How are these two equal? Which of these terms is greater ?
$2x-6y+1$ or $1$
if $x^4 + 3y^2=0$

According to the text they are equal ?How is that ?

What number system are $x,y$ in? If $x,y\in\mathbb{R}$, then it is simple, since $x^4$, $3y^2$ $\ge 0$, we must have that $x,y=0$, hence $2x-6y+1=1$.


What are the components of this force in x-z direction I'm not able to visualize the components of $F$ in x and z direction. As I can tell $F_y=300\cos(30^\circ)$.

I don't think the drawing is clear.  If you want a real answer you should have a three view drawing, which will show where the angles are measured more clearly.  When I first looked, the line $30^\circ$ from the force seemed to be perpendicular to the plate, the direction a drill bit would go through to drill the hole.  If that is true the force is purely in the $x$ direction.  On a second look, the line $30^\circ$ from the force seems to be parallel to $y$.  That would support your claim that $F_y=300 \cos(30^\circ)$.  I don't understand what the blue line is trying to tell us.  I also can't tell whether $F$ has a component in the $z$ direction or not.

You obiously need to do two projections.
See the axis which starts at point A and is perpendicualr to the y axis. Let's call it $v$.
Then $F_v=300 \cdot \sin (30º)$
Now project that one into the x and z components.
$F_x=F_v\cdot cos(150º) = - 300\ N\sin (30º)\cos (30º)$


The best way of integrating irrational functions Ok, so, here is the example integral:
$$I=\int\frac{x-2-\sqrt{-x^2-4x+4}}{x^2-\sqrt{-x^2-4x+4}}dx$$
I always solve these types of integrals using Euler's substitutions, but, recently, I came across some more difficult integrals like the one above which, after using Euler's substitution, become the integral of a rational function which is too complex to calculate, or at least to me.
So, what I did with the integral above is use second Euler's substitution:
$$\sqrt{ax^2+bx+c}=x\cdot t \pm \sqrt{c}$$
which gives:
$$\sqrt{-x^2-4x+4}=x\cdot t-2$$
Then, we have:
$$x=\frac{4-2t}{1-t^2},\ \ \ dx=\frac{2(t^2-1)+2t(4-2t)}{(1-t^2)^2}dt$$
After the substitution, the integral becomes:
$$I=\int \frac{\frac{4-2t}{1-t^2}-2-\frac{4-2t}{1-t^2}\cdot t+2}{\frac{(4-2t)^2}{(1-t^2)^2}-\frac{4-2t}{1-t^2}\cdot t+2}\cdot \frac{2(t^2-1)+2t(4-2t)}{(1-t^2)^2}dt$$
Which, after some calculation is:
$$I=\int\frac{8-40t+24t^3-8t^4}{2t^3-6t^2+14t-14}dt$$
Then, I can divide them and get two integrals, one from the table and the other one of a rational function with numerator's degree lower than denominator's.
Then, I should integrate that remaining rational function, but I'm not sure how, since I can't factor the polynomial in denominator.
My question is, is there a simpler way to calculate integrals like this one and if there isn't, what do I do with the remaining integral of rational function? What if I get a polynomial of degree $5$, $6$, $7$ etc. in denominator?
Thank you for your time.

There is a mistake in your computation. I added my comments here since there is a limited number of characters I can put into the comment window.
\begin{equation}
x=\frac{4-2t}{1-t^2},\ \ \ dx=\frac{-2(t^2-4 t+1)}{(1-t^2)^2}dt
\end{equation}
Then after the substitution, the integral becomes:
\begin{equation}
I=\int \frac{\frac{4-2t}{1-t^2}-2-\frac{4-2t}{1-t^2}\cdot t+2}{\frac{(4-2t)^2}{(1-t^2)^2}-\frac{4-2t}{1-t^2}\cdot t+2}\cdot \frac{-2(t^2-4 t+1)}{(1-t^2)^2}dt
\end{equation}
Which, after some calculation is:
\begin{equation}
I=\int \frac{2 (t-2) ((t-4) t+1)}{(t+1) (t (t-2) (2 t+5)+9)} dt
\end{equation}
Then, I can divide them and get two integrals:
\begin{equation}
I=\int \frac{-2}{(t+1)} dt + \int \frac{6 t^2-16 t+14}{2 t^3+t^2-10 t+9} dt =-2 \log (t+1) +\int \frac{6 t^2+2 t -10}{2 t^3+t^2-10 t+9}+\int \frac{24- 18 t}{2 t^3+t^2-10 t+9}=-2 \log (t+1)+\log (2 t^3+t^2-10 t+9)+2 \int \frac{12- 9 t}{2 t^3+t^2-10 t+9}
\end{equation}


Analytic proof of area probability $X,Y$ are i.i.d. $unif(-1,1)$ random variables.
Prove that
$$P(X^2+Y^2\leq 1)=\frac{π}{4}$$

Geometrically, I understand how that happens. $(X,Y)$ is a random point in square having centre at origin and vertices $(-1,-1),(-1,1),(1,-1),(1,1)$. Probability that a random point in this square lies in the unit circle is the ratio of their areas.

But, is there any analytic proof ?

Let $D = \{(x, y): x^2+y^2 \leq 1\}$. Let $g(x, y)$ be the density of $(X, Y)$. It exists because $X$ and $Y$ are independent, and equals to multple of their densities. (Fubini theorem)
$$P(X^2+Y^2\leq1) = \iint_D g(x, y)dxdy = \frac{1}{4}\iint_D dxdy = \frac{\pi}{4}$$


Name for "3D quadrilateral" shape? I am interested to know the name of the following solid construction - kind of a deformed cube... but I don't know the name, or even a general name:

Left and right faces parallel with the $yz$ axes
Bottom face parallel with the $xz$ axes
Back face parallel with the $xy$ axes.
Front and top faces not necessarily parallel,

assuming $x=(1, 0, 0),y=(0,1,0),z=(0,0,1)$. If there is no such name, then is "3d quadrilateral" correct ?

This is a right-quadrilateral-based right prism. Say prism for short.

It is called Triclinic. It is one of the 7 crystal systems.
Wiki


How to solve $Ax=y$ according to minimal linear error? I have an overconstrained problem $Ax = y$ that I’m solving for x with a least squares solver (eigen library).
I would like to have the problem solved however such that not the sum of squared errors is minimized, but the sum of errors.
Is there a way to modify the original problem so that I can still use my least-squares solver to solve for x?
If not, what algorithms can solve for that?

As the comments mentions you want to look into L1 regression but first give yourself some time to thinkt if that would make sense for your problem, maybe explaint why you would want to use L1.

If you're using MATLAB or Python with a convex solver such as CVX you shouldn't have any problem. Instead of solving $$\min_\mathbf{x}\,\|A\mathbf{x}-\mathbf{y}\|_2^2 $$ just solve the problem $$\min_\mathbf{x}\,\|A\mathbf{x}-\mathbf{y}\|_2 $$ which is still a convex program. Convergence will be slower than the squared version, but it's still a problem that can be efficiently solved.


probability space over the coin flips I recently read the definition of differential privacy, which is as follows:

  a randomized function $K$ gives $\epsilon$-differential privacy if for all data sets $D$ and $D'$ differing in at most one row, and all output space $S$, we have $Pr[K(D) \in S] \leq \exp(\epsilon).Pr[K(D') \in S]$ where the probability space in each case is over the coin flips of $K$

Could somebody explain what it means when saying "probability is over the coin flip" ? an example would be greatly appreciated

Cheers

the probability space in each case is over the coin flips of K

Simply put, $K$ has values in a discrete set $\mathfrak K$ with $\Pr(K=k)=p_k$ for every $k$ in $\mathfrak K$, and
$$
\Pr(K(D)\in S)=\sum_{k\in\mathfrak K}p_k\cdot\mathbf 1_{k(D)\in S}.
$$
Similarly for $\Pr(K(D')\in S)$.

I am also confused by this expression when reading Prof. Dwork's book. I guess maybe authors only want to highlight that the algorithms discussed in differential privacy are not deterministic but probabilistic with randomization.


Reverse of Deduction Theorem Why is it "easy to see" that if $S \vdash (A\to B)$ then $S \cup\{A\} \vdash B$?

Let $\langle\varphi_1,\varphi_2,\ldots,\varphi_n\rangle$ be a proof of $(A\to B)$ from $S$: $\varphi_n$ is $(A\to B)$, and each $\varphi_k$ with $k<n$ either is in $S$, is an axiom, or follows from earlier statements in the list by one of your inference rules. Then $\langle\varphi_1,\ldots,\varphi_n=(A\to B),A,B\rangle$ is a proof of $B$ from $S\cup\{A\}$: the presence of $A$ in the list is justified by the fact that $A\in S\cup\{A\}$, and $B$ follows from $(A\to B)$ and $A$ by modus ponens, which is surely one of your rules of inference.

It is not so easy to see!

The only axiom we have is the original single axiom DDpDqrDDtDttDDsqDDpsDps, and our only rule of inference other than uniform substitution is

{D$\alpha$D$\beta$$\gamma$, $\alpha$} $\vdash$ $\gamma$.

This "if S⊢(A→B), then S∪{A}⊢B" involves $\rightarrow$.  I'll rewrite that as:
"if S$\vdash$Cab, then S ∪ {a} $\vdash$b."  Thus, using the definition

Cpq := DpDqq, we want to show that:

"if S$\vdash$DaDbb, then S ∪ {a} $\vdash$b."

Suppose that we've proved DaDbb.  This means that in the rule above we can substitute $\alpha$ with "a", $\beta$ with "b", and $\gamma$ with "b" yielding:

{DaDbb, a} $\vdash$ b.  Now from that since we have $\vdash$DaDbb, and "a" consists of a hypothesis, $\vdash$b.


What's the difference between early transcendentals and late transcendentals? Anton, Bivens, and Davis have written calculus books with late

transcendentals and Stewart has a calculus book with early transcendentals.

What's this all about?

edit 1: (Both terms show up in the titles of the books.)

Here's the Anton, Bivens, Davis book:
https://www.wileyplus.com/WileyCDA/Section/Calculus-Late-Transcendentals-10e.id-813274.html

Here's a link to an amazon page selling the Stewart calculus book:

https://www.google.com/shopping/product/7326832019121784605?q=james+stewart+calculus+early+transcendentals&prds=hsec:specs&ved=0CAUQ4Ss&ei=ovplUsfDNceBiwLSm4DYDA

The difference is entirely a pedagogical one. Both approaches ultimately cover the same calculus.

“Late transcendentals” is the traditional approach to teaching calculus where the treatment of logarithmic and exponential functions is postponed until after integration is introduced.  In the traditional method, the natural logarithm is defined by the formula
$$
\ln x \;\underset{\scriptscriptstyle\mathrm{def}}{=}\; \int_1^x \frac{1}{t}\,dt,
$$
and $e^x$ is then defined to be the inverse function of $\ln x$.  Important properties of $\ln x$, such as that $\ln(xy) = \ln x + \ln y$, are proven from the integral definition using $u$-substitution, and the properties of the exponential function follow from these.  Arbitrary powers $a^b$ where $a>0$ and $b\in\mathbb{R}$ are then defined by the formula
$$
a^b = e^{b \ln a}
$$
and it is shown that this agrees with the existing definition when $b$ is rational.

In the “early transcendentals” method, the logarithmic and exponential functions are introduced shortly after the definition of the derivative.  Exponentiation where the power is an arbitrary real number is defined by the formula
$$
a^b \;=\; \lim_{q\to b}\, a^q
$$
where the limit is taken over the rational numbers. (There are some technical issues in proving that this limit exists, which are often ignored.)  An argument is then given that there exists a number $e$ for which
$$
\frac{d}{dx}\bigl[e^x\bigr]\biggr|_{x=0} \;=\; 1,
$$
though again the presentation of this argument is sometimes less than completely rigorous.  It follows that the derivative of $e^x$ is $e^x$, and the natural logarithm is defined as the inverse of the exponential function.

The traditional method has the advantage of being cleaner, and the proofs are simple enough that they can be presented to starting calculus students in a mathematically rigorous way.  However, it has several distinct disadvantages:

It is not very intuitive, since the natural logarithm and exponential function are essentially summoned out of thin air by what looks like magic. The “early transcendentals” approach, on the other hand, corresponds much more closely with how we actually think of exponentials and logarithms.
Students who take only a single semester of calculus, which includes most biology majors at some colleges in the United States, do not see the exponential function and natural logarithm.  This is a serious problem, because these are among the most important functions for applications in biology.
Even students who take two semesters of calculus learn about exponential and logarithmic functions fairly late, which means they don't have time to get used to these functions.

These arguments are generally considered persuasive enough that most universities in the United States have now adopted calculus books that use the “early transcendentals” approach.

Of course, not everyone agrees with this change, and further arguments can be presented on both sides of the debate.  Books using the traditional approach continue to be used at some universities and in some calculus classes.  In particular, some universities offer an honors calculus sequence designed specifically for math majors and similar students, and there's a relatively strong argument for using the traditional approach in such a course.  My impression is also that the “late transcendentals” approach remains quite popular outside the United States, though my evidence for this is purely anecdotal.

Early transcendentals means it has review at the beginning and late transcendentals or if the book doesn’t say anything me and that it has no review and jumps right in. Early transcendentals books will usually cost more than others because it has that extra review part in it. This is based on a recent google I just did. I hope this helps and is not outdated for you.


