Find a subset of columns to maximize the number of rows whose sum of entries in selected columns is equal or larger than a given number Given a real m*n matrix A (m rows, n columns), a real positive number b. Each entry in A can be positive or negative.

Find a subset of columns to maximize the number of rows whose sum of entries in those selected columns is at least b.

The equivalent mathematical formulation:
$$
\max_{x\in\{0,1\}^n} \sum_{i=1}^{m} I(A_{i.}x-b)
$$
where

$I(x) = 1 \text{ if } x\geq0, 0$ otherwise.
$A_{i.}$ is the row vector of the i-th column of A
$x$ is a binary column vector of size n.

What I have tried so far:

When m=1, the problem is equivalent to the optimization version of the subset sum problem, assuming entries of A are non-negative numbers (Even if the assumption is violated, my solution is to add the absolute value of the most negative value in that row to all the entries and b) The reference to the optimization of subset sum is here (https://courses.cs.washington.edu/courses/csep521/05sp/lectures/sss.pdf)

Current I am stuck when $m\geq2$. I have looked at generalized assignment problem (https://en.wikipedia.org/wiki/Generalized_assignment_problem), weighted bipartite graphs (https://en.wikipedia.org/wiki/Matching_(graph_theory)#In_weighted_bipartite_graphs). However, I cannot reduce my problem to any of the above two.

Please kindly advise.

Did you manage to solve this problem? I am trying to solve exactly the same problem... I think for m = 1, it is not subset sum sine subset sum is looking for a subset whose sum equals precisely b, while not at least b. If it is at least b, as in your problem, the best way is to just pick all the non-negative columns and sum them up -- this the largest sum you could get.

It is not clear how to solve this problem for large m.


Where to multiply & divide/plus & minus a number? So, here's an example:

$$\int \frac{x^2}{\sqrt{1 - x^2}}\, \mathrm dx$$

My math's teacher just say Add and Subtract 1 like this

$$\int \frac{1 + x^2 - 1}{\sqrt{1 - x^2}}\, \mathrm dx.$$

This is just one example. This thing is in almost half of the problems.

I get the next steps but what disturbs me is WHERE should one multiple/divide or add/subtract a number to make it easier. Of course there must be some concept behind it which I don't know so it will be a great help. I have seen this thing last year while studying trigonometry and this year while studying Calculus. If someone can explain how does this happen. That would be really helpful :)

The whole idea is to simplify fractions, since monomials are easier to integrate than fractions in general. E.g.: $$\int \frac{x}{x+a} \ {\rm d}x =\int \frac{x\color{red}{+a-a}}{x+a} \ {\rm d}x =\int \left(\frac{x+a}{x+a} -\frac{a}{x+a}\right) \ {\rm d}x \\ =\int \left(1 -\frac{a}{x+a}\right) \ {\rm d}x =  x-a\ln|x+a|+c$$ rather than using integration by parts with $u=x$ and ${\rm d}v = {\rm d}x/(x+a)$.


Probability of an event in a chain of events with given conditional probabilities Meredith Watts, police chief of Shorewood, knows that the probability of any assault that occurs in the city being reported to the police is .5.  If an assault is reported, the probability that someone will be arrested is .6.  If someone is arrested, the probability of a conviction is .9.  If a person is convicted of assault the probability of actually serving jail time is .7.  Given that an assault occurs, what is the probability of a person serving jail time for the crime?

I knew that by the question stating "given" that an assault occurs that I need to use the rule of conditional probability. I know the formula for that is $Pr(A |B) = Pr (A \cap  B)/Pr(B)$.

So I thought the best way to get started would be to use a probability tree, but I have never done that before and I am not exactly sure how to get started.

Hint:

Let (symbol) denote the event of (event)

As - Assault

R - Report

A - Arrested

C - Conviction

S - Serving

We know ( Note that | means given)

P(R|As) = 0.5

P(A|R) = 0.6

P(C|A) = 0.9

P(S|C) = 0.7

and

We want to know:

P(S|As) = ?

Before computing this mathematically, through conditional probability, let us think about this intuitively using probability tree diagrams.

anything with "^c" means complement as in the event did not occur e.g. R^c ($R^c$) means that the assault was not reported.

It seems that the only way to arrive at S from As is through R, A, C and S. This is correct as unreported assaults do not result in arrests, not being arrested cannot result in a conviction and not being convicted means not serving.

When going through the probability tree this way, we multiply the probabilities (eg the probability of getting double 6's when rolling a pair of fair dice is (1/6)*(1/6) = 1/36).

Thus, the answer would be 0.5*0.6*0.9*0.7, but how do we state this formally? More importantly, why does going through the probability tree in such a way call for multiplication?

Let us note that if we serve a sentence, it follows that we have been convicted.

If we were given that a conviction happened and were asked to compute the probability for a sentence, the answer would be 0.7, and we are done.

Is being convicted given? No. Go up the tree.

Being convicted means we were arrested.

Is being arrested given? No. Go up the tree.

Being arrested means we were reported. Given? No. Up.

Being reported means we committed an assault. This one is given.

Oh also, note that $S = R \cap A \cap C \cap S$, that is the event S is equal to the events R, A, C and S all happening together since there is no way someone can be sentenced if they are not first convicted, arrested, reported and are guilty of assault.

Recall we noted earlier: It seems that the only way to arrive at S from As is through R, A, C and S. This is correct as unreported assaults do not result in arrests, not being arrested cannot result in a conviction and not being convicted means not serving.

Thus, $P(S | As) =  P(R \cap A \cap C \cap S | As)$.

By definition of conditional probability, we have

$P(R \cap A \cap C \cap S | As) = \frac{P(R \cap A \cap C \cap S \cap As)}{P(As)}$.

If we could say that the events R, A, C, S and As are independent, then we would have $\frac{P(R \cap A \cap C \cap S \cap As)}{P(As)} = \frac{P(R)P(A)P(C)P(S)P(As)}{P(As)}$.

Unfortunately, they are not independent, I think. If not, there is a rule here we can use called the chain rule of probability.

We compute $P(R \cap A \cap C \cap S \cap As)$ as follows:

$P(R \cap A \cap C \cap S \cap As) = P(S | C \cap A \cap R \cap As) P(C | A \cap R \cap As) P(A | R \cap As) P(R|As) P(As)$

Hey look something we know, P(R|As) = 0.5.

What about the others?

We could use the extremely long and tedious process of doing things like:

$P(A | R \cap As) = \frac{P(A \cap R \cap As)}{P(R \cap As)}$

or we could take note of the fact that

$R \cap As = R$

$A \cap R \cap As = A$

$C \cap A \cap R \cap As = C$

This is true by similar reasoning to $S = R \cap A \cap C \cap S$.

Thus, we have

$P(R \cap A \cap C \cap S \cap As) = P(S|C) P(C|A) P(A|R) P(R|As) P(As)$

Hey wait, we don't know P(As). All we know is that it happened.

That's okay, we don't want to compute $P(R \cap A \cap C \cap S \cap As)$. We instead want to compute

$P(R \cap A \cap C \cap S | As) = \frac{P(R \cap A \cap C \cap S \cap As)}{P(As)} = \frac{P(S|C) P(C|A) P(A|R) P(R|As) P(As)}{P(As)}$.

Alternatively, you can write $P(R \cap A \cap C \cap S \cap As| As)$ and then note that $P(As | As) = 1$.

To understand chain rule of probability more, you may want to check out The Birthday Problem. It's a popular probability problem.


How to prove $\liminf_{n \to \infty} \prod_{k=1}^n |1-\exp(2\pi k i\alpha)|=0$ I wonder how to show the following limit inferior ($\alpha$ is an irrational number)
$$\liminf_{n \to \infty} \prod_{k=1}^n |1-\exp(2\pi k i\alpha)|=0.$$
Intuitively this is quite true since $\{\exp(2\pi k i\alpha)\}$ is uniformly distributed on the unit circle at origin (since elementary proofs are known, please feel free to use this result). When $n$ is large, $|1-\exp(2\pi k i\alpha)|$ is at most $2$ but can frequently be close to zero. So the product will be frequently close to zero (MatLab experiment also shows the same phenomenon). But how do we make this into a rigorous argument?
Another result we can consider using is the Huiwitz's theorem, which states that there are infinitely many relatively prime integers $a, q$ such that
$$|\alpha-\frac{a}{q}|<\frac{1}{\sqrt{5}q^2}.$$
Let $(a_n),(q_n)$ be such a sequence with $q_n \to \infty$ and $\alpha=\frac{a_n}{q_n}+\beta$ and $|\beta|<\frac{1}{\sqrt{5}q^2}.$ I speculate that
$$\lim_{n \to \infty} \prod_{k=1}^{q_n} |1-\exp(2\pi k i\alpha)|=0$$
(of course you don't have to use my idea to answer the question I asked)
Here is the reasoning for the result of limsup, which might help.

Let $a_n = \prod_{k=1}^n|1 - \exp(2\pi i \alpha k)|$
So
$$ a_n = \prod_{k=1}^n\sqrt{(\cos(2\pi\alpha k) - 1)^2 + \sin^2(2\pi\alpha k)}$$
$$ a_n = \prod_{k=1}^n\sqrt{2 - 2\cos^2(2\pi\alpha k)} $$
$$ a_n = 2^{\frac n2} \prod_{k=1}^n|\sin(2\pi\alpha k)|$$
Does this lead anywhere?


Modulo operation: Why does -1 = 59 (mod 60), whereas 1 = 1(mod 60)? The latter makes perfect sense to me, but what's the skinny with the former?  I researched but couldn't understand the drift. TIA

Consider the Number line marked with integers. Now, Paste the integer $60$ over $0$ and wrap around the entire number line over the circle formed as above.

Now, we see that when we count on the above number line constructed by the above procedure,  the integer $-1$ coincides with the integer $59$. This is called Modular Arithmetic.

Remarks:
$1$. The above construction can be generalized to any positive integer $n$ by replacing $60$ with $n$.
$2$. In general, while doing Modular arithmetic, we look for a solution inside the set $\{0,1,2,...,n-1\}$. But we can consider any set containing "$n$" consecutive integers.

Note: You can refer to any Elementary Number Theory Book. It will be more helpful.

This video Link might be useful for the above-said construction

I hope this answers your questions.

See here:

  The % (modulo) operator yields the remainder from the division of the first argument by the second. The numeric arguments are first converted to a common type. A zero right argument raises the ZeroDivisionError exception. The arguments may be floating point numbers, e.g., 3.14%0.7 equals 0.34 (since 3.14 equals 4*0.7 + 0.34.) The modulo operator always yields a result with the same sign as its second operand (or zero); the absolute value of the result is strictly smaller than the absolute value of the second operand [2].

So -1 % 60 must return a nonnegative (the same sign as 60) integer that is less than 60 in absolute value, and is congruent to -1 modulo 60. The unique integer satisfying that is 59. If you instead did -1 % -60 you would get -1.


Find generating function expressing number of rectangles I have a 3 x n rectangle. I need to find a generating function expressing the number of little 1 x 3 rectangles inside this bigger one. How can I do this? I have no idea how to even begin.

Hint: $a_{n+3} = a_n+a_{n+2}$.

Expanded: $a_n$ obviously, is the number of tilings of a $3\times n$ rectangle. Now, consider a $3\times (n+3)$ rectangle. In any tiling, the bottom left corner is either tiled by a vertical tile, or a horizontal tile. In the first case, the rest of the is $3 \times (n+2)$ and can be tiled $a_{n+2}$ ways. Otherwise, the cell just above the bottom left has to be tiled by a horizontal tile, and same for the one above it. In which case the rest is a $3 \times n$ rectangle, which can be tiled $a_n$ ways. These cases are obviously mutually exclusive.

The answer seems to be $$\sum_{n=0}^\infty a_n x^n = \dfrac{1}{1-x-x^3}$$.

This is in particular equation (9) in Tilings of Rectangular Regions by Rectangular Tiles: Counts Derived from Transfer Matrices, ArXiv:1406.7788  by Richard J. Mathar.


Prove $A\cap A = A$. How can $A\cap A = A$ be proved?
Here is my approach-

$A \cap A$
$= \{x : x \in A \cap A \}$
$= \{x : x \in A \wedge x \in A \}$
$= \{x : x \in A \}$
$= A$

I curious to know if there exists more ways to prove the aforementioned statement apart from my approach ?
A detailed proof would be helpful.

You can just look at what intersection, i.e. $\cap$, produces for a set. Let $X,Y$ be sets, then

$$X\cap Y=\{z\mid z\in X\text{ and }z\in Y\}$$

Thus

$$A\cap A=\{x\mid x\in A\text{ and }x\in A\}=\{x\mid x\in A\}=A$$

The set $A \cap B$ is definied as $\{x: x \in A \text{ and } x \in B\}$

The set $A \cap A$ is, by consequence, $\{x:  x \in A \text{ and } x \in A\}$.

