Liar paradox because there is no Liar sentence. See the entries on Tarski and Tarski's truth definitions for more on Tarski's views of truth.

Statements such as "this statement is false" and "Window is half open" are those which cannot have fixed truth values. Hence these are not propositions.
Reference:
http://mathworld.wolfram.com/Proposition.html

P.S. By not fixed truth values I mean its debatable. The glass is half full ;)


Euler–Lagrange equation has no solutions Show that the Euler–Lagrange equation for the functional:

$$I(y) =  \int_{0}^{1}y dx$$

subject to y(0) = y(1) = 0 has no solutions. Explain why no extremum for I exists.

When forming the E-L equation I get 1=0.
How would I go about doing this question?

If you consider the function $f_n$, where $f_n(x) = n, x \in [\frac{1}{n}, 1 -\frac{1}{n}]$, $f_n(x) = n^2x, x\in [0, \frac{1}{n}],$ and $f_n(x) = n^2(1-x), x\in [1 -\frac{1}{n}, 1],$ then clearly $I(f_n)$ can be as large as you want. $I(-f_n)$ can also be as large as you want.

So you've answered your own question. You obtain 1=0 which obviously doesn't have any solutions. There is no reason why this should have an extremum either. Here is a proof (ish) Take y to be any function from [0,1] to the real line satisfying the condition. Then if we add a function delta to y (with the right boundary conditions) it will change I up to first order in the integral of delta so if y where a minimum then adding a delta with a small negative integral would make I smaller and therefore it can't be a minimum. The same happens with a maximum.


Number of mixed doubles pairs such that no one plays with his/her spouse? Can you help me with this problem?

  There are $7$ married couples. What will be the number of mixed double pairs in tennis such that no one plays with his/her spouse?

Can some one help me with this? The answer is $840$.

Thanks!

EDIT : The question asks for the number of teams possible in which no wife is teamed up with her husband.

If no married couple can play together, but are allowed to oppose one another, you can pick the first pair in $7\cdot 6=42$ ways.  Then you can pick the second woman in one way as the wife  of the first man, in which case she can pair with six other men, or five ways as not the wife of the first man, in which case she can pair with five other men.  In total we have $7\cdot 6 (1\cdot 6+5\cdot 5)=1302$ possibilities if the order of pairs matters.  If the order of pairs does not matter, it is half this, $651$ possibilities.

882-210-21 = 651

ttl mixed double pairs possible - no. of one married couple on either side - no. of exactly one married couple & one mixed couple


bug on the box problem A bug walks on the surface of a box ($L=B=1$, $H=2$), starting at a corner, $A$. You want to feed the bug but you also want the bug to walk the longest distance.  The bug takes the shortest path possible.  Where do you put the food to make it walk the longest?

The box floats in the air and the bug can walk on any face of box it wants.
Hint: its not $B$ Refer the image

The two existing answers provide good intuition with their net diagrams and go a long way towards a solution, but they can be improved by combining them.

Oscar Lanzi’s answer is wrong in that the golden path should travel along the face, as in David G. Stork’s answer, not along the edge.

David G. Stork’s answer is missing the symmetry argument from Oscar Lanzi’s answer that reduces the number of coordinates to be optimized from $2$ to $1$.

Combining the two, we can consider a point on the diagonal of the opposite square face with coordinates $(x,x)$ measured from $B$. The path that uses one rectangular face has squared length

$$
(2+(1-x))^2+(1-x)^2=2x^2-8x+10\;,
$$

whereas the path that uses two rectangular faces has squared length

$$
(2+x)^2+(2-x)^2=2x^2+8\;.
$$

These are equal for $-8x+10=8$, that is, for $x=\frac14$. Their derivatives have opposite sign, so increasing one will decrease the other; thus this point $\frac14$ along the diagonal from $B$ towards the corner above $A$ is the point at maximal distance. The maximal distance is

$$
\sqrt{2\left(\frac 14\right)^2+8}=\frac{\sqrt{130}}4\approx2.850\;,
$$

compared to the distance $\sqrt8\approx2.828$ to $B$, a difference of less than one percent.

We can also find the minimal height of the box at which this phenomenon appears. Replacing $2$ by $h$ above, we have

$$
(h+(1-x))^2+(1-x)^2=2x^2-4x-2hx+(h+1)^2+1
$$

and

$$
(h+x)^2+(2-x)^2=2x^2-4x+2hx+h^2+4\;.
$$

These are equal for

$$
-2hx+(h+1)^2+1=2hx+h^2+4
$$

and thus

$$
x=\frac{h-1}{2h}\;.
$$

Thus, for $h\lt1$ there is no solution, so $B$ is the point at maximal distance. For $h=1$, the two paths to $B$ have the same length, which makes sense, since in this case they are related by symmetry. For $h\gt1$ the point $D$ at equal distance along the two paths is on the diagonal of the square face, and as $h\to\infty$ it goes to the centre of the square face, which also makes sense, since in this limit the distance on the rectangular faces is $h$ no matter where you enter the square face. For $h\ge1$ the distance to $D$ is

$$
\sqrt{2\left(\frac{h-1}{2h}\right)^2-4\left(\frac{h-1}{2h}\right)+2h\left(\frac{h-1}{2h}\right)+h^2+4}=\sqrt{\frac{\left(h^2+1\right)\left(2h^2+2h+1\right)}{2h^2}}\;.
$$

Interestingly, this distance is less than the distance $\sqrt{h^2+4}$ to $B$ for $h\lt h^*=\frac{3+\sqrt{17}}4\approx1.781$ and only becomes the maximal distance as $h$ increases beyond $h^*$. This is possible because for $h\lt2$ the angle between the path $AB$ and the face diagonal $BC$ is acute after unfolding, so that a displacement of the food from $B$ along the face diagonal initially shortens the path. (Thanks to Oscar Lanzi for pointing this out.) At $h=h^*$ the triangle formed by the unfolded paths from $A$ to $B$ and to $D$ and the face diagonal is isosceles, and beyond that the distance to $D$ is greater than the distance to $B$ even though the angle at $B$ remains acute up to $h=2$.

