Is there any intuition behind the statement $E[X_\tau \mid \mathcal{F}_\sigma]=X_\sigma$ Is there any intuition behind the statement $E[X_\tau \mid \mathcal{F}_\sigma]=X_\sigma$ a.s.
I mean I know that the interpretation of the conditional expectation and how to visualize it somehow but I kind of lose track when I see stopped processes as they are way abstract for me to make any real sense of it. I mean I can use the definitions and prove a few things but then I would really like to understand the relevance of these stopping theorems.

P.S: In my course on continuous time finance we did a lot of stopping theorems and their proofs which I think I understood but then I have no intuition
Sorry if this question is not very precise but I have been struggling with this for quite some time and I felt compelled to ask

It is relatively intuitive. It says that your information of the past up to now doesn't help you to predict the future. Your expectation is neither higher is smaller of current's value.
You mentioned studying finance. Those traders who follow trend for example , think it is not martingale and they follow the trend because they think the expected price in the future is not equal of today's value.


Integrating the derivative of a multivariate function If $u=u(x,t)$ how do I compute $\displaystyle \int \frac{du}{dt}dt$ ?

Would I be correct in saying it is not simply a case of cancelling the $dt$'s and getting an answer of $u$+ constant, seeing as $u$ is a multivariate function?

What does the integration mean?

Integration with one variable $t$ is an operation acts on a function with the variable $t$

So $\int \frac{d}{dt}u(x,t) dt$ means nothing more than we integrate the function $\frac{d}{dt}u(x,t)$ with respect to $t$ with $x$ fixed. If $x$ was a function  of a variable $t$, then it must have been noted before one uses the integrations. However, it's not the case.

Hence, it is just $F(x)=\int \frac{d}{dt}u(x,t) dt$. And to emphasize that $x$ is fixed under the integration, set $u_x(t):=u(x,t)$

Then, $F(x)=\int \frac{d}{dt}u_x(t) dt = u_x(t) + C = u(x,t)+ C$.

This is just the fundamental theorem of calculus.

I don't have time to write more so I hope you understand it now.

\begin{equation}
\frac{du}{dt}dt=du
\end{equation}
So
\begin{equation}
\int \frac{du}{dt}dt= \int du
\end{equation}
Where
\begin{equation}
du=u_{x}dx+u_{t}dt
\end{equation}
Where
$$u_{x_{k}}=\frac{\partial u}{\partial x_{k}}$$


Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!

  Problem:

  For any natural number $n , n^3 + 2n$ is divisible by $3.$
  This makes sense

  Proof:

  Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
  $2 \times 0 = 0.$ So it is divisible by $3.$

  Induction: Assume that for an arbitrary natural number $n$,
  $n^3+ 2n$ is divisible by $3.$

  Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
  the induction hypothesis. Got it

  $$( n + 1 )^3+ 2( n + 1 ) = (  n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$

  $$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
 and regrouping}\}$$
  $$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$

  which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
  by the induction hypothesis. What?

Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$

In the inductive hypothesis, you assumed that $n^3 + 2n$ was divisible by 3 for some $n$, and now you're proving the same for $n+1$. It's like knocking down dominoes: if you can prove that the first domino falls over (base case) and each domino knocks over the next (inductive step), then that means that all of the dominoes get knocked down eventually.

You know that $(n^3 + 2n) + 3(n^2 + n + 1)$ is divisible by 3 because $n^3 + 2n$ is (because of the inductive hypothesis) and $3(n^2 + n + 1)$ is (because it's 3 times an integer). So, their sum is as well.

RTP : $a^3+3a$  is in the form of $3k$

Proof : $a^3+2a=
            a^3-a+3a=
            a(a^2-1)+3a=
            a(a+1)(a-1)+3a$
We know that sum of three consecutive numbers is divisible by 3           $$3k+3a=3(k+a)=3k$$


Intro to Probability A person randomly places 9 rooks on a 9 by 9 grid (the 8 by 8 case corresponds to a chess board). What is the probability that none of the rooks can capture any of the other rooks?

I have a good start on the problem, but I think that I am possibly missing a step or something. Any hits would be apprciated. I know that each time you place a rook, you lose spaces that you can place the next rook. It goes 9^2 spaces to 8^2 spaces to 7^2 spaces...down to 1 space at the end. Thats for the chances to satisfy the problem. For total options it goes 81 spaces to 80 spaces down to 72 when you place the last rook. So my answer is (81*64*49*36*25*16*9*4*1)/(81*80*79*78*77*76*75*74*73*72) Am I missing anything here?Thank you

If you think about it, placing 9 rooks on a 9x9 chessboard so that they don't attack each other is equivalent to asking how many permutations there are of the numbers $1,2,\cdots,9$. This is because each allowed rook configuration corresponds to a $9\times 9$ permutation matrix $M$, with $M_{ij}=1$ if there's a rook at $(i,j)$ and 0 otherwise: there cannot be two 1's on the same row or column. So the answer is immediately $9!/\binom{81}{9}$.

Don't have enough points for a comment, but you seem to be contradicting yourself when you first say that for the first rook you have 9^2 , for the second you have 8^2 and then for the total options you cite 81, 80, etc.


An n×n matrix B is define e^B by e^B=∑(B^j/j!) . Let P be the char polynomial of B.then the matrix e^P(B) is An $n \times n$ matrix $B$ defines $e^B$ by $e^B=\sum\left(B^j/j!\right)$. Let $P$ be the char polynomial of $B$. Then the matrix $e^{P(B)}$ is

$I(n \times n)$
$O(n \times n)$
$eI(n \times n)$
$πI(n \times n)$

By the Cayley-Hamilton theorem, every matrix satisfies it's own characteristic polynomial.  Thus

$P(B) = 0; \tag 1$

it follows that

$e^{P(B)} = e^0 = I. \tag 2$


Let $G$ be a group with $33$ elements acting on a set with $38$ elements. Prove that the stabilizer of some element $x$ in $X$ is all of $G$. I'm trying to figure out this old qualifying exam question:
Let $G$ be a group with 33 elements acting on a set with 38 elements. Prove that the stabilizer of some element $x \in X$ is all of $G$.
I think I'm supposed to use the orbit-stabilizer theorem to prove that the orbit of any $x\in X$ must be trivial, i.e. $orb_G(x)=\{x\}$. This is what I know:
$|G|$ and $|X|$ are relatively prime.
Since $|orb_G(x)| $ divides $|G|$ we must have that $|orb_G(x)|=1, 3, 11 $ or 33.
The orbit of each $x\in X$ partitions $X$.
If $|orb_G(x)|=1$ then by the orbit-stabilizer theorem: $|G|=|orb_G(x)||stab_G(x)| \implies |stab_G(x)|=33$.
I just don't see how to put this together in the right way. I wondered if $|orb_G(x)| $ necessarily needs to divide $|X|,$ but I didn't find anything to support that.

For a fixed $x \in X$, define a group action $f: G\times X\to X  $ by $(g,x) \mapsto x$ where $g\in G_x\subset G$, this map is not necessarily injective because $G_x$ can have more than one element. Since $1\in G_x$, this map is surjective.
Now by the orbit-stabilizer theorem, if we have for some $x\in X$ that is not equivalent to other elements in X,
$$1=|{x}|=|\{f(x)\}|=\vert G\cdot x|=|G|/|G_x| $$
This forces $|G_x|=|G|$, i.e., stabilizer of x in $X$ is all of G.
Now we only need to show the existence of such nonequivalent element;
Assume that there is an equivalent element y$\in X$ of x, then $x\dot g=y$ for all x$\in X$ for some $\dot g\in G$ and $G\cdot x=G\cdot y$. By orbit-stabilizer equation, we have $|G_x|=|G_y|$. Since $\cup_x G_x = G$ and the existence of equivalence of every element forces $|G|=33$ is divisible by 2, which is a contradiction.
Hence there exist a nonequivalent element x and we complete the proof.


Where does the guassian function/normal or bell curve come from? I am confused as to where the function for the normal distribtuion comes from.  Where does the e and pi come from?  In my textbook I am presented with the function,but I am unsure about where it came from.  I have spent hours on google and have yet to find a good proof that i can understand.  Come someone provide a source that shows an elementary proof of the normal curve function.

The Gaussian (normal or bell curve) arises, when you look at a huge collection of random variables and whose distribution you are not aware of.

You might want to look up Central Limit Theorem (CLT). The CLT explains why when you have a huge collections of random variables, they appear to follows the Gaussian.

A version of the CLT goes as follows.

Consider $n$ independently and identically distributed random variables, $\{X_1,X_2,\ldots,X_n\}$, with mean $\mathbb{E}(X_k) = \mu$ and variance $\mathbb{E}((X_k-\mu)^2) = \sigma^2 < \infty$. Now look at the sample mean $$S_n = \dfrac{X_1 + X_2 + \cdots + X_n}{n}$$
The central limit theorem states that $$\sqrt{n}(S_n - \mu) \to_{d} \mathcal{N}(0,\sigma^2)$$ as $n \to \infty$.

A proof for a weak version goes like this. Consider the characteristic function of the random variable $X$, say $\phi_X(t)$. Expanding using Taylor series about $0$, we get that $$\phi_{X-\mu}(t) = \phi_{X-\mu}(0) + \phi_{X-\mu}'(0) t + \dfrac{\phi_{X-\mu}''(0)}2 t^2 + \mathcal{O}(t^4/n^4) = 1 - \dfrac{\sigma^2}2 t^2 + \mathcal{O}(t^4/n^4)$$
The characteristic function for $(S_n - \mu)$ is $$\phi_{S_n - \mu}(t) = \phi_{X - \mu}(t/n)^n = \left( 1 - \dfrac{\sigma^2}2 t^2/n^2 + \mathcal{O}(t^4/n^4)\right)^n$$
Hence, the characteristic function for $Z_n = \sqrt{n}(S_n - \mu)$ is $$\phi_{Z_n}(t) = \phi_{\sqrt{n}(S_n - \mu)}(t) = \phi_{S_n - \mu}(\sqrt{n}t) = \left( 1 - \dfrac{\sigma^2}2 nt^2/n^2 + \mathcal{O}(t^4/n^2)\right)^n = \left( 1 - \dfrac{\sigma^2 t^2/2}n + \mathcal{O}(t^4/n^2)\right)^n$$
Hence, $$\lim_{n \to \infty} \phi_{Z_n}(t) = \lim_{n \to \infty} \left( 1 - \dfrac{\sigma^2 t^2/2}n + \mathcal{O}(t^4/n^2)\right)^n = \exp(-\sigma^2 t^2/2)$$

Note that the proof above is true for independent, identically distributed random variables arising from any nice probability distribution. With a little more effort, you can show that the convergence holds for independent (but not necessarily identically distributed) random variables. This is why the normal distribution is fundamental and appears everywhere in probability and statistics. The number $e$ appears since $$e = \lim_{n \to \infty} \left(1+\dfrac1n \right)^n.$$ The $\pi$ you are reffering to appears as a normalizing constant since $$\int_{-\infty}^{\infty} \exp(-x^2) dx = \sqrt{\pi}$$

I'm not aware of an elementary derivation of the functional form itself, but I can give you my favorite bit of intuition on where it comes from.

Suppose I told you I had a function $f : \mathbb{R} \to \mathbb{R}$, and that $f(0) = 0$.  Based on this extremely limited information, I then ask you to find me $g(a \, | \, x)$, a probability density function for the value of $f(x)$.

There is one crucial observation you can make that allows you to find a partial solution.  That observation is: for any $a, \Delta \in \mathbb{R}$, intuitively speaking, it must be true that $P(f(x) = a \, | \, f(0) = 0) = P(f(x + \Delta) = b + a \, | \, f(\Delta) = b)$.  In other words, you have no reason to believe that the point $(0, 0)$ is special: you might as well rephrase the problem as what are the odds that $f$ increases by $a$ when its parameter is changed by $\Delta$? and assume that these odds are the same at any point of $f$.

We can use this to set up an integral equation.  When $f$ travels from $(0, 0)$ to $(x, ?)$, imagine it first travelling to some intermediate point $x'$ and then on to $x$ from there.  Then for any $x'$ between $0$ and $x$, it must be true that:
$$g(a \, | \, x) = \int_{-\infty}^{\infty} \, g(a' \, | \, x') g(a - a' | x - x') \, da'$$

The solution to this integral equation is:
$$\frac{1}{\sqrt{2\pi c_1x}} \, e^{-\frac{1}{2}\frac{(a - c_2 x)^2}{c_1 x}}$$

This equation is commonly referred to as the Gaussian with drift.  $c_1 x$ is the variance and $c_2 x$ is the mean.


What is the value of $\angle ABD$ in the following figure? In the given figure $\angle BAE, \angle BCD$ and $\angle CDE$ are
  right angles and $AB = 4, BC=3, CD=4$  and $DE=5$. What is the value
  of $\angle ABD$?

I found this problem in a sheet of contest math problems. However the problem stated in the sheet was "Find the value of $AE$ ". I proved the problem using $BE^2 = 8^2+4^2$. Then I noticed that it could be proved by making rectangle $ABED'$. I tried to prove why $\angle ABD$ is a right angle only with the given conditions but failed. Is the $\angle ABD$ even a right angle (then how to prove it) or it can have different values?

Continue with $BE = \sqrt{8^2+4^2} = 4\sqrt5$ and recognize that the triangle BDE is isosceles with $BD =DE=5$,

$$\cos \angle ABE =\frac{AB}{BE} = \frac1{\sqrt5}, \>\>\>\>\>
\cos \angle DBE =\frac{BE/2}{BD} = \frac2{\sqrt5}
$$

Thus,

$$\angle ABD = \angle ABE + \angle DBE = \arccos \frac1{\sqrt5}
+ \arccos \frac2{\sqrt5}=90^\circ$$

BD = 5.

