Handshake/pigeonhole principle problem? There a number of people in a party. Each person shakes hands with exactly 20 people. For each pair of people that shakes hands with each other, there is exactly 1 other person who shakes hands with both of them; while for each pair of people that don’t shake hands with each other, there are exactly 6 other people who shake hands with both of them. I am tasked to find the total number of people in the room. I am very confused as to how to invoke 3rd condition, could someone guide me in this respect? Thank you!

For reference, the two complicated rules are

For every pair of people who shook hands, the is exactly $1$ person who shook hands with both of them.
For every pair of people who did no shake hands, there are exactly $6$ people who shook hands with both of them.

Let $n$ be the number of people. There are $20n/2=10n$ handshakes.

Call a triple of people $\{a,b,c\}$ a "two-chain" if exactly two pairs of them shook hands. Let's count the number of two-chains in two different ways:

Let $N$ be the number of pairs of people who did not shake hands. Per rule $(2)$, there are $6N$ two-chains.
Let $H$ be the number of handshakes. For each handshake $a\leftrightarrow b$,
by rule $(1)$, there is another person $c$ which shook both their hands. Let $A$ be the set of 18 other people $a$ shook hands with, besides $b$ and $c$. Same for $B$. Per rule $(1)$ again, no one in $A$ shook hands with $b$ (because only $c$ shook hands with both $a$ and $b$), so for every person $a'\in A$, we have $\{a,a',b\}$ is a two-chain.  Similarly for $\{a,b,b'\}$ for $b'\in B$. This leads to $18+18=36$ two-chains per handshake. However, this double-counts the number of two chains since each two-chain has two handshakes, so the number is actually $36H/2=18H$.

This shows that $6N=18H$, or $N=3H$.
But we also must have $N+H=\binom{n}2$, the total number of pairs of people. It follows that $$4H=\binom{n}2.$$ But we already had that $H=10n$. Substituting this into the above, you get $n=81$.

I'm pretty sure its 90. So I shake 20 peoples hands. And for each set of handshakes means that there is 1  not shook. Meaning 20*.5=10. 20+10=30. But there are 6 extra people whom I don't shake hands with. Meaning 6*10=60. QED 30+60=90.


rigorous proof that : $E[X] = \sum_{k \in \mathbb{N}} \mathbb{P}(X \geq k)$ Let $X$ be a random variable : $\mathbb{N} \to \mathbb{N}$. Then if $X$ has an expectation we have :
  $$E[X] = \sum_{k \in \mathbb{N}} \mathbb{P}(X \geq k)$$

It's quite logical that its true. The problem is that I don't know how to prove this result rigourously just by manipulating sums. For example I can do the following : $\sum_{k \in \mathbb{N}} \mathbb{P}(X \geq k) = \sum_{k \in \mathbb{N}} \sum_{i = k}^\infty \mathbb{P}(X = i)$. It would be nice if I can isolate the $\mathbb{P}(X = i)$ for single $i$, but I don't know how to do this. Maybe I should introduce indicator functions somehow ? and then use Fubini to interchange the summations ?

Thank you !

For non-negative numbers $a_{nm}$ we always have $\sum\limits_{n=1}^{\infty} \sum\limits_{m=1}^{\infty} a_{nm}=\sum\limits_{m=1}^{\infty} \sum\limits_{n=1}^{\infty} a_{nm}$ even without the assumption of convergence. Fubini's Theorem is not required for this. You can just use partial sums and take limits to show that each side is less than or equal to the other. Apply this result with $a_{nm}=P(X= n)$ for $n \geq m$ and $0$ for $n<m$ to get the expression for $EX$.

Here is how you can "discover" the desired result. Begin with
$$P(X\geq 1)=P(X=1)+P(X\geq 2)$$
Add $P(X\geq 2)$ to both sides:
$$P(X\geq 1)+P(X\geq 2)=P(X=1)+2P(X\geq 2)=P(X=1)+2P(X=2)+2P(X\geq 3)$$
in the r.h.s equality the fact that $X$ takes only integer values is used.
Add $P(X\geq 3)$ to both sides:
$$\sum_{k=1}^3P(X\geq k)=\sum_{k=1}^3kP(X=k)+3P(X\geq 4)$$
And by induction, you get that for every natural number $n$
$$\sum_{k=1}^nP(X\geq k)=\sum_{k=1}^nkP(X=k)+nP(X\geq n+1)\quad (*)$$
So now your problem is to prove that
$$\lim_{n\to\infty}nP(X\geq n+1)=0$$
But you are given that the expectation $E(X)$ exists, which by definition means that the series
$$\sum_{k=1}^{\infty}kP(X=k)$$
is convergent, which implies that $nP(X=n)$ tends to zero as $n\to\infty$. Since
$$nP(X\geq n+1)\leq nP(X\geq n)$$
we deduce that $nP(X\geq n+1)$ tends to zero as well, and letting $n\to\infty$ in (*) gives the desired result.


Is it wrong to say that $100$ is solution of $\sqrt x+10=0$? Is it wrong to say that 100 is solution of $\sqrt x +10=0$?

I know that range of $\sqrt{x}$ is $[0, \infty)$ by convention. This convention is because of definition of a function. So if I consider $\sqrt x+10=0$ as a simple equation then can I say hundred is a solution to the equation?

I think I can say that because if I square both sides I get $x=100$ and also if I consider $\sqrt x+10=0$ just as a equation then there is no need of thinking about functions and the convention of omitting negative roots.

The solutions to $f(x) = k$ are a SUBSET of the solutions to $f(x)^2 = k^2$ but not all the solution to $f(x)^2 = k^2$ are solutions to $f(x) = k$.  Squaring both sides of an equation add extraneous solutions.

$\sqrt{x} + 10 = 0$

$\sqrt{x} = -10$ Doesn't just mean that $(\sqrt{x}^2 = (-10)^2$.  It ALSO means that $\sqrt{x} = -10 < 0$.

When we square both sides we LOSE information.

$\sqrt{x}^2 = (-10)^2$

$x = 100$ but we have completely LOST that $\sqrt{x} < 0$.

Any  $\sqrt{100} + 10 = 10 + 10 = 20 \ne 0$.  SO it simply DOESN'T work.

Consider this:

$x =2 $ has one solution. Square both sides and you get $x^2 = 4$.  Which has TWO solutions!!!  Where did that solution $x = -2$ come from?

It came because when we squared both sides we added invalid extraneous solutions.

Solutions to $x= 2$ is $\{2\}$.  Solutions to $x^2 = 4$ has solutions $\{2,-2\}$ and $\{2\} \subset \{2,-2\}$.  But it doesn't go the other way.  It only goes one way.


Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$ Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$
Let $d=\gcd(4n^2+1,24)$ then we have:
$$d|24n^2+6,24n^2\ \Rightarrow\ d|6\ \Rightarrow\ d|6n^2,4n^2+1\ \Rightarrow\ d|12n^2,12n^2+3\ \Rightarrow\ d|3\ \Rightarrow\ d=1\ or\ 3$$
Using modular equivalence it's very easy to show that $d$ can't be 3,but how can I show it WITHOUT using  modular equivalence???

Theorem. $4n^2 + 1$ is not a multiple of $3$.

Proof. Since $n^2 = (-n)^2$, we need only prove this for
$n \in \mathbb W = \{0, 1, 2, \dots\}$.

Let $T = \{k\in \mathbb W : 3 | 4k^2 + 1\}$.

We need to show that $T$ is the empty set.

If we assume that $T$ is not the empty set, then it must contain a smallest member, say $s$. (This is called "The Well-Ordering Principle".) Then $s$ is the smallest member of $\mathbb W$ such that $3 \mid 4s^2 + 1$.

Note that
\begin{align}
    4\cdot 0^2 + 1 &= 1 \\
    4\cdot 1^2 + 1 &= 5 \\
    4\cdot 2^2 + 1 &= 17 \\
\end{align}
So $s-3 \in \mathbb W$ and $s-3 \not \in T$. But if $3 \mid 4s^2 + 1$ and
$3 \not \mid 4(s-3)^2 + 1$, then

\begin{align}
    3 &\not \mid (4s^2 + 1) - (4(s-3)^2 + 1) \\
    3 &\not \mid 4(s^2 - (s-3)^2)\\
    3 &\not \mid 4(6s - 9)\\
\end{align}

But clearly $3 \mid 4(6s - 9)$. So by contradiction, $T$ is the empty set.

Finally I could find a desired solution:
We consider 3 cases for $n:n=3k,3k+1,3k+2$
$n^2=9k^2,9k^2+6k+1,9k^2+12k+4\ \Rightarrow\ 4n^2+1=3k'+1,3k'+2,3k'+2$
So $4n^2+1$ is never divisible by 3,thus $d=1$,that's it!


Proof verification related to a rotation of an operator and convexity Let $(\mathcal{H}, \langle\cdot,\cdot\rangle)$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ is the algebra of all bounded linear operators on $\mathcal{H}$.

For ${\bf T} = (T_1,\cdots,T_n)\in \mathcal{B}(\mathcal{H })^n$, we consider the following subset $\mathbb{C}^n$
\begin{eqnarray*}
JtMaxW({\bf T})
&=&\{(\lambda_1,\cdots,\lambda_n)\in \mathbb{C}^n:\;\exists\,(x_i)\subset \mathcal{H}\;\;\hbox{such that}\;\|x_i\|=1,\displaystyle\lim_{i\rightarrow+\infty}\langle T_k x_i,x_i\rangle=\lambda_k,\\
&&\phantom{++++++++++}\;\hbox{and}\;\displaystyle\lim_{i\rightarrow+\infty}\|T_kx_i\|\rightarrow \|T_k\|,\; 1\leq k \leq n\;\}.
\end{eqnarray*}

  Why the following facts which is taken from this reference(theorem 3.8) hold?

  Note that Lemma 3.4 indicates that $\pi_k(JtMaxW_A({\bf T}))$ is convex for all $k\in\{1,\cdots,n\}$, where $\pi_k$ is the projection from $\mathbb{C}^n$ to $\mathbb{C}$ i.e. $\pi_k(x_1,\cdots,x_n)=x_k$, for all $k\in\{1,\cdots,n\}$.

  If the above claim is false. Is it true if $n=1$?

My attempt: Suppose now that $0\notin JtMaxW({\bf T})$, then $\forall\{x_i\}\in\mathcal{H}$ such that $\|x_i\|=1$ we have,
$$\langle T_k x_i,x_i\rangle\nrightarrow 0 \quad \mbox{or}\quad\|T_kx_i\|\nrightarrow \|T_k\|,\,\, 1\leq k \leq n.$$
Let us rotate ${\bf T}$, that is we consider ${\bf \widehat{T}}=(e^{i\alpha_1}T_1,...,e^{i\alpha_n}T_n)$. Moreover, if $\lambda=(\lambda_1,...,\lambda_n)\in JtMaxW({\bf  \widehat{T}})$, then there exists a sequence $\{y_i\}$ in $\mathcal{H}$ such that
$$\|y_i\|=1\quad \mbox{and}\quad\|\widehat{T}_ky_i\|\rightarrow \|\widehat{T}_k\|,\,\, 1\leq k \leq n.$$
i.e.
$$\|y_i\|=1,\,\,e^{-i\alpha_k}\langle T_k y_i,y_i\rangle\rightarrow \lambda_k \quad \mbox{and}\quad\|T_ky_i\|\rightarrow \|T_k\|,\,\, 1\leq k \leq n.$$
Since, $\displaystyle\lim_{i\rightarrow\infty}\langle T_k y_i,y_i\rangle\neq0$, then $\lambda_k\neq0$ for all $1\leq k \leq n.$ Assume that there exists $\delta_k$ such that $\Re e(\lambda_k)\geq\delta_k>0$

  Why for all $x\in \mathcal{H},\;\|x\|=1$ we have $\Re e\langle \widehat{T}_k x,x\rangle\geq\delta_k>0$)?

The claim as it stands is false because $JtMaxW(T)$ might be empty. Then clearly, $(0,\ldots,0)\notin JtMaxW(T)$, but also $JtMaxW(T') = \emptyset$ for any "rotation" $T'$ of $T$.

An example for which $JtMaxW(T)$ is empty is given by
$$
T_1 = \begin{pmatrix}1&0\\0&0\end{pmatrix}\quad\text{and}\quad T_2 = \begin{pmatrix}0&0\\0&1\end{pmatrix}.
$$


Why is a Riemannian metric $g$? A Riemannian metric on a manifold is universally denoted with the letter $g$, but unlike many other mathematical objects like a function $f$, a distance $d$, a manifold $M$ or a group $G$ there does not seem to be a connection to the name or meaning of the actual quantity, neither in English nor German.

Questions:

When and where has this notation been introduced?
Is there an evident reason why the letter $g$ was chosen and accepted by everyone?

Some Remarks:

Of course $g$ carries a lot of information about the geometry of a manifold, but this seems to be a little too unspecific to me.
In his book on Riemannian geometry, Do Carmo only uses $g_{ij}$ but never $g$ for the metric. On the other hand at some point he uses $g$ for the Gauss-map of a surface, but that also does not directly relate to the metric tensor.
Do Carmo also points out that Riemann introduced the concept of a quadratic form assigned to each point of a manifold in his paper "Über die Hypothesen, welche der Geometrie zugrunde liegen." But in there I did not find the notation $g$ as well.

I always thought it was $g$ for the Gram matrix, which, after picking a basis, specifies the component of an inner product:

$g_{ij} = \langle \hat{e}_i, \hat{e}_j \rangle$

On a Riemannian manifold, the metric is just an inner product-valued field.

The symbol $\mathbb Z$ is used because the German word for number is Zahlen.

In German, "gestalt" means "shape", "form", "figure", etc.

Perhaps this is the reason behind the choice of $g$.


How many ways can seven people sit around a circular table? How many ways seven people can sit around a circular table?

For first, I thought it was $7!$ (the number of ways of sitting in seven chairs), but the answer is $(7-1)!$.

