Prove by contradiction that every integer greater than 11 is a sum of two composite numbers I have thought a lot but am failing to arrive at anything encouraging.

First try: If this is to be proved by contradiction, then I start with the assumption that let $n$ be a number which is a sum of two numbers, of which at least one is prime. This gives $n = p + c$, where $p$ is the prime number and $c$ is the composite number. Also, any composite number can be written as a product of primes. So I can say, $n = p + p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$. From this, I get $n - p = p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$, but I have no clue what to do next.

Second try: For an instant let me forget about contradiction. Since $n > 11$, I can say that $n \geq 12$. This means that either $p \geq 6$ or $c \geq 6$. Again I'm not sure what to do next.

Finally, consider that the number 20 can be expressed in three different ways: $17+3$ (both prime), $16+4$ (both composite), and $18+2$ (one prime and one composite). This makes me wonder what we are trying to prove.

The textbook contains a hint, "Can all three of $n-4$, $n-6$, $n-8$ be prime?", but I'm sure what's so special about $4, 6, 8$ here.

Spoiler #1

   You can write $n = (n - \varepsilon) + \varepsilon$, where $\varepsilon \in \{4, 6, 8\}$.

Spoiler #2

   $n - \varepsilon > 3$, as $n > 11$.

Spoiler #3

   One of the three numbers $n - \varepsilon$ is divisible by $3$, as they are distinct modulo $3$.

Only 9 even numbers greater than 4 can't be expressed as the ORDERED sum of two ODD composites, namely 6, 8, 10, 12, 14, 16, 22, 32, 38.

Look at the 4 identities:
1. pp(2n)=pr[2,n]-pc(2n)
2. cc(2n)=c[2,n]-cp(2n)
3. pp(2n)=pr[n,2n-2]-cp(2n)
4. cc(2n)=c[n,2n-2]-pc(2n)

where pp(2n)=number of ordered sum of 2 primes = 2n, cc(2n)=# of ordered sums of 2 composites=2n, cp(2n)=number of ordered sums of 1 composite and 1 prime (in that order)=2n, and pc(2n)= number of ordered sums of 1 prime and 1 composite (in that order)=2n, and a+b is an ordered sum iff a< or = to b, pr[a,b] = number of primes in[a,b], c[a,b] = number of composites in [a,b]

Lots of other identities to construct from the 4 above - have fun playing with.


Prove that $e^x \tan x$ is strictly monotonically increasing on $(-\pi/2, \pi/2)$ I want to prove that $f: (-\pi/2,\pi/2) \to \mathbb{R}: x \mapsto e^x \tan x$ is strictly monotonically increasing on the given interval $D = (-\pi/2,\pi/2)$. I'd do this with the condition that
$$
f'(x) > 0 \, \forall x \in D \Rightarrow f \text{ strictly mono. incr.}
$$
See Q&A answer below. It's a trivial question, really, but I already wrote it all out, so here it goes.

So, after writing my question down, I already found the answer myself, so this is gonna be a Q&A answer.

Taking the derivative, we have
$$
\frac{d}{dx}f(x) = e^x \left( \tan x + \frac{1}{\cos^2 x}\right)
$$
Now, this product is obviously bigger than 0 if both factors are positive or both are negative. $e^x$ is always positive for $x \in \mathbb{R}$, so that means we must show that $(\tan x + \frac{1}{\cos^2 x}$) is positive.

This is where I was stuck. I then did the following steps:
$$
\begin{align}
\tan x + \frac{1}{\cos^2 x} &>0 \\
\tan x &> -\frac{1}{\cos^2 x} \\
\frac{\sin x}{\cos x} &> -\frac{1}{\cos^2 x} \\
\sin x \cdot \cos x &> -1
\end{align}
$$
Now we know that for $x \in D, 0<\cos x \leq 1$ and $-1 < \sin x < 1$, and the product of these will always be $>-1$. Q.E.D.

by Maclaurin Series we have:
  e^x=1+x/1!+x^2/2!+x^3/3!+⋯,-∞

  from this if 0 < x < y then e^x < e^y and it follows that e^(-y) < e^(-x). hence e^x is strictly increasing on the whole real axis.


Discriminant of the polynomial $f(x)=4x^3-ax-b$ Definition. The discriminant of the polynomial $f(x)=4(x-x_1)(x-x_2)(x-x_3)$ is the product $16\{(x_2-x_1)(x_3-x_2)(x_3-x_1)\}^2$.

How to prove that the discriminant of $f(x)=4x^3-ax-b$ is $a^3-27b^2$.

Any help would be appreciated.

We know the discriminant of a cubic polynomial of the form $ax^3 + bx^2 + cx + d$ is $\Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$. For $f(x)=4x^3-ax-b$, we just plug in the coefficients and get $\Delta = -4(4)(-a)^3 -27(4^2)(-b)^2$ which simplifies to  $\Delta = 16(\color{blue}{a^3 - 27b^2})$.

Edit: I realize now we are going by the definition. In that case, how I'd do it is start with the definition of the discriminant for a polynomial and find that it is $\Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$ as I mentioned earlier. Then plug in the numbers. You'll then get the right answer, up to a constant factor.

This document is a nice derivation of the discriminant of a monic cubic polynomial (highest degree term is $1$). You can  follow the same approach in your case (where the highest degree term is $4$). It's a roundabout way of doing it, but it'll work.

The discriminant shows the behavior of the roots (or zeros in the graph). It shows whether the graph has positive, negative, or imaginary roots. You can prove it by solving the quadratic  and plugin $a,b,c$ into the discriminant to see your results.

In other words, since $a=4$, $b=a$, $c=-b$, we can say say that $a=4$, $b=a=4$, $c=-b=-(a)=-4$. Thus, we can plug it into the discriminant and figure out if the polynomial has either positive, negative, or imaginary roots. And then you can plug it into the quadratic formula.

If you need more, just let me know.


Proving a polynomial $f(x)$ composite for infinitely many $x$ Let $f(x)=a_0+a_1x+ \ldots +a_nx^n$ be a polynomial with integer
  coefficients, where $a_n>0$ and $n \ge 1$. Prove that $f(x)$ is
  composite for infinitely many integers $x$.

I can easily show that there are infinitely many composite numbers of the form $a_0+a_1x+ \ldots +a_nx^n$ if $a_0 \ge 2$, we just note that $f(x)$ is composite for every $x$ being a multiple of $a_0$. But I can't find a way to prove this in the case $a_0=1$.

Choose $m$ such that $f(m)\ne\pm1$, then choose any prime $p$ dividing $f(m)$, and think about $f(m+pk)$ for $k=1,2,\dots$.

Let $x_k=(\text{lcm}(a_0,a_1,\dots,a_n))^k$ for some $k$. Then all $x_k$ with $k\in\mathbb N$ are values which produce composite numbers.


A question about partially ordered sets and their subsets I was reading about partially ordered sets and in the book, a theorem was proven. The theorem was that, given an poset, $(X, \le)$ there exists a set $Y$ of subsets of $X$ such that $(X, \le) \cong (Y, \subset)$. The proof went as follows:

"For each $a \in X$, let $Z_a = \{b \in X : b \le a\}$, and let $Y = \{Z_a : a \in X\}$. Define a map $\pi$ from $X$ to $Y$ by $\pi(a) = Z_a$. Clearly $\pi$ is a bijection. Moreover $a_1 \le a_2 \iff Z_{a_1} \subset Z_{a_2}$, so $\pi$ is an isomorphism between $(X, \le)$ and $(Y, \subset)$."