Drawing a perpendicular from D to AE at G,
DG = 4 so EG = 3
and AG = 5 so AE = 8.

As to the angles,
EDG = CDB (both 3-4-5)
so CDE = AED = CBD
so BDG = 90
so ABD = 90.


Is there a derivation of the linear regression equation from statistical principles? Let us say that we have the equation $y = ax + b$ where $a = \frac{rs_{x}}{s_{y}}$ and $b = \bar{y} - a\bar{x}$. Where does $a$ and $b$ come from? How do statisticians reason their "derivation" if there is one?

See Ordinary Least Squares and Simple Linear Regression at Wikipedia.


I've been told that the use of ellipsis in "$S = x_1 + x_2 + x_3 + x_4 + \dots$" is ambiguous and meaningless. Is it? What else can the definition of the ellipsis symbol, "$\dots$", mean in this context?

$$S = x_1 + x_2 + x_3 + x_4 + \dots$$

All I can see is that you have an infinite sum of $x$s, where the first one is $x_1$, the next is $x_2$, then $x_3$, and so on forever, for as many natural numbers as exist and in order. But, for some reason, I am being told that such a definition is ambiguous and meaningless compared to formal mathematics.

Stating ambiguity of this summation is bad faith. The pattern is obvious and the ellipsis clearly indicates an unlimited sequence of terms.

I would be more critical towards a sum like

$$1+2+4+\cdots\ ?$$

In common practice, if the first few terms/indexes (as little as $3$) follow an arithmetic progression, it can be considered implied.

To give you an idea of what can go wrong with arbitrary entries in the summation the following example shows these complications. Begin with $S=1 + 2 + 4 + 8 + ...$ and so $S=1+2(1+2+4+8+...)$ and so $S=1+2S$ giving us finally that $S=-1$ and we now have a sum of positive numbers equal to a negative number, which is absurd. These complications must be addressed in order to deal with sums of this kind meaningfully.


Proving $f'$=$g'$ has some c such that $g=f+c$ Suppose that $f$ and $g$ are differentiable functions on $(a,b)$ and suppose that $g'(x)=f'(x)$ for all $x \in (a,b)$. Prove that there is some $c \in \mathbb{R}$ such that $g(x) = f(x)+c$.

So far, I started with this:

Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then $h'(c)=0 \implies h(c)=c$

After this i'm not sure where to go, or if this is correct at all, any hints?
This is also my first post in Latex so sorry if there's any mistakes!

Fixing your proof a little bit, I can extend your proof.

Suppose that your domain of function is (a,b). And let's choose any $\beta\in (a,b)$.

So for any $x\in (a,b)$ $\exists \alpha\in (\beta,x)$ (or $\alpha \in(x,\beta)$ if $x<\beta$) such that $h'(\alpha)=\frac{h(x)-h(\beta)}{x-\beta}=0$.

Thus,

$$\frac{h(x)-h(\beta)}{x-\beta}=0 \implies h(x)-h(\beta)=0\\ \implies f(x)-g(x)=f(\beta)-g(\beta)\\ \implies f(x)=g(x)+f(\beta)-g(\beta)$$

Thus, the $c$, we were finding is $f(\beta)-g(\beta)$.

Assume f-g is non-constant. Then f'-g' is not identically zero. But f'=g' by hypothesis. Contradiction. Hence f-g is constant.


Simplify a set (A ∪ B)\(A \ (A ∩ B)) I'm trying to simplify a set and I'm stuck.
The set is: $(A \cup B) \setminus (A \setminus (A \cap B))$.

If someone can help with some instructions and explanations it would be great!

If $x \in (A \cup B) \setminus (A \setminus (A \cap B))$, then $(x \in A$ or $x \in B)$ and ($x \not \in A$ and $(x \in A$ and $x \in B$)).
Since $x \in A$, $x \not \in A$ is a contradiction, it must be that $x \in B$, meaning that the expression simplifies into $B$.
You could also draw Venn diagrams to get a visual as to what's actually happening.

Since this is a simple set expression involving two sets, we can check it by just using $A=\{1,3\}$ and $B=\{2,3\}$ (see here for details).

Thus we have

\begin{align}
(A\cup B)\setminus(A\setminus(A\cap B))
&= (\{1,3\}\cup\{2,3\})\setminus(\{1,3\}\setminus(\{1,3\}\cap\{2,3\}))\\
&= \{1,2,3\}\setminus(\{1,3\}\setminus\{3\})\\
&= \{1,2,3\}\setminus\{1\}\\
&= \{2,3\} = B
\end{align}

Therefore the set expression simplifies to $B$.


What is the value of an infimum if it's taken over an empty set? I just got to this theorem in Royden where, a set function, $\mu$, over an arbitrary collection of subsets, S, of a set X can induce an outer measure.  But the outer measure is defined as an infimum of a sum with the infimum taken over countable collections that cover any set.  If $E\subseteq X$, then define

$\mu ^* E= \inf \sum_{k=1}^\infty \mu E_k$ over collections $\left\{E_k\right\}$ in S that cover E.

I'm just confused when there are no collections in S that cover E.  Or should I not bother thinking about this because useful set functions should at least be defined on X and evaluate $\mu X$.

The infimum is the greatest lower bound.  It is as "great" as it can get subject to the constraints imposed by the members of the set whose infimum you're taking.  For each such member, the constraint is merely that the infimum is not allowed to be greater than that number.  If there are no such constraints, then the infimum is $+\infty$.

PS (later edit):  Here's a practical (?) example.  Suppose the distance between two points on a manifold is defined as the infimum of the set of all lengths of paths connecting the two points.  Now suppose the two points are in different connected components of the space.  Then, since the greatest lower bound of the empty set is infinite, the distance between two points not connected by any path is infinite.  I seem to recall seeing some proposition somehere, in which this way of looking at it obviated the need for a hypothesis that the space is path-connected.

The null set is covered by any subset.  The infimum you see is not a void infimium.


Counting sets - Addition principle Theorem: If A and B are non-empty sets, and A and B are disjoint, then
$$ |A \bigcup B| = |A|+|B|$$
If I have n sets and all of them are disjoint, then
$ |A_1 \bigcup A_2 \bigcup...\bigcup A_n| = |A_1|+|A_2|+...+|A_n|$

If I want to prove that, can I do like this?

1) By the theorem we know that is it true fpr n=2

2) we suppose that it is $ A_1, A_2, ... , A_k $ is disjoint then
$$ |A_1 \bigcup A_2 \bigcup...\bigcup A_k| = |A_1|+|A_2|+...+|A_k|$$

3)  we suppose that $ A_1, A_2, ... , A_k, A_{k+1} $ is disjoint sets, then it follows that $A_1 \bigcup A_2 \bigcup...\bigcup A_k$ and $A_{k+1}$ are disjoint sets. The theorem above then applies that
$$ |A_1 \bigcup A_2 \bigcup...\bigcup A_k\bigcup A_{k+1}| = |A_1|+|A_2|+...+|A_k|+|A_{k+1}|$$
and by 2)
$$ |A_1 \bigcup A_2 \bigcup...\bigcup A_k| = |A_1|+|A_2|+...+|A_k|$$
so
$$ |A_1 \bigcup A_2 \bigcup...\bigcup A_k\bigcup A_{k+1}| = |A_1|+|A_2|+...+|A_k|+|A_{k+1}|$$
The result follows now from induction principle.

What you have is fine.  I've tidied it up a little, exploiting the $\bigcup$ notation a little more here:

Theorem: If $A$ and $B$ are non-empty sets, and $A$ and $B$ are disjoint, then
$$ \left\vert A \cup B\right\vert = |A|+|B|$$

Proposition: If we have $n$ pairwise disjoint sets, then
$ \left\vert\bigcup_{j=1}^{n} A_j\right\vert = \sum_{j=1}^{n}|A_j|$

Proof of Proposition:

The case of $n=1$ is trivial.

Base Case: By the theorem we know that the proposition holds for $n=2$

Induction hypothesis: Assume that for some $k$, whenever $ A_1, A_2, \ldots, A_k $ are disjoint then
$$ \left\vert\bigcup_{j=1}^{k} A_j\right\vert = \sum_{j=1}^{k}|A_j|$$

We now suppose that $ A_1, A_2, \ldots , A_k, A_{k+1} $ are pairwise disjoint. It follows that $\bigcup_{j=1}^{k}A_j$ and $A_{k+1}$ are disjoint sets. The Theorem tells us that
$$ \left|\bigcup_{j=1}^{k+1} A_j\right| = \left\vert \left(\bigcup_{j=1}^{k} A_{j}\right) \cup A_{k+1}\right\vert= \left\vert \bigcup_{j=1}^{k} A_{j} \right\vert + \left\vert A_{k+1}\right\vert.$$

Now applying the induction hypothesis to this, we get
$$\left|\bigcup_{j=1}^{k+1} A_j\right|=\left(\sum_{j=1}^{k}|A_j|\right)+|A_{k+1}|=\sum_{j=1}^{k+1}\vert A_{j}\vert$$
concluding our proof.

I've tried to follow your reasoning as closely as I could here. It is actually possible to be slightly more efficient.  The base case really should be $n=1$.  The proof for $n=2$ follows from this base case and the argument for the inductive step (which is really just an invocation of the Theorem).


If $\sin A$, $\sin B$, $\sin C$ are in AP, then $a$,$b$,$c$ are in If $\sin A$, $\sin B$, $\sin C$ are in AP, then $a$,$b$,$c$ are in..

$1$ AP, $2$.GP, $3$. HP, $4$.none
My attempt:
Let
$$\begin {align*}
\sin A&=x\\
A&=\sin^{-1} (x)\\
\sin B&=y\\
B&=\sin^{-1} (y)\\
\sin C&=z\\
C&=\sin^{-1} (z).
\end{align*}
$$
So, what's next?

If sin = 0, 1/2, 1,
then angles = 0, $\pi/6, \pi/2$.

Therefore not 1
($\pi/2-\pi/6 \ne \pi/6-0$),
not 2
(can't have a zero term),
not 3
(can't have zero).

Therefore 4.

Note:
Same applies to cos,
with the angles being
$\pi/2, \pi/3, 0$.

We know $a/\sin A=b/\sin B=c/\sin C=2R$.
       So, by instruction $\sin A$, $\sin B$, $\sin C$ are in AP.
     So, $2\sin B=\sin A+\sin C$.
         Therefore $2b=a+c$.
       So $a$, $b$, $c$ are in also AP.


why is $2.2250738585072014\text{e}{-308}$ not a number? In programming the min value of a float is: $$2.2250738585072014\text{e}{-308}$$

but when I type this into a calculator, it says Not a Number. what I am wondering is why this is an invalid number?

Note: I am only just finishing grade 8 math and haven't learned fully what the 'e' in an equation does. I think it has something to do with putting the number to the power of ten times the number after the e.

You are correct about your interpretation of the $e$ stuff. Indeed, this value is

$2.2250738585072014 \cdot 10^{-308}$

Which is, to be sure, a number. It's just that, in order to properly store this number, the calculator will need a fair bit of memory to store it accurately, and you've essentially attempted to exhaust it.

In the so called "scientific notation" of numbers the e stands for 10 to the power of x. So in your case 2.22... times 10 to the power of -308.
Your number happens to be the smallest number representable by the very common data type "double", a 64 bit binary floating point representation. So your calculator is unable to represent it because it is too small and wrongfully tells you it was not a number when in fact it is a number, simply too small.


Is composition of surjective continuous function with discontinuous function discontinuous? Let $I_1,I_2,I_3$ be intervals $\subset \mathbb{R}$. Suppose $f:I_1 \to I_2$ is a surjective continuous function and $g: I_2 \to I_3$ is a discontinuous function. Must the composition $g \circ f$ be discontinuous?

There are some easy counter-examples if $f$ is not assumed to be surjective, e.g. taking $f$ to be a constant function, or in a way that "dodges" the discontinuous point(s) of $g$.

However if such "dodging" is prohibited, I fail to construct such functions nor find an answer from many similar questions on this site. So I am interested to know whether counter-examples exist? If not, is there a proof? Does it have something to do with intermediate value theorem?

If $g$ is discontinuous at $x=a$, then $gf$ cannot be continuous. Suppose $gf$ is continuous. For any sequence $\{x_n\}\in I_2, x_n\to a$, due to the surjectivity of $f$, $\exists \{y_n\}\in I_1, f(y_n)=x_n,\forall n$.

$y_n\in I_2$ which is compact, so there must be a subsequesce $y_{n_k}$ that converges. Let $\lim_{k\to \infty}y_{n_k}=L$.

Since $f$ is continuous, $f(L)=\lim_{k\to \infty} f(y_{n_k})=\lim_{k\to \infty}x_{n_k}=\lim x_n=a$. Also, $gf(L)=g(a)$.

So, $\lim g(x_n)=\lim gf(y_{n_k})=gf(L)=g(a)$, thanks to the continuity of $gf$.

The result is that we establish the continuity of $g$ from the continuity of $f$ and $gf$. So $gf$ cannot be continuous.

Given that $g$ is discontinuous there exists $x_0 \in I_2$ and a sequence $(x_n)_{n \in \mathbb{N}} \subseteq I_2$ such that $\lim\limits_{n \to \infty}x_n = x_0$ and $g(x_0) \neq \lim\limits_{n \to \infty}g(x_n)$. Now let $y_0 \in f^{-1}(x_0)$ and for any $n \in \mathbb{N}$ let $y_n \in f^{-1}(x_n)$ be given. It follows
\begin{align*}
(g \circ f)(y) = g(x_0) \neq \lim\limits_{n \to \infty} g(x_n) = \lim\limits_{n \to \infty} (g \circ f)(y_n).
\end{align*}
We have therefore shown that $g\circ f$ is discontinuous at $y$.