I don't understand how sitting around a circular table and sitting in seven chairs are different. Could somebody explain it please?

In a circular arrangement we first have to fix the position for the first person, which can be performed in only one way (since every position is considered same if no one is already sitting on any of the seats), also, because there are no mark on positions.

Now, we can also assume that remaining persons are to be seated in a line, because there is a fixed starting and ending point i.e. to the left or right of the first person.

Once we have fixed the position for the first person we can now arrange the remaining $(7-1)$ persons in $(7-1)!= 6!$ ways.

Are we dealing with chairs or a circular bench? If we are using chairs, then the number of combinations is $(7-1)!$ as previously noted. However if we are using a circular bench, then the location of the first two sitters is irrelevant. Accordingly the next five sitters can be placed in $(7-2)!=$$720$ ways.


Does Every Continuous Function Have an Antiderivative? I was thinking about integrals over continuous function and wanted to know if every continuous function has an antiderivative. Generally, the only way I can think of proving this is through a linear algebra approach.
I wanted to first prove or disprove

Let $D_{x}$ be the differential operator, $C^{1}$ be the set of differentiable functions thats derivative is one time continuous, and $C$ be set of continous functions.
$$C\subseteq D_{x}(C^{1})$$

and then if this were true then wanted to prove or disprove

$D_{x}$ is surjection from $C^{1}$ to $C$

The problem is first I don't know if truth of the propositions would imply that every continuous function has an antiderivative. Next, Im not really comfortable with infinite subspaces. I know that if this did happen to be finite we could use rank nullity theorem and since constant function derivative would be 0 thus in nullspace thus not injective thus not surjective, but I'm pretty sure this type of logic doesn't apply to infinite subspaces.
Also If I didn't state these propositions correctly feel free to edit them

No, e.g., the Cantor function: http://en.wikipedia.org/wiki/Cantor_function is a.e., constant, so that its derivative is a.e. $0$, although it is not constant, i.e., you cannot recover the function from its derivative. It is a canonical example of a function thatis integrable, but without an antiderivative.  I think what you are looking for is for the function to be absolutely-continuous, which will guarantee the existence of an antiderivative.


“$ x $ is a brother of $ y $.” Why is this not transitive? I am working on a problem set at the moment, and while checking my answers I realized that I have listed "x is a brother of y" as a transitive relation, while the answers say that it is not.

EDIT: I realized I forgot to put some assumptions from the question:
i) Domain is people in general.
ii) Assume that all siblings share both parents.

But if x is a brother of y, and y is a brother of z, then surely x is a brother of z? Am I missing something here?

Please help a student out. Thanks!

If Fred is Bob's brother, and vice versa, transitivity would imply that Fred is his own brother.

What if Y is female.  So X is a brother of Y, but Y is a sister of X.


Missing $1/r$ factor in gradient of polar function I'm trying to derive the gradient in polar coordinates using the chain rule.

So the idea is that when we have a function $f(x,y)$ and we switch to polar coordinates, we're really composing $f$ with $P(r,\theta) = (r\cos(\theta),r\sin(\theta))$.  So then the gradient of $f$ in polar coordinates should just be $\nabla (f\circ P)(r,\theta)$.

From my last question I know that $$\nabla(f\circ P)(r,\theta) = \begin{bmatrix}(\partial_1f\circ P)(r,\theta) & (\partial_2f\circ P)(r,\theta)\end{bmatrix}\begin{bmatrix}\partial_1P_1(r,\theta) & \partial_2P_1(r,\theta) \\ \partial_1P_2(r,\theta) & \partial_2P_2(r,\theta)\end{bmatrix} \\ = \begin{bmatrix} (\partial_1f\circ P)(r,\theta)\cdot\partial_1P_1(r,\theta) + (\partial_2f\circ P)(r,\theta)\cdot\partial_1P_2(r,\theta)  \\ (\partial_1f\circ P)(r,\theta)\cdot\partial_2P_1(r,\theta) +(\partial_2f\circ P)(r,\theta)\cdot\partial_2P_2(r,\theta)\end{bmatrix}^T$$

But I also know that $$\partial_1(f\circ P)(r,\theta) = (\partial_1f\circ P)(r,\theta)\cdot\partial_1P_1(r,\theta) + (\partial_2f\circ P)(r,\theta)\cdot\partial_1P_2(r,\theta) \\ \partial_2(f\circ P)(r,\theta) = (\partial_1f\circ P)(r,\theta)\cdot\partial_2P_1(r,\theta) +(\partial_2f\circ P)(r,\theta)\cdot\partial_2P_2(r,\theta)$$

by the regular chain rule.

So, putting that together I get $$\nabla(f\circ P)(r,\theta) = \begin{bmatrix}\partial_1(f\circ P)(r,\theta) & \partial_2(f\circ P)(r,\theta)\end{bmatrix}$$

i.e. $$\nabla(f\circ P) = \frac{\partial (f\circ P)}{\partial r}\mathbf {\hat r} + \frac{\partial (f\circ P)}{\partial \theta}\mathbf {\hat \theta}$$

If I were to then write this out in more traditional notation, where $f$ and $f\circ P$ are not distinguished, it should look like $$\nabla f = \frac{\partial f}{\partial r}\mathbf {\hat r} + \frac{\partial f}{\partial \theta}\mathbf {\hat \theta}$$

But comparing this to the correct formula for the gradient in polar coordinates, which for reference is usually written as $$\nabla f = \frac{\partial f}{\partial r}\mathbf {\hat r} + \frac 1r\frac{\partial f}{\partial \theta}\mathbf {\hat \theta},$$ I see that I'm missing a factor of $\frac 1r$ on the second term.  Where does that come from?

Edit: BTW, I notice something interesting, though it may have nothing to do with the problem I'm having.  But $$\begin{bmatrix}\partial_1P_1(r,\theta) & \partial_2P_1(r,\theta) \\ \partial_1P_2(r,\theta) & \partial_2P_2(r,\theta)\end{bmatrix} = \begin{bmatrix}\cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r\cos(\theta)\end{bmatrix}$$ would be an orthogonal matrix (in fact, it'd be a rotation) if we multiplied the second column by $\frac 1r$.  But then that's the column that goes into $\partial_2(f\circ P)(r,\theta)$.  So if I normalized that column, then somehow my formula would have come out correctly with the $\frac 1r$.  But I see no reason why I should do that.  Is this just a weird coincidence?

Polar coordinates:
$$
\begin{cases} x = r\cos(\theta) \\ y = r \sin(\theta) \end{cases} \quad \Longrightarrow \quad
\begin{cases} \mathbf {\hat r} = \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix} \\
\mathbf {\hat \theta} = \begin{bmatrix} -\sin(\theta) \\ \cos(\theta) \end{bmatrix} \end{cases}
$$
Chain rules:
$$
\frac{\partial f}{\partial r} =
\frac{\partial f}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r}
= \cos(\theta) \frac{\partial f}{\partial x} + \sin(\theta) \frac{\partial f}{\partial y} \\
\frac{\partial f}{\partial \theta} =
\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}
= - r \sin(\theta) \frac{\partial f}{\partial x} + r \cos(\theta) \frac{\partial f}{\partial y}
$$
Matrix format:
$$
\begin{bmatrix} \Large \frac{\partial f}{\partial r} \\ \Large \frac{1}{r} \frac{\partial f}{\partial \theta} \end{bmatrix} =
\begin{bmatrix} \cos(\theta) & \sin(\theta) \\ - \sin(\theta) & \cos(\theta) \end{bmatrix}
\begin{bmatrix} \Large \frac{\partial f}{\partial x} \\ \Large \frac{\partial f}{\partial y} \end{bmatrix}
$$
Inverse transform:
$$
\begin{bmatrix} \Large \frac{\partial f}{\partial x} \\ \Large \frac{\partial f}{\partial y} \end{bmatrix} =
\begin{bmatrix} \cos(\theta) & - \sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}
\begin{bmatrix} \Large \frac{\partial f}{\partial r} \\ \Large \frac{1}{r} \frac{\partial f}{\partial \theta} \end{bmatrix}
= \frac{\partial f}{\partial r} \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix}
+ \frac{1}{r} \frac{\partial f}{\partial \theta} \begin{bmatrix} - \sin(\theta) \\ \cos(\theta) \end{bmatrix}
$$
Conclusion:
$$
\nabla f = \frac{\partial f}{\partial r}\mathbf {\hat r} + \frac 1r\frac{\partial f}{\partial \theta}\mathbf {\hat \theta}
$$

Surely the extra 1/r term in the expression for $\nabla f$ is because unit vectors are being used. If the base vectors are not normalised to unit vectors then grad f takes the simpler form:
$$
\nabla f = \frac{\partial f}{\partial r}\mathbf {r} + \frac{\partial f}{\partial \theta}\mathbf{\theta}
$$


if $a,b,c$ are positive prove that $a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$ if $a,b,c$ are positive prove that $$a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$$

I have tried to use the  p-q-r method.
let $p=a+b+c,q=ab+bc+ca,r=abc$
we have to prove $$p^2-3q+r+2-p\ge 0$$
as $p^2\ge 3q$ we have to prove $$r+2-p\ge 0$$.
but i don't think this is true. Also i know the following inequalities but i can't figure out how to use them.

$pq\ge 9r$
$2p^3+9r\ge 7pq$
$q^3\ge 27r^2$ etc. etc.

Please note that i am looking for a solution only using pqr.I would be grateful if anyone can explain their logic behind the proof as it is the first time i am using 'pqr' method.

The inequality $r+2-p \geqslant 0$ does not true, example $a = 2, b = 1, c = \frac 12.$
We need to prove
$$p^2-p-3q+r+2 \geqslant 0,$$
equivalent to
$$(p^2+2r+1-4q)+(p^2-2p-2q+3) \geqslant 0.$$
Because
$$p^2-2p-2q+3 = (a-1)^2+(b-1)^2+(c-1)^2 \geqslant 0.$$
It's remain to prove that
$$p^2+2r+1 \geqslant 4q.$$
Using the AM-GM inequality, we have
$$2r+1 \geqslant 3\sqrt[3]{r^2} \geqslant \frac{9r}{p}.$$
Therefore, we will show that
$$p^2+ \frac{9r}{p}\geqslant 4q.$$
Which is Schur inequality.
Note. See here (page 4, Vietnamese).

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$a^2+b^2+c^2+abc+2\geq a+b+c+ab+ac+bc$$ or $f(v^2)\geq0,$ where
$$f(v^2)=9u^2-6v^2+w^3+2-(3u+3v^2)=9u^2-9v^2+w^3+2-3u.$$
But $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $v^2$.
Now, $a$, $b$ and $c$ are roots of the equation $$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$-x^3+3ux^2+w^3=3v^2x.$$
Now, let $u$ and $w^3$ are constants and $v^2$ is changed. In this situation $a$, $b$ and $c$ are changed.
Since the last equation has three roots $a$, $b$ and $c$, we see that a graph of $g(x)=-x^3+3ux^2+w^3$ and the line $y=3v^2x$ have three common points and $v^2$ gets a maximal value for equality case of two numbers between $a$, $b$ and $c$.
Actually, if $a\leq b\leq c$ we can see that it happens for $b=c$.
Indeed, easy to draw a graph of $g$: $(0,w^3)$ it's a minimum point of $g$, $x_{max}=2u$ and the line $y=3v^2x$ goes across the origin.
If $v^2$ increases, so the line $y=3v^2x$ is  rotated and we can see, that $v^2$ may be maximal, if we want that our equation would have three positive roots, when $b=c$.
Id est, it's enough to prove our inequality for equality case of two variables.
Since our inequality is symmetric, it's enough to assume $b=c$ and we obtain a quadratic inequality of $a$ and the rest is smooth, which you can see here: show that $a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$


Minimum value of the integral I am struggling to find the solution to this problem. If anyone could help to explain how to solve this problem to me, it would be really appreciated.

What is the minimum value of the integral? y(x)=?

$\displaystyle\int_0^4 \left[y^2+\left(\frac{dy}{dx}\right)^2\right]\,dx$

$y(0)=0$ and $y(4)=1$

Euler-Lagrange equations: If

$$I = \int_0^4 dx \ F(y,y')$$

then $I$ is stationary with respect to $y$ and $y'$ when

$$\frac{d}{dt} \frac{\partial F}{\partial y'} = \frac{\partial F}{\partial y}$$

In this case, $F(y,y') = y^2+y'^2$.  Then the E-L equation implies that $y''=y$, so that

$$y(x) = A e^x + B e^{-x} $$

Given the boundary conditions, the solution is

$$y(x) = \frac{\sinh{x}}{\sinh{4}} $$

For this solution in $y$ the value of $I$ is $\coth{4}$.  Comparing this with another function that satisfies the BC's, say, $y=x/4$, at which $I=19/12 > \coth{4}$, we may conclude that the solution is a minimum.


How to prove that complex numbers $(\mathbb{C})$ are linear space over real numbers $(\mathbb{R})$ field? How to prove that complex numbers $\mathbb{C}$ are linear space over real numbers field $\mathbb{R}$ ?

Verify that it satisfies the axioms, of course!

The sum of two complex numbers is a complex number.
Multiplying a complex number by a real number yields a complex number.
Addition of complex numbers is associative.
Addition of complex numbers is commutative.
The complex number $0$ is such that $z+0 = z$ for all complex numbers $z$.
Given a complex number $z$, there is a complex number (namely, $-z$) such that $z+(-z) = 0$.
Multiplication by reals numbers is associative: $\alpha(\beta z) = (\alpha\beta)z$ for all $\alpha,\beta\in\mathbb{R}$, $z\in \mathbb{C}$.
Multiplication by reals distributes over the sum of complex numbers: $\alpha(z+w) = \alpha z + \alpha w$ for all $\alpha\in\mathbb{R}$, $z,w\in\mathbb{C}$.
Multiplication by a complex number distributes over the sum of real numbers: $(\alpha+\beta)z = \alpha z + \beta z$ for all $\alpha,\beta\in\mathbb{R}$, $z\in\mathbb{C}$.
Multiplication by the real number $1$ is the identity mapping: $1z = z$ for all $z\in\mathbb{C}$.