I understand why these two sets are isomorphic, but I don't understand why $Y$ is a subset of $X$. If $(X, \le)$, then there is a relation on $X$ and a relation is defined to be a subset of the Cartesian product. If thats the case, then the relation set must be a set of ordered pairs. The set $Z_a$ is the set of all elements which have a partial order on $a$. $Y$ is the set of all $Z_a$'s. But, if every $Z_a$ is based on only that which has a relation on $a$, doesn't that break the ordered pairs (since they are in the form $(a,b)$, and with any $a$ considered, only the $b$ elements would be in the set) and imply that ordered pairs cant be in any $Z_a$ set?

The only other interpretation I can think of is that $a$ itself is an ordered pair, because its a member of $X$, but then, I don't see how its possible for any $Z_a$ to have elements, given its definition.

Am I misunderstanding something?

Indeed, the $Z_a$ do not consist of ordered pairs. But they needn't; the $Z_a$ are simply the elements on which $\subset$ becomes the relation.

As an example, consider $(\Bbb N, \le)$ with $1$ and $2$. Then $Z_1 = \{0,1\}$, and $Z_2 = \{0,1,2\}$, and both are in $y$.

Now $\subset$ becomes a relation on $y$, so that we have $Z_1 \subset Z_2$, or more formally, $(Z_1, Z_2) \in {\subset}$ (and $\subset$ itself is then to be considered, well, a subset of $y \times y$).

This theorem means : if $X$ be any nonempty paritially ordered set and $Y$ is the power set of $X$ ( power set which is the set of all subset of $X$)

Then $Y$ is partially ordered set with the relation "subset or equal"
The proof doesn't depends on $X$ (if was it poset or not)

You can prove it as follows:
Every $A$ belong to $Y$ then $A$ is subset of $A$
Then the relation is reflexive...


Under what condition are $U$ and $V$ uncorrelated? Let $X$ and $Y$ be independent random variables with finite variances, and let $U = X + Y$ and $V = XY$. Under what condition are $U$ and $V$ uncorrelated?

MY ATTEMPT

We say that two variables are uncorrelated if their covariance equals zero. Bearing this in mind, we have
\begin{align*}
\textbf{Cov}(U,V) & = \textbf{E}(UV) - \textbf{E}(U)\textbf{E}(V)\\\\
& = \textbf{E}(X^{2}Y + XY^{2}) - \textbf{E}(X+Y)\textbf{E}(XY)\\\\
& = \textbf{E}(X^{2})\textbf{E}(Y) + \textbf{E}(X)\textbf{
E}(Y^{2}) - \textbf{E}(X)^{2}\textbf{E}(Y) - \textbf{E}(X)\textbf{E}(Y)^{2}\\\\
& = \textbf{E}(Y)[\textbf{E}(X^{2}) - \textbf{E}(X)^{2}] + \textbf{E}(X)[\textbf{E}(Y^{2}) - \textbf{E}(Y)^{2}]\\\\
& = \textbf{E}(Y)\textbf{Var}(X) + \textbf{E}(X)\textbf{Var}(Y) = 0
\end{align*}

Is there a specific name for this last expression? Any contribution is appreciated. Thanks :)

First of all, there is no specific name of what you've got. Second of all, assuming r.v. $X$ is not degenerated we have the following condition
$$
\mathbf{E} Y = - \mathbf{E} X \frac{\mathbf{Var} Y}{\mathbf{Var} X}.
$$


Does anyone know a reference to best-fitting lines with integral coefficients? I'm writing up a manual on how to generate "nice" Linear Algebra problems; that is, where the solutions tend to be integral. I "discovered" the following fact about the best-fitting line:

Theorem. Let $x_1,x_2,\ldots,x_n$ be integers such that (a) $\displaystyle\sum_{i=1}^n x_i=0$, and (b) $x_i\not=x_j$ for some $i,j$. Then the best-fitting line $y=ax+b$
for the data points $(x_1,y_1), (x_2,y_2), \ldots, (x_n,y_n)$
has integral coefficients iff
$$\sum_{k=1}^n x_i y_i \equiv 0 \pmod {\sum_{i=1}^n {x_i}^2}
\quad\quad\quad\quad{\rm and}\quad\quad\quad\quad
\sum_{k=1}^n y_i \equiv 0 \pmod {n}.$$

This is the sort of result which is nice, was probably discovered before, and which I cannot find any reference to. Has anyone seen this in a paper anywhere?

The trial function is
$$
 y(x) = a_{0} + a_{1} x.
$$
The data will consist of a sequence of measurements $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}.$
The goal is to create restraints to insure that $a\in\mathbb{Z}^{2}$.

Derivation

The linear system is
$$
\begin{align}
  \mathbf{A} a  &= y \\
%
\left[ \begin{array}{cc}
 \mathbf{1} & x
\end{array} \right]
%
\left[ \begin{array}{c}
 a_{0} \\ a_{1}
\end{array} \right]
%
&=
\left[ \begin{array}{c}
 y
\end{array} \right]
%
\end{align}
$$
The normal equations are
$$
\begin{align}
  \mathbf{A}^{*} \mathbf{A} a  &= \mathbf{A}^{*} y \\
%
\left[ \begin{array}{cc}
 \mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
 x \cdot \mathbf{1} & x \cdot x
\end{array} \right]
%
\left[ \begin{array}{c}
 a_{0} \\ a_{1}
\end{array} \right]
%
&=
\left[ \begin{array}{c}
 \mathbf{1} \cdot y \\
 x \cdot y
\end{array} \right].
%
\end{align}
$$
The solution for the normal equations is
$$
\begin{align}
  a &= \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} y \\
%
&=
%
\left(
 \det \left( \mathbf{A}^{*} \mathbf{A} \right)
\right)^{-1}
%
\left[ \begin{array}{rr}
 x \cdot x & -\mathbf{1} \cdot x \\
 -\mathbf{1} \cdot x & \mathbf{1} \cdot \mathbf{1}
\end{array} \right]
%
\left[ \begin{array}{c}
 \mathbf{1} \cdot y \\
 x \cdot y
\end{array} \right].
%
\end{align}
$$
The task is simplified with the a priori constraint that the $x$ values have zero mean:
$$
  \mathbf{1} \cdot x = 0.
$$
Now the solution is
$$
\begin{align}
  a
%
&=
%
\left(
 \det \left( \mathbf{A}^{*} \mathbf{A} \right)
\right)^{-1}
%
\left[ \begin{array}{cc}
 x \cdot x & 0 \\
 0 & \mathbf{1} \cdot \mathbf{1}
\end{array} \right]
%
\left[ \begin{array}{c}
 \mathbf{1} \cdot y \\
 x \cdot y
\end{array} \right].
%
\end{align}
$$