The largest relative difference between the maximal distance and the distance to $B$ occurs at $h\approx5.083$, at which point $x\approx0.4016$ and the relative difference is about $4.6\%$.

Given that the 2D topology of the 3D box is as shown in the figures in this answer and previous answers, there is a simple geometric solution that clearly shows the max distance is the distance to B if you can only put food in one location of the box. If you draw a circle centered at A that completely encloses the 2D topology of the box, the only place that circle touches the box is at B. The radius is the longest straight line distance within the circle and that radius is the $sqrt{10}$Box inscribe in a circle


Find the derivative of $y=(\tan (x))^{\log (x)}$ Find the derivative of $y=(\tan (x))^{\log (x)}$

I thought of using the power rule that:
$$\dfrac {d}{dx} u^n = n.u^{n-1}.\dfrac {du}{dx}$$
Realizing that the exponent $log(x)$ is not constant, I could not use that.

Note that $$\log(y) = \log(x).\log(\tan(x))~.$$ Now, apply chain rule on both sides.

$(\tan(x))^{\ln(x)}=\exp(\ln x(\ln(\tan x)))$ by apply the formula off the derivative of a composition:

$f'(x)=(\ln(x)\ln(\tan(x))'\exp(\ln(\tan(x))$.

$(\ln(x)(\ln(\tan(x)))'={1\over x}\ln(\tan(x))+\ln(x){1\over {\tan(x)}}{1\over{cos(x)^2}}$.


Expand a function in a Fourier Series Expand the function
$$ f(x) = \sin^6{x} \cdot \cos^3{x} $$
in a Fourier series on the segment $[-\pi, \pi]$.

I think we should start from the fact that this function is even, but then the $ a_n $ coefficient turns out to be very big

Repeatedly use the identities $\sin^{2}x=1-\cos^{2}x$, $\cos^{2}x=\frac {1-\cos (2x)} 2$ and $\cos^{3}x=\frac {\cos (3x)+3\cos x} 4$ to write  the function as finite Fourier series.

This is not an answer to your question because I assume you want to calculate your Fourier series manually. Nevertheless it may be useful to know how you can do this using a CAS.

Using Maxima 5.40.0 http://maxima.sourceforge.net I get

(%i1) f1:sin(x)^6*cos(x)^3;
                                   3       6
(%o1)                           cos (x) sin (x)
(%i2) f2:trigreduce(f1);
               (- cos(9 x)) + 3 cos(7 x) - 8 cos(3 x) + 6 cos(x)
(%o2)          -------------------------------------------------
                                      256
(%i3) f3:multthru(f2);
                   cos(9 x)    3 cos(7 x)   cos(3 x)   3 cos(x)
(%o3)           (- --------) + ---------- - -------- + --------
                     256          256          32        128

when I use the basic commands to manipulate expressions.

But simpler si the following

(%i4) poissimp(f1);
                   cos(9 x)    3 cos(7 x)   cos(3 x)   3 cos(x)
(%o4)          (- --------) + ---------- - -------- + --------
                     256          256          32        128

The commands are described in the Maxima Reference Manual.

The following numerical test should increase your confidence in this solution:

(%i5) sum(abs(ev(float(f1)-float(f3),\
    x=-%pi+k/N*2*%pi,numer=true,tr_numer=true)),k,0,N),N=100;
(%o5)                       1.665336423076783E-15

We calculate the absolute value of the difference of $f1$ and $f3$ at 100 equidistant points between $-\pi$ and $\pi$ and sum them up. These sum is smaller than $2\cdot 10^{-15}$. The values of $\sin$ and $\cos$ in these interval vary between $-1$ and $1$.


Use the half-angle formula for cosine to compute $\cos(\theta/2)$ given $\cos(\theta)=63/68$ where $0\lt\theta\lt\pi/2$ Use the half-angle formula for cosine to compute $\cos(\theta/2)$ given $\cos(\theta)=63/68$ where $0\lt\theta\lt\pi/2$.

I know that $\cos(\theta/2)= \pm\sqrt{\frac{\cos(\theta)+1}{2}}$.  Therefore the answer would be $\pm\sqrt{\frac{(63/68)+1}{2}}$.  However, I don't know whether or not its negative or positive.

Given that $0 < \theta < \pi/2$, it follows that the cosine function is positive, allowing you to determine a sign. Always go back to the graph when in doubt.

T. Bongers is right: When an angle is on the interval  $0 < \theta < 2\pi$, the cosine of the angle is positive. More generally, the six basic circular trig functions { $\sin(x)$, $\cos(x)$, $\tan(x)$, $\csc(x)$, $\sec(x)$, $\cot(x)$ } follow these patterns:

Quadrant I: All six functions are positive.
Quadrant II: $\sin(x)$ and $\csc(x)$ are positive.
Quadrant III: $\tan(x)$ and $\cot(x)$ are positive.
Quadrant IV: $\cos(x)$ and $\sec(x)$ are positive.