If all 10 are true, then $\mathbb{C}$ is a vector space/linear space over $\mathbb{R}$.

(More generally, if $\mathbf{F}$ is a field, and $\mathbf{K}$ is a field that contains $\mathbf{F}$, then $\mathbf{K}$ is a vector space over $\mathbf{F}$).

$\mathbb{C}=\{a+b\mathrm{i}\mid a,b\space \mathrm{from}\space \mathbb{R}\}$  and $u, v$ vectors from $\mathbb{C}$. $u=(a_1+b_1\mathrm{i})$ and $v=(a_2+b_2\mathrm{i})$.
To show that $\mathbb{C}$ is a linear space over $\mathbb{R}$, we can prove it by this:
$u+v=((a_1+b_1\mathrm{i})+(a_2+b_2\mathrm{i}))=(a_1+a_2)+(b_1\mathrm{i}+b_2\mathrm{i})=(a_1+a_2)+(b_1+b_2)i$ so the new vector formed by summary of two other vectors has the form $A+B\mathrm{i}$.
This means that the new vector is part of $\mathbb{C}$, and by this we prove that $\mathbb{C}$ is a linar space over $\mathbb{R}$ field.


Sum of infinite cosine series I'm sure this has been asked before, but I'm not sure how to search for this.

  Evaluate the sum $$\sum_{k=1}^{\infty} \frac{\cos(k)}{2^k}.$$

I thought about expanding $\cos(x)$ as a Taylor series, but that didn't go anywhere. Any hints?

HINT:

$$\cos x =\text{Re}\left(e^{ix}\right)$$

Then, the series of interest is

$$\text{Re}\left(\sum_{k=1}^{\infty}\left(\frac{e^{i}}{2}\right)^k\right)$$

Amusement Time

Firstly I thought about a joke, namely: let's take the series, and white it like:

$$S = \frac{e^i}{2} + \frac{e^{2i}}{4} + \frac{e^{3i}}{8} + \frac{e^{4i}}{16} + \ldots $$

Now:

$$S = \frac{e^i}{2}\cdot\left[1 + \frac{e^i}{2} + \frac{e^{2i}}{4} + \frac{e^{3i}}{8} + \frac{e^{4i}}{16} + \ldots \right] = \frac{e^i}{2}\cdot (1 + S)$$

Then re-arranging the terms to get S:

$$S = \frac{e^i}{2-e^i}$$

Now we have to take the Real part:

$$\Re \frac{e^i}{2 - e^i} = \Re \frac{i\sin(1) + \cos(1)}{2 - i\sin(1) -\cos(1)} = \frac{\cos(1)}{2-\cos(1)} $$

Unfortunately this is not the sum of that series.

But I guess that a similar road shall be undertaken because the sum of that series is (thanks to Mathematica)

$$S = \frac{1 - 2\cos(1)}{4\cos(1) - 5}$$


Graphs being plotted despite being undefined. I've not really understood why this happens. I'm quite new to the idea of limits, etc. But I've seen that when you use a grapher, it plots undefined points. For example in the function $f(x) = \frac{x^2+x-6}{x-2}$, the value $f(2)$ would be undefined, so why does a grapher not show that $f$ is undefined at $2$ and instead carries on through it? It seems to be plotting the $\lim\limits_{x \to 2} \frac{x^2+x-6}{x-2}$.

I hope this makes sense, apologies if it doesn't.

A simple way a computer can make a graph of a function $f$ on an interval $[a,b]$ is to divide $[a,b]$ up in $n$ ascending points $x_1,\dots x_n$ with $x_1=a$ and $x_n=b$ and then draw straight lines between $\left(x_i,f(x_i)\right)$ and $\left(x_{i+1},f(x_{i+1})\right)$. The number $n$ and the ways the partition $x_1,\dots x_n$ is chosen depends on the software and is often adjustable. If I use the following Mathematica code

Plot[(x^2 + x - 6)/(x - 2), {x, 0, 4}]

to plot your function I get the following image.

If I now ask Mathematica which points it used to plot I get a long list of which these are a few:

$$\dots,(1.88078, 4.88078), (1.96018, 4.96018), (2.03801,5.03801), (2.12244, 5.12244),\dots.$$

So we see that the point $x=2$ wasn't sampled and as a result we might as well have asked it to plot $f(x)=x+3$. The shortcomings of this naive way of plotting are felt stronger when we for instance plot $f(x)=\tan(x)$. The code

Plot[Tan[x], {x, 0, Pi}]

produces the following image.

The software does not know that the function is not defined at $x=\frac{1}{2}\pi$. Sophisticated software often knows how or has a way to deal with this. In Mathematica this

Plot[Tan[x], {x, 0, Pi}, Exclusions -> {Cos[x] == 0}]

produces

That graph is wrong. There should be a hole at $x=2$, as you correctly guessed. The correct plot are two lines with arrows at the ends where they touch the point $(2,5)$.


Multiplicative group of prime order greater than 2 For any multiplicative group $\mathbb{Z}_n$, the order is $\phi(n)$. We know that $\phi(n)$ is always even for $n\ge3$.

Then how can I get a prime order multiplicative group.

Is the order of a multiplicative group means the multiplicatiive order of the generator element in that group?

In Builinear pairing, it is given that $e: \mathbb{G} \times \mathbb{G}_T$ , where both are of same prime order.

Please provide an example for $\mathbb{G}$ and $\mathbb{G}_T$.

I think you're confusing two senses of the phrase "multiplicative group". Often that phrase is used to mean any group, where the group operation is written using product notation: the identity is written $1$, inverses are written $x^{-1}$, products are written using concatenation $xy$, and powers are written using superscripted exponents like $xx=x^2$.

One particular kind of multiplicative group is the group of units modulo $n$, written $(\mathbb Z/n\mathbb Z)^*$. As you note, for $n\geq 3$, the order of this group is even. When discussing this group, one often emphasizes the word multiplicative to distinguish it from the additive group $\mathbb Z/n\mathbb Z$.
There are many other groups that are written multiplicatively. These include the multiplicative group of complex numbers $\mathbb C^*$ and its subgroups. As Adelafif notes in the comments, the group of third roots of unity is such a group of order $3$.

When a group is cyclic, yes, its order is the same as the order of any of its generator elements. When a group is not cyclic, it doesn't have a generator element, so of course this doesn't apply.

$Z_n$ is not a multiplicative group, inverse does not exist for 0. The order is not \phi(n), please dont get confused, multiplicative group means "a set together with multiplication operation becomes group".for example $U(n)=\{x \in N | x < n$ and $gcd(x,n)=1\}$ is a multiplicative group.


How much work is done to pump all of the water out of the outlet pipe? A water tank has the shape of a cylinder with base radius 2 meters and height 6 meters.
The tank is positioned vertically, so that its circular base is on the ground, and it has an outlet pipe
at the top. It is half full of water. How much work is done to pump all of the water out of the
outlet pipe? (Recall that the acceleration due to gravity is g = 9.8 meters per second per second
and the mass density of water is ρ = 1000 kilograms per cubic meter.) Your answer should be in
Newton-meters. Feel free to leave your answer in terms of g and ρ.

My attempt

Since, the water is only half-filled, I took the integral from $0$ to $3$.

$\int _{ 0 }^{ 3 }{ (1000)(9.8)(4\Pi )(6-y)dy } $

Can anyone verify this answer

Since the tank has a uniform cross section, you can also get the answer (or check your integration) by thinking about how far you need to move the center-of-mass.

The tank is full from 0-3 meters, so the center of mass is as 1.5 m and needs to be moved up 4.5 meters.

$$W = mg\Delta h = (1000*4\pi*3)*9.8*4.5$$

It's no mistake that $$3*4.5 = \int_0^3 (6-y)dy$$ which is the only way in which the two methods differ.

Your calculation is correct. But note that you can empty the tank with almost zero work using a thin hose of length $>12$ meters. Throw this hose over the rim of the tank such that it falls to level $0$ on both sides. Then use suction on the outside until the hose is full of water. Finally just let the water spill on the ground outside until the tank is empty.


correspondence between meromorphic functions on complex torus vs. $\mathbb{C}$ itself I have an integration to do along the path which is the border of a parallelogram,
namely, I need to show using integration the sum of the orders on the complex torus is $0$.

I reparameterized it correctly, and solved the problem, by showing that $\gamma_1,\gamma_3$ cancel and so do the other two when orientation is chosen correctly. Letting $\omega_1,\omega_2$ be my linearly independent elements of $\mathbb{C}$, I need to show that I can shift my problem over to $\mathbb{C}$ as we have not learned integration on Riemann Surfaces yet, then bring it back to the torus. In doing so I concluded that there's a 1-1 correspondence between meromorphic functions defined on the complex torus and meromorphic defined on $\mathbb{C}$ itself.

Let

$f: \mathbb{C} / L \rightarrow \mathbb{C}$

be meromorphic, this corresponds to

$F: \mathbb{C} \rightarrow \mathbb{C}$

which is meromorphic and doubly periodic, that is $F(\omega_i+z)=F(\omega)$ for $i=1,2$

and is defined via $F(z)=f(z)$ for when $z \in \mathbb{C} / L$, but what do I defined $F(z)$ to be for when $z \in L$?

Then I run the integration on $F$ instead of $f$ which is the same exact work shown, then I conclude by saying that

$0=\sum_{z \in \mathbb{C}}$ord$_z(F)$ = $\sum_{p \in X}$ord$_p(f)$

Any help or hint on my correspondence between $F$ and $f$? Since if $z \in L$, we are $0$ mod $L$, do you send $F(z)$ to $0$? I wanna say if $z \in L$ we just map $F(z)$ to 0.

Your question is unclear. With $h$ meromorphic $1,t$ periodic with no zeros/poles on $a+n+mt+\Bbb{R},a+n+mt+\Bbb{R}t$ then $$\int_{[a,a+1]\cup[a+1,a+1+t]\cup[a+1+t,a+t]\cup[a+t,a]}h(z)dz$$ $$=\int_{[a,a+1]}-\int_{[a,a+1]}+\int_{[a,a+t]}-\int_{[a,a+t]}h(z)dz=0$$
When doing so with $h(z)=f'(z)/f(z)$  we get that $f$  meromorphic $1,t$ periodic
 has the same number of zeros/poles on the fundamental paralellogram.


Is there a constructive discontinuous exponential function? It is well-known that the only continuous functions $f\colon\mathbb R\to\mathbb R^+$ satisfying $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb R$ are the familiar exponential functions.  (Prove $f(x)=f(1)^x$ successively for integers $x$, rationals $x$, and then use continuity to get all reals.)

The usual example to show that the identity $f(x+y)=f(x)f(y)$ alone doesn't characterize the exponentials requires the axiom of choice.  (Define $f$ arbitrarily on the elements of a Hamel basis for $\mathbb R$ over $\mathbb Q$, then extend to satisfy the identity.)

Is there an explicit construction of a discontinuous function satisfying the identity?  On the other hand, does the existence of such a function imply the axiom of choice or some relative?

Let $g(x)=\ln f(x)$, so that $g(x+y)=g(x)+g(y)$. The solutions to this equation are precisely the ring homomorphisms from $\mathbb R\to\mathbb R$ and they are in bijection with solutions to your original equation. Since $\mathbb R$ is an infinite dimensional $\mathbb Q$-vector space, there are infinitely many such ring homomorphisms. They are determined by a Hamel basis for $\mathbb R$.

Does this answer your question?


Prove that $\gcd(a, b) = \gcd(b, a-b)$ I can understand the concept that $\gcd(a, b) = \gcd(b, r)$, where $a = bq + r$, which is grounded from the fact that $\gcd(a, b) = \gcd(b, a-b)$, but I have no intuition for the latter.

Let $d = \text{gcd}(a,b)$, then $d|a$, and $d|b$, so $d|(a-b)$, and $d|\mbox {gcd}(b,a-b)= d'$. Also $d'|b$, and $d'|(a-b)$, so $d'|(b+(a-b)) = a$, and $d'|\mbox {gcd}(a,b) = d$. So $d = d'$.

First of all,

definition: $b|a \iff \exists k \in \mathbb{Z}: a = k \cdot b$
Lemma 1: $a|b \wedge a|b \Rightarrow a = \pm b$
Lemma 2: $d|a \wedge d|b \Rightarrow d|(ax + by)$, for any $x,y \in \mathbb{Z}$
Theorem 1: If a and b are any integers, not both zero, then $gcd(a,b)$ is the smallest positive element of the set $\{ax + by: x,y \in \mathbb{Z}\}$ of linear combinations of a and b.

Thus, we need to prove $gcd(,)|gcd(,−)$ and $gcd(,−)|gcd(,)$, and as $gcd$ is always positive, they must be equal by Lemma 1.

1. $gcd(,)|gcd(,−)$

Let $d = gcd(,)$, thus $d|a \wedge d|b$ and, also, $d|(a-b)$ as it is a linear combination of $a$ and $b$ (Lemma 2).

$d|b$, $d|(a-b)$, and $gcd(,−)$ is a linear combination of $b$ and $a-b$ (Theorem 1). Therefore, $d|gcd(,−)$ by Lemma 2.

2. $gcd(,−)|gcd(,)$

Let $d = gcd(,−)$, thus $d|b \wedge d|(a-b)$.

$a = b + (a - b)$, thus $a$ is a linear combination of $b$ and $a - b$.

Therefore, $d|a$ by Lemma 2.

Again, $gcd(a,b)$ is a linear combination of $a$ and $b$ by Theorem 1, thus $d|gcd(a,b)$.

Finally, using Lemma 1:

$gcd(,)|gcd(,−) \wedge gcd(,−)|gcd(,) \Rightarrow gcd(,) = gcd(,−)$

  Note that if $a = b = 0$, then $gcd(a,b) = gcd(b, a - b) = gcd(0,0) = 0$ by definition.