Fix the denominator

Turn attention to the denominator
$$
\det
\left( \mathbf{A}^{*} \mathbf{A} \right)
%
=
%
\left( \mathbf{1} \cdot \mathbf{1} \right) \left( x \cdot  x \right) -
\left( \mathbf{1} \cdot x \right)
%
=
%
\left( \mathbf{1} \cdot \mathbf{1} \right) \left( x \cdot  x \right).
$$
The solution is
$$
 a_{LS} = \left( \left( \mathbf{1} \cdot \mathbf{1} \right) \left( x \cdot  x \right)\right)^{-1}
%
\left[ \begin{array}{c}
 \left( x \cdot x \right) \left( \mathbf{1} \cdot y \right) \\
 \left( \mathbf{1} \cdot \mathbf{1} \right) \left( x \cdot y \right)
\end{array} \right].
$$
The explicit form for intercept $a_{0}$ and slope $a_{1}$ makes the constraints apparent:
$$
a_{0} =
\frac{\mathbf{1} \cdot y} {\mathbf{1} \cdot \mathbf{1}}, \quad
%
a_{1} =
\frac{ x \cdot y} { x \cdot x }
%
$$


Both $0^0$ and $0^1$ are undefined? Following up on my previous question at What values of $0^0$ would be consistent with the Laws of Exponents? I am still trying to get a handle on $0^0$. It now seems to me that not only is it undefined, but so is $0^1$. What mistake, if any, am I making here? (Yes, I know, you can simply define values for both and hope no contradictions arise, but I don't find such an approach to be very satisfying.)

In the natural numbers, exponents greater than 1 are naturally defined as follows:

$x^2=xx$

$x^3=xxx=x^2x$

$x^4=xxxx=x^3x$

$x^5=xxxxx=x^4x$

and so on.

Therefore $x^{n+1}=x^nx$ for $n\gt 1$.

What about $x^0$ and $x^1$?

Suppose $x^1x=x^2$. Then $x^1=x^2/x = x$ for $x\ne 0$.

Suppose $x^0x=x^1$. Then $x^0=x^1/x = 1$ for $x\ne 0$.

Thus $x^0$ and $x^1$ are both undefined for $x\ne 0$.

CONCLUSION:

For proof that $0^1=0$ assuming the usual Laws of Exponents, see comments by Henry Swanson below.

EDIT:

See FOLLOW-UP in my answer below.

In set theory and combinatorics, and for purposes of expansions in power series, $0^0$ is $1$.

It is $1$ because the number of factors multiplied is $0$.  When you multiply no numbers, you get $1$, just as when you add no numbers at all you get $0$.  Not multiplying by anything is the same as multiplying by $1$, just as adding no numbers at all is the same as adding $0$.  Google the term "empty product" and read about it.

"Everybody knows" that
$$
e^z = \sum_{n=0}^\infty \frac{z^n}{n!}.
$$
But when $z=0$, then the first term is $\dfrac{0^0}{0!}$.  But the first term must be $1$.

One the other hand $0^0$ is an indeterminate form in that if $f$ and $g$ approach $0$ as $x$ approaches something, then $f^g$ could approach any positive number at all or $0$ or $\infty$, depending on which functions $f$ and $g$ are.  (But if $(f,g)$ approaches $(0,0)$ from within a sector bounded by two lines of positive slope, then the limit is $1$.)

In the natural numbers, exponents greater than 1 are naturally defined as follows:
$x^2=xx$
$x^3=xxx=x^2x$
$x^4=xxxx=x^3x$
$x^5=xxxxx=x^4x$
and so on.

More formally, exponentiation can be defined as a binary function on $N$ such that
(1) $\forall x\in N:x^2=xx$
(2) $\forall x,y\in N:x^{y+1}=x^y x$
It can then be shown that
(1) $\forall x\in N:(x\ne 0 \implies x^1=x)$
(2) $\forall x\in N:(x\ne 0 \implies x^0=1)$
(3) $\forall x,y,z\in N:(x^{y+z}=x^y x^z) \implies 0^1=0 \land (0^0=0 \lor 0^0=1)$
Formal proof in DC Proof format (126 lines) at http://dcproof.com/Ambiguity-of-0-to-the-0.htm
FOLLOW-UP
Down vote quite justified! My thinking on this topic has evolved considerably in the weeks since I posted the above, although I still have exponentiation being defined as a binary function on $N$ such that
(1) $\forall x\in N: (x\ne 0 \implies x^0=1)$, or equivalently $\forall x\in N:x^2=xx$
(2) $\forall x,y\in N:x^{y+1}=x^y x$
$0^0$ is assumed to be a natural number, but no is actual value is assigned to it. Contrary to my previous claim, however, it can be shown that we must have $0^1=0.$
See details and formal proofs at "Oh, the Ambiguity!" at my math blog.


Why do we define quotient groups for normal subgroups only? Let $G \in \mathbf{Grp}$, $H \leq G$, $G/H := \lbrace gH: g \in G \rbrace$. We can then introduce group operation on $G/H$ as $(xH)*(yH) := (x*y)H$, so that $G/H$ becomes a quotient group when $H$ is a normal subgroup.
But why do we only work with quotient groups by normal subgroups? If we introduce the notion of left quotient group in the above manner, how much good properties of a quotient group do we lose?

Added. So, what is the problem? Let's look at the simplest example of a non-normal subgroup. Take $G=S_3$, and $H=\{e,(1,2)\}$. If we compose permutations right to left, the left cosets of $H$ in $G$ are:
\begin{align*}
eH = (1,2)H &= \{ e, (1,2)\};\\
(1,2,3)H = (1,3)H &= \{(1,2,3), (1,3)\};\\
(1,3,2)H = (2,3)H &= \{(1,3,2), (2,3)\}.
\end{align*}
If we try multiplying cosets term-by-term, we run into problems. Multiplying by $eH$ is not a problem, but take $(1,2,3)H$ multiplied by itself. The products are:
$$\{ (1,2,3)(1,2,3), (1,2,3)(1,3), (1,3)(1,2,3), (1,3)(1,3)\} = \{(1,3,2), (2,3), (1,2), e\}$$
which is not a coset. If we multiply using representatives, as in the original question,we also run into problems: if we multiply $(1,2,3)H\cdot(1,3,2)H$ as $(1,2,3)(1,3,2)H$, we get $eH$. But $(1,2,3)H = (1,3)H$, and $(1,3,2)H = (2,3)H$, and if we multiply them by looking at these alternative representatives/names, we get $(1,3)H\cdot (2,3)H = (1,3)(2,3)H = (1,3,2)H\neq eH$. That is, the multiplication rule depends on the name we give the coset, rather than on what the coset is. This means that the rule is not well-defined.

We can let $G$ act on the set of left cosets of $H$, by $g(xH) = gxH$; this is not a group, but a group action. We can also, separately, put group structures on the set of cosets of $H$ in $G$, since it will be a set, and every nonempty set can be given a group structure (assuming the Axiom of Choice; in fact, equivalent to the Axiom of Choice). But that structure will generally have nothing to do with the operation in $G$.

If we want an operation on the set of cosets, and have the operation in the cosets to be induced by the operation in $G$, then we have:

THEOREM. Let $G$ be a group and $H$ a subgroup. Let $G/H = \{ gH\mid g\in G\}$ be the set of left cosets of $H$ in $G$. The operation
$$\cdot \colon G/H \times G/H \to G/H\qquad\text{defined by}\qquad (xH)\cdot(yH) = (xy)H$$
is well defined (that is, if $xH = x'H$ and $yH=y'H$ then $xyH = x'y'H$) if and only if $H$ is normal in $G$.