Bounded Set for Rational Number $$A = \{x \in \Bbb Q: x^2 < 11\}$$

I am looking for the upper and lower bounds of set $A$. I know they should be the greatest and the smallest rational number within $(- \sqrt{11}, \sqrt{11})$

So the difficulty is how to get those rational numbers in exact value.
Thanks

As you correctly noted, $A = (-\sqrt{11}, \sqrt{11}) \cap \mathbb Q$. However $\pm\sqrt{11}\notin \mathbb Q$ and $\mathbb Q$ is dense in $\mathbb R$. Therefor $A$ is dense in $(-\sqrt{11}, \sqrt{11})$ and we can find a sequence in $A$ converging to any of $\pm\sqrt{11}$. This proves that no upper or lower bound exists in the set (i.e. $\min A$ and $\max A$ are undefined). However, $\sup A$ and $\inf A$ are defined and are least upper and greatest lower bounds for $A$. Since we already have a convergent sequence to $\pm\sqrt{11}$ and both are bounds on $A$, we can see that
$$\sup A = \sqrt{11}\\
\inf A = -\sqrt{11}$$


"Are there finitely or infinitely many Fermat primes?": decidable? Has anyone ever proven that there exists a proof or disproof that there are finitely many Fermat primes. I know that it's an unsolved problem whether there are finitely or infinitely many Fermat primes but my question is only whether it has been proven to be possible to prove or disprove it. If so, how can I access such a proof?

It's a $\Large{\Pi_2}$ statement of an incomplete theory (Peano arithmetic) so we can't a priori say "there exists a proof or disproof of this theorem".

If it was $\Large \Sigma_1$ statement then there would certainly exist a proof of it were true - but not necessarily if it were false. If it was a statement from a complete theory we could say for sure there was a proof without having one.

Personal opinion: On the other hand it's clearly true that there are only 5 Fermat primes, and proving must be possible (since it doesn't do any horrible self-reference or encoding into it) but by ideas that have not been discovered.

http://en.wikipedia.org/wiki/Arithmetical_hierarchy

Currently no such proof exists. I doubt that one ever will.

A simple search for "Fermat Prime" will tell you a lot.


What makes it legitimate to multiply both sides? Having the proof of the cancelation law for multiplication:

$$cb=ab$$

$$(cb)b^{-1}=(ab)b^{-1}\tag{Inverse}$$

$$cbb^{-1}=abb^{-1}\tag{Associativity}$$

$$c\cdot 1=a\cdot 1\tag{Indentity}$$

$$c=a$$

This may seem stupid, but exactly at the second step: I can multiply both sides by the inverse, but is this a consequence of the field axioms or of the equality defined as an equivalence relation? I guess it's the second one, but I'm not sure if one could prove it with the field axioms alone.

Because if there were some operation you could perform on the two quantities to get different results, they would not be equal.

If two things are equal, the results of multiplying them each by $b^{-1}$, or of performing any other operation on them, must be equal, because that is what it means for two things to be equal.

Yes, you can prove it from the properties of real numbers.

Proof:
If $a$, $b$, and $c$ are any real numbers, and $a = b$.

$ac$ is a unique real number $\iff$ closure property of multiplication

$ac = ac$  $\iff$reflexive property of equality

$a = b$ given

$ac = bc$ $\iff$ substitution principle


Abstract algebra homomorphism question This was a question on my abstract algebra test. I didn't know what to do with it.

Let $\phi:\mathbb{Z}\rightarrow \mathbb{Z}_8$, such that $\phi(1) = 6$. For what values of $n$ does $\phi(n) = 0$?
This is a homomorphism.

I listed the multiples of $8$, since they are all equal to $0\bmod 8$. That's obviously wrong, though, and I should be ashamed.

We have $\phi(n)=\phi(1+1+\cdots+1)=n\phi(1)=6n$.

To find $\ker \phi$, we must solve $6n \equiv 0 \bmod 8$.

This equation reduces to $3n \equiv 0 \bmod 4$, or $n \equiv 0 \bmod 4$.

Therefore, $\ker \phi = 4\mathbb Z$.

f(1)=6. f(4)=24=0 . 4 is the least in absolute value that maps to 0. 4Z maps to 0.


Is the zero vector not being a possible eigenvector just an arbitrary decision? I understand that if 0 is allowed to be an eigenvector, then it would be an eigenvector of every matrix (corresponding to some λ).

But why is that a bad thing?  Does allowing this cause any contradiction? Or this is really just some convention for convenience? In that case, what is this convenience?

Thanks.

[Made my comment an answer]

It's just not useful. Eigenvectors tell you something about the operator (many times, they tell you everything). The zero vector tells you absolutely nothing about it - not even an eigenvalue, since $L0=\lambda 0$ for any $\lambda$ and any $L$.

Because it would be an eigenvector corresponding to EVERY $\lambda$, so the notion of eigenvalue would be quite useless if every matrix could have any possible $\lambda$ as eigenvalue.


Evaluate $\log_{5} x^2(y-3)$ Given that $\log _{5} x=-2$ and $\log_{2} y=3$

Evaluate $\log_{5} x^2(y-3)$

My attempt,

$=\log{5} x^2+ \log_{5} (y-3)$

I'm stuck here. Can anyone explain how to proceed? Thanks in advance.

HINT:

$\log _{5} x=-2 \implies x=5^{-2}$ and $\log_{2} y=3 \implies y=8$

Can you evaluate it now?


Is a function useful that has zeros on the prime integers? So I want a function that is zero on the reals only on the prime integers and which doesn't depend on knowing the primes. I construct:

$$f(x) = e^{-x^2} - \sum\limits_{n=2}^\infty e^{-n^2} \frac{ \sin(\pi x)^2 }{ n^2\sin(\pi x/n)^2}$$

Which has zeros on the real line only on the positive and negative prime integers.

( $c_n(x)=\frac{\sin(\pi x)^2}{n^2\sin(\pi x/n)^2}$ has a well defined Taylor series and can be defined everywhere. It is $1$ when $x$ is a multiple of $n$ and 0 otherwise. The exponentials are just to make the whole thing converge.)

So my question is, is a function like this useful? As in, would it tell us anything about the primes?

Edit: if it is analytic, as well as zeros on the real axis it will have many complex zeros.

I also note that it is quite "easy" to convert this into a series of the form:
$f(x)=\sum\limits_{k=0}^\infty a_{k} x^{k}$  where the coefficients $a_{2k} =\frac{(-1)^k}{k!} - \sum\limits_{n=2}^\infty c^{(2k)}_n(0)e^{-n^2}$. It begins $f(x)=0.981-0.97722x^2+...$ although you would need a lot of terms when $x$ is big!!! But we could say $f^{-1}(0)\subset$primes.

Edit 2: I think the function actually also has zeros at points very close to the primes and only the zeros where $f'(x)<0$ are primes. (Basically every second zero.) This works equally well replacing the exponentials with $1/x^2$.

Yes: In case all zeroes have same order, in doubt consider f/f', one could derive a function with simple poles and residual 1 in primes; then a circular integral would deliver the number of enclosed primes.
Concerning convergence rate, just an idea: May be there is a representation with elliptic functions, then fast Jacobi elliptic functions could serve.

I know that in computer science, a lot of cryptography algorithms rely on prime numbers because no one has ever found such an analytic function.

But yes, a function that has say, three or four zeros which happen to land on the prime might be useful to someone.  I'll let others chime in on who might use it and why.


How do I find the error of nth iteration in Newton's Raphson's method without knowing the exact root In our calculus class, we were introduced to the numerical approximation of root by Newton Raphson method. The question was to calculate the root of a function up to nth decimal places.

Assuming that the function is nice and our initial value does lead to convergence. Our teacher terminated the algorithm when the two successive iterations had the same first n digits and told us that the the approximation was correct up to the nth digit!

I feel that the termination step is valid if $f(x_n)$ and $f(x_{n+1})$ has different signs but my teacher disagrees. How do I sort this out??

Futhermore how do I find the error in the  nth iteration without knowing the exact root?

Your teacher's example cannot work in general as I present a counter example below.
Nonetheless, I think that your teacher's approach is a reasonable way to explain the intuition behind what happens in a typical case, provided that the proper caveats are given.

I think a more reasonable stopping condition, for programming purposes, is to iterate until the value of $f$ is very small.  If the first derivative is relatively large in a neighborhood of the last iterate, this might be enough to prove that there is definitively a root nearby.  Of course, Christian Blatter has already provided sufficient conditions.

For a counter example, let's suppose that
$$f(x) = x(x-\pi)^2 + 10^{-12}.$$
Then, the Newton's method iteration function is
$$N(x) = x-f(x)/f'(x) =  x-\frac{x (x-\pi)^2+10^{-10}}{3 x^2-4\pi  x+\pi ^2}$$
and if we iterate $N$ 20 times starting from $x_0=3.0$, we get
$$
3., 3.07251, 3.10744, 3.12461, 3.13313, 3.13736, 3.13948, 3.14054, \
3.14106, 3.14133, 3.14146, 3.14153, 3.14156, 3.14158, 3.14158, \
3.14159, 3.14159, 3.14159, 3.14159, 3.14159, 3.14159
$$
Thus, your teacher's method implies there is a root at $x=3.14159$ when, of course, there is no root near here.  There is, however, a root near zero to which the process eventually converges after several thousand iterates.

To place this in a broader context, let's examine the basins of attraction for this polynomial in the complex plane.  There are three complex roots, one just to the left of zero and two at $\pi\pm\varepsilon i$ where $\varepsilon$ is a small positive number.  In the picture below, we shade each complex initial seed depending on which of these roots Newton's method ultimately converges.

Now, it is a theorem in complex dynamics that, whenever two of these basins meet, there are points of the third basin arbitrarily near by.  As a result, there is definitely a number whose decimal expansion starts with $3.14159$ that eventually converges to the root near zero under iteration of Newton's method.

By the quadratic convergence of the algorithm, the corrections decrease very quickly, so that the error is much smaller than the last correction. Usually, you can stop iterations when the latter falls below your tolerance.


Constraint length in a convolutional code Regarding convolutional error correcting code with $k=1$ (single input), I'm puzzled with different definitions for the constraint length.

In some books [4, 5], the constraint length is defined as the number of shift-registers.
But in [1, 2, 3], the constraint length is defined as the number of shift-registers plus one.

What about this difference in definition? What can I follow?

References:

[1] Page 231 of the book "Andrew J. Viterbi, Jim K. Omura - Principles of Digital Communication and Coding"

[2] https://www.mathworks.com/help/comm/ref/poly2trellis.html

[3] https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-02-introduction-to-eecs-ii-digital-communication-systems-fall-2012/readings/MIT6_02F12_chap07.pdf

[4] Page 459 of the book "Shu Lin, Daniel Costello - Error Control Coding - 2nd edition"

[5] Page 272 of the book "Richard E. Blahut - Algebraic codes for data transmission"

Each message bit influence a span of n(m+1) successive output bit the quantity n(m+1) is called as constraint length


Values of $\sum_{n=0}^\infty x^n$ and $\sum_{n=0}^N x^n$ Why does the following hold:

\begin{equation*}
\displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ?
\end{equation*}

Can we generalize the above to

  $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?

Are there some values of $x$ for which the above formula is invalid?

What about if we take only a finite number of terms? Is there a simpler formula?

  $\displaystyle \sum_{n=0}^{N} x^n$

Is there a name for such a sequence?

This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.

and here: List of abstract duplicates.

By definition, a "series" (an "infinite sum")
$$\sum_{n=k}^{\infty} a_n$$
is defined to be a limit, namely
$$\sum_{n=k}^{\infty} a_n= \lim_{N\to\infty} \sum_{n=k}^N a_n.$$
That is, the "infinite sum" is the limit of the "partial sums", if this limit exists. If the limit exists, equal to some number $S$, we say the series "converges" to the limit, and we write
$$\sum_{n=k}^{\infty} a_n = S.$$
If the limit does not exist, we say the series diverges and is not equal to any number.

So writing that
$$\sum_{n=0}^{\infty} 0.7^n = \frac{1}{1-0.7}$$
means that we are asserting that
$$\lim_{N\to\infty} \sum_{n=0}^N0.7^n = \frac{1}{1-0.7}.$$

So what your question is really asking is: why is this limit equal to $\frac{1}{1-0.7}$? (Or rather, that is the only way to make sense of the question).