Let $x$, $y$, $z$ be positive. Prove an inequality... Let $x$, $y$, $z$ be positive. Prove the inequality
$$4(x+y+z)^3 \ge 27(yx^2+zy^2+xz^2+xyz)$$

I have no idea of how the proof should look like, tried to get rid of the braces but it seems to be the wrong way, as the equation becomes very long and has very different positive and negative coefficients, so doing a factorization is quite impossible.

I guess some kind of theorem or something can be applied here, but have no idea which one

Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.

Hence, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x+xyz=xy\cdot x+yz\cdot y+zx\cdot z+xyz\leq ab\cdot a+ac\cdot b+bc\cdot c+abc=$$
$$=b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq4\left(\frac{b+2\cdot\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}(x+y+z)^3$$
Done!

Let z is parameter, x+y = t is parameter.
Try to proof what right side is maximal if x = y.
And simple  Weierstrass  works


Hire purchase problems A bank offers loans at interest of $12\%$ per annum , compounded monthly. Jack took a $\$50,000$ loan and makes a monthly installment payments of $\$4200$ at the end of each month . Find his outstanding loan after his third installment payment.

I'm confused with what to substitute into the compound interest formula:

$$ A = 50,000 \left( 1 + \frac{0.12}{12} \right)^n $$

Since I do not know what's the total number of interest periods, which number should I substitute into $n$ ?

While the solutions to approach this through "brute force" are fine for a small number of periods, I thought I'd show a more general approach when you're asked to do the computation for longer durations. If you find the following heavy-going, then just do the brute force computation (I indicate where you can immediately do this in the working below).

Approach this problem from first principles. An annual interest rate of $12\%$ compounded monthly is a compound monthly rate of $1\%$. A monthly compounding means the amount owed will grow by a factor of $(1 + 0.01) = 1.01$ with each month.

Let the initial amount at the beginning of the first month (the principal) be represented by $A_0$.

Then the amount at the end of the first month $A_1 = 1.01A_0 - 4200$.

The last term is because he does a partial repayment by $\$4200$.

In fact, this is the general formula for the recursion:

$A_{n+1} = 1.01A_n - 4200$.

At this point, you can immediately apply this recursive calculation to find $A_3$, which is the simple "brute force" approach. Use $A_0$ to find $A_1$, then $A_1$ to find $A_2$, then finally $A_2$ to find $A_3$.

But if you want a more general solution, you can work through this to find a closed form for $A_n$.

$A_n = 1.01A_{n-1} - 4200 = 1.01^2A_{n-2} - 4200(1.01 + 1) \\= 1.01^3A_{n-3} - 4200(1.01^2 + 1.01 + 1) \\= \dots \\= 1.01^nA_0 - 4200(1.01^{n-1} + 1.01^{n-2} + \dots + 1.01^0)$

the last term of which can be further simplified by the geometric sum formula.

$$4200(1.01^{n-1} + 1.01^{n-2} + \dots + 1.01^0) = 4200\frac{1.01^n-1}{1.01-1} = 4.2 \times 10^5(1.01^n-1)$$

After a little rearrangement, you will get:

$A_n = 4.2 \times 10^5 - 1.01^n(4.2 \times 10^5 - A_0)$

This closed form can be proved using induction for rigour.

From this, you can find out $A_3 = 4.2 \times 10^5 - 1.01^3(4.2 \times 10^5 - 50000) = \$38,788.63$. This is the answer to your question.

If you pursue this to the end (by setting $A_n = 0$), you will find he will have fully repaid his loan by the $13$th month.

Let a month be your period, so the answer to your actual question is that you should substitute 3 into n, because 3 is the number of periods.  However, the interest rate is not $0.12\%/12$

To find the monthly interest rate you must first use $n=12$ and put $12\%$ on the left hand side of your equation in order to calculate the monthly interest rate $x$:

$112\%=(1+x)^{12}$

$x=\sqrt[12]{1.12}-1 = 0.9489\%$ per month$

Now just use brute force to calculate what he owes each month by applying this algorithm 3 times to the principal $\$50,000$:  Add $x\%$ interest and deduct $\$4,200$.

After 1 month he owes $\$46,274.44$

After 2 months he owes $\$42,513.53$

After 3 months he owes $\$38,716.93$


What does it mean to say breaking RSA generically is equivalent to factoring? I am giving a one hour presentation on the RSA crypto-system as one of the requirements for Masters degree. I just want to get some facts straight here. I was told casually by a professor that RSA is equivalent to factoring, but I am having a hard time verifying this with resources online.

So, is breaking RSA generically equivalent to factoring as in the title of this paper? https://eprint.iacr.org/2008/260.pdf

What does it mean to use the word "generically" to describe the difficulty of breaking RSA.

Any other resources would be helpful.

I am turning my comment into an answer.

RSA is secure under the RSA Assumption, which basically states that RSA is secure. This assumption is not known to be the same as the assumption that factorization is hard. While others have pointed out that factorization is sufficient to break RSA, it has not been proven that factorization is necessary to break RSA.

Supporting the position that breaking RSA is easier than factoring, if factoring is easy then the RSA secret key may be obtained from the public key, but breaking RSA encryption only requires distinguishing between the encryptions of two adversarially-chosen plaintexts, which is intuitively an easier task. Similarly, breaking RSA signing does not require obtaining the secret key, but instead forging the signature of an adversarially-chosen message, which is again intuitively easier. Of course I can't offer any proofs because the problem is open.

As for the meaning of the word "generically", it appears to mean "only by performing operations in the algebraic structure being analyzed" as opposed to exploiting additional structure, such as the bit representation of numbers. Note that proving the equivalence of factoring and the RSA assumption "generically" is a more modest result than proving their equivalence universally because the computational models that they consider have been restricted to just perform certain operations, and it's possible that by allowing other operations, the two problems are no longer equivalent.

I think the idea is this: the RSA algorithm is powerful because it relies on the fact that for large numbers, say numbers with 200 digits, it is very difficult to factor them or even find out if they are composite. If we were somehow able to figure a way to factor large numbers relatively easily, then the RSA algorithm would no longer be that powerful, thus it would be "broken".

Edit: As for what "generically" means, it might explain that in the paper but my guess is that the author is using the word "generic" to mean that this is characteristic of the RSA algorithm, i.e. that is why we can say it is equivalent to factoring.


Proving that ${n}\choose{k}$ $=$ ${n}\choose{n-k}$ I'm reading Lang's Undergraduate Analysis:

Let ${n}\choose{k}$ denote the binomial coefficient,
$${n\choose k}=\frac{n!}{k!(n-k)!}$$
where $n,k$ are integers $\geq0,0\leq k\leq n$, and $0!$ is defined to be $1$. Prove the following assertion:
$${n\choose k}={n\choose n-k}$$

I proceded by making the adequate substitutions:
$${n\choose n-k}=\frac{n!}{\color{red}{(n-k)}!(n-\color{red}{(n-k)})!}$$
And then I simplified and achieved:
$$\frac{n!}{(n-k)!-k!}$$
But I'm not sure on how to proceed from here, I've noticed that this result is very similar to $\frac{n!}{k!(n-k)!}$. What should I do? I guess it has something to do with the statement about the nature of $n$ and $k$:

$n,k$ are integers $\geq0,0\leq k\leq n$

So should I just change the minus sign to plus sign and think of it as a product of $(n-k)!$?
$$\frac{n!}{(n-k)!-k!}\Rightarrow\frac{n!}{(n-k)!+k!}\Rightarrow \frac{n!}{k!(n-k)!}$$
I'm in doubt because I've obtained the third result on Mathematica, but obtained the first with paper and pencil. I'm not sure if there are different rules for simplification with factorials. I'm not sure if this $(n-k)!+k!$ mean a sum or a product in this case.

The proof you've suggested, and the other answers have corrected, is the purely arithmetic way to see the desired equality.  The result is fairly clear to see intuitively as well, just ponder the following statement:

"To form a collection of $k$ objects from $n$ total objects, one may choose which $k$ objects to include in the collection, or equivalently, one may choose which $n-k$ objects to exclude from the collection."


uncountable well-ordered chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$ without $AC$ If we assume $AC$ we can construct an uncountable well-ordered chain of subsets of $\mathbb{N}$ by well-ordering the reals and then using the same Dedekind's cut construction as in this question.

edit: as pointed out in the comments and Asaf's answer the construction above doesn't work

What if we don't assume $AC$? Can we show the existence of an uncountable, well-ordered chain of subsets of $\mathbb{N}$ in $ZF$ alone?

You're confusing well-ordered with well-orderable.

There is no uncountable well-ordered chain in $\mathcal P(\Bbb N)$, because if there was such $\{A_\alpha\mid\alpha<\omega_1\}$, map $\alpha$ to the least $n$ in $A_{\alpha+1}\setminus A_\alpha$, and you got yourself an injection from $\omega_1$ into $\omega$. Something which you cannot get with or without choice.

But it is also consistent without the axiom of choice that there is no uncountable well-orderable chain in $\mathcal P(\Bbb N)$, as it is consistent that there is no injection from $\omega_1$ into $\mathcal P(\Bbb N)$ which means every well-orderable subset is countable. So we cannot prove in $\sf ZF$ the existence of an uncountable well-orderable subset of $\mathcal P(\Bbb N)$, chain or otherwise.

There is none since the set of integers is countable. The dedekind cut produces an uncountable chain which isn't well ordered.


What's the Lebesgue measure of this set? The following citation is from Folland's Real Analysis.

  Let $m$ denote the Lebesgue measure on $\mathbb{R}$ and $\{ r_j \}$ be an enumeration of the rational numbers in $[0,1]$, and given $E > 0$, let $I_j$ be the interval centered at $r_j$ of length $\epsilon \, 2^{-j}$. Then the set $U = (0,1)\cap \left( \bigcup_{1}^{\infty} I_j \right)$ is open and dense in $[0,1]$, but $m(U) < \sum_{1}^{\infty}\epsilon\, 2^{-j} = \epsilon$;

Since $U$ is a non-empty open set, $m(U)>0$. But why does the above argument show that $m(U)=0$?

It seems that you're arguing that $m(U) < \varepsilon$ for all $\varepsilon > 0$, so $m(U) = 0$.

But $U$ depends on the choice of $\varepsilon$, so the above argument doesn't work!

All this argument shows is that there are open and dense sets of arbitrarily small positive Lebesgue measure.

In other words, this is the difference between

$\qquad\exists U$ open and dense $\forall \varepsilon > 0 : m(U) < \varepsilon$ (which is not true)

and

$\qquad \forall \varepsilon > 0 \exists U$ open and dense $: m(U) < \varepsilon$ (which is what Folland shows).


Var$(X) = \mathbb{E}((X - \mathbb{E}(X))^2) = \mathbb{E}(X^2) - \mathbb{E}(X)^2$ I have a question about something my teacher told us:

let $\mathbb{E}$(X) donate the expected value of a certain random variable $X$.

Then Var$(X) = \mathbb{E}[(X - \mathbb{E}(X))^2] = \mathbb{E}(X^2) - \mathbb[{E}(X)]^2$. I can't understand why the last equation holds. Can you explain this to me?

It's a simple calculation. First,
$$
\mathbb{E}[(X-\mathbb{E}[X])^2] =
\mathbb{E}[X^2 + \mathbb{E}[X]^2 - 2X\mathbb{E}[X]].
$$
Linearity of expectation implies
$$
\mathbb{E}[X^2 + \mathbb{E}[X]^2 - 2X\mathbb{E}[X]] =
\mathbb{E}[X^2] + \mathbb{E}[X]^2 - 2\mathbb{E}[X]^2 = \mathbb{E}[X^2] - \mathbb{E}[X]^2,
$$
using the fact that $\mathbb{E}[X]$ is a constant.

Expand $\mathbb{E}[(X−\mathbb{E}(X))^2]$ and use linearity of $\mathbb{E}$.


maximal girth $g(n,m)$ of a graph with $n$ vertices and $m$ edges The girth of a graph is the length of its smallest cycle.

For $m \ge n \ge 3$, let $g(n,m)$ denote the maximal girth over all graphs with $n$ vertices and $m$ edges.

Is it true that if $m > c n $ and $c > 1$, then the girth is $O( \log n)$? (I.e. is there a constant $C$ which depends on $c$ but not on $n$ such that $g(n,m) < Cn$?)

Here is a sketch of a proof if $m > cn$ and $c > 2$. Such a graph has average degree $$d = 2m/n > 4.$$ Than a standard fact in graph theory (e.g. by repeatedly deleting vertices of small degree) gives that there is a subgraph $H$ with minimum degree $\delta [H] \ge d/2 > 2$. Since $\delta$ is an integer, $\delta [H] \ge 3$. Now consider a vertex $v \in H$ as a root, consider neighbors of $v$ and their neighbors, etc. Until there is a cycle, every vertex at distance $d$ from $v$ has at least two neighbors at distance $d+1$. So the number of vertices doubles at each step, and the diameter is at most $\log_2 n$. The girth is at most $2 \log_2 n + 1$.

Take a graph $G$ with $n$ vertices and $m$ edges, satisfying $m > cn$. By deleting vertices with degree at most $1$, we get a graph $G'$ with $n' = n-k$ vertices and $m' \ge m-k$ edges, which means $m' > cn'$.