The way my Calc teacher tells me to remember this pattern (or maybe it's called a mnemonic?) is cycling through the Quadrants in order: "All Students Take Calculus!" The first letter of each word corresponds to the basic function(s) that produces positive values in that Quadrant. A: all functions; S: $\sin(x)$; T: $\tan(x)$; C: $\cos(x)$. Because the inverse of a number does not change its sign, the reciprocal functions of those mentioned above follow the same positive negative patterns as the 3 functions I just mentioned. With any Quadrantal angle (when $x$ is a multiple of 0, 90, 180, or 270 degrees), each of the aforementioned functions has its own set of special cases that are more easily understood by practicing to make the graphs of the functions.


does this trigonometric series vanish? I have a question regarding a trigonometric series:

Shouldn't

$$\sum_{n,m}^\infty s^{n+m}sin\left((n-m)\phi\right)=0$$

where $s \in \mathbb{R}$ because the sinusoidal function is antisymmetric in $n$ and $m$ and vanishes for $n=m$?

$$\begin{align}
& &\sum_{n,m}^\infty s^{n+m}\sin\left((n-m)\phi\right)\\&=&
\text{Im}\sum_{n,m}^\infty s^{n+m}\text{e}^{(n-m)\text{i}\phi}\\&=&
\text{Im}\sum_{n}^\infty (s\text{e}^{\text{i}\phi})^n \sum_{m}^\infty (s\text{e}^{-\text{i}\phi})^m\\&=&
\text{Im}\frac{1}{1-s\text{e}^{\text{i}\phi}}\cdot\frac{1}{1-s\text{e}^{-\text{i}\phi}}\\
&=&\text{Im}\frac{1}{1-2\cos\phi s +s^2}\\
&=&0.
\end{align}
$$

Well,

\begin{align}
\sum_{n,m}^\infty s^{n+m}\sin\left((n-m)\phi\right)&=\\
\sum_{n<m}^\infty s^{n+m}\sin\left((n-m)\phi\right)+\sum_{n>m}^\infty s^{n+m}\sin\left((n-m)\phi\right)+\sum_{n=m}^\infty s^{n+m}\sin\left((n-m)\phi\right)&=\\
\sum_{n<m}^\infty s^{n+m}\sin\left((n-m)\phi\right)+\sum_{n<m}^\infty s^{m+n}\sin\left((m-n)\phi\right)+\sum_{n=m}^\infty s^{n+m}\sin\left(0\cdot\phi\right)&=\\
\sum_{n<m}^\infty s^{n+m}\sin\left((n-m)\phi\right)-\sum_{n<m}^\infty s^{m+n}\sin\left((n-m)\phi\right)+0&=\\
0&
\end{align}


Are there scalars which makes $u_1$ to $u_3$ linearly independent? There are 4 vectors:
$v_1 = (1,4,2,-3) , v_2 = (-2,4,-4,27) , v_3 = (1,8,2,4) , v_4 = (2,-12,4,-41)$

Are there any scalars $a_i, b_i, c_i ( 1 \le i \le 4)$ that the vectors

$u_1 = \sum\limits_{i=1}^4 a_iv_i$

$u_2 = \sum\limits_{i=1}^4 b_iv_i$

$u_3 = \sum\limits_{i=1}^4 c_iv_i$

be independent?

Well, I put $v_1,v_2,v_3,v_4$ on a matrix and got 3 vectors which are linearly independent.

Which means that if I want $u_1,u_2,u_3$ to be linearly independent I will need to put for every $u_i$ vector two scalars to be zero.

for example:

$u_1 = a_1v_1$

$u_2 = b_2v_2$

$u_3 = c_3v_3$

and like that I will have about 6 options I guess to make linearly independency. What do you think? Did I solve this all wrong and there's a better way to solve it?

After putting v1, v2, v3, v4 into a matrix and obtain its row echelon form, obtain the linear independent set from the corresponding pivotal ones. Suppose v1, v2 and v3 are linearly independent. Then just choose u1=v1, u2=v2 and u3=v3 by assigning appropriate values for ai's.


Find an upper bound for $f(x) = \sin(\sin(x))$. I've run into this hard calculus problem that I can't seem to solve.  The question is:

  If $f(x) = \sin(\sin x)$, use a graph to find an upper bound for $|f^{(iv)}(x)|$.

I am not sure what I have to do.  Should I find the fourth derivative and then find the maximum of that graph.  Or do I have to do something completely different?

Thanks in advance.

To solve the problem, I agree with you that you must find the 4th derivative of the given functio then graph it using any graphing tools. But this is tedious. However, you can use Wolfram Alpha To help you on answering your problem. Wolfram Alpha gives the result below for the 4th derivative of $f$ ogether with the graph:


10 red cars, 10 blue cars, 10 green cars are distributed randomly in a line. What is the expected number of times a red car precedes a blue car? We can apply linearity of expectation to this to make it easier. This means we reduce the problem to finding the expectation that the red car i precedes the blue car j. This is just the probability of this event. I'm having trouble computing this probability. I appreciate the help.

We find the expected number of times that a red car immediately precedes a blue car. Imagine that the red cars are labelled $1$ to $10$.

Let $X_i=1$ if red car $i$ immediately precedes a blue car, and let $X_i=0$ otherwise. We want to find the expectation of $X_1+\cdots+X_{10}$. By the linearity of expectation and symmetry this is $10E(X_1)$.

We have $E(X_1)=\Pr(X_1=1)$. For $X_1$ to be $1$, red car $1$ must be in positions $1$ to $29$ (probability $\frac{29}{30}$). Given that it is in one of these positions, the probability there is a blue immediately behind it is $\frac{10}{29}$. Thus $\Pr(X_1=1)=\frac{10}{30}$.

Well, you don't care much about the colors of the cars at this point. In fact, you don't even care about the other cars. In half the cases the red car will be before the blue car, and in the other half it'll be after. By linearity of expectation, you just count the number of pairs of red-blue cars and divide that by $2$ to get your answer.