The set $A$ is obviously $\{x: x \in A\}$

Now, given any element $G \in A$, $G \in A
\cap A$, as the definition suggests.

Also, for any element $G \in A \cap A$, $G \in A$ has to be satisfied.

Thus, every element of set $A$ is also in the set $A \cap A$ and vice versa

That should be enough to declare the sets equal.


prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction How to prove
 $$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$

by Mathematical induction，n$\ge $1

As far as I can tell, none of the answers so far gives a proof by induction. Here's one.

Let

$$S_n=\sum_{k=n+1}^{2n}\frac1k\;.$$

By explicit computation

$$\frac{25}{36}-\frac1{60}-S_{16}=\frac{238793353}{20629078984800}\gt0\;.$$

As others have pointed out, the sequence monotonically increases, so this establishes the inequality up to $n=16$. Now we can prove

$$
S_n\lt\frac{25}{36}-\frac1{4(n-1)}
$$

for $n\ge16$ by induction. The base case is handled above. The induction step is

$$
\begin{align}
S_{n+1}
&=
S_n+\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}
\\
&=
S_n+\frac1{(2n+1)(2n+2)}
\\
&\lt
S_n+\frac1{4n(n-1)}
\\
&\lt
\frac{25}{36}-\frac1{4(n-1)}+\frac1{4n(n-1)}
\\
&=
\frac{25}{36}-\frac1{4n}\;.
\end{align}
$$

The general insight to be gained here is that a proof by induction sometimes becomes easier when we make the conclusion stronger. That's because this not only strengthens the conclusion of the induction step but also its premise. In the present case, the original claim can't be proved directly by induction, since knowing that the inequality holds doesn't help in proving that it holds if we add something to the lesser side; but by strengthening the conclusion slightly more for $n$ than for $n+1$ we can create some space to work with in the induction step.

EDIT: This approach is unfortunately wrong, as pointed out in the comments.

Let $S(n) = \frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}.$

(1) Base step. Let $n=1$. Then $n+1 = 2n$, so the series only has one term. $S(1) = \frac{1}{2} < \frac{25}{36}$, hence the statement is true for $n=1$.

(2) Inductive step. Suppose the induction hypothesis is true. Then

$S(n+1) = \frac{1}{ (n+1)+1}+\frac{1}{ (n+1)+2}+\cdots+\frac{1}{2(n+1)}$

$= \frac{1}{ n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}$

$= - \frac{1}{n+1} + \left(\frac{1}{n+1} + \frac{1}{ n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}\right) + \frac{1}{2n+1} + \frac{1}{2n+2}$

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

$< - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} + \frac{25}{36}$

where the inequality comes from the induction hypothesis.

Now $S(n+1) < \frac{25}{36}$, as we want to show, requires
$- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} < 0.$

As $- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} < - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+1}$, this is certainly true if

$- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+1} < 0$, or $\frac{1}{n+1} < \frac{2}{2n+1}$. The last inequality is trivial to show.


Factor Rings of Polynomial Rings. Let $\mathbb F$ be a field and $\mathbb F[x]$ the ring of polynomials with coefficients in  $\mathbb F$. Let $p(x)$ be an irreducible polynomial in $\mathbb F[x]$. Let $k$ be a positive integer and consider the vector space $V$, over the field $\frac{\mathbb F[x]}{(p(x))}$with basis

$$1, p(x), p(x)^2, \ldots, p(x)^{k-1}.$$

That is, $$V=\left\{q_0+q_1p(x)+\cdots +q_{k-1}p(x)^{k-1}\;\;:\;\;q_i\in\frac{\mathbb F[x]}{(p(x))}\right\}.$$

Define in $V$ a product modulo $p(x)^k$. That is, if

$$v=v_0+v_1p(x)+\cdots +v_{k-1}p(x)^{k-1}\;\;\text{and}\;\;u=u_0+u_1p(x)+\cdots +u_{k-1}p(x)^{k-1},$$
then we multiply $v$ by $u$ the obvious way, using the distributivity and assuming that $p(x)^k=0$. This makes $V$ an algebra. My question is:

  Is this algebra $V$ isomorphic to $\frac{\mathbb F[x]}{(p(x)^k)}$?

If I understand well $V=L^k$, where $L=\mathbb F[X]/(p)$ is a field. On $V$ you define a multiplication by $$(a_0,a_1,\dots,a_{k-1})(b_0,b_1,\dots,b_{k-1})=(a_0b_0,a_0b_1+a_1b_0,\dots,a_0b_{k-1}+\cdots+a_{k-1}b_0).$$

It's easy to see that $V\simeq L[Y]/(Y^k)$ and thus your problem reduces to the following:

  Let $\mathbb F$ be a field and $p\in\mathbb F[X]$ irreducible. Set $L=\mathbb F[X]/(p)$. Is it true that $L[Y]/(Y^k)$ is isomorphic to $\mathbb F[X]/(p^k)$?

Denote $L$ by $\mathbb F(t)$, where $t$ is the residue class of $X$ modulo $(p)$. (Obviously $p(t)=0$.) I'll prove that
$$L[Y]/(Y^k)\text{ is isomorphic to } \mathbb F[X]/(p^k)$$
whenever $p'(t)\neq 0$. (This happens, for instance, if the characteristic of $\mathbb F$ is $0$.)

Define a ring homomorphism $\varphi:\mathbb F[X]\to L[Y]/(Y^k)$ by sending $X$ to $y+t$, where $y$ denotes the residue class of $Y$ modulo $(Y^k)$. We have $\varphi(p(X))=p(y+t)=p(t)+yh(y)$, $h$ being a polynomial with coefficients in $L$ and having the property that $h(0)=p'(t)$. Since $p(t)=0$ we get $\varphi(p(X))=yh(y)$ and using that $y^k=0$ we obtain $\varphi(p^k(X))=0$, that is, $p^k(X)\in\ker\varphi$. But $\ker\varphi$ is a principal ideal (of $\mathbb F[X]$), so $\ker\varphi=(f(X))$ for some $f\in\mathbb F[X]$. Since $f(X)\mid p^k(X)$ we must have $f(X)=p^i(X)$ for $1\le i\le k$. Suppose $i<k$. Then $\varphi(p^i(X))=0$ which is equivalent to $y^ih^i(y)=0$, that is, $Y^ih^i(Y)\in(Y^k)$. We thus get $Y\mid h(Y)$, a contradiction (with $h(0)\neq 0$). As a consequence we have proved that $\ker\varphi=(p^k(X))$. It remains to prove that $\varphi$ is surjective. But this follows easily observing that $\varphi$ is a homomorphism of $\mathbb F$-vector spaces of the same dimension (equal to $k\deg p$).

No, the algebra you've defined is isomorphic to $\mathbb F[x]/x^k$, where $x$ is represented by $p(x) \in V$.  It has dimension $k$ whereas $\mathbb F[x]/p(x)^k$ has dimension $k\cdot\deg p$.


Quaternion to Euler angles conversion I have written the following MATLAB code for transforming Quaternion to Euler angles based on the mathematical formula from wikipedia:

function [phi theta psii]=quat(q0,q1,q2,q3)

q_length=sqrt(q0*q0+q1*q1+q2*q2+q3*q3);

q0=q0/q_length;
q1=q1/q_length;
q2=q2/q_length;
q3=q3/q_length;

phi=atan2(2*(q0*q1+q2*q3),1-2*(q1*q1+q2*q2));
theta=asin(2*(q0*q2-q3*q1));
psii=atan2(2*(q0*q3+q1*q2),1-2*(q2*q2+q3*q3));

end

I tested the code:

[phi theta psi]=quat(0.99619,0,0,0.08716)
psi*180/pi
ans =

   -115.0435

The result is different compared to this website which suggests:

$\theta=0,~~~\phi=0,~~~\psi=-90$

Be careful when you convert between quaternions and euler angles. One of the main source of confusions are the conventions adopted to represent angles. The same quaternion can represent a rotation or it's inverse based on the adopted convention.

Your equations seems to be correct at first glance. As a comparison write some code to compute the conversion to the rotation vector. I do not know what the website does and I believe that when it comes to quaternions it is safe not to use software from different sources because you do not know the conventions that are adopted.

By the way, quaternions are provided by matlab in the aerosym toolbox (included in the complete version), but they follow the JPL convention and not the Hamiltonian.


How to multiply two numbers in base 3 using Karatsuba I want to multiply $a =  11201$ and $b =  1022$ which are in base $3$, so I have a problem I do not know how to start, normally I should split the numbers into other number of two digits, but I do not know how since a has $5$ digits.

Any help please ?

Sorry I did not read the question ... Use the Karatsuba algorithm. I use the notation in the wiki page.
So our base $B=3$ and the best split of the numbers is into three digit numbers thus $m=3$. (All numbers will be understood to be in base 3 from here (apart from the exponent of $B$))
\begin{eqnarray*}
x=11B^3+201 \\ y=1B^3+022
\end{eqnarray*}
Now the theory says $xy= z_2 B^6+z_1 B^3 +z_0$
\begin{eqnarray*}
z_2=x_1 y_1 \\ z_1=(x_1+x_0)(y_1+y_0)-z_2-z_0 \\ z_0=x_0y_0
\end{eqnarray*}
The finesse,of course is in the second equation, only one multiplication (computationally expensive) needs to be performed at the expense of a few more additions & subtractions. We have
\begin{eqnarray*}
z_2=11 \\ z_1=(212)(100)-11-12122=1220 \\ z_0=12122
\end{eqnarray*}
Now put six zeros on the end of $z_2$  three zeros on the end of $z_1$ and add these onto $z_0$ and we have $20002122$.

You don't need the messy lattice. Just multiply exactly the same as in base 10, remembering the base 3 multiplication table:

$$0* 0 = 0 \ \ \ 0*1 = 0 \ \ \  0*2 = 0$$
$$1*0 = 0 \ \ \ 1*1 = 1 \ \ \  1*2 = 2$$
$$2*0 = 0 \ \ \  2 * 1 = 2 \ \ \ 2*2 =11_3$$

The only carrying you need to do is when you have $2*2$
Then remember base 3 addition, noting $2 + 1 = 10_3$


Logical question with a balance and flour The problem is: I have a balance of $9$ kgs of flour and two weights; $250$g and $50$g. In the matter of $3$ steps I have to divide them into $2$ bags of $7$ and $2$ kg, respectively. I know that after I have measured a bag with $250$g I can combine it with the $250$ weight, so $500$g in total for example. Using the balance counts as a step! So you can use the balance only 3 times.
I hope you understand my English :]

The problem is really to measure out $2$ kg of flour in one bag.
The second bag simply receives all the remaining flour.

If you are allowed to put two bags on the scale and then transfer flour
between them until the scale balances, you can use the scale to
divide a bag of flour in half in one step.
Put the bag to be divided on one side of the scale,
put an empty bag on the other side,
and transfer flour between the bags until they balance.

Assuming you can divide a bag as described above,
here is a hint about how to measure $2$ kg
from the original $9$ kg in three steps:
$$ 9 = 2.25 \times 4. $$

I'm not sure if I am allowed to answer on my question but I solve it.
First I combine the 250 and 50g weight and I got a bag of 300g.
Then I combine the bag of 300g and the weights of 250g and 50g - I get a bag of 600g. Then lastly I combine the 600g and 300g bag the 250 weight and put the 50g weight on the other side of the balance. I got a bag with 1100g.
1100g + 600g + 300g = 2000g = 2kg.


Getting 0 solving Schrodinger equation with Dirac delta by Fourier transform I am attempting to solve the Schrödinger equation with the potential $V = - \delta (x)$. This leads to a differential equation

$$
\alpha \psi''(x) + (E + \delta(x)) \psi(x) = 0
$$

where

$$
\alpha \equiv \frac{\hbar^2}{2m}
$$

My instinct to solve the equation was to use an integral transform as I didn't know what else to do with the delta function. Since I am concerned with all space, I chose the Fourier transform. To get the Fourier transform, I transform each term, $\alpha \psi''(x)$, $E \psi(x)$, $\delta(x) \psi(x)$, and $0$. I get these transforms:

\begin{align}
&\mathcal{F}\{ \alpha \psi''(x) \} = -4\alpha \pi^2 p^2 \hat{\psi}(p)\\
&\mathcal{F}\{ E \psi(x) \} = E \hat{\psi}(p) \\
&\mathcal{F}\{ \delta(x)\psi(x) \} = \hat{\psi}(p) \\
&\mathcal{F} \{0 \} = 0
\end{align}

Where $\hat{\psi}(p)$ is the Fourier transform of $\psi(x)$. Unfortunately, when I combine this with the differential equation above, I get

$$
(1 + E -4 \alpha \pi^2 p^2) \hat{\psi}(p) = 0
$$

which results in the $\hat{\psi}(p)$ term going away and leaving me with a function only of $p$, so I am not able to solve for $\psi(x)$ via an inverse Fourier transform like I expected. I did replace the right hand side with an arbitrary function and work through it, but that results in $\psi(x) = 0$ is that function is $0$.