The resulting graph $G'$ has minimum degree $2$ (and some vertices of higher degree), which means we can think of it as a subdivision of a multigraph $H$ with minimum degree $3$. We'd like to pass to that multigraph by smoothing out all degree $2$ vertices, but before we do that, we'd like to make sure that no edges of $H$ correspond to very long paths in $G'$.

Suppose $G'$ contains a path with $t > \frac{c}{c-1}$ edges, all of whose vertices, except the endpoints, have degree $2$. Deleting this path reduces the number of vertices by $t-1$ and the number of edges by $t$, producing a graph $G''$ with $n''=n'-(t-1)$ vertices and $m''=m'-t$ edges. If $t > \frac{c}{c-1}$, then $t < c(t-1)$, which means $m'' > cn''$.

Once no such paths are left, we can smooth out $G''$ to get a multigraph $H$ with minimum degree $3$, all of whose edges correspond to paths in $G''$ of length at most $\frac{c}{c-1}$. We can actually assume $H$ is a graph: it might have loops or parallel edges, but if it does, then they correspond to a cycle in $G''$ of length at most $\frac{2c}{c-1}$.

Now, by the argument in the question, $H$ has a cycle of length at most $2\log_2n + 1$. This corresponds to a cycle in $G''$ (and therefore a cycle in $G$) of length at most $\frac{c}{c-1}(2\log_2n +1)$, because each edge of the cycle in $H$ corresponds to at most $\frac{c}{c-1}$ edges in $G$.

In a cycle of n vertices, the girth is of n. Hence, an upper bound of O($log_{2}n$) on girth size won't exist. The problem with answer given by Misha Lavrov is that when we remove vertices with degree equal to 1, there might be a decrease in degree of remaining vertices.


Problem about the subgroup generated by the commutators. I want to make a question about this exercise.

  Let $G$ be a group, and let $a$, $b$, be elements of $G$. We define the commutator of $a$ and $b$ as follows:

  \begin{equation}
[a,b]:=aba^{-1}b^{-1}.
\end{equation}

  Let $C:= \langle \{[a,b] \mid a,b\in G\} \rangle$ be the subgroup generated by all the commutators in $G$.

  ($i$) Prove that $\forall\ a,b,c\in G$, $[a,b]^{-1}=[b,a]\ $, and $c[a,b]c^{-1}=[cac^{-1},cbc^{-1}].$

  ($ii$) Using ($i$), prove that $C \lhd G$.

I proved ($i$), and I'm now stuck on $(ii)$. My try has been this:

  We have to see that $\forall\ g\in C, [a,b]\in C$ it happens that
  \begin{equation}
g^{-1}[a,b]g\in C.
\end{equation}

  But this is true, because by ($i$) we have that
  \begin{equation}
g^{-1}[a,b]g= [g^{-1}ag,g^{-1}bg]
\end{equation}

Now I would say that it's over, this element is always in $C$ and hence $C \lhd G$. Is this correct?

Thanks.

There's something missing from your argument.  $C$ is normal if for all $c \in C$ and all $g \in G$ we have $g^{-1}cg \in C$.  But $C$ is the subgroup generated by all commutators: it is (in general) not just the set of commutators.  Elements of the subgroup generated by a set $S$ can be written as products $s_1,\ldots,s_n$ where each $s_i$ is either an element of $S$ or the inverse of an element of $S$.

This isn't a big problem for your argument, but you need something extra.  Let $c$ be any element of $C$. Then $c$ equals a product $c_1 c_2 \cdots c_n$ where $c_i = [g_i,h_i]^{\epsilon_i}$ where $\epsilon_i= \pm 1$ Then $g^{-1}cg = \cdots$

Try to prove this first:

   any element of $C$ is a finite product of integral powers of commutators of $G$

Thus it is enough  to show for any element $g\in G$ $g(aba^{-1}b^{-1})g^{-1}\in C$

Now $g(aba^{-1}b^{-1})g^{-1}=((ga)b(ga)^{-1}b^{-1})bgb^{-1}g^{-1}\in C$


Reference to angular distance metric properties. I know that the Angular Distance is a proper metric but I'm struggling to find a reference that states that and proves all the properties for that distance. In the book Mining of Massive Datasets (Page 95) the authors briefly argue that the properties hold for what they call "Cosine Distance", which is just the angle between two vectors.
However, I've read that the Angular Distance must be normalised in order to be considered a metric.

Does the Angular Distance have to be normalised for it to be a metric?
Can you point me to a reference that proves that the Angular Distance is a proper metric?

Metric

Let $X$ be an arbitrary set. A function $d: X \times X \rightarrow \mathbb{R}$ is called a metric on $X$, iff
        the following properties hold true for all $x, y, z \in X$:

identity of indiscernibles: $\forall x, y \in X: d(x, y) = 0 \Leftrightarrow x = y$
symmetry: $\forall x, y \in X: d(x, y) = d(y, x)$
triangle inequality: $\forall x, y, z \in X:d(x, z) \leq d(x,y) + d(y, z)$

Cosine Distance

Cosine distance is defined as:

$$d_{cos}(x, y) := 1 - \cos(\theta) = 1 - {\mathbf{A} \cdot \mathbf{B} \over \|\mathbf{A}\| \|\mathbf{B}\|} = 1 - \frac{ \sum\limits_{i=1}^{n}{A_i  B_i} }{ \sqrt{\sum\limits_{i=1}^{n}{A_i^2}}  \sqrt{\sum\limits_{i=1}^{n}{B_i^2}} }$$

Cosine distance is not a metric in $\mathbb{R}^n$, because the identity of indiscernibles does not hold true:

$$d(0.5, 1) = 1 - \cos(0°) = 1 - \frac{1}{1} = 0\text{, but } 0.5 \neq 1$$

To the question if cosine distance is a metric on $S_n = \{x \in \mathbb{R}^n: ||x|| = 1\}$:

\begin{align}
x &= y\\
\Rightarrow d(x,y) &= d(x,x)\\
&= 1 - \sum_{i=1}^n x_i^2\\
&= 0
\end{align}

So that looks good.

We also know that $\cos(\theta) = 1 \Leftrightarrow \theta \in \{0°, 180°\}$. This means it still does not have the identity of indiscernibles attribute. We have to reduce the space to the non-negative unit sphere $S_n^+ = \{x \in \mathbb{R}^n: ||x|| = 1 \land x^{(i)} \geq 0\}$.


Proof that a stopped continuous-time martingale is a martingale. The proof for a stopped discrete-time martingale is shown as follows.

  Let $M=(M_n)_{n\ge0}$ be a discrete-time martinglae w.r.t. the filtration $(\mathcal F_n)_{n\ge0}$, and let $M^T=(M_{n\land T})_{n\ge0}$ be the stopped martingale, where $T$ is a stopping time w.r.t. $(\mathcal F_n)_{n\ge0}$. Since:

  We have
  $$\begin{align}
M_{n\land T}&=1_{T\ge n+1}M_n+1_{T\le n}M_T\\
&=1_{T\ge n+1}M_n+ \sum_{k=1}^n 1_{T=k}M_k\\
\end{align}$$
  where $1_{T\ge n+1}, M_n, 1_{T=k}$, and $M_k$ are all $\mathcal F_n-measurable$, hence $M_{n\land T}$ is $\mathcal F_n-measurable$;
  ${\Bbb E}|M_{n\land T}| \le \underbrace{{\Bbb E}|M_1|+…+{\Bbb E}|M_n|}_{{\Bbb E}|M_i| \text{ is integrable, }\forall i\ge0}<\infty$, i.e., $M_{n\land T}$ is integrable $\forall n \ge 0$;
  We have
  $$\begin{align}
{\Bbb E}(M_{n+1 \land T}|{\mathcal F_n}) &= {\Bbb E}(M_{n \land T}+1_{T \ge n+1}(M_{n+1}-M_n)|{\mathcal F_n})\\
&=M_{n \land T}+1_{T \ge n+1}{\Bbb E}(M_{n+1}-M_n)\\
&=M_{n \land T}\\
\end{align}$$

  Therefore, the stopped martingale satisfies the definition of a discrete martingale. Proof complete.

However, I am not able to extend these three parts of proof to a continuous version, because I cannot devide the time into separate spots with one next to another as the discrete version did. So I really wonder how to give a proof for a continuous-time martingale.

We are going to prove the following proposition:

If $M$ is a continuous martingale and $T$ a stopping time, the stopped process $M^{T}$, i.e., $\{M(t \wedge T), t \geq 0\}$ is a martingale with respect to $\left(\mathcal{F}_{t}\right)$.

Here we go.

The process $M^{T}$ is obviously continuous and adapted.
Firstly we use a weak form of optional stopping theorem, saying that a martingale has equal expectation at any bounded stopping time. If $S$ is a bounded stopping time, so is $S \wedge T$; hence $$ E\left[M_{S}^{T}\right]=E\left[M_{S \wedge T}\right]=E\left[M_{0}\right]=E\left[M_{0}^{T}\right]. $$
    Then we use this conclusion twice to get our desired equality. If $s<t$ and $A \in \mathcal{F}_{s}$ the r.v. $T=t \mathbf{1}_{A^{c}}+s \mathbf{1}_{A}$ is a stopping time and consequently
    $$ E\left[X_{0}\right]=E\left[X_{T}\right]=E\left[X_{t} \mathbf{1}_{A^{c}}\right]+E\left[X_{s} \mathbf{1}_{A}\right] .$$
    On the other hand, $t$ itself is a stopping time, and $$ E\left[X_{0}\right]=E\left[X_{t}\right]=E\left[X_{t} \mathbf{1}_{A^{c}}\right]+E\left[X_{t} \mathbf{1}_{A}\right]. $$

Comparing the two equalities yields $X_{s}=E\left[X_{t} | \mathcal{F}_{s}\right].$

With dyadics, page 2 : http://www.dam.brown.edu/people/huiwang/classes/Am264/Archive/Brownian.pdf. The idea is to discretize the time, build discrete stopping times on these time slots, and conclude using backward martingales.

Optional sampling theorem : If $X = (X_t,\mathcal{F}_t)$ is a supermartingale and $T$ is an arbitrary stopping time, then the stopped process $X^T=(X_{T\wedge t},\mathcal{F}_t)$ is also a supermartingale.

Proof : Here we give a sketch of the proof. Assume $X$ is a supermartingale. First, for a fixed $n\ge 1$, let
$$
D_n = \left\{\frac{k}{2^n}, k= 0,1,2,\ldots\right\}\subset D_{n+1}\subset \cdots
$$
be the set of non-negative dyadic rationals of order no greater than $n$. It follows that
$$
X = (X_t,\mathcal{F}_t; t\in D_n)
$$
is a super martingale (discrete time).

Second, we construct a stopping time $T_n$ such that $T_n\ge T$ and $T_n$ only take values in $D_n$. Indeed, let
$$
T_n(\omega) = \inf\left\{t\in D_n; t\ge T(\omega)\right\}.
$$
Then $T_n\ge T_{n+1} \ge \cdots $ and $T_n$ is a stopping time. Fix $0\le s\le t$, we wish to show that
$$
\mathbb{E}\left[X_{t\wedge T}|\mathcal{F_s}\right]\le X_s
$$
almost surely. Similarly define
$$
t_n = \inf\left\{u\in D_n; u\ge t\right\} \ge t_{n+1}\ge\cdots
$$
and
$$
s_n = \inf\left\{u\in D_n; u\ge s\right\} \ge s_{n+1}\ge\cdots
$$
It follows from the discrete time optional sampling theorem that
$$
\mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s_m}\right] \le X_{s_m \wedge T_n}
$$
for any integers $m\ge n$. Letting $m\to\infty$, we have $s_m \to s$ decreasing and $\mathcal{F}_{s_m}\to\mathcal{F}_s$. By Lévy's Downward theorem and the continuity of process $X$, we have
$$
\mathbb{E}\left[X_{t_n\wedge T_n}|\mathcal{F}_{s}\right] \le X_{s_m \wedge T_n}
$$
Observe that $(X_{t_n\wedge T_n},\mathcal{F}_{t_n\wedge T_n})$ is a backward martingale, whence it is uniformly integrable. Letting $n\to\infty$, we arrive at
$$
\mathbb{E}\left[X_{t\wedge T}|\mathcal{F}_{s}\right] \le X_{s\wedge T}.
$$
This completes the proof.


Utility of the Derivative of Laplace Transforms for ODE's Many texts discuss the derivative of Laplace transform $dF(s)/ds$. In general, differentiation of a Laplace is equivalent to multiplying the original function by $t$, and vice versa. So, if $\mathscr{L}\{f(t)\} = F(s)$, then:
$$\mathscr{L}\{tf(t)\}=-\frac{dF(s)}{ds} $$
Similarly, if $\mathscr{L}\{f'(t)\} = sF(s) - f(0)$, then:
$$\begin{split}\mathscr{L}\{{tf'(t)}\}&=-\frac{d}{ds}\left\{sF(s) - f(0)\right\}\\
                                       &=-s\frac{dF(s)}{ds} - F(s)\end{split}$$
The problem (my problem) with the last equation is that a derivative with non-constant coefficient $(tf'(t))$ yielded a derivative with non-constant coefficient $(sF'(s))$. How then would this transform be useful? Are there any examples of the utility of this property of Laplace transforms.

