Cauchy Criterion divergence I have problem checking of this series converge or diverge using Cauchy Criterion:

$$ \sum \frac {\sin(n^n+e^{n!})}{4^n} $$

I managed to show that the sum is smaller then $ \frac{p}{4^n}$ since $ -1\le \sin(x) \le 1 $ but I think I have to get rid of $p$ somehow.

Any advice?

Cauchy Criterion for convergence:

For every $\enspace\epsilon>0\enspace$ exists $\enspace N>0\enspace$ with $\enspace\displaystyle |\frac{\sin(m^m+e^{m!})}{4^m}-\frac{\sin(n^n+e^{n!})}{4^n}|<\epsilon$

for all $\enspace m,n\ge N$ .

Proof:

For $\enspace N\leq \min(m,n)\enspace$ and $\enspace\displaystyle \epsilon:=\frac{1}{4^{N-1}}\enspace$ we get

$\displaystyle |\frac{\sin(m^m+e^{m!})}{4^m}-\frac{\sin(n^n+e^{n!})}{4^n}|\leq \frac{1}{4^m}+\frac{1}{4^n}\leq \frac{2}{4^{\min(n,m)}}<\frac{1}{4^{N-1}}=\epsilon\enspace$ .

We have $|\frac {sin(n^n+e^{n!})}{4^n}| \le \frac {1}{4^n}$ for all $n$ and $\sum\frac {1}{4^n}$ is convergent ......


How to prove a matrix property? Let $$A=\begin {bmatrix}
                    1 & 3 &  1\\
                    1 & 1 & -2\\
                    1 & 2 & -1\\
                   \end{bmatrix}
                  $$

I  need to demonstrate this relation:
$$\forall n \in \mathbb{N}, A^n \neq I_3$$
where $I_3$=identity matrix.

If you have seen eigenvalues, you know that if $\lambda$ is an eigenvalue of $A$,
$\lambda^n$ is an eigenvalue of $A^n$. But all eigenvalues of $A$ are such that $|\lambda| \neq 1$ ($\lambda \approx 1.839$ and $\lambda \approx -0.420 \pm 0.606 i$), thus it is impossible, whatever $n$, to have $|\lambda|^n=1$ (the eigenvalue of $I_3$).

Proof using mathematical induction:

The base case: for $n=1$ we have $$A^1 = \begin {bmatrix}
                    1 & 3 &  1\\
                    1 & 1 & -2\\
                    1 & 2 & -1\\
                   \end{bmatrix}$$
So $A^1 \ne I$ is true for the base case.

Induction hypothesis: Assume that $A^n \ne I$ is a correct mathematical proposition and let proof that $A^{n+1} \ne I$ is also correct.
$$A^{n+1}=A^{n}\cdot A$$
According to the base case we have $A \ne I$ and according to the induction hypothesis we have $A^n \ne I$ , so $$A^n\cdot A \ne I \cdot I$$
$$\Leftrightarrow A^{n+1} \ne  I$$

Hence $A^{n+1} \ne  I$ is true. So $$\forall n \in \mathbb{N}: A^n\ne I$$


Understanding sum rule for marginal probability If $p(x,y)$ is the joint distribution of two discrete random variables $x, y$. The sum rule states that:
$$p(x) = \sum_{y \in T} p(x,y)$$

Where $T$ are that states of the target space of random variable $Y$

As per my understanding, this is basically the law of total probability. If events associated with target space of Y are a partition of the outcome space $\Omega$. We can calculate the probability of $x$ (marginal) regardless of $y$ (please correct me if there is something not accurate).

Now, my issue is with the other form of the sum rule (for continuous random variables):
$$p(x) = \int_{T} p(x,y) dy$$

It seems logical to me, but I want to understand how can we end up with this form for continuous r.v., so any pointers?

It seems to me to follow from the definition of a joint density $p(x,y)$, which is

$$P(X \in A, Y \in B) = \int_A \int_B p(x, y) \, dy \, dx$$

and the definition of the marginal density $p(x)$, which is

$$P(X \in A) = \int_A p(x) \, dx$$

If $T$ are all the possible values of $Y$, then, since $P(X \in A) = P(X\in A, Y \in T)$, and using the definition of a joint density:

$$P(X \in A) = \int_A \int_T p(x, y) \, dy \, dx$$

Notice that this is just the definition of the marginal density, but with $p(x)$ replaced with $\int_T p(x, y) dy$. So, that must be the marginal density.

The keyword is the probability measure you define on the space $T$. The last notation is the most general one in probability theory. When $T$ is discrete and has the Dirac point measure, it reduces to a sum.
The measure tells you how elements of the set $T$ are to be "counted in" when computing probabilities. The measure here is $dy$ i.e. an uniform democratic probability for all elements in $T$.


Are there any interpretations for the Gronwall's inequality in view of comparison theorem? One form of the Gronwall's inequality is that

If $\alpha(x),u(x)$ are non-negative continuous functions on $[0,1]$, and $$\forall x\in [0,1], u(x)\leq C+\int_{0}^{x}[\alpha(s)u(s)+K]ds\;(C,K\geq0),$$ then we have that $u(x)\leq[C+Kx]e^{\int_{0}^{x}\alpha(s)ds}$.

One form of comparison theorem is the following.

  Assume that $f(x,y),F(x,y)$ are continuous on a domain $\Omega\supset [0,1]\times\mathbb{R}$ and $f(x,y)<F(x,y)(\forall (x,y)\in\Omega)$, $y=\phi(x),y=\varphi(x)$ are solutions to $y'=f(x,y),y'=F(x,y)$ (respectively), and $\phi(0)=\varphi(0)$. Then we have that $\phi(x)<\varphi(x)\,(\forall x\in(0,1])$.

Are there any interpretations for the Gronwall's inequality in view of the comparison theorem? I am not sure if they have any connections besides the fact that Gronwall's inequality can be used to prove the comparison theorem.

Will someone be kind enough to give some comments on this?
Thank you very much!

Maybe this is a remark, but I'm not qualified to make remark, so I wrote here.
For simplicity, we could consider $K$ to be $0$. Now we are lead to $u(x)\le C+\int_0^x a(s)u(s)ds$, "differentiate" on both sides, $u'(x)\le a(x)u(x)$, now consider another ODE, $u'(x)=a(x)u(x)$, with initial value $u(0)=C$, we could apply comparison principle, right?


Why does it matter to approach zero from the left or right in 1/0? I was surprised to see that 1/0 is undefined. One answer mentions that $1/0$ can be +$\infty$ or -$\infty$ depending on whether $0$ is approached by the left or the right:

But why does this make a difference? Aren't both numbers equal by all measurable accounts?

Also, if zero is neither positive nor negative, why $1/0$ does not equal "unsigned infinity", which would be infinity in its own dimension (like imaginary numbers), and hence, $0$ in the real dimension?

Also, if we keep dividing 1 in parts of zero size, won't we have +$\infty$ number of parts?

Changed to an answer and expanded at asker's request.

The problems with extending the definition of division to make $\frac{1}{0}$ meaningful appear when you treat it like any other real number. The real numbers have certain operations defined on them, notably addition, subtraction, multiplication and (except when the denominator is $0$) division. These satisfy certain properties, such as $b\cdot \frac{a}{b}=a$ and $0\cdot a=0$ for any real numbers $a,b$ such that the expressions are defined. If we want these to still hold after defining $\frac{1}{0}$, we would have to let $1=0\cdot \frac 1 0 = 0$! This is absurd, so whatever we build by defining $\frac{1}{0}$ is very different from the real numbers, and lacks at least one of its fundamental properties.

You have to be careful here; you're poking at some advanced mathematics. Before delving into the discussion at hand, concerning your "infinity in [its] own dimension" remark, have a look at this. Now, onto why $\frac{1}{0}$ is (for the most part) problematic. We shall begin with a brief discussion of the concept of number and division.

We may only consider the concept of a non-negative integer, as the rest follows :) While there are a number of good, and technical, definitions of this type of mathematical object, it will suffice for us to interpret it as a set of sets. So, for example we may say $0 = \varnothing = \{\}, 1 = \{0\} = \{\{\}\},$ $2 = \{0, 1\} = \{\{\},\{\{\}\}\},$ and so on (Reference: Advanced Calculus, Shlomo Sternberg). Now, we may conveniently define $\mathbb{Z}^*$ (the set of non-negative integers) to be the set containing all objects describable in the form "$\{\{\}_1,\{\{\}\}_2,..., \{...\{\}...\}_{n - 1}\}$", where $n \in \mathbb{Z}^+$ and the indices have only been shown for convenience. Notice, that this disallows the number "$\infty$", since "$\infty$" is defined as (for our purposes) the object such that no number is greater than or equal to it. Indeed, this is problematic given our current framework, hence the link at the beginning. Anywho, we can now also define the binary operators "$+$", "$-$", "$\div$", and "$\times$". I'll leave the details to you, and only focus on "$\div$".

$$
\mathbf{\tag{1} Def:}\  \forall x, y \in \mathbb{Z}, \ {x} \div {y} = \frac{x}{y} = r \in \mathbb{Z}^* : (r \times y = x ) \iff (x \div y = r : y \text{ can be subtracted } r \text{ times from } x) \text{ (i.e., } x - r \times y = 0).
$$

Certainly, $x \geq y$ and $y \neq 0$ or $x < y$ and $x,\ y \neq 0$ is a necessary condition for $x \div y$ to be defined. As a corollary to what we have covered so far, $\forall x \in \mathbb{Z}^*,\ x \div 0$ is undefined, for there is simply no integer, $r$, such that $r \times 0 = x$ if $x > 0$; if $x = 0$, $x \div 0$ is still undefined, though, for a subtly different reason which I'll allow you the pleasure of working out for yourself.

The fact that "$\lim_{x \rightarrow \infty} x^{-1} = \pm \infty$" is merely a subtlety of the real number system, the notion of limit, and mathematicians rejecting the idea of defining infinity with coarse machinery.


Question on proof of no rational number whose square is 2 in the proof that there is no rational number whose square is 2:

Suppose x is a rational number whose square is 2.

Then x can be written in lowest terms as $\frac{a}{b}$ , where a is an integer and b is a positive integer.

Since $x^2 = 2$, $\left (\frac{a}{b}  \right )^2$, so $\frac{a^2}{b^2}$
But then a is even, so $a = 2n$ for some integer n.

Then $(2n)^2$ = $2b^2$, so $4n^2 = 2b^2$.

Then $2n^2 = b^2$ , so $b^2$ is even, and thus b is even.

Then a and b both have 2 as a common factor, so $\frac{a}{b}$ cannot be in lowest terms, a contradiction.
Thus x cannot be rational.

The part I am struggling with is why we can assume that x can be written in lowest terms. Or in other proofs, they say that the gcd(a,b) must be 1, but I do not understand why this is important to the proof. Any help is appreciated

You can assume that $\dfrac{a}{b}$ is in lowest terms because any fraction can be written in lowest terms, so we just take $\dfrac{a}{b}$ to be that lowest terms representation to make the proof easier. This is exactly the same as assuming $\gcd(a,b)=1$. This assumption is crucial, as it allows us to go through the proof, and eventually conclude that $2\mid a$ and $2\mid b$, which gives a contradiction because of our construction that $\gcd(a,b)=1$. Since $\gcd(a,b)=1$ there cannot be any common factors of $a$ and $b$ other than $\pm 1$. In other words, if $\dfrac{a}{b}$ is in lowest terms, the numerator and denominator cannot both be even, because then they would both be divisible by two, and thus the fraction is not in lowest terms.

You might think of the problem in another way. It just makes the proof finite.

Say you don't consider a & b to be the lowest terms. Follow it through...
You will find something along the lines of:
a = 2c
c = 2d
d = 2e
e = 2f
f = 2g
...

This is analogous to numerical methods whereby each iteration increases accuracy but you never get to 100% accurate - wherever you stop you only have an approximation.

This is the nature of irrational numbers compared to rational numbers - they cannot be defined precisely with integers. Resting upon the lowest factors argument is shorthand for saying "this iteration shouldn't occur infinitely for rational numbers".


Derivative of $\operatorname{Log}(\operatorname{Log}(z^2))$ Please help me with this question: (i don't know how to start)

Suppose that $f(z)$ = $\operatorname{Log}(\operatorname{Log}(z^2))$. Find $f'(z)$ where it exists, and determine the set of points at which $f$ is analytic.

Well, $log(log(z^2))$ is a composite function. More in general the derivative of a composition of 3 functions $f(g(h(x)))$ is $f'(g(h(x)))\cdot g'(h(x))\cdot h'(x)$. So, in this case
$$D[log(log(z^2))] =$$
$$\frac{1}{log(z^2)}\cdot \frac{1}{z^2}\cdot 2z =$$
$$\frac{1}{log(z^2)}\cdot \frac{2}{z} $$

As regards the $differentiability$ of the function, you need to study the continuity of $f'(x)$:
the only critical points are when
\begin{cases}
log(z^2) = 0 \to  z^2 = 1 \to z=\pm 1\\
z = 0
\end{cases} so the function is differentiable if $ z \neq \pm 1$ and $z \neq 0$


$\mathbb{Z}_{(2)}$ has one maximal ideal My lecture notes state that the set $\mathbb{Z}_{(2)}$, defined as

$$\mathbb{Z}_{(2)}:=\left\{\frac{a}{b}\in\mathbb{Q}\mathrel{}\middle|\mathrel{}\gcd(a,b)=1\text{ and } 2\nmid b\right\}$$ has a unique maximal ideal generated by $\left<2\right>$. Is this example trying to say that the set of even integers is the maximal ideal, and if so, why is this maximal. Can anyone prove that this is indeed a maximal ideal. Also in the set $\mathbb{Z}$, both $2\mathbb{Z}$ and $3\mathbb{Z}$ are maximal ideals, but I thought a maximal ideal is an ideal which contains all other ideals.