Is there something that I have done incorrectly in my approach? Is there a better approach to solving this equation? My professor and book both solved this question by solving on $(-\infty, 0)$ and $(0, \infty)$ and determining the condition at the origin by integrating the Schrödinger equation over $(-\epsilon, \epsilon)$, but I was hoping to be able to solve it in a more general way and plan to move to a Dirac comb next, which should be very similar using Fourier transformations.

Fourier transforms is a perfectly fine tool to solve the system. The advantage of using it is that you avoid having to solve two equations, one for $x<0$ and one for $x>0$, and then matching at $x=0$. The drawback is that you will have to compute the inverse transform which is usually a bit more work than solving and performing the matching if you have not done it before. However if you can use Fourier transform tables then this is imo the simplest way to solve it. The only mistake in your approach so far is not using the correct formula for the Fourier transform of $\delta(x)\psi(x)$. If you correct this then the derivation is straight forward and below I show an example of such a derivation for completeness.

For a general delta-function sink, $V(x) = -\lambda\delta(x)$, the Schrödinger equation reads

$$-\alpha\frac{d^2\psi(x)}{dx^2} - \lambda \delta(x)\psi(x) = E\psi(x)$$

where $E<0$ for a bound state (which is the case we are interested in). Taking the Fourier transform, using the convention $\int {\rm d}x\, e^{ipx}$ for simplicity, we obtain

$$\alpha p^2\hat{\psi}(p) - \lambda\psi(0) = E\hat{\psi}(p) \implies \hat{\psi}(p) = \frac{A}{p^2 + p_0^2}$$

where $p_0 = \sqrt{\frac{-E}{\alpha}}$ and $A= \frac{\lambda\psi(0)}{\alpha}$. Transforming back to real space we get the wavefunction

$$\psi(x) = \frac{A}{2\pi}\int_{-\infty}^\infty\frac{e^{-ipx}}{p^2+p_0^2}{\rm d}x = \frac{A}{2 p_0} e^{-p_0|x|}$$

where I simply used a Fourier transform table to compute the integral. If you want to know how to solve it analytically see this answer. The final thing left is to compute the energy which is determined by imposing wave-function normalization

$$\int_{-\infty}^\infty |\psi(x)|^2{\rm d}x = 1\implies \psi(0) = \sqrt{p_0} \implies E = -\frac{\lambda\alpha}{2}p_0 = -\frac{\lambda^2}{4\alpha}$$

so the full solution can be written

$$\psi(x) = \sqrt{p_0}e^{-p_0|x|}~~~\text{where}~~~p_0 = \frac{\lambda}{2\alpha}$$

From the Schrodinger equation
$$\alpha \psi''(x) + (E + \delta(x)) \psi(x) = 0$$
and
\begin{align}
&\mathcal{F}\{ \alpha \psi''(x) \} = -4\alpha \pi^2 p^2 \hat{\psi}(p)\\
&\mathcal{F}\{ E \psi(x) \} = E \hat{\psi}(p) \\
&\mathcal{F}\{ \delta(x)\psi(x) \} = \psi(0) \\
&\mathcal{F} \{0 \} = 0
\end{align}
then it is seen that:
$$(1 + E -4 \alpha \pi^2 p^2) \hat{\psi}(p) = \psi(0).$$
Now, since $\hat{\psi}$ is on $(-\infty, \infty)$ then the only way to determine $\psi(0)$ is to consider the two regions, $p < 0$ and $p > 0$, and use $\psi(0)$ as the matching condition. In general the delta function acts like a discontinuity and needs to be treated as a discontinuity.


Gaussian Elimination - What am I doing wrong? I'm trying to solve a system of equations using Gaussian Elimiation but I'm really having some trouble as I seem to be getting too many 0's. Here are the equations:

 +  +  =

 +  −  =

 +  +  = 4

To get my first 0 in the bottom left corner, I have performed R3 - R2 + R1:

[1 1 1 = 1

4 3 -1 = 6

0 3 5 = -1]

Then to get the "0" on the second row, I perform R2 * R1 - R2:

[1 1 1 = 1

0  0  0 = 0

0 3 5 = -1]

Already I can see this is wrong as I have way too many 0's and can't see how I'll be able to get my diagonal 1's. My issue with this method is that I'm aware of what I'm allowed to do, but I'm not aware of what I'm not allowed to do.

Could someone please step me through a better way of performing Gaussian Elimitation with these equations?

Thank you in advance!

To obtain the RREF matrix we are allowed to do three types of elementary row operations:

1) Swapping two rows

2) Multiplying a row by a non-zero number

3) Adding a multiple of one row to another row

For example we can obtain

$$\begin{bmatrix}1&1&1&1\\4&3&-1&6\\3&5&3&4 \end{bmatrix}
\stackrel{R3-3R1 + R2-4R1}
\to\begin{bmatrix}1&1&1&1\\0&-1&-5&2\\0&2&0&1 \end{bmatrix}
\stackrel{R3+2R2}
\to\begin{bmatrix}1&1&1&1\\0&-1&-5&2\\0&0&-10&5 \end{bmatrix}$$

R2 * R1 - R2

This is not a valid elementary row operation. You can't multiply two rows together. You can only multiply them by (nonzero!) constants, or add and subtract them from each other, or swap them around.

Try something like $R_2 - 4R_1$ instead.


A small college has 1095 students. What is the approximate probability that more than five students were born on Christmas day? A small college has 1095 students. What is the approximate probability that more than five students were born on Christmas day? Assume that the birthrates are constant throughout the year and that each year has 365 days.

I tried doing
$X \sim Pn(3) $ and calculating $P(X\gt5)$. My calculation turned out to be 0.22....., which was wrong. (What was wrong with my approximation?). The solution given used a Normal approximation to get an answer of 0.0735. When I tried using a Normal approximation, I was still unable to get that answer. Here is how I attempted it.

$ N\sim(3,1092/365)$.
$ P(X\gt5)=P(N\gt4.5) $  #continuity correction
$= P\left(Z\gt\frac{4.5-3}{\sqrt{\frac{1092}{365}}}\right)$
=$ P(Z>0.86721)$.
=$0.193$.

Any help is much appreciated.

poisson probability distribution $\lambda = 1095 / 365$
then calculate

using

$P(N) = \frac{\lambda^n}{n!} \exp(-\lambda)$

1 - P(0) - P(1) - P(2) - P(3) - P(4) - P(5) =

$1 - \exp(-\lambda)(1 + \lambda + \lambda^2 / 2! + \lambda^3 / 3! + \lambda^4/4! + \lambda^5/5!) = .084$

note this is an approximation,

also assuming that 5 is NOT included

N should be 5.5 not 4.5

P(X>5) = P(N >5.5) // continuity correction

N~(μ,σ^2)

N~(3,2.99) <--- σ^2 should not be (1092/365)

μ = np(amount of students)(probability of person born on Christmas Day)

=1095 *(1/365)

=3

σ = sqrt(n*p(1-p))

= sqrt(1095*(1/365)(1-(1/365)))

= sqrt(3*(364/365))

= sqrt(2.99)

P(X > 5) = P(X > 5.5)

      = P(Z >(5.5-3)/(sqrt(2.99)))         //Z =((X-μ)/σ)

      = P(Z > 1.45)      //using the Standard Normal Probabilities table

      = 0.0735

enter image description here


Ring with 10 elements is isomorphic to $\mathbb{Z}/10 \mathbb{Z} $ How do I prove that a finite ring $R$ of order 10 is isomorphic to the ring $\mathbb{Z}/10 \mathbb{Z}$?

I know that as a group under addition, $(R,+)$ is isomorphic to the group $(\mathbb{Z}/10 \mathbb{Z}, + )$, but the multiplication is rather mysterious to me.

I assume your definition of "ring" requires a unit element, which I'll write as $1$ (without that requirement, the statement is false: you could make all products be $0$).  Now if $1+1=0$, we'd have
$r+r = (1+1)\cdot r = 0$ for all $r \in R$, but then the order of the additive group of $R$ couldn't be $10$.  Similarly, $1+1+1+1+1$ can't be $0$.  So $1$ must have order $10$ in the additive group of $R$, and all members of $R$ are $0, 1, 2=1+1, \ldots, 9=1+1+1+1+1+1+1+1+1$.  Expanding it out, any $i \cdot j$ is the sum of $i j$ $1$'s, and this is $k$ where $k \equiv i j \mod 10$.

Note that if $1+1=0$ then $\mathrm{char}R =2$ and the additive group if it will have order $10$ then it should have a subgroup of order $5$ which is cyclic and generated by $x$. But $x^2=x+x=0$ contradiction.


Proving two paths are not path homotopic If I wanted to prove that two paths $f:[0,1] \rightarrow X$ and $g:[0,1] \rightarrow X$ are not path homotopic, I can prove that $f * \bar{g}$ is not path homotopic with the constant path that stays at $f(0)$. Here $\bar{g}$ denotes the reverse of $g$.
I'm not sure why showing $f * \bar{g}$ is not nulhomotopic tells us that $f$ and $g$ is not path homotopic. Can someone explain this?

Commenter: That is what the term "path-homotopic" means in every reference.
OP: If f and g are path-homotopic, then f * (g-bar) is based homotopic to g * (g-bar) which is based homotopic to the identity, as you saw when you introduced fundamental groups. Contradiction.


Showing an automorphism I am trying to show that $Aut(Z_4+Z_2)$ is the dihedral group $D_4$ or the Quaternions $Q$. The Quaternions have 1 element of order 2, while $D_4$ has 5 elements of order 2. So if I show two automorphisms, then its $D_4$. So far I have defined one automorphism as $\phi_1: (1,0)\rightarrow (3,0)$ and $(0,1)\rightarrow (0,1).$ This works because $\phi_1 \circ \phi_1=e.$ I need another one, but I cant seem to find it. I have tried sending $(1,0)$ to its other generators, $(1,1),(3,1), and (1,0)$, but the only one that works is $(1,0)$, but I know that that one is too simple as in obvious. Is it really the only one that works?


Is $\pi$ more transcendent than $e$? I’ve often wondered about this, and I conjecture the affirmative, based mainly on that it is so much easier to prove the transcendence of $e$ than that of $\pi$.

I would be surprised if, just as numbers form a linearly ordered classes ranging from algebraic of degree $n$ and culminating in the class of transcendental numbers, the transcendental numbers themselves are not further divided into a linearly ordered classes, with $\pi$ belonging to the top class, and $e$ belonging to some class below it, so this is really a reference request. Could someone please cite chapter and verse where this is done?

This question at MathOverflow has several answers discussing several senses in which one can say that one real number is more irrational than another, including irrationality measures and other hierarchies of complexity for real numbers.

This might be a subset of the previous answers but I think this is still worth writing it as a separate answer.

One reason to believe $e$ is less transcendental than $\pi$ is that there is a nice pattern in the continued fraction of $e$ as opposed to that of $\pi$.

$e = [2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,1,14,1,1,16,\ldots]$ whereas there is no such pattern for $\pi$. (We can find patterns in $\pi$ as well if we look at generalized continued fractions where the "numerators" in the continued fraction need not be $1$ always)


Different way of finding square roots of numbers? this is my first post here and I wanted to run something by people who are more knowledgeable than I. In the past couple days I've decided that I need to teach myself more advanced mathematics as it is of great importance to my studies in physics, astronomy and biology. In school we never went past algebra 1 and geometry and that has proven to be a hindrance. So I decided to brush up on the basic math that I learned in school and then move on to more advanced maths. I decided to make up a bunch of problems to solve while waiting for my books to arrive and I ran into my old nemesis, finding square roots of large numbers. I was taught prime factorization, which didn't work for this number, and the Babylonian Method, which can yield accurate results depending on the accuracy of your initial estimate and how many times you're willing to run the equation. What I wanted was an equation that used simple math and a known number instead of an assumed estimate (that must be high in the case of the Babylonian Method) that didn't have to be high or low that would yield an equally/greater accurate estimate of the square as does the BM, and would only have to be run once. I realize this is simple mathematics compared to what is normally discussed here, so I apologize, and also please forgive me if my terminology isn't quite correct. What I came up with was

√S≈((S/2)/(e/2)+e)/2

in this S is the number we are trying to find the square of and e is the closest perfect square root regardless of weather it is high or low.