Proof of lemma 1

By definition, $b|a \Rightarrow |b| \leq |a|$. Therefore, $a|b \wedge b|a \Rightarrow |a| = |b| \Rightarrow a = \pm b$.

Proof of lemma 2

As $d|a \wedge d|b$, there are some integers $k_1$ and $k_2$ which meets $\frac{a}{d} = k_1$ and $\frac{b}{d} = k_2$.

Now, we need to prove $(ax + by) = k \cdot d$, for some $k \in \mathbb{Z}$.

$(ax + by) = k \cdot d \iff k = \frac{a}{d}x + \frac{b}{d}y = k_1x + k_2y$

As the multiplication of two integers is an integer, and this also follows for the sum of two integers, thus $k_1x + k_2y \in \mathbb{Z}$.

Proof of Theorem 1

This is proved at: Cormen, T.H., Leiserson, C.E., Rivest, R.L., Stein, C., Introduction to algorithms, MIT Press, 3a.
ed., 2009.

page 930, Theorem 31.2.


Formula for calculating the total interest payable over the life of a loan wondering if someone can help a non-mathematician out.

I am looking for the formula for calculating the total interest payable over the life of a loan.

Given that we know:

P: Principal (amount) of loan

R: The monthly repayments on the loan

T: The term of the loan (i.e. the number of repayments)

r: The annual interest rate

Assume that interest is accrued monthly and that repayments are made in arrears (at the end of each loan period).

If you know the monthly payment and the number of payments, the total of payments is $RT$, so the total interest you pay is $RT-P$.  The fact that the interest is accrued monthly and payments are made in arrears only go into calculating the payment from the interest rate, which has already been done for us.

You said Given that we know:
P: Principal (amount) of loan
R: The monthly repayments on the loan
T: The term of the loan (i.e. the number of repayments)
r: The annual interest rate

On a \$50,000 loan at an interest rate of 8% for 5 years, then your monthly payment is \$1013.82 each month for 60 months.

You then multiply \$1013.82 x 60 payments to determine you will pay \$60,829.20 over the life of the loan.

By subtracting your original loan amount of \$50,000 (P) from \$60,829.20 (RT) you will see that over the life of the loan you will pay \$10,829.20 in total interest.


$I_n=\frac{1}{\pi}-\frac{n(n-1)}{\pi^2}I_{n-2}$ I have to calculate $I_n=\int_0^1 x^n \sin{(\pi x)} dx$.
By double partial integration I got $I_n=\frac{1}{\pi}-\frac{n(n-1)}{\pi^2}I_{n-2}$ and $I_0=\frac{2}{\pi}$, $I_1=\frac{1}{\pi}$.
How to solve this recurrence, in order to find $\lim_{n\to\infty} I_n$ and to deduce convergence of $\sum_{n=1}^\infty I_n$?
Any help is welcome. Thanks in advance.

$$(\forall n\in \Bbb N)\;\;\big|I_n\big|=\left|\int_0^1x^n\sin(\pi x)dx\right|$$
$$\le \int_0^1\big|x^n\sin(\pi x)\big|dx$$
$$\le \int_0^1x^ndx=\frac{1}{n+1}$$
$$\implies \lim_{n\to+\infty}I_n=0$$
The recursive relation you got by double partial integration, can also be written as
$$\boxed{I_{n+2}\pi^2=\color{red}{\pi}-(n+2)(n+1)I_n.}$$
$$\lim_{n\to+\infty}I_{n+2}=0\implies$$
$$\implies \lim_{n\to+\infty}(n+2)(n+1)I_n=\color{red}{\pi}$$
$$\implies I_n\sim \frac{\pi}{(n+2)(n+1)}\sim \frac{\pi}{n^2}$$
$$\implies \sum I_n \text{ converges}$$

$\;\sum\limits_{n=1}^\infty I_n=\text{Si}(\pi)-\dfrac2\pi\;$
and as far as I know there is not a simplier  way to write the sum of your series.


What is wrong with my attempt of $0 \lt w \lt \frac{\pi}{6}$ then $w \csc w$ is? For $0 \lt w \lt \frac{\pi}{6}$,  the value of  $ w  \csc w $ is
A) less than $ \frac{\pi}{6}$
B) greater than $ \frac{\pi}{6}$
C) less than $\frac{\pi}{3}$
D) greater than $ \frac{\pi}{3}$

I know from L'hospital rule that $ \lim_{h\to 0} \dfrac{h}{\sin h} = 1 $
and $\sin \frac{\pi}{6} = 0.5 $, thus
$ w  \csc w $ goes from $1$ to $\frac{\pi}{3}$. (Open interval)
Thus answer should be (B and C) greater than $\frac{\pi}{6}$ and less than $\frac{\pi}{3}$.
But the answer is only C.

Per the comments, this appears to be a case of looking for the "best answer," whatever that this. If this was given to you by an instructor rather than taken from a textbook, you may wish to bring it up with them.


What is the principal cubic root of $-8$? According to my book it should be a real number, and according to WolframAlpha it should be $1+1.73i$

What is the correct answer?

The cubic roots of $-8$ are:
$$
2e^{i\pi/3}=2(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}) \qquad 2e^{i\pi}=-2 \qquad 2e^{i2\pi/3}=2(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2})
$$

it seems that WolframAlpha consider as principal root the root with the minimum value of the argument, i.e $e^{i\pi/3}$, but usually, if there is a real root this is considered the principal root.

Consider $x^3 =  - 8 $. It "seems" we can just say $x = -2$ because $(- 2)^3 = -8$ but $- 2$ is not the official Primary Solution!
Using De Moivre's theorem again:
$$x^3 =  - 8$$
$$( r \operatorname{cis}(\theta) )^3 = -8$$
$$r^3 \operatorname{cis}(3\theta) = 8\operatorname{cis}( 180^\circ + 360^\circ n)$$
$$r^3  = 8 \text{ and } 3\theta = 180^\circ + 360^\circ n $$
$$r = 2 \text{ and } \theta = 60^\circ + 120^\circ n = 60^\circ, 180^\circ, 300^\circ $$
$$x_1  =  2\operatorname{cis}( 60^\circ ) = 1 + i\sqrt3$$
$$x_2  = 2\operatorname{cis}(180^\circ) =  - 2$$
$$x_3  = 2\operatorname{cis}(240^\circ) =  1 - i\sqrt3$$
The Primary Solution is $x_1 = 1 + i\sqrt3   ≈ 1 + 1.732i$
So the cube root of $-8$ or $(-8)^{\frac13}=1+1.732i$ and NOT $-2$.


How to apply law of large numbers for this problem Let $X_1,...,X_n,...$ be independent variable satisfying $P(X_i=0)=P(X_i=1)=\frac{1}{2}$ for all i then denote $Z_i=X_iX_{i+1}$ for all i .I want to show that $\lim_{n\to \infty}\frac{Z_1+Z_2+...+Z_n}{n}=\frac{1}{4}$ a.s by using law of large number. I'm new to this area and struck immediately since $\{Z_i\}$is not independent.
Perhaps Borel-Cantelli lemma can be used here?

really thanks for your help

Hint: $Z_n$'s are not independent but

what about $Z_{2n}$'s or $Z_{2n+1}'s$?

I am not quite sure the answer is 1/4 considering $Z_n$'s are not independent. I think it is 1/2

Got the idea from here, w/c involves Borel-Cantelli Lemmas, but I don't think you need those.


If $a+b=2$ so $2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$ Let $a$ and $b$ be non-negative numbers such that $a+b=2$. Prove that:
  $$2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$$

In the left inequality the equality occurs also for $a=2$ and $b=0$, which makes additional problems.

For $b=2-a$ we get inequalities of one variable, but derivatives are very ugly and I think it can't help (I mean here $f'(x)=0$...).

Since the equality occurs for $a=b=1$, we can try the Taylor series, but I  failed in this way.

Thank you!

It's easy to prove
$$2+\frac{ab(a-b)^2}{4}\leq 2+\frac{(a-b)^2}{4}$$
AS $$0\leq a,b \leq 2$$
Since by AM-GM inequality
$$ab \leq 1$$

So above inequality is true.
For the rest of the inequality this graph proves everything....

Here $a=2-b$


If $a|(b+c)$ and $\gcd(b,c)=1$, prove that $\gcd(a,b)=1$ and $\gcd(a,c)=1$. If $a|(b+c)$ and $\gcd(b,c)=1$, prove that $\gcd(a,b)=1$ and $\gcd(a,c)=1$.

I started with:
Suppose $a|(b+c)$ and $\gcd(b,c)=1$.
This means that $ak=b+c$, for some integer $k$.
And $1|b$ and $1|c$.
I know I can solve this using the theorem that if $\gcd(a,b)=1$ then there exists integers $u$ and $v$ such that $au+bv=1$.
But I was wondering how else I can solve this, without using this theorem.

The fact that $\gcd(b,c)=1$ implies there are $k,l \in \Bbb Z$ such that $kb + lc= 1$ (Bézout identity). $a|(b+c)$ tells us $ma = b+c$ for some $m \in \Bbb Z$.
Can you find a Bezout identity for the pairs $(a,b)$ and $(a,c)$ from that?

One thing I like to do is figure if $k$ is a common (positive) divisor of $a$ and $b$, and $a|b+c$ then $k|b+c$.  But we also have $k|b$ so $k| (b+c)-b = c$.  So $k$ is a common (positive) divisor of $b,c$.  But $\gcd(b,c) = 1$ so the only common (positive) divisor of $b,c$ is one so $k=1$.  So the only common divisor of $a$ and $b$ is $1$.  So $\gcd(a,b)=1$.
And doing similar to prove $\gcd(a,c) =1$ is nearly entirely the same.  If $d$ is a common divisor of $a$ and $c$ then $d|b+c$ and $(b+c)-c=b$ and so $d$ is a common divisor of $b$ and $c$ which are relatively prime.
....
Or maybe even easier: $a|b+c$.  So $\gcd(a,b)|a$ so $\gcd(a,b)|b+c$ but $\gcd(a,b)|b$ so $\gcd(a,b)|(b+c)-b=c$.  So $\gcd(a,b)|b$ and $\gcd(a,b)|c$ so $\gcd(a,b)|\gcd(b,c)=1$.  So $\gcd(a,b) = 1$.


Proof of replacement rule in conditional probability The answer to this question gives a replacement rule for conditional probability. But how do you prove this? I tried integrating both sides w.r.t. $P_X$ and fiddling around but it didn't get me anywhere.

  Rule: Let $X$ and $Y$ be as above random variables, and let $\xi:\mathbb{R}^2\to\mathbb{R}$ be $(\mathcal{B}(\mathbb{R}^2),\mathcal{B}(\mathbb{R}))$-measurable. Then
  $$
P(\xi(X,Y)\in D\mid X=x)=P(\xi(x,Y)\in D\mid X=x),\quad D\in\mathcal{B}(\mathbb{R}),
$$
  holds for $P_X$-a.a. $x$. This is saying that "conditional on $X=x$ we may replace $X$ by $x$".

[EDIT]

Managed to prove this without $Y$, for measurable $g$ we have: $\int_C \mathbf{P} (g (X) \in B \mid X = x) P_X \left( \text{d} x \right)
=\mathbf{P} (g (X) \in B, X \in C) = \int_C {\bf 1}_{g (x) \in B} P_X
\left( \text{d} x \right)$. Also $\int_C \mathbf{P} (g (x) \in B \mid X = x)
P_X \left( \text{d} x \right) = \int_C {\bf 1}_{g (x) \in B} P_X \left(
\text{d} x \right)$.

But this no longer works with $Y$ as $g(X,Y)$ is no longer a function of $X$.

[EDIT2] Many thanks to Stefan Hansen who posted an answer here to the same question.

Probability of an event is the expectation of an indicator function that the event occurs, and
$\mathsf E(g(X,Y)\mid X=x) = \mathsf E(g(x,Y)\mid X=x)$

Thus:
$$\begin{align}
\mathsf P(\xi(X,Y)\in D\mid X=x)
 & =\mathsf E(\mathbf 1_{\xi(X,Y)\in D}\mid X=x)
\\ & = \mathsf E(\mathbf 1_{\xi(x,Y)\in D}\mid X=x)
\\ & = \mathsf P(\xi(x, Y)\in D\mid X=x)
\end{align}$$


Variance and Expected value of internet connection I am working on a probability/statistics problem!

The problem is as follows:

Your internet connection is very poor. It constantly alternates between being functional for x minutes and being down for y minutes. If you try to check your email at a random time, how long do you have to wait, on average, to get internet connection? What is variance of this waiting time?

I don't even know how I would approach this problem because the random time of x and y minutes do not have a probability that I would need to compute the expected value (which is summation of probability * value). I guess these are expected value of continuous variables, but then again I am very stuck.

Could someone guide me through this problem? thanks!

What is the probability that you try to check your email during the first $x$ minutes of the $(x+y)$-minute cycle?  If you do that, then your waiting time is $0$.

What is the probability that you try to check your email when more than $w$ minutes, but fewer than $y$ minutes, remain of the $(x+y)$-minute cycle, where $0<w<y$?  In that case, you need to wait more than $w$ minutes.

This gives you a cumulative probability distribution function of the waiting time.


If the normalizer of a subgroup in a group is equal to the subgroup then is the subgroup abelian? If $H$ is a proper subgroup of $G$ such that $H=N_G(H)$ ( the normalizer of $H$ in $G$ ) , then is it true that $H$ is abelian ?

Let $A \ast B$ be any non-trivial free product, that is $A,B \neq \{1\}$. Then the normalizer of $A$ (resp. $B$) is $A$ (resp. $B$). However, there is no restriction on the groups $A$ or $B$.

Here is a hint: Consider $H=S_{3}$ and $G=S_{4}$. Then (why?) $N_{G}(H)=H$. The previous hint that I posted was completely wrong. Here is a better hint.