You can show that of the complement of $2 \mathbf{Z}_{(2)}$ is $\mathbf{Z}_{(2)}^{\times}$, showing thereby that $2 \mathbf{Z}_{(2)}$ is the unique maximal ideal of $\mathbf{Z}_{(2)}$. This show that $\mathbf{Z}_{(2)}$ is a local ring.

"Alternatively", consider the "map" from $\mathbf{Z}_{(2)}$ to $\mathbf{F}_2$ defined by sending an element $x= \frac{a}{b}$ with $a,b$ integers and $2$ not dividing $b$ to $\overline{a} \overline{b}^{-1}$, where $\overline{x}$ is the image in $\mathbf{F}_2$ of the integer $x$. Show that this map is well defined, that is, that it does not depend on the writing $x= \frac{a}{b}$ with $a,b$ integers and $2$ not dividing $b$, and that it is a ring morphism. Show that its kernel is the ideal $2 \mathbf{Z}_{(2)}$. As this morphism is surjective, this implies passing to the quotient that the induced map $\mathbf{Z}_{(2)} / 2 \mathbf{Z}_{(2)} \to \mathbf{F}_2$ is an isomorphism. As $\mathbf{F}_2$ is a field, this shows that $2 \mathbf{Z}_{(2)}$ is a maximal ideal of $\mathbf{Z}_{(2)}$.

Intuition. More generally instead of $2$, let $p$ be a prime. What is $\mathbf{Z}_{(p)}$ ? The ring $\mathbf{Z}$ is a ring where only $\pm 1$ are invertible. We want to construct a bigger ring, containing $\mathbf{Z}$, and in which every guy from $\mathbf{Z}$ that is not divisible by $p$ will become invertible. This ring is $\mathbf{Z}_{(p)}$. (For any ring, $A^{\times}$ denotes the set of invertible elements of $A$.) Then, $\mathbf{Z}_{(p)}^{\times} = \mathbf{Z}_{(p)} \backslash p\mathbf{Z}_{(p)}$.

Remark. More generally, if $A$ is a commutative ring with unit and if $\mathfrak{p}$ is a prime ideal of $A$, with the same arguments, you have an isomorphism $A_{\mathfrak{p}} / \mathfrak{p} A_{\mathfrak{p}} \to \textrm{Frac}(A/\mathfrak{p})$, where $A_{\mathfrak{p}}$ is the localisation of $A$ with respect to the multiplicative subset $S = A \backslash \mathfrak{p}$ of $A$. The ring $A_{\mathfrak{p}}$ is also called by abuse of language the location of $A$ at the prime ideal $\mathfrak{p}$, or also more simply, the localization of $A$ a $\mathfrak{p}$. This vocabulary applies of course to $A = \mathbf{Z}$ and $\mathfrak{p} = 2 \mathbf{Z}$.

The maximal ideal is $2\mathbb{Z}_{(2)}$.

Show it is maximal by showing that $\mathbb{Z}_{(2)}/2\mathbb{Z}_{(2)}$ is a field.

To show that it is the only maximal ideal, there are many ways :

Show that $J(\mathbb{Z}_{(2)})=2\mathbb{Z}_{(2)}$ where $J$ is the Jacobson ideal.
Use the following theorem : If $R$ is a commutative ring and $M$ is a maximal ideal of $R$, then the following are equivalent :

(1) $R$ has a unique maximal ideal (such a ring is called local)

(2) $R\backslash M=R^{\times}$

(3) $\forall a,b\not\in R^{\times},a+b\not\in R^{\times}$.


inverse system vs inverse sequence I am wondering about such problem. Let $\{X_i,\phi_{ij},I\}$ be an inverse system, where the directed set $I$ has such property that there exists a sequence $i_1 \leq i_2\leq\cdots\subset I$ such that for every $j\in I$ there exists $i_n$ with $i_n\geq j$.

So it gives us inverse sequence indexed by the elements of the sequence. Are there any connections between inverse limits of these inverse systems? Maybe they are the same sometimes? I am especially interested in inverse limits of topological spaces.

Write the inverse limit of the full system as $A$ and of the subsystem as $B$. We have a map from $A$ to $B$ given on $X_{i_j}$ by the composition of the identity on $X_{i_j}$ with the projection of $A$ onto $X_{i_j}$ and a map $B\to A$ given similarly on the $X_{i_j}$ and on the other $X_i$ by composing the identity on $X_i$ with the given map $X_{i_n}\to X_i$ and the projection $A\to X_{i_n}$. You can check directly that both compositions are the identity, since their components are the projections $A\to X_i$ and $B\to X_{i_j}$.

The sequence is also directed, and cofinal in I. It's a theorem that the inverse limits of the inverse systems are isomorphic.


You are flipping n fair coins, putting aside those that come up heads after each flip. What's the expected number of rounds? This question is related to https://www.quora.com/You-are-flipping-100-coins-putting-aside-those-that-come-up-heads-after-each-flip-How-many-flips-do-you-expect-to-make-before-running-out-of-coins-to-flip

At each stage, you flip all your coins and only keep the ones that are tail. You stop when you ran out of coins. Some Quora answers mention that the expected number of rounds is logarithmic with respect to $n$. Any suggestions on how you would prove this fact?

Since the probability of getting a head is $1/2$ after every round the number of coins will be probably halved. So let's call $C_j$ the number coins at the $j$-th round:

$$C_j=\frac{n}{2^j}$$

The game ends when:

$$C_j<1$$
$$2^j>n$$
$j>\log_2(n)$

So the expected value $x(n)$ searched is:

$$x(n) \sim \log_2(n) $$

:)


Why do parabolas' arms eventually become parallel? Here is what this site states

  All parabolas are the same shape, no matter how big they are. Although they are infinite, meaning that the arms will never close up, the arms will eventually become parallel.

Now, I have an argument against it. Let $f(x) = ax^2 + bx + c$ be a quadratic polynomial with $a , b$ and $c$ being real numbers and $a \ne 0$. So, its graph will give us a parabola ($\because$ graph of a quadratic polynomial is a parabola).

Now,

$$\dfrac{d (ax^2 + bx + c)}{dx}$$$$ = 2ax + b$$

i.e. the slope of a quadratic polynomial is given by $g(x) = 2ax + b$. Now, differentiating the equation for slope of the quadratic

$$\dfrac {d (2ax + b)}{d x}$$
$$ = 2a$$

So, if $a \gt 0$ then the slope of $g(x)$ will be increasing. This means that the slope of $f(x)$ will also be increasing. Similarly, if $a \lt 0$ then the slope of $f(x)$ will be decreasing.

This means that for all $x$ the slope of $f(x)$ will be different. So, this contradicts the fact (according to me) that the arms of a parabola will eventually be parallel. Where am I going wrong?

By "eventually" they mean "in the limit". As $a$ goes to positive infinity, the slope at $a$ goes to positive infinity, which means the arm is becoming more and more vertical. As $a$ goes to negative infinity, so does the slope at $a$ - but a slope of "negative infinity" also means that the line is vertical, so the left arm is also becoming closer and closer to vertical. If both arms are getting closer to closer to vertical, then they must be getting closer and closer to being parallel to one another.

Of course, your argument shows that they will never actually be parallel. Just very close.

Obviously, the tangents become vertical. Verticals are parallel.


Alternative model of Euclidean geometry I'm planning to teach high-school geometry. As usual, this will be by building from axioms. (The axioms used are AFAICT particular to the book I've been assigned, but they're some combination of Hilbert's, SMSG's, and God knows what.) I'm considering demonstrating that geometry's axioms need not have their usual model by presenting an alternative model of at least a few basic axioms. Can anyone recommend such a model? I'd need it to be accessible to high schoolers (so, for example, not this).

The rational plane $\mathbb{Q}^2$ is a model for Euclid's five axioms, and I would think (hope?) that it is accessible to high-schoolers. Many common geometric constructions don't work as expected for it; for example, here's an excerpt from Explanation and Proof in Mathematics, p.66:

A geometric model for euclidean geometry should be presented as follows:

Euclidean plane be the open unit disk
A point of E_plane be a point inside the unit circle and vice versa
A line of E_plane be part of circle inside the unit circle which passes through a fixed point on the unit circle.

It seems that any student can easily prove that sum of internal angles is equal to $180^\circ$ for triangles.

A.V.Farimani


expected value of failures before success in n experiments let $Y \sim B(15, p)$ be the number of successful experiments out of 15 and let $X$ be the number of failed experiments before the first success. I need to calculate $E[X|Y=1]$ and get a final answer without $p$.
I defined $W$ as the number of experiments until the first success so $W \sim G(p)$ hence $X=W-1$ so $E[X|Y=1] = E[W|Y=1 - 1] = E[W|Y=1] - 1$ but when I try to calculate $E[W|Y=1]$ I get
$$E[W|Y=1] = \sum_{i=1}^{15}ip(1-p)^{i-1}=\sum_{i=1}^{15}\binom{i}{1}p(1-p)^{i-1}$$ and I don't know how I can get a solution without p from this


Binomial and Uniform joint distribution statistics question. I was just wondering if someone could explain the solution to this example in my textbook for me. I am not really understanding it. The question is,
X follows Binomial(n,p) and p follows Uniform(0,1). Show, P(X =0) = P(X = 1) = · · · = P(X = n). You are given that the integral of the binomial distribution over (0,1) is equal to  [i!(n-1)!]/(n+1)!.

$P(X=k)=\binom{n}{k}\int_0^1 p^k(1-p)^{n-k}dp$.
Integrate by parts $du=(1-p)^{n-k}dp,\ v=p^k$.
You will get$P(X=k)=\binom{n}{k} \frac{k}{n-k+1}\int_0^1 p^{k-1}(1-p)^{n-k+1}dp=P(X=k-1)$

Therefore all P(X=k) are equal.


Difference between the difference operator in recursion equations vs the derivation opertor. The derivation operator is $D=d/dx$, where we have: $D(f(x)g(x)) = fDg+gDf$.

Now while reading a book on difference equations, it occured to me that for the discrete operator, i.e $\Delta(y_n) = y_{n+1}-y_n$, we don't have the same identity, i.e. we have: $\Delta(z_n y_n) = z_{n+1}y_{n+1}-z_n y_n$, and $z_n\Delta y_n + y_n \Delta z_n = z_n(y_{n+1}-y_n)+y_n(z_{n+1}-z_n) = z_n y_{n+1}+y_nz_{n+1}-2z_ny_n$.

What other differences are there between these operators?
I prefer those that aren't obvious, although if you were to ask me before calculating I would have said the same identity should be also true for the discrete case.

They are pretty similar but at the same time slightly different due to the fact that:

$$\Delta y=y(x+1)-y(x)=\frac{y(x+1)-y(x)}{(x+1)-x}$$

The solution to the equation:

$$y=Dy$$

Is as we know $ce^x$.

The solution to the equation:

$$y=\Delta y$$

Is on the other hand,

$$y=c2^x$$

Given $f(x)=Dy$ then to find $y$ we integrate (a constant also appears when we integrate):

$$y= \int f(x) dx$$

Given $f(x)=\Delta y$ then to find $y$ we sum and add a corresponding constant for example,

$$\sum_{i=0}^{x-1} f(x)=\sum_{i=0}^{x-1} (y(i+1)-y(i))=y(x)+y(0) \implies y(0)+\sum_{i=0}^{x-1} f(x)=y(x)$$

We also have a pretty similar but different taylor series:

$$y(x)=\sum_{i=0}^{\infty} \Delta^i(0){x \choose i}$$

If we take $\Delta^0(0)=1$ and $\Delta^i(0)$ to represent $\Delta^i(f(x))$ when $x=0$.

The product rule  similar as mentioned in the comments illustrates similarity but slight difference , but "the chain rule" is very different and more complex: Chain rule for discrete/finite calculus.

The product rule can also be written as,

$$\Delta(f(x)g(x))=g(x)\Delta(f(x))+f(x)\Delta(g(x))+\Delta(f(x))\Delta(g(x))$$

Here is the quotient rule which is again slightly different:

$$\Delta(\frac{f(x)}{g(x)})=\frac{g(x)\Delta f(x)-f(x)\Delta g(x)}{g(x)(g(x)-\Delta g(x))}$$

The Chain Rule is trivial for differences: $$\Delta (g\circ f)(c)= \Delta g\, (f(c)).$$


Proof that $a(a+1)(2a+1)$ is divisible by $6$ for every integer a This is from the book Elementary Number Theory by Jones & Jones
Example 3.6
Let us prove that a(a+1)(2a+1) is divisible by 6 for every integer a
By taking least absolute residues mod(6) we see that $a \equiv 0,\pm1,\pm2 or 3$.
If $a \equiv 0$ then $a(a+1)(2a+1) \equiv 0 \cdot 1 \cdot 1 \equiv 0$, if $a \equiv 1$, then $a(a+1)(2a+1) \equiv 1 \cdot 2 \cdot 3 = 6 \equiv 0$, and similar calculations (which you should try for yourself) show that $a(a+1)(2a+1) \equiv 0$ in the other 4 cases, so $6 \vert a(a+1)(2a+1)$ for all a.

I don't understand the proof at all starting with the first line - By taking least absolute residues mod(6) we see that $a \equiv 0,\pm1,\pm2 or 3$. - How does taking absolute residues mod(6) give $a \equiv 0,\pm1,\pm2 or 3$?

The remainder when you divide a number by $6$ can only be $0,1,2,3,4,5$.
Furthermore, $4 \equiv -2 \pmod{6}$ and $5 \equiv -1 \pmod{6}$.
After that, you just have to enumerate the $6$ cases to verify that it is true.