For example, using 1,863 the number I was initially trying to find the square of, we know that the closest perfect square root is 43. So it would look like this

√1,863≈((1863/2)/(43/2)+43)/2

√1,863≈((931.5)/(21.5)+43)/2

√1,863≈(43.325+43)/2

√1,863≈ 86.325/2

√1,863≈ 43.163

Actual square of 1863=43.1624

so the estimate it yielded was very close. I've run a bunch of numbers through it both large and small and so far it appears to give better estimates than the BM (depending on the accuracy of your initial high estimate and number of times you run it through the equation) and without the requirement that the estimate is high, and also we don't have to make an initial estimate ourselves as the starting estimate is fixed for us by the nature of the equation.

What do you guys think? Is there something I've missed or perhaps is there an easier equation like it that I am not familiar with?

Thanks!

The relation $\sqrt S\approx\dfrac{1}{2} \left(\dfrac{\frac{S}{2}}{\frac{ e}{2}}+e\right)$ can be simplified to

$\sqrt S\approx\dfrac{1}{2} \left(\dfrac{S}{e}+e\right) $

if you square both sides you get (trust me)

$S\approx\dfrac{1}{4} \left(\dfrac{S^2}{e^2}+e^2+2S\right)$

the right side can be written as (trust me again)

$S\approx S+ \dfrac{\left(S-e^2\right)^2}{4 e^2}$

$S$ can be cancelled from both sides and we have

$\dfrac{\left(S-e^2\right)^2}{4 e^2}\approx 0$

Numerator is very very little because we choose $e $ close to $\sqrt S$ while denominator is about $4S$ therefore the fraction is very little and the approximation works

The larger the $S$ the more accurate is the approximation

I tried $\sqrt{123456789}$ and with you formula I got $11111.1110611$

while the actual value is $11111.1110605$

The only problem is to know that the  square closest to $123456789$ is $11111^2$

Hope this helps :)

If you allow yourself to use a table of square roots, the best method is to lookup the closest entry in the table and start the Babylonian method from there. It is hards to beat that (I mean in terms of choosing an initial guess).

But what you describe is strictly the Babylonian method, nothing new. Note that it is wrong that the initial estimate must be by excess.


Have I proved that $f$ is not continuous correctly? Consider the function

$f:(\Bbb R, T) \rightarrow(\Bbb R,T)$

Where $T$ is generated by the basis $ \{(−q,q)|q ∈ \Bbb Q , q > 0\},$ and $f(x) = x^2$

I want to decide if f is continuous.

Here is my attempt

Consider an open set $U\subset (\Bbb R, T)$

There are 3 possibilities for what $U$ will be

$U$ is the empty set
$U$ is all of $\Bbb R$
$U$ is of the form $(-q,q)$

for 1. inputting the empty set in any function will output the empty set , which is in $T$ so 1. checks out

for 2. we can choose q large enough so that the set $(-q,q)$ is all of $\Bbb R$, so we need only look at 3.

for 3. consider the inverse of $f, f^{-1} $, we know that $f^{-1}=\sqrt{x}$

$f^{-1}(q)=+_-\sqrt{q}$

$f^{-1}(-q)=+_-i\sqrt{q}$

Therefore $f^{-1}((-q,q))=(-i\sqrt{q},-\sqrt{q})\cup (i\sqrt{q},\sqrt{q})$

But $+_-\sqrt{q}$ and $+_-i\sqrt{q}$ are not elements of $\Bbb Q$ so neither set in the above union can be open in T, therefore f can not be continuous .

Does this proof look about right ?

Your $f$ is indeed continuous, we can prove this with sequences since $(\mathbb{R},T)$ is second countable.

Assume that $x_n \to x$ in $(\mathbb{R},T)$. We wish to prove that $x_n^2 \to x^2$ in $(\mathbb{R},T)$.

Let $q \in \mathbb{Q}, q > 0$ such that $x^2 \in (-q,q)$. Then $x \in (-\sqrt{q}, \sqrt{q})$. Pick $r \in \mathbb{Q}$ such that $0 \le |x| < r < \sqrt{q}$. Then $x \in (-r,r)$ so, since $x_n \to x$, there exists $n_0 \in \mathbb{N}$ such that $n\ge n_0 \implies x_n \in (-r,r)$. Hence for $n \ge n_0$ we have $x_n^2 \in (-r^2,r^2) \subseteq (-q,q)$ which proves $x_n^2 \to x^2$.

Therefore $f$ is continuous.

No, it is not right. The set $f^{-1}\bigl((-q,q)\bigr)$ is equal to $\left(-\sqrt q,\sqrt q\right)$, which is an open set, since, if $(q_n)_{n\in\mathbb N}$ is an increasing sequence of rational numbers such that $\lim_{n\to\infty}q_n=q$, we have$$\left(-\sqrt q,\sqrt q\right)=\bigcup_{n\in\mathbb N}(-q_n,q_n)\in T.$$Therefore, $f$ is indeed continuous.


Mathematical syntax when defining domain and range What is the correct syntax when writing about the domain and range of a function?

For example, let's say:

$f(y)=\cos{y}$ and $y(x)=\arcsin{x}$

In order to simplify the function $f(y(x))=\cos{\arcsin{x}}$ I want to write down the domains and ranges of all functions involved in a thorough way.

So I would start with the domain of $\arcsin{x}$ which is $[-1,1]$ then the range, which is $[-\frac{\pi}{2},\frac{\pi}{2}]$, which is also the domain of $\cos{y}$, which implies that the range of $\cos{y}$ is always positive.

How to write the above using the right mathematical syntax? I mean, using the right symbols?

Further question: what does the "element of" $\epsilon$ symbol mean and how to use it? Is that symbol useful in this case?

Furthermore I also sometimes see symbols like: $\mathfrak{D}$ and $\mathfrak{R}$. How are they correctly used?

There are several mistakes which are probably just the result of sloppy writing, but can also lead to some wrong results:

The function $f(x)$ is not equal to $\cos \arcsin x$. If $f(y)=\cos y$, then $f(x)=\cos x$. I believe what you want to write down is $f(y(x))$.
Second of all, you write that the range of $\arcsin$ is $\left[-\frac\pi2, \frac\pi2\right]$ which is true, but then you say "which is also the domain of $\cos y$", and this is not true. The domain of $\cos y$ is $\mathbb R$.

Other than that, it's hard to give you a full answer because I don't understand your question. Do you want to write down the domain of the function $\cos\arcsin x$? Then that domain is equal to the domain of $\arcsin x$, which is $[-1,1]$. Or do you also want to calculate the range?

You have two functions.

First function:

$f(x)=\cos x$

Domain: $x \in \mathbb R$

Range: $-1 \le f(x) \le 1$

Second function:

$g(x)=\arcsin x$

Domain: $-1 \le x \le 1$

Range: $-\frac {\pi}2 \le g(x) \le \frac {\pi}2$

Now consider $h(x)=f(g(x))$

The domain of $g(x)$ is $-1 \le x \le 1$.

The range of $g(x))$ is $-\frac {\pi}2 \le g(x) \le \frac {\pi}2$, which is all suitable as a domain for $f(x)$.

However, since the domain of $f(x)$ has now been restricted to $-\frac {\pi}2 \le g(x) \le \frac {\pi}2$, the range of $f(x)$ is restricted to $0 \le h(x) \le 1$

In conclusion:

Domain of $h(x)$ is $-1 \le x \le 1$.

Range of $h(x)$ is $0 \le h(x) \le 1$


Representations of finite groups over $\mathbb{C}[t]$ Let $G$ be finite group. I want to find simple and projective modules over the group algebra $A=\mathbb{C}[t][G]$. First observation is $A \cong \mathbb{C}[t]\otimes_{\mathbb{C}} \mathbb{C}[G] \cong \mathbb{C}[t] \otimes \oplus_{i=1}^r \operatorname{End}(S_i)$, where $S_i$ are simple $\mathbb{C}[G]$ modules.

Then we have the following simple modules $\mathbb{C}[t]/(t-\lambda) \otimes_\mathbb{C} S_i$. Is it true that there are no other simple modules over $A$?

Modules $\mathbb{C}[t] \otimes_\mathbb{C} S_i$ are projective because they are direct summands of $A$. Is it true that there are no other indecomposable projective modules over $A$?

Are there any books/papers where such representation theory of finite groups is discussed?

Since the algebra is a direct sum of matrix algebras over $\mathbb C[t]$, both irreducible and projective modules over this algebra are just modules over the summands. Moreover, modules over $\text{Mat}_n(R)$ correspond bijectively to modules over $R$ preserving simplicity and projectivity (Morita equivalence). Thus, we just need to classify simple and finite projective modules over $\mathbb C[t]$. And this is easy (any finite projective $\mathbb C[t]$-module is free).

Any textbook on graduate algebra (which talks about Morita equivalence) should be a good reference.


Every ideal of $K[x_1,\ldots,x_n]$ has $\leq n$ generators? Is this true: Every ideal of $K[x_1,\ldots,x_n]$ is generated by some subset with $\leq n$ elements?

It is true when $n=1$, since $K[x]$ is a PID.

I'm trying to prove it is not true for $n\geq2$, via the example $I:=\langle x^2,xy,y^2\rangle\unlhd K[x,y]$.

Does the SINGULAR code below confirm that $I$ is not generated by $1$ or $2$ polynomials?

ring R=0,(x,y),ls;
ideal I=x2,xy,y2;
minbase(I);

From the SINGULAR manual:

  5.1.76 minbase
  Syntax:  minbase ( ideal_expression )  minbase ( module_expression )
  Type:  the same as the type of the argument
  Purpose:  returns a minimal set of generators of an ideal, resp. module, if the input is either homogeneous or if the ordering is local.

What does a "minimal set of generators" mean? Minimal w.r.t. cardinality, or w.r.t. $\subseteq$?

Expanding on my comment above:

Let $J=\langle x^3,x^2y,xy^2,y^3\rangle$.  Since $J\subset I$, if $I$ is generated by two elements in $K[x,y]$ then its image, $I'\subset K[x,y]/J$, is generated by two elements as a $K[x,y]$-module.

But $I'$ in $K[x,y]/J$ is just $\{ax^2+bxy+cy^2: a,b,c\in K\}$.  As a $K[x,y]$-module, $x$ and $y$ act as zero on $I'$, and, as a vector space over $K$, $I'$ is 3-dimensional.  So there cannot be two generators for $I'$, and hence there cannot be two generators in $K[x,y]$.

In general for a graded ring, $R=\oplus_{i\geq 0} R_i$ with $R_0=K$ a field, any generating set for the ideal $I_j=\oplus_{i\geq j} R_i$ has at least as many elements as the dimension of $R_j$ as vector space over $K$.

Edit: I finally remembered how to do algebra, I hope. Below is a completely different answer than my original (and hopefully a correct one).

Your example works just fine. In fact, if we let $R = K[x_1,\ldots,x_n]$ and $\mathfrak{m}=\langle x_1,\ldots,x_n\rangle$ (which is maximal) then we get that $\mathfrak{m}^2$ requires more than $n$ generators for any $n\geq2$. To see this, note that we do not change the generators by localizing at $\mathfrak{m}$, so we can assume $\mathfrak{m}^2 = \langle f_{1}\ldots f_{n}\rangle$ for $f_{1},\ldots,f_{n}\in R_\mathfrak{m}$ to derive a contradiction. Any polynomial $p\in \mathfrak{m}^2$ must have degree at least $2$, so each $f_{1},\ldots,f_{n}$ must have degree at least $2$ and any linear combination (over $R_\mathfrak{m}$) of these with coefficients outside $K$ must have degree more than $2$, thus of the $\binom{n+1}{2}$ monomials of degree $2$ or less, at most $n$ are contained in $\mathfrak{m}$. Hence $R_\mathfrak{m}/\mathfrak{m}^2$ has dimension at least $\binom{n+1}{2}-n$. Yet $K\cong (R_\mathfrak{m}/\mathfrak{m}^2)/(\mathfrak{m}/\mathfrak{m}^2)$ has dimension $1$ and $\mathfrak{m}/\mathfrak{m}^2$ has dimension $n$ because $R_\mathfrak{m}$ is a regular local ring, hence $R_\mathfrak{m}/\mathfrak{m}^2$ has dimension $n$. But for $n\geq 2$ we have $\binom{n+1}{2}-n>n$, a contradiction, hence $\mathfrak{m}^2$ has more than $n$ generators.

I believe this can actually be generalized further to $\mathfrak{m}^k$ for all $k\geq 2$, but this requires computing the dimension of $\mathfrak{m}^k/\mathfrak{m}$ inductively using the exact dimensions of $R/\mathfrak{m}^j$ for $j < k$ in order to establish the desired inequality, which seems a pain at best.


$2d^2=n^2$ implies that $n$ is multiple of 2 I'm reading a proof of the irrationality of $\sqrt 2$. In a step it states that $2d^2=n^2$ implies that $n$ is multiple of 2. How?

$\;2\;$ is a prime and divides the left side in $\,2d^2=n^2\;$ ,
 so by the fundamental theorem of arithmetic it also divides the right hand side...

square of even numbers are even, so $n$ is even


Question on convergence in probability I have this sequence: $X_n=(-1)^nX+\frac{1}{n}$ where $X$ is $U(-1,1)$ random variable. Does this converge in probability?