In order to figure out the limit, it is useful (but not strictly necessary) to have a formula for the partial sums,
$$s_N = \sum_{n=0}^N 0.7^n.$$
This is where the formulas others have given come in. If you take the $N$th partial sum and multiply by $0.7$, you get
$$\begin{array}{rcrcrcrcrcrcl}
s_N &= 1 &+& (0.7) &+& (0.7)^2 &+& \cdots &+& (0.7)^N\\
(0.7)s_N &= &&(0.7) &+& (0.7)^2 &+&\cdots &+&(0.7)^N &+& (0.7)^{N+1}
\end{array}$$
so that
$$(1-0.7)s_N = s_N - (0.7)s_N = 1 - (0.7)^{N+1}.$$
Solving for $s_N$ gives
$$s_N = \frac{1 - (0.7)^{N+1}}{1-0.7}.$$
What is the limit as $N\to\infty$? The only part of the expression that depends on $N$ is $(0.7)^{N+1}$. Since $|0.7|\lt 1$, then $\lim\limits_{N\to\infty}(0.7)^{N+1} = 0$. So,
$$\lim_{N\to\infty}s_N = \lim_{N\to\infty}\left(\frac{1-(0.7)^{N+1}}{1-0.7}\right) = \frac{\lim\limits_{N\to\infty}1 - \lim\limits_{N\to\infty}(0.7)^{N+1}}{\lim\limits_{N\to\infty}1 - \lim\limits_{N\to\infty}0.7} = \frac{1 - 0}{1-0.7} = \frac{1}{1-0.7}.$$
Since the limit exists, then we write
$$\sum_{n=0}^{\infty}(0.7)^n = \frac{1}{1-0.7}.$$

More generally, a sum of the form
$$a + ar + ar^2 + ar^3 + \cdots + ar^k$$
with $a$ and $r$ constant is said to be a "geometric series" with initial term $a$ and common ratio $r$. If $a=0$, then the sum is equal to $0$. If $r=1$, then the sum is equal to $(k+1)a$. If $r\neq 1$, then we can proceed as above. Letting
$$S = a +ar + \cdots + ar^k$$
we have that
$$S - rS = (a+ar+\cdots+ar^k) - (ar+ar^2+\cdots+a^{k+1}) = a - ar^{k+1}$$
so that
$$(1-r)S = a(1 - r^{k+1}).$$
Dividing through by $1-r$ (which is not zero since $r\neq 1$), we get
$$S = \frac{a(1-r^{k+1})}{1-r}.$$

A series of the form
$$
\sum_{n=0}^{\infty}ar^{n}
$$
with $a$ and $r$ constants is called an infinite geometric series.
If $r=1$, then
$$
\lim_{N\to\infty}\sum_{n=0}^{N}ar^{n}
= \lim_{N\to\infty}\sum_{n=0}^{N}a
= \lim_{N\to\infty}(N+1)a
= \infty,
$$
so the series diverges. If $r\neq 1$, then using the formula above we have:
$$
\sum_{n=0}^{\infty}ar^n = \lim_{N\to\infty}\sum_{n=0}^{N}ar^{N} = \lim_{N\to\infty}\frac{a(1-r^{N+1})}{1-r}.
$$
The limit exists if and only if $\lim\limits_{N\to\infty}r^{N+1}$ exists. Since
$$
\lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll}
0 &\mbox{if $|r|\lt 1$;}\\
1 & \mbox{if $r=1$;}\\
\text{does not exist} &\mbox{if $r=-1$ or $|r|\gt 1$}
\end{array}\right.
$$
it follows that:
$$
\begin{align*}
\sum_{n=0}^{\infty}ar^{n} &=\left\{\begin{array}{ll}
0 &\mbox{if $a=0$;}\\
\text{diverges}&\mbox{if $a\neq 0$ and $r=1$;}\\
\lim\limits_{N\to\infty}\frac{a(1-r^{N+1})}{1-r} &\mbox{if $r\neq 1$;}\end{array}\right.\\
&= \left\{\begin{array}{ll}
\text{diverges}&\mbox{if $a\neq 0$ and $r=1$;}\\
\text{diverges}&\mbox{if $a\neq 0$, and $r=-1$ or $|r|\gt 1$;}\\
\frac{a(1-0)}{1-r}&\mbox{if $|r|\lt 1$;}
\end{array}\right.\\
&=\left\{\begin{array}{ll}
\text{diverges}&\mbox{if $a\neq 0$ and $|r|\geq 1$;}\\
\frac{a}{1-r}&\mbox{if $|r|\lt 1$.}
\end{array}\right.
\end{align*}
$$

Your particular example has $a=1$ and $r=0.7$.

Since this recently came up (09/29/2011), let's provide a formal proof that
$$
\lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll}
0 &\mbox{if $|r|\lt 1$;}\\
1 & \mbox{if $r=1$;}\\
\text{does not exist} &\mbox{if $r=-1$ or $|r|\gt 1$}
\end{array}\right.
$$

If $r\gt 1$, then write $r=1+k$, with $k\gt0$. By the binomial theorem, $r^n = (1+k)^n \gt 1+nk$, so it suffices to show that for every real number $M$ there exists $n\in\mathbb{N}$ such that $nk\gt M$. This is equivalent to asking for a natural number $n$ such that $n\gt \frac{M}{k}$, and this holds by the Archimedean property; hence if $r\gt 1$, then $\lim\limits_{n\to\infty}r^n$ does not exist. From this it follows that if $r\lt -1$ then the limit also does not exist: given any $M$, there exists $n$ such that $r^{2n}\gt M$ and $r^{2n+1}\lt M$, so $\lim\limits_{n\to\infty}r^n$ does not exist if $r\lt -1$.

If $r=-1$, then for every real number $L$ either $|L-1|\gt \frac{1}{2}$ or $|L+1|\gt \frac{1}{2}$. Thus, for every $L$ and for every $M$ there exists $n\gt M$ such that $|L-r^n|\gt \frac{1}{2}$ proving the limit cannot equal $L$; thus, the limit does not exist. If $r=1$, then $r^n=1$ for all $n$, so for every $\epsilon\gt 0$ we can take $N=1$, and for all $n\geq N$ we have $|r^n-1|\lt\epsilon$, hence $\lim\limits_{N\to\infty}1^n = 1$. Similarly, if $r=0$, then $\lim\limits_{n\to\infty}r^n = 0$ by taking $N=1$ for any $\epsilon\gt 0$.

Next, assume that $0\lt r\lt 1$. Then the sequence $\{r^n\}_{n=1}^{\infty}$ is strictly decreasing and bounded below by $0$: we have $0\lt r \lt 1$, so multiplying by $r\gt 0$ we get $0\lt r^2 \lt r$. Assuming $0\lt r^{k+1}\lt r^k$, multiplying through by $r$ we get $0\lt r^{k+2}\lt r^{k+1}$, so by induction we have that $0\lt r^{n+1}\lt r^n$ for every $n$.

Since the sequence is bounded below, let $\rho\geq 0$ be the infimum of $\{r^n\}_{n=1}^{\infty}$. Then $\lim\limits_{n\to\infty}r^n =\rho$: indeed, let $\epsilon\gt 0$. By the definition of infimum, there exists $N$ such that $\rho\leq r^N\lt \rho+\epsilon$; hence for all $n\geq N$,
$$|\rho-r^n| = r^n-\rho \leq r^N-\rho \lt\epsilon.$$
Hence $\lim\limits_{n\to\infty}r^n = \rho$.

In particular, $\lim\limits_{n\to\infty}r^{2n} = \rho$, since $\{r^{2n}\}_{n=1}^{\infty}$ is a subsequence of the converging sequence $\{r^n\}_{n=1}^{\infty}$. On the other hand, I claim that $\lim\limits_{n\to\infty}r^{2n} = \rho^2$: indeed, let $\epsilon\gt 0$. Then there exists $N$ such that for all $n\geq N$, $r^n - \rho\lt\epsilon$. Moreover, we can assume that $\epsilon$ is small enough so that $\rho+\epsilon\lt 1$. Then
$$|r^{2n}-\rho^2| = |r^n-\rho||r^n+\rho| = (r^n-\rho)(r^n+\rho)\lt (r^n-\rho)(\rho+\epsilon) \lt r^n-\rho\lt\epsilon.$$
Thus, $\lim\limits_{n\to\infty}r^{2n} = \rho^2$. Since a sequence can have only one limit, and the sequence of $r^{2n}$ converges to both $\rho$ and $\rho^2$, then $\rho=\rho^2$. Hence $\rho=0$ or $\rho=1$. But $\rho=\mathrm{inf}\{r^n\mid n\in\mathbb{N}\} \leq r \lt 1$. Hence $\rho=0$.

Thus, if $0\lt r\lt 1$, then $\lim\limits_{n\to\infty}r^n = 0$.

Finally, if $-1\lt r\lt 0$, then $0\lt |r|\lt 1$.  Let $\epsilon\gt 0$. Then there exists $N$ such that for all $n\geq N$ we have $|r^n| = ||r|^n|\lt\epsilon$, since $\lim\limits_{n\to\infty}|r|^n = 0$. Thus, for all $\epsilon\gt 0$ there exists $N$ such that for all $n\geq N$, $| r^n-0|\lt\epsilon$. This proves that $\lim\limits_{n\to\infty}r^n = 0$, as desired.

In summary,
$$\lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll}
0 &\mbox{if $|r|\lt 1$;}\\
1 & \mbox{if $r=1$;}\\
\text{does not exist} &\mbox{if $r=-1$ or $|r|\gt 1$}
\end{array}\right.$$

The argument suggested by Srivatsan Narayanan in the comments to deal with the case $0\lt|r|\lt 1$ is less clumsy than mine above: there exists $a\gt 0$ such that $|r|=\frac{1}{1+a}$. Then we can use the binomial theorem as above to get that
$$|r^n| = |r|^n = \frac{1}{(1+a)^n} \leq \frac{1}{1+na} \lt \frac{1}{na}.$$
By the Archimedean Property, for every $\epsilon\gt 0$ there exists $N\in\mathbb{N}$ such that $Na\gt \frac{1}{\epsilon}$, and hence for all $n\geq N$, $\frac{1}{na}\leq \frac{1}{Na} \lt\epsilon$. This proves that $\lim\limits_{n\to\infty}|r|^n = 0$ when $0\lt|r|\lt 1$, without having to invoke the infimum property explicitly.

Using the fact that $$x^n - y^n=(x-y)\sum_{k=1}^nx^{k-1}y^{n-k}$$
Set $x=1$ and $y \in \mathbb{N}$ to get the desired result.
$$1 - y^n=(1-y)\sum_{k=1}^ny^{n-k} \iff \sum_{k=1}^ny^{n-k}=y^n\sum_{k=1}^ny^{-k}=\frac{y^n-1}{y-1} \iff \sum_{k=1}^n \frac{1}{y^k} = \frac{y^n-1}{y^{n+1}-y^n}$$
Now, using the limit of the result we have $$\large{\lim_{n\to \infty}\frac{y^n-1}{y^{n+1}-y^n} = \frac{1}{y-1}}$$.
Some examples,
$$\sum_{k=1}^\infty \left(\frac{1}{2}\right)^k = 1$$
$$\sum_{k=1}^\infty \left(\frac{1}{3}\right)^k = 1/2$$
$$\sum_{k=1}^\infty \left(\frac{1}{4}\right)^k = 1/3$$
$$\sum_{k=1}^\infty \left(\frac{1}{5}\right)^k = 1/4$$
Also, if $y \to \infty \implies $
$$\sum_{k=1}^\infty \left(\frac{1}{y}\right)^k = 0$$


Problem 8 (a) in Exercises after Sec. 18 in Munkres' Topology, 2nd ed.: How to show this set is closed? Let $X$ be an arbitrary topological space, let $Y$ be an ordered set in the order topology, and let the maps $f, g \colon X \to Y$ be continuous. Then how to show that the set $S$ given by
$$S \colon= \{ \ x \in X \ \colon f(x) \leq g(x) \ \}$$
is closed in $X$?

I'll try to give an elaborate hint: It's very similar to the following proof

Let $h:X \rightarrow \mathbb{R}$ be defined as $h(x)=g(x)-f(x)$

$h$ is continuous, so $S=h^{-1}([0,\infty))$ is closed since $[0,\infty)$ is closed in $\mathbb{R}$.


Does $\lim \frac{a_n}{e^{\delta n}}=0$ for every positive $\delta$ imply that $\lim \frac{a_n}{\sum\limits_{k=1}^n a_k} =0$ Let's say we have an increasing sequence $a_n$ such that $\underset{n\rightarrow\infty}{\lim} a_n=\infty$. Now it's fairly clear to me, though I haven't proven this yet, that:
$$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=K>0
\Rightarrow \exists\ \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=1$$
In fact, I think $\delta=1/K$ but I could have that mixed up. And as a corrollary:
$$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=0
\Rightarrow \forall\  \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=0$$

But here's my question: is the converse also true? Specifically is it the case that:
$$
\forall\ \delta>0, \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=0
\Rightarrow \underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=0
$$

I think that this is true but when I try to show it, the problem comes up that I can show that $a_n$ is dominated by the same sequence that dominates $\sum\limits_{k=1}^n a_k$, but I can't show that $\sum\limits_{k=1}^n a_k$ is large relative to $a_n$. In my mind though, I know that if the RHS is not true and this limit is positive, then at some point the sequence is proportional to its own rate of change, and that means it should be asymptotically equivalent to an exponential function. But the details of the proof are killing me.

Any help would be greatly appreciated :-)

Here is a counterexample to your first conjecture, stating that for every unbounded increasing sequence $(a_n)$,
$$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=K>0
\Rightarrow \exists\ \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=1$$
Take $a_n=n2^n$. Then $\{a_n\}$ is increasing and $a_n\to\infty$. Then
$$
a_1+\cdots+a_n=(2-1)(a_1+\cdots+a_n)=2^2+2\cdot 2^3+\cdots+n2^{n+1}-2^1-2\cdot 2^2-\cdots-n2^{n}=n2^{n+1}-2-(2^2+2^3+\cdots+2^n)=n2^{n+1}-2-2^{n+1}+2^2=(n-1)2^{n+1}+2.
$$
Also
$$
\frac{a_n}{a_1+\cdots+a_n}=\frac{n2^n}{(n-1)2^{n+1}+2}\to \frac{1}{2}
$$
But, there is no $\delta>0$, such that
$$
\frac{a_n}{\mathrm{e}^{n\delta}}=\frac{n2^n}{\mathrm{e}^{n\delta}}=
\exp(\log n+n(\log 2-\delta))
$$
converges to $1$, as $\log n+n(\log 2-\delta)\to \infty$ or $\log n+n(\log 2-\delta)\to -\infty$,
if $\delta\ge\log 2$ or $\delta<\log 2$, respectively.

Not true.