Hint:
It's much better to use:
$$a(a+1)(2a+1)=a(a+1)(2(a-1)+3)=2\underbrace{(a-1)a(a+1)}_{\text{ The product of } 3\text{ consecutive integers }}+3a(a+1)$$
Use The product of $n$ consecutive integers is divisible by $n$ factorial


Calculating the are between 2 functions, by y-axis So, I have these 2 equations:
$$
y_1=e^{2x}\\y_2=-e^{2x}+4
$$
And I need to calculate the area they have limited right of the y-axis
this is how it looks:
I have converted the formulas to x:
$$
x_1 = \frac{\log(y_1)}{2}\\x_2 = \frac{\log(4-y_2)}{2}\\
$$
Calculated the limits:
$$
\frac{\log(y_1)}{2}= \frac{\log(4-y_2)}{2}\\y = 2 \Rightarrow x = \frac{\log(2)}{2}
$$

And I got to the point of integrating:
$$
\int_0^{\frac{log(2)}{2}}(\frac{\log(4-y)}{2}- \frac{\log(y)}{2})dy = \\=\int_0^{\frac{log(2)}{2}}(\frac{\log(4-y)}{y})dy
$$
And I don't know how to integrate $\log(x)$ can you please help me?

You don't need to inverse the function. Instead do the following:

Find the intersection by solving: $y_1=y_2$

Integrate $y_2-y_1$ from 0 to the intersection. This will give you the area.

Full calculation:

Solving $y_1=y_2$:
$$e^{2x}=-e^{2x}+4$$
$$2e^{2x}=4$$
$$e^{2x}=2$$
$$2x=\ln(2)$$
$$x=\frac12\ln(2)=\ln(\sqrt2)$$
$$y=e^{2x}=e^{2(\frac12\ln(2))}=2$$
Integrating the difference:
$$\int_{0}^{\frac12\ln(2)}(4-e^{2x}-e^{2x})dx=\int_0^{\frac12\ln(2)}4dx-\int_0^{\frac12\ln(2)}2e^{2x}\\
=2\ln(2)-[e^{\ln(2)}-e^{2*0}]=2\ln(2)-2+1=2\ln(2)-1$$

When in doubt go to integral tables:

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


Proving two different expressions of non-centrality parameters are equivalent I am stuck in proving $$\sum_{i=1}^{K}\xi_i(\mu_i - \bar{\mu})^2 = \sum_{i,j}\xi_i\xi_j(\mu_i - \mu_j)^2,$$
where $\bar{\mu} = \sum_{i=1}^{K}\xi_i\mu_i$ and $\sum_{i=1}^{K}\xi_i = 1$.
I am not sure whether it is true or not.


How do I perform a change of variables? Use a change of variables to evaluate:
  $$\iiint\limits_{D}xy\,\mathrm{d}V$$$D$ is bounded by the planes $y-x=0$, $y-x=4$, $z-y=0$, $z-y=1$, $z=0$, $z=5$.

A similar question was asked here. It was never answered, because the OP demonstrated that he/she understood how to perform the change of variables. In that question, the OP sets $u$, $v$, and $w$ equal to expressions involving $x$, $y$, and $z$ and uses those equations to solve the problem. I do not understand how to come up with equations for $u$, $v$, and $w$. Can someone please walk me through that process?

The idea is to choose coordinates such that the limits of integration are somehow 'simpler'. Note that the $0 \leq z \leq 5$ limits are already 'nice' - the issues are with $0 \leq y - x \leq 4$ and $0 \leq z - y \leq 1$.

We would like to have something like $0 \leq u \leq 4$ and $0 \leq v \leq 1$ where $u$ and $v$ are some new coordinates - so do precisely that, and choose $u = y - x$, $v = z - y$, and $w = z$.

With respect to these new coordinates, the domain $D$ becomes $D' = \lbrace (u, v, w) : u \in [0, 4], \,v \in [0, 1], \, w \in [0, 5] \rbrace$. Then we solve this for $(x, y, z)$ by observing that $u + v = z - x$ so $x = w - u - v$, which leads to $(x, y, z) = (w-u-v, w-v, w)$.

You can do the choice $u = y-x,v=z-y$ and $w=z$. Then the bounds in the integral for These new variables are easy, e.g. integral over $u$ goes from 0 to 4. Solve above equation System by $x,y$ and you will get $x(u,v,w),y(u,v,w)$, i.e. the "old coordinates" in dependence of the "new coordinates".

The next step is calculating the Jacobi Matrix $J$ (by computing derivatives of old coordinates by new coordinates) and then its determinant. Because $w=z$ this calculation simplifies a lot.

After this you can compute

$\int \int \int_D xy dV = \int_0^5 \int_0^1 \int_0^4 x(u,v,w)y(u,v,w) det J du dv dw$


Norms Induced by Inner Products and the Parallelogram Law Let $ V $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$.

It's not hard to show that if $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ for some (real) inner product $\langle \cdot, \cdot \rangle$, then the parallelogram equality
$$ 2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u - v\rVert^2 $$
holds for all pairs $u, v \in V$.

I'm having difficulty with the converse.  Assuming the parallelogram identity, I'm able to convince myself that the inner product should be
$$ \langle u, v \rangle = \frac{\lVert u\rVert^2 + \lVert v\rVert^2 - \lVert u - v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u\rVert^2 - \lVert v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 - \lVert u - v\rVert^2}{4} $$

I cannot seem to get that $\langle \lambda u,v \rangle = \lambda \langle u,v \rangle$ for $\lambda \in \mathbb{R}$.  How would one go about proving this?

Since this question is asked often enough, let me add a detailed solution. I'm not quite following Arturo's outline, though. The main difference is that I'm not re-proving the Cauchy-Schwarz inequality (Step 4 in Arturo's outline) but rather use the fact that multiplication by scalars and addition of vectors as well as the norm are continuous, which is a bit easier to prove.

So, assume that the norm $\|\cdot\|$ satisfies the parallelogram law
$$2 \Vert x \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y \Vert^2 + \Vert x - y \Vert^2$$
for all $x,y \in V$ and put
$$\langle x, y \rangle = \frac{1}{4} \left( \Vert x + y \Vert^2 - \Vert x - y \Vert^2\right).$$ We're dealing with real vector spaces and defer the treatment of the complex case to Step 4 below.

Step 0. $\langle x, y \rangle = \langle y, x\rangle$ and $\Vert x \Vert = \sqrt{\langle x, x\rangle}$.

Obvious.

Step 1. The function $(x,y) \mapsto \langle x,y \rangle$ is continuous with respect to $\Vert \cdot \Vert$.

Continuity with respect to the norm $\Vert \cdot\Vert$ follows from the fact that addition and negation are $\Vert \cdot \Vert$-continuous, that the norm itself is continuous and that sums and compositions of continuous functions are continuous.

Remark. This continuity property of the (putative) scalar product will only be used at the very end of step 3. Until then the solution consists of purely algebraic steps.

Step 2. We have $\langle x + y, z \rangle = \langle x, z \rangle + \langle y, z\rangle$.

By the parallelogram law we have
$$2\Vert x + z \Vert^2 + 2\Vert y \Vert^2  = \Vert x + y + z \Vert^2 + \Vert x - y + z\Vert^2 .$$

This gives
$$\begin{align*}
\Vert x + y + z \Vert^2 & = 2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 - \Vert x - y + z \Vert^2 \\
& = 2\Vert y + z \Vert^2 + 2\Vert x \Vert^2 - \Vert y - x + z \Vert^2
\end{align*}$$
where the second formula follows from the first by exchanging $x$ and $y$. Since $A = B$ and $A = C$ imply $A = \frac{1}{2} (B + C)$ we get

$$\Vert x + y + z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x + z \Vert^2 + \Vert y + z \Vert^2 - \frac{1}{2}\Vert x - y + z \Vert^2 - \frac{1}{2}\Vert y - x + z \Vert^2.$$

Replacing $z$ by $-z$ in the last equation gives
$$\Vert x + y - z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x - z \Vert^2 + \Vert y - z \Vert^2 - \frac{1}{2}\Vert x - y - z \Vert^2 - \frac{1}{2}\Vert y - x - z \Vert^2.$$

Applying $\Vert w \Vert = \Vert - w\Vert$ to the two negative terms in the last  equation we get
$$\begin{align*}\langle x + y, z \rangle & = \frac{1}{4}\left(\Vert x + y + z \Vert^2 - \Vert x + y - z \Vert^2\right) \\
& = \frac{1}{4}\left(\Vert x + z \Vert^2 - \Vert x - z \Vert^2\right) +
\frac{1}{4}\left(\Vert y + z \Vert^2 - \Vert y - z \Vert^2\right) \\
& = \langle x, z \rangle + \langle y, z \rangle
\end{align*}$$
as desired.

Step 3. $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{R}$.

This clearly holds for $\lambda = -1$ and by step 2 and induction we have $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{N}$, thus for all $\lambda \in \mathbb{Z}$. If $\lambda = \frac{p}{q}$ with $p,q \in \mathbb{Z}, q \neq 0$ we get with $x' = \dfrac{x}{q}$ that
$$q \langle \lambda x, y \rangle = q\langle p x', y \rangle = p \langle q x', y \rangle = p\langle x,y \rangle,$$
so dividing this by $q$ gives
$$\langle \lambda x , y \rangle = \lambda \langle x, y \rangle \qquad\text{for all } \lambda \in \mathbb{Q}.$$
We have just seen that for fixed $x,y$ the continuous function $\displaystyle t \mapsto \frac{1}{t} \langle t x,y \rangle$ defined on $\mathbb{R} \smallsetminus \{0\}$ is equal to $\langle x,y \rangle$ for all $t \in \mathbb{Q} \smallsetminus \{0\}$, thus equality holds for all $t \in \mathbb{R} \smallsetminus \{0\}$. The case $\lambda = 0$ being trivial, we're done.

Step 4. The complex case.

Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$, observe that $\langle ix,y \rangle = i \langle x, y \rangle$ and $\langle x, y \rangle = \overline{\langle y, x \rangle}$ and apply the case of real scalars twice (to the real and imaginary parts of $\langle \cdot, \cdot \rangle$).

Addendum. In fact we can weaken requirements of Jordan von Neumann theorem to
$$
2\Vert x\Vert^2+2\Vert y\Vert^2\leq\Vert x+y\Vert^2+\Vert x-y\Vert^2
$$
Indeed after substitution $x\to\frac{1}{2}(x+y)$, $y\to\frac{1}{2}(x-y)$ and simplifications we get
$$
\Vert x+y\Vert^2+\Vert x-y\Vert^2\leq 2\Vert x\Vert^2+2\Vert y\Vert^2
$$
which together with previous inequality gives the equality.

(sorry, I'm not familiar with LaTex, hope you have no difficutly reading my answer).

I have another method to prove this result for $\ x,y \in \mathbb{R^n}$, $\lambda \in \mathbb{R}$.

step.1.
prove that $\langle \lambda x,\lambda y \rangle = \lambda^2 \langle x, y \rangle$, use polarisation identity to expand inner product, it's easy to prove it.

step.2.
prove that $\langle \lambda x, y \rangle = \langle x, \lambda y \rangle$, i.e.prove $\langle \lambda x, y \rangle - \langle x, \lambda y \rangle=0$, the proof is similar to proving $\langle x+y, z \rangle = \langle x, z \rangle+\langle y, z \rangle$, you may use $\langle x+y, z \rangle = \langle x, z \rangle+\langle y, z \rangle$ and $\langle -x, y \rangle = -\langle x, y \rangle=\langle x,-y \rangle$ in your proof and the latter is easy to prove.

step.3.
use the result of step.2., we have $\langle \lambda x,\lambda y \rangle = \langle \lambda^2 x, y \rangle$, compare to the result in step.1., we get that $\langle \lambda^2 x , y \rangle = \lambda^2 \langle x, y \rangle$, for $\lambda^2 >0$.

step.4
when $\lambda <0$, $\langle \lambda x, y \rangle = \langle-(-\lambda x),  y \rangle=-\langle(-\lambda x),  y \rangle$, use the result in step.3., we get $\langle(-\lambda x),  y \rangle=-\lambda\langle x,  y \rangle$, which means the result holds for $\lambda <0$, and it's easy to prove when $\lambda =0$.

It's a bit more complex to prove for $\ x,y \in \mathbb{C^n}, \lambda \in \mathbb{C}$, while the method is general, you just have to seperate it into two parts, the real and the imaginary.


Multiples that are one less than Squares I was inspired to ask this problem after trying to find all $(x,y,u,v)$ for which $xy+1,xu+1,xv+1,yu+1,yv+1,uv+1$ are all sqaure.

After some basic calculation I was easily able to find
$x=n, y=n+2, u=4n+4$. However, I had some difficulty finding $v$.

It took some time, but I was able to find $v=4(n+1)(2n+1)(2n+3)$.

Are there more easier ways of finding $(x,y,u,v)$?

And is the following generlization possible: There exists infinitely many sets $X_{ n }=\{ x_{ i }|1\le i\le n \quad i\in \mathbb{N}\}$ and ${ X }_{ n }\subset  \mathbb{N}$ for any finite natural number $n$ for which ${ x }_{ i }x_{ j }+1$ are all perfect squares?

This problem is well-known in the literature, where such sets are called Diophantine $4$-tuples.  Dujella has made an extensive study of them: see his web page at https://web.math.pmf.unizg.hr/~duje/dtuples.html for plenty of references.

In particular, Dujella proved that there are no Diophantine $6$-tuples and that any $5$-tuples, if they exist, are bounded by an absolute constant so there are at most finitely many (none have been found).  So the conjecture in your last paragraph is false for every $n \ge 6$.

Your question about extending a triple of integers to a $4$-tuple has also been considered (see https://web.math.pmf.unizg.hr/~duje/quint.html for the reference).  We may always take

$$v = x + y + z + 2xyz + 2\sqrt{(xy+1)(yz+1)(xz+1)}.$$

It's conjectured that the above value of $v$ is unique (which would immediately imply that $4$ is the largest possible Diophantine tuple).