I have $P(\left|(-1)^nX+\frac{1}{n}-X\right|>\epsilon=0\ \forall$ even $n$, as I can choose a large $N$ corresponding to any $\epsilon$ such that $P(1>N\epsilon)=0$.

But for odd $n$, I have the condition that $P(\left|-2X+\frac{1}{n}\right|>\epsilon)$ should go to zero. As $n\to \infty$, this is the probability that $P(2X>\epsilon)$ which is nerly 0.5 for $\epsilon\to 0$.

So I conclude this sequence does not converge in probability to a uniform random variable. Is my argument correct? Could someone please confirm this?

Assume that $(X_n)$ converges in probability, say to $Y$. Then, $Z_n=|X_n-X_{n+1}|$ converges in probability to $|Y-Y|=0$. Since $Z_n\to2|X|$ almost surely, $Z_n\to2|X|$ in probability. The limit in probability is unique up to a null set hence $2|X|=0$ almost surely. This is absurd hence $(X_n)$ does not converge in probability.

On the other hand, the distribution of $X$ is symmetric and $\frac1n\to0$ hence $X_n$ converges in distribution to $X$.

Edit: Here is a rewriting of (what I understand from) OP's argument.

First, for every positive $\varepsilon$, $[|X_{2n}-X|\geqslant\varepsilon]=[1\geqslant n\varepsilon]=\varnothing$ for every $n$ large enough hence $\mathrm P(|X_{2n}-X|\geqslant\varepsilon)\to0$ when $n\to\infty$, that is, $X_{2n}\to X$ in probability.

Thus, if $(X_n)$ converges in probability, $X_{2n+1}\to X$ in probability. But $X_{2n+1}-X=-2X+\frac1n$  hence that would imply that $\mathrm P(|2X-\frac1n|\geqslant\varepsilon)\to0$. If $n\geqslant\frac2\varepsilon$, $[|2X-\frac1n|\geqslant\varepsilon]\supseteq[|X|\geqslant\varepsilon]$, hence one would have $\mathrm P(|X|\geqslant\varepsilon)\to0$. For $\varepsilon\lt1$, this is absurd since $\mathrm P(|X|\geqslant\varepsilon)=1-\varepsilon$.

If the question is, do $(-1)^n X + \frac{1}{n}$ converge in probability to $X$, which it looks like based on your answer, yeah it looks good to me.

BTW i think you mixed up the even and odd cases?


Why are the differences between consecutive squares equal to the sequence of odd numbers? I was playing around with the squares and saw an interesting pattern in their differences.

$0^2 = 0$

      + 1

$1^2 = 1$

      + 3

$2^2 = 4$

      + 5

$3^2 = 9$

      + 7

$4^2 = 16$

      + 9

$5^2 = 25$

      + 11

$6^2 = 36$

      etc.

(Also, in a very related question, which major Math Research Journal should I contact to publish my groundbreaking find in?)

A nice thing to notice.

It's basically because

$(n + 1) ^ 2 - n ^ 2 = 2n + 1$

The difference between two consecutive cubes x^3 and (x+1)^3 will be 3x^2+3x+1.
It gets confusing after that... for consecutive hypercubes, it's 4x^3-6x^2+4x-1.


Sketching Inverse Functions I have to find the inverse of the function $f(x) = \frac{1}{x}$, where $x \geq 3$. However in the answer, the coordinates seem to be given as $(3,1)$. Shouldn't it be $(3,\frac{1}{3})$? Or is it that since it's a sketch, it's just an estimation of how the functions would look like and so it doesn't have to be 'accurate'?
I've just started doing these and I'm a bit confused on how I'm supposed to sketch these reflections.

When doing the inverse of a function the x-values and y-values are swapped. This would make sense why your answer would be (3,1) rather than (3,1/3).

Another thing that is important to remember is that the inverse is relfected about the line y=x. Just something to keep in mind!


There are four hundred students in a class and their total age is 8000 years. Show that the total age of the oldest hundred is at least 2000 years. I'm pretty sure I'm supposed to use the pigeonhole principle, but not exactly sure how. It's evident that since 2000 is the average, it must be greater but I'm not sure how to show it.

Split the $400$ students into $4$ groups of $100$. By the pigeonhole principle, at least one of them must have a total age of at least $8000/4 = 2000$. But the oldest $100$ will have an age that is greater than or equal to any group of $100$, and you are done.

A sketch of a proof that $T_N,$ the total age of the $N$ oldest students is at least $20N.$

Consider the age of the $N$th oldest student, $A_N.$ There are two cases:

$A_N \geq 20.$ Conclude that $T_N \geq 20N.$

$A_N < 20.$ Conclude that $8000 - T_N < 20(400 - N)$ and therefore $T_N > 20N.$

Set $N=100$ for the proof required in the question.

Note that this is a proof sketch. I do not claim it to be a complete proof. But if the idea is fundamentally unworkable (that is, if there is something that prevents doing a complete proof along these lines), I would appreciate if someone would point out specifically why it cannot work.


Syntax of an epsilon delta proof/why is this version incorrect So we have the regular $\delta$-$\epsilon$ definition of continuity as:

(1) For all $\epsilon>0$, there exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.

My question is why is the following definition incorrect?

(2) There exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$ for all $\epsilon>0$.

The obvious response is that $\forall x$ $\exists y$ $\neq$ $\exists y$ $\forall x$ (or rather, they are not always equal), but look harder at the grammar: is that necessarily what is going on?

Let us define $p:=$ "There exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$"

So we now have the statement:

"$p$ is true for all $\epsilon>0$"

Isn't that sentence identical to the English sentence: "for all $\epsilon>0$, $p$ is true"?

In which case you would have (1):

For all $\epsilon>0$, there exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.

The counterargument that $\forall x$ $\exists y$ $\neq$ $\exists y$ $\forall x$ makes sense if you look at it immediately from a logical view, but because of the way English sentences work (and their vagueness, in a case like this), the "for all $\epsilon>0$" clause can be placed anywhere without changing the meaning of the statement in English.

To illustrate better what I'm talking about, let us imagine that we write our definition as follows:

There exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$ (for all $\epsilon>0$).

Or, maybe:

There exists a $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$ (N.B. we're talking about all $\epsilon>0$ here!).

Aren't these parentheticals sort of "overlying" the whole statement? Is the fault in my reasoning that $\epsilon$ is being "bound" to an if-then statement before it has been defined? Or am I just blatantly incorrect and this is a case of $\forall x$ $\exists y$ $\neq$ $\exists y$ $\forall x$.

The essential point of all of this is that, if you have the statement: $$ \forall A \exists B : {B \subseteq A} $$
Then in English, this is equivalent to saying, both "For all A, there exists a B such that if B then A", and, "There exists a B such that if B then A, for all A".

The English sentence

  There exists $A$ such that $P(A,B)$ for all $B$.

is ambiguous -- it can either mean $\exists A\forall B\,P(A,B)$ or $\forall B\exists A\,P(A,B)$, and there is no generally observed convention about which of them it ought to be understood as.

Therefore, for the sake of clarity sentences of this shape should be avoided in mathematical writing.

In symbolic logic, there is no ambiguity because the quantifier should always come first. Some beginning students (and, sadly, some teachers) write things like "$P(x) \,\forall x$", as a kind of shorthand for the English wording "$P$ of $x$ holds for all $x$", but this is not proper syntax for predicate logic. If you use symbols, you should stick to the grammar that goes with them.

On the other hand, symbolic logic regrettably has a different ambiguity, namely whether $\forall x\;P(x)\Rightarrow Q$ means $(\forall x\;P(x))\Rightarrow  Q$ or $\forall x\;(P(x)\Rightarrow Q)$. There are conventions for which one to choose here, but unfortunately they give different results! Explicit parentheses are recommended for both of these meanings, unless you're really really sure that your readers know and follow the same convention that you do.

The definition is incorrect because the order in which the values are picked matters. I'm glad you asked this because this is not nearly hammered as much as it should be to get people to understand and appreciate what is going on the first time around.

First you pick $\epsilon$. THEN you pick $\delta$. The $\delta$ depends on the $\epsilon$. This is hugely important because if you were to pick another $\epsilon$ you would get a different $\delta$. (Caveat: if you're talking about the case of uniform continuity, then no matter what $\epsilon$ you pick, the same $\delta$ works for any of them).

For me, the easiest way to grasp the intuition behind this is to draw it out. Pick a continuous function and draw it. Then draw 2 lines parallel to the y-axis a certain distance apart (say half an inch, or whatever distance you want). At the points they intersect the curve (i.e. function) mark the corresponding values on the y-xis. So now you have your first ($\epsilon_1$, $\delta_1$) pair. Repeat this process but this time around change the distance between the 2 parallel lines, so that you get another ($\epsilon_2$, $\delta_2$) pair.

Remember that $\epsilon_1$ =/= $\epsilon_2$ by design. Now look at $\delta_1$ and $\delta_2$. If you picked a continuous function (but not uniformly continuous) $\delta_1$ won't be the same as $\delta_2$. If you picked a uniformly continuous function, they will be the same.

For these kinds of long-winded explanations that articulate the intuition you're supposed to grasp, I highly recommend the book "Introduction to Real Analysis" by Michael J. Schramm. It is not nearly as popular as the terse Rudin, but I found it invaluable in helping me to understand how to do analysis. (I don't mean to bash Rudin. His book is great for exercises once you've had some hand-holding from Schramm.)


Riemann Zeta Function's Analytic Continuation I am doing some independent learning on the Riemann Zeta function just out of curiosity.  I have searched the site, and I know that there are plenty of questions on the Riemann Zeta function.  However, this particular one has not been answered for me nor have I been able to find it easily on google.  I teach high school math and adjunct at a local community college in lower level classes such as Trig and College Algebra.  I say that so that you would keep in mind the level of mathematics that I have been exposed to and that you will word your answers accordingly (at an elementary level with little sophistication).  I understand that it is claimed that the zeta function is only valid for Re(s)>1 which is why negative arguments require analytic continuation.  But why is the function as it is written not defined for Re(s)<1?  Is this because the series diverges when Re(s)<1?  And if so, why may we claim that the function is undefined for those values just because it is divergent for those values?  It will really take some convincing for me to believe that we can change a function just because it is divergent at some values.

Let's begin by presenting a simple example of representing a function with a series.  Let $f(z)=\frac{1}{1-z}$ for $z\ne 1$.  Recall that $f(z)$ can be represented by the geometric series
$$f(z)=\sum_{n=0}^\infty z^n \tag 1$$
for $|z|<1$.  So, although $f(z)$ exists for all $z\ne 1$, its representation as given in $(1)$ is valid only when $|z|<1$.
We can represent also represent $f(z)$ by the series
$$f(z)=-\sum_{n=1}^\infty \left(\frac{1}{z}\right)^n \tag 2$$
for $|z|>1$.  So, we have two representations for the same function that are valid in distinct regions of the complex $z$-plane.

Now, suppose that we represent the function denoted $\zeta(s)$ by the series
$$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$$
for $\text{Re}(s)>1$.  We can easily extend the definition by writing
$$\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^s}}_{\zeta(s)}=\underbrace{2\sum_{n=1}^{\infty}\frac{1}{(2n)^s}}_{2^{1-s}\zeta(s)} -\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}\tag 3$$
for $\text{Re}(s)>1$.  Upon rearranging $(3)$ we find
$$\zeta(s)=(1-2^{1-s})^{-1}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\tag 4$$
But notice that the series on the right-hand side of $(4)$ converges for $\text{Re}(s)>0$.  So, we have just developed another representation for $\zeta(s)$ that is valid in a larger region of the complex $s$-plane.
There are other series representations of the Riemann Zeta function, such as its Laurent series,
$$\zeta(s)=\frac1{s-1}+\sum_{n=1}^\infty \frac{(-1)^n\gamma_n}{n!}(s-1)^n$$
which converges for all $s\ne 1$.
And there are integral representation of $\zeta(s)$ such as
$$\zeta(s)=\frac1{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x-1}\,dx$$
which converges for $\text{Re}(s)>1$ and (see 25.5.11 this reference)
$$\zeta(s)=\frac12 +\frac{1}{s-1}-2 \int_0^\infty \frac{\sin(s\arctan(x))}{(1+x^2)^{s/2}(e^{2\pi x}+1)}\,dx$$
which is valid for all $s\ne 1$.

When the real part is more than one, the total length of the pieces $|n^{-k}|$ is finite.  So there is a well-defined sum.
When the real part is between zero and one, the pieces shrink to zero, but the total length of the pieces is infinite.  As a result, you can rearrange the terms and form any sum you like.
For example,
$$1-\frac12+\frac13-\frac14+...=\ln 2\\
(1+\frac13-\frac12)+(\frac15+\frac17-\frac14)+(\frac19+\frac1{11}-\frac16)+...>(1+\frac13-\frac12)=5/6$$

When $k<0$, so the pieces are $n^{-k}$ with a positive exponent, the pieces get bigger and bigger, so as you say can't settle to anything.