Proof. Suppose $H$ is normal. Then $gH=Hg$ for every $g\in G$. If $xH=x'H$ and $yH=y'H$, then
$$xyH = x(yH) = x(Hy) = x(Hy') = (xH)y' = (x'H)y' = x'(Hy') = x'(y'H) = x'y'H$$
so the operation is well defined.

Conversely, suppose the operation is well defined so that whenever $xH=x'H$ and $yH=y'H$, then $xyH = x'y'H$. We want to show that $g^{-1}Hg\subseteq H$ for all $g\in G$. For each $h\in H$, since $hH = eH$, we have $eHgH = egH=gH$ is equal to $hHgH = hgH$. So $gH = hgH$, hence $H=g^{-1}hgH$, so $g^{-1}hg\in H$ for each $h\in H$. Thus, $g^{-1}Hg\subseteq H$. This holds for any $g\in G$, so $g^{-1}Hg\subseteq H$ for all $g\in G$, hence $H\triangleleft G$, as claimed. QED

And now I'm going to hijack your question a bit and go on a long excursion to a closely related subject which is not necessary to answer your question, but which I think is interesting. (Sorry!)

This is really connected to the notion of a congruence on an algebra (in the sense of universal algebra). A subgroup $H$ defines an equivalence relation on $G$ by $x\sim y$ if and only if $xH = yH$ (it also defines the equivalence relation of right congruence, but ignore that for now). Being an equivalence relation, it corresponds to a subset $\Phi$ of $G\times G$, with $(x,y)\in\Phi$ if and only if $x\sim y$, if and only if $xH=yH$.

But $G\times G$ also has a natural group structure inherited from $G$. Since $\Phi$ is a subset of the group $G\times G$, it makes sense to ask whether $\Phi$ is a subgroup of $G\times G$ or not.

THEOREM. $\Phi$ is a subgroup of $G\times G$ if and only if $H$ is normal in $G$.

Proof. Note that $\Phi$ always contains $(e,e)$ (in fact, the entire diagonal). Assume first that $H$ is normal. If $(x,y)\in \Phi$, then $xH=yH$. Hence, $Hx = Hy$, and taking inverses of each element we get $x^{-1}H = y^{-1}H$, so $(x^{-1},y^{-1})\in\Phi$; thus, $\Phi$ is closed under inverses. If $(x,y),(z,w)\in\Phi$, then $xH=yH$, $zH=wH$, so by what we proved above we have $xzH = ywH$, so $(xz,yw)=(x,y)(z,w)\in\Phi$. Thus, $\Phi$ is closed under products, so $\Phi$ is a subgroup of $G\times G$.

Conversely, suppose that $\Phi$ is a subgroup. If $xH=yH$ and $zH=wH$, then $(x,y),(z,w)\in \Phi$, hence $(x,y)(z,w)=(xz,yw)\in\Phi$; thus, $xzH = ywH$. This proves that the operation $xH\cdot zH \mapsto xzH$ is well-defined, so by the theorem above $H$ is a normal subgroup of $G$. QED

Added. (Stuff between the two horizontal lines was added in this edit).

Okay, how about this? Can we define some equivalence relation $\sim$ on a group $G$ so that we can multiply equivalence classes through representatives, have it be well-defined, but not have $\sim$ "come from" a normal subgroup?

That is: suppose $\sim$ is an equivalence relation on the group $G$ (any equivalence relation; all we require is that it be reflexive, symmetric, and transitive on the underlying set of $G$), and let $[x]$ denote the equivalence class of $x$ for each $x$ in $G$. We want to define a multiplication on the quotient set $G/\sim$ by $[x][y] = [xy]$. When is this well defined?

THEOREM. Let $G$ be a group, and let $\sim$ be an equivalence relation on $G$. Then the multiplication on $G/\sim$ defined by $[x][y]=[xy]$ is well defined if and only if $\sim$ is a subgroup of $G\times G$.

Proof. If $\sim$ is a subgroup of $G\times G$ and $[x]=[z]$, $[y]=[w]$, then $(x,z),(y,w)\in \sim$, so $(x,z)(y,w) = (xy,zw)\in \sim$, hence $[xy]=[zw]$, so multiplication is well defined.

Conversely, suppose the multiplication is well defined. Since $\sim$ is an equivalence relation, $(e,e)\in\sim$. Also, if $(x,z), (y,w)\in\sim$, then $[x]=[z]$, $[y]=[w]$, so $[x][y]=[xy]$ equals $[z][w]=[zw]$, so $(xy,zw)=(x,y)(z,w)\in\sim$. Thus, $\sim$ is a submonoid of $G\times G$. It only remains to show that $\sim$ is closed under inverses. Suppose that $(x,y)\in\sim$. Multiplying by $(x^{-1},x^{-1})$ on the left and $(y^{-1},y^{-1})$ on the right (both elements of $\sim$) we have $(y^{-1},x^{-1})\in\sim$. Since $\sim$ is an equivalence relation, it is reflexive, so $(x^{-1},y^{-1})\in\sim$. Thus, $\sim$ is a subgroup of $G\times G$, as claimed. QED

Okay, so we can define the multiplication on the quotient set if and only if the equivalence relation is a subgroup of $G\times G$. Do all such equivalence relations "come" from subgroups? Yes:

THEOREM. Let $\sim$ be an equivalence relation on $G$ that is a subgroup of $G\times G$, and let $H=\{g\in G\mid g\sim e\}$. Then $H$ is a normal subgroup of $G$, and $x\sim y$ if and only if $xH=yH$.

Proof. We can identify $G$ with the subgroup $G\times\{e\}$ of $G\times G$; then $H$ corresponds to the intersection of the subgroup $\sim$ and the subgroup $G\times\{e\}$, hence is a subgroup of $G$.

To see that $\sim$ is in fact left congruence modulo $H$, simply note that $xH=yH$ if and only if $y^{-1}x\in H$, if and only if $(y^{-1}x,e)\in \sim$, if and only if $(y,y)(y^{-1}x,e)\in\sim$, if and only if $(x,y)\in \sim$, if and only if $x\sim y$. That $H$ is normal now follows because multiplication of cosets (equivalently, of equivalence classes modulo $\sim$) is well defined. QED

When we generalize this to an arbitrary algebra (in the sense of universal algebra; examples include magmas, monoids, semigroups, groups, rings, $K$-algebras, lattices, etc), we get the notion of congruence:

Definition. Let $A$ be an algebra (in the sense of universal algebra). A congruence on $A$ is an equivalence relation $\Phi$ on $A$ such that $\Phi$ is a subalgebra of $A\times A$.

THEOREM. Let $A$ and $B$ be algebras of the same type, and let $f\colon A\to B$ be a homomorphism; define $\Phi_f = \{(x,y)\in A\times A\mid f(x)=f(y)\}$. Then $\Phi_f$ is a congruence on $A$. Conversely, if $\Psi$ is a congruence on $A$, then there exists an algebra $\mathcal{B}$ and a homomorphism $\pi\colon A\to \mathcal{B}$ such that $\Psi=\Phi_{\pi}$.

Special case. Let me do the proof with semigroups. First, let $f\colon A\to B$ be a semigroup homomorphism; then $\Phi_f$ is an equivalence relation on $A$. To show it is a subsemigroup of $A\times A$, suppose $(x,y),(z,w)\in \Phi_f$. Then $f(xz) = f(x)f(z) = f(y)f(w) = f(yw)$, so $(xz,yw)\in\Phi_f$. Thus, $\Phi_f$ is a subsemigroup of $A\times A$, hence a congruence.