Stars and bars to find "how many $x$ digit numbers are there with sum of digits $y$"? This question poses a seemingly very simple method to solve problems of the sort "how many $x$ digit numbers are there with sum of digits $y$?", but I don't understand it. Why are the "bad" solutions in correspondence to the solutions of $y_1 + 10 + x_2 + x_3 + x_4 + x_5 = 23$? What's the idea?

For instance, what "correct solution" corresponds to taking $y_1=13$? How to solve this problem?

The referred question asks for the number of non-negative integer solutions of
\begin{align*}
  x_1+x_2+x_3+x_4+x_5=23
  \end{align*}
with only one additional restriction $0\leq x_1\leq 9$. In this case we do not need the inclusion-exclusion principle.

  In order to determine all integer solutions with
  \begin{align*}
  x_1+x_2+x_3+x_4+x_5=23\qquad\qquad0\leq x_1\leq 9,0\leq x_2,x_3,x_4,x_5\tag{1}
  \end{align*}

  we look at all integer solutions of

  \begin{align*}
  x_1+x_2+x_3+x_4+x_5=23\qquad\qquad\qquad\qquad 0\leq x_1,x_2,x_3,x_4,x_5\tag{2}
  \end{align*}

  and subtract all solutions of

  \begin{align*}
  x_1+x_2+x_3+x_4+x_5=23\qquad\qquad\qquad x_1\geq 10, 0\leq x_2,x_3,x_4,x_5\tag{3}
  \end{align*}

The solutions of (3) are the so-called  bad solutions in the referred question, meaning the invalid solutions which are to subtract when determining the solutions of (2).

  In order to calculate (2) with the nice range $x_1,x_2,x_3,x_4,x_5\geq 0$ we can use the stars-and-bars
  technique and obtain
  \begin{align*}
  \binom{23+4}{4}=\binom{27}{4}=17550
  \end{align*}

In order to calculate (3) we transform the range by a proper substitution which enables us to apply the stars-and-bars technique again. Instead of $x_1\geq 10$, we substitute $x_1=y_1+10$ and the range $x_1\geq 10$ can then be transformed to $y_1+10\geq 10$ or equivalently $y_1\geq 0$.

  This way we obtain from (3)

  \begin{align*}
  x_1+x_2+x_3+x_4+x_5&=23\qquad\qquad\qquad x_1\geq 10, 0\leq x_2,x_3,x_4,x_5\\
  (y_1+10)+x_2+x_3+x_4+x_5&=23\qquad\qquad\qquad 0\leq y_1,x_2,x_3,x_4,x_5\\
  y_1+x_2+x_3+x_4+x_5&=13
\end{align*}

  We are now in the same situation as in (2) and can apply the stars-and-bars technique again.

  \begin{align*}
  \binom{13+4}{4}=\binom{17}{4}=2380
  \end{align*}

$$ $$

  We finally obtain the number of wanted solutions as the total number of solutions of (2) minus the number of solutions of (3)
  \begin{align*}
  \binom{27}{4}-\binom{17}{4}=17550-2380=15170
  \end{align*}

I'm also not great at combinatorics, but I'll give this a shot to give you a start. This isn't going to be an answer, but an extended comment.

First, we need to establish the number of nonnegative numbers satisfying \begin{equation}\sum_{i=1}^5 x_i=23\end{equation} (call this (*)). This is a standard stars and bars type problem, where we have $23+(5-1)=27$ stars and $5-1=4$ bars. But we've overcounted, because some of these solutions will have some of the numbers being greater than or equal to 10, which should not be allowed because we wants a 5 digit number. For this, we can use the inclusion-exclusion principle (we want the second quoted equation).

Let $A$ be the set of solutions to (*) where none of the integers is greater than or equal to 10, $B$ the set of solutions where 1 integer is greater than or equal to 10, and $C$ where two of the integers are greater than or equal to 10. After this, we're done, because at most 2 of the $x_i$ in the solutions we counted can be greater or equal to $10$. Then
\begin{align} |A\cup B\cup C|= |A| + |B| + |C| - |B\cap C|\end{align} (I excluded the parts of the sum that were empty). We want to know $|A|$, we know $|A\cup B \cup C|$, so we need to compute $|B|,|C|,|B\cap C|$. How one would compute $|B|,|C|$ is given, more or less, in the answer I quoted in the comments.

How one would compute $|B\cap C|$, I'm not sure off the top of my head. We also need to exclude the solutions where $x_1=0$. In the quoted method, the number is small enough to count by hand. For this case we would need a more robust method.

Edit: See also here for ideas. I'm out the door, but best of luck to you!

Edit 2: Back, and I messed up a little in haste. $B\cap C$ is obviously equal to the number of solutions where exactly 2 numbers are $\geq 10$. So we can just take $|A\cup B\cup C|=|A|+|B|+|C|$, where $A$ is as before, $B$ is the number of solutions with exactly one number $\geq 10$, and $|C|$ the number of solutions with exactly two numbers $\geq 10$. Then we'll need to refine by excluding the case where $x_1=0$, this can be done with the trick $x_1'=x_1-1$, though some care will need to be taken.


In the triangle, AB = 4, AC = 5 and BC = 7. Calculate angle A to the nearest degree/ In the triangle, AB = 4, AC = 5 and BC = 7. Calculate angle A to the nearest degree/

Hint You can either use the cosine formula and find the $\cos$ of $A$;or use the Heron formula of the area and find the $\sin$.

Note that $\angle BAC=cos^{-1}(\frac{AC^2+AB^2-BC^2}{2 \cdot AC \cdot AB})$. You can use a calculator to get $\boxed{102}$ as the answer. It's about $101.537$
Hope this helped!


Proper map on $n$-sphere Proper functions are defined over topological spaces.
For functions $f:\mathbb{R}^d\rightarrow \mathbb{R}^d$ the conditions "simplify" to $\lim_{x\rightarrow \infty}|f(x)|=\infty$.
Are similar "simplications" know for functions over the n-sphere $S^{d}=\{x\in\mathbb{R}^{d+1} \mid \|x\|=1\}$, i.e., $f:S^{d}\rightarrow S^{d}$?

A function $f$ from $\mathbb{S}^d$ to itself is proper iff it is
continuous.

Note that
$\mathbb{S}^d$ is compact and so every closed subset of it is compact and viceversa

Continuous$\rightarrow$ proper.
Every continuous function $f:\mathbb{S}^d\to \mathbb{S}^d$ is proper, since $f^{-1}$ sends closed (=compact) sets to closed (=compact) sets.

Proper $\rightarrow$ continuous. Every proper function $f:\mathbb{S}^d\to \mathbb{S}^d$ is such that $f^{-1}$ sends compact (=closed) sets to compact (=closed) sets. This is one of the possible definition of continuity

The result is true in a more general form: given an Hausdorff compact space $X$, $f:X\to X$ is proper iff it is continuous,

$f: S^d \to S^d $ is proper if an only if it is continuous.


Basic probability - either event occurs but not both I'm taking a graduate course in probability and statistics using Larsen and Marx, 4th edition and I'm struggling with a seemingly basic question.

If A and B are any two events, not mutually exclusive: $$P((A \cup B) ^\complement) = 0.6, P(A \cap B) = 0.2$$

What is the probability that A or B occurs, but not both?  Or in other words:  $$P((A \cap B)\ ^\complement \cap (A \cup B)) = ?$$

So far, I've been able to infer the following: $$ P(A \cup B) = 1 - P((A \cup B) ^\complement ) = 1 - 0.6 = 0.4 $$ and $$P((A \cap B) ^\complement) = 1 - P(A \cap B) = 1 - 0.2 = 0.8$$

Can someone kindly give a hint as to how to approach from here?  Am I heading in the right direction?  How should I think about these types of problems in general?  I feel like the text gives you the basic set of axioms to define things but then neglects the showing  of solutions for slightly more complicated examples of compound probability equations such as the above.

Hint: Find the probability of $A$ only plus the probability of $B$ only.

Let $x= P(A~ \text{only})$, $y=P(B ~\text{only})$. Then $x +y + 0.2 =0.4$. Thus $x+y= 0.2$

$P((A \cup B)\setminus(A \cap B)) = P(A\setminus B) + P(A\setminus B) = P(A) + P(B) − 2P(A \cap B)$
  you will get the answer!


How to find the maximum of the value $\frac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$ Find the maximun of the value
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$$
I use wolframpha this found this maximum is $\dfrac{4\sqrt{2}}{5}$,But How to prove  and how to find this value?(without derivative)
idea 1
let $\tan{\dfrac{x}{2}}=t$,then we have
$$f=\dfrac{(t+1)^2}{\sqrt{t^4+2t^3+6t^2+2t+5}}$$
Therefore,it suffices to prove that
$$\dfrac{(t+1)^4}{t^4+2t^3+6t^2+2t+5}\le\dfrac{32}{25}$$
or$$32(t^4+2t^3+6t^2+2t+5)-25(t+1)^4\ge 0$$
or
$$ (t-3)^2(7t^2+6t+15)\ge 0$$
But this method if we without derivative,we don't known the maximum is $\dfrac{4\sqrt{2}}{5}$.
idea 2
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{(\sin{x}+1)+2(\cos{x}+1)}}$$
Let $u=\sin{x}+1,v=\cos{x}+1$,then $(u-1)^2+(v-1)^2=1$,find the maximum of the
$$\dfrac{u}{\sqrt{u+2v}}$$

Change variables so that $\sin(x)$ is $x$ and $\cos(x)$ is $\sqrt{1-x^2}$. The range of $x$ is $[0,1]$, of course.
Square the function to get rid of the radical, so that now your function is:
$(1+x)^2/(3+x+2\sqrt{1-x^2})$.

Try to prove by direct manipulation that $(1+x)^2/(3+x+2\sqrt{1-x^2})]\leq1$ in $[0,1]$ , i.e. get all but the radical on one side, then get rid of the radical and arrive at a polynomial inequality. You will be able to factor it and show that it is strictly negative.

Hint:

Assume that $f(x)\leq 4\sqrt2/5$

Now cross multiply, do squares on both sides and ultimately if it's the maximum, you will get an inequality where the equality condition can hold, then write the whole sum backwards!


How do I reverse the order of integration for this double integral? $$\int_0^{\pi/2}\int_{\sin(x)}^2dydx$$ This is the integral in question. My closest was $$\int_0^{2}\int^{\sin^{-1}(y)}_2dxdy$$
but I believe this is wrong.
Sorry for the bad formatting, It's my first time using it properly

As saulspatz pinted out in the comment, you will have to split the integral into two if you switch the order:
$$
\int_0^1\int_0^{\sin^{-1}(y)}dxdy+\int_{1}^2\int_0^{\pi/2}dxdy
$$


How to win at roulette? I know that the answer to my attention-grabbing question is : "You can't win at roulette, it's a negative Expected Value game".

Yes, you're right, long term speaking.

Let's imagine a medium-short term situation (just an after-dinner at the casinò) where there will be <100 bets.

What's the best strategy to use to win some money?

I imagine a hypothetical formula that considers our "money target" and percentage of success as inversely proportional (the more we want to win the less likely we are going to succeed).

What do you think? Are those "red-black double if lose" systems useful?

If you want to maximise your probability of increasing your fortune to some target amount of money $\$a$ (greater than the amount you already have) then the correct strategy is the "bold strategy". If you have an amount of money $\$b$, then you should bet $\min\left(a-b,b\right)$ (i.e. bet everything you have unless you are near to your target, in which case bet just what you need to reach the target). This is because roulette is a negative expectation game, every time you bet you are on average going to lose money. So you want to achieve your target as quickly as you can. The Martingale strategy (double-if-you-lose) is bad because you make lots of bets, each one of which is negative expectation.

The bold strategy is proved to be optimal in this paper: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC223086/pdf/pnas00211-0067.pdf.

Of course, while this strategy does maximise the chance of hitting the target, it also has a large probability of losing all your money. In real life, "the only winning move is not to play".

I think, best strategy is this:

Choose an event with double choice, let's say red/black.

Bet in this way: always on the same color, doubling bet at each round if you do not win.

I mean:

first round: bet 1 on red (total loss: 1, win: 2)
second round: bet 2 on red (total loss: 3, win: 4)
3rd round: bet 4 on red (total loss: 7, win: 8)
....

When you win, you start again from beting 1.

At the n-round you can loose only if $n$ not-red (black  or zeros) are consecutively drawn: $P("n-not-red") =(\frac{19}{37})^n$ , so with $n = 5$ round you have 96% of probability of win 1 (but also 4% probability of loose 31)


Which of the statements are true if $ABCDE = I$? Let $A,B,C,D,E$ be five real square matrices of the same order such that $ABCDE = I$,
where $I$ is the unit matrix. Then which of the following are true?

(A) $B^{−1}A^{−1}= EDC$

(B) $BA$ is a nonsingular matrix

(C) $ABC$ commutes with $DE$

(D) $ABCD = {1\over detE}$ Adj $E$.

Hint: the multiplicativity of the determinant implies that all $5 $ mattices are invertible,  now it easily follows that (B), (C) and (D) are true. About (A), you know that $ B^{-1} A^{-1}=(AB)^{-1}=CDE $, so just find an example where $ CDE\neq EDC $ and you'll have a counterexample to (A).