Hope that helps,


Not sure how to start proving that $\int_E f\,d\lambda$ = $\lim_{n \to \infty} \int_{E_n} f\,d\lambda$ Here is the following problem:

Suppose that $f$ is Lebesgue integrable over E and that $\{E_{n}\}_{n=1}^{\infty}$ is a increasing sequence of lebesgue measurable sets, that is, $E_1$ $\subset$ $E_2$ $\subset \cdots$ and $\cup_{n=1}^{\infty} E_{n}$ := E.  Prove that
$$\int_E f\,d\lambda = \lim_{n \to \infty} \int_{E_n} f\,d\lambda$$

Any suggestions/ hints on how to start this problem would be appreciated.

Hint: Let $f_n := f\chi_{E_n}$, where $\chi$ is the characteristic function. Then $f \chi_E = \lim_{n\to\infty} f_n$. Hence you would like to move the limit inside the integral, which theorem would be useful for that?

Case 1: Let $ f \ge 0$.

$$\lim_{n \to \infty} \int_{E_n} f\,d\lambda$$

$$\lim_{n \to \infty} \int_{E_n} f\,d\lambda$$

$$ = \lim_{n \to \infty} \int_{\Omega} f1_{E_n}\,d\lambda$$

$$ = \int_{\Omega} \lim_{n \to \infty}  f1_{E_n}\,d\lambda \tag{1}$$
$$ = \int_{\Omega} f 1_{E}\,d\lambda \tag{2}$$
$$ = \int_{E} f \,d\lambda$$

$(1)$ Now, try out: DCT? MCT? ICT? VCT? Scheffe's Lemma?

$(2)$ Good ol' $\varepsilon-N$ definition.

Case 2: Now, write $f = f^+-f^-$

Replace all $f$'s in Case 1 with $f^+-f^-$.


What is $dx$ in integration? When I was at school and learning integration in maths class at A Level my teacher wrote things like this on the board.

$$\int f(x)\, dx$$

When he came to explain the meaning of the $dx$, he told us "think of it as a full stop".  For whatever reason I did not raise my hand and question him about it.  But I have always shaken my head at such a poor explanation for putting a $dx$ at the end of integration equations such as these.  To this day I do not know the purpose of the $dx$.  Can someone explain this to me without resorting to grammatical metaphors?

The motivation behind integration is to find the area under a curve. You do this, schematically, by breaking up the interval $[a, b]$ into little regions of width $\Delta x$ and adding up the areas of the resulting rectangles. Here's an illustration from Wikipedia:

Then we want to make an identification along the lines of

$$\sum_x f(x)\Delta x\approx\int_a^b f(x)\,dx,$$

where we take those rectangle widths to be vanishingly small and refer to them as $dx$.

An integral gives you the area between the horizontal axis
and the curve.  Most of the time this is the x axis.

                         y

                         |                    |
                       --|--              ----|---- f(x)
                     /   |   \          /     |
                    /    |     --------       |
          |        /     |                    |
     -----|-------       |                    |
          |              |                    |
          |              |                    |
----------|--------------+--------------------|----- x
        a                                   b

And the area enclosed is:

Area =  $\int^b_a f(x) dx$

But say you didn't want to use an integral to measure the
area between the x axis and the curve.  Instead you just
caclulate the average value of the graph between a and b
and draw a striaght flat line y = avg(x)   (the average
value of x in that range).

Now you have a graph like this:

                         y

                         |                    |
                       - | -              - - | - - f(x)
          |          /   |   \          /     |
     -----|-----------------------------------|---- avg(x)
          |        /     |                    |
     - - -|- - - -       |                    |
          |              |                    |
          |              |                    |
----------|--------------+--------------------|----- x
        a                                   b

And the area enclosed is a rectangle:

Area = avg(x) w          where w is the width iof the section

The height is avg(x) and the width is w = b-a or in English,
"the width of a slice of the x axis going from a to b."

But say you need a more accurate area.  You could break the
graph up into smaller sections and make rectangles out of them.
Say you make 4 equal sections:

                           y

                           |                    |
                      |----|---|        |-------|---- f(x)
                      |    |   |        |       |
                      |    |   |--------|       |
            |         |    |   |        |       |
       -----|---------|    |   |        |       |
            |         |    |   |        |       |
            |         |    |   |        |       |
  ----------|---------|----+---|--------|-------|----- x
            a                                   b

And the area is:

Area = section 1  + section 2  + section 3  + section 4

= avg(x,1) w + avg(x,2) w + avg(x,3) w + avg(x,4) w

where w is the width of each section.  The sections are all
the same size, so in this case w=(b-a)/4 or in English,
"the width of a thin slice of the x axis or 1/4 of the
width from a to b."

And if we write this with a sumation we get:

Area =   $\sum^4_{n=1}avg(x,n) w$

But it's still not accurate enough.  Let's use
an infinite number of sections.  Now our area
becomes a summation of an infinite number of sections.
Since it's an infinite sum, we will use the
integral sign instead of the summation sign:

Area = $\int avg(x) w$

where avg(x) for an infinitely thin section will be
equal to f(x) in that section, and w will be "the width of
an infinitely thin section of the x axis."

So instead of avg(x) we can write f(x), because they are
the same if the average is taken over an infinitely small width.

And we can rename the w variable to anything we want.
The width of a section is the difference between the
right side and the left side.  The difference between
two points is often called the delta of those values.
So the difference of two x values (like a and b) would
be called delta-x.  But that is too long to use in an
equation, so when we have an infinitely small delta,
it is shortened to dx.