Take $a_n = n * (\frac{e}{2})^{n}$. It's easy to verify that $\lim \frac{a_n}{e^n} = 0$

On the other hand, $\frac{a_n}{\sum a_n} = \frac{1}{\sum a_i/a_n}$, where i runs from 1 to n

Now consider the denominator:

$\sum \frac{a_i}{a_n} = \sum \frac{i}{n}*(\frac{e}{2})^{i-n} $ < $\sum (\frac{e}{2})^{i-n}$

Let $k = \frac{e}{2}$. Then, the RHS can be written as: $\frac{1}{k^n}*\sum k^i$ = $\frac{k-1/k^n}{k-1} < \frac{k}{k-1} = \frac{e}{e-2}$

Thus, $\frac{a_n}{\sum a_n} = \frac{1}{\sum a_i/a_n} > \frac{e-2}{e} = constant > 0 $.

Then it can't go to $0$.


$\pi = 0$! It can't be! Keep in mind: I'm still in high-school so forgive my poor maths. Also remember that I'm in HIGH-SCHOOL so nothing to complex
I like to mess around with equations and I find it quite fascinating the results I can somehow come up with. Recently, I found Euler's Identity and started to mess around with this. Somehow I can up with the following result: $π = 0.$ I thought to myself, as anyone would, how is this possible?
I then decided to show my proof to my maths teacher and my parents and neither one could show me where I went wrong so I was hoping someone here could.
This is my proof:
$$e^{iπ} + 1 = 0$$
$$e^{iπ} = -1$$
$$e^{2iπ} = 1$$
$$2iπ = \ln(1) = 0$$
$$4\times-1\timesπ = 0$$
$$-4π^2 = 0$$
$$0 = 4π^2$$
$$0 = 2π$$
$$0 = π$$
Edit
As I pointed out above, my maths knowledge isn't that advanced, compared to people on this site. Therefore the answer(s) I accept and/or upvote will be based on understandability as well as how well they answer the question.

If $x$ and $y$ are real, then $e^x=e^y$ implies $x=y$.

But if $x$ and $y$ are complex, $e^x=e^y$ only implies that
$$\frac{x-y}{2\pi i}\in\Bbb Z$$

Therefore, from $e^{2\pi i}=e^0$ we can't deduce that $2\pi i=0$.

The problem is simple. Note that if $0=\log(1)=\log(e^{2\pi I})=2\pi i$ than the logarithm is no mapping. Canonically one give a solution to this type of equations by restricting to the main branch of the complex logarithm.


Why do mathematicians sometimes assume famous conjectures in their research? I will use a specific example, but I mean in general. I went to a number theory conference and I saw one thing that surprised me: Nearly half the talks began with "Assuming the generalized Riemann Hypothesis..." Almost always, the crux of their argument depended on this conjecture.

Why would mathematicians perform research assuming a conjecture? By definition, it is not known to be true yet. In the off-chance that it turns out to be false, wouldn't all of the papers that assumed the conjecture be invalidated? I may be answering my own question, but I speculate that:

There is such strong evidence in support of the particular conjecture (Riemann Hypothesis in particular) and lack of evidence against it, that it is "safe" to assume it.
It's not so much about result obtained, but the methods and techniques used to prove it. Perhaps by assuming the conjecture, in the case of the Riemann Hypothesis, it leads to development of new techniques in analytic number theory.

For four main reasons:

If the famous conjecture is proven true, the demonstrated results are proven true too.
If the reasoning is correct but demonstrated results are proven false, the famous conjecture is proven false too.
Others may be able to prove further results based on the demonstrated results which may themselves be proven false, thus again proving the famous conjecture false.
Showing interesting consequences of the famous conjecture and surprising connections into other areas of research generates interest in proving that the famous conjecture is true.

You can't expect a mathematician to find proof of a conjecture directly. Assuming that a conjecture is true is a creative approach to understanding the conjecture. There is no pretty systematic way to solve a problem.


Why should $u_1-v_1\in U_1,\ldots,u_m-v_m\in U_m$ imply that each $u_j-v_j=0$? Linear Algebra Done Right -Sheldon Axler, Third Edition, Section 1.C, Page 23.

Condition for a direct sum:

  Suppose $U_1,U_2,\ldots,U_m$ are subspaces of $V$. Then $U_1+\cdots+U_m$ is
  a direct sum if and only if the only way to write $0$ as a sum
  $u_1+\cdots+u_m$, where each $u_j$ is in $U_j$, is by taking each
  $u_j=0$.

I don't seem to understand their proof.

They first write $v=u_1+\cdots+u_m$ (where $u_1\in U_1,u_2\in U_2,\ldots,u_m\in U_m$)
and also suppose that $v=v_1+\cdots+v_m$ (where $v_1\in U_1,\ldots,v_m\in U_m$).

Next, they subtract the two equations and write:

$$0=(u_1-v_1)+\cdots+(u_m-v_m)$$

"Because $u_1-v_1\in U_1,\ldots,u_m-v_m\in U_m$, the equation above implies that each $u_j-v_j=0$. Thus $u_1=v_1,\ldots,u_m=v_m$." I didn't understand how they deduced this. Why should $u_1-v_1\in U_1,\ldots,u_m-v_m\in U_m$ imply that each $u_j-v_j=0$ ?

They are proving that $U_1 + \ldots + U_m$ is a direct sum if the only way to write $0 = z_1 + \ldots + z_m$ with $z_i \in U_i$ is when all $z_i = 0$. To prove that, they assume that the only way to write $0 = z_1 + \ldots + z_m$ is when all $z_i = 0$, and then they prove that $U_1 + \ldots + U_m$ is indeed a direct sum.

To prove that something is a direct sum, one needs to prove that each vector $v \in U_1 + \ldots + U_m$ has a unique representation of the form $ v = u_1 + \ldots + u_m$ with $u_i \in U_i$. So they assume that they have some vector $v$ has two representations $v = u_1 + \ldots + u_m = v_1 + \ldots + v_m$, and they show that these representations are actually equal, that is, there's no way to have two different representations, so that the representation must be unique.

The way they prove is to note that if $v = u_1 + \ldots + u_m = v_1 + \ldots + v_m$, then $0 = v - v = (u_1 + \ldots + u_m) - (v_1 + \ldots + v_m) = (u_1 - v_1) + ... + (u_m - v_m)$. But since $u_i, v_i \in U_i$, we also have $u_i - v_i \in U_i$. But now, the assumption is that the only way to write $0 = z_1 + \ldots + z_m$ is when all $z_i = 0$, thus $u_i - v_i = 0$ for all $i$.

The sum $U_1 + \cdots + U_m$ is direct if $$U_j \cap (U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m) = \{0\}$$ for all $j \in \{1, \ldots, m\}$.

Assume that the intersections are trivial.
If $$0 = u_1 + \cdots + u_m$$

is a representation of $0$ with $u_i \in U_i$ then $$-u_j = u_1 + \cdots + u_{j-1} + u_{j+1} + u_n$$

The left side is in $U_j$ and the right side is in $U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m$. Thus, $$-u_j \in U_j \cap (U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m) = \{0\}$$

so $u_j = 0$. Since this is valid for all $j \in \{1, \ldots, m\}$, we conclude that $0$ has only the trivial representation $$0 = 0 + \cdots + 0$$

Conversely, assume that $0$ has only the trivial representation.

Take $x \in U_j \cap (U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m)$ for some $j \in \{1, \ldots, m\}$.

Since $x \in U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m$ there exists a decomposition

$$x = \underbrace{v_1}_{\in U_1} + \cdots + \underbrace{v_{j-1}}_{\in U_{j-1}} +  \underbrace{v_{j+1}}_{\in U_{j+1}} + \cdots + \underbrace{v_m}_{\in U_m}$$

We have

$$0 = x - x = \underbrace{v_1}_{\in U_1} + \cdots + \underbrace{v_{j-1}}_{\in U_{j-1}} + \underbrace{-x}_{\in U_j} + \underbrace{v_{j+1}}_{\in U_{j+1}} + \cdots + \underbrace{v_m}_{\in U_m}$$

which is a representation of zero. Hence, it must be trivial so $x = 0$. We conclude that all intersections

$$U_j \cap (U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m)$$

for $j \in \{1, \ldots, m\}$ are trivial.


Does edge weight of a undirected weighted graph must be a metric? In a undirected weighted graph, each edge is assigned with a weight $w: E\rightarrow R$. Does $w(\cdot)$ must be a metric ? That is, $w$ satisfies the following three conditions

(1) $w(x, y) > 0 $ for any edges pair $x$ and $y$;

(2) $w(x, y) = w(y, x)$;

(3) $w(x, y) + w(y, z) \geq w(x, z)$.

No, it doesn't have to be a metric.

I haven't found a complete and fully explicit definition of a weighted graph anywhere, but I've found several informal descriptions of weighted graphs, and none of them say that the weight function has to be a metric.

In particular, none of the descriptions I've found say that the weight function has to satisfy the triangle inequality $w(x, y) + w(y, z) \geq w(x, z)$. For example, these lecture notes by William D McQuain of Virginia Tech show a weighted graph which includes three vertices $b$, $h$ and $i$, such that $w(b,h) = 5$ and $w(h,i) = 15$, but $w(b,i) = 25$.

It's obvious that $w$ cannot be a metric in general, because $w(x,y)$ is undefined if $(x,y)$ is not an edge. But:

Suppose $(V,E)$ is a connected undirected graph and $w(x,y)>0$ for every $(x,y)\in E$ with $x\ne y$; if edges of the form $(x,x)$ are allowed, assume $w(x,x)=0$ whenever $(x,x)\in E$. For $x,y\in V$ define $d(x,y)=0$ if $x=y$ and otherwise say $d(x,y)$ is the infimum of $\sum w(x_j,x_{j-1})$ over all paths $x=x_0,\dots, x_n=y$ (such that each $(x_j,x_{j-1})$ is an edge). Then it's easy to show $d$ is a metric. (Here I'm assuming that "undirected" means that if $(x,y)\in E$ then $(y,x)\in E$ and $w(y,x)=w(x,y)$.)


The monodromy representation $\pi_1(E)\to Mod(F)$ Given a fiber bundle $F\to X\to E$, with $F,E$ are Riemann surfaces. I know the monodromy gives a permutation of the fiber. But how to see we have the monodromy representatio: $\pi_1(E)\to Mod(F)$? Here Mod($F$) is the mapping class group of the fiber.

Update: Now I think by locally trivialization, we can get the monodromy gives a homeomorphism from the fiber to itself. But how do we know it preserves the orientation?

If you define the mapping class group as isotopy classes of orientation preserving homeomorphisms, then the bundle needs to be oriented.


Using the Limit Comparison Test on $\sum_{n=1}^{\infty} \frac{n^2} {n!}$ is this right ?
$$
\sum_{n=1}^{\infty} \frac{n^2} {n!}
$$
i need to use quotient criterion

$$
\lim_{n\to\infty} \frac{\frac{n^2}{n!}}{ \frac{1}{n!}} =    \lim_{n\to\infty} {\frac{n^2}{n!}} { \frac{n!}{1}} = \lim_{n\to\infty} {n^2} = \infty
$$

so $$ \lim_{n\to\infty} \frac{\frac{n^2}{n!}}{ \frac{1}{n!}} $$ equals $\infty$ and $$\sum_{n=1}^{\infty} \frac{1} {n!} $$ is divergent  it means that
$$ \sum_{n=1}^{\infty} \frac{n^2} {n!}$$ is divergent

Can you show that $\dfrac{n^2}{n!}<\dfrac{1}{n^2}$ for sufficiently large $n$?

Then use the monotonic property of the sum from that $n$ onwards. The other $n$'s less than that particular $n$ contribute only to a finite sum.

EDIT: Since the question used the words "Limit Comparison", consider comparing with $\dfrac{1}{(n-2)!}$.

Notice that $$\lim_{n\to\infty}\dfrac{\dfrac{n^2}{n!}}{\dfrac{1}{(n-2)!}}=\lim_{n\to\infty}\dfrac{n}{n-1}=1$$

Note that $\dfrac{1}{n!}<\dfrac{1}{n^2}$ for all $n>3$. Thus $\sum_{n=3}^\infty\dfrac{1}{(n-2)!}$ converges. Hence $\sum_{n=3}^\infty \dfrac{n^2}{n!}$ converges, and hence $\sum_{n=1}^\infty\dfrac{n^2}{n!}$ converges.

This doesn't work!

In the Limit Comparison Test, you need
$$
\lim_{n\to \infty} \frac{a_n}{b_n}
$$
to be a positive real number (and not infinity) when comparing $\sum a_n$ and $\sum b_n$.

Consider the example where
$$
a_n = \frac{1}{n} \quad\text{and}\quad b_n = \frac{1}{n^2}.
$$
Here $\sum a_n$ is divergent and $\sum b_n$ is convergent, but
$$
\lim_{n\to \infty} \frac{a_n}{b_n} = \infty.
$$

I don't know how you would use the Limit Comparison Test here. I see now that Landon does know. So you should see his answer!

As a sidenote, note that
$$
\sum_{n=1}^{\infty} \frac{1}{n!}
$$
is convergent and not divergent.


Evaluating $\int \frac{5-e^{x}}{e^{2x}}dx$ without Partial Fractions I am trying to evaluate

  $$\int \frac{5-e^{x}}{e^{2x}} \mathrm dx$$

I tried rewriting the integral by throwing $e^{2x}$ up on top and using $$u=e^{x}$$ $$du = e^{x} dx$$

I then tried another substitution where $v = 5-u$
and $dv = -1 du$ but then I can only simplify the integral to

$$\int \frac{v}{(v-5)^{3}} \mathrm dv$$

Which would then require partial fractions, which my class has not gotten to quite yet (so I'm not allowed to use the method for homework, sadly).

Is there a simple substitution I am overlooking from the beginning or something?

Thanks.

Rewrite the integrand $5e^{-2x}+e^{-x}$. You know how to integrate $ae^{bx}$ right?