Conversely, suppose that $\Psi$ is a congruence. Let $\mathcal{B}=A/\Psi$, the set of equivalence classes of $\Psi$ in $A$. Let $[x]$ denote the class of $x$. We define a multiplication on $\mathcal{B}$ by $[a][b]=[ab]$. We must prove that this is well defined: suppose $[x]=[y]$ and $[z]=[w]$. Then $(x,y),(z,w)\in\Psi$, so $(xz,yw)\in\Psi$ (since $\Psi$ is a congruence). Therefore, $[xz]=[yw]$, so multiplication is well-defined. It is now easy to verify that multiplication is associative:
$$[x]([y][z]) = [x][yz] = [x(yz)]=[(xy)z] = [xy][z] = ([x][y])[z].$$
Thus, $\mathcal{B}$ is a semigroup. The map $\pi\colon A\to \mathcal{B}$ given by $\pi(a)=[a]$ is easily verified to be a homomorphism, and $\Phi_{\pi}=\Psi$.

The proof for an arbitrary algebra with operations $f_1,\ldots,f_k$ of arities $a_1,\ldots,a_k$ (or even an infinite number of operations, each with finite arities) is essentially the same.
QED

One can then prove the Isomorphism Theorems for "congruences" in exactly the analogous manner as for groups and rings, for any class of algebras.

In Groups and Rings, we can "get away" with not studying congruences because if $\Phi$ is a congruence on $G$, then we can let $N = \{ g\in G\mid (g,e)\in\Phi\}$, and it turns out that $\Phi$ is the equivalence relation induced by $N$. Essentially, a congruence tells you that certain things are equal, but in groups (and rings), instead of saying "$x$ is equal to $y$", we can say "$xy^{-1}$ is the identity". The normal subgroup (resp. the ideal) is the set of elements equivalent to the identity. In more general classes, such as semigroups, we don't have the luxury of expressing any identity in the form $w=e$, so we must actually keep track of both sides of the equality.

This is all part of Universal Algebra. I cannot recommend George Bergman's book highly enough if you want a great introduction to the subject (with a Category Theory flavor).

To elaborate on a comment made by another user in the above, if the operation $(xH)\ast(yH):=(xy)H$ is well-defined, then $(G/H,\ast)$ is a group. Hence $f:G\to G/H$ defined by $f(g)=gH$ is a homomorphism and $H=\ker f$. However, the kernel of a group homomorphism is always normal. Therefore $H$ has to be normal.

(As shown by Arturo Magidin's answer, actually $(xH)\ast(yH):=(xy)H$ is well-defined if and only if $H$ is normal.)


Little confused about the constraint of Injective Functions and Surjective. From my understanding,

A Function is called to be Injective, if different elements of the first set are mapped to different elements of the second set.

Let set A = {a,b,c} and set B = {1,2,3}

Are the following points correct,

if:

1) f(a) = 1 , f(b) = 2 , f(c) = 3    > The function F: A -> B is injective.

2) f(a) = 1 , f(b) = 1 , f(c) = 3    > The function F: A -> B is not       injective.

Did I cover all the possible possibilities?

A Function is called to be Surjective, if all elements of the second set are mapped to some element of the first set.

Let set A = {a,b,c, d, e} and set B = {1,2,3}

Are the following points correct,

if:

1) f(a) = 1 , f(b) = 2 , f(c) = 3 > The function F: A -> B is surjective.

2) f(a) = 1 , f(b) = 1 , f(c) = 3 > The function F: A -> B is not surjective.

Did I cover all the possible possibilities?

Thank you!

Your definitions are wrong.

Your second example says $f(a) = 1$, $f(a) = 2$, $f(c) = 3$.  That is not even a function.  $f(a)$ cannot be more than one thing.

"Injective" means no two elements of the domain are mapped to the same element of the codomain.

"Surjective" means for every element of the codomain, some element of the domain is mapped to it.

Yes, this is all correct. Of course you can construct more functions $f: A \to B$, that are not injective / not surjective.

Your definition of surjective sounds a little strange, but I think you mean the right thing. A function $f: A \to B$ is called surjective, if for every element $b \in B$ there exists an element $a \in A$, such that $f(a) = b$.


Line intersection in 3D I have 4 points. Only three are known:

$P1(a,b,c)$
$P2(x,y,z)$

$P3(d,e,f)$

$P4(g,h,i)$

$L1=(P1P2)$, $L2=(P2P3)$ and $L3=(P3P4)$ are lines. $L1$ and $L2$ intersect with each other at point $P2$. $L2$ and $L3$ intersect with each other at point $P3$.

Length of the lines are also known as:

$L1=K$

$L2=L$

$L3=M$

On the other hand, plane's equations which contain $L1$ and $L3$ are also known.

I would like to know $x,y,z$ intersection point of $L1$ and $L2$.

Thanks in advance.
If there is any misinformation in order to be able to slove the problem, please let me know.

Isn't there any solution related this topic. This is a mechanism and every line is an arm of the mechanism. I need to find every position of the mechanism related to motion.


Why are we allowed to cancel fractions in limits? For example:

$$\lim_{x\to 1} \frac{x^4-1}{x-1}$$

We could expand and simplify like so:

$$\lim_{x\to 1} \frac{(x-1)(x^3 + x^2 + x + 1)}{x-1} = \lim_{x\to 1} (x^3 + x^2 + x + 1) = (1^3 + 1^2 + 1^1 + 1) = 4$$

In this case we divided out $x-1$ on top and bottom even though technically, at $x=1$, we have $\frac{0}{0}$ that we're just tossing aside.

But what allows us to do this?

Simply because we are dealing with values $x\neq 1$ in this case, thus for algebraic rule we are allowed to cancel out

$$\lim_{x\to 1} \frac{x^4-1}{x-1}=\lim_{x\to 1} \frac{\color{red}{(x-1)}(x^3 + x^2 + x + 1)}{\color{red}{x-1}}$$

Remember indeed that by the definition of limit we are demanding that
$$\forall \varepsilon>0 \quad \exists \delta>0 \quad \text{such that}\quad \color{green}{\forall x\neq1}\quad|x-1|<\delta \implies|f(x)-L|<\varepsilon$$

Note also that the same cancellation is used to prove the basic derivatives case, for example for $f(x)=x^2$

$$\lim_{x\to x_0}\frac{x^2-x_0^2}{x-x_0}=\lim_{x\to x_0}\frac{\color{red}{(x-x_0)}(x+x_0)}{\color{red}{x-x_0}}=\lim_{x\to x_0}(x+x_0)=2x_0$$

Well, the simple answer is: Because limits never depend on values in single points, so we are allowed to modify the limited function in finitely many points without changing the limit.

Either the point is the one in which we are taking the limit; but we know that it's excluded.
Or it's any other point, and than it's "far away" from the limit point and can be ignored altogehter.

However, you of course cannot do crazy stuff like claiming that
$$ \lim_{x\to0-} \frac{x\sqrt{x}}{\sqrt{x}} = \lim_{x\to0-} x = 0$$
because here you modify the domain of the function in more than finitely many points.