If we replace avg(x) and w with these equivalent things:

Area =  $\int f(x) dx$

So what the equation says is:

Area equals the sum of an infinite number of rectangles
that are f(x) high and dx wide (where dx is an infinitely
small distance).

So you need the dx because otherwise you aren't summing
up rectangles and your answer wouldn't be total area.

dx literally means "an infinitely small width of x".

It even means this in derivatives.  A derivative of
a function is the slope of the graph at that point.
Slope is usually measured as the y difference of two
points divided by the x difference of those points:

Slope = (y2 - y1) / (x2 - x1)

But the closer these points get the smaller these
differences get.  Let's start calling them deltas,
because the difference between two points is often
called the delta of those values.

Slope = delta-y / delta-x

The deltas get smaller and smaller as these two x,y
points get closer and closer.  When they are an
infinitely small distance apart, then the delta-y
and delta-x is shortened to dy and dx:

Slope = dy / dx

The slope is still Slope = (y2 - y1) / (x2 - x1)
but these points are infinitely close together, so
we use dy and dx to tell ourselves that they are
infinitely close or "differential distances."


How can we produce another geek clock with a different pair of numbers? So I found this geek clock and I think that it's pretty cool.

I'm just wondering if it is possible to achieve the same but with another number.

So here is the problem:

We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \mathbb{N}^+$ times in any mathematical expresion to produce results in range $[1, 12]$. No rounding, is allowed, but anything fancy it's ok.

If you're answering with an example then use one pair per answer.

I just want to see that clock with another pair of numbers :)

Notes for the current clock:

1 o'clock: using 9 only twice, but it's easy to use it 3 times with many different ways. See comments.

5 o'clock: should be $\sqrt{9}! - \frac{9}{9} = 5$

For $n=12$ and $k=12$ here is a solution:

$1=\frac{12}{12+12+12+12+12+12-(12+12+12+12+12)}$

$2=\left(12 \times \frac{12}{12-12+12-12+12+12+12+12+12+12}\right)$

$3=\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)\right)}\right)$

$4=\left(12-\frac{12}{\left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$5=\left(12 \times \frac{12}{\left(12 \times \left(12 \times \frac{12}{\left(12-\left(12-\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)}\right)\right)}\right)$

$6=\left(12+\frac{12}{\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$

$7=\frac{12}{\left(12 \times \frac{12}{\left(12-\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$8=\left(12+\left(12 \times \frac{12}{\left(12+\left(12+\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)}\right)\right)$

$9=\frac{12}{\left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$

$10=\left(12 \times \frac{12}{\left(12-\left(12 \times \frac{12}{\left(12-\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)\right)}\right)\right)}\right)$

$11=\left(12+\frac{12}{\left(12-\left(12 \times \left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)}\right)\right)\right)}\right)$

$12=\left(12+\left(12+\left(12-\left(12 \times \left(12 \times \frac{12}{\left(12+\left(12+\left(12+\left(12+\left(12+12\right)\right)\right)\right)\right)}\right)\right)\right)\right)\right)$


Prove that if the speed of a particle is constant, $\vec a$ is perpendicular to $\vec v$ My train of reasoning (well, not even that, more like "translating") so far:

Let $C$ be some constant.
$\|\vec {v(t)}\|=C$, so $\frac {d\|\vec {v(t)}\|} {dt}=0$.

But, where to go from here? I don't know how to express $\frac {d\|\vec {v(t)}\|} {dt}=0$ in terms of $\vec a$, aside from just flatly saying "$\mathrm{comp}_va=0$", which I feel doesn't sound very smooth or explanatory.

With some help from Semiclassical, I figured it out:

Let $C$ be some constant; we know
$||\vec {v(t)}||=C$ and $||\vec{v(t)}||^2=\vec v • \vec v$ so $\vec v • \vec v=C^2$ and $$\frac {d(\vec v • \vec v)} {dt}=2(\vec v • \frac {d \vec v} {dt})=0$$Therefore, $$\vec v • \vec a=0$$

a in direction of v = $\frac{d|v|}{dt}=0$
a in perpendicular direction of v = $\frac{|v^2|}{R}\ne0$ where R is radius of curvature


Intuitively understanding the probability that one of the two coins is a head given that there is a tail Suppose that I flip two coins without showing you the result. Now, I tell you that one of the coins is T. The probability that the other coin is H is
$$
\mathbf{P}[HT, TH| \text{at least one } T] = \frac{\mathbf{P}[HT, TH]}{\mathbf{P}\text{[at least one } T]} = \frac{2}{3}
$$

But the thing is, a fair coin should always yield $50\%$ for each flip. I understand that the two coins are no longer independent, but can someone provide me of an intuitive explanation of why this is?

Just list the possibilities.  Initially you have $HH,HT, TH, TT.$  You have ruled out $HH$, so you have three possibilities left.  In two of them the non-tails coin is heads.

The paradox arises simply because there are two ways to flip one head and one tail (the most likely outcome of flipping two coins) but only one way to flip two tails.

As a generalisation, consider flipping 24 coins and then revealing that there was a tail. What is the probability that at least one other coin was a head? Out of the $2^{24}-1=16777215$ 24-coin flips that contain a tail, only one of them contains 24 tails, leaving the conditional probability that at least one coin is a head as $\frac{16777214}{16777215}$ – in other words, virtually certain.


Does $\sum\limits_{x=1}^\infty\sin(x)$ converge? I received a task to find out whether the following series converges:

$$\sum_{x=1}^\infty\sin(x)$$

On first look it seems simple, but as I keep thinking about it, there's not a single lemma or criterion that I can use to tackle the problem.