Option B and D is correct. Note that $det(AB)=det(A)det(B)$. Take determinant of both side of $ABCDE=I$. Since $det(I)=1\neq0$, none of the matrix in left hand side is singular. So does $BA$

For Option D, Since $ABCDE=I$, $ABCDEE^{-1}=IE^{-1}=E^{-1}=Adj(E)/det(E)$. It is correct also (since E is non singular)

For Option A, $ABCDE=I$, $$A^{-1}ABCDE=A^{-1}I=A^{-1}$$$$BCDE=A^{-1}$$ $$B^{-1}BCDE=B^{-1}A^{-1}$$$$B^{-1}A^{-1}=CDE\neq EDC$$

EDIT: Option C is is also correct. Note that $(AB)^{-1}=B^{-1}A^{-1}$ and $(AB)C=A(BC)$. From DKal's answer, one knows that $DE=(ABC)^{-1}$. So $ABC(ABC)^{-1}=(ABC)^{-1}ABC$


Actually
$\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}=$
$=\pi+\tan^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}}\right)$
We can notice that
$\frac{\pi}{2}<\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}<\pi$.
I am going to prove that
if $\frac{\pi}{2}<\alpha+\beta<\frac{3\pi}{2}$ then $\alpha+\beta=\pi+\tan^{-1}\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}\right).$
Proof:
$\tan(\alpha+\beta-\pi)=\frac{\tan\alpha+\tan(\beta-\pi)}{1-\tan\alpha\cdot\tan(\beta-\pi)}=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}$
As $|\alpha+\beta-\pi|<\frac{\pi}{2}$ and the function tangent is invertible in $\left]-\frac{\pi}{2},\frac{\pi}{2}\right[$, it follows that
$\alpha+\beta-\pi=\tan^{-1}\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}\right)$, therefore:
$\alpha+\beta=\pi+\tan^{-1}\left(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}\right)$.


Is there any difference between mapping and function? I wonder if there is any difference between mapping and a function. Somebody told me that the only difference is that mapping can be from any set to any set, but function must be from $\mathbb R$ to $\mathbb R$. But I am not ok with this answer. I need a simple way to explain the differences between mapping and function to a lay man together with some illustration (if possible).

Thanks for any help.

I'm afraid the person who told you that was wrong. There is no difference between a mapping and a function, they are just different terms used for the same mathematical object. Generally, I say "mapping" when I want to emphasize that what I am talking about pairing elements in one set with elements in another set, and "function" when I want to emphasize that the thing I am talking about takes input and returns output. But that's just a personal preference, and there is no convention I'm aware of.

By Nii: To my best understanding, mapping is just a process of matching elements of one set to elements of another set. Mapping is not a function unless some conditions are defined. Thus every mapping is a retation but not necessary a function.


Adding always the half of the previous number to a number beginning with 1... This question blows my mind.

Let's say we start at 1 and keep adding always 0.5 * the previous number.

1 + 0.5 + 0.25 + 0.125 + .....

The question is: will it reach infinity?

I really don't know, kept thinking about this alot.

Rewrite it with fractions and it might be a bit clearer:

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} ...$

Now calculate the partial sums:

$1$, $1 \frac{1}{2}$, $1 \frac{3}{4}$, $1 \frac{7}{8}$, $1 \frac{15}{16}$, ...

See a pattern?  Can you see this going over $2$?  How close do you think it might get?

So what you're looking for is the value of this series, which will call it $S$.

$$S=\sum_{n=0}^{\infty} \left( \frac{1}{2} \right)^n=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$$

If we divide this value $S$ by $2$ we get

$$\frac{S}{2}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=S-1 \Longrightarrow S=2S-1$$

so now with simple algebra we get the result that the sum of these numbers $S$ tends to $2$, which is not $\infty$.


Is it true that $R[x]+R[y]=R$ where $R=\mathbb Z_n$? Consider the ring $R=\mathbb Z_n$ where $n=pq$, $p,q$ are primes. $[x]\in \mathbb Z_n$
Assume $\gcd(p,q)=1$.
Let $[x]=a[p], [y]=b[q]$ where $1\le a\le q-1, 1\le b\le p-1$.
Is it true that $R[x]+R[y]=R$ where $R=\mathbb Z_n$?

My try:
Clearly $R[x]=\langle [x]\rangle $, $R[y]=\langle [y]\rangle $ where $\langle [x]\rangle ,\langle [y]\rangle $ is the ideal generated by $[x],[y]$.
Since $\gcd(p,q)=1\implies \langle [p]\rangle +\langle [q]\rangle =\mathbb Z_n$.
Also  $\gcd(p,q)=1\implies pt+sq=1\implies t[p]+s[q]=[1]$ for some $t,s\in \mathbb Z$.
I am stuck here. How to show that $R[x]+R[y]=R$ from here?
Can anyone please help?
I need to show that $[1]\in \langle [x]\rangle +\langle [y]\rangle $

First of all, you should convince yourself that $a[p]=[a][p]$ for all $a$ in $\mathbb{Z}$.

As $q$ is prime and $1\leq a<q$ then $gcd(a,q)=1$; hence there exist $u,v\in\mathbb{Z}$ such that $ua+vq=1$. Multiplying $p$ on this last relation we get $uap+vpq=p$ and taking class we have $ua[p]=[p]$, i.e., $u[x]=[p]$. Therefore $\langle [x]\rangle=\langle [p]\rangle$ since $[x]=a[p]\in\langle [p]\rangle$ and $[p]=u[x]\in\langle [x]\rangle$.

Using  an analogue reasoning you will get $\langle [y]\rangle=\langle [q]\rangle$. But we know that $\langle [p]\rangle+\langle [q]\rangle=\mathbb{Z}_n$ hence
 $$\langle [x]\rangle+\langle [y]\rangle=\mathbb{Z}_n$$


Prove that series $\sum \frac{a^n}{n^b(\log n)^c}$ is convergent for $|a|<1,b,c \in \mathbb{R}$ Prove that series $$\sum \frac{a^n}{n^b(\log n)^c}$$ is convergent for $|a|<1,b,c \in \mathbb{R}$

when $b,c >0$  Then $\ \ \frac{|a|^n}{n^b(\log n)^c} < |a|^n$ and hence series is  converge by comparison test.

but how to prove for other values of $b$ and $c$. Any hint ??

$\sum{\frac{a^n}{n^b(\log n)^c}}<\sum {\frac{a^n}{n^b(n)^c}}$
since

$a<1$ we can observe that the graph of $f(x)={\frac{a^x}{x^{b+c}}}$ is similar to that of $f(x)=\frac{1}{x}$ and hence we can approximate the sum as the integral

$\int_1^\infty\frac{a^x}{x^{b+c}}$ <$\int_0^\infty \frac{a^x}{x^{b+c}}$

which is convergent  because it is basically the Gamma function Of (b+c+1) multiplied by some constant and although the gamma function does not converge for values less than 0, our integral does because it does not take values between (0,1).


What are stress-free boundary conditions? (Continuum Mechanics) I've been doing some work on continuum mechanics and I can't find a good explanation for "stress free boundary conditions". I'm doing the question below:

  A solid linearly elastic isotropic circular cylinder of length $h$, radius $a$ and constant density $r$ is suspended with the axis vertical. The cylinder is slightly deformed due to its own weight. The $X_3$ axis corresponds to the axis of the cylinder. The lower end of the cylinder lies in the plane $X_3 = 0$ with the origin at its centre. The lateral surface of the cylinder $X^2_1 +X^2_2 - a^2 =0$ and the lower face $X_3 =0$ are stress free. State the stress free boundary conditions on these surfaces.

I can draw the cylinder but I'm very lost on what to write for the boundary conditions? I'd prefer to understand it over just knowing the answer flat out please (:

Thank you!

consider a hallow cylinder, such that it is opened in upper surface and closed in lower surfaces, now we can see what is stress free boundary? the opened upper surface of cylinder has imaginary boundary so that we are only assume, but sensation is not possible, that is called stress free boundary, and now the lower surface has a boundary so that possible to sense, that is called rigid boundary


Surjective, injective and bijection problem I have a homework problem that is to find the range that allows the function below to be bijective. $$\ f(x) = x^3 -x^2 -x+1  $$
 My problem is I don't know how to go about showing this function is either surjective or injectieve. So far we have worked with simple stuff like $\ x+1 $ or $\ x^2 $ I understand that injective functions are found like this:

$$ \begin{align}
& f(x) = x^3\\
& f(a) = f(b)\\
& a^3 = b^3\\
& a = b\\
\end{align}$$

and for surjective functions I need  $\ b = f(a) $ so,

$$ \begin{align}
f(a) = b\\
a^3 = b\\
a=\sqrt[3]{b}
\end{align}$$

I am quite confident about finding the domain, but have no idea how to tackle this function.

Thanks for looking.

Since the function is a third degree polynomial, you will have a harder time proving injectivity and surjectivity, but it will be possible. I will first only give you one hint:

Try to plot the function as accurately as possible. Once you have plotted it, you can usually guess if the function is surjective and if it is injective. Once you guess the right answer, it is easier to prove it (since you are not proving the "wrong" answer.

To plot the function, you should find its zeros and the zeros of the derivative, which should not be hard.

A function that is continuous and monotonous increasing or decreasing is injective.
Every polynomial function is continuous everywhere.

In order to be surjective, just watch out that the codomain of $f$ is equal to its range. As an example, assume that $f(x)$ has domain $D$ and range $R \neq \mathbb{R}$. In that case, $f:D \rightarrow R$ is surjective, but $f: D \rightarrow \mathbb{R}$ is not.


Prove $a \in\mathbb I, b \in \mathbb I \implies a^b \in \mathbb I$ Let $\mathbb I$ be the set of Irrational numbers and  $\mathbb Q$ be the set of Rational numbers. Show $a \in \mathbb I, b \in\mathbb I\implies a^b \in\mathbb I$

Using a counter example I can prove this:

$a = \sqrt{2}^\sqrt{2}$
$b = \sqrt{2}$

$\sqrt{2}^{\sqrt{2}\times\sqrt{2}}  = 2$

since $Q \subset I$ we can say $2 \in I$

is this correct? Or is there a rigorous proof? Thanks. :)

Your proof is a sketch of the proof that there are irrational numbers $a,\,b$ such that $a^b$ is rational - that is, assuming that you accept the law of excluded middle (that a proposition must be either true or false) so this is not a proof to an intuitionist: but I digress.

You argue as follows. Let $a = \sqrt{2}^\sqrt{2}$. Now either $a$ is rational or it is not (this is where an intuitionist would reject the proof).

If it is irrational, then raising to the irrational power $\sqrt{2}$ yields a rational number, to wit $2$. Therefore we have an example of an irrational ($a$) to an irrational power that is rational.

On the other hand, if $a$ is not irrational, it is rational, and so $a$ itself stands as an example of an irrational ($\sqrt{2}$) raised to a irrational power ($\sqrt{2}$) which yields a rational result.

Either way, we have a counterexample. Curiously, there is no way from this proof to tell which one is the counterexample: we simply know we have one, disproving the proposition.

To show that
$a^b \in I$,
you have to use induction.

Assumptions:
$I$ means the natural (positive) integers,
$a \in I$ and $b \in I$.

Needed results:
Inductive definition of exponentiation
and the product of
two elements of $I$
is also in $I$.

Base step:
For $b=1$,
$a^1 = a \in I$.

Suppose true for $b$
(i.e., $a^b \in I$).
Then

$\begin{array}\\
a^{b+1}
&=a^b\cdot a
\quad\text{(by the inductive definition of exponentiation)}\\
&\in I
\quad\text{(because $a^b \in I, a \in I$, and product of elements in $I$ is in $I$)}\\
\end{array}
$


What is meant by "evenly divisible"? "What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?"
Is it different from divisible?

Evenly divisible means that you have no remainder.
So, 20 is evenly divisible by 5 since 20 / 5 = 4.
Though, 21 is not evenly divisible by 5 since 21 / 5 = 4 R 1, or 4.2.

Means the same as "divisible". Answer is $2\times3\times5\times7\times11\times13\times17\times19$.


A question about finite dimensional function vector space. Let $V$ be a nontrivial finite-dimensional subspace of continuous functions on $[0,1]$, prove that there exists $g \in V$, such that $\int^1_0g^2=1$ and $\max g^2\ge\dim V$.

(Assuming we're talking about real-valued functions:)

Say $f_1,\dots, f_n$ is an orthonormal basis for $V\subset L^2([0,1])$. Since $$\int_0^1\sum_{j=1}^n f_j^2=n$$there must exist $t\in[0,1]$ with $$\sum_{j=1}^n f_j(t)^2\ge n.$$Hence there exist scalars $a_1,\dots a_n$ with $\sum a_j^2=1$ and $$\sum a_jf_j(t)\ge\sqrt n.$$


Probability that coupon cannot be used to spell a string Each packet of blades sold contains a coupon which is equally likely to bear the letters $A,B,or C$.

If $m$ packets are purchased,then the probability that the coupons cannot be used to spell $BAC$ is-----

The total possible ways in which the letters can be asssigned to the blades are $3^m$.

The total unfavouable cases are$$3^{m-3}+3^{m-6}+\cdots +3^{m-3 [{m/3}]}$$

I couldnot proceed ahead of this.

Please help me in this regard.

Thanks.

So in this problem the following are the favourable cases

Either all A,or all B or all C

Both A and B and not C

Both B and C and not A

Both C and A and not B

Note that  case 1 is obtained in 3 ways,case 2 in ($2^m-2$)ways.similarly for others.

Hence the required probability is

$$\frac{3+3*(2^m-2)}{3^m}$$ which simplifies to

$$\frac{2^m-1}{3^{m-1}}$$

I hope this helps.


Find a formula of sequence: $\frac{-1}{2}, 0, \frac{1}{10}, 0, \frac{-1}{26}, 0, \frac{1}{50}, 0, \frac{-1}{82}, 0, \frac{1}{122}, 0, \dots$ I'm working on a discrete math homework that finding a formula for the following sequence:

$$\frac{-1}{2}, 0, \frac{1}{10}, 0, \frac{-1}{26}, 0, \frac{1}{50}, 0, \frac{-1}{82}, 0, \frac{1}{122}, 0, \dots$$

I've tried but can't find the same difference between those numbers.

I come up with something as: $\frac{1}{2}((-1)^{(x-1)}-1)\times0.5\times0.2$
But it's not anywhere close as I mentioned, I don't know the difference.

Also, is there way to apply to find formula for this kind of problem.

Thank you so much.