It is better to use a more General approach. We write the system.

$$\left\{\begin{aligned}&xy+T=a^2\\&xz+T=b^2\\&xq+T=c^2\\&yz+T=d^2\\&yq+T=k^2\\&zq+T=n^2\end{aligned}\right.$$

If the number $T$, lay at the multipliers. We find then the desired settings.

$$T=3(p-t-s)(p+t-s)(p+s)^2$$

Then the solution can be written as.

$$x=t^2+2s^2+2ps-p^2$$

$$y=t^2-s^2+2ps+2p^2$$

$$z=4t^2-(p-s)^2$$

$$q=3(p+s)^2$$

$$a=p^2+ps+s^2-t^2$$

$$b=2t^2+s^2+ps-2p^2$$

$$c=3(p+s)s$$

$$d=2t^2-2s^2+ps+p^2$$

$$k=3(p+s)p$$

$$n=3(p+s)t$$


A logarithimic simplification How could I simplify this? It must give $\dfrac{9}{2}+\ln (2)$.

$$\left( \dfrac{4}{2}+6+\ln (2)\right) - \left( \dfrac{1}{2}+3+\ln (1)\right) =8+\ln (2)-\dfrac{7}{2}-\ln (1)=\dfrac{9}{2}+\ln (2)-\ln (1)$$


Generalization of an Integral Trick? There is an interesting trick that can be used to evaluate integrals in the form
$$I=\int_{-a}^a \frac{E(x)}{b^x+1}dx$$
where $E$ is an even function. Notice that, by substituting $x\to -x$,
$$I=\int_{-a}^a \frac{E(-x)}{b^{-x}+1}dx=\int_{-a}^a \frac{b^xE(x)}{b^{x}+1}dx$$
and so
$$I+I=\int_{-a}^a \frac{E(x)+b^xE(x)}{b^x+1}dx=\int_{-a}^a E(x)dx$$
and so
$$I=\frac{1}{2}\int_{-a}^a E(x)dx=\int_{0}^a E(x)dx$$
For example, this trick can be used to evaluate the intimidating integral
$$\int_{-1}^1 \frac{x^{100}}{e^x+1}dx=\frac{1}{101}$$

QUESTION: Is there some way to generalize this trick to integrals of the form
$$I=\int_{-a}^a \frac{E(x)}{(b^x+1)^2}dx$$
or will this type of integral just have to be done the hard way?

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

  Lets
  $\ds{{1 \over b^{x} + 1} =
{1 \over \expo{\beta x} + 1} \equiv \mathrm{f}\pars{x}:\ Fermi\!-\!Dirac\ Distribution\ Function.\quad\beta \equiv \ln\pars{b} > 0}$

\begin{align}
&\mbox{It satisfies the identity}\quad
\mrm{f}\pars{x} =
\left\{\begin{array}{lcrcl}
\ds{\Theta\pars{-x} + \,\mrm{sgn}\pars{x}\mrm{f}\pars{\verts{x}}} & \mbox{if} &
\ds{x} & \ds{\not=} & \ds{0}
\\[2mm]
\ds{1 \over 2} & \mbox{if} & \ds{x} &  \ds{=} & \ds{0}
\end{array}\right.
\\[5mm]
&\mbox{where}\ \Theta\ \mbox{is the}\ Heaviside\ Step\ Function.
\end{align}

Then, with $\ds{a > 0}$:
$$
\bbx{\int_{-a}^{a}\mrm{E}\pars{x}\mrm{f}\pars{x}\dd x =
\int_{-a}^{0}\mrm{E}\pars{x}\dd x + \int_{0}^{a}\bracks{\mrm{E}\pars{x} - \mrm{E}\pars{-x}}\mrm{f}\pars{x}\dd x}
$$

  Moreover

\begin{align}
\int_{-a}^{a}{\mrm{E}\pars{x} \over \pars{b^{x} + 1}^{2}}\,\dd x & =
\int_{-a}^{a}\mrm{E}\pars{x}
\bracks{-\,{1 \over \beta}\expo{-\beta x}\mrm{f}'\pars{x}}\,\dd x
\\[1cm] & \stackrel{\mrm{IBP}}{=}\,\,\,
-\,{\mrm{E}\pars{a}\expo{-\beta a}\mrm{f}\pars{a} -
\mrm{E}\pars{-a}\expo{\beta a}\mrm{f}\pars{-a} \over \beta}
\\[2mm] & +
\int_{-a}^{a}\bracks{%
{1 \over \beta}\,\mrm{E}'\pars{x}\expo{-\beta x} -
\mrm{E}\pars{x}\expo{-\beta x}}\mrm{f}\pars{x}\,\dd x
\end{align}

  which is reduced to the previous case !!!.


Simplifying sum with factorial in denominator I am trying to to find the sum of this series: $\sum_{x=0}^{\infty}\frac{x^2(1/2)^xe^{-1/2}}{x!}$, but I am stuck because I don't know how to deal with the factorial in the denominator. Is this perhaps somehow related to a Taylor series expansion?

$$
\frac{x^2}{x!} = \frac{x(x-1)}{x!} + \frac{x}{x!} = \frac 1 {(x-2)!} + \frac 1 {(x-1)!}
$$

Next, break the sum into two sums: one with the first of these two fractions and one with the second.  And notice that the cases $x=0$ and $x=1$ require some special attention.


Understanding use of factor theorem to factor polynomial based on possible factors from constant I'm reviewing my algebra in prep for Linear Algebra, using the review of Algebra in Stewart's Single Variable Calculus, here: http://stewartcalculus.com/data/CALCULUS_8E_ET/upfiles/6et_reviewofalgebra.pdf

On p.4, they use the Factor Theorem to factor $$x^3 - 3x^2 -10x + 24$$

They state the following:

  If P(b) = 0, where b is an integer, then b is a factor of 24.

They then use this fact to test possible integer values of b as a factor of 24 (+1, -1, +2, -2, etc.).  What I'm not understanding is how they arrive at the fact that b must be a factor of 24 based on the the Factor Theorem.  I understand from the Factor Theorem that if P(b) = 0, then x - b will be a factor of P(x), but how did they discern that b would be a factor of 24?  Thanks in advance.

Because, by the rational root theorem, every rational root $q_0$ of a polynomial $a_0+a_1x+\cdots+a_nx^n$ with integer coefficients can be written as $\frac ab$, where $a$ and $b$ are integers such that $b$ divides $a_n$ and $a$ divides $a_0$. In your case, $a_n=1$ and $a_0=24$. So, $q$ must be an integer that divides $24$.

The product of all roots of this polynomial is $24$ as you can see. Also $b$ is a root of this polynomial as given $P(b)=0$. Therefore $b$ must be a factor of $24$.


Can every transcendental number be expressed as an infinite continued fraction? Every infinite continued fraction is irrational.  But can every number, in particular those that are not the root of a polynomial with rational coefficients, be expressed as a continued fraction?

The infinite continued fractions are precisely the irrational numbers; you will find a proof here, along with a proof that the expansion is unique.

The set of non-computable numbers is a strict subset of the transcendental numbers. If infinite continued fractions are computable, then I believe that they would not include all the transcendental numbers. Chaitin's constant is transcendental, but cannot be represented by an infinite continued fraction. Chaitin's constant cannot be represented through an algorithmic process.


Prove that if each row of a matrix sums to zero, then it has no inverse. Could anyone help me with this proof without using determinant? I tried two ways.

  Let $A$ be a matrix. If $A$ has the property that each row sums to zero, then there does not exist any matrix $X$ such that $AX=I$, where $I$ denotes the identity matrix.

I then get stuck. The other way was to prove by contradiction, and I failed too.

Hint: You can sum the elements of a row by multiplying this row with a vector of $1$'s. Can you find now a matrix $X$ (with appropriate columns) such that $AX=Ο$?

I'm not too sure but my reasoning is that if the sum of a row is 0, then the rows of the matrix A are linearly dependent because they are a linear combination. If the rows of A are linearly dependent then the columns of A transpose is linearly dependent. Therefore matrix A is not invertible by the Invertible Matrix Theorem and the determinant is equal to 0.


68-95-99.7 Rule and Normal Distribution Question. I need help solving this question. It is on my final exam review. I was given the answer but i do not know how to solve it.

1) Not everyone pays the same price for the same model of a car. Suppose the price of a car is normally distributed. The mean price of a particular model of a new car is 22,000 and the standard deviation is 750. Use both the 68-95-99.7 Rule and Normal Distribution to find the percentages of buyers who paid more than 22,750?

ANS: P(x>22,750) = 0.16

 Normal Distribution
 P(x>22,750) = 0.1587

Since $22,750-22,000=750$, and $750$ is the standard deviation, the cutoff of $22,750$ is one standard deviation above the mean. Thus, you want to know what percentage of the data is more than one standard deviation above the mean

The $68$-$95$-$99.7$ rule says that about $68\%$ of the data in a normally distributed data set lie within one standard deviation of the mean. That leaves $100\%-68\%=32\%$ of the data more than one standard deviation away from the mean. The normal distribution is symmetric about the mean, so half of that $32\%$ is more than one standard deviation below the mean, and half of it is more than one standard deviation above the mean. Half of $32\%$ is $16\%$, so the $68$-$95$-$99.7$ rule says that about $16\%$ of the data are above $22,750$.

To get a more accurate value you’ll need to use the normal distribution itself. This requires either a calculator with the appropriate function or access to a table of the normal distribution. These tables come in several forms. The one on this web page gives the area under the normal curve to the left of a cutoff value. In your case the cutoff value is $1.00$, one standard deviation above the mean, and the table says that $0.8413$, or $84.13\%$, of the area under the curve is to the left of that value. That leaves $100.00\%-84.13\%=15.87\%$ of the area above the cutoff, meaning that about $15.87\%$ of the buyers will have paid more than $\$22,570$.

Trivial problem if you remember what the rule means: 68% is the proportion of buyers who pay within 1 standard deviation of the mean. 32% pay more or less, so 16% pay more. 22750 is 1 standard deviation above the mean.


Polynom equality modulo p I found these two equations:

(a) $$X^4 + 1 \equiv (X + 1)^4 \mod \ 2$$
(b) $$X^4 + 1 \equiv (X^2 - X - 1)(X^2 + X - 1) \mod \ 3$$

I would like to understand the concept of modulo for Polynoms.

How were they made? And how to verify these equality ?

Thanks in Advanced!

Note that $\;2k=0\pmod 2\;,\;\;\forall\,k\in\Bbb Z\;$ , and thus

$$(x+1)^4=x^4+\overbrace{4x^3+6x^2+4x}^{=0\pmod 2}+1=x^4+1\pmod 2$$

Try now something similar with the other case

For $(a),$

$y^{m+1}-y=y(y^m-1)$ is divisible by $y(y-1)$ which being the product of two consecutive integers is even

$\implies y^{m+1}\equiv y\pmod2$ for $m\ge1$

$\implies y^4\equiv y\pmod2$ set $y=X, X+1$

For $(b),$

$(x^2-1-x)(x^2-1-x)=(x^2-1)^2-x^2=x^4-3x^2+1\equiv x^4+1\pmod3$


Find the sum and product of $f(x)=3x-5, g(x)=2x^2-4x+3$ in $Z_8$. Find the sum and product of $f(x)=3x-5, g(x)=2x^2-4x+3$ in $Z_8$.

Does using high school style of solving this yields a different answer?

What would be the answer?


How to show that two normally distributed density functions are independent Show that $U$ and $V$ are independent, where $U=X+Z$ and $V=X-Z$.

I´m given that $X\sim N(0,1)$, $Z\sim N(0,1)$ and X and Z are independent.

First I find U and V.
This leads to $U\sim N(0,2)$ and $V\sim(0,2)$

From here I'm stuck, I know that $independence \implies Cov(V,U)=0$ but i'm not sure if $Cov(V,U)=0 \implies independence.$

The suggestion in the comment is correct. If you can show that U and V are jointly gaussian with covariance zero then you have their independence. There are details about this lemma online, for example this page.

In this case, as X and Z are independent it's easy to show that all linear combinations of U and V are gaussian RVs. From there you can show the independence.

You are also correct that zero covariance does not imply independence. For example, $A$ is a RV with $\mathbb{E}A =  0$ and $\mathbb{E}[A^2] = 0$, and let $B=A$. Then we have $Cov(AB) = 0$ but $A$ and $B$ are clearly dependent.


Is it possible to prove everything in mathematics by theorem provers such as Coq? Coq has been used to provide formal proofs to the Four Colour theorem, the Feit–Thompson theorem, and I'm sure many more. I was wondering - is there anything that can't be proved in theorem provers such as Coq?

A little extra question is if everything can be proved, will the future of mathematics consistent of a massive database of these proofs that everyone else will build on top of? To my naive mind, this feels like a much more rigorous way to express mathematics.

One of the computability results is that there must exist provable theorems in number theory that require very long proofs relative to their statements.

If $n$ is the encoding of a statement in number theory which can be proven, and $f(n)$ is the length of the shortest proof of that statement, then the function $$G(n)=\max \{f(k): k\leq n, k \text{ encodes a provable theorem}\}$$ is not only not computable, but it cannot be bounded above by any computable function, or else we could write a program that could essentially "find" all decidable statements - the set of decidable theorems would be recusrive, not just recursively enumerable. (I initially had an argument here about why that would be a problem, but that argument was flawed. I'm still pretty sure it is wrong for the decidables to be recursive, but it will take more work.)

In particular, then, for any total computable function $h$, there must be a provable theorem encoded by some $n$ such that the shortest proof, when encoded, is greater than $h(n)$.

So, in a practical sense, there will always be harder theorems, where we need more storage and compute time to verify the entire proof.

It is unlikely that Archimedes thought about theorems in category theory. In the future there will likely be new areas of mathematics. While a theorem prover might have encodings of results in these new areas they might not be readable for humans.