A real function can change its form as you cross a point.  There is a function that is zero if $x\leq0$, but $\exp(-1/x^2)$ when $x>0$  This function is quite smooth at zero.

A complex function can't change suddenly.  When a complex function is differentiable, there is only one way to extend it.  That is second-year calculus.

We have $f(x)=\sum n^{-k}$ defined for $\Re(k)>1$ and $g(x)$ defined everywhere.  $f(x)=g(x)$ when $\Re(k)>1$, and there is only one way to extend it.  So we call the sum $g(x)$, even though the sum doesn't converge.


Evaluation of $\int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{1+\tan(x)}}dx$ Evaluate the given integral:

  $$\int_{\pi/6}^{\pi/3} \frac{1}{\sqrt{1+\tan(x)}}dx$$

After using $\int_{a}^{b}f(x)dx= \int_{a}^{b} f(a+b-x)dx$, we get $\int_{\pi/6}^{\pi/3} \sqrt{\sin(x)+\cos(x)} dx$ but I am not able to solve this integral. Could someone help me with this?

Hint:  Let $u=\sqrt{1+\tan x}$, so $du=\frac{\sec^2x}{2\sqrt{1+\tan x}}dx\;$.  Then

$\displaystyle\int\frac{1}{\sqrt{1+\tan x}}dx=2\int\frac{1}{1+\tan^2x}\frac{\sec^2 x}{2\sqrt{1+\tan x}}dx=2\int\frac{1}{1+(u^2-1)^2}du=2\int\frac{1}{u^4-2u^2+2}du$.

Next use partial fractions:

Since $u^4-2u^2+2=[(u^2)^2+2\sqrt{2}u^2+(\sqrt{2})^2]-(2+2\sqrt{2})u^2=(u^2+\sqrt{2})^2-\left(\sqrt{2+2\sqrt{2}}u\right)^2$

$\hspace{1.32 in}=\left(u^2-\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)\left(u^2+\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)$,

$\hspace{.4 in}\displaystyle\frac{1}{u^4-2u^2+2}=\frac{Au+B}{u^2-\sqrt{2+2\sqrt{2}}u+\sqrt{2}}+\frac{Cu+D}{u^2+\sqrt{2+2\sqrt{2}}u+\sqrt{2}}$ so

$\hspace{.3 in}1=(Au+B)\left(u^2+\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)+(Cu+D)\left(u^2-\sqrt{2+2\sqrt{2}}u+\sqrt{2}\right)$.

1) $u=0\;$ gives $B+D=\frac{\sqrt{2}}{2}$

2) the $u^3$ coefficient gives $0=A+C$

3) the $u$ coefficient gives $0=\sqrt{2}(A+C)+\sqrt{2+2\sqrt{2}}(B-D)$, so $\color{blue}{B=D=\frac{\sqrt{2}}{4}}$

4) the $u^2$ coefficient gives $0=\sqrt{2+2\sqrt{2}}(A-C)+(B+D)=\sqrt{2+2\sqrt{2}}(A-C)+\frac{\sqrt{2}}{2}$,

so $A-C=-\frac{1}{2\sqrt{1+\sqrt{2}}}=-\frac{\sqrt{\sqrt{2}-1}}{2}$ and $\color{blue}{A=-\frac{\sqrt{\sqrt{2}-1}}{4}}$ and $\color{blue}{C=\frac{\sqrt{\sqrt{2}-1}}{4}}$.

(Now integrate each term.)

hint: $\sin x + \cos x = \sqrt{2}\sin(x+\frac{\pi}{4})$, and you get an Elliptic Integral of Second Kind. This part has been posted here at MSE.


Beast ball league ways to get 10 points I know we can start by making the choices +1, +2, and +3. We can find how many ways to get 1 point, 2 points, 3 points, 4 points... But, I can’t find a pattern to find out how many ways you can get 10 points.

Let $B_{k, W}$ be the number of ways to get $k$ points ending with a Win.
Let $B_{k, L}$ be the number of ways to get $k$ points ending with a Lose.

We have the recurrence relations:

   $ B_{k, W } = B_{k-3, W } + B_{k-2, L} $
 $ B_{k, L} = B_{k-1, W} + B_{k-1, L}$
 The starting conditions are $B_{1, W} = 0 , B_{1, L } = 1, B_{2,W} = 1, B_{2, L} = 1, B_{3,W} = 1, B_{3,L} = 2$.

From here, we can find $ B_{10, W} + B_{10, L}$, which is left to the reader.

We are essentially trying to find the number of nonnegative solutions to $x+2y+3z=10$. To do this, consider the $\textbf{generating function}$ $$G(x)=(1+a+a^2+a^3+\cdots)(1+a^2+a^4+a^6+\cdots)(1+a^3+a^6+a^9+\cdots).$$
You are trying to find the coefficient of $x^{10}$ in this expression, and I'll leave for you to do. ^_^


Finding, not verifying, the limit of a sequence. If I'm told to prove that $\lim \limits_{n \to \infty}{\frac{1}{n^2}} = 0$, I would do the following:

$|\frac{1}{n^2} - 0| = \frac{1}{n^2} < \epsilon \implies N(\epsilon) = \frac{1}{\sqrt{\epsilon}} < n$

Every $\epsilon > 0$ is defined, so $\frac{1}{n^2} \to \infty$.

However, what if I'm not told $L = 0$? How would I find that $L = 0$, without looking at graphs or testing increasingly larger $n$?

I used $a_n = \frac{1}{n^2}$ as a simple example. I'd prefer some method that works for other sequences, too.

Finding/evaluating a limit requires the use of limit theorems like

algebra of limits
Squeeze theorem
Cesaro Stolz theorem
standard limits like $1/n \to 0$ as $n\to\infty$ which are proved using definition of limits.

Apart from these techniques sometimes we need to use theorems on existence of limits like a monotone and bounded sequence tends to a limit and then use the this fact to find the limit. Such solutions are often tricky. I show one example.

Let $a_n=x^{n}$ where $x$ is real. I deal with  the case when $0<x<1$. Clearly $a_n$ is monotone and bounded and hence it tends to some limit $L$.  Now we know that $a_{n+1}=xa_{n}$ and taking limits as $n\to \infty$ we see that $L=xL$ and hence $L=0$.

Your question about sequence $1/n^{2}$ can be solved by using product rule and the standard limit $1/n\to 0$. In fact we can generalize this and use the result $1/n^{a}\to 0$ for $a>0$ as a standard limit. Sometimes a combination of the techniques mentioned above may be required.

The above techniques exclusively deal with theorems related to sequences. Sometimes we also need results which deal with limits of functions of a real variable. Two such results which are popular are as follows:

If $a_n\to L$ and a function $f$ is continuous at $L$ then $b_n=f(a_n) \to f(L) $.
If $f(x) \to L$ as $x\to\infty$ then $f(n) \to L$ and if $g(x) \to L$ as $x\to 0$ then $g(1/n)\to L$.

Thus for example $n\sin(1/n)\to 1$ as $n\to\infty$.

Prove that $1/n^2$ cannot converge to $L$ when $L\ne 0.$ As follows:

(1).  If $L<0$ and $0 <\epsilon <|L|/2$ then $|L-1/n^2|>\epsilon$ for ALL $n$.

(2). If $L>0$ there exists $n_0\in \mathbb N$ with $n_0>\sqrt {2/L}.$ Then with $\epsilon  = 2/L$ we have :  $\quad |L-1/n^2|>\epsilon$ for all $n\geq n_0$.


Suppose the pdf of $X$ is $f(x) = \frac{-3}{4} (x-3)(x-5)$. Find the pdf of $X^2/8$. Suppose that $X$ is a continuous random variable with a pdf $f(x) =
 \frac{-3}{4} (x-3)(x-5)$ with $3\leq x \leq 5$. What is the pdf of the random variable $Y$,
  where $Y = X^2/8$.

My attempt:
First I notice that if $X\in[3,5]$, then $Y\in[9/8, 25/8]$.

I derived the cumulative distribution function as follows:
$$P(Y<y) = P(X^2<8y)=P(-\sqrt{8y}<X<\sqrt{8y})=\int_{-\sqrt{8y}}^{\sqrt{8y}}{f(x)dx}=-\sqrt{2y}(8y+45)$$

Then differentiate this cdf to obtain the pdf of $Y$:$$\frac{d}{dy}(-\sqrt{2y}(8y+45)) = -\frac{3(8y+15)}{\sqrt{2y}}$$

However, when I tried to integrate this pdf:  $\int_{9/8}^{25/8} {-\frac{3(8y+15)}{\sqrt{2y}}dy} =-94\neq1$

It turns out that the integral does not equal to 1, which suggests that there is something wrong with my answer. Any help would be highly appreciated.

$$\int_{-\sqrt{8y}}^{\sqrt{8y}} f(x)\ dx = \int_{-\sqrt{8y}}^{3} f(x)\ dx +\int_{3}^{\sqrt{8y}} f(x)\ dx$$

The first term is zero because $f(x)=0$ in that range. If $3\le\sqrt{8y}\le5$, then you can go on and integrate the second term.


Evaluate $\iint_{R}(x^2+y^2)dxdy$ $$\iint_{R}(x^2+y^2)dxdy$$

  $$0\leq r\leq 2 \,\, ,\frac{\pi}{4}\leq \theta\leq\frac{3\pi}{4}$$

My attempt :

Jacobian=r

$$=\iint_{R}(x^2+y^2)dxdy$$

$$x:=r\cos \theta \,\,\,,y:=r\cos \theta$$

$$\sqrt{x^2+y^2}=r$$

$$\int_{\theta=\pi/4}^{\theta=3\pi/4}\bigg[\int_{r=0}^{r=2}\bigg(r^2\bigg)dr\bigg]d\theta$$

$$.....=\boxed{\frac{4\pi}{3}}$$

  Is it correct?

>

Multiply Jacobian in integrand

$$\int_{\theta=\pi/4}^{\theta=3\pi/4}\bigg[\int_{r=0}^{r=2}\bigg(r^3\bigg)dr\bigg]d\theta$$

while converting in polar co-ordinate you have directly taken dxdx=drd(theta),
you have to use polar transformation Jacobin to do so, that is dxdx=rdr*d(theta), then integrate normally

answer will be (ip/2).


Which sequences converge in the topological space $\left(\mathbb{N}\times\mathbb{N}, \mathcal{J}_{\text{Lexicographic}}\right)$? Which sequences converges in the topological space $\left(\mathbb{N}\times\mathbb{N}, \mathcal{J}_{\text{Lexicographic}}\right)$?

That i have tried:

For each $(p,q)\in \mathbb{N}\times\mathbb{N}$ we have that the sequence $\{(p,q),(p,q),(p,q),\cdots,(p,q)\cdots$} coverges to $(p,q)$ for all $(p,q)\in \mathbb{N}\times\mathbb{N}$.

Am I ok?

Is it correct? Do you have a different example approach?

HINT: Apart from the eventually constant sequences, note that every point which is not a successor, or the minimum, is the limit of a sequence of smaller points. Can you characterize these points? If so, then you can characterize the sequences which converge.

ALERT: sorry the next answer is not correct:

I think that you are right.

As we have that $\left(\mathbb{N}\times\mathbb{N}, \mathcal{J}_{\text{Lexicographic}}\right)$ is a discrete space, it has the trivial convergence. By definition, a space $ (X, \mathcal{T})$ has trivial convergence if the only sequences that converges are the constant sequences.


Addition of two Binomial Distribution What is the distribution of the variable $X$ given $$ X = Y + Z, $$where $Y \sim $ Binomial($n$, $P_Y$) and $Z\sim$ Binomial($n$, $P_Z$)?

For the special case, when $P_Y = P_Z = P$, I think that X~Binomial($2n$, $P$) is correct. If $P_A ≠ P_B$, the distribution might eventually just be Binomial$\left(2n, \frac{P_A + P_B}{2}\right)$ but I can't prove it.

If the problem is more complicated than I expect and we can't derive the whole distribution, can we tell something about the mean and the variance of $X$?

It will be a special case of the Poisson Binomial Distribution.

In the limit as $n \to \infty$, your binomials become Gaussian and since it seems you are implicitly assuming your two binomials are independent, a sum of two independent Gaussians is Gaussian with mean and variance parameters given by the sum of the parameters for the two Gaussians, so yes, in the limit as $n \to \infty$, your distribution will converge to Binomial$(2n, (P_A+P_B)/2)$.

Also you are right, if $P_A = P_B = P$ and you assume independence, then the distribution is precisely Binomial$(2n,P)$.