This can be used to solve differential equations having a derivative term multiplied by $t$. e.g. : $$t\frac{\mathrm dy}{\mathrm dt}=t.$$


Does probability of picking something change if we repick from a part of original objects? Here's an illustration , suppose I have of deck of 52 cards(well shuffled) and I pick 10 cards at random and now again I pick a card from those 10 cards,

(I)what is the probability that it's ace of diamonds?

(II)what is the probability that a cards drawn is a face card.
By deck of cards I mean normal playing cards

I feel the answer should be 1/52.
I have tried searching this over web but no success.

The probability of drawing out an ace of diamond from those 10 cards depend on whether it is in those 10 cards or not. The probability of the ace of diamond  being in the first 10 cards that were drawn is ${10\over 52}$ and probability of drawing an ace of diamond out of those 10 card is ${1\over 10}$. So in total the probability of drawing an ace of diamond is
$${10\over 52}*{1\over 10}={1\over 52}$$
So yes, your answer is right. The probability is not changed, its still ${1\over 52}$.
I think we can use proportion for the part ii). As saying I have got 5 marks out of 10 is equivalent to I have got 50 out of 100. Similarly there are 12 face cards in 52 cards is equivalent to we have ${30\over 13}$ face cards in those 10 cards. So the probability of choosing a face cards from those 10 cards is $${{30\over 13}\over 10} = {3\over 13}$$ The first part can be solved in the same way.

Its wrong  . This is a classical probability question , if 10 cards are selected from 52 , the number of diamond , aces , queens must be know before you can take the probability , the no o diamonds in the 10 cards was not give, so you can't take the probability


Does the operation of $G/A$ on $A$ by conjugation just give a trivial homomorphism? I'm having doubts about an exercise, since I reach a trivial conclusion. I hope it's not an issue to see if I'm screwing up.

Let $G$ be a group, and $A$ a normal abelian subgroup. Show $G/A$ operates on $A$ by conjugation and in this way get a homomorphism of $G/A$ into $\text{Aut}\ (A)$.

So the action is $gA\bullet a=gAa(gA)^{-1}=gAag^{-1}A=gAaAg^{-1}=gAg^{-1}=A$. Also $(gAhA)\bullet a=gA\bullet(hA\bullet a)$ and $A\bullet a=A$. So is this just a trivial homomorphism which maps everything to $A$?

I think I'm applying things wrong since I thought it's supposed to be $A\bullet a=a$ and the action should return elements of $A$, not $G/A$. How to interpret correctly?

Remember that any action of a group $G$ on a set $X$ induces a homomorphism $G\to S_X$, the group of permutations on $X$; and conversely, that any group homomorphism $G\to S_X$ induces an action of $G$ on $X$. Given an action $G\times X\to X$, the homomorphism $\varphi\colon G\to S_X$ maps $g$ to $\varphi(g)$, the function such that $\varphi(g)(x) = g\cdot x$. Conversely, given a homorphism $\psi\colon G\to S_X$, $\psi$ defines an action by $g\cdot x = \psi(g)(a)$.

With that preliminary, let let $G$ act on $A$ by conjugation in the usual way: $g\cdot a=  gag^{-1}\in A$ (it's in $A$ because $A$ is normal). This is an action of $G$ on $A$.

This induces a homomorphism $G\to S_A$, where we consider $A$ as a set.

However, because $A$ is abelian, for every $a\in A$ and every $b\in A$ we have $b\cdot a = a$. That is, the action of $A$ on itself is the identity permutation, so the homomorphism $G\to S_A$ factors through $G/A$. This gives a homomorphism $G/A \to S_A$. And of course, any homomorphism from a group $H$ to a permutation group $S_X$ induces an action of $H$ on $X$, so $G/A$ acts on $A$.

The action of $G/A$ on $A$ is as follows: given $gA\in G/A$ and $a\in A$, $gA\cdot a = gag^{-1}$. This is well defined, because if $gA=hA$, then $h^{-1}g\in A$, so $(h^{-1}g)a(g^{-1}h) = a$, hence $gag^{-1}=hah^{-1}$.

The action is not trivial unless $A$ is central.

Let $\alpha$: G-->Aut(A) is a homomorphism, then if $\alpha$(g)(a)=ga$g^{-1}$=a, we can get that g belong to A, so ker$\alpha$=A, from this point we obtain $\alpha^{'}$ which is induced by $\alpha$:G/A-->Aut(A), which is a injective inner aut.


According to math-quiz.co.uk the $x^3+12x^2−19x=−20$ factorised is $(x−4)(x+5)(x+1)$ My answer in red under "Your answer" isn't my answer I just had to chose something since it was a multiple choice

My working to check if it's correct

$(x−4)(x+5)(x+1)$

$(x^2-4x+5x-20)(x+1)$

$x^3+x^2-4x^2-4x+5x^2+5x-20x-20$

$x^3+2x^2-19x-20$

Am I wrong or is math-quiz.co.uk wrong?

The given expression has a real root $\approx  -13.5153$ and two complex roots.

The question was set cleverly to start with, but the setter seems somehow upset.

Edit=I hadn't checked your answer, I just saw their answer is wrong.

Maths quiz is wrong, but your answer is wrong too. The real answer is a lot more complicated, and includes a factor with no real solutions.

Your proof that their answer is wrong is correct though.


sliding a shape with area= 5 on a grid so it covers at least 6 of it's points - riddle place a shape on an integer coordinates grid, which is continuous without holes, and that its area is 5. Explain why you can slide it (without twisting or warping it), so that it will cover at least six points of the grid?

Let $A$ be any compact set with area 5. Fix a point $P$ in $A$, and define the function $f:[0,1]\times[0,1] \to \mathbb Z_{\ge0}$ by

$(x,y) \mapsto$ the number of lattice points covered by $A$ when $P$ is translated to $(x,y)$

Then the integral of $f$ over $[0,1]\times[0,1]$ is 5. So unless $f = 5$ almost everywhere, $f(x,y)$ must be $\ge$ 6 for some $(x_0,y_0)$.

Now, suppose that $f(x,y) < 5$ for some $(x,y)$. $A$ is compact, so its complement is open. Thus, when $P$ is at $(x,y)$, all lattice points that lie outside $A$ are bounded away from $A$. Hence there is a neighbourhood of $(x,y)$, with positive area, in which $f$ remains less than 5. (This uses the fact that $A$ is bounded, so only a finite number of lattice points can be close to $A$.) Therefore $f$ is not equal to 5 almost everywhere.

So now all we have to show is that $f$ can't be equal to 5 everywhere. Suppose otherwise, and imagine sliding $P$ from $(0,0)$ to $(1,0)$. At every point there are exactly 5 lattice points covered by $A$. But the set of lattice points covered when $P=(0,0)$ is not the same as the set of lattice points covered when $P=(1,0)$. So at some point $P=(a,0)$, a lattice point $L_1$ must get left behind by $A$ as it moves to the right. To compensate, a new lattice $L_2$ point must be consumed by $A$. But then at $P=(a,0)$, both $L_1$ and $L_2$ are in $A$, and we have $f(a,0) \ge 6$. QED

Note: It may be that more than one lattice point leaves $A$ at $P=(a,0)$. But then the same number of lattice points must be consumed by $A$, and the same argument applies.

Start with a shape that is 100 x 1/20 rectangle, and stick it on the x coordinate line. That covers 100 points


Linear Algebra Matlab matrix loop? I'm new to both linear algebra and Matlab, so naturally I'm probably asking an obvious question. If someone could attempt to explain to me in layman's terms how to go about answering this question in Matlab it'd be great, thanks in advance.

Loops are for suckers.

function x_new = matrix_recursion(A, endp, x_start)
 if (endp == 0)
  x_new = x_start
 else
  x_new = matrix_recursion(A, endp-1, A*x_start);
 end

I don't have MatLab on this computer so I can't test it, but it should work fine.


Is any open subset $A≠∅$ of $R^n $ with the usual topology uncountable? I think the answer is: yes, it is!. Due to all the points of an open set are interior points, the set contains a ball of radius r>0 , I tried to translate the question to demonstrate that any ball is an infinity uncountable set, but I don't see how to do it.
I would like to receive some suggestion.

Yes !!. If a non-empty finite set A, with cardinality 'n', was open. Take an element x$\in$A.  Then, there must be a ball around 'x', say B$_{\epsilon}$(x) completely in the parent set.

i.e. B$_{\epsilon}$(x)$\subseteq$A which will imply $|$ B$_{\epsilon}$(x)$|$$\leq$$|$A$|$ i.e. $\infty$$\leq$n. A contradiction.

Similarly, you can do the same for any countably infinite open set.


Proof that $a^2 + b^2 + ab$ and $a^2 + b^2 - ab$ can not both be perfect squares If $a$, $b$ are co-prime integers, how can I prove that $a^2 + b^2 + ab$ and $a^2 + b^2 - ab$ cannot be perfect squares? I know that perfect squares should be capable of expression in the form $a^2 + b^2 + 2ab$ and $a^2 + b^2 - 2ab$, but it is not immediately clear to me that there might not be other integers $x$, $y$ that would work even if $a$, $b$ do not. Thanks if anybody can help.

I should have said that $a$, $b$ are distinct integers and are both positive so that they could not both be equal to 1.

Please note that the premise of this question has now been disproved with a straightforward counter example, which is all the answer I require. Thanks to everyone who has taken the trouble to contribute. I am grateful for the help but won't be continuing to monitor further responses.

Above simultaneous equations shown below:

$(a^2+ab+b^2)=p^2$

$(a^2-ab+b^2)=q^2$

I don't think there are integer solution's to above equation.

But if "OP' is interested there is solution in irrational's

and is given below:

$(p,q)=((m^2+mn+n^2),(m^2-mn+n^2))$

where,  $(m,n)$=$((2+5^{1/2}),(1))$

$(a,b))$=$((8+4*5^{1/2}),(5+2*5^{1/2}))$

$(p,q)$=$((12+5*5^{1/2}),(8+3*5^{1/2}))$


Finding a closed form solution for recurrence Consider the following recurrence:

$P(1) = 3$
$P(n) = 3nP(n-1)$

I have to find a closed form for this recurrence. Expanding it a bit, we get:
$$P(n) = 3nP(n-1) = 3n(3(n-1)P(n-2)) = 3n(3(n-1)3(n-2)P(n-3)) = 3^3n(n-1)(n-2)P(n-3)$$
The pattern looks something like $P(n) = 3^kP(n-k)$, however it is missing those terms in between. How to genelarize them in this case? It reminds me of factorial, but it is still blurry to me.
Any help is appreciated.

Your enumeration suggests, after backwards-iterating $k$ times to the $P(n-k)$ term,

$$P(n) = 3^k n^{\underline k} P(n-k)$$

Here, $n^{\underline k}$ is the Pochhammer symbol - it denotes a falling factorial when used like this, which is like a factorial that doesn't finish. You can express it in several equivalent ways depending on preference, experience, and handiness:

$$n^{\underline k} = \underbrace{n(n-1)(n-2)\cdots (n-k+1)}_{\text{total of k factors}} = \prod_{i=0}^{k-1} (n-i) = \frac{n!}{(n-k)!} = P(n,k) = \; _nP _k$$

You can read up more on the Pochhammer symbol on Wikipedia.

In any event, the pattern is clear, and suggests, once you set $k=n-1$ to get to your initial condition of $P(1)=3$, the above expressions simplify and you would see

$$P(n) = 3^{n-1} \cdot n! \cdot P(1) = 3^n \cdot n!$$

What I'll leave up to you: verify that this does in fact work. You can prove this by induction on $n$.

You can utilize factorial $n!$ or permutation $_nP_r$


If $\sum_{n=1}^\infty a_n$ is convergent, then $\sum_{n=1}^{\infty} a_n\sin(n)$ If the series $\sum_{n=1}^\infty a_n$ is convergent (absolutely convergent or conditionally convergent), then
  $$ \sum_{n=1}^\infty a_n \sin(n) $$
  is also convergent.

Any hint? today I'm not at my best. Thanks in advance.

When the series is absouletely convergent, the proof is fairly straightforward as we have
$$
\sum_{n=1}^\infty |a_n| \geq \sum_{n=1}^\infty |a_n \sin(n)|
$$

as $ |\sin(n) | \leq 1 $ for all n. For conditionally convergent series, I'm not too sure on how it can be proven. Indeed, for a lot of conditionally convergent series, doing anything to the $a_n$ terms can cause the new series to not converge.

Hint: Using the alternating series test we know that for each monotonically decreasing sequence $a_n$ with nonnegative $a_n$ and $a_n\to 0$ the series $\sum_{n=0}^{\infty}(-1)^na_n$ converges.


Simulation of 2-dimensional Brownian motion I am trying to simulate (for the first time) a 2-dimensional SDE, in Matlab.

$$dX(t)=F(t,X(t))\,dt + \sigma(t,X(t))\,dBt$$

I have no problem using the Euler-Maruyama method in the one dimensional case, but I am not sure if I am correct with the 2-dimensional Brownian motion $B=(B_1,B_2)$.

I need to simulate two sequences $dB_1$ and $dB_2$. Can I just simulate 2 sequences in Matlab $\sqrt{dt}*randn$? Will they be independent ?

I tried to search for tutorials on simulation of two independent Brownian motions. Can you help me with this? What is the easiest way of generating two independent sequences of normally distributed random numbers?