Show $f(x)=x^ 2 - 14$ can't be factorized as $(ax + b )(c x +d)$ in $\mathbb Z_{2014}[x]$ show $f(x)=x^ 2 - 14 $ can't be expressed as $(ax + b )(c x +d)$ in $\mathbb Z_{2014}[x]$
I tried this by different  way comparing with it in my book .
Here is my attempt
I've got some idea from factorization of 2014.
Take the case, equation $f(x)= 0 $ for mod ${2014}$
If so, $f(x)\equiv 0 \pmod 2$, $f(x)\equiv 0 \pmod{ 19}$and $f(x)\equiv 0 \pmod {53}$ holds.
plus, put $f_2 \equiv f \pmod 2$,  $f_{19} \equiv f \pmod   {19} $ and vice versa $f_{53}$
Suppose the $f$ can  be factorized by those two polynomials(I.e. $f = (ax+b)(cx+d)$)
Since $ac= 1$ for mod $2014$, $gcd(a,2014) = gcd(c,2014) =1$
Therefore $ deg f_2 =deg f_{19}= deg f_{53}= 2 $ the reason why $(ax + b )(c x +d) \equiv ( [ ax ]_p + [b]_p)( [ cx]_p + [d]_p)$ for mod $p$ (Here the $p$ is a prime number)
Therefore$ f_2 , f_{19} $and $f_{53}$ are reducible over  fields $\mathbb Z_{2}$,  $\mathbb Z_{19}$ and  $\mathbb Z_{53}$  respectively. Plus their degree are $2$, so only we left examine the existence of the roots in each fields. But the $f_{19}$ doesn't have any roots in $\mathbb Z_{19}$ by Legendre symbol. (I.e. $14$ isn't quadratic residue for mod $19 $ . ) So the $  f_{19}  $ is irreducible over $\mathbb Z_{19}$. It is contradiction.
Is  my solution  is right? please verify my solution. Thanks.

You could reduce Parcly Taxel's verbiage and increase clarity as follows: Assume the factorization holds mod 2014. Then it holds mod 19. Then $(-b/a)$ is a root of $f$ mod 19, contradicting your computation of Legendre symbol QED.

You could reduce your proof's verbiage (and increase its clarity) as follows: if $x^2-14$ factors as specified over $\mathbb Z_{2014}$, since $a$ and $c$ are coprime to $2014$ there must exist exactly two values $x$ where $x^2\equiv14\bmod2014$. By the Chinese remainder theorem this is equivalent to $14$ being a quadratic residue modulo $2$, $19$ and $53$, but $14$ is a non-residue modulo $19$ and hence there is no factorisation into the given form.


Congruence between Bernoulli numbers I fell on the following fact as a by-product:

If p is a prime number, then the sum of the Bernoulli numbers, from index 0 to p - 2, is congruent with - 1 modulo p.

Do you know a simple proof ? Or a reference to the literature ? Thanks beforehand.
Edit. It was not a very smart question. The congruence is an easy consequence of Bernoulli's formula for the sum of the $(p-2)$-powers of the natural numbers from 1 to $p-2$.
Edit. As suggested by barto, I give the proof. For $p = 2$, it is trivial, thus assume $p \geq 3$. Bernoulli's summation formula is (for $m \geq 1$) :
$$
(m+1) \sum_{k=1}^{n-1} k^m = \sum_{k=0}^m \binom{m + 1}{k} B_k n^{m + 1 - k}
$$
or in more detail
$$
(m+1)(1^m + 2^m + \cdots + (n-1)^m) = \sum_{k=0}^m \binom{m + 1}{k} B_k n^{m + 1 - k}.
$$
(I use the definition with $B_1 = -1/2$. The Wikipedia article :
https://en.wikipedia.org/wiki/Bernoulli_number#Sum_of_powers
uses the other definition, hence a slight difference.)
With $n = p-1$ and $m = p-2$, this becomes
(1) $(p-1) \sum_{k=1}^{p-2} k^{p-2} = \sum_{k=0}^{p-2} \binom{p-1}{k} B_k (p-1)^{p-1-k}$,
where the involved Bernoulli numbers are $p$-integral rational numbers.
Now, by Fermat's little theorem, $\sum_{k=1}^{p-1} k^{p-2}$ (with one more term than in the preceding sum) is modulo $p$ the sum of the inverses of the $p-1$ nonzero elements. Since inversion is a bijection, this sum is equal modulo $p$ to $\sum_{k=1}^{p-1} k = p(p-1)/2$, which is zero modulo $p$. Thus (1) gives
$-(p-1)(p-1)^{p-2} \equiv \sum_{k=0}^{p-2} \binom{p-1}{k} B_k (p-1)^{p-1-k} \pmod{p}$
$-1 \equiv \sum_{k=0}^{p-2} \binom{p-1}{k} B_k (p-1)^{p-1-k}\pmod{p}$.
But $\binom{p-1}{k} \equiv (-1)^{k} \pmod{p}$, thus the right member is congruent modulo $p$ to $\sum_{k=0}^{p-2} B_k$, and we are done.
Edit : here is a simpler proof. It is well known that, for $m \geq 2$,
$\sum_{i=0}^{m-1} \binom{m}{i} B_{i} = 0.$ Putting $m=p-1$, we find
$\sum_{i=0}^{p-2} \binom{p-1}{i} B_{i} = 0.$
Since $\binom{p-1}{k} \equiv (-1)^{k} \pmod{p}$ and the involved Bernoulli numbers are $p$-integral rational numbers, this gives
$\sum_{i=0}^{p-2} (-1)^{i} B_{i} \equiv 0 \pmod{p}.$
Now, $B_{1} = -1/2$ and for $i$ odd and $> 1$, $B_{i} = 0$, hence the statement.

What you have is incorrect. $\displaystyle \sum_{k=0}^{p-2} B_k$ is not guaranteed to be an integer in the first place, let alone being $-1 \pmod{p}$. For instance, $$\sum_{k=0}^{3-2} B_k = \dfrac12$$
$$\sum_{k=0}^{5-2} B_k = \dfrac23$$
$$\sum_{k=0}^{7-2} B_k = \dfrac{19}{30}$$


Are sorted i.i.d still i.i.d? I need help with that question.
We have i.i.d. Let us sort the values in ascending order.Is new series is i.i.d?
From my opinion the answers is yes because function y that sort the x is only mapping from x, but im not sure. Can anybody correct my understanding?

No, unless $X$ is a constant (non-random). First of all, the $y_i$ are clearly dependent: you already know that $y_2\geq y_1$. Furthermore, they are not identically distributed: the later values are more likely to be large.

It depends whether you know the values of the $\{y_i\}$ or not. The knowledge of the values makes the random variables not IID anymore. If you are curious about the case when you do know the values at ordering, check out the "order statistic". Deriving the laws is a bit cumbersome in my opinion but completly doable by any students knowing probability.
https://en.wikipedia.org/wiki/Order_statistic
However, if you do not know the values at ordering time, then yes you can rearrange the IID random variables as you wish yourself :)


A cubic diophantine Equation While reading Diophantine equations I came across the following equation $$x^3+cy^3-3yx=0$$
Is there any known method to solve this equation for any $c$?

We have $y\mid x^3$ and $x\mid cy^3$. If we assume $y=ax$ for some integer $a$ then
substituting this in $x^3+cy^3=3xy$ yields
$$
x=a(3-ac), \quad y=a^2(3-ac).
$$
Conversely this is a solution of $x^3+cy^3=3xy$ for all integers $a$. This is a first step to solve this Diophantine equation (which is not homogeneous, i.e., not given by a binary cubic).

first we have (0,0,0),(0,a,0),(0,0,c) are solutions ∀ y=a  ,a,c∈Z

now,

by multiplying by xy we have

$yx^4-3x^2 y^2+cxy^4=0$
let,
$x^2=s   ,y^2=r ,   y=a   ,cx=k$

then
$as^2-3sr+kr^2=0$

the discriminant  ,     $d=9-4ak=9-4cxy$

as d should be greater than or equal to zero and for Diophantine equations d is a perfect square

$d>=0  , d=u^2$

$cxy<=2$, if we considered that cxy>=0

implies  $cxy=2$ , or $cxy=0$  because 1 makes d non-square

so both x,y >=0  or x,y<=0 together

take c=∓2 ,

$xy=∓1,  x^3∓2y^3∓3=0 ,(∓1,1) -is-the-only-solution $

take c=∓1 ,

$xy=2,  x^3∓y^3=∓6 , -has-no-solution $

take c=0 ,

$  x^3-3xy=0 , -has-infinite-solutions $

if cxy<0 ,

$  x^3+cy^3-3xy=0 , -has-infinite-solutions $

hint: d=u^2 and d is odd then if cxy<0    2-cxy=m(m-1), $m∈z^+$

(m-2)(m+1)=-cxy  ,∀ m>2 & either one of c,x,y is negative or all


Drawing Balls Without Replacement A box contains $3$ red balls, $4$ blue balls, $6$ green balls. Balls are drawn one-by-one without replacement until all the red balls are drawn. Let $D$ be the number of draws made. Calculate $P(D \le 9)$.

Since we are drawing without replacement, this is a hypergeometric distribution.

Minimum of $D$ is $3$.

$P(D \le 9) = \sum_{i=3}^{9}\frac{\binom{3}{3}\binom{10}{i-3}}{\binom{13}{i}}=0.734265734$

But the answer is just

  $\frac{\binom{3}{3}\binom{10}{9-3}}{\binom{13}{9}}=0.293706293$

Why does it only consider the case $P(D=9)$?

Edit 1: Pictures of the question and solution from the textbook

It's just a counting argument.

Choose the three red balls, and choose six others. Divide by the total number of possible ways to draw nine balls.

When you get the three red balls doesn't matter, as long as they're part of the nine. You just keep drawing until you get nine balls if you already have the three red ones.

Or, just grab nine balls from the box all at once, if that's conceptually easier.

You're doing a drawing balls without replacement, what conduct us to supose that is hypergeometric distribution as follows:

$P(x = k) = \frac{\binom{M}{k}\binom{N-M}{n-k}}{\binom{N}{n}} $

Where N is our population (all balls), M are the sample we are interested, and N-M the others kind of balls collors in our box.

Balls are drawn until the three red balls are drawn and will happen 9 drawn. So we are intersting in the probability to find all the three balls with minimum trials three and max trials nine.

Doing this to three trials we obtain:

$P(x = 3) = \frac{\binom{3}{3}\binom{10}{0}}{\binom{13}{3}} = 0.0034 $

Doing this to four trials we obtain:

$P(x = 4) = \frac{\binom{3}{3}\binom{10}{1}}{\binom{13}{4}} = 0.013$

Until the last trial:

$P(x = 9) = \frac{\binom{3}{3}\binom{10}{6}}{\binom{13}{9}} = 0.29$

If we sum all the equations  $ P(D \le 9) = \sum_{i=3}^{9}\frac{\binom{3}{3}\binom{10}{i-3}}{\binom{13}{i}}=0.734265734$.

Is correct your calculus, maybe the question if badly formulated or is missing something else.


What is the smallest positive multiple of 450 whose digits are all zeroes and ones? What is the smallest positive multiple of 450 whose digits are all zeroes and ones?

I tried guess and check but the numbers grew big fast.  Thanks in advance!

Can we agree that it must be an even multiple of 450?  Otherwise the last two digits will be 50.

What is the smallest positive multiple of 900 such that all the digits are 0s or 1s?

A rule of multiples of 9:  the sum of the digits of a multiple of 9 is a multiple of 9.

This rule goes both ways. If the sum of the digits is a multiple of 9, the number is a multiple of 9.

That makes  11111111100 our winner.

one more way. let x be the required number.

for a number to be multiple of 450, it should be multiple of 10.  so 10.y =x
now y should be multiple of 45. it should be first multiple of 5. so a number with only 1, o will be multiple of 5 if last digit is 0.
 so y = 2. w/9...dividing 10 by 5 gives 2 in numerator.

2.w should be mulitple of 9 and should contain only 1,0.

using the fact that sum of digits should be multiple of 9 for a number to be multiple of 9,

2.w should be 111111111.

adding two zeros extra.
11111111100.

thus 11111111100 is the required number.

lengthy process but easy to understand.


Analog of $(a+b)^2 \leq 2(a^2 + b^2)$ Is there an inequality such as
$$(a+b)^2 \leq 2(a^2 + b^2)$$
for higher powers of $k$
$$(a+b)^k \leq C(a^k + b^k)?$$

The generalized mean inequality states that
$$\dfrac{a+b}2\leq \left(\dfrac{a^k+b^k}2\right)^{1/k},$$
with equality if and only if $a=b$, from which it follows that
$$(a+b)^k\leq 2^{k-1}(a^k+b^k),$$
with equality if and only if $a=b$. Thus $C=2^{k-1}$ works and no smaller $C$ works.

The first inequality is true because : $2ab ≤ a^2+b^2$

There is no specific inequality for higher powers of k, although you can always give a very higher value for 'C' to make the second inequality true.


The interval [0,1] andd [3,5] are equivalent. Is my proof correct? The intervals $[0,1]$ and $[3,5]$ are equivalent.

My proof goes like this.

Proof. To show that the two sets are equivalent, we should show a bijection between them. Consider the function $f:[0,1] \to [3,5]$ such that $f(x)=2x+3$.
Since $f$ is  linear, it is bijective.
Therefore, the intervals $[0,1]$ and $[3,5]$ are equivalent.

I am not sure if this is right. Assuming that this is correct, is it better to show that $f$ is reflexive, symmetric, and transitive?
If this is wrong, what is the right way of proving?
Sorry if this is so simple.