However if $P_A \neq P_B$ and you assume independence, then the exact distribution is different from Binomial$(2n,(P_A + P_B)/2)$. If you let $X = X_A + X_B$ be the random variable which is the sum of your two binomials, then $P(X = k)$ is the summation over all the ways that you get $X_A = k_A$ and $X_B = k_B$ where $k_A + k_B = k$. It is easy to write down this summation formula if you know the formulas for binomial distribution, and summation notation. However I'm inclined to believe there is no closed form formula for it, unless it's something crazy like hypergeometric.


Can we find a function $f(t)≥1$ such that the function $t^{a}/(f(t)exp(t^{2/3+b/loglog t})$ is increasing. Let $a,b$ be two positive real numbers.

Then my question is: Can we find a function $f(t)≥1$ such that the function

$$
\frac{t^{a}}{f(t)\exp\left(t^{2/3+\frac{b}{\log\log t}}\right)}$$ is increasing for all $t≥2$.

The answer is no. Put
$$g(t)=\frac{t^{a}}{f(t)\exp\left(t^{2/3}+\frac{b}{\log\log t}\right)}.
$$

If $f(t)\geq 1$,  then $$g(t)\leq \frac{t^{a}}{\exp\left(t^{2/3}+\frac{b}{\log\log t}\right)}.
$$
But the RHS goes to zero as $t\to \infty$, so $g(t)\to 0$, and $g$ is increasing and positive, so it must be zero. Which is nonsense.

You can pick your favorite function $g(t)$ which is both increasing and at least $1$ (say, pick $g(t) = e^t + 1$), set that expression equal to $g(t)$, and solve for $f(t)$ algebraically.


Proving $(C_b(X,Y),d_{\infty})$ is complete if $(Y,d_y)$ is complete Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. We say a function $f: X
\rightarrow Y$ is bounded if $f(X)$ is a bounded set of $Y$. Consider
  $$C_b(X,Y) = \left\{f: X \rightarrow Y \mid f \ \text{is continuous
and bounded} \right\} $$ with the metric $d_{\infty}(f,g) =
 \sup\left\{d_y(f(x),g(x)) \mid x \in X \right\}. $ Prove that
  $(C_b(X,Y),d_{\infty})$ is complete if $(Y,d_Y)$ is complete.

Attempt: In my course, we already proved that if $(X,d_X)$ is a metric space, then $$C_b(X, \mathbb{R}) = \left\{f: X \rightarrow \mathbb{R} \mid f \ \text{is continuous
and bounded} \right\}$$ is a complete space for the $d_{\infty}(f,g)= \sup\left\{|f(x) - g(x)| \mid x \in X \right\}$ metric. I want to use this result now. I know that if $Y$ is a complete metric space, and $X$ is a bounded subspace of $Y$, then $X$ is also complete.

So in this case I said $C_b(X,Y)$ is a subspace of the complete space $(C_b(X,\mathbb{R}),d_{\infty})$. So I need to prove that $C_b(X,Y)$ is closed in $C_b(X, \mathbb{R})$ ? Then let $f_n \in C_b(X,Y)$ be a convergent sequence. Denote the limit by $f \in C_b(X,\mathbb{R})$. I need to prove that $f \in C_b(X,Y)$.

I'm not sure how to prove the continuity of $f$. Also, I haven't used the fact yet that $(Y,d_Y)$ is complete so I'm not sure if this is the right direction. Any help is appreciated.

To use the result shown in your class, take $y\in Y$ and consider $f_n^y(x)=d(y,f_n(x))$. It is a family of functions of $C_b(X,R)$ which is continuous and converges towards $f^y(x)=d(y,f(x))$ where $f(x)=lim_nf_n(x)$. You conclude that $f^y$ is bounded, this implies that $f$ is bounded.

To show that $f$ is continuous at $x$, take $y=f(x)$ and consider $f^{f(x)}(z)=d(f(x),z)$ it is continuous at $x$ i.e for every $c>0$, there exists $d>0$ such that $d(z,x)<d$ implies that $|f^{f(x)}(z)-f^{f(x)}(x)|=d(f(z),f(x))<c$. This equivalent to say that $f$ is continuous at $x$.


Uniform distribution on $\mathbb Z$ or $\mathbb R$ I was assisting once the course in Probability Theory where students learnt quite quickly that there are ways to assign the uniform distribution to any finite set - or even subsets of $\mathbb R$ of a finite measure. They also clearly understtod the proof that there are no ways to put a uniform distribution (as a probability measure) on $\mathbb Z$ or a uniform distribution in  a sense of a constant density on $\mathbb R$. But then I was asked the question:

  Why is it impossible? What is intuitive meaning of this?

The only answer I could find that time refers to $\mathbb R$: well, the density is just a density w.r.t. Lebesgue measure - so why do we want the uniform distribution to have a constant density w.r.t. precisely Lebesgue measure? (as for me, if we omit the algebraic structure on $\mathbb R$, it would be just a measure).

However, I couldn't answer the question why there is no a uniform distribution on $\mathbb Z$ where the sentence "each value comes with equal probability" does have sense, I hope. Now this question bothers me myself and any help is appreciated.

If you have a uniform distribution $\mu$ on $\mathbb{R}$, it should assign the same value $v$ to $[z,z+1)$ for every $z\in\mathbb{Z}$. But if $v=0$, you have $\mu(\mathbb{R})=\sum_{z\in\mathbb{Z}}0=0$. On the other hand, if $v>0$ then $\mu(\mathbb{R})=\sum_{z\in\mathbb{Z}}v=\infty$. So $\mu(\mathbb{R})=1$ can not be when $\mu$ is countably additive, which is the usual assumption in probability. The same argument applies to the nonexistence of a uniform distribution on $\mathbb{Z}$, when you look at the probability assigned to each point.

Let's say that a uniform distribution on $\mathbb{Z}$ is a function $\mu:\mathcal{A}\to[0,1]$, with $\mathcal{A}$ being an algebra, such that, $\mu(\emptyset)=0$, $\mu(\mathbb{Z})=1$, $A,B\in\mathcal{A}$ and $A\cap B=\emptyset$ implies that $\mu(A\cup B)=\mu(A)+\mu(B)$ (so we require $\mu$ to be only finitely additive). Lastly, the uniformity is expressed by the property that for all $A\in\mathcal{A}$, $A+z=\{a+z:a\in A\}\in\mathcal{A}$ and $\mu(A)=\mu(A+z)$.

Under these assumptions, a uniform distribution on $\mathbb{Z}$ exists. Let $\mathcal{A}$ consist of all subsets of $\mathbb{Z}$ that are finite or have a finite complement. This is an algebra. Let $\mu(A)=0$ if $A$ is finite and let $\mu(A)=1$ if $A$ has a finite complement. You can verify that this satisfies all requirements we made on a uniform distribution. It is even possible to generate such a uniform distribution with $\mathcal{A}=\mathcal{P}(\mathbb{Z})$, by using the Hahn-Banach theorem. You can find some details here, together with other notions of uniformity.

So the non-existence of a uniform distribution on $\mathbb{Z}$ is essentially a matter of accepting countable additivity. There is no "elementary" reason.

Think of it this way. You start dividing Z and R into N equal slices over a histogram. The overall probability has to sum to 1. So to know the probability to assign to each slice of the histogram you need to know the finite number of slices you are going to make. But Z and R are infinite in size. So you can't slice Z and R into any particular finite number of equal slices.

It is the equal size & equal probability of each slice that is the problem. Of course there are functions which have probability of 1 over infinity. For instance the normal distribution is defined for all values of X from -infinity to +infinity, and it sums to 1.

As an example consider dividing the set of integers into groups of 10. So if I have integers between 1 and 100, then I have 10 groups. Each group thus has a probability of 1/10 of being randomly chosen when I choose a random integer between 1 and 100.

But what happens if I extend set of integers to Z? I end up with an infinite number of groups and 1/infinity is undefined as far as probability goes. So I can't define a probability for a particular group of 10 being chosen.


Maximal chain in the collection of all invariant subspaces for compact operator $K$ Let $X$ be a Banach space over ${\Bbb C}$, and $K\in K(X)$ ($K(X) = $ compact operators space). Show that if ${\cal L}$ is a maximal chain in the $Lat K$ ($Lat K = $ the collection of all invariant subspaces for $K$), then ${\cal L}$ is a maximal chain in the lattice of all subspaces of $X$.

I do not have any idea about it. Please give me a hint. Thanks in advance.

Let $L$ not be a maximal chain in the lattice of all subspaces of $X$. $M \notin $ Lat K for some M subspace of X, i.e. $Kx \notin M$ for some $x \in M$.
Suppose that $\{x_n\}$ is a sequence in $M$ such that tends to $x$ in $M$ weakly. By Prop. VI.3.3 [A course in Functional Analysis, John B. Conway] $K$ is completely cont., so that $\{ K{x_n}\}$ converges in norm to $Kx \in M$; this contradicts that $Kx \in M$.


How does the Maclaurin series for $e^x$ converge for negative values? It really doesn't seem to do so. It just oscillates wildly while increasing in magnitude. wtf?
$e^{-\infty }$ should approach 0, but that doesn't seem to be the case here.
https://www.wolframalpha.com/input/?i=sum+from+0+to+111+of+%28-10000%29%5En%2Fn%21

The problem is to know how many terms you need to sum for a given accuracy.
In other words, adding $(n-1)$ terms you want to know $n$ such that
$$\frac {x^n}{n!}  < 10^{-k}\implies n!>x^n\, 10^k$$
If you look at this question of mine, you will see a magnificent approximation by @robjohn for the inverse of the factorial (have a look here).
$$\bbox[5px,border:2px solid #C0A000]{n\sim ex\exp\left(\operatorname{W}\left(\frac k{ex}\log(10)-\frac1{2ex}\log(2\pi x)\right)\right)-\frac12}$$ where $W(\cdot)$ is Lambert function.
Using this approximation for $x=10^4$ and $k=6$ would give $n=27190.6084224$ while the exact solution is $n=27190.6084239$
Checking
$$\frac{(10^4)^{27190}}{27190!}=1.84\times 10^{-6} > 10^{-6}$$
$$\frac{(10^4)^{27191}}{27191!}=6.76\times 10^{-7} < 10^{-6}$$

The Taylor series around zero is $e^z=\sum_{n\ge0}\dfrac {z^n}{n!} $.
By Cauchy-Hadamard, the radius of convergence is  $r=1/\limsup_{n\to\infty}\sqrt [n]{1/n!}=1/0=\infty$.


Some Philosophical Questions About Mathematics and Logic The following questions may seem very philosophical and I guess that you guys will tell me that this is not the right place for asking them.
But for me it is important to get answers from mathematicians and not from philosophers.

Note: The questions are subjective. I am asking for YOUR opinion. So if you tell me your opinion I would be very happy ;-D

Is "5 is not a set" a mathematical statement? I don't want to know, if you think it is a true statement, or a false statement, but I want to know whether you think it is a statement at all.
In type theory, one can't express "5 is not a set". However, should a foundation for mathematics be able to express this sentence?
Is "second-order logic" a logic at all? Or is it just "set theory in disguise"?
Do we use the language of higher-order logic in ordinary mathematics? Or are we always working with first-order logic?
Comment: Maybe you are going to say that the definition of a topological space is higher-order because we quantify over sets of sets of ... individuals. But that is not what I mean with "using higher-order logic". The example "definition of topological space" can be expressed in set theory which is a first-order theory. Thus with "using higher-order logic" I do not mean quantification over sets, which is a first-order concept, but I mean "quantification over properties".
Is "working in formal systems (like the predicate calculus)" mathematics? That is: Does "proving of logical tautologies" belong to mathematics?
Is it a subfield of mathematics? Is there research in this area?

For 1. : Is "5 is not a set" a mathematical statement?

Yes, of course; it is expressible in set theory (like $\mathsf{ZF}$) and set theory, much more than a foundational theory, is a mathematical theory.

  For 2. : Is "working in formal systems (like the predicate calculus)" mathematics?

The "mathematical part" of mathematical logic is mainly the meta-theory. In terms of mathematical knowledge, a formal derivation in predicate calculus is quite uninteresting (nor more interesting that performing a sum). But Gödel Completeness Theorem for predicate calculus is an outstanding "piece" of mathematics.


Does this statement represent a set? Does the following statement represent a set?
"The collection of all the stars in the universe?". I think it represents a set. Please explain if someone agrees or disagrees.

Whether or not it's a set really depends on the context in which you are asking the question.

The answer is "no" if you're doing mathematics, since there's no mathematical definition of a star (in the sky).

But questions like this frequently appear at the beginning of a class where you are first learning informally about sets in mathematics. A set might be described as "a collection of things". The answer would be "yes",  as long as you're willing to agree that everyone knows what a star is (something astronomers might argue about). Then the informal collection of all of them is a set. The sun is a member, so is alpha centauri. The moon is not. Your computer is not.