Expected number of tosses to get either $n$ heads or $n$ tails in a row? I have solved the problem for the expected number of tosses to get $n$ heads in a row, which is $2^{n+1} - 2$. To get $n$ heads OR $n$ tails, I think the problem is substantially more difficult, and I can't figure out a way to generalize this for any $n$.
Is there a straightforward approach to generalizing this? The lack of information on this online seems to suggest that it isn't so straightforward.

Suppose the answer to the expected time to  get $n$ heads or $n$ tails was $f(n)$.
Then the expected time to  get $n$ heads would be $2f(n)$  and similarly the the expected time to  get $n$ tails would be $2f(n)$: you try to get one of them and with probability $\frac12$ it is the one you want while with probability $\frac12$ you start again and it is like the expected time to get a single head.
But you already know $2f(n)=2^{n+1}-2$, so $f(n)$ is half this, namely $2^{n}-1$

For $n$ heads, you start with $0$ heads and when you have $k$ heads, there is a 50:50 chance that you change to either $k+1$ heads or to $0$ heads.
For $n$ heads-or-tails, after one toss you definitely have $1$ head-or-tail, and when you have $k$ heads-or-tails, there is a 50:50 chance that you change to either $k+1$ heads-or-tails or to $1$ (other) head-or-tail.
Hence the answer to the second problem for $n$ is the same as the answer to the first problem with $n-1$, plus $1$.

Put differently, replace the original sequence of heads and tails by (the first toss and) XOR results of consecutive tosses (i.e., either $=$ or $\neq$). These occur like coin tosses, i.e., independently and with probability $\frac 12$. Your goal is to collect $n-1$ consecutive "$=$" (after the initial toss).  So again, we arrive at $(2^{n-1}-2)+1$.


Expected time between successive visits to state $i$ in a Markov chain conditionally on no visit to state $j$ Say I have a Markov chain $\{X_n: n \geq 1\}$ with state space $E = \{1,2,3,4,5\}$ and transition matrix,

$$ P = \begin{bmatrix} 0 & 1/2 & 1/2 & 0 & 0 \\\ 1/2 & 0 & 1/2 & 0 & 0 \\\ 1/4 & 1/4 & 0 & 1/4 & 1/4 \\\ 0 & 0 & 1/2 & 0 & 1/2 \\\ 0 & 0 & 0 & 1/2 & 1/2\ \end{bmatrix} $$

With a stationary distribution,

$$ \pi^T =  [1/6 \quad 1/6 \quad 1/3 \quad 1/6 \quad 1/6]$$

Assuming that $X_1 = 1$, it is easy to use this information to calculate the expected number of transitions between successive visits to state 1 (the answer is $1/\pi_1 = 6$}.

What is not clear to me, however, is how to calculate the expected number of transitions between successive visits to state 1 conditional on the fact that state 5 is never visited.

Two potential approaches to answer this are:

1) Constructing the Markov chain graph without state 5, recalculating the transition matrix and the stationary distribution. This approach yields a transition matrix of

$$ P = \begin{bmatrix} 0 & 1/2 & 1/2 & 0 \\\ 1/2 & 0 & 1/2 & 0 \\\ 1/3 & 1/3 & 0 & 1/3 \\\ 0 & 0 & 1 & 0\ \end{bmatrix} $$

a stationary distribution of

$$ \pi^T =  [1/4 \quad 1/4 \quad 3/8 \quad 1/8]$$

and an answer of 4.

2) Calculating the answer as $(1/\pi_1)$ after conditioning the stationary distribution to account for the fact that state $5$ is not visited. In other words, we normalize $\pi_1 \ldots \pi_4$ by a factor of $1 - \pi_5$ to account for the fact that state $5$ is not visited. This yields a stationary distribution

$$ \pi^T =  [1/5 \quad 1/5 \quad 2/5 \quad 1/5 \quad 0]$$

and an answer of $5$.

Some friends have argued that approach #1 is the correct approach, but cannot explain why approach #2 is wrong.

Any insight is appreciated.

The reason number 2 is wrong is that each number is predicated upon being able to either go to or come from 5 in each state, which does not uniformly affect each of the states. The degree to which each state's steady state varies depends upon the column corresponding to 5 in the transition matrix, which is not the same for each.

I found this document which support Jikai's comment. Here is a link. In this document, it tells that 'the ratio of two limiting state probabilities represents the expected number of visits to state j between two successive visits to i which is $\pi_j/\pi_i$'. So for your problem, the answer is that 1/$\pi_1$ - $\pi_5/\pi_1$ which is '1/(1/9) - (1/3)/(1/9) = 6'.

the stationary distribution i got is $\pi=[1/9, 1/9, 2/9, 2/9, 1/3]$ which is different from your expression. The $\pi_3$ is not satisfied in your result.

hope this helps.


Inverse of right engel element is a left engel element. Heineken shows that inverse of a right engel element is a left engel element. For this he showed the commutator identity $[x,_{n+1} g]=^g [^xg^{-1},_n g]$.
I am studying the textbook "A course in the theory of groups" by Robinson.
I am not able to conclude the claimed result from this commutator identity.
Please help me.

First I apologize for such question. The answer of the above question, if g is an right engel element then so $^xg$ also. Now by using the given commutator identity we can have the desired result.


Proving that all integers are even or odd I know that $\mathbb{Z}$ is a group under addition with a multiplication defined. I have just the definition of even and odd integers: $n$ is even if $n = 2k$ for some integer $k$ and $n$ is odd if $n = 2k+1$ for some integer $k$.

Using just this I am wondering how to prove that all integers are either even or odd. That is, how can I prove that given an integer $n$, $n$ must be even or odd?

My problem with this is that it seems so simple. I know that one can divide an integer by $2$ and the remainder will be $0$ or $1$. Using this, it is clear that the even and odd integers make up everything. But how can I prove it without using this fact about remainders and such? I guess one could also use facts about prime numbers, but I am looking for a proof that just uses the definition of odd and even.

If you really want to avoid using general facts about division with remainder, you can use mathematical induction on $n$ to prove it for nonnegative integers:

Base case: $0$ is even because $0=2\cdot 0$.

Induction step: Assume that $n$ is odd or even; then we must prove that $n+1$ is also either odd or even.

First subcase: $n$ is even, so $n=2k$ for some $k$. Then $n+1=2k+1$ and so it is by definition odd.

Second subcase: $n$ is odd, so $n=2k+1$ for some $k$. Then $n+1=2k+1+1=2k+2=2(k+1)$, and so $n+1$ is even.

That completes the induction proof, and now we just need to know that negative integers are also all either odd or even. But if $n$ is negative, then $-n$ is positive. One easily sees that if $-n$ is even, then $n$ is even too, and if $-n$ is odd, then $n$ is odd too.

Whether this satisfies your requirements is a bit debatable, because the induction part of the proof is essentially the same as how you prove that division with remainder works in general.

However, you CANNOT prove that every integer is either odd or even solely from the fact that the integers form "a group under addition with a multiplication defined", that is a commutative ring with unit. The reason for this is that there are some rings where not all elements are either odd or even -- such as $\mathbb Z[X]$, the ring of polynomials with integer coefficients. Here the polynomial $X+1$ is neither twice something, nor twice something plus one.

If you know or believe that every integer can be expressed in decimal, then this is easy: A number is even or odd according to whether the units digit in its decimal representation is even or odd. This is because $10$ is even and every single-digit number is either even or odd.

Of course, proving that every integer can be expressed in decimal (or any base) requires induction.