The function is not symmetric since there is the point (0,3) but not the point (3,0)  because f is not even defined in 3.
Your argument is correct, if you're talking about equivalence as topological spaces (showing there's a continuous bijection between both spaces). May be you could argument that it is bijective because it has a left and right inverse: g(x)=(x-3)/2 and continuity is clear because it is a composition of continuous functions.


last 4 digits of $7^{65}$ As the name implies, what are the last 4 digits of $7^{65}$

I can't find anything that does not involve bashing, I tried $(8-1)^{65}$, or maybe  $7^{65}$ mod 625 and 16,

If you calculate the first powers of 7, you will see quickly see the pattern that develops - for example, the last digit is going 7, 9, 3, 1, 7, 9, 3, 1, ... - I can calculate that in my head.
Simply try the first ten or fifteen powers and write them down, and you'll see the pattern.


Why are the irreducible components $T$-stable? I'm having trouble with part of a proof (7.1.5) in Springer's Linear Algebraic Groups.  Let $r: G \rightarrow \textrm{GL}(V)$ be a rational representation of a linear algebraic group $G$, $B$ a Borel subgroup of $G$, and suppose that there exists a nonzero $v \in V$ such that $V$ is spanned by $g \cdot v : g \in G$, and $$B = \{ g \in G: gv \in kv\}$$ Let $T$ be a maximal torus of $G$ which is one dimensional, $\lambda: k^{\ast} \rightarrow T$ an isomorphism, $e_1, ... , e_n$ a basis for $V$, such that for some integers $m_1 \geq \cdots \geq m_n$, not all equal, we have $$\lambda(a)e_i = a^{m_i}e_i$$ Now $r$ induces an isomorphism of $G/B$ onto a closed subvariety $Z$ of $\mathbb{P}(V)$, the set of one dimensional subspaces of $V$, via $gB \mapsto [g\cdot v]$.  Let $U_1$ be the affine open set in $\mathbb{P}(V)$ consisting of points with nonzero first coordinate (with respect to the basis $e_i$).  Then $\mathbb{P}(V) \setminus U_1$ is closed, and $$[\mathbb{P}(V) \setminus U_1] \cap Z$$ is closed and clearly $T$-stable.  Springer claims that the irreducible components of $[\mathbb{P}(V) \setminus U_1] \cap Z$ are also $T$-stable, but gives no indication of why.  Why should this be the case?

$G$ (and hence $T$) acts by projective automorphisms on $\mathbb P(V)$ and they are automorphisms of $\mathbb P(V)$ in algebro-geometric sense thus they preserve irreducible components of any subvariety of $\mathbb P(V)$ they stabilize.

Since the basis of $V$ are chosen to be eigenvectors of $T$, the set with first coordinate zero is obviously $T$-stable.

Let $T$ act on $X$, the stabilizer of a component $X_1$ equals to $\cap_{x \in X_1} \{t \in T | t.x \in X_1\}$, is a closed subgroup of $T$, and is of finite index ($t \in T$ permutes components of $X$, easily seen since $T$ is a group, $t^{-1}$ exists). Its identity component is contained in $T$, both are closed irreducible and have the same dimension, they must be equal.

This is true for any connected algebraic group $T$.


Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$ Could any one give an example of a bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$?

Thank you.

First, note that it is enough to find a bijection $f:\Bbb R^2\to \Bbb R$, since then $g(x,y,z) = f(f(x,y),z)$ is automatically a bijection from $\Bbb R^3$ to $\Bbb R$.

Next, note that since there is a bijection from $[0,1]\to\Bbb R$ (see appendix), it is enough to find a bijection from the unit square $[0,1]^2$ to the unit interval $[0,1]$. By constructions in the appendix, it does not really matter whether we consider $[0,1]$, $(0,1]$, or $(0,1)$, since there are easy bijections between all of these.

Mapping the unit square to the unit interval

There are a number of ways to proceed in finding a bijection from the unit square to the unit interval.  One approach is to fix up the "interleaving" technique I mentioned in the comments, writing $\langle 0.a_1a_2a_3\ldots, 0.b_1b_2b_3\ldots\rangle$ to $0.a_1b_2a_2b_2a_3b_3\ldots$.  This doesn't quite work, as I noted in the comments, because there is a question of whether to represent $\frac12$ as $0.5000\ldots$ or as $0.4999\ldots$. We can't use both, since then $\left\langle\frac12,0\right\rangle$ goes to both $\frac12 = 0.5000\ldots$ and to $\frac9{22} =  0.40909\ldots$ and we don't even have a function, much less a bijection.  But if we arbitrarily choose to the second representation, then there is no element of $[0,1]^2$ that is mapped to $\frac12$, and if we choose the first there is no element that is mapped to $\frac9{22}$, so either way we fail to have a bijection.

This problem can be fixed.

(In answering this question, I tried many web searches to try to remember the fix, and I was amazed at how many sources I found that ignored the problem, either entirely, or by handwaving. I never did find it; I had to remember it. Sadly, I cannot remember where I saw it first.)

First, we will deal with $(0,1]$ rather than with $[0,1]$; bijections between these two sets are well-known, or see the appendix. For real numbers with two decimal expansions, such as $\frac12$, we will agree to choose the one that ends with nines rather than with zeroes. So for example we represent $\frac12$ as $0.4999\ldots$.

Now instead of interleaving single digits, we will break each input number into chunks, where each chunk consists of some number of zeroes (possibly none) followed by a single non-zero digit.  For example, $\frac1{200} = 0.00499\ldots$ is broken up as $004\ 9\ 9\ 9\ldots$, and $0.01003430901111\ldots$ is broken up as $01\ 003\ 4\ 3\ 09\ 01\ 1\ 1\ldots$.

This is well-defined since we are ignoring representations that contain infinite sequences of zeroes.

Now instead of interleaving digits, we interleave chunks.  To interleave $0.004999\ldots$ and $0.01003430901111\ldots$, we get $0.004\ 01\ 9\ 003\ 9\ 4\ 9\ldots$.  This is obviously reversible. It can never produce a result that ends with an infinite sequence of zeroes, and similarly the reverse mapping can never produce a number with an infinite sequence of trailing zeroes, so we win.  A problem example similar to the one from a few paragraphs ago is resolved as follows: $\frac12 = 0.4999\ldots$ is the unique image of $\langle 0.4999\ldots, 0.999\ldots\rangle$ and $\frac9{22} = 0.40909\ldots$ is the unique image of $\langle 0.40909\ldots, 0.0909\ldots\rangle$.

This is enough to answer the question posted, but I will give some alternative approaches.

Continued fractions

According to the paper "Was Cantor Surprised?" by Fernando Q. Gouveâ, Cantor originally tried interleaving the digits himself, but Dedekind pointed out the problem of nonunique decimal representations. Cantor then switched to an argument like the one Robert Israel gave in his answer, based on continued fraction representations of irrational numbers.  He first constructed a bijection from $(0,1)$ to its irrational subset (see this question for the mapping Cantor used and other mappings that work), and then from pairs of irrational numbers to a single irrational number by interleaving the terms of the infinite continued fractions.  Since Cantor dealt with numbers in $(0,1)$, he could guarantee that every irrational number had an infinite continued fraction representation of the form $$x = x_0 + \dfrac{1}{x_1 + \dfrac{1}{x_2 + \ldots}}$$

where $x_0$ was zero, avoiding the special-case handling for $x_0$ in Robert Israel's solution.

Cantor-Schröder-Bernstein mappings

The Cantor-Schröder-Bernstein theorem takes an injection $f:A\to B$ and an injection $g:B\to A$, and constructs a bijection between $A$ and $B$.

So if we can find an injection $f:[0,1)^2\to[0,1)$ and an injection $g:[0,1)\to[0,1)^2$, we can invoke the CSB theorem and we will be done.

$g$ is quite trivial;  $x\mapsto \langle x, 0\rangle$ is one of many obvious injections.

For $f$ we can use the interleaving-digits trick again, and we don't have to be so careful because we need only an injection, not a  bijection. We can choose the representation of the input numbers arbitrarily; say we will take the $0.5000\ldots$ representation rather than the $0.4999\ldots$ representation. Then we interleave the digits of the two input numbers.  There is no way for the result to end with an infinite sequence of nines, so we are guaranteed an injection.

Then we apply CSB to $f$ and $g$ and we are done.

Appendix

There is a bijection from $(-\infty, \infty)$ to $(0, \infty)$.  The map $x\mapsto e^x$ is an example.
There is a bijection from $(0, \infty)$ to $(0, 1)$.  The map $x\mapsto \frac2\pi\tan^{-1} x$ is an example, as is $x\mapsto{x\over x+1}$.
There is a bijection from $[0,1]$ to $(0,1]$.  Have $0\mapsto \frac12, \frac12\mapsto\frac23,\frac23\mapsto\frac34,$ and so on. That takes care of $\left\{0, \frac12, \frac23, \frac34,\ldots\right\}$.  For any other $x$, just map $x\mapsto x$.
Similarly, there is a bijection from $(0,1]$ to $(0,1)$.

First, note that the exponential function is a bijective map of $\mathbb R$  to $(0,\infty)$.
Now let $G$ be the irrationals in $(0,\infty)$. I'd like a bijective map of $(0,\infty)$ to $G$.  It can be done as follows: if $x = r \pi^n$ for some nonnegative integer $n$ and rational $r$, let $f(x) = \pi x$, otherwise $f(x) = x$.

Finally, it suffices to find a bijective map of $G^3$ to $G$.  This can be obtained using continued fractions.  Each $x \in G$ can be expressed in a unique way as an infinite continued fraction $x = x_0 + \dfrac{1}{x_1 + \dfrac{1}{x_2 + \ldots}}$ where $x_0$ is a nonnegative integer and $x_1, x_2, \ldots$ are positive integers.  Denote this as $[x_0; x_1, x_2, \ldots]$.  We then map $(x,y,z) \in G^3$ to
$[x_0; y_0+1, z_0+1, x_1, y_1, z_1, \ldots]$.


Determine $\phi(2^{399}+1)$ This is a question I had on an exam so there is no access to calculators or software. I checked the answer using Maple after and as expected, the answer is very large and some of the prime factors are also very large, with over 10 digits.

Is there any reasonable way to break this down and get an answer of some sort?

After getting feedback on this question in class, it appears that there is no feasible way to do this by hand and it can be easily computed using software such as Maple.


$x^2-2mx+m^2-1=0$ Find the range of m when one root lies in (-2, 4) Let: $x^2-2mx+m^2-1=0$. For all $m$ there exist real roots for the above equation. If one root lies in between $(-2,4)$ Find the value range for $m$

My Work

Since coefficient of $x^2$ is 1 the graph is a maximum. Also the discriminant is positive. How can I use this fact to solve this problem ? Please Help me. Thanks a lot

The roots are $m+1$ or $m-1.$ Since one of them must stay within $(-2,4),$ we must have $-2<m+1<4,$ or $-2<m-1<4,$ which give $-3<m<3,$ or $-1<m<5.$ Thus, $m$ must satisfy the conditions $$-3<m<5.$$


The $P(X) = 0.7$ and $P(Y) = 0.6$. What is the probability of $X$ or $Y$ occurring? So let's say we have two independent events. The $P(X) = 0.7$ and $P(Y) = 0.6$. What is the probability of $X$ or $Y$ occurring? I know how to calculate an or probability but this question is confusing me.

Recall inclusion exclusion
$$P(X\cup Y) = P(X)+P(Y)-P(X\cap Y)$$
and recall something special about
$$P(X\cap Y)$$
from the fact that they are independent.
By the way, this problem is an "or" problem.

Given that P(x) = 0.7
P(Y) = 0.6

Then P(X or Y) = P(X) + P(Y) - P(X and Y)(If you assume independence for X and y)

= 0.7 + 0.6 - (0.7*0.6)
= 1.3 - 0.42

P(X or Y) = 0.88


Notation for set of equivalence classes Suppose we define an equivalence relation $\sim$ on a set $S$. The exact details of the relation and checking that it verifies reflexivity, symmetry, and transitivity is less important, so let's assume it satisfies those properties. Then $S/\sim$, read "S mod $\sim$ denotes the set of equivalence classes.
I am trying to make sense of this notation, specifically the use of "mod." In a quotient $R/S$, we say that $r_1, r_2 \in R$ are equivalent if $r_1 - r_2 \in S$. Similarly, in $\mathbb{Z}/2\mathbb{Z}$, $a$ is equivalent to $b$ if $a - b \in 2\mathbb{Z}$. I do not see any parallel in general to $S/\sim$. The only similarity is that they are equivalent "with respect to $\sim$."
Am I missing something? Is there a better way to think of this notation or even what it means to "quotient by $\sim$"?

When you defined $a \equiv b \pmod n$ we use the notation $\pmod n$ BECAUSE $a- b\in n\mathbb Z$ is an equivalence relation that means that the relation $a-b\in n\mathbb Z$ "partitions" $\mathbb Z$ into equivalence classes and $\mathbb Z/n\mathbb Z$ is the set of equivalence classes it is partitioned into.
$\mathbb Z/n\mathbb Z = \{[0],[1],......., [n-1]\}$ where $[k]=\{m\in \mathbb Z| m\equiv k \pmod n\} = \{m \in \mathbb Z| m-k \in n\mathbb Z\} = \{....,k-n, k, k+n, k+2n,.....\}$.
But there is nothing magical about the relation $a\equiv b$.  EVERY equivalence relation will partition a set, $S$ into equivalence classes $U_\alpha$ where each and every $s\in S$ is in exactly one, and only one, $U_\alpha$ and all the elements in $U_\alpha$ are related to each other.  And $S/\sim$ is the set of all these equivalence classes.
So for instance; take any equivalence relation $\sim$ an a set $S$.  For example let say $S= \mathbb N$ and $a \sim b\iff $ the highest power of $3$ that divides $a$ is the same highest power of $3$ that divides $b$.
for example $3\sim 6 \sim 12 \sim 51 etc. $ because $3|6, 12, 51$ but $3^2 \not\mid 6,12,51$.
If we let $U_1 = \{3, 6, 12, 15, 21, 24, .....\}=\{k\in \mathbb Z| 3\mid k, 9\not \mid k\}$ then that is one equivalence class.  If we let $U_2=\{9,18,36,....\} = \{k\in \mathbb Z| 9|k, 27\not \mid k\}$ that's another equivalence class.  And we can let $U_m =\{k\in \mathbb Z| 3^m|m, 3^{m+1}\not \mid k\}$ can be any other.
Also note $\{0\} = U_\infty = \{k\in \mathbb N| 3^j\mid k$ for all $k\in \mathbb N\}$ is the final equivalence class..
Then $\mathbb Z/\sim = \{U_\infty, U_1, U_2, U_3......\}$ is the set of all equivalence classes.
That is entirely equivalent to $\mathbb Z/(\pmod n) = \{[0],[1],.....[n-1]\}$ where $[k] = \{.... k-2n, k-n, k, k+n,...\} = \{k+mn|m\in \mathbb Z\} = \{z\in \mathbb Z| a\equiv z\pmod n\} = \{z\in \mathbb Z| a-z\in n\mathbb Z\}$ are the set of all equivalence classes.
=====

I do not see any parallel in general to S/∼. The only similarity is that they are equivalent "with respect to ∼."