Hint: Consider the sequence
\begin{eqnarray*}
2(n^2+(n+1)^2).
\end{eqnarray*}


How does eve find the message m that Bob encrypted? The following is given:

  $p=13$, $g=7$, $A=5$, $(c1,c2)=(10,8)$

I know $A=g^a \to a=3$  the message $m$ should be $c2*c1^{-a}$, which is $8\times 10^{-3} \mod 13$ but I don't know how to calculate the inverse part.

So what I get from your first computation is the following: you have looked for $a$ such that
$A = g^a \mod 13$ and have found that $a = 3$. In your second step, you want to compute $$c_2\cdot c_1^{-a} = 8 \cdot 10^{-3} \mod 13.$$

Since $\text{gcd}(10,13) = 1$, we have that $10$ has a multiplicative inverse modulo $13$. This means that there is some $x$ such that
$$10\cdot x \equiv 1 \mod 13$$
and we denote $x = 10^{-1}$. Therefore, your question becomes: 'find $x$ which satisfies this equation'.

Okey, so we have the following equations:
$$\begin{align}
13 &= 1 \cdot 10 + 3\\
10 &= 3 \cdot 3 + 1\\
3 &= 3 \cdot 1
\end{align}$$
this is Euclids division algorithm. From it , you can see that (starting from the second equation):
$$\begin{align}
1 &= 10 - 3 \cdot 3\\
  &= 10 - 3\cdot (13 - 1\cdot 10)\\
  &= 4 \cdot 10 - 3 \cdot 13
\end{align}$$
and if we look at this modulo $13$, we see that
$$1 = 4 \cdot 13 - 3 \cdot 13 \equiv 4 \cdot 10 \mod 13.$$
Therefore, we have shown that $10^{-1} \equiv 4 \mod 13$. Note that $10^{-3} \equiv 4^3 = 64 \equiv 12 \equiv -1 \mod 13$.

Using all of this, we find that
$$c_2 \cdot c_1^{-a} = 8\cdot 10^{-3} \equiv 8 \cdot (-1) \equiv -8 \equiv 5 \mod 13.$$

I hope this helps :)

To find $10^{-3}\mod 13$ we first note that it is equal to $\left(10^{-1}\right)^3\mod 13$.

We can find $10^{-1}\mod 13$ by the definition of inverses - we want to find an $x$ such that $x\times 10\equiv 1\mod 13$ where $x$ can also be denoted by $10^{-1}$

I used WolframAlpha to find  $10^{-1}=4\mod 13$, however you can do this by trying all values of $x$ from $2$ to $12$ until you find the desired value (there are far better methods for doing this btw, but beyond the scope of this answer). We note that $10$ will have an inverse modulo $13$ as $13$ is prime.

Therefore, \begin{align}10^{-3}\mod 13&\equiv\left(10^{-1}\right)^3\mod 13\\
&\equiv 4^3\mod 13\\
&\equiv 64\mod 13\\
&\equiv 12\end{align}


What is the $\lim_{t\to 0^+}u(1,t)$? Consider $u_t=u_{xx}$ $x\in \mathbb{R}, t>0$

$
u(x,0) =
     \begin{cases}
       \text{1,} &0\leq x\leq1 \\
       \text{0,} &\quad\text{otherwise.} \\
     \end{cases}
$

What is the $\lim_{t\to 0^+}u(1,t)$

Is the question complete? I mean no boundary conditions are given.

I know how to solve the problem when the boundary conditions are given. Here is the source from which I am studying http://tutorial.math.lamar.edu/Classes/DE/SolvingHeatEquation.aspx

Can somebody help me in finding the solution to this problem.

Thanks a lot.

Actually the boundary values are given via the function $u(x,0)$. So what you have is a standard heat equation whose solution is obtained by the standard methods. It is:

$$u(x,t)=\frac{1}{\sqrt{\pi}}\left(\int_0^{\frac{1-x}{2\sqrt{t}}}e^{-s^2}\,ds+\int_0^{\frac{x}{2\sqrt{t}}}e^{-s^2}\,ds\right)$$
Therefore
$$u(1,t)=\frac{1}{\sqrt{\pi}}\int_0^{\frac{1}{2\sqrt{t}}}e^{-s^2}\,ds$$
so that
$$\lim_{t\to 0^{+}}u(1,t)=\frac{1}{\sqrt{\pi}}\int_0^{\infty}e^{-s^2}\,ds=\frac{1}{2}$$


Chinese remainder Theorem in polynomials in one variable. In a question I was  answering, I needed to solve these congruences to proceed, and find the least $k<1000$

$$2k+k^2 = 0 \pmod 3$$

$$ 2k^3 + 6k = 0 \pmod 7$$

$$ k = 0 \pmod 2$$

My try:

due to the third statement

$$ k = 2j$$ for  some integer $j$,

due to the second statement
$$ k  = 7x ; k= 7x+2$$ for some integer $k$

due to the first statement

$$ k = 3y+1; k = 3y$$

can I just exhaust all of the $4$ possible cases due to these statements and use the Chinese remainder theorem? I don't know if I'm doing it right or not.

As pointed out in the comments, the solutions for the second congruence are $k\equiv0,2,5\bmod7$. Now combine the congruences one-by-one: the first and third give $k\equiv0,4\bmod6$, which when combined with the second gives
$$k\equiv0,12,16,28,30,40\bmod42$$
So the least positive $k$ is $12$, and the greatest $k<1000$ is $996$.

Least non-negative $k$ is $0$. If you are looking for non-zero solutions then, essentially $k = 2j$; $j = 7i+1, 7i-1$ or $7i$; $i = 3b, 3b+1$;


How can I prove this equation? My prof put this equation on the board, without any kind of explanation or proof. When I asked him for one, he didn't really give me a solid answer.

$$w = r(\cos \theta + i\sin \theta)$$

Then

$w^n = r^n(\cos n\theta + i\sin n\theta)$ for $n \in \mathbb{R}$

I would try to prove this, but it's a bit beyond me considering I learn't what a complex number is a week ago. Please don't use Euler's Theorem.

EDIT: I ended up using induction since the formula is not valid for non-integers.

I don't know if you've been taught the form $e^{i\theta} = cos(\theta) + i sin(\theta$).
If yes, then $$w = re^{i\theta}$$ and $$w^n = r^ne^{in\theta} = r^n(cos(n\theta) + i sin(n\theta))$$

Maybe your prof did not give you the explanation because he haven't taught you this form yet. I don't remember very well what comes first in the math program between complex numbers and exponentials (plus it can depend on the country).


How find this maximum $f(x)=\sqrt{\frac{1+x^2}{2}}+\sqrt{x}-x$ let $x\ge 0$,Find the follow maximum
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x$$

I find
$$f'(x)=\dfrac{1}{2}\left(\dfrac{\sqrt{2}x}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{x}}-2\right)$$
so
$$f'(x)=0\Longrightarrow x=1$$
so I think
$$f(x)_{max}=f(1)$$

But I this methods is ugly, and I think this have AM-GM inequality or Cauchy-Schwarz inequality to solve this problem
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x\le\sqrt{\dfrac{1+x^2}{2}}+\dfrac{1}{2}(1+x)-x=\sqrt{\dfrac{1+x^2}{2}}+\dfrac{1}{2}(1-x)$$

Then I can't,Thank you

Simply consider that $\sqrt{x}$ is a concave function, hence, for any $a,b\geq 0$, $\sqrt{a}+\sqrt{b}\leq \sqrt{2(a+b)}$. By taking $a=x$ and $b=\frac{1+x^2}{2}$ you get $f(x)\leq 1$ with equality only when $x=\frac{1+x^2}{2}$, i.e. $x=1$, as wanted.

Find f''(x)

if  f''(x_0) is -ve than x_0 is point of maximum

else f''(x_0) is  +ve than x_0 is point of minimum


Evaluate nested summation of a function I'm trying to relearn summation simplification, I haven't touch math in a while.  I'm having trouble simplify this nested summation down and I don't even know where to start. Could anyone please give me a detail break down of what I need to do here? Thank you.

$\sum_{i=1}^n \sum_{j=i}^n c*(n-j)$

It may be helpful to swap order of summation. Note change in index limits.

$$\begin{align}
\sum_{i=1}^{n}\sum_{j=i}^n c(n-j)&=\sum_{j=1}^n\sum_{i=1}^j c(n-j)\\
&=c\sum_{j=1}^{n}j(n-j)\\
&=c\left[n\sum_{i=1}^{n}j-\sum_{i=1}^n j^2\right]\\
&=c\left[n\cdot \frac12n(n+1)-\frac16n(n+1)(2n+1)\right]\\
&=\frac c6(n-1)n(n+1)
\end{align}$$

Alternatively:

$$\begin{align}
\sum_{i=1}^nj(n-j)&=(n-1)\sum_{i=1}^nj-\sum_{i=1}^nj(j-1)\\
&=(n-1)\sum_{i=1}^n\binom j1-2\sum_{i=1}^n\binom j2\\
&=(n-1)\binom {n+1}2-2\binom {n+1}3\\
&=3\binom {n+1}3-2\binom{n+1}3\\
&=\binom{n+1}3\\
c\sum_{i=1}^nj(n-j)&=c\binom{n+1}3\\
&=\frac c6 (n-1)n(n+1)\end{align}$$


Example of infinite field of characteristic $p\neq 0$ Can you give me an example of infinite field of characteristic $p\neq0$?

Thanks.

One very important example of an infinite field of characteristic $p$ is
 $$\mathbb{F}_p(T)=\left\{\,\frac{f}{g}\,\Bigg|\,\,\,f,g\in\mathbb{F}_p[T], g\neq0\right\},$$

the rational functions in the indeterminate $T$ with coefficients in $\mathbb{F}_p$ (the symbol $\mathbb{F}_p$ is just a synonym for $\mathbb{Z}/p\mathbb{Z}$). In other words, these are ratios of polynomials in $\mathbb{F}_p[T]$; this is the same construction as the one we use to make $\mathbb{Q}$ from $\mathbb{Z}$. The field $\mathbb{F}_p(T)$ is infinite because, for example, it contains $1$, $T$, $T^2$, $\ldots$, and it is of characteristic $p$ because it contains $\mathbb{F}_p$ (alternatively, because the kernel of the unique ring homomorphism $\mathbb{Z}\to\mathbb{F}_p(T)$ is $p\mathbb{Z}$.)

Another important example is $\overline{\mathbb{F}_p}$, the algebraic closure of the finite field $\mathbb{F}_p$. If you accept, for the moment, that every field has an algebraic closure (which is certainly not an obvious statement), then the fact that there are no finite algebraically closed fields means that the algebraic closure of a field of characteristic $p$ will have to be an infinite field of characteristic $p$.

Michael Hardy raises some good questions below.

Is one of $\mathbb{F}_p(T)$, $\overline{\mathbb{F}_p}$ a subfield of the other?
Is $\mathbb{F}_p(T)$ algebraically closed?
What is the relationship between these two fields?

These questions are all related. First, we should look at the definitions of algebraic element, algebraic extension, algebraically closed field, and algebraic closure.

Given a field $K$, and another field $L$ containing $K$ (i.e., $L\supseteq K$), we say that $\alpha\in L$ is an algebraic element over $K$ (or, for short, just that "$\alpha$ is algebraic over $K$") when there exists some (non-zero) $f\in K[x]$ such that $f(\alpha)=0$. If $\alpha\in L$ is not algebraic over $K$, we say it is transcendental over $K$. Wikipedia presents the following (standard) examples:

$\sqrt{2}\in\mathbb{R}$ is algebraic over $\mathbb{Q}$, because there is a non-zero polynomial $f\in\mathbb{Q}[x]$ (i.e., $f$ has rational coefficients) such that $f(\sqrt{2})=0$; we could take $f=x^2-2$, or $f=3x^3-6x$, or any other polynomial in $\mathbb{Q}[x]$ with $\sqrt{2}$ as a root.
$\pi\in\mathbb{R}$ is transcendental over $\mathbb{Q}$, because there is no non-zero polynomial in $\mathbb{Q}[x]$ with $\pi$ as a root; in other words, $\pi$ satisfies no algebraic relation with the rational numbers.
However, $\pi$ is algebraic over $\mathbb{R}$, because there are many easy examples of non-zero polynomials in $\mathbb{R}[x]$ with $\pi$ as a root - first and foremost, $x-\pi$. (Given any field $K$, any $\alpha\in K$ is algebraic over $K$ because $x-\alpha$ is a polynomial in $K[x]$ having $\alpha$ as a root. This demonstrates the importance of specifying algebraic (or transcendental) over what field.)

The setup of two fields $K$ and $L$, with $L\supseteq K$, is referred to as a field extension. We refer to the extension as a single entity by the expression $L/K$ (this is not a quotient like this or this, though). An extension $L/K$ is an algebraic extension when every $\alpha\in L$ is algebraic over $K$.
A field $K$ is algebraically closed when any non-constant $f\in K[x]$ (i.e., $f$ is a polynomial of degree $\geq 1$) has a root in $K$.
The requirement that the root is in $K$ is the key property.  For example, any non-constant polynomial $f\in\mathbb{R}[x]$ has a root, but some of them (e.g. $x^2+1$) don't have any roots that are actually in $\mathbb{R}$; thus $\mathbb{R}$ is not algebraically closed. (The fact that $\mathbb{C}$ is algebraically closed is often referred to as the Fundamental Theorem of Algebra).
Given any field $K$, there exists an algebraic extension $L/K$ such that $L$ is algebraically closed; such an $L$ is called an algebraic closure of $K$. It is unique up to isomorphism (so we often talk about "the" algebraic closure of $K$), and we write $L=\overline{K}$, or sometimes $L=K^{\text{alg}}$.

Now we have the concepts necessary to compare and contrast $\mathbb{F}_p(T)$ and $\overline{\mathbb{F}_p}$.