Thank you :)
Bilka

The following code

>> X = cumsum(0.01 * randn(10000,2));
>> plot(X(:,1), X(:,2))
>> axis equal

generates this plot. Adjust parameters to taste.

"Can I just simulate 2 sequences ..."

Yes you can just do it in the straightforward way, and it will do the right thing.


Upper bound on the expectation of a random variable indexed by a random variale Imagine that we are given $X(u)$, a random variable indexed by some parameter $u \in I \subset \mathbb{R}$. Furthermore, assume that we know that:
\begin{equation*}
\mathbb{E}(|X(u)|)\leq 1 \quad \text{$\forall$ $u\in I$.}
\end{equation*}
Now, let $\theta$ be a random variable such that $\mathbb{P}(\theta \in I) = 1$. We do not suppose that $\theta$ and $X(u)$ are independent.
Does it hold that:
\begin{equation*}
\mathbb{E}(|X(\theta)|)\leq 1 \quad \text{?}
\end{equation*}

At least one declared downvote for extramathematical reasons, by user @stuartstevenson. This sucks.

The result does not hold without independence, as the simple example below shows.

Consider the parameter space $I=[0,1)$ and the probability space $(\Omega,\mathcal F,P)$ defined by $\Omega=[0,1)$, $\mathcal F=\mathcal B(\Omega)$, and $P$ the Lebesgue measure restricted to $\Omega$.

For every $u$ in $I$ and $\omega$ in $\Omega$, let $X(u)(\omega)=2\cos^2(2\pi(\omega+u))$. For every $\omega$ in $\Omega$, let $\theta(\omega)=1-\omega$.

Then, $E(X(u))=1$ for every $u$ in $I$ but the random variable $Y=X(\theta)$ is, by definition, such that $Y(\omega)=X(1-\omega)(\omega)=2$ for every $\omega$ in $\Omega$ hence $E(Y)=2$.

Yes, it holds.

If $\theta \in I$ with probability $1$ then $\theta \in I$.

The rest follows from definition.

The only issue I might see is if this is in some kind of limiting sense in which case you might use the suffix, "in the limit" or "almost surely" depending on the context.


Problem on total probability Let us conisder the following question from the textbook named Introduction to Probability

  Alice is taking a probability class and at the end of each week she
  can be either up-to-date or she may have fallen behind. If she is
  up-to-date in a given week, the probability that she will be
  up-to-date (or behind) in the next week is 0.8 (or 0.2, respectively).
  If she is behind in a given week, the probability that she will be
  up-to-date (or behind) in the next week is 0.6 (or 0.4, respectively).
  Alice is (by default) up-to-date when she starts the class. What is
  the probability that she is up-to-date after three weeks?

My solution:

Therefore my answer is $$0.8 \times 0.8 \times 0.8+0.8 \times 0.2 \times 0.6+0.2 \times 0.6 \times 0.8+0.2 \times 0.4 \times 0.6 = \bf{0.752}$$

but according to textbook, it is $\bf{0.688}$.

Where I went wrong?

Let U be uptodate, B be behind.

The four possible paths are from U to {UUU,UBU,BUU,BBU}

The required probability $= 0.8^3+(0.8)(0.2)(0.6) + (0.2)(0.6)(0.8) + (0.2)(0.4)((0.6) = 0.752$

The book answer is wrong.

Way accepted answer is arrived at is little incorrect. Looking at Reference, it is clear that solution arrived in original question is correct - http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.706.2591&rep=rep1&type=pdf

This is a question of total probability. So here our possible paths are only two - {UUU, UBU}. Given first week is U (by default up-to-date). We need to calculate probability of week2 and week3 given week1

Let Ui be up-to-date in given week and Bi behind in given week. Given week1 is by default up-to-date, represented as (U1).

So calculation will be -

P(U3) = P(U3|U2)*P(U2) + P(U3|B2)*P(B2)
P(U3) = 0.8*P(U2) + 0.6*P(B2)

Now, we need to find values of P(U2) and P(B2)
P(U2) = P(U2|U1)*P(U1) + P(U2|B1)*P(B1)
P(U2) = 0.8*0.8 + 0.6*0.2 = 0.76

Also,
P(B2) = P(B2|U1)*P(U1) + P(B2|B1)*P(B1)
P(B2) = 0.2*0.8 + 0.4*0.2 = 0.24

So, replacing values in our equation 1 -
P(U3) = 0.8*P(U2) + 0.6*P(B2)
P(U3) = 0.8*0.76 + 0.6*0.24 = 0.752


Peano Induction Axiom This is a typical rendition the Peano Axiom of Induction:

If subset $S \subseteq \mathbb{N}$ contains $1$ and is closed under the successor function (i.e., $n \in S$ implies $\sigma\text{n} \in S$ where $\sigma$ is the successor function) then that subset $S$ is all of $\mathbb{N}$.

How do we make the leap that $S$, which might not be all of $\mathbb{N}$ eventually does cover all of $\mathbb{N}$? I can only assume $\sigma\text{n}$ takes things outside of $S$?

How do we make the leap that $S$, which might not be all of $\mathbb{N}$ eventually does cover all of $\mathbb{N}$? I can only assume $\sigma n$ takes things outside of $S$?

When you say that "$S$..might not be all of $\mathbb{N}$", you are actually trying to find out whether $S$ and $\mathbb{N}$ are identical or not. In other words, you need a proof of the question whether $S=\mathbb{N}$ or not. But you can only prove the equality between two sets when you know that definition of each set. So, you need to answer first,

  What is the definition of $\mathbb{N}$?

The Peano Induction Axiom can be thought of as one of the 5 essential properties of the natural numbers. See Peano's Axioms. (There are other, less widely used formulations.)

The Induction Axiom says that, if you have a subset $P$ of $\mathbb{N}$, then we can show that all elements of $\mathbb{N}$ will be in $P$ if

(a) $1 \in P$, and

(b) if $x \in P$, then the successor of $x$ will also be in $P$


Why are the diagonals of a parallelogram not equal? Imagine a parallelogram and draw its diagonals. Now the areas of the two triangles on one of the bases is equal. But by Heron's formula, the areas are not equal. So what is the explanation for it.

Your reasoning is a little murky, but it looks like you're thinking:

  Here are two triangles with the same $a$ and $b$ but different $c$. Since the area can be computed from $a$, $b$ and $c$ using Heron's formula, different $c$s lead to different areas, and so the triangles can't have the same area.

But it is not true that different $c$ necessarily lead to different areas. Heron's formula is
$$ \sqrt{s(s-a)(s-b)(s-c)} \qquad\text{where }s=\frac{a+b+c}2$$
Increasing $c$ while holding $a$ and $b$ constant will make $s$ larger, and therefore the $s(s-a)(s-b)$ factors become larger. But the $s-c$ factor becomes smaller by an amount that can dominate the increase in the three other factors -- or, in appropriate cases, can cancel it out exactly!

Indeed, setting $c$ to either $a+b$ or $|a-b|$ will make the area $0$, and for some $c$ between those extremes the triangle will have maximal area, varying continuously with $c$. Every possible area less than the maximum is created by two different $c$, namely one above and one below the $c$ that gives maximal area.

I just encountered this question myself and after looking through some answers online I wasn't satisfied with any since I found them all to be over complicated and didn't really get to the point. So I'll post my conclusion here :

Imagine a parallelogram with horizontal lines AB and CD and vertical lines AC and BD.
Now imagine dropping a perpendicular line from C onto AB and another perpendicular line from B onto CD. You'll see that these two perpendicular lines and the horizontal lines of the parallelogram form a rectangle.

The diagonal CB of the parallelogram is also one of the diagonals of this rectangle as C and B are 2 of 4 vertices that make up the rectangle.
A rectangle's diagonals are equal, so a diagonal equal to CB would be defined by the points where the perpendicular lines dropped from B and C intersect the horizontal lines of the parallelogram and define the remaining 2 vertices of the rectangle.

Try drawing this and you'll see that the 2nd diagonal of the parallelogram, AD, is clearly not equal to this and therefore not equal to the diagonal CD either.


Differential forms and smooth functions Let $f_1$ and $f_2$ two smooth functions of the real variables $x$ and $y$. My questions is about the existence of a function $g$ non-zero such that $gf_1dx+gf_2dy$ is the differential of a smooth function. Is this equivalent to find a $g$ such that $(\partial/\partial y)(gf_1)=(\partial/\partial x)(gf_2)$? If this is true, how can I prove the existence of such $g$? Thanks!

Hint: Existence of differential form is one of the Goal of  De_Rham_cohomology which investigate for differential form with prescribed properties. and the best appraoch which presented above by Akaba comment is  g(x)=0 since it is  smooth function and satisfies the titled differential form .


Analysis, Limit Points of the alternating sequence (-1)^n I've recently learnt the definition of limit point in the context of metric spaces. I'm comfortable with the definition geometrically, but am having trouble showing that $$A=(-1)^{n}, n \in \mathbb{N} $$

Doesn't have any limit points.

I initally thought it would do, because for example, could I not take $$U(-1,4)$$ where $$U(a,r)$$ denotes the open ball center a radius r, r>0.  Then under the usual metric this would give an open interval from (-5, 3). Then looking at all points in A not including -1, we only have the point 1. But the point 1 is also in the open interval, so the intersection is non-empty...

I'm clearly misunderstanding the concept.

Any help appreciated,

Thanks.

You need to show that it holds for all balls centered at the supposed limit point, not just one.

For this problem, you must let $x$ be a real number, and find a deleted ball around it which does not intersect $A$.

The range of the sequence is the set E={1,-1} and it does not converge to some point. 1 and -1 are not limit point because the ball $U(1,1/4)$ does not contain a point of E different from 1, and the ball $U(-1,1/4)$ does no contain a point of $E$ different from -1. So the sequence have no Limits Points.


Stuck in problem involving combination/ permutation How many ten-digit binary numbers have exactly three 1’s?

What I have find is that some common patterns are they have three 1s and seven 0s. I have an idea by common sense. But, how to prove it mathematically.

This is what I got

C(7,3) = 35

Not sure if go for permutation or combination

$\frac{10!}{7!3!}=120$  Sorry I am on mobile so I cannot write much. But there $3$ 1's and $7$ 0's. So by the permutation formula. Permutations:$\frac{n!}{r_{1}!r_{2}!...r_{n}!}$

There are 10 ways to put the first 1 in a 10 bit binairy number. Pick any position for the first 1. Then there are 9 slots left for the second 1. Again pick any position... Now there are 8 slots left for the a 1.

So I think the answer could be 10x9x8=720 combinations of binairy numbers containing exactly three ones.


boundary of any finite set of points in the plane The problem is sketch "boundary of any finite set of points in the plane" and determine its boundary. I'm totally confused since there'are infinitely finite sets in the plane, such as one point alone. How can I sketch its picu

There isn't one possible picture, but they are all about equally interesting. The boundary is the set of points in the plane which have both points in and out of your finite point-set 'near' them. Since a finite point set is exclusively made up of separated points (the interior of the set is empty) the boundary is empty.


Characterize all $m \in \mathbb N$ such that $\frac{m^2-1}{m^2-n} \in \mathbb N$ and $1<n<m^2-1$ This just came when I was considering when $m^2-1$ can be divisible by more than two numbers $1 \leq n \leq m^2-1$.

Since $n=1$ and $n=m^2-1$ are always solutions the question is about when it is possible to have $$\dfrac{m^2-1}{m^2-n} \in \mathbb N$$ for $1<n<m^2-1$

How to characterize all such $m$?

There are solutions for all $m > 2$: simply let $n = m^2 - m + 1$, then
$$
\frac{m^2 - 1}{m^2 - n} = \frac{m^2 - 1}{m - 1} = m + 1.
$$
Essentially your question is: when is $m^2 - 1$ prime? And the answer is: essentially never, because $m^2 - 1 = (m-1)(m+1)$, and unless $m = 2$ (or the unusual case $m = 1$) this gives two non-trivial factors.

To start with $$\frac{m^2-1}{m^2-n}-1=\frac{n-1}{m^2-n}$$
So $$n\lt m^2\lt 2n$$

For the first $10000$ values of $n$, 8537 have no solutions for $m$ in this range,
1244 have one solution, 170 have two, 48 have three and one, namely 4621,
has four solutions, in this case $m=68,69,71,76$.  All have $m=1$ as well.  $4621$ also has a negative solution for $m=67$.


Finding Pythagorean triplet given the hypotenuse I have a number $c$ which is an integer and can be even or odd. It is the hypotenuse of a right angled triangle. How can I find integers $a,b$ such that

$$ a^2 + b^2 = c^2 $$

What would be the complexity of the calculation?

You may go by trial and error, or through the following lines. Consider the factorization of $c^2$ and separate its members according to them being $4^n$, $p^{2m}$ with $p\equiv 1\pmod{4}$ and $q^{2h}$ with $q\equiv 3\pmod{4}$. Every prime $p\equiv 1\pmod{4}$ splits in $\mathbb{Z}[i]$ and so it can be represented as a sum of two squares in a essentially unique way, $p=a_p^2+b_p^2$. That representation can be recovered with a finite descent that exploits Lagrange's identity:
$$ (a_1^2+b_1^2)\cdot(a_2^2+b_2^2) = (a_1 a_2+b_1 b_2)^2 + (a_1 b_2-a_2 b_1)^2. $$
In order to compute $a_p$ and $b_p$ from $p$, we need about $\log(p)$ steps.
Have a look at the second part of my answer to this question in order to understand how the descent really works.