And if you go that route, its nicer to prove directly that every integer can be expressed in binary and the look at the units digit, which is $0$ or $1$.


If set $A_n$ converges to $A$ in $L^1$, then is it also true for the $r$-neighborhood? Let $A$ be a bounded simply connected set in $\mathbb{R}^2$, and $A^r$ denote the $r$-neighborhood of $A$, that is, $A^r:=\{x: d(x,A) \le r\}$. Suppose there exists $A_n \rightarrow A$ in $L^1$, that is, $|A \Delta A_n| \rightarrow 0$, then prove $A_n^r \rightarrow A^r$ in $L^1$.

It seems to be very intuitive based on the picture, but strangely I've no idea how to make a rigorous proof. Any comments will be really appreciated.

Thanks to the answer provided by @zhw, I realized I had asked a trivial question. Later I asked another question here On the $r$-neighborhood of a set in $\mathbb{R}^2$. to fully make sense of the background and my concern.

No, here's a counterexample: Let $A = D(0,1),$ and $A_n = D(0,1) \cup \{(x,y): |y| < (e^{-x^2}) /n\}.$ Then $A_n \to A$ in $L^1$ (in your sense), but $|A_n^r| = \infty$ for every $r$ and $n.$


Finite dimensional subspaces Let $X$ be a complex Banach space of infinite dimension. Does there exist a finite dimensional subspace of $X$ of arbitrary (finite) dimension which is complemented by a projection of norm 1?

I am not entirely sure what you mean by "which is complemented by a projection of norm 1", but I suppose you mean: If $Y$ is a subspace of $X$, is there a non-trivial projection $P$ such that $Y\oplus P(X) = X$. If this is not what you mean, then I am sorryt for the misunderstanding.

There exists such a projection if , and only if, the is a projection $\hat P$ such that $\hat P (X) = Y$, because then $P$ can be chosen as $1-\hat P$.

To find out for which subspaces $Y$ this is true, take the map $x \mapsto y$, where $y$ is the unique vector out of $Y$ such that $x$ can be written as $x=y+u$, for some $u$ in a complementary subspace. This map is a projection if and only if it is continuous, but now it is easy to see that this is true if and only if $Y$ is a closed subspace.

Hence for all finite dimensional subspaces and for all infinite dim. subspaces which are closed the answer to your question is yes, otherwise no.


On the theorem $(p \cdot q) \supset (p \supset q)$ In the book Gödel's Proof, by Nagel and Newman, the authors write (p. 119):

  There is a theorem in logic which reads:
  $$(p \cdot q) \supset (p \supset q)$$

  or when translated, 'If both $p$ and $q$, then (if $p$ then $q$)'.

As the authors' translation at the end of the quote shows, they use $\cdot$ to denote logical conjunction (aka $\wedge$ or $\&$), and $\supset$ to denote implication (aka $\Rightarrow$).  Furthermore, elsewhere in the text, they use $\sim$ for logical negation (aka $\lnot$), and the $\vee$ for logical disjunction.  To minimize confusion, I will adopt the authors' notation for the rest of this post.

If I convert the implications to disjunctions, the theorem becomes

$$ \sim\! (p \cdot q) \vee (\sim\! p \vee q)$$

...and applying de Morgan's law to the first disjunct above, we are left with

$$ \sim\! p \; \vee \sim\! q\; \vee \sim\! p \vee q $$

This is clearly a tautology, but the role of $p$ in the whole thing seems to me entirely superfluous.  In other words, the theorem here looks to me like an unnecesarily cluttered rendition of $\sim\! q \vee q$.

I find this very confusing.

Part of my confusion stems from two senses of the word "theorem".  For example, according to one of these senses, the equation

$$139 - 31 = 3^3 \times 2^2$$

is "a theorem of number theory", but according to the other sense it isn't, since nobody calls this equation "a theorem of number theory"; that designation is reserved to things like "The Infinitude of Primes", "Fermat's Last Theorem", "The Chinese Remainder Theorem", etc.

I am left wondering which of the two senses of "theorem" did Nagel and Newman intend when they described $(p \cdot q) \supset (p \supset q)$ as "a theorem in logic".

Either way, why invoke a theorem that includes a clearly superfluous reference to $p$?  Or to put it differently, why invoke such an obfuscated form of $\sim\! q \vee q$?  It's as if, in a number theory proof, one were to invoke the "theorem"

  Either there are infinitely many primes or $\pi^{\sqrt {2.6 + e}} < 1\frac{3}{32}$.

It is true, but why would anyone bother to use such a thing in a proof?

Clearly I'm missing the authors' point, and would appreciate a cluebrick or two.

"Theorem" in propositional calculus means a formula derivable form propositional axioms : see Nagel & Newman, page 49 for the propositional axioms used into Principia Mathematica.

The axiom may vary but we want that they are enough to derive all tautologies usung the modus ponens rule of inference (see page 54).

A theorem is number theory is a formula expressible in the language of arithmetic that is derivable from the axioms of logic and the specific axioms of arithmetics : Peano's axioms.

Consider the tautology $p ∨ ¬p$, which is derivable from logical axioms alone.

We have that $∀x(x=0) ∨ ¬∀x(x=0)$ is a theorem of arithmetic because we can derive it from axioms (in this case without using the specific arithmetical ones).

Of course, we need the specific arithmetical axioms to derive more "interesting" arithmetical theorems, like e.g. commutativity of $+$.

The formula in page 119 is used in the context of the formalization of Euclid's argument.

It is used to derive from the intermediate result : (6) $Pr (x) \land (\exists z) [Pr (z) \land Gr (z, x)]$ the final arithmetical theorem : (7) $(x) [Pr (x) \to (\exists z)(Pr (z) \land Gr (z, x))]$.

Thus, the authors use the tautology : $(p \land q) \to (p \to q)$ and modus ponens (the Rule of Detachment) to derive from (6) :

  (6') $Pr (x) \to (\exists z)[Pr (z) \land Gr (z, x))]$

and then "generalize" it to get (7).

The gist of the example is to produce an "exercise in formalization" : in the context of a standard mathematical proof, having proved that ‘$x$ is a Prime and there is at least one other Prime $z$ such that $z$ is Greater than $x$’ ( i.e. $Pr (x) \land (\exists z) [Pr (z) \land Gr (z, x)]$) it is enough to conclude, without further steps, that ‘for every $x$, if $x$ is Prime, then there is at least one other Prime $z$ that is greater than $x$’ ( i.e. $(x) [Pr (x) \to (\exists z)(Pr (z) \land Gr (z, x))]$).

I am left wondering which of the two senses of "theorem" did Nagel and Newman intend when they described $(p \cdot q) \supset (p \supset q)$ as "a theorem in logic".

This statement corresponds to usual first row of the truth table for implication: If  the antecedent and consequent are both true, then the implication is true. It is theorem that can be derived using only (1) a premise rule to introduce assumptions, (2) a conclusion rule to discharge assumptions and (3) a rule to eliminate conjunctions. See my formal proof.


Analytic continuation commuting with series Suppose $f_1,f_2,...$ are entire functions, and there is an open subset $U \subseteq \mathbb{C}$ such that the series $F(z) = \sum_{n=1}^{\infty} f_n(z)$ converges normally on $U$. Also suppose that $F$ can be analytically continued to an entire function.