D'alembert ? Doesn't work: The following is meaningless IMHO: $\lim\limits_{x \to \infty}{}\frac{\sin(x+1)}{\sin(x)}$

That series isn't monotonic... you can't understand the rules for when will a member of the series be negative or positive.

All I know is that $\sin(x)$ is blocked between (-1) and 1.
Though it's easy to see that $ \sum\limits_{x=1}^\infty\lvert\sin(x) \rvert$ diverges.

May I use Leibniz formula for $\pi$ in order to construct  2 subseries:

One that shows that $\sin(x)$ converges to the limit 1
Another one that shows that $\sin(x)$ converges to the limit 0

And we know that a series can't converge to 2 different numbers, hence it diverges?

The sequence $(\sin n)$ doesn't converge to $0$ so the given series is divergent.

Strictly speaking, the sum does not exist. That is because the sin(n) terms do not converge to zero, but keep oscillating in the interval (-1, +1). Hence all the partial sums also keep oscillating.

On the other hand, it is a close affair. The amplitude of the terms do not converge to zero, but they also do not diverge. More importantly, the fact that we are dealing with an oscillating sequence means that a lot of cancellation takes place. If one can combine neighbouring terms in a clever way, the cancellation might be such that the sum is convergent!

In my opinion this is an example of a (non-convergent) sum that can be made convergent by relative simple mathematical tricks. Instead of combining terms (which might work, I haven't verified it), there is an even simpler trick. Simply multiply each term with a convergence factor exp(-epsilon*n). Since epsilon is very small, the exponential factor is very close to 1 but just a little bit smaller; this is sufficient to make the summation convergent. If we proceed this way, we obtain:

S = (1/2i) * Sum(n=1 to inf) {exp(n*(i-eps)) - exp(n*(-i-eps))}

It is straightforward to perform the summations over n, which are now convergent.

S = (1/2i) * {exp(i-eps)/(1 - exp(i-eps)) - exp(-i-eps)/(1 - exp(-i-eps))}

We are already in the comfortable position that we can take the limit epsilon -> 0. Rearranging terms and using exp(ix) = cos(x) + i*sin(x), we get

S = (1/2i) * {1/(1 - cos(1) - i * sin(1)) - 1/(1 - cos(1) + i * sin(1))}

S = (1/2) * sin(1) / (1 - cos(1))

S = (1/2) * cos(1/2) / sin (1/2)

The calculation has resulted in a clear, unambiguous result for the sum S !


How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$? How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$?

Hmmm... So I didn't know what to do here so I tested some cases for a rule.

If a number had the factors $3^2$ and $2^1$, you can make $5$ distinct factors: $2^1$, $3^1$, $3^2$, $2^1 \cdot 3^1$, $2^1 \cdot 3^2$... I don't see a pattern yet.

How does one go about this?

I don't think the answer is $5!$....

If $\begin{equation}x = a^p \cdot b^q\cdot c^r+...\end{equation}$
then there are $(p+1)(q+1)(r+1)...$ numbers that divde $x$.
Any number that divides $x$ will be of the form $a^\alpha\cdot b^\beta\cdot c^\gamma$ .
So we have p p+1 options for $\alpha$ because we need to consider $\alpha = 0$ also. Similarly, we have $q + 1$ options for $\beta$ and $q+1$ options for $\gamma$.
Therefore, we multiply these out.

To get the intuition of why this is so, let us take the example of the number $24$.
$24 = 2^3\cdot3^1$
Any number that divides $24$ will be of the form $2^\alpha\cdot3^\beta$.
We have 4 choices for $\alpha:0,1,2,3$ and 2 choices for $\beta:0,1$.
So we have $4\cdot2 = 8$ numbers that divide 24. These can be listed out as:

$$(2^03^0,2^13^0,2^23^0,2^33^0),(2^03^1,2^13^1,2^23^1,2^33^1)$$

So, $2^5*3^4*5^3*7^2*11^1$ has $6\cdot5\cdot4\cdot3\cdot2 = \boxed{720}$

Your approach with $3^2$ and $2^1$ is very close. You only forgot to account for the factor $1 = 3^0\cdot2^0$. This brings us to $6$ distinct factors for this smaller example. You can calculate this by multiplying the number of all the different exponents used for each factor. $3$ can occur either $2$, $1$ or $0$ times, while $2$ can only occur $1$ or $0$ times. This brings us to $3\cdot2 = 6$ combinations of these exponents.

Now, the "real" question has another little trick to it because the factors aren't all that nice just yet. $2^5\cdot3^4\cdot4^3\cdot3^2\cdot2^1 = 2^6\cdot3^6\cdot4^3$
Much better.
Now, we can have between $0$ and $6$ occurences of $2$, making it $7$ possibilities.
Another $7$ possibilities for occurences of $3$ and finally $4$ possibilities for $4$.
This makes a total of $7\cdot7\cdot4 = 196$ possible distinct factors of $2^6\cdot3^6\cdot4^3 = 2^5\cdot3^4\cdot4^3\cdot3^2\cdot2^1$.

Edit
Alright, I didn't pay attention either. As Konstantinos Gaitanas stated, $4^3 = 2^6$, making the whole thing $2^{12}\cdot3^6$.
So the solution should indeed be $13\cdot7$ = 91.


Proof for the following nascent delta function Let $f(x)$ be a (nonzero) rapidly decreasing function in $\mathbb{R}$ (i.e. $f(x)$ and all its derivatives go to zero as $x\to\pm\infty$ faster than any negative power of $x$). The sequence of functions
$$
\eta_\epsilon(x)=\frac{\epsilon f(x)}{[\epsilon \pi  f(x)]^2+x^2}
$$
approximates the delta distribution $\delta(x)=\lim_{\epsilon\to 0}\eta_\epsilon(x)$.