Given the representations of every prime $p\equiv 1\pmod{4}$ dividing $c$, we may also recover, through Lagrange's identity, every representation of $c^2$ as a sum of two squares.

c can be found as

$c=2 r^2 +s^2+2 r s$
$a=2r^2+2 r s$
$b=s^2+2 r s$

Edit
Was using x^2+y^2=z^2 variables changed to a^2+b^2=c^2


How to find a counter example for non convexity? Consider a simple function $f(x,y)=\frac{x}{y}, x,y \in (0,1]$, the Hessian is not positive semi definite and hence it is a non convex function. However, when we plot the function using Matlab/Maxima, it "appears" convex. For the sake of clarity we want to find points which violate the definition of convexity, in other words identify the region (in the plot) where the function is non-convex. Can you suggest a procedure to do the same ?

The Hessian is $$\nabla^2 f(x) = \begin{bmatrix} 0 & -y^{-2} \\ -y^{-2} & 2xy^{-3} \end{bmatrix}=\frac{1}{y^3}\begin{bmatrix} 0 & -y \\ -y & 2x\end{bmatrix}.$$
For nonzero $y$, the determinant $-y^{-4}$ is strictly negative. This means that there must be exactly one positive eigenvalue and one negative eigenvalue. The function is nonconvex over the entire domain. Indeed, it is nonconvex at every point $(x,y)$, $y\neq 0$.

The function is affine for fixed $y$, and convex for fixed positive $x$. So it may be that when you look at the MATLAB plot you're seeing the convexity of these cross sections. At any given point, however, it is actually concave along the direction defined by the eigenvector corresponding to the negative eigenvalue.

I cannot tell you how many people I have encountered that insist that this function is convex for positive $y$! Good for you for not falling into that trap.

EDIT: That suggests a way to graphically demonstrate non-convexity, by the way. Pick a point, say $x$, and find the negative eigenvector, $v$, or at least a vector satisfying $v^T\nabla^2f(x) v<0$. Plot the function $g(t)=f(x+tv)$. It will be concave, probably visually so, at $t=0$.

EDIT 2: Here you go: $x=0.25+t$, $y=0.75\pm t$:

I think if you only look at one quadrant of $f(x)=\frac{x}{y}$ then it appears to be convex but if you look at all the quadrants then you should see that its non-convex.


Second degree multi variable taylor polynomial Let f (x, y ) = x cos(πy ) − y sin(πx)  point: 1,2

I am following the standard formula, which starts with taking the partial of f with regards to x twice, which gives me:

ysin(πx)π  But plugging in 1,2 gives me 0 because sin(π) = 0. This isn't right, but I am not sure what I am doing wrong.

Example of formula: http://www.maa.org/press/periodicals/loci/resources/calcplot3d-an-exploration-environment-for-multivariable-calculus-taylor-polynomials-of-a-function-of

Thanks.

The Taylor series out to second-order terms for a function of two variables is $$
f(x,y)=f(x_0,y_0)+(x-x_0)f_x(x_0,y_0)+(y-y_0)f_y(x_0,y_0)+\frac12((x-x_0)^2f_{xx}(x_0,y_0)+2(x-x_0)(y-y_0)f_{xy}(x_0,y_0)+(y-y_0)^2f_{yy}(x_0,y_0))+O(\|(x,y)-(x_0,y_0)\|^3).
$$ Just as with the Taylor series for $\cos x$ at $0$, in which every other coefficient is $0$, it’s perfectly fine for some of these partial derivatives to be $0$. In this case, $f_y=-\sin{\pi x}-\pi x\sin{\pi y}=0$ and $f_{xx}=\pi^2y\sin{\pi x}=0$ at $(1,2)$, so the series won’t have a $(y-2)$ or $(x-1)^2$ term.


How can I find ALL complex roots of $z^4 + 1 = 0$? For $z \in \mathbb{C}$, how can I find all roots of the equation
$$
z^4 + 1 = 0
$$
Obviously, this equation implies
$$
z^4 = -1 = e^{i \pi}
$$
and thus, one of our roots must be
$$
z_1 = e^{\frac{i \pi}{4}}
$$

However, I am aware that this is not the only root. How can I find all other roots of this equation?

Actualy you should write

$$z^4=-1=e^{i(\pi+2k\pi)}\to z_k=e^{\frac{i\pi(2k+1)}{4}}$$

Now make $k=0,1,2,3$ and get your roots.

If $z=\mathrm e^{i\theta}$, you just have to solve
$$4\theta\equiv \pi\mod2\pi\iff\theta\equiv \frac\pi4\mod\frac\pi 2.$$

Alternatively, once you have one $4$th root of a complex number, you obtain all the roots multiplying it by the $4$th roots of unity, i.e. by $1,-1, i,-i$.


Is sample variance always less than or equal to population variance I was reading this wikipedia article on Bessel's correction:  https://en.wikipedia.org/wiki/Bessel%27s_correction.  The article says that sample variance is always less than or equal to population variance when sample variance is calculated using the sample mean.  However, if I create a numpy array containing 100,000 random normal data points, calculate the variance, then take 1000 element samples from the random normal data, I find that many of my samples have a higher variance than the 100,000 element population.

import numpy as np

rand_norm = np.random.normal(size=100000)

# save the population variance
pop_var = np.var(rand_norm, ddof=0)

# execute the following 2 lines a few times and and I find a variance of the
# sample that is higher than the variance of rand_normal
samp=np.random.choice(rand_norm, 1000, replace=True)

# calculate the sample variance without correcting the bias (ddof = 0)
# I thought that the variance would always be less than or equal to pop_var.
np.var(samp,ddof=0)

Why am I getting sample variance which is greater than the population variance?

You have misinterpreted the article. The passage you are looking at never says anything about the actual population variance.

The passage literally says:

  Now a question arises: is the estimate of the population variance that arises in this way using the sample mean always smaller than what we would get if we used the population mean?

The pronoun "what" refers to an estimate of the population variance. To spell it out more explicitly, the article compares two ways of estimating the population variance:

Subtract the sample mean from each observed value in the sample. Take the square of each difference. Add the squares. Divide by the number of observations.
Subtract the population mean from each observed value in the sample. Take the square of each difference. Add the squares. Divide by the number of observations.

The article then says that Method 1 always gives a smaller result except in the case where the sample mean happens to be exactly the same as the population mean, in which case both methods give the same result.

This is a simple consequence of the not-so-simple fact that if you take any finite list of numbers $(x_1, x_2, \ldots, x_n)$ and consider the function $f(m)$ defined by
$$ f(m) = (x_1 - m)^2 + (x_2 - m)^2 + \cdots + (x_n - m)^2, $$

the smallest value of $f(m)$ occurs when $m$ is the mean of that list of numbers,
that is, when $m$ is the sample mean.

Notice that none of the preceding statements compared anything with the actual population variance. The actual population variance could be unknown.
All the above statements are concerned only with estimates of the variance.

All of this does not mean that every sample will underestimate the population variance. We might draw a sample in which the data values are unusually far from the population mean. But in that case the variance we would compute using Method 2 would overestimate the variance by even more than Method 1.
And while this can happen, it is not the usual thing to happen. More often the variance computed by Method 2 is nearly correct or smaller than the true population variance, and the variance computed by Method 1 is simply too small.

That's the thing about statistics like this.
You can use a bad method and yet it sometimes will give you a correct result just by luck.

From the quote, I think it may means that the expectation value of the sample variance is always less than or equals the expectation value of population variance. As we know, sample variance and population variance are both random variables.


Example of a very simple math statement in old literature which is (verbatim) a pain to understand As you know, before symbolic notations were introduced and adopted by the mathematical community, even simple statements were written in a very complicated manner because the writer had (nearly) only words to describe an equation or a result.

What is an example of a very simple math statement in old literature which is, verbatim, a pain to understand?

Here's Proposition 2 from Book 5 of Euclid's Elements:

  If a first magnitude and a third are equal multiples of a second and a fourth, and a fifth and a sixth are equal multiples of the second and fourth, then the first magnitude and fifth, being added together, and the third and the sixth, being added together, will also be equal multiples of the second and the fourth, respectively.

Or in modern notation: $a(x + y) = ax + ay$.

Euclid - "a point is that which has no part"


Question on irreducible polynomial I know the standard definition of an irreducible polynomial. But, in the book "a course in abstract algebra" by authors "Khanna and Bhambri", irreducible polynomial is defined as following:

  Let $R$ be an integral domain with unity. A polynomial $f(x)\in R[x]$ of positive degree(i.e. deg≥1) is said to be an irreducible polynomial over $R$ if it can not be expressed as product of two polynomials of positive degree. In other words if whenever $f(x)=g(x)h(x)$, then $\deg g(x)=0$ or $\deg\text h(x)=0$

According to usual definition of an irreducible polynomial, $f(x)=2x^2+4=2(x^2+2)$ is irreducible over $\mathbb{Q}$ but reducible over $\mathbb{Z}$(reducible over $\mathbb{Z}$ because neither $2$ is unit nor $x^2+2$ is unit in $\mathbb{Z}[x]$) . Hence according to our usual definition, above polynomial is reducible over smaller set $\mathbb{Z}$ but irreducible over bigger set $\mathbb{Q}$

Whereas, if I use above definition in the book then, $f(x)=2(x^2+2)$ then $f(x)$ is irreducible over $\mathbb{Q}$ and over  $\mathbb{Z}$ too. I think this looks more appropriate!  Irreducible in bigger set and hence it must be irreducible in smaller set!

So which definition is more correct? and is there is disadvantage of using above definition in book?

I think that the definition that $a$ is irreducible in $R$ if $a=xy$ in $r$ means one of $x$ and $y$ is a unit is more correct.
$x^4 +1$ is irreducible in  $\mathbb{Z}$ but is reducible in $\mathbb{Z}/{n\mathbb{Z}}$ for all $n$ >$1$


How to solve: $\frac{x}{\log_2(x) }= y$ For example, I can solve: $x \log_2(x) = y$

$x \log_2(x) = x \log_e(x) / \log_e(2) = e^{\log_e(x)} \log_e(x) / \log_e(2)$

$e^{\log_e(x)} \log_e(x) = y\log_e(2)$

$e^{W(z)} W(z) = z$, where W(z) is the Lambert W-function

$log_e(x) = W(y\log_e(2))$

$x = e^{W(y\log_e(2))}$

But how to (find $x$) solve: $\frac{x}{\log_2(x)} = y$

Answer:

$$\displaystyle\begin{array}$x&=& \frac{1}{e^{W(ln(\frac{1}{2})^{1/y})}} = \frac{1}{e^{W(−ln(2)/y)}}
\end{array}$$

$$\begin{array}{rcl}
\displaystyle\frac{x}{\log_2(x)}&=&y\\
\displaystyle\frac{x}{\ln(x)}&=&\displaystyle\frac{y}{\ln(2)}\\
\displaystyle\frac{\ln(x)}{x}&=&\displaystyle\frac{\ln(2)}{y}\\
\displaystyle\frac{-\ln(x)}{x}&=&\displaystyle-\frac{\ln(2)}{y}\\
\displaystyle\frac{1}{x}{\ln(\frac{1}{x})}&=&\displaystyle-\frac{1}{y}\ln(2)\\
\end{array}$$
Now, from Wikipedia, $W(t\ln(t))=\ln(t)$ so
$$\begin{array}{rcl}
\displaystyle W\biggl(\frac{1}{x}\ln\left(\frac{1}{x}\right)\biggr)&=&\displaystyle W\biggl(-\frac{1}{y}\ln(2)\biggr)\\
\displaystyle\ln\left(\frac{1}{x}\right)&=&\displaystyle W\biggl(-\frac{1}{y}\ln(2)\biggr)\\
\displaystyle\ln\left(x\right)&=&\displaystyle -W\biggl(-\frac{1}{y}\ln(2)\biggr)\\
\Rightarrow x&=&\displaystyle\mathrm{e}^{-W\biggl(-\frac{1}{y}\ln(2)\biggr)}
\end{array}$$


What does it mean when a number 'y' is pseudoprime to base 'x' I am self learner so I don't really understand about pseudoprime to base $x$

  For example, $91$ is a pseudoprime to base $3$ then is $91$ also a
  pseudoprime to base $2$?

thank you please explain.

edit** I did a bit research

  For example, an odd composite integer $N$ will be called a Fermat
  pseudoprime to base $a$, if $\gcd(a, N) = 1$ and $a^{N−1} \equiv 1 \pmod{N}$.

My question is what about base $2$? Do we use the same formula like $2^{90}$ and divide by $91$? If I don't get a remainder as $1$, it is not a pseudoprime, right? But when I plugged in on the calculator, the number is too big so how can I find a remainder on calculator?

Yes. If nothing else is said, it is assumed that pseudoprime means Fermat's pseudoprime, i.e., a number $n$ such that $a^{n-1}\equiv1\bmod{a}$. The number $a$ is the base here, so pseudoprime w.r.t. base $3$ is a number such that $3^{n-1}\equiv1\bmod 3$.

We call it a pseudoprime, because all primes are pseudoprimes w.r.t. all bases (but not vice versa, as you realized).