Thats all the similarity that there is. but that's that's enough similarity for anyone.  That is a HUGE similarity.

There is in fact a much more general answer to your question that covers quotient structures in general. First, read this post, which explains how we can form quotient structures using a suitable partition of the larger structure into parts that respects the structure (the operations and predicates). Notice that it uniformly explains the isomorphism theorems for all kinds of structures. Next, observe that we write $S/\sim$ when $\sim$ is an equivalence relation that partitions $S$, but we also write $S/T$ when $T$ is an ideal of $S$, meaning that special part referred to in the linked post that we can use to construct a structure-respecting partition $S$. That is precisely why we write $ℤ/nℤ$, and $G/H$ for groups $G,H$ with $H ◁ G$, and $V/W$ for vector spaces $V,W$ with $W ⊆ V$, and $R/I$ for ring $R$ and ideal $I$ of $R$, and so on...


To Solve $\displaystyle q+xp=p^2$ using Charpit's Method p = ∂z/∂x
q = ∂z/∂y

My Attempt:

$\displaystyle \frac{dx}{-(x-2p)}=\frac{dy}{-1}=\frac{dz}{-p(x-2p)-q}=\frac{dp}{2p}=\frac{dq}{0}$

So, do I get.. $\displaystyle \log p=\log c_1 \implies p=c_1 $

$\displaystyle q=c_2 $

Going to $\displaystyle dz=pdx+qdy $

This won't give the desired solution

Now, I can't think ahead ..

The given answer is: $\displaystyle z=axe^{-y}-\frac{1}{2}a^2e^{-2y}+b$

$\dfrac{\partial z}{\partial y}+x\dfrac{\partial z}{\partial x}=\left(\dfrac{\partial z}{\partial x}\right)^2$

$\dfrac{\partial^2z}{\partial x\partial y}+x\dfrac{\partial^2z}{\partial x^2}+\dfrac{\partial z}{\partial x}=2\dfrac{\partial z}{\partial x}\dfrac{\partial^2z}{\partial x^2}$

$\dfrac{\partial^2z}{\partial x\partial y}+\left(x-2\dfrac{\partial z}{\partial x}\right)\dfrac{\partial^2z}{\partial x^2}=-\dfrac{\partial z}{\partial x}$

Let $u=\dfrac{\partial z}{\partial x}$ ,

Then $\dfrac{\partial u}{\partial y}+(x-2u)\dfrac{\partial u}{\partial x}=-u$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{du}{dt}=-u$ , letting $u(0)=u_0$ , we have $u=u_0e^{-t}=u_0e^{-y}$

$\dfrac{dx}{dt}=x-2u=x-2u_0e^{-t}$ , we have $x=u_0e^{-t}+f(u_0)e^t=u+f(ue^y)e^y$


Evaluate $\int ( 3x + \frac{6x^2 \sin^2(\frac{x}{2})}{x - \sin x} ) \frac{(x-\sin x)^{3/2}}{\sqrt{x}} \mathrm{d}x$ I have a really vague integration problem. It's some substitution and then integration by parts maybe.
I got this from a friend, but it seems it's unlikely to be solved.

 Now my question is, how to approach such big problems?

I reduced it to $$\displaystyle{\int \bigg(3x(2x-x\cos x - \sin x)\sqrt{1-\frac{\sin x}{x}}} \bigg)\mathrm{d}x$$

How to proceed from here?

I think I got it! Consider your simplified version and try to regroup terms to make it look more organized:

$$\int 3x(2x-x\cos x-\sin x )\sqrt{1-\frac{\sin x}{x}}dx$$
$$=\int 3\sqrt{x}(x(1-\cos x) +(x-\sin x))\sqrt{x-\sin x}dx$$
Now start recognition of similar terms. Let's define
$$u=x-\sin x$$
Observe $du=(1-\cos x)dx$. The integral becomes:
$$=\int 3(x(du/dx) +u)\sqrt{ux}dx=\int 3(xdu+udx)\sqrt{ux}$$
But now you see how the parenthesis looks like a total derivative of $ux$. This leads us to the conclusion, that $u$ is not a good variable substitution, but $ux$ is! Define $t=ux=x^2-x\sin x$ and you get
$$\int 3 \sqrt{t}dt$$
You can take it from here.


Solve $(y+ x^3y + 2x^2)dx + (x + 4xy^4+ 8y^3)dy = 0$ Basically I tried to group the terms
ydx + xdy is a perfect differential but I couldn't think a method for  ($x^3$y)dx + (4x$y^4$)dy . Also dividing by xy didn't help me.

The implicit solution is given by Wolfram Alpha:

$$
\frac13 x^3 + y^4 + \log(x y + 2) = c
$$

Since this cannot be solved for $y$, I guess further progress will be difficult.


Singularity of the power function I am reading some complex analysis and I am confused with the power function.

So I understand that $z^a$ has a branching point at $0$ if $a$ is not integer and that the number of branches can be infinitely many if $a$ is irrational. Then this means that $0$ is an essential singularity of the function and I actually computed few coefficients of the Laurent series of $z^\sqrt{2}$ as an example and it seems true.

So then, the big Picard theorem states that the function will take all possible values except possibly one infinitely many times. But on the other hand I find that $|z^a|=|z|^a$ for $a>0$, which means that not only there is a strict bound but also the limit to $0$ exists and it is $0$.

So I am confused as these cannot be both true. Where is my mistake?

If you consider $z^\alpha$ (for $\alpha\notin\Bbb Z$) or $\log z$ as functions on $\Bbb C$ they do not have isolated singularities at $z=0$: there is no way to define them as holomorphic, or even continuous, functions on a punctured neighbourhood of the origin.

If you prefer, you can view those functions on (possibly infinitely sheeted) Riemann surfaces, and if you do so, the functions will indeed be holomorphic everywhere, but in that case, the origin will not lie on the Riemann surface in question, and it doesn't make sense to talk about the "singularity" at $z=0$.

In either way, $z=0$ is not an essential singularity, at least not with the common usage of the term. I have no idea how you "computed the Laurent series" for $z^\sqrt2$; that function does not admit a Laurent series around $0$.

You are talking about $z^a$ which is  $0$ at $z = 0$.  When a function is zero at a point, that is not a singularity of any kind , even if it is also a branch point.  All the branches of this function will be $0$ at $z = 0$; otherwise for the same z the different branches have different values.

A singularity is when the function goes to infinity as you approach the point, at least along some path.   If the singularity is isolated the Laurent series will tell you what kind of singularity you have, by showing you which of the negative coefficients are $0$.  If only a finite number of them are nonzero, you have a simple or multiple pole.  If infinitely many of them are non-zero you have an essential singularity and can then apply the big Picard theorem.

Log z is a different issue.  It does indeed have a singularity at $0$, and a branch point as well; but it is not an isolated singularity.  You can read more about that here. Type of singularity of $\log z$ at $z=0$


Is the product of two complete TVS a complete TVS? I'm reading about Topological Vector Spaces (using Treves's book) and I failed to answer the question on title. I'll be more precise:

1) We say that a TVS $E$ is complete if every Cauchy filter in $E$ is convergent.

2) Given $E_1$ and $E_2$ two TVS, we consider $E=E_1 \times E_2$ with the product topology. With this topology and the natural structure of vector space, $E$ is a TVS.

In the Bourbaki's book is proved that it's true when we use the notion of uniform structure. But looking at page 140, ex. 5 of Topological Vector Spaces and Distributions (by John Horvath, where he uses just the definition 1), he states just the converse.

So, is it true that $E$ is complete when $E_1$ and $E_2$ are complete? If it's not true, are there some example?

Yes the product of tow complete topological vector space is complete. The very simple example is considering $\mathbb R $ and $\mathbb R^2$


Show that $\lim_{j\to\infty}\int_{\mathbb{R}^n}f_jd\mathcal{L}^n=\int_{\mathbb{R}^n}f\mathcal{L}^n$ Let $g_j\le f_j\le h_j$ be three sequences of Lebesgue-integrable functions with $g_j\to g, f_j\to f, h_j\to h$ such that $g,h\in L^1(\mathbb{R}^n)$ and $\lim_{j\to\infty}\int_{\mathbb{R}^n}h_jd\mathcal{L}^n=\int_{\mathbb{R}^n}hd\mathcal{L}^n,\;\lim_{j\to\infty}\int_{\mathbb{R}^n}g_jd\mathcal{L}^n=\int_{\mathbb{R}^n}gd\mathcal{L}^n $.

Show that $f\in L^1(\mathbb{R}^n)$ and $\lim_{j\to\infty}\int_{\mathbb{R}^n}f_jd\mathcal{L}^n=\int_{\mathbb{R}^n}fd\mathcal{L}^n$.

My work so far: $f$ is integrable by the definition of the Lebesgue integral since both the negative (which I will denote by $f_{\_}$) and positive part of $f$ (which I will denote by $f_{+}$) are bounded by $g_{\_}$ and $h_{+}$.
My problem is the second part: I want to obtain something like $f_j\ge h_j+\epsilon$ and $f_j\le g_j-\delta$ for some $\epsilon,\delta> 0$ in order to be able to use the fact that we can switch integral and limit for $g$ and $h$. I have already tried to make the convergence uniform by using Egorov locally and then gluing the result together but I don't know how to continue from here.

I don't think you should have uniform convergence. Here is my idea:
Let $f_{n-}$ denote the negative part of $f_n$ for every $n$. $f_{n-}$ is always a positive function. Thus, by Fatou's Lemma $$\int_{\mathbb{R}^n}f_-\;d\mathcal{L}^n\leq\liminf_{n\to\infty}\int_{\mathbb{R}^n}f_{n-}\;d\mathcal{L}^n\leq\limsup_{n\to\infty}\int_{\mathbb{R}^n}f_{n-}\;d\mathcal{L}^n\leq\int_{\mathbb{R}^n}f_-\;d\mathcal{L}^n.$$ The same follows analogously for $f_+$. The proof follows from the definition of the Lebesgue Integral.


Prove that if $n^2$ is divided by 3, then also $n$ can also be divided by 3. $n\in \Bbb N$
Prove that if $n^2$ is divided by 3, then also n can also be divided by 3.

I started solving this by induction, but I'm not sure that I'm going in the right direction, any suggestions?

Hint: $p|ab \implies p|a$ or $p|b$ if $p$ is a prime

nnn + 2*n = n*(nn+2) = n{(n*n -1) + 3} = n*3 + {n*(n-1)*(n+1)} ==> P + Q

Here
  P = n*3 ==> divisible by 3
  Q = (n-1)n(n+1) ==> this is multiplication of three consecutive number and hence it is divisible by 3
==> P + Q is divisible by 3


Max and min value of $f(x,y)$ $$f(x,y) = 2\cdot x\cdot y$$
$$x^2+y^2 \leq 4$$
I have no idea about this question.There is a region.How can I solve?

At first make two cases:
$$x^2+y^2 <4$$
this one is open hence you need the gradient to be zero there.

The second case is the boundary so
$$x^2+y^2=4$$
Here take langrange multipliers

Hint :Use A.M - G.M inequality for $|x|,|y|$.

$\Rightarrow x^2+y^2\ge2|xy|$

$\Rightarrow 4\ge x^2+y^2\ge 2|xy|$

$\Rightarrow-4\le2xy\le 4$

Equality holds when $|x|=|y|=\sqrt{2}$


Is $x_0$ local minimum/maximum or inflection point if $f'(x_0)=0$? Is the following statement true or false? (I think it is true, and I tried proving it by contradiction and failed.)

Let $f:(a,b)\to \mathbb{R}$ be differentiable, and let $x_0\in (a,b)$ such that $f'(x_0)=0$. Then $x_0$ is a local minimum/maximum or inflection point.

This is false. Let $$f(x)=\begin{cases}x^2 \sin (1/x) & x \neq 0 \\ 0 & x=0\end{cases}$$
Then $f$ is differentiable and $f'(0)=0$, but $0$ is neither a local extremum nor an inflection point.

If $f'(x_0) = 0$, then the tangent to the curve $y = f(x)$ at $x = x_0$ is horizontal. Thus it is as you say.


Base three numbers and expanded form I need help understanding this. Write each of the following base three numerals in expanded notation.

$22_3$
$212_3$
$12110_3$

Surely this means to expand them into base ten ...

So 22 in base three is actually 8 in base ten (this is 2 + (2*3)).
212 in base three is actually 14 in base ten (this is 2 + 1*3 + 2*9).
12110 in base three is actually 147 in base ten.