Any idea for proving of this?

It is similar to the well-known example $\frac{\epsilon}{\pi(\epsilon^2+x^2)}$, but not the same.

Use the practical definition of the Dirac function below:

$$ \delta(x)=\{ \begin{matrix}
 0, & x\ne0\\ \infty, & x=0
\end{matrix}
$$
$$\int_{-\infty}^{\infty}\delta(x)dx=1$$

It is straightforward to verify the first condition and it remains to show that

$$ \lim_{\epsilon\to 0}\int_{-\infty}^{\infty} \eta_\epsilon(x)dx=\lim_{\epsilon\to 0}I_\epsilon= 1$$

where,

$$I_\epsilon = \int_{-\infty}^{\infty} \eta_\epsilon(x)dx =\int_{-\infty}^{\infty}\frac{\epsilon f(x)dx}{[\epsilon \pi  f(x)]^2+x^2}$$

With the variable transformation,

$$ u = \frac{x}{\epsilon \pi  f(x)}$$

The integral $I_\epsilon$ becomes

$$I_\epsilon = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{du}{1+u^2}
+\epsilon \int_{-\infty}^{\infty} \frac{xf'(x)dx}{[\epsilon \pi  f(x)]^2+x^2}$$

The second integral simply vanishes in the limit $\epsilon \rightarrow 0$. Therefore,

$$\lim_{\epsilon\to 0}I_\epsilon=\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{du}{1+u^2}
=\frac{1}{\pi}\tan^{-1}(u)|_{-\infty}^{\infty}=1$$

Thus, $\lim_{\epsilon\to 0}\eta_\epsilon(x)=\delta(x)$ indeed approximates the delta function.


Is it common to notate a probability density function with Pr(..)? My understanding is - and my internet-research reassured me - that one uses lower-case letters for the probability density function and upper-case letters for absolute probabilities. 'p' vs 'Pr' for example.

Now, I'm reading an established textbook ("Computer Vision: Models, Learning, and Inference" the PDF-textfile is available here: http://www.computervisionmodels.com/) and the author uses 'Pr' to notate probability density functions. See bottom of page 40 for example, where one can find the following formula:

Hence, I'm wondering whether this notation is common. Because I'm thinking about adapting it.

You are right. It is very common to denote absolute probabilities with uppercase letters and density functions with lower case letters. The letters "P" and "p" are the most commonly used but, of course, you can find other letters as well. In the textbook you are referring to they are just using a less standard notation. This is also very common.


conversion of distribution function Find distribution function  $F_Y(y)$

$$F_X(x) = \left\{ \begin{array}{ll}
         \frac{x}{\theta} &  \  0\leq x \leq \theta;\\
        1 & x \geq \theta \end{array} \right.  $$

Given that $Y=\min(X,\frac{\theta}{2})$

Can anyone please explain me how we convert distribution function?

Edit: I don't want to solve this particular problem, I want to know how these types of problems are solved.

Edit 2:  If necessary one can change the question to another random variable $Y$ or to a different distribution function $F_X(x)$ so that I can understand more about my problem.

They're solved mostly by definition-chasing.

$F_Y(y) = \Pr(Y\le y)$.  That is a definition.

$\Pr(Y\le y) = \Pr(\min\{X,\theta/2\}\le y)$.  That is a definition.

$\min\{X,\theta/2\}\le y$ if and only if $X\le y\text{ or }\theta/2\le y$.  That is an immediate corollary of a definition.  Thus we have
$$
F_Y(y) = \begin{cases} 1 & \text{if }y\ge\theta/2, \\ ? & \text{otherwise}. \end{cases}
$$
So suppose $0\le y < \theta/2$.  Then $\min\{X,\theta/2\}\le y$ if and only if $X\le y$, and the probability of that is $y/\theta$.  Si
$$
F_Y(y) = \begin{cases} 1 & \text{if }y\ge\theta/2, \\ y/\theta & \text{if }0\le y <\theta/2,  \\ ? & \text{otherwise}. \end{cases}
$$
It's not hard to see why the "otherwise" value must be $0$.

Don't try to solve problems like this by applying an algorithm.  Use logic.

For $y < \theta/2$, $F_Y = F_X$ is the same since $Y = X$. We see that for $y \geq \theta/2$, $F_Y = 1$ since all values of $Y$ are less than or equal to $\theta/2$.


How to determine if $z(x,y)=\ln(x^2 + y^2)$ is a harmonic function So I know that a function is harmonic if it satisfies Laplace's equation:

$\frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} = 0$

But I'm just not sure how I should put it in to action for a practice question I have, which is:
Determine if $z(x,y) = \ln(x^2 + y^2)$ is a harmonic function.

By the Chain Rule, $\partial z/\partial x=2x/(x^2+y^2)$. Now by the quotient rule, $$\partial^2 z/\partial x^2=\frac{2(x^2+y^2)-2x(2x)}{(x^2+y^2)^2}=\frac{2(y^2-x^2)}{(x^2+y^2)^2}$$

Similarly $\partial^2 z/\partial y^2=\frac{2(x^2-y^2)}{(x^2+y^2)^2}$. But this is the negative of the previous answer, so they add to zero.

Any real function $u(x,y)$ with continuous second partial derivatives which satisfies Laplace's equation,

$$\nabla ^2u(x,y)=0$$is called a harmonic function.