Trying to show that limit does not exist Consider the function, with domain $\mathbb{R}$, defined by

$$f(x)
= \begin{cases}
      x & \text{if $x$ is rational,} \\
      -x & \text{if $x$ is irrational.}
   \end{cases}
$$

When referring to above.

Q) Prove that for any $a \neq 0$, $f$ does not have a limit at $a$.

Is it sufficient for me to say:

Leading towards a contradiction, for $a=0$, $f$ does not have a limit at a. By the above, we see that this is not true, because when $a=0$, there exists a limit which has been proven. Hence, for any $a\neq0$, $f$ does not have a limit at $a$.

Okay, so the problem here, is that the contradiction of the statement "for any $a \neq 0$, $f$ does not have a limit  at $a$" , is not "for $a = 0$, $f$ does not have a limit at $a$". The answer is : "there is some $a \neq 0$ for which the limit exists", Therefore, behaviour of $f$ at $0$, is not required to answer this question.

In words : The limit of $f(x)$ at $x = a$ exists, if bringing $x$ close enough to $a$ means that $f(x)$ can be brought close enough to "some number", which we call the limit at that point.

Coming to the answer itself, we have to show that $\lim_{x \to a} f(x)$ does not exist. Why could this be true? Well, let $a$ be a rational point. There are plenty of rational and irrational points near $a$. Roughly, if $x$ a rational point close to $a$, then we know that $f(x) = x$ is very close to $f(a) = a$. But, if $x$ is an irrational point close to $a$ then is $f(x) = -x$ close to $f(a) = a$? We are trying to say that if $x$ is close enough to $a$, then $f(x)$ should be close enough to $f(a)$. But here, we are landing up, with both $x$ and $-x$ being close to $a$. How can this happen?

Well, it can happen if $x$ and $-x$ are close to each other i.e. $x$ is close to $0$, and therefore at $a =0$ we see that the limit exists intuitively. at other points, $x$ and $-x$ simply wouldn't be close to each other: if one of them is close to $a$, the other is going to be far away, and that's the problem.

A rigorous proof: See, in the limit definition, for each $\epsilon$ we find a $\delta$, right? Here, I will show you an $\epsilon$ for which no $\delta$ will work! And therefore, the limit wouldn't exist!

  Let $a \neq 0$. Put $\epsilon = \frac{|a|}{2}$. Note that $\epsilon > 0$ (this is important, right!) Finally, let $\delta > 0$ be any positive real number which is such that $|x - a| < \delta \implies |f(x) - f(a)| < \epsilon$. This means, by the triangle inequality, that $f(a) + \frac {|a|}{2} > f(x) > f(a) - \frac{|a|}{2}$. Since we know that $|f(a)| = |a|$, the above shows that $f(x)$ has the same sign for all $|x - a| < \delta$ i.e. it is either positive for all such $x$, or negative for all such $x$. However, if I pick $x$ such that $|x - a| < \delta$, that is opposite to $a$ i.e if $a$ is irrational, then $x$ is irrational , and vice-versa, then $f(x)$ and $f(a)$ would definitely have opposite sign, which contradicts our earlier statement. Hence, we see  that no $\delta$ exists for this $\epsilon > 0$, which finishes the arugment that the limit does not exist.

Vitally, the fact that $a \neq 0$ was used in the first line itself in this argument.

Again, the usual comment : read the proof carefully, and you can get back for further queries.

If $a$ rational, $f(a)=a$ and $f(a+{e\over n})=-a-{e\over n})$. We have $lim_na+{e\over n}=a$ and $lim_nf(a+{e\over n})=lim_n-a-{e\over n}=-a\neq f(a)$.

If $a$ is irrational, $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists $a_n, lima_n=a, a_n\in\mathbb{Q}$. We deduce that $lim_nf(a_n)=lim_na_n=a$ and $f(a)=-a$.


On group automorphism of subgroup a group $G$ Let $G$ be a group and $H$ be a subgroup of $G$. When is $\rm{Aut}(H)$ a subgroup of $\rm{Aut}(G)$?

The restriction of an automorphism of G to H is an automorphism of H requires that H is a characteristic  sub-groupe  in G, in this case the restriction morphism of Aut (G)   to Aut (H) has a sense and is an epimorphism. In  this cas it is hoped that Aut (G) is a direct product  of Aut (H)  ie  the exacte sequence

I said that a sufficient condition for the $Aut(H)$
to be  injected into $Aut(G)$ is that $H$  a characteristic
subgroup in $G$ and the  exact  short sequence:  $ ker(rest)
\hookrightarrow Aut(G)\rightarrow Aut(H)$(this last morphism is
the epimorphism restriction) split. Then a partial answer that I
deduced from the last comment given to this issue as similar
question in terms of exact short sequence of groups is this: under
the condition H to be a direct factor of G and simultaneously a
 characteristic subgroup in G, then $Aut(H)$ is injected into
$Aut(G)$. This injection may in no case be canonical (as Aut (.)
Is not a functor), but the restriction is canonical and has a
sense here because H supposed characteristic.

Prove: Suppose $G=H\times K$
and $H$ characteristic in $G$, so the restriction morphism
$Aut(G)\rightarrow Aut(H)$ is an epimorphism $f\mapsto f_H $, it
can viewed in this situation as the composite morphism $p_H\circ
f\circ i_H : H\hookrightarrow H\times K \rightarrow H\times
K\times \rightarrow H $ where $i_H$ the canonical
monomorphism,$f\in Aut(H\times K)$ and  $p_H$ the canonical
projection. Since $H$ characteristic in $H\times K$ then  $K$ is
also characteristic in $H\times K$ because for any $f\in
Aut(H\times K)$ the composite morphism $p_K\circ f$ left via the
naturel projection $p_K:H\times K\rightarrow K$. and so we
obtained   the direct product $Aut(H\times K)\simeq Aut(H)\times
Aut(K)$ with the isomorphism restriction $f\mapsto (f_H,f_K)$ and
of course the already exact short sequence split.


One event of a r.r.v as a function of another r.r.v Let $(X_n)_{n \in \mathbb{N}}$ a sequence of r.r.v (real-value random variable) such as

$\exists p \in [0,1] : \forall n \in \mathbb{N}$

$P(X_{n+1} = 0) = p$, and

$P(X_{n+1} = X_n + 1) = 1 -p$.

It was written in my example that $\forall n \in \mathbb{N}, E[X_{n+1}] = (1 -p)\ E[X_{n}]$. However, in the measure-theory perspective, I do not understand how that is possible. Indeed, by integrating $X_{n+1}$ over its set of events according to its discret probability measure, $E[X_{n+1}]$ should be equal to $0 \times p + (1 -p) X_{n}$, but $X_{n}$ is not here a real but a r.r.v. and therefore a measurable function.

1) Is this formulation of the distribution of $X_{n+1}$ correct?

2) If it is, is $E[X_{n+1}] = (1 -p)\ E[X_{n}]$ correct and why?

3) Else, what is the right result and formulation of $X_{n+1}$ as a distribution?

Thank you in advance

A rigorous formulation of the hypothesis (which is ambiguously stated) might be that $$X_{n+1}=B_{n+1}\cdot(X_n+1)$$ where $B_{n+1}$ is independent of $(X_k)_{0\leqslant k\leqslant n}$ and $$P(B_{n+1}=0)=p\qquad P(B_{n+1}=1)=1-p$$ Then, $$E(X_{n+1})=E(B_{n+1})\cdot(E(X_n)+1)=(1-p)\cdot(E(X_n)+1)$$ Thus, unless you copied something incorrectly, your source is wrong.

In terms of distributions (but this is not really useful), one would translate the above as $$P_{X_{n+1}}=p\,\delta_0+(1-p)\,\delta_1\ast P_{X_n}$$ where, as usual, $P_Y$ denotes the distribution of the random variable $Y$ and $\delta_x$ denotes the Dirac measure at $x$.


Prove that the limit exists and then find its limit We are given a function $h(x)$ which strictly positive. The function $h$ is defined on $\mathbb{R}$ and that satisfies the following:\

$$\lim_{x\to 0}(h(x)+\frac{1}{h(x)})=2$$

The question is to prove that the limit of $h$ exists at$ 0$ and then find its limit as $x\to 0$

If we assume that the limit exists (say it's equal to $\lambda$, then we have $\lambda+\frac{1}{\lambda}=2$ and by solving this equation, we get $\lambda=1$. However, I am completely clueless on how to prove the existence of the limit of $h(x)$ when $x\to 0$.

We have $h(x) +\dfrac 1{h(x)} \to 2$, so $$f(x) :=h(x) +\dfrac 1{h(x)}-2=\frac{h(x)^2 -2h(x) + 1}{h(x)}=\frac{(h(x) - 1)^2}{h(x)}\to 0$$as $x \to 0$.

Show that for $x$ sufficiently small $h(x)$, we have $\frac12<h(x) < 2$ (any upper bound will work). So $$f(x)>\frac12\big(h(x)-1\big)^2$$

Now can you use the squeeze theorem to derive the result?

Note that if either $h(x)$ or $\frac{1}{h(x)}$ is not bounded in $(-\varepsilon, \varepsilon)$ then the original limit cannot exist. We'll use that property further, refering it as $(*)$

$$\lim_{x\to 0}\left(h(x) + \frac{1}{h(x)}\right) = 2 \Leftrightarrow \lim_{x\to 0}\left(h(x) -2 + \frac{1}{h(x)}\right) = 0 \Leftrightarrow \lim_{x\to 0}\left(\sqrt{h(x)} - \frac{1}{\sqrt{h(x)}}\right)^2 = 0 \Leftrightarrow \lim_{x\to 0}\left(\frac{h(x) - 1}{\sqrt{h(x)}}\right)^2 = 0 \Leftrightarrow \lim_{x\to 0}\frac{(h(x) - 1)^2}{h(x)} = 0 \stackrel{(*)}{\Rightarrow} \lim_{x\to 0}(h(x)-1)^2 = 0$$

Wich means that $$\lim_{x\to 0} h(x) = 1$$


Solving linear congruence (modular inverse or fraction) via gcd Bezout equation Isn't finding the inverse of $a$, that is, $a'$ in $aa'\equiv1\pmod{m}$ equivalent to solving the diophantine equation $aa'-mb=1$, where the unknowns are $a'$ and $b$? I have seem some answers on this site (where the extended Euclidean Algorithm is mentioned mainly) as well as looked up some books but there is no mention of this. Am I going wrong somewhere or is this a correct method of finding modular inverses? Also can't we find the Bézout's coefficients by solving the corresponding diophantine equation instead of using the extended Euclidean Algorithm?

Yes, it's well-known and  occurs here many times, e.g. here where it is special case $\,b\! =\! 1\,$ below
$\quad\ \, \exists\, x\in\Bbb Z\!:\  ax\equiv b\pmod{\! m}\!\iff\! \exists\, x,y\in\Bbb Z\!:\ ax\!+\!my = b\!\overset{\rm\ Bezout}\iff{\color{#c00}{\overbrace{\gcd(a,m)}^{\large d}}}\mid b$
Proof $\ \ (\Rightarrow)\ \ ax\equiv b\pmod{\!m}\Rightarrow ax\!+\!my = b\,$ for $\,y\in\Bbb Z\,$ so $\,\color{#c00}{d\mid a,m}\Rightarrow\,d\mid \color{#c00}ax\!+\!\color{#c00}my = b.\ $
$(\Leftarrow)\ \ $ By Bezout: $\,\exists\,\bar x,\bar y\in\Bbb Z\!:\,$ $\,a\bar x\!+\!m\bar y = d\,$ $\Rightarrow\, a(c\bar x)\!+\!\color{#0a0}m(c\bar y) = b,\,$ via scaling by $\,c = b/d,\ $ hence reducing the prior equation $\!\bmod \color{#0a0}m\,$ yields $\,a(\color{#90f}{c\bar x})\equiv b\pmod{\!\color{#0a0}m},\,$ so let $\,x=\color{#90f}{c\bar x}$.
Remark $ $ Switching back-and-forth between a linear congruence and its associated linear Diophantine equation is something so basic and ubiquitous  that it is rarely explicitly mentioned - just as for  use of other basic laws (associative, commutative, distributive etc.)
By the above arrows, computing modular fractions (= inverses when $\,b=1)\,$ is equivalent to solving the associated linear Diophantine equation.
There are various ways to solve such congruences and equations, e.g. see here & here for a handful of methods applicable when the solution is unique.
See here for the general method when the solution is not unique, which includes a handy  fractional view of above, and its use in the extended Euclidean algorithm. As is often the case, use of fractions may yield  significant simplification and conceptual insight.
See here for the equivalence of the above and CRT = Chinese Remainder Theorem.

Sure you can do it that way. If low enough though, you can use long division on negative numbers. and then simple remainder knowledge ex.

$$\;\; -17\\7\overline{)-121}\\\underline{-(-70)}\\\; -51\\\;\underline{-(-49)}\\\;\; -2$$

which by $-121$ being $-1$ mod $120$ and $1-(-1)=2$ implies $-17$ mod $120$ is 7's  inverse multiplicatively. $-17\equiv 103\bmod 120$  is then it's least positive form.


GRE Quant Combinatorics problem The University of Maryland, University of Vermont, and Emory University have each $4$ soccer players. If a team of $9$ is to be formed with an equal number of players from each university, how many number of ways can the selection be done?