I have a situation where all $f_n$ vanish at some point $z_0$, but unfortunately $z_0 \notin U.$ Can we still say that $F(z_0) = 0$?

I would guess not, since it feels like bending the rules of analytic continuation in a way that shouldn't be allowed. But I didn't think of a counterexample.

The answer is NO.

Example. Let $U$ be the unit disk, and
$$
f_n(z)=(1-z)z^n.
$$
Then $F(z)=\sum f_n(z)=z$, and while $f_n(1)=0$, we have that $F(1)=1$.

No. Let $f_n(z)=\frac{z^n}{n!}$ , $z_0=0$ and $U= \mathbb C \setminus \{0\}$. Then $ \sum_{n=0}^{\infty} f_n(z)$ converges normally on $U$ to $F(z)=e^z$. But $F(z_0)=1 \ne 0$.


Finding the nth derivative of $y=\frac4{(6x+8)^3}$ Find the $n$-th derivative of $y=\frac{4}{(6x+8)^3}$

\begin{align}
y' ={} & 4(-3)(6) \frac{1}{(6x+8)^4}
\\
y''={} & 4(-3)(-4)(6)^2 \frac{1}{(6x+8)^5}
\\
y'''={} & 4(-3)(-4)(-5)(6)^3\frac{1}{(6x+8)^6}
\end{align}
I recognise the pattern but can't interpret that into a formula.

$y=\frac{4}{(6x+8)^3} y'$ so $\frac{\rm d}{\rm dx}(\ln y)=\frac{(6x+8)^3}{4}$.
Now you can find $\ln y$ and hence $y$, and then take the derivatives.


Recursive definition of this sequence I'm having some trouble finding a recursive definition for the following sequence:

$x_0 = \sqrt{1+n}$

$x_1 = \sqrt{1+n\sqrt{1+(n+1)}}$

$x_2 = \sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)}}}$

$x_3 = \sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)}}}}$

and so on (where $n \in \mathbb{N}$). I tried using something like $x_{i+1} = \sqrt{1 + n f(x_i)}$ where $f$ would be a function that sends $n$ to $n+1$ wherever $n$ appears in $x_i$. So for instance $f(x_0) = \sqrt{1 + (n+1)}$ and $f(x_1) = \sqrt{1+(n+1)\sqrt{1+(n+2)}}$. My problem is that I can't figure out an explicit form for $f$. If it's any help, the limit of the sequence is $n+1$.

One has $x_k(n) = f_{k+1}(n)$, where $(f_k)_{k \geqslant 0}$ is the sequence of functions defined recursively as follows:

$f_0(n) = 1$

$f_{k+1}(n) = \sqrt{1+nf_k(n+1)}$

Indeed,

$f_0(n) = 1$

$f_1(n) = \sqrt{1+n}$

$f_2(n) = \sqrt{1+n\sqrt{1+(n+1)}}$

$f_3(n) = \sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)}}}$

etc.

Recall that a sequence $(x_k)$ can be defined recursively if there exists some function $g$ such that $x_{k+1}=g(x_k)$ for every $k$.

Of course, every given strictly increasing sequence, such as the sequences considered in the question, can be defined recursively, using any function $g$ defined on $X=\{x_k\mid k\geqslant0\}$ by $g(x_k)=x_{k+1}$ and, say, by $g(x)=0$ for every $x\notin X$.

This version of the question is rather shallow since every sequence $(x_k)$ can be defined recursively in this sense except if there exists $k<\ell$ such that $x_k=x_\ell$ and $x_{k+1}\ne x_{\ell+1}$, which never happens for strictly monotonous sequences.

A more interesting version would be to ask whether all the sequences $(x_k)$ in the question can be simultaneously defined recursively, that is, whether there exists a common function $g$ for every value of $x_0$, or, equivalently, for every value of $n$.

Then the answer is negative.

To wit, for $n=4$, $x_0=\sqrt5$ and $x_1=\sqrt{1+4\sqrt6}$ while, for $n=2$, $x_1=\sqrt5$ and $x_2=\sqrt{1+2\sqrt{1+3\sqrt5}}$, thus, one would need that $$\sqrt{1+4\sqrt6}=g(\sqrt5)=\sqrt{1+2\sqrt{1+3\sqrt5}}$$ which is absurd.


Power set of Cartesian product I'm trying to see the differences between a Power set of a cartisian product and the cartisian product of two power sets. I have these 2 examples:

$P(\{1, 2\} \times \{3, 4\})$

$P(\{1, 2\}) \times P(\{3, 4\})$

With the first one I can arrive here, but I'm not sure how to continue to compute the power set.

$P(\{(1,3), (1,4), (2,3), (2,4)\})$

The second one I'm stuck on the cartisian product between the to result of the power set.

$\{\{\}, \{1\}, \{2\}, \{1,2\}\} \times \{\{\}, \{3\}, \{4\}, \{3,4\}\} $

How can I keep going from there? And the notation is correct?

For finite sets they are definitely equal. $A \cap \phi = \phi$.


If $x>0$ is such that $x^{n}+\frac{1}{x^n}$ and $x^{n+1}+\frac{1}{x^{n+1}}\in \mathbb{Q} \implies x+\frac{1}{x}\in\mathbb{Q}$? Let $n \in \mathbb{N}$. If $x>0$ is such that $x^{n}+\frac{1}{x^n}$ and $x^{n+1}+\frac{1}{x^{n+1}}\in \mathbb{Q} \implies x+\frac{1}{x}\in\mathbb{Q}$?

Any thoughts on how to solve the above problem. Working for $n=2$ says that this result is true, but not sure if one can generalize

Proffering the following argument based on elementary properties of algebraic numbers.

From $x^n+x^{-n}=q_1$ it follows that $x$ satisfies the polynomial equation $x^{2n}-q_1x^n+1=0$. Furthermore, it is obvious that all the zeros of this polynomial are $x^{\epsilon}\zeta_n^k$, where $\epsilon=\pm1, \zeta_n=e^{2\pi i/n}$ and $k=0,1,2,\ldots,n-1$.

Therefore the zeros of the minimal polynomial $m(T)$ of $x$ (over $\Bbb{Q}$) are among those numbers.

But, from $x^{n+1}+x^{-(n+1)}=q_2$ it similarly follows that the zeros of the minimal polynomial of $x$ are among the numbers $x^{\epsilon}\zeta_{n+1}^\ell, \ell=0,1,2,\ldots,n$.

Therefore the zeros of $m(T)$ are either just $x$, or both $x$ and $x^{-1}$.
In the former case $x$ is rational, and the claim is immediate. In the latter case $x+\dfrac1x$ is rational because it is the coefficient of the linear term of the minimal polynomial $m(T)=(T-x)(T-1/x)\in\Bbb{Q}[x]$.

This argument also proves that if $x^n+x^{-n}$ and
$x^{n+1}+x^{-(n+1)}$ are both integers, then $x+1/x$ must also be an integer. This is because in this case $x$ is an algebraic integer, and hence the coefficients of $m(T)$ are all integers. This old trick then implies that $x^k+x^{-k}\in\Bbb{Z}$ for all $k\in\Bbb{Z}$.

Well, if $x^n\in\mathbb{Q}$, then $\frac{1}{x^n}\in\mathbb{Q}$, so we can ask the question if $x^n\in\mathbb{Q}$ for all $m$ in $n+m$ if $n$ and $n+1$