$z(x,y)=\ln(x^2+y^2)$
$$\nabla ^2 z(x,y)=\frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial y^2}$$
 you can solve the above expression and it must equal to $0$ to be satisfied for  the harmonic function


Fulton, Algebraic Curves, Exercise 1.37 I am stuck trying to prove this:

  Let $k$ be any field, $F\in k[X]$ a monic polynomial of degree $n>0$.
  Then the residues $\bar{1},\bar{X},...,\overline{X^{n-1}}$ form a
  basis for $k[X]/(F)$ over $k$.

I thought I had done it correctly but my proof isn't convincing me, so back to square one.

Any help will be greatly appreciated!

EDIT: The question does say "any field" but it is in the section that says $k$ is algebraically closed for the whole section.

Here's the attempt.

Let $G$ be any polynomial with residue $\bar{G}\in k[X]/(F)$, so $G-aF^m=0$ for some $a\in k$ and $m>0$. Since $k$ is algebraically closed, we may divide $G$ by the irreducible factors of $F$ so that $G$ does not vanish for any $P\in V(F)$. Moreover, since $F$ is monic, $\deg(G)=\deg(G-aF^m)+1$ for some $a\in k$ and $m>0$. It follows then that $\{\bar{1},\bar{X},...,\overline{X^{n-1}}\}$ is a basis for $k[X]/(F)$.

The fact that $F$ is monic is useful it says that
$$x^{n}=a_1x^{n-1}+\cdots +a_n.$$


Primes of the form $x^2+ny^2$ where $n\equiv 1\pmod{4}$ is a squarefree number Let $n\equiv1\pmod{4}$ be a squarefree number and $p\equiv1\pmod{4n}$ be a prime number. Does there exist $x,y\in\mathbb{N}$ such that $p=x^2+ny^2$?


Flirting Sequences (Real Analysis) I need help with a homework problem and am pretty sure my real analysis teacher made the following definition up:

In a metric space, a sequence $\{P_n\}_n$ $ \ $ flirts with $p$ iff for each $\epsilon > 0$, there is a $n \in \mathbb{N}$ and $m > n$ such that $0 < d(p_n,p) < \epsilon$ and $d(p_n,p) < d(p_m,p)$. The sequence $\{P_n\}_n$ $ \ $ is a flirting sequence if there is a point $p$ such that $\{P_n\}$ flirts with $p$.

Give an example of a sequence that flirts with $1$.

I am trying to $(1)$ understand the concept of "flirting" and $(2)$ figure out an example of a sequence that flirts with $1$. Actually, I would be happy to see an example of a any sequence that flirts with something.

I translated the definition like this:

In a metric space, the sequence $\{P_n\}$ flirts with $p$ iff
$$
\forall \epsilon > 0 \quad \exists(m,n \in \mathbb{N}, m >n): 0 < d(p_n,p) < \epsilon \quad  \text{and} \quad  d(p_n,p) < d(p_m,p).
$$

I have concluded that in $\mathbb{R}$ with the usual metric, the sequences {$2-\frac1n$} and {$1-\frac1n$} do not flirt with $1$.

Can anyone help me understand this concept and/or provide an example of any sequence that flirts to some point?

Consider the following real-valued sequence $(p_n)_n$ which flirts with $0$:
$$p_n = \begin{cases} 1 & \text{ if } n \text{ is even}\\
\frac{1}{n} &\text{ if } n \text{ is odd}
\end{cases}$$ can you see why this is the case?

Note that

a sequence can flirt with $p$ but not converge to $p$: as in the above  (it will have a subsequence converging to $p$, though)
a sequence can flirt with $p$ and converge to $p$:
$$p_n = \begin{cases} \frac{1}{2^n} & \text{ if } n \text{ is even}\\
\frac{1}{n} &\text{ if } n \text{ is odd}
\end{cases}$$

Remember that a easy example of a flirting sequence is $~-1, ~1/2,~ -1/3,~ 1/4, ~-1/5, ~1/6,~$ etc.

Mainly if it is even it is positive, if it is odd it is negative, just like $~(-1)^n~.~$

I think the formula is  $-1/n,~~ +1/(n+1)$

Even better, $-1/(2*(n)),~~ +1/(2*n+1)$.


If $n>k$, $n \mid km$, and $m \nmid n$, does this imply that $k \mid n$? I think I am missing something very obvious here, but I've got the following question:

  Be $k,m,n \in \mathbb{Z}$ with $n \ge m$, $n > k$, $n \mid k \cdot m$ and $m \nmid n$

Does this imply $k \mid n$?

No. For a counterexample, you can take $k = 6, m = 10$, and $n = 15$.

It is possible you are misremembering the following similar statement.

If $a \mid bc$ and $\gcd(a,b) = 1$, then $a \mid c$. This is stated and proved on MSE elsewhere.

the statement shows us that $k*m = r*n$ for some $r$. if $gcd(k,r) = d$, $d > 1$and $d$ doesn't divide $n$, then clearly $k$ doesn't divide $n$


Infinite geometric series and convergence I have been given the infinite geometric series-:
$$
2-\frac 4x + \frac8{x^2}
-\frac{16}{x^3}+\dots
(x\neq 0)$$
I am asked to

Find the convergence area to the sequence
Find the sum $S(x)$ and the domain

I'm completely stuck as to how I should proceed!

When I use $X$ values greater than $0$ and less than $1$ it seems as though the series alternates between $-\infty$ and $+\infty$  as I approach infinity...

With $X$ values less than $0$ it seems to approach $+ \infty$?

I have no idea, could some kind soul shine a light?

The series is $\sum_{n=0}^\infty a_n$, for  $a_n=2(\frac{-2}{x})^n$.

A good first step is to compute $\frac{a_{n+1}}{a_n}$ and simplify.  If this is constant, you have a geometric series you can sum with the formula.  If not, you can apply the ratio test.