Possible answers are: $3, 4, 12, 16, 25$.

I have encountered that problem in GRE Quant.

Here is my solution: From each university we should take $9:3=3$ players and this choice could be done with $C_4^3=4$ ways. Hence the selection can be done in $4\times 4 \times 4=4^3=64$ ways. But such answer does not exist.

Can anyone explain this moment please?

I'd frame exactly the same computation you use as "the number of ways to pick the one player that is excluded from each University ...".  Of course, I'd get the same answer...

As a quick enumeration of at least 26 different ways to make some of these teams...

Let $a, b, c, d$ represent the four players from the University of Maryland.  Let $f, g, h, i$ represent the four players from the University of Vermont.  Let $p, q, r, s$ represent the four players from Emory University.  The question is this ungrammatical mess "If a team of 9 is to be formed with an equal number of players from each university, how many number of ways can the selection be done?"  In particular, "how many number of ways" is not grammatically correct English.  Nevertheless,  we will represent a "selection" by who from each University is not selected (since listing the three selected players is equivalent to listing the one non-selected player from each University).  \begin{align*}
afp && afq && afr && afs  \\
agp && agq && agr && ags  \\
ahp && ahq && ahr && ahs  \\
aip && aiq && air && ais  \\  \\
bfp && bfq && bfr && bfs  \\
bgp && bgq && bgr && bgs  \\
bhp && bhq && &&\text{+2 more}  \\
&&&&\text{+4 more} \\  \\
&&&&\text{+16 more} \\  \\
&&&&\text{+16 more} \\  \\
\end{align*}

I cross check with other answer is 12=4+4+4

The question is asking selection for each university hence Principal of Addition will apply. Principal of Multiplication will be applied when asked about number of possible combination of team.
The question specifically asks selection from each university not selection of team


How do I solve delta epsilon proofs for quadratic equations? For the $\lim\limits_{x\rightarrow 2} (x^2 + 5x - 2) = 12$ I need to show how to find a $\delta$ such that $|f(x) - L| < \varepsilon$ for all $x$ satisfying $0 < |x - a| < \delta$.

Help is appreciated.

$\forall\epsilon >0\ \exists\delta>0\ $ such that $|x-a|<\delta\implies |f(x)-L|<\epsilon$.  Then you have, $|x-2|<\delta\implies |x^2+5x-2-12|<\epsilon$.  Factoring the right side of $\implies$
$$|(x-2)(x+7)|<\epsilon\implies |x-2|<\frac{\epsilon}{|x+7|}$$. Now we need to get the fraction on the right hand side only in terms of $\epsilon$ and some number.  You can set $|x-2|<1,$ because the limit only applies as $x\to 2,$ so you can restrict $\epsilon$ to being $<1$ so $|x-2|<1$.  Then this gives $-1<x-2<1\implies 1<x<3\implies 8<x+7<10.$  Since the maximum of $|x+7|$ makes $\frac{\epsilon}{|x+7|}$ a miniumum, then $$|x-2|<\frac{\epsilon}{|x+7|}<\frac{\epsilon}{10}.$$  Now you can let $\delta=\min(1,\frac{\epsilon}{10}).$  Then, just to show you if you let $\delta=1$, then $|x-2|<\delta =1\implies |x-2||x+7|<|x+7|.$  Since you know that $1<\frac{\epsilon}{10}$, then this means that $\epsilon<10.$  Therefore, $|x-2|x+7|<|x+7|<10<\epsilon\implies |(x-2)(x+7)|<\epsilon\implies|x^2-5x-14|<\epsilon.$ The proof for $\delta=1$ is done.  Now you can prove it for $\delta=\frac{\epsilon}{10},$ and once you do that the proof is done in entirety.  I'll leave that last part up to you for when $\delta=\frac{\epsilon}{10}$.


Geometric interpretation for median formula $m_c = \sqrt{\frac{2(a^2 + b^2)- c^2}{4}}$ for $a + b < c$? For a triangular with side lengths $a, b, c$, the length of the median of $c$ is $m_c = \sqrt{\frac{2(a^2 + b^2) - c^2}{4}}$.

If $a + b < c$, the configuration is obviously not a triangle anymore.
However, there are values for $a, b, c$ with $a + b < c$ (e.g. $a = 7, b = 2, c = 10$) where the root is still positive. Is there any geometric interpretation for that?

If $a+b>c$, then we can draw circles of radius $a$ and $b$ about endpoints of segment $\overline{AB}$ of length $c$, and these circles will meet at distinct points $C$ and $C^\prime$. The circle about $M$, the midpoint $\overline{AB}$, with radius $m:=\frac12 \sqrt{2(a^2+b^2)-c^2}$, meets those points, as well. Thus, $\overleftrightarrow{CC^\prime}$ is a common secant line, and therefore also the common "radical axis" or "power line", of the circles. (When $a+b=c$, we can replace "common secant line" with "common tangent line".) This allows us to characterize the length $m$ thusly:

  The circle coaxal with $\bigcirc A$ and $\bigcirc B$, centered at the midpoint of $\overline{AB}$, has radius $m$.

(The diagram illustrates the property that, for any point $P$ on the radical axis of a pair (or family) of circles, the tangent segments from $P$ to each of those circles are all congruent.)

As it turns out, the same description holds for $a+b<c$.

Here's a quick coordinate proof (valid whether $a$, $b$, $c$ make a triangle or not) that leverages a convenient property of radical axis equations.

Take $A = (-c/2, 0)$, $B = (c/2, 0)$, $M= (0,0)$. Then the equations of the circles are
$$
\bigcirc A: \left(x + \frac{c}{2}\right)^2 + y^2 = a^2 \qquad
\bigcirc B: \left(x - \frac{c}{2}\right)^2 + y^2 = b^2 \qquad
\bigcirc C: x^2 + y^2 = m^2
$$

Now, the equation of the radical axis of two circles is obtained by eliminating the $x^2$ (and $y^2$) terms from the equations of those circles. With our equations, a simple subtraction will do:
$$\begin{align}
\text{radical axis of $\bigcirc{A}$ and $\bigcirc{B}$}\; &\;:\; \bigcirc{A} - \bigcirc{B}\;\quad\to\quad \phantom{-}2cx = a^2 - b^2 \\[4pt]
\text{radical axis of $\bigcirc{A}$ and $\bigcirc{M}$} &\;:\; \bigcirc{A} - \bigcirc{M} \quad\to\quad \phantom{-2}cx = a^2 - m^2 - \frac{c^2}{4} \\[4pt]
\text{radical axis of $\bigcirc{B}$ and $\bigcirc{M}$} &\;:\; \bigcirc{B} - \bigcirc{M} \quad\to\quad \;-cx = b^2 - m^2 - \frac{c^2}{4}
\end{align}$$

These three (vertical) lines will coincide precisely when
$$m^2 = \frac{1}{4}\left(\; 2 ( a^2 + b^2 ) - c^2 \;\right)$$

$\square$

For the median, what we need is not just $m_c\ge0$. We actually need $\displaystyle m_c\ge\frac{|a-b|}{2}$.

When $a=7$ and $b=2$, $\displaystyle \frac{|a-b|}{2}=\frac{5}{2}$.

Now $\displaystyle \sqrt{\frac{2(a^2+b^2)-c^2}{4}}=\frac{\sqrt{6}}{2}<\frac{5}{2}$.


How to show $f$ is uniformly continuous if it is continuous Let $X$ be a metric space such that every real-valued continuous function on $X$ is uniformly continuous on $X$. How to show that, given an arbitrary metric space $Y$ and a continuous function $f:X\to Y,$ $f$ is uniformly continuous on $X?$

Please help! I am just clueless to start.

A function $f:X\to {Y}$ is continuous at $a$ if and only if it is sequentially continuous at $a$.

Let $f$ is continuous at $a$. Let $\epsilon$ be given. Suppose $x_n\to {a}$ then there exists $\delta>0$ such that $d(f(x),f(a))< \epsilon$ for $d(x,a)<\delta$ and there exists $M\epsilon \mathbb{N}$ such that $d(x_n,a)\delta$ for $n>M$. It follows that $d(f(x_n),f(a))<\epsilon$ for $n>N$. So, $f(x_n)\to f(a)$ is sequentially continuous at $a$.

Conversely, suppose that $f$ is not continuous at $a$. Then there exists $\epsilon_0>0$ such that for every $n\epsilon N$ there exists $x_n \epsilon X$ with $d(x_n,a)<\frac{1}{n}$ and $d(f(x_n),f(a))>\epsilon_0$ then $x_n\to a$ but $f(x_n)$ doesn't approaches $a$. So, $f$ is not sequentially continuous at $a$

Uniform continuity can be proved in the same way.


Arranging cards so that no two consecutive values remain consecutive Let us say we have 52 cards with values ranging from 1-13 (4 sets of cards from 1-13). Assume that you wanted no two consecutive values to be next to each other in the pile of cards. For example, a 3 cannot be next to a 2 or a 4.
How many ways can I arrange these cards that there are no consecutive values next to each other?

Can someone suggest a permutation that fulfills these requirements or suggest a computer program to solve the problem?

total ways are 52!. now let us assume we have two cards consecutive . consider them as a single group we can have 104 ways . they can be arranged in 2! ways between themselves.We have 1 set of 13 so number of ways of 1 pair of two consecutive cards is 2!.11! such arrangements in 104 ways can be done and of four different sets and these four sets can be arranged in 4! ways.  The total ways that cards are not together =52!-(2!.11!.4!.104.4)


find $\lim_{x \to \infty} \frac{\ln(x+1)}{x}$ I need to find: $$\lim_{x \to \infty} \frac{\ln(x+1)}{x}$$
without using L'Hospital, integrals or derivative!
I would like your help in order to know i'm not doing something invalid.
I was thinking of the following way:
$$
\lim_{x \to \infty } \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty } \frac{1}{x} \ln(x+1)  =  \lim_{x \to \infty } \ln(x+1)^\frac{1}{x}
$$
and because $\frac{1}{x}$ tend to $0$ as $x$ approaches $\infty$, than $(x+1)^\frac{1}{x}$ tends to $1$ and $\ln(1) = 0$
and the limit is $0$.
But this way sounds off. I was trying also to express $x$ with $t$ and use $e$ (euler) somehow, but it didn't work.
Is my way valid? If not, do you have a better way without using L'Hospital's rule?
Thanks!

Edit:
While reading answers I was thinking of another way,
$$\lim_{x \to \infty} \frac{\ln(x+1)}{x}  =  \lim_{x \to \infty} \frac{\ln(x(1+\frac{1}{x}))}{x} = \lim_{x \to \infty} \frac{\ln(x)}{x} + \frac{\ln(1+\frac{1}{x})}{x} = $$ $$\lim_{x \to \infty} \ln(\sqrt[x]{x}) + \frac{\ln(1+\frac{1}{x})}{x}$$
now $\sqrt[x]{x}$ tends to $1$ as x approaches to $\infty$, hence $\ln(\sqrt[x]{x})$ tends to $0$ (is it valid to say so?)
and $\frac{\ln(1+\frac{1}{x})}{x}$ tends to $0$ is pretty obvious.
Am on missing something in this one?

Without L'Hospital Rule (or series for that matter!) you are left with using some basic properties of functions. For example set $\ln(x+1)=t$ so that $x=e^t-1$ and now you have an expression $\frac{t}{e^t-1}$ with $t $ going to infinity. To what extend does your teacher want you to prove that an exponential functions grows much faster than a linear term?

It is easy to show (See Here) that $\log(x)<\sqrt{x}$ for all $x>0$.  Therefore, we have
$$\frac{\log(1+x)}{x}\le \frac{\sqrt{x+1}}x$$
And you can finish now.


Normal distribution: standard deviation given as a percentage. Book question: The concentration of salt in a cell, $X$, can be modelled by a normal distribution with mean $\mu$ and standard deviation $2$%. Find the value of $\alpha$ such that $P(\mu-\alpha< X < \mu+\alpha) = 0.9 $.
What does it mean, standard deviation $2$%? I assumed it meant $X \sim N(\mu, 0.02^2)$, and I then did:
$P(X<\mu-\alpha) = 0.05$. So the $Z$-value is: $z=\frac{x - \mu}{\sigma} = \frac{(\mu-\alpha) - \mu}{\sigma} = \frac{-\alpha}{0.02} = -50\alpha$, where $Z \sim N(0,1)$ is the standard normal deviation. And using the inverse normal distribution function, I get: $-50-\alpha = -1.6448... \implies \alpha =0.0329\ (3sf)$.
But the answer in the back is $3.29$.
Ah but this is $100$ times more than my answer, so perhaps the % sign is just a typo in the question? Standard deviation $2$ instead of $0.02$ would get me the correct answer I think.

The textbook answer is correct. There's just a misunderstanding because the concentration is always misured in %. Thus the result is 3.29 (%, obviously).
Your error is due to the fact that you transformed in % a number that originally was already expressed in %
The correct solution was
$$\mathbb{P}[Z<\frac{a}{2}]=0.95 \rightarrow a\approx3.29$$
Anyway also your reasoning is correct. It is enough that you realize that your result is a % espressed with respect to 1 and not to 100

It means the standard deviation is $0.02 \mu$.  So you should be looking at $N(\mu, (0.02 \mu)^2)$.