Did you try integrating by parts where $dv = \frac{3}{e^{3x}}$ and $u = 5 - (e^x)$ ? It may be a bit laborious of a calculation,but that's what I'd try if I was instructed NOT to use partial fractions. This is where knowing as many techniques of integration as possible comes in very handy in baby calculus.


Methods to solve $x^2+(x')^2=1$ and similar nonlinear differential equations How do I solve the following:

$x^2 + \left(\frac{dx}{dt}\right)^2 = 1$

Please refer me to an online resource that has worked examples that I can work through to practice.

I have the answer : $ x = \cos{t} $

I need to understand how to get to the answer.

$x^2 + (\frac{dx}{dt})^2 = 1 \Leftrightarrow \frac{dx}{dt} = \sqrt{1-x^2} $ or $\frac{dx}{dt} = -\sqrt{1-x^2} $

Now you have :

$\int \frac{dx}{\sqrt{1-x^2}} = \int dt$ or $ \int -\frac{dx}{\sqrt{1-x^2}} = \int dt $

Which yield you the solutions : $\arcsin(x) = t+c$ or  $-\arcsin(x) = t+c$

Which means : $x(t) = -\sin(t+c_1)$ or $x(t) = \sin(t+c_1)$

Now, depending on your initial conditions, you will be leaded to the answer you mentioned : $x(t) = \cos t$


Why must n be even? Skew Symmetric matrices Let B be any real, nonsingular $nxn$ matrix, where n is even, and set $A=B-B^T$. Show that A does not admit an LU decomposition without pivoting.

I know that A is a skew symmetric matrix. The 2x2 case is

0 -a

a 0

It is easy to see we can't eliminate a so we must pivot. What I don't understand is why n must be even? I think even odd cases show this... Thanks.

I don't think any invertible skew-symmetric real matrix admits an LU decomposition without pivoting.

From $A^T=-A$, if $A=LU$ we get $$LU=-(LU)^T=-U^TL^T.$$ We are assuming that $A$ is invertible, so $L$ and $U$ are also invertible. Thus $$(U^T)^{-1}L=-L^TU^{-1}.$$ Here the left-hand-side is lower triangular, and the right-hand-side is upper triangular; this implies that both are diagonal. So $(U^T)^{-1}L=D$ for an invertible diagonal matrix $D$. We can write $L=U^TD$. We also have $-L^TU^{-1}=D$, and we obtain
$$
L^T=-DU,
$$
which we can write as $L=-U^TD.$ It follows that $U^TD=-U^TD$, which would imply that $U^TD=0$, a contradiction since $U$ and $D$ are invertible.

If $n$ is odd then the characteristic polynomial for $A$ emits at least one real root. And zero happens to be a root which means $A$ is singular.

As an exercise, try to see if you can prove that $A$ is singular when $n$ is odd.

Hint: Consider $\det(A) = \det(A^T) = \det(-A)$.


Mean, mode and median equal for a random variable Suppose for a random variable mean, mode and median are equal. What is the intuitive meaning of this ?

Notice that the mean is the average of the values the random variable can take, while median is just the middlemost value. Thus if the mean is larger than the median then the distribution is heavy-tailed, that is it assigns higher probability mass to large enough values, and so the the distribution must be positively skewed. Similarly If the mean is less than the median then the distribution is negatively skewed intuitively (although there may may be exceptions).

If mean=median, then obviously the distribution is symmetric. Moreover if mean=mode=median then the distribution is both symmetric and unimodal.


what can we say about $f(x)= |\cos x| + |\sin x|$ If function $f:\mathbb R \to  [0,2]$ is defined by $f(x)= |\cos x| + |\sin x|$ then

$f$ is one one
$f$ is onto
$f$ is differentiable on $\mathbb R$
The minimum value of $f$ is $1$.

I countered option 3 by fact that $f (x)$ is not differentiable at $x=\pi$ so it is not differentiable on $\mathbb R$.

I countered option 4 by inequality that $|\cos x|+|\sin x|\ge|\cos x +\sin x | \ge \sqrt{2}$ so min value is $\sqrt{2}$

I have doubt in deciding option 1 and 2. And also what is correct option
thanks. Btw book says answer is option 4

Since $\cos$ and $\sin$ are periodic with the same period, so is $f$, so $f$ being one to one is out of the question. It should be easy to see that $1$ is false.
The second point is false because $f(x)$ cannot reach $0$ at any value of $x$.
Your answer for 3 is correct.
Your answer for $4$ is incorrect. Your inequality, $\cos x + \sin x \geq \sqrt 2$, is not true at all. For example, $\cos 0 + \sin 0 = 1 < \sqrt 2.$

A function $f:a \to B$ is 1-1 iff $\forall a,b \in A$, $f(a)=f(b) \implies a=b$

But your function is periodic. You can easily provide a counter example by picking some value and then 'moving up' the number line by a full period and showing that $f(a)=f(b)$ but $a \not= b$

A function $f:A \to B$ is onto means that: $ \forall b \in B$ $ \exists  a \in A$ such that $f(a)=b$

There does not exist any $x \in \mathbb{R}$ such that $f(x)=o$. This proves 2 is false.

you're correct
consider f(0): |cos(0)| + |sin(0)| = |1| + |0| = 1 < $\sqrt2$


Scientific notation and negative numbers My daughter is learning scientific notation in school, and her textbook says something to the effect of this:

Scientific notation is a method of writing numbers as the product of
two factors where the first factor is a number greater than or equal
to 1 but less than $10$ and the second factor is a power of $10$.

The teacher is taking this to mean that you cannot express a negative number in scientific notation. So that e.g.
$$-4 \times 10^{50}$$
would not be valid scientific notation because $-4$ is less than $1$.
Is there such a view of scientific notation? It certainly doesn't jive with my memory (or wikipedia), or is that description just deficient, and should better read:

Scientific notation is a method of writing numbers as the product of
two factors where the first factor is a number whose absolute value is greater than or equal
to 1 but less than 10 and the second factor is a power of $10$.

And if it is a legitimate view, how do you express negative numbers in scientific notation?

You are right. The textbook and teacher are wrong.

Scientific notation is where numbers are written in the form $a × 10^b$ where $a ∈ ℝ$ and $b ∈ ℤ$.

Normalized scientific notation also stipulates that $1 ≤ |a| < 10$.

$\therefore \;\; -4 × 10^{50}$ is correct normalized scientific notation, as common sense would dictate.

You would write the problem as: $|-4| \times 10^{50}$, or $4\times 10^{50}$. Your teacher is right, you cannot express a scientific notation as a negative number. It has to be a number greater than or equal to one, but less than 10. The absolute value would be taken, therefore, giving you $4 \times 10^{50}$. I just did one like this in college Chemistry. Thanks! MB


Abstract Algebra topic: Equivalence relations If R1 is reflective and not transitive, R2 is transitive but not symmetric and R3 is symmetric but not reflexive. We need to find an example of a set S and the three relations R1 R2 R3.

$R_1$ use the $\delta_{1}$ relation "made" reflective ($=1 \vee =$)
$R_2$ use the implication relation ($\Rightarrow$) in any fashion;
$R_3$ use the inequality relation ($\neq$) in any fashion.
$S$ could be $\{1,2,3\}$ for example.

For the specific $S$ the relations could be
$$\begin{align*}R_1 = & \{(1,1), (1,2), (1,3), (2,1), (3,1) , (2,2), (3,3)\} & = \{1\} \times S \cup S \times \{1\} \cup {\rm id}\\
R_2 = & \{(1,2), (2,3), (1,3)\} \\
R_3 = & \{(1,2), (1,3), (2,3), (2,1), (3,1), (3,2)\}
\end{align*}$$

Suppose $S$ is the set of all integers and $n$ be some fixed integer. Take the relation on $S$, that $x$ is related to $y$ if $y$ when divided by $n$ leaves the same remainder as $x$ leaves when divided by $n$. i.e.
$$x \sim y :\Leftrightarrow x \equiv y \quad ({\rm Mod}\ n)$$


Proof by induction with a prime number I'm having trouble wrapping my head around proofs by induction with this problem:

Prove by careful induction on $k$ that, if $p$ is a prime number dividing the product $m_1m_2...m_k$ of (ordinary) integers $m_1,...,m_k$ where $k\geqslant1$, then $p$ divides at least one of the integers $m_i, i =1,...,k.$

Trying again based on feedback

So I start with the base case, that is:

$k = 1$

If $p$ divides  $m_1m_2...m_k$, that gives us $1/p$, meaning $p|1$ which supports the hypothesis.

So the base case holds.

Now for the Inductive step, we have $k +1$ which would equate to $2$

Then we have $m_1m_2...m_k = 1*2$ and then $1*2/p$.

Since $p∣ab \implies p∣a$ or $p∣b$, we know that $p|1$ or $p|2$.

So the inductive step holds, and so does the hypothesis.

Base Case:  m$_1$...m$_k$ is m$_1$.  The hypothesis implies that p thus divides m$_1$.  Thus, p divides at least one of the integers of the product m$_1$.

Recursive Step:  Suppose that for at least one m$_k$, p divides m$_1$...m$_n$.  By Bill Dubuque's hint, p divides m$_1$...m$_n$m$_j$ where j = (n + 1) implies that p divides m$_1$...m$_n$ or m$_j$.  If it divides m$_1$...m$_n$, then by assumption for at least one m$_k$ it divides m$_1$...m$_n$.  If it divides m$_j$, then for at least one m$_k$ it divides m$_j$.  Thus, in either case for at least one m$_k$, p divides m$_1$...m$_n$m$_j$.  Therefore, if for at least one m$_k$, p divides m$_1$...m$_n$, then for at least one m$_k$, p divides m$_1$...m$_n$m$_j$.


Mass and center of mass using double integrals Disclaimer: This was given as a homework from college but the teacher didn't teach us anything about density or mass or anything related.

A lamina has the form of the region limited by the parabola $ y = x^2 $ and the straight line $ y = x $. The density varies as the distance from the $ X $ axis.

Find the mass and center of mass.

what i could find however is that the formula of mass is the following
$$M = \int\int_R \rho(x,y)dA $$

so i tried doing something like this
$$ \int_0^1\int_y^{\sqrt(y)} ? dxdy $$

the thing is that they say the density varies as the distance from the x axis, so i don't know what to replace for the density.. is it $ x + y $?

The mass density varies as the distance from the x-axis implies that $\rho =Ky$ where $K$ is a constant.

Now, the total mass is given by

$$M=\int_0^1 \int_y^{\sqrt{y}} (Ky) dx dy$$

The moment about x is

$$\frac{\int_0^1 \int_y^{\sqrt{y}} x(Ky) dx dy}{M}$$

$$\frac{\int_0^1 \int_y^{\sqrt{y}} y(Ky) dx dy}{M}$$

Can you complete?  Notice that the moments are independent of $K$.

Mass of lamina is K/15
(Density=KY)
On the vertical strip on the region of integration-
Y varies from parabola to straight line i.e. x^2 to x
And X varies from 0 to 1
Thus after double integration by the mass formula, we get mass of lamina=K/15


As $x \to 0$ this approaches zero since the zero of $e^{\frac{-1}{x^2}}$ beats out the pole of $\frac{1}{Q_k(x)}$? In the Robert Strichartz's book "A Guide to Distribution Theory and Fourier Transforms" at the page $4$, we have an interesting exercise :

$$ \psi(x) =
  \begin{cases}
    e^{\frac{-1}{x^2}}      & x>0\\
    0 & x \leq 0\\
  \end{cases}
$$

has continuous derivatives of all orders $$\frac{d^k (e^{\frac{-1}{x^2}})}{dx} = \frac{P_k(x)}{Q_k(x)} e^{\frac{-1}{x^2}},$$ whre $P_k(x)$ and $Q_k(x)$ are two polynomials such that deg($P_k(x)$) $<$ deg ($Q_k(x)$). Is there anyone could explain (at least a hint) to me why as $x \to 0$ this approaches zero since the zero of $e^{\frac{-1}{x^2}}$ beats out the pole of $\frac{1}{Q_k(x)}$ and what is its utility in the problem??

Let us first prove (by induction) a more explicit formula for the derivatives of $f = \textrm e ^{- \frac 1 {x^2}}$ ($x>0$): if $k \ge 1$, then $f^{(k)} = \dfrac {P_k (x)} {x^{3k}} \ \textrm e ^{- \frac 1 {x^2}}$ with $\deg P_k \le 2k-2$ and $P_k (0) = 2^k$.

Checking it for $k=1$ is easy. Next, assume the formula valid for $k$ and obtain it for $k+1$ (use Leibniz's formula):

$$f^{(k+1)} = \left( \frac {P_k (x)} {x^{3k}} \ \textrm e ^{- \frac 1 {x^2}} \right)' = \frac {P_k ' (x) x^3 - 3k x^2 P_k (x) + 2 P_k (x)} {x^{3(k+1)}} \ \textrm e ^{- \frac 1 {x^2}} .$$

Notice that the three terms in the numerator have degrees $2k$, again $2k$ and finally $2k-2$, therefore the degree of the numerator $P_{k+1}$ is at most the largest of them, i.e. $\deg P_{k+1} \le 2k$. Also, a quick look at the formula of the numerator shows that $P_{k+1} (0) = 2 P_k (0) = 2^{k+1}$.

It is clear then that when computing $\lim \limits _{x \to 0^+} f^{(k)} (x)$ we may safely ignore the numerator (which will only contribute a $2^k$) and focus on $\lim \limits _{x \to 0^+} \dfrac {\textrm e ^{- \frac 1 {x^2}}} {x^{3k}}$, which upon making the change of variable $t = \frac 1 {x^2}$ becomes $\lim \limits _{t \to \infty} \dfrac {t^{\frac 3 2 k}} {\textrm e ^t} = 0$ (the exponential is stronger than any power). This shows that $\lim \limits _{x \to 0^+} f^{(k)} (x) = 0$, from which it follows easily that all the derivatives of $\psi$ in $0$ exist and are continuous. Since $\psi$ obviously has continuous derivatives of any order on $\Bbb R \setminus \{0\}$, the above shows that $\psi$ has continuous derivatives of any order on the whole $\Bbb R$ (i.e. $\psi \in C^\infty (\Bbb R)$) and $\psi ^{(k)} (0) = 0$.