Of course defining a set as a "collection of things" is OK for  informal understanding but won't do as a mathematical statement unless you had previously defined "collection" ...

To define a set, you just need to define its elements.

here,  elements are stars or suns.

the elements are well-defined, so we can speak about the set $S $ of these elements.

for example, $($vega, the  sun$) \in S^2 $.


Convex Functions and their maximums? Show that the maximum of a convex function on a convex polytope occurs at one of the extreme points of the polytope. I am really stuck here and don't know where to begin.

Hint: For the following reasoning use

The definition of convexity of a function
The fact that the convex polytope over which the function is maximized contains all convex combinations of points in it. In fact every point in the convex polytope can be expressed as a convex combination of it's extreme points.
The definition of an exterme point (that it cannot be written as a linear combination of two distinct points).

General idea of proof (you need to write it in more detail): Consider the point $$x_0=λx_1+(1-λ)x_2 \quad \in C$$ where $λ \in [0,1]$ and $x_1, x_2 \in C$. Since $f$ is convex (I named your convex function $f$) you know that $$f(λx_1+(1-λ)x_2)\le λf(x_1)+(1-λ)f(x_2)$$ for every $x_1, x_2 \in C$ (I named the convex polytope $C$), $λ \in [0,1]$. But this implies that the inner point $x_0:=λx_1+(1-λ)x_2$, for $λ \in (0,1)$ (which is in $C$, since $C$ is convex) cannot have a maximum since it's value under $f$ is dominated by the RHS of the above expression (where $x_1$ and $x_2$ can be chosen to be extreme points).

(Note of course, than when the inequality holds with a equality, then the maximum is attained in the inner point $x_0$ as well as in the extreme points $x_1, x_2$ so the theorem is again valid.)

for example:
Now, note that for any fixed $t\geq0$, the function $g_{t}:R^{r}\rightarrow R$ defined by $g_{t}(x)=-\frac{1}{2}\sum_{k=1}^{r}ln(1+2tx_{k})$ is convex. Hence, its maximum over the simplex $\{x\in R^{r}: \sum_{k=1}^{r}x_{k}=1,x\geq 0\} $ is attained at a vertex of the simplex.
 so $g_{t}(x)=-\frac{1}{2}\sum_{k=1}^{r}ln(1+2tx_{k})\leq-\frac{1}{2}ln(1+2t)$

I want to ask how to understand this?


subring of 2x2 matrices with real entries. Let $S = \left\{\begin{bmatrix}a&b\\0&0\end{bmatrix} :a,b \in \mathbb {R} \right\}$. Prove that the set $S$ is a subring of $M_2 (R)$ with its usual addition and multiplication. Is $S$ a ring with identity? Is it a commutative ring? Describe the units of this ring.

yes this is a subring of $M_2 (R)$, $S$ has an identity but its not the identity in $M_2 (R)$ it is commutative and it has no units.
Does the fact that the identity is not the same matter?

$ \begin{bmatrix}1&0\\0&0\end{bmatrix} \in S$ so S is non empty.

$ \begin{bmatrix}a&b\\0&0\end{bmatrix} \begin{bmatrix}c&d\\0&0\end{bmatrix}= \begin{bmatrix}ac+0&ad+0\\0&0\end{bmatrix}$ $ac,ad \in \mathbb {R}$ closed under multiplication

$ \begin{bmatrix}a&b\\0&0\end{bmatrix} + \begin{bmatrix}c&d\\0&0\end{bmatrix} = \begin{bmatrix}a+c&b+d\\0&0\end{bmatrix}$ $(a+c),(b+d) \in \mathbb {R}$ closed under addition

$ \begin{bmatrix}a&b\\0&0\end{bmatrix} \begin{bmatrix}c&d\\0&0\end{bmatrix}= \begin{bmatrix}ac+0&ad+0\\0&0\end{bmatrix}$

$ \begin{bmatrix}c&d\\0&0\end{bmatrix} \begin{bmatrix}a&b\\0&0\end{bmatrix}= \begin{bmatrix}ca+0&cb+0\\0&0\end{bmatrix}$

so not a commutative ring.

$\begin{bmatrix}1&0\\0&1\end{bmatrix} $ is not in S so S has no unity.

Since $ \forall s \in S $ the det(s)=0 nothing is a unit.


Is there a specific terminology for numbers which are nontrivial multiples of triangular numbers? (Note: Please see this new question for the motivation.)

A number $T$ is said to be triangular if it could be written in the form
$$T=\frac{n(n+1)}{2},$$
where $n$ is a positive integer.

Here is my question:

  Is there a specific terminology for numbers which are nontrivial multiples of triangular numbers?

That is, is there a specific name for numbers $T'$ which are of the form
$$T' = dT = d\cdot\bigg(\frac{n(n+1)}{2}\bigg),$$
for some integer $d > 1$?

I tried searching for the relevant sequence in OEIS, but all I am seeing are references to $T'$ being triangular as well.

I'm going to answer based on the confirmation in the comment.

Note that if $T'>1$, then taking $d=T'$ and $n=1$ (so that
$T=1*2/2=1$) works.
Also, since $d>1$, $T'=1$ is impossible.
And since $n,d\ge1$, $T'<1$ is impossible as well.

Therefore, a name for the possible values of $T'$ is "integers greater than 1".


Confidence band for Brownian Motion with uniformly distributed hitting position Let $(B_t)$ denote the standard Brownian motion on the interval $[0,1]$. For a given confidence level $\alpha \in (0,1)$ a confidence band on $[0,1]$ is any function $u$ with the property that
$$
P(\omega; |B_t(\omega)| < u(t), \quad \forall t\in [0,1])=\alpha.
$$
In other words, the probability that a path of the Brownian motion stays within a confidence band is $\alpha$. Additionally the boundary hitting position for those paths leaving the band must be uniformly distributed on $[0,1]$. This condition can be stated using the stopping time
$$\tau(\omega) = \inf [ t \in [0,1], |B_t(\omega)|=u(t) ].
$$
Then $\tau $ is the time of the first hitting, and one asks that $\tau$ is uniformly distributed on $[0,1]$ conditionally on the event that $\tau$ is finite.

I am interested in

References and links to literature or papers considering this or similar problems
Thoughts, ideas, discussion

The context of the problem is a rather boring one, so will not state it here. The problem itself seem to be non-trivial and interesting.

If I understand you correctly, you are looking for a curve $u(t)$ with $t \in [0,1]$ so that the probability the absolute value of a standard Wiener process does not cross the curve is $\alpha$ and that the probability density of the first crossing is a constant $1-\alpha$.

The following simulation in R may help indicate the shape of $u_\alpha(t)$:

##simulated boundary for standard Wiener process
##time for absolute value to cross boundary first time
##uniformly distributed on [0,1] given crosses boundary
steps     <- 100                      #how many steps in (0,1]
cases     <- 100000                   #how many processes to simulate
alpha     <- 0.00                     #probability  does not cross boundary
normmat   <- matrix(rnorm(steps*cases), ncol=steps)
brown     <- normmat/sqrt(steps)      #for var=1 after all steps
for (i in 2:steps){brown[,i] <- brown[,i-1] + brown[,i]}      #cumulative sum
absbrown  <- abs(brown)
boundary  <- rep(0,steps)
for (i in 1:steps){
     boundary[i] <- quantile(absbrown[,i],
                     probs = (steps-i*(1-alpha))/(steps-(i-1)*(1-alpha)),
                     names = FALSE)
     absbrown    <- absbrown[!(absbrown[,i] > boundary[i]), ] #del crossed
    }
plot( c(0,(1:steps)/steps), c(0,boundary), type="l", xlab="t",
      ylab="boundary", main=paste("simulated boundary for alpha =",alpha) )
abline(h=0)
abline(v=0)

Here are an example with $\alpha =0$.  The actual curve will be smoother.

Here is another.  If George Lowther is correct then this is simply the first half of the previous curve stretched upwards.

Added for comment: Taking the left hand half of the first graph (black below) and taking a shrunken version of the second (red below, dividing $t$ by $2$ and the boundary by $\sqrt{2}$), there is a very good match, except for the $y$ axis which may be a rounding effect in the simulation. So George Lowther looks correct.

Thoughts, ideas and discussions (not answers):

Consider the random variable $X(\omega)= sup_{t \in [0,1]} |B_t(\omega)|$. Let $c^*$ be the upper $1-\alpha$ quantile of this distribution; i.e, $P(\{\omega \in \Omega | X(\omega)>c^*\})=1-\alpha$.

The first $u(t)$ that came to my mind was the following: $u(t)=c^*$ for all $t \in [0,1]$; so that the confidence band is a constant. This works, right? Also, this is not conservative as

$\{\omega | |B_t(\omega)|\leq c^* \quad \forall \quad t \in [0,1]\} \subseteq \{\omega | X(\omega) \leq c^*  \}$.

but the other inclusion also holds.

Seems like an interesting problem. Presumably there are different confidence regions that you can build. Any ideas on what are the properties that you would like to have? I'd be interested in the context of this inference problem!


Quick question about contravariant and covariant tensors I have seen many different notations to denote contravariant/covariant and mixed tensors.  For example, I think the notation $\omega^{v}_{\,\,\,\mu}$ stands for a mixed tensor, where one index transforms contravariantly and the other covariantly.   What then would the notation $\omega^{v}_{\mu}$ denote?

A covariant vector transforms like $a'_{\mu} = \beta_{\mu}^{\,\,\,v}a_v$, but what is the difference between this and writing $a'_{\mu} = \beta_{\mu}^{v}a_v$ for example?

The only instance where I have seen an equivalence between the two notations is for the Kronecker delta, $\delta^v_{\,\,\,\mu} = \delta_{\mu}^{\,\,\,v} = \delta^v_{\mu}$, but the reason for this is not quite clear.

Many thanks.

In Einstein summation convention, repeated indices represent summation. Depending on the range of indices ($\nu=1$ to n), you would have a sum of n terms, and the contravariant index would cancel out the covariant index, leaving just a covariant $a'_\mu$


Why isn't $-2$ solution for $x$? I came across an logarithm problem recently. I don't know why solution to this problem cannot be $-2$. Now, don't downvote now because you don't know why I'm asking this. I know that logarithms' domains must be greater than $0$, can't take a negative number. Just read the whole post and wait until I get to the point to see why I'm asking this "ridiculous question".

This is the problem:
$$\log_{10}x=1-\log_{10}(x-3)$$
You can solve it like this:
$$\log_{10}x=\log_{10}10-\log_{10}(x-3)$$
$$\log_{10}x=\log_{10}\frac{10}{x-3}$$
$$x=\frac{10}{x-3}$$
$$x(x-3)=10$$
$$x^2-3x-10=0$$
$$(x+2)(x-5)=0$$
$$x\stackrel{?}{=}-2,\,\,x\stackrel{?}{=}5$$
If you plug in $5$, it will work just fine. You will get when simplified $\log_{10}5=\log_{10}5$. But if you plug in $-2$, you get something like this:
$$\log_{10}-2=\log_{10}\frac{10}{-2-3}$$
$$\log_{10}-2=\log_{10}-2$$
I know that negative logarithms with any base don't exist, but I thought of something. There's a logarithm property which states:
$$\log_by=\log_bx \Rightarrow y=x$$
Using this property, we can say:
$$\log_{10}-2=\log_{10}-2 \Rightarrow -2=-2$$
And, now (I hope) everyone will agree that $-2$ is indeed equal to $-2$ (itself).

So why besides $5$, isn't $-2$ solution to this problem? Even if the property that I stated doesn't apply to this case, could I theoretically invent a new imaginary number unit of $v_b=\log_b(-1)$ like someone back in the days did with $i=\sqrt{-1}$ and state that $\log_{10}(-2)=v_{10}+\log_{10}2$ ?

Okay we've talked a lot about complex logarithms, let's try and solve the problem over the complex numbers. (I think that is what the OP is really interested in; it has been reiterated that the only solution over the reals is 5)

We have:

$\log_{10} (-2)$

The change of base formula is consistent for all complex numbers, of which 10 is one. ($0i+10$). But we must be careful about branch selection. In this exercise we will denote $L(x)$ as the principal branch of the natural complex logarithm.

Therefore:

$\log_{10} (-2)$ =$ \frac{L(-2)}{L(10)}$

By the definition of the principle branch:

$\log_{10} (-2)$ =$ \frac{\ln(2)+iπ}{\ln(10)}$

As @uqtredd1 has noted, this solution is not one that should be submitted in the context of the presented problem, but is completely extraneous. The complex logarithm is a multivalued function, which is why branch selection was so important. There are many more possible "answers" to what $\log_{10} (-2)$ is in the complex sense.

If $-2$ is a solution, you must calculate $\log_{10}-2$, but doesn't exist.