If you are being asked to write numbers in the expanded notation for Base $3$ such as $112201_3$ (just a random number I made up) then you would need to write it like so (starting from the $3^5$th place, the 5th place of the digit)

$$112201_3 = 1.3^5 + 1.3^4 + 2.3^3 + 2.3^2+ 0.3^1 + 1.3^0$$

Use this video to help yourself understand
https://www.youtube.com/watch?v=vOyiHMa-mtQ


How to prove that $\mathbb{N}$, the set of natural numbers, is well ordered? I am reading classic set theory by D.C. Goldrei. I am stuck on understanding an argument for proving $\mathbb{N}$ is well ordered (every non-empty subset is of a least element). The author assumed that $B$ is a nonempty subset with no least element. Then the following subset of $\mathbb{N}$ is constructed:
\begin{equation}
A = \left\{\boldsymbol{n} \in \mathbb{N}: \boldsymbol{m} \not\in B\ \textrm{for all}\ \boldsymbol{m} \leq \boldsymbol{n}\right\},
\end{equation}
and the author proved that $A = \mathbb{N}$, so that $B$ is empty, which leads to a contradiction. I am not quite sure about how this argument leads to the fact that $B$ is empty, and what is the relation between the assumption that $B$ is nonempty with no least element and the construction of set $A$. I have no idea why $A$ has to be constructed as such. Could anyone provide hints?

Assume $k \in B$.  Then $k+1 \notin A$ because there is a natural number ($k$) that is less than $k+1$ that is also in $B$.  If $A= \Bbb N$, then $k+1 \notin A$ can't happen, so your initial assumption that $k \in B$ must be false.  But $k$ was an arbitrary natural number, so that means no natural numbers can be elements of $B$; i.e., $B$ is empty.

Okay.
Let's do a sort of but not quite proof by contradiction.
The  well-ordered principal says:  Every non-empty subset of $\mathbb N$ has a least element.
Let's assume $B\subset \mathbb N$ without a least element.
If we can prove that $B$ can't be non-empty we are done.  Because if we can prove for $B$ to not have a least element we must have $B=\emptyset$ than that means for any subset that isn't empty, it can't be possible that it doesn't have a least element (because if it doesn't have a least element we have proven it must be empty.).
Okay, let's prove $B$ must be empty.
If we can prove that for any $k \in \mathbb N$ that $k \not \in B$ we will be done.  $B\subset \mathbb N$ but none of the $k \in \mathbb N$ can actually be in $B$ than nothing can be in $B$.  So $B$ would have to be empty.
So let's do that.  Let's prove that if $k \in \mathbb N$, then $k\not \in \mathbb N$.
So your text considers the set $A = \{n\in \mathbb N| $ if $m \le n$ then $m\not \in B\}$.
We can prove $1\in A$.  If $m\le 1$ then $m = 1$.  ANd if $m \in B$ then $m$ is the least element in $B$. But $B$ doesn't have a least element. So $1\not \in B$. so all $m\le 1$ are not in $B$.  So $1 \in A$.
Suppose we have a $k$ so that if $n \le k$ then $n\in A$.  We can prove $k+1\in A$.
If $m \le k+1$ then either $m \le k$ or $m = k+1$.  If $m\le k$ then $k \in A$ by our assumption.  If $m=k+1$ the either $k+1 \in B$ or $k+1\not \in B$.  Because $k\in A$ then all $n< k+1$ then $n \le k$ so $n\not \in B$.  So if $k+1 \in B$ then $k+1$ is the least element of $B$.  But $B$ has no least element. So $k+1 \not \in B$.  So for all $n \le k+1$ we have $n \not \in B$.  So $k+1 \in A$.
Now the natural numbers have the induction principal.  So if $1 \in A$ and if $k\in A \implies k+1 \in A$ we can conclude all $n \in \mathbb N$.
So for all $n \in \mathbb N$ we have $n \in A$.
But that means $B$ is empty.  Why?  Because if $m \in \mathbb N$ then $m \in A$.  And if $m \in A$ then all $n\le m$ are not in $B$.  And $m \le m$ so $m \not \in B$.  So no $m\in \mathbb N$ are in $B$.
So $B$ is empty.
And we have proven everything we needed to prove.
We have proven that if $B$ is any subset of $\mathbb N$ without a least element then $B$ must be empty.  That means for any non-empty set, the set can't not have a least element.  So every non-empty set of natural numbers has a least element.
Note, this will hold for any set that has the induction principal.  Note $\mathbb Z, \mathbb Q, \mathbb R$ etc. don't have the induction principal so $\mathbb Z, \mathbb Q, \mathbb R$ need not be well-ordered.
=====
Actually if it were me I wouldn't bother with the set $A$.
I'd just show

$1 \not \in B$.  (Because otherwise $1$ is the minimal element in $B$)

If $n \le k$ not in $b$ then $k+1$ is not it $B$ (else $k+1$ would be the least element.)

So by induction there are no $n \in \mathbb N$ that are in $B$.  So $B$ is empty.


Determine $a$, $b$ and $n$ in the following expression (binomial theorem backwards) You are given the following expression:

$$(ax + by)^{n} = -15120x^{4}y^{3}$$

Determine the constants $a$, $b$ and $n$.

My attempt to solve this problem is by trying to use the binomial theorem backwards.

The binomial theorem: $(ax + by)^{n} = \displaystyle\sum_{k = 0}^{n} {\binom{n}{k}(ax)^{n-k}(by)^{k}}$

And so if you compare the expressions, you get:

$$\binom{n}{k}(a)^{n-k}(b)^{k} = -15120$$

$n - k = 4$ and $k = 3$ so that $n = 7$

Here is where I get stuck, because now I have:

$$\binom{7}{3}(a)^{4}(b)^{3} = -15120$$

Two unknowns... How to solve it? Am I even doing it correctly?

I think, that the exercise is

  You are given the following expression:

  $(ax + by)^{n} = \color{blue}{\ldots} -15120x^{4}y^{3}+
 \color{blue}{\ldots} $

  Determine the constants $a$, $b$ and $n$.

In this case you know, that the 5th summand is ${7 \choose 4} \cdot (ax)^4\cdot (by)^3=-15120x^{4}y^{3}$

$15120=2^4\cdot3^3\cdot5\cdot7$.

${7 \choose 4}=5\cdot 7$

Thus $b=-3$ and $a=2$

You get $b= (\frac{-15120}{35 \, a^{4}})^{\frac{1}{3}}$ , $a$ can be any value as we only have one term.


Integral with a Dirac Delta and a Heaviside Theta function in statistical mechanics $$I=\frac{4\pi L }{h^2}\int_{-L}^{L} dy \int_{0}^{+\infty} dp \delta(\epsilon-\frac{p^2}{2m}-ky)$$ I put $\epsilon-ky=\epsilon_1$ so the integral:
$$I=\frac{4\pi L }{h^2}\int_{-L}^{L} dy \int_{0}^{+\infty} dp \delta(\epsilon_1-\frac{p^2}{2m})$$ now from the identity $\delta(f(p))=\frac{\delta(p-p_0)}{f'(p_0)}$, here $p_0=\pm\sqrt{2m\epsilon_1}$ so the integral becomes:
$$I=\frac{4\pi L m}{p_0h^2}\int_{-L}^{L} dy \int_{0}^{+\infty} dp [\delta(p-p_0)+\delta(p+p_0)]\theta(\epsilon_1)$$ with the choice of this theta, the integral becomes:
$$I=\frac{4\pi Lm }{h^2}\int_{-L}^{L} dy \theta(\epsilon_1)=\frac{4\pi Lm }{h^2}\int_{-L}^{L} dy \theta(\epsilon-ky)$$ and the integral solution is:
$$I=\frac{4 \pi Lm}{h^2k}[(\epsilon+kL)\theta(\epsilon+kL)+2kL\theta(\epsilon-kL)]$$ but the solution should be:
$$I=\frac{12 \pi Lm}{h^2k}[(\epsilon+kL)\theta(\epsilon+kL)-(\epsilon-kL)\theta(\epsilon-kL)]$$ could you please tell me where I'm wrong?
The $\theta$ function is the Heaviside function.

Using the notation for positive and negative parts, OP's integral becomes
$$\begin{align}\frac{h^2}{4\pi L }I~=~&\int_{[-L,L]}~\mathrm{d}y \int_{\mathbb{R}_+} \!\mathrm{d}p~ \delta(\epsilon-\frac{p^2}{2m}-ky)\cr
~\stackrel{k,m>0}{=}&\frac{\sqrt{2m}}{k}\int_{\mathbb{R}_+} \!\mathrm{d}p\int_{[-kL,kL]}~\mathrm{d}y ~ \delta(\epsilon-p^2-y)\cr
~=~&\frac{\sqrt{2m}}{k}\int_{\mathbb{R}_+} \!\mathrm{d}p~1_{\{-kL\leq \epsilon-p^2\leq kL\}}(p)\cr
~=~&\frac{\sqrt{2m}}{k}\int_{\mathbb{R}_+} \!\mathrm{d}p~1_{\{\epsilon-kL\leq p^2\leq \epsilon+kL\}}(p)\cr
~=~&\frac{\sqrt{2m}}{k}( \sqrt{(\epsilon+kL)^+}-\sqrt{(\epsilon-kL)^+} ).
\end{align} $$


Let $f(x)=x^3 - 3x + 1$ Let $f(x)=x^3 - 3x + 1$.

a) Expand $(x+h)^3$.
Would this simply be $x^3 + h^3$?

b. Use the formula $f’(x) =\lim_{h\to 0}\big(f(x+h)-f(x)\big)/h$ to show that the derivative of $f(x)$ is $3x^2-3$.
So, don’t we plug in $x^3 - 3x + 1$ every time we see $f(x)$ in that equation?

c. find the intervals over which the graph of $f$ is increasing and decreasing. State the coordinates of any relative extreme values for the graph of $f$.

for this, do we find the derivative of $f(x)=x^3 - 3x + 1$ which is $3x^2 - 3$ then solve for x and do a chart to find the positive and negative values.

d. Find the intervals over which the graph of $f$ is concave up and concave down. State the coordinates of any points of inflection.

For this, do we take the second derivative and do the same?

Based on the comments we need to begin at the beginning.

$(x+h)^3 = (x+h)(x+h)(x+h)$

Multiplication distributes over addition.

Lets start with one set of multiplications.

$(x+h)^2 = (x+h)(x+h)\\
(x+h)(x+h) = (x+h)x + (x+h)h\\
x\cdot x+h\cdot x  + x\cdot h+h\cdot h\\
(x+h)^2 = x^2 + 2xh + h^2$

Now lets work in the next factor.

$(x+h)^3 = (x+h)^2(x+h)\\
(x^2 + 2xh +h^2)(x+h)\\
(x^2 + 2xh +h^2)(x) +(x^2 + 2xh +h^2)(h)\\
x^3 + 2x^2h +xh^2+x^2h + 2xh^2 +h^3\\
(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$

$f(x+h)$

We plug in $(x+h)$ everywhere we see $x$ in the definition of  $f(x)$

$f(x+h) = (x+h)^3 - 3(x+h) + 1$

I will let you multiply that out using the information above.

Next you will need to work out:

$f(x+h) - f(x) = (x+h)^3 - 3(x+h) + 1 - x^3 + 3x - 1$

c) what you describe is correct.  But without a) b) you can't find $f'(x)$

d) You are going to need, a second derivative!

Using the same processes you applied to $f(x)$ you will apply to $f'(x)$ to find

$\lim_\limits{h\to 0} \frac {f'(x+h) - f'(x)}{h}$

a) $(x+h)^3 = (x+h)(x+h)^2 = ? $

b) $f(x+h) = (x+h)^3-3(x+h)+1$.

Use a) to expand and reduce. Then, compute $\frac{f(x+h)-f(x)}{h}$. When $h\to 0$, the expression you'll find should tend to $3x^2-3$.

c) and d) : yes and yes !


Question about conditional expectation on finite space Suppose $X_i$'s are Bernoulli random variables with success probability $p_i$ and $f$ is a function of $X_1,...,X_n$. Is that true $E[f\mid X_1,\dots,X_n]=\mathbf{1}_{\{X_n=1\}}E[f\mid X_1,\dots,X_n=1]+\mathbf{1}_{\{X_n=0\}}E[f\mid X_0,\dots,X_n=0]$ and
$E[f\mid X_1,\dots,X_{n-1}]=p_i E[f\mid X_1,\dots,X_n=1]+(1-p_i)E[f\mid X_1,\dots, X_n=0]$?
I think both are correct but not sure how to check the requirement that if $Y=E[X\mid\mathcal{F}]$, then $\int_AYdP=\int_A XdP$ for all $A\in \mathcal{F}$.
Revision: we assume $X_1,\dots,X_n$ are independent.

Yees, if you mix the notation for expectation conditioned by a random variable with the notation for expectation conditioned by an event, which are confusingly too similar in appearance.

Otherwise, for a sequence of Bernoulli random variables, then indeed:

$$\bbox[1ex,lemonchiffon]{\mathsf E(f\mid X_1,\ldots,X_n)~{=~ {\mathbf 1_{X_n=1}\mathsf E(f\mid X_1,\ldots,X_{n-1},\mathbf 1_{X_n=1}) }+{ \mathbf 1_{X_n=0}\mathsf E(f\mid X_1,\ldots,X_{n-1},\mathbf 1_{X_n=0}) } } }$$

And since $\mathsf E(f\mid X_1,\ldots, X_{n-1})=\mathsf E(\mathsf E(f\mid X_1,\ldots,X_n)\mid X_1,\ldots, X_{n-1})$ and also $\mathsf E(\mathbf 1_{X_n=1}\mid X_1,\ldots,X_{n-1})\overset{(1)}=\mathsf E(\mathbf 1_{X_n=1})\overset{(2)}=p_i$ (only if (1) mutually independent and (2) identically distributed) then too:

$$\bbox[1ex,lemonchiffon]{\mathsf E(f\mid X_1,\ldots,X_{n-1})~{=~{ p_i~\mathsf E(f\mid X_1,\ldots,X_{n-1},\mathbf 1_{X_n=1})} + {(1-p_i)~\mathsf E(f\mid X_1,\ldots,X_{n-1},\mathbf 1_{X_n=0})}}}$$