Now, it $\psi$ analytic? It is clearly so on $(-\infty, 0)$ (being constant) and on $(0, \infty)$ (being a composition of the analytic functions $\exp$ and $-\frac 1 {x^2}$). Nevertheless, it cannot be analytic on any neighbourhood of $0$ because if it were, then it would be equal to its own Taylor series in $0$, which is $0$ (because $\psi ^{(k)} (0) = 0 \ \forall k \ge 0$), so we would have $\psi = 0$ around $0$ - clearly not true.

To conclude, what this (classic) example constructs is a smooth function that is not analytic - clearly not a trivial thing.

As
$x \to 0$,
$\dfrac1{x^2} \to +\infty$
so
$e^{\frac1{x^2}} \to \infty$
so
$e^{-\frac1{x^2}}
=\dfrac1{e^{\frac1{x^2}}}
\to 0$
and does so very strongly.


High school math question $f(x)=ax+2$, $g(x)=a^2x^2-\ln x+2$ $a\in \mathbb{R}$, $x>0$. Q: Is there a negative $a$, for any positive $x$, $f(x)\le g(x)$? If $a$ exist, solve it, else, show the reason.

I want to know how to solve this problem, and are there any soft wares to show the figure of $ax-a^2x^2+\ln x\le 0$?

It is easy enough to see that you can find an $a$ and an $x$ to satisfy this by looking at what happens around $x = 0$. As $x$ approaches $0$, $f(x)$ goes towards $2$ and $g(x)$ goes towards infinity (due to the $\ln(x)$ component). That makes it easy to find a negative a that for small $x$ makes the $f(x) \le g(x)$.

Sorry I can't comment yet, but why don't you just try graphing it in mathematica or even on WolframAlpha if you don't have access to mathematica. Especially useful is mathematica's variable plot function which allows you to graph functions and then watch the graphs change as you vary a parameter, such as a or x. You could even make two variables that can be manually changed and check if f(x) is ever less than g(x).

Mathematica is often your best bet for graphing functions and stuff


Prove the equation has unique class of solutions Find the solutions of equation:
$$
x^y + y^x = 1 + xy \quad x,y \in \mathbb{R} \quad x,y >0
$$

My quest

First, $x=1$ or $y=1$ gives us obvious solutions, so let's suppose $x \not =1$ and $y \not= 1$.

Using Bernoulli's inequality I was able to show the inequality:
$$
x^y + y^x > 1 + xy \quad \forall x,y>0 \quad and \quad (x-1)(y-1) \gt 0
$$
Therefore there is no solution with $x,y\gt1$ or $x,y\lt1$.

My goal is to show that there is no solution having $x \not =1$ and $y \not= 1$ and the only case I was not able to cover is $(x-1)(y-1)<0$.

Any help will be appreciated.

you have given no source and no indication of your background. The conclusion that the solution set is the lines (in the first quadrant only) $y=1$ and $x=1$ seems quite likely given these plots; if you know multivariable calculus this gives some directions to pursue.

By the fundamental theorem on solutions to non linear equations, there are exactly n - 1 solutions, where n is the number of variables


The set of homomorphisms of modules is a module Suppose $M,N$ are $R$-modules. Let $Hom_R(M,N)$ be the set of all $R$-module homomorphisms from $M$ to $N$ with operations defined by $(\phi+\psi)(m)=\phi(m)+\psi(m)$ and $(r\phi)(m)=r(\phi(m))$. How can I prove that $Hom_R(M,N)$ is an $R$-module?

I just started studying modules and have been very confused, so explanations would be really appreciated!

just prove these for all r , s $\in$ R and $\phi , \psi $ in $*Hom_R(M,N)*$

$Hom_R(M,N)$ is an abelian group under the addition $
$r(\phi + \psi)$ = $r.\phi +r.\psi$.
$(r+s).\phi = r.\phi +s.\phi$
$(rs).\phi = r.(s.\phi)$
$ 1_R. \phi = \phi$ for all $phi$  in  R


Finite product of compact subspaces is compact I am working through an old qualifying exam and I have run into what I think is a very simple question but the way it is posed is confusing to me and I am worried I a missing the point.  The question is:

Assume as known that a finite product of compact spaces is itself compact.  Let $K_i$ be a compact subspace of the space $X_i$, $i=1,...,n$.  Show that $K_1\times...\times K_n$ is a compact subspace of $X_1\times...\times X_n$.

This seems like a silly question.

It is my understanding that a subspace of a space is compact iff it is a compact space in the subspace topology.  So since each $K_i$ is a compact subspace, each $K_i$ is a compact space in the subspace topology.  But we are to assume that finite products of compact spaces are compact.  Hence, $K_1\times...\times K_n$ is a compact space in the subspace topology, so a compact subspace of $X_1\times...\times X_n$.

Is there something more subtle going on here that I am missing?

I would assume they want you to prove that the subspace topology induced by the product topology is the same as the product topology of the induced subspace topologies.

Something does seem rather strange in the question as stated.  It might be more convincing to actually prove what the statement is telling you to assume.  It's actually not hard to prove that $K_1 \times \ldots \times K_n$ is compact in $X_1 \times \ldots \times X_n$ provided that each $K_i \subseteq X_i$ is compact.  Let $\mathcal{U} = \{U_i \; | \; i \in I\}$ be some open cover for $K_1 \times \ldots \times K_n$, and note that this induces an open cover for each $K_j$, namely we can write $\mathcal{U}_j = \{U_i \cap X_j \; | \; i \in I\}$.  This induces a finite subcover over $K_j$ by picking out some finite set of indices $I_j \subseteq I$.  We can therefore let $\widetilde{\mathcal{U}}_j = \{U_i \; | \; i \in I_j\}$.  The union

$$
\bigcup_{j=1}^n \widetilde{\mathcal{U}}_j
$$

induces a finite subcover over $K_1 \times \ldots \times K_n$.


What is the square root of $i^4$? What is the $\sqrt{i^4}$?

$i^4$ = $(i^2)^2$ So is $\sqrt{i^4}$ = $\sqrt{(i^2)^2}$ = $i^2$ = $-1$?

Or is $\sqrt{i^4}$ = $\sqrt{1}$ = $1$?

When I plug it into my TI-89 Titanium, I get $1$.

Edit: I now see why $i^4 \neq -1$. $$\sqrt{x^2} = \left|{x}\right| $$ So$$\sqrt{(i^2)^2} = \left|i^2\right| = \left|-1\right| = 1$$

The answer is $1$.  The reason it is not $-1$ is because, to get there, you inherently split up the square root into a product of square roots as

$$\sqrt{(i^2)^2} = \sqrt{i^2} \sqrt{i^2}$$

This is not valid for nonreal numbers.

I think it should be $-1$

note that $$i=e^{\left(\frac{\pi}2+2k\pi\right)i}$$ $$i^4=e^{4\left(\frac{\pi}2+2k\pi\right)i}=e^{\left(2\pi+8k\pi\right)i}$$
$$\sqrt{i^4}=e^{\frac12\left(2\pi+8k\pi\right)i}=e^{\left(\pi+4k\pi\right)i}=-1$$


Why are there only 2 solutions for $x^n=1$? (where $n>0$)

I have been taught that an equation with the highest power $n$ will always have $n$ solutions. This does not appear to be the case with:
$$x^n=1 \implies x=\pm1$$
Where $n$ is even, the solutions are $1$ and $-1$; where $n$ is odd, the only solution is $1$.

Why? Is it a special case, or is there something I'm missing?

It has $n$ roots, but only if you consider roots $x_i\in \Bbb C$. These "special" ones are calleds the roots of unity.

As an example, consider $X^4=1$; try to check that $X_1=i$ is a root of said equation.

Notice, we have $$x^n=1$$ $$x=(1)^{1/n}$$ $$=(\cos 0+i\sin 0)^{1/n}$$ $$=(\cos 2k\pi+i\sin2k\pi)^{1/n}$$ $$=\left(\cos \left(\frac{2k\pi}{n}\right)+i\sin\left(\frac{2k\pi}{n}\right)\right)$$ Thus there are $n$ roots which can be determined by setting $k=0, 1, 2, \ldots (n-1)$


Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$ $$P(x)=x^{4}-4x^{3}+12x^{2}-24x+24$$
I'd like to prove that $P(x)=|P(x)|$. I don't know where to begin. What would be the first step?

.In this question, you have to attempt to complete two squares, namely:
$$
x^4-4x^3 + 12x^2-24x+24 = x^4 -4x^3+6x^2-4x+1 + 6x^2-20x+23
\\  = (x-1)^4 + 6x^2 -20x+23
$$

Now, as it turns out, we can complete the second square, and:
$$
6x^2-20x+23 = \frac{2}{3}(3x-5)^2 + \frac{19}{3}
$$

Hence, we can rewrite the whole expression as:
$$
x^4-4x^3 + 12x^2-24x+24 = (x-1)^4 + \frac{2}{3}(3x-5)^2 + \frac{19}{3}
$$

It is a sum of positive expressions, hence is always positive, hence $P(x) = |P(x)|$.

The first step is definitely recalling a modulus function's definition:
$$|x| := \begin{cases}x & \text{if } x\ge 0 \\ -x & \text{if } x\le 0\end{cases}$$
So the requirement $P(x) = |P(x)|$ is equivalent to $P(x) \ge 0$.

Hope you can continue from that.


How to multiply Roman numerals? How to multiply Roman numerals? I need an algorithm of multiplication of numbers written in Roman numbers. Help me please.

Make a table with two columns, and enter the two numbers to be multiplied into the first row.
Make the next row by halving the first number (discarding remainders) and doubling the second. Continue until there is nothing left to halve.
Cross out all the rows where the left number is even.
Add the remaining numbers in the second column. The result is the product of the first two numbers

Examples

Source

It is conversion to Hindu numerals, Not Arabic. Just because Europeans learnt from Arabs does not mean the founders change!

https://rbutterworth.nfshost.com/Tables/romanmult is one place multiplication is explained


How do you find the sine of the angle between two vectors? I do not know what the sine of the angle between two vectors is. I think it may be the vector created by connecting the tips of the two vectors but I am not sure.

How do you find the sine of the angle between two vectors?

In general for $\mathbf{a},\mathbf{b}\in\mathbb{R}^3$, we have
$$\|\mathbf{a}\times \mathbf{b}\|=\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta,$$
where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.

Translate your two vectors so that their tails are at the origin.  Then draw a line through each of those two vectors.  The smaller of the two angles is the called the "angle between the two vectors".

Hint on how to find it:  The angle $\theta$ between two vectors $\vec u$ and $\vec v$ is given by the formula $$\theta = \arccos\left(\frac {\vec u\cdot\vec v}{|\vec u||\vec v|}\right)$$

Or -- here's a slightly harder way (because it involves calculating a cross product).  You know that $|\vec u\times \vec v| = |\vec u||\vec v|\sin(\theta)$, so find the cross product and all of the norms and then plug in.


Prime and Maximal Ideals of $\mathbb{Z}_2 \times \mathbb{Z}_4$ I'm having some trouble finding ideals in general. The problem I'm stuck on is:

Find all prime and maximal ideals of $\mathbb{Z}_2 \times \mathbb{Z}_4$.

I know that a finite integral domain is a field, which means the prime and maximal ideals are the same. That means I just have to find one or the other, right? But I don't really know how to go about finding ideals for $\mathbb{Z}_n \times \mathbb{Z}_m$. Any help would be great. Thank you!

An ideal of $Z_2 \times Z_4$ must have the form $I \times J$ where $I$ is an ideal of $Z_2$ and $J$ is an ideal of $Z_4$, both of which are not necessarily proper. So clearly there's not so much ideals to try!


The definition of function series Regarding the definition of a "function series"

$$\sum_{n=0}^{\infty} f_n = f_0+f_1+...$$

such as:

$$\sum_{n=0}^{\infty} \frac{nx}{1+n^4x^2}$$

Does the function series term-wise (as a sequence of terms) look like

$$(0, \frac{1x}{1+1^4x^2}, \frac{2x}{1+2^4x^2}, ...)$$

then?

Be aware of the difference between a map $f$ and the value $f(x)$ of $f$ at some point $x$; knowing this, you know that the series
$$
\sum_{n=0}^{\infty} \frac{nx}{1+n^4x^2}
$$
is the series whose general term is the value $f_{n}(x)$ of the map $f_{n}: \xi \mapsto \frac{n\xi}{1+n^{4}\xi^{2}}$ at some point $x$.

Yes, the terms of your function series would be the $f_i$, just as you list them. But presumably you are really interested in the sequence of partial sums:

$$\sum_{0 \le k \le n} \frac{k x}{1 + k^4 x^2}$$


How to solve $x_{100}$ with $x_{n+1}=\frac{2x_n}{2+x_n}+1$ and $x_{1}=3$ How to solve $x_{100}$ with $x_{n+1}=\frac{2x_n}{2+x_n}+1$ and $x_{1}=3$?

Can anybody shed light on this? regards.

tl;dr: considering a related sequence will help you obtain (very easily) the closed-form for $x_n$, from which you can plug in $n=100$ and get
$x_{100} = 2 + \frac{3}{2^{200}-1} \simeq 2 + 2\cdot 10^{-60}.$

Details: It is pretty clear that the sequence converges to $2$ (if you do not know how to show it, prove by induction it is monotone decreasing and positive, hence convergent; and show the only possible limit is then $\ell=2$).