First, note that $T\in\mathbb{F}_p(T)$ is transcendental over $\mathbb{F}_p$ - there's no non-zero $f\in\mathbb{F}_p[x]$ such that $f(T)=0$. This is really what we originally meant when we said $T$ is an "indeterminate" - it stands in no relation to $\mathbb{F}_p$, we have only added it in as a formal symbol, so the only way we can get
$$a_nT^n+\cdots+a_1T+a_0=0\text{ for }a_i\in\mathbb{F}_p$$
is if every $a_i=0$, so there is no non-zero polynomial with coefficients in $\mathbb{F}_p$ having $T$ as a root. In fact, any element of $\mathbb{F}_p(T)$ that isn't itself an element of $\mathbb{F}_p$ (i.e., anything having a $T$ in it) is transcendental over $\mathbb{F}_p$, by the same argument - so the extension $\mathbb{F}_p(T)/\mathbb{F}_p$ is really super-duper non-algebraic.

However, every element of $\overline{\mathbb{F}_p}$ is algebraic over $\mathbb{F}_p$, because part of the definition of algebraic closure includes that the extension $\overline{\mathbb{F}_p}/\mathbb{F}_p$ is algebraic.

So, if we had $\mathbb{F}_p(T)\subseteq\overline{\mathbb{F}_p}$, then we'd have transcendental elements inside our algebraic extension $\overline{\mathbb{F}_p}/\mathbb{F}_p$, which is a contradiction.

On the other hand, if we had $\overline{\mathbb{F}_p}\subseteq\mathbb{F}_p(T)$, then we would have that there were some $\frac{f}{g}\in \mathbb{F}_p(T)$ such that $\frac{f}{g}\notin\mathbb{F}_p$ and $\frac{f}{g}\in\overline{\mathbb{F}_p}$ (because $\overline{\mathbb{F}_p}$ is infinite and $\mathbb{F}_p$ is finite), and they would have to be algebraic over $\mathbb{F}_p$, which we've also seen is a contradiction.

Thus, neither $\mathbb{F}_p(T)\subseteq\overline{\mathbb{F}_p}$ nor $\overline{\mathbb{F}_p}\subseteq\mathbb{F}_p(T)$.

It is also easy to see that $\mathbb{F}_p(T)$ is not algebraically closed; for the sake of simplicity, let's set $K=\mathbb{F}_p(T)$. If $K$ were algebraically closed, then every non-constant $f\in K[x]$ would have to have a root in $K$; but there are many such $f$'s that do not, for example $f=x^2-T$ (remember, $T\in K$, so don't be thrown by the existence of two indeterminates; this is just like $x^2-2$ in $\mathbb{Q}[x]$). If the polynomial $x^2-T$ had a root $\frac{f}{g}\in K=\mathbb{F}_p(T)$, then
$$\left(\frac{f}{g}\right)^2-T=0$$
$$f^2=Tg^2$$
$$2\cdot\deg(f)=\deg(f^2)=\deg(Tg^2)=1+2\cdot\deg(g)$$
which is a contradiction (can't have even = odd). This mirrors the usual proof that there is no $\frac{a}{b}\in\mathbb{Q}$ such that $\left(\frac{a}{b}\right)^2=2$, i.e. that $\sqrt{2}$ is irrational; the $\deg$ function is analogous to the 2-adic order function. (As  Pierre-Yves Gaillard points out in the comments, this actually shows that $K(T)$ is not algebraically closed, for any field $K$.)

So, in summary, what is the relationship between the fields $\mathbb{F}_p(T)$ and $\overline{\mathbb{F}_p}$? Other than the fact that they are both extensions of $\mathbb{F}_p$, not much. The extension $\mathbb{F}_p(T)/\mathbb{F}_p$ is not algebraic, while the extension $\overline{\mathbb{F}_p}/\mathbb{F}_p$ is; in fact the only elements $\mathbb{F}_p(T)$ and $\overline{\mathbb{F}_p}$ have in common are $\mathbb{F}_p$ itself. They are both extremely important in algebra, number theory, algebraic geometry, and they both fundamental examples of characteristic $p$ fields.

Another construction, using a tool from the formal logic: the ultraproduct.

The cartesian product of fields
$$P = {\Bbb F}_p\times{\Bbb F}_{p^2}\times{\Bbb F}_{p^3}\times\cdots$$
isn't a field ("isn't a model of..." ) because has zero divisors:
$$(0,1,0,1,\cdots)(1,0,1,0\cdots)=(0,0,0,0,\cdots).$$
The solution is taking a quotient: let be $\mathcal U$ a nonprincipal ultrafilter on $\Bbb N$. Define
$$(a_1,a_2,\cdots)\sim(b_1,b_2,\cdots)$$
when
$$\{n\in\Bbb N\,\vert\, a_n=b_n\}\in\mathcal U.$$
The quotient $F=P/\sim$ will be a infinite field of characteristic $p$.


Order Limit Theorem for Functional Limits Let $f$ and $g$ be functions defined on a set $A$ and assume $f(x) \le g(x)$ for all $x \in A$. Further, let $c$ be a limit point of $A$ and assume that the limits for $f$ and $g$ exist at $c$. Show that $$\lim_{x\to c} f(x) \le \lim_{x\to c} g(x)$$.
I know that it makes sense I'm just not sure how to prove it

This is for interpreting the question as having $f,g$ functions defined on a set $A$ of reals including all $x$ near but not (necessarily) equal to $c.$ Let the limits as $x \to c$ of $f,g$ be respectively denoted $F,G,$ and recall one is assuming $f(x) \le g(x)$ for all $x \in A.$

We wish to show the claim that $F \le G.$ Setting things up for a contradiction proof, we assume to the contrary that $F>G.$ Now define $H=(F+G)/2,$ and note since $f(x) \to F$ there is $\delta_1>0$ for which $f(x)>H$ as long as $|x-c|<\delta_1.$ [Apply here that $f \to F >H.$] Similarly there is $\delta_2$ for which $g(x)<H$ as long as $|x-c|<\delta_2.$ Then if $\delta = \min(\delta_1,\delta_2)$ and $|x-c|<\delta,$ we have both $f(x)>H$ and $g(x)<H.$ Combining these gives $g(x)<H<f(x),$ i.e. $g(x)<f(x),$ going against our assumption that $f(x) \le g(x)$ for all $x \in A.$


How to prove that $\sum_{n=1}^{\infty}\frac{a_{n}}{1+a_{n}}$ converges absolutely If $\sum_{n=1}^{\infty}a_{n}$ converges absolutely, show that
$$\sum_{n=1}^{\infty}\dfrac{a_{n}}{1+a_{n}}$$ converges absolutely.

My try: since $\sum_{n=1}^{\infty}a_{n}$ converges absolutely, then
$$\sum_{n=1}^{\infty}|a_{n}|$$ converges, then there exsit $M>0$, such
$$\sum_{n=1}^{N}|a_{n}|<M$$

then, how to prove than
$$\sum_{n=1}^{N}\dfrac{|a_{n}|}{|1+a_{n}|}<cM?$$
where $c$ is a constant?

Hint: Prove that if $\sum a_n$ converges absolutely and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges absolutely.

It is well-known that
$$
\frac{x}{x+1}<x
$$
for positive x. Therefore we have for each term that $|a_n|\geq \frac{|a_n|}{|1+a_n|}$. Therefore
$$
\infty >\sum_{n=0}^\infty |a_n| \geq \sum_{n=0}^\infty \frac{|a_n|}{|1+a_n|}
$$
And the sum converges absolutely.


Show that two matrices with the same eigenvalues are similar First assume that $A$ and $B$ are $p \times p$ matrices and that $\lambda_1,\ldots , \lambda_p$ are distinct eigenvalues of $A$ and $B$. I want to show that $A$ and $B$ are similar.

Here is my approach:
The goal is to show that there is a nonsingular $p \times p$ matrix  $P$ such that $B = P^{-1}A P$.
We know the following is satisfied for each eigenvalue
$$ A \textbf x_j = \lambda_j \textbf x _j  \text { where } j = 1,\ldots,p$$
and $\textbf x_j$ are the eigenvectors. Since $A$ and $B$ share the same eigenvectors, $B$ must satisfy the following
$$ B \textbf x_j = \lambda_j \textbf x _j  \text { where } j = 1,\ldots,p$$
Thus we have
$$ B \textbf x_j = A \textbf x_j$$
We can take the $\textbf x_j$ as the columns of the matrix $P$ and get
$$BP=AP$$
we know that the columns of $P $ are linearly independent so the matrix must be non singular. Unfortunately this means that
$$A=B$$
but this is not what we wanted to show. Am I on the right track? Is there an obvious mistake I am making?

The idea in your approach almost works. But you cannot assume that $A$ and $B$ have the same eigenvectors! So you will have
$$
Ax_j=\lambda_jx_j,\ \ \ By_j=\lambda_jy_j.
$$
As $\lambda_1,\ldots,\lambda_p$ are distinct, $x_1,\ldots,x_p$ and $y_1,\ldots,y_p$ are each linearly independent, so they are bases. Let $P$ be the change of basis $y\to x$, i.e. $P$ is the matrix such that $Py_j=x_j$, $j=1,\ldots,n$. It is clearly nonsingular: if $P(c_1y_1+\cdots+c_py_p)=0$, then
$$
0=c_1P(y_1)+\cdots+c_pP(y_p)=c_1x_1+\cdots+c_px_p,
$$
so $c_1=\cdots=c_p=0$. Now, for each $j$,
$$
PBy_j=\lambda_j Py_j=\lambda_jx_j=Ax_j=APy_j.
$$
As the $y$ are a basis, we get that $PB=AP$, i.e. $PBP^{-1}=A$.

Two matrices with same eigenvalues are similar if they are non-singular! If the matrices are singular they may or may not be similar!

eg:
  $$M_1=
\begin{bmatrix}
   1 & 0 & 0 \\
   0 & 0 & 1 \\
   0 & 0 & 0 \\
  \end{bmatrix}$$
 $$ M_2=
\begin{bmatrix}
   1 & 0 & 0 \\
   0 & 0 & 0 \\
   0 & 0 & 0 \\
  \end{bmatrix}$$

here $M_1$ & $M_2$ have the same set of eigenvalues! but they are not similar.


Discretization of the Anisotropic Diffusion Operator for Finite Difference Method I have to derive and apply a Finite Difference scheme to solve a steady state, anisotropic, diffusion equation. So I have to find a discretization of the following equation

$$\nabla \cdot ( \mathbf{K}(x,y) \mathbf{\nabla} u(x,y) ) + f = 0$$

where

$$\mathbf{K}(x,y) = \left(\begin{matrix} A(x,y) && B(x,y) \\ C(x,y) && D(x,y) \end{matrix}\right)$$

are the components of the space-dependent diffusion/permeability tensor.

Using some standard approach found for example on "Morton K. W., Mayers D.F. - Numerical Solution of Partial Differential Equations, An Introduction (Cambridge University Press, 2005" I translated the local, differential, equation, into an integral form and I wrote a flux conservation law for each rectangular cell of my domain.

If $\mathbf{K}$ is diagonal all is fine, I end un with something like a 5-point stencil for the laplacian with adjusted "weigths".

But if $B \ne 0$ (usually $B = C$) I end-up with an issue, because, I think, in the contribution to the total flux from a generic edge of the cell, appear the derivative of $u$ along the direction of the edge.

I'm not sure how to estimate this derivative, because the edge is at half way between grid nodes...

$$
0 \approx \Delta x_i \Delta y_j f_{ij} + \Delta y_j \left(A u_x + B u_y\right)_{x_{i-1/2},y_j}^{x_{i+1/2},y_j} + ...
$$

$$
0 \approx \Delta x_i \Delta y_j f_{ij} + \Delta y_j \left\{ A_{i+1/2,j} \times 2 \frac{ u_{i+1,j} - u_{i,j} }{\Delta x_i + \Delta x_{i+1}} - A_{i-1/2,j} \times 2 \frac{ u_{i,j} - u_{i-1,j} }{\Delta x_{i-1} + \Delta x_{i}} + B_{i+1/2,j} \times ? \right\} + ...
$$

The only ideas that comes to my mind are:

Use some near grid nodes, f.i:

$$
(u_y)_{i+1/2,j} \approx (u_y)_{i+1,j} \approx 2 \times \frac{u_{i+1,j+1} - u_{i+1,j-1}}{\Delta y_{j-1} + 2 \Delta y_j + \Delta y_{j+1}}
$$

Use some weighted mean / interpolation between nearest nodes:

$$
(u_y)_{i+1/2,j} \approx \frac{\Delta x_{i+1} (u_y)_{i,j} + \Delta x_{i} (u_y)_{i+1,j}}{\Delta x_{i} + \Delta x_{i+1}}
$$

Use a non-compact, bigger, 13-point (?) stencil and write an equation every $2\times2$ cells of the discretization "mesh"

Unfortunately I didn't find any reference on this topic.

Maybe this would help https://www.mia.uni-saarland.de/weickert/Papers/book.pdf#page=107

It gives the stencil for a laplacian of the form $\text{div}(D\nabla)$ where :
$D = \left(\begin{matrix} a & b \\ b & c \end{matrix}\right)$

Look this:
https://www.sciencedirect.com/science/article/pii/S0021999114003155

This causes diffusion, heat conduction, and viscous momentum loss, to effectively be aligned with the magnetic field lines. This alignment leads to different values for the respective diffusive coefficients in the magnetic field direction and in the perpendicular direction, to the extent that heat diffusion coefficients can be up to 1012 times larger in the parallel direction than in the perpendicular direction. This anisotropy puts stringent requirements on the numerical methods used to approximate the MHD-equations since any misalignment of the grid may cause the perpendicular diffusion to be polluted by the numerical error in approximating the parallel diffusion. Currently the common approach is to apply magnetic field-aligned coordinates, an approach that automatically takes care of the directionality of the diffusive coefficients. This approach runs into problems at x-points and at points where there is magnetic re-connection, since this causes local non-alignment. It is therefore useful to consider numerical schemes that are tolerant to the misalignment of the grid with the magnetic field lines, both to improve existing methods and to help open the possibility of applying regular non-aligned grids. To investigate this, in this paper several discretization schemes are developed and applied to the anisotropic heat diffusion equation on a non-aligned grid.