So let us consider instead the positive sequence $(a_n)_{n\geq 1}$ defined by $a_n \stackrel{\rm def}{=} x_n -2$ for all $n\geq 1$, which now converges to $0$ and satisfies the recurrence relation (from the one that $(x_n)_n$ satisfies)
$$
a_{n+1} = \frac{a_n}{a_n+4} = \frac{1}{1+\frac{4}{a_n}}, \qquad n\geq 1 \tag{1}
$$
with $a_1 = 1$.

Mmh, there are a lot of reciprocals there, and they are not that nice. Let us go further defining the (new) positive sequence $(b_n)_{n\geq 1}$ by
$b_n \stackrel{\rm def}{=} \frac{1}{a_n} = \frac{1}{x_n -2}$, so that
$$
b_{n+1} = 1+{4}{b_n}, \qquad n\geq 1 \tag{2}
$$
with $b_1 = 1$.

This looks much nicer. Solving this should not be too hard: there are various standard methods, and they will lead to
$$
b_n = \frac{4^n-1}{3}, \qquad n\geq 1
$$
from which you eventually get the closed-form expression for $x_n$:
$$
x_n = \frac{1}{b_n}+2 = \frac{3}{4^n-1}+2, \qquad n\geq 1. \tag{3}
$$
(Sanity check: it does converge to $2$, is positive, and decreasing with $x_1=3$.)

And finally: from (3), you get
$$
x_{100} = 2 + \frac{3}{2^{200}-1}.
$$

you are trying to find the fixed point of the function f(x)=(2x/(1+x))+1 with a derivative 2/(1+x)^2<1 on the domain [1,10],say. The fixed point is 2. you can find the term you are after by consulting any book on numerical analysis ((Mendelson's book)ang go from  there to the fixed point algorithm.


Prove that $(a_n)$ converges if and only if there is a natural number $N$ big enough and an integer $m$ such that for any $n \geq N$, ${a_n} = m$ Suppose that $(a_n)$ is a sequence of integers. Prove that $(a_n)$ converges if and only if
there is a natural number $N$ big enough and an integer $m$ such that for any $n \geq N$,
${a_n} = m$. I thought this is the definition of convergence of a sequence. Anyone can guide me on how to prove this ?

Take any $0<\epsilon<\frac12$, and use the fact that convergent sequences are Cauchy to conclude that a convergent sequence of integers must be (eventually) constant. The other direction should be trivial.

That is wrong. Consider the sequence $a_n = 1 - 2^{n + 1}$, which converges to $1$. No integer $a_n$ in sight, no constant value for $a_n$ starting at some $N$ whatsoever.


On the definition of locally connected spaces I have seen two a priori different definitions of locally connected space :

1) For all point $x$, and neighborhood $V$ of $x$, there is a connected neighborhood $C$ of $x$ such that $C\subseteq V$

2) For all point $x$, and open set $U$ containing $x$, there is an open connected neighborhood $O$ of $x$ such that $O\subseteq U$.

I imagine those two definitions are identical, but I don't see why.

The same thing for locally path-connected spaces, we can either use open neighborhoods or just neighborhoods, are the two definitions the same?

By definition a neighborhood of a point $x$ is a set $I$ that contains an open set $A$ such that $x\in A$. So 1) $\Rightarrow$ 2) by considering $U$ as your neighborhood; 2) $\Rightarrow$ 1) by considering the open set of the definition of neighborhood.


Trapezoid area proof by dividing it into two triangles? I am trying to figure out how the formula for the area of a trapezoid with exactly two parallel sides is deduced. In my textbook it says that the formula for the area of a trapezoid is deduced by dividing the trapezoid into two triangles, one with base a and height h, and one with base b and height h.

$A={a\times h\over 2} + {b\times h\over 2} = {a\times h + b\times h\over 2} = {h(a+b)\over 2}$

I have drawn this diagram in GeoGebra. It may not be drawn to scale (it is not congruent) with the diagram in my textbook, but it is actually similar. I actually scanned the diagram from my textbook and then made this diagram on top of it. In the textbook, they did not mark the vertices. But I have named the vertices (blue) to simplify explanation, so that we have something to refer to.

The formula above is the one that's given in the textbook.

From just the looks of it, does this even make any sense to you? They drew this diagram and named the sides of it as a, b, c and d. They also drew the altitude h and the diagonal DB.

I can't make the connection with the formula. And I have seen an actual proof of the area of a trapezoid like this on a website. There are at least two different proofs for the area of a trapezoid. Perhaps the most common proof is where you would divide the trapezoid into two triangles and a rectangle. But two triangles??...

So what I am asking for is for someone to give me a proof that the formula for the area of a trapezoid can be deduced by dividing the trapezoid into two triangles, as indicated by this diagram.

I know that the area of a triangle is base times the height divided by two, or 1/2 times the base times the height. It is essentially half of the area of the rectangle. So if I consider that first part of the formula above I get this.

(Image upload to Imgur doesn't work at the moment. I'll get back to it.)

Update:

$[DBE]={a\times h\over 2}$

I have constracted the altitude DF. The area for triangle DBE is half of the area of DFBE.

$[BGA]={b\times h\over 2}$

I have constructed the altitude AH. The area for triangle BHA is half of the area of BEHA.

But this gives me the overlapping triangle GBH. It overlaps with triangle DBE. Is the area of GBH equal to the area of AGD?

And what about the triangles EBC and DFA?

Update:

I think I got it right now. So here's that second diagram again.

And here is the third diagram once again.

They are not the same size this time. I think I got the scale wrong on export to PNG image. But here you can see that I have shaded and measured the areas of the rectangles and triangles to show how it interacts with the area of the trapezoid.

I know that this is not really a formal proof of the area of a trapezoid. But I think this makes sense to me now. I was getting confused by the fact that the triangle ABD had no height or altitude. Or it was not within the triangle itself, it was outside of the triangle. And also I had hard time seeing how that overlapping part GHB "transforms" (or however you want to call it) into that other empty space. But I now see it more clearly.

So this really comes from the formula of the area of a triangle then? Or maybe we can say that the area of a triangle is used as a postulate to prove the area of a trapezoid?

When you divides the trapezoid in two triangles, the area of the trapezoid is the sum of the aeras of the triangles.
The triangle $DBC$ has an area of $\frac{a*h}{2}$ ($\frac{\text{basis }*\text{ height}}{2}$) and the triangle $ABD$ an area of $\frac{b*h}{2}$, hence the area of the trapezoid : $\frac{a*h}{2} + \frac{b*h}{2} = \frac{(a+b)*h}{2}$

Given Trapezoid ABCD with AB || CD.  Let AB be $B_1$ and CD be $B_2$

First draw in auxiliary diagonal AC to make $\triangle ADC$ and $\triangle ABC$.

Find the midpoint of segments AD, AC, and BC, Label these points E, F, and G respectively.
Next, draw in the 'midsegments' of the two triangles.

By the midsegment of a triangle theorem, $EF || DC$ and $FG|| AB$.   Thus, by the transitive property, $ AB ||EG||DC$.  Additionally, the midsegment of a triangle guarantees that $EF=\frac{1}{2}DC$  (use substitution to say) $EF=\frac{1}{2}B_2$.

The area of $\triangle ADC=\frac{1}{2}B_2*h$, where $h$ is the height of the triangle and the trapezoid.  Additionally by the midsement of a triangle theorem, $FG=\frac{1}{2}AB$ (use substitution to say) $FG=\frac{1}{2}B_1$.  The area of $\triangle ABC = \frac{1}{2}B_1*h$.

By area addition, the area of  $\triangle ADC + \triangle ABC$ = the area of Trapezoid ABCD.

Trapezoid ABCD has an area of(½B2*h) + (½B1*h).  We can factor this to get Area = ½h*(B1 + B2).

WHAT!!!  BTW, when I took the Praxis exam years ago, we had to prove the trapezoid area formula in two different ways.  I used the rotated trapezoid to make a parallelogram as the other way.


Equilateral triangle and 120° angle P is the angle's vertex. With A,B we form an equilateral triangle on the same direction the angle's rays open up.

How can I  prove the triangle's third vertex C is always on the bisector's line?

As pointed out by some comments, the problem has been originally misstated. It should be: we are given a 120 degrees angle with vertex P and an equilateral triangle ABC inscribed in it, such that the vertices A and B lie on the sides of the angle and C is in its interior, as in the drawing. Show that the line PC bisects the angle APB (i.e., that the angles APC and CPB are both  60 degrees).

Proof. Consider  the quadrangle ACBP. It  is cocyclic (its vertices lie on the same circle) since the sum of the angles at its  opposite vertices  P and C is 180 degrees. Thus the angles APC and CPB are equal, since they are inscribed angles in the same circle that are facing equal segments. QED

From the proof you can see that it is enough to assume that ABC is isosceles, i.e. AC=BC, and that the sum of the angles and P and C is 180 degrees.

Call $\;M\;$ to the intersection point of $\;PC\,,\,AB\;$ , and observe that

$$\Delta APM\sim\Delta BPM\;\;\text{by s.a.s., since applying the angle bisector theorem we get}$$

$$\frac{AM}{MB}=\frac{AP}{PB}\;,\;\;\text{on the triangle}\;\;\Delta APB$$

This means $\;PC\perp AB\;$ (why?) , so both $\;\Delta CBP\;,\;\Delta CAP\;$ are right triangles (and, in fact, golden ones $\;30-60-90\;$).

Now just remember that an angle's bisector is the locus of all points on the plane that are equidistintant from both angle's rays.


Non-existent expectation and variance I have a problem the answer for which is intuitively obvious but is more difficult than anticipated to actually show.
Claim: If X is a random variable such that the expectation of X does not exist, then the variance of X does not exist either.
This seems true on an intuitive level, as the variance is a measure taken relative to the mean. My issue is in showing that, for a discrete random variable X with E[X] and E[X^2] both infinity, V[X] does not exist. In this case, using the definition V[X] = E[X^2]-E[X]^2, I have a sum which diverges to infinity take (the square of) another sum which diverges to infinity, which I don't really know how to treat.
Any tips would be appreciated.

Please see if the below is useful
\begin{align*}
\int\limits_{x=-\infty}^\infty x^2 f_X(x) dx &= \int\limits_{x=-\infty}^0 x^2 f_X(x) dx + \int\limits_{x=0}^1 x^2 f_X(x) dx+\int\limits_{x=1}^\infty x^2 f_X(x) dx\\
&\geq \int\limits_{x=-\infty}^0 x^2 f_X(x) dx + \int\limits_{x=0}^1 0 f_X(x) dx+\int\limits_{x=1}^\infty x^2 f_X(x) dx\\
&\geq \int\limits_{x=-\infty}^0 x f_X(x) dx + \int\limits_{x=0}^1 0 f_X(x) dx+\int\limits_{x=1}^\infty x f_X(x) dx\\
&= \int\limits_{x=-\infty}^\infty x f_X(x) dx +  \int\limits_{x=0}^1 (-x) f_X(x)  \\
&\geq \int\limits_{x=-\infty}^\infty x f_X(x) dx -  \int\limits_{x=0}^1 1 f_X(x)  \\
&\geq \int\limits_{x=-\infty}^\infty x f_X(x) dx -  \int\limits_{x=-\infty}^\infty f_X(x)  \\
&= \left[\int\limits_{x=-\infty}^\infty x f_X(x) dx\right]-1\\
&= \mathbb E [X]-1
\end{align*}
Therefore $\mathbb E [X^2] \geq \mathbb E [X]-1 = \infty$.
Let's call $\mathbb E[X] = \mu$. Then, $Var(X)=\mathbb E[(X-\mu)^2]$.
\begin{align*}
\lim\limits_{\mu\to\infty}Var [X] &=\lim\limits_{\mu\to\infty}\int\limits_{x=-\infty}^\infty (x-\mu)^2 f_X(x) dx\\
&=\lim\limits_{\mu\to\infty}\int\limits_{x=-\infty}^\infty [x^2 + \mu(\mu-2x)] f_X(x) dx\\
&=\int\limits_{x=-\infty}^\infty x^2f_X(x) dx +\lim\limits_{\mu\to\infty} \int\limits_{x=-\infty}^\infty\mu(\mu-2x) f_X(x) dx\\
&=\int\limits_{x=-\infty}^\infty x^2f_X(x) dx +\lim\limits_{\mu\to\infty} \mu\int\limits_{x=-\infty}^\infty(\mu-2x) f_X(x) dx\\
&\geq\int\limits_{x=-\infty}^\infty x^2f_X(x) dx +\lim\limits_{\mu\to\infty} \mu\int\limits_{x=-\infty}^\infty f_X(x) dx\\
&=\int\limits_{x=-\infty}^\infty x^2f_X(x) dx +\lim\limits_{\mu\to\infty} \mu\cdot 1\\
&=\mathbb E [X^2] +\lim\limits_{\mu\to\infty} \mu\cdot 1\\
&=\infty
\end{align*}


a Problem about Sequence Let $a_1$ be an integer. Then we assume

  $$
a_{n+1} =
\begin{cases}
3a_n+1,&\text{$a_n$ is odd}\\
\frac{a_n}{2},&\text{$a_n$ is even}
\end{cases}
$$

Now we prove that

  for any $a_1\in\mathbb N$, there exists $N$ which satisfy: $a_n=1,2$ or $4$,$n\geq{N}$.

At first I want to give it a suitable category for the problem: analysis. And I want to use the basic method: evaluate the upper bound for $a_n$, however I find it's not easy because the iteration is rely on the odd or even property of $a_n$. So I attempt the method of number theory. But I failed to find any way to go over it. Can anyone have idea? Thank you.

It's false because if $a_n=1$ then $a_{n+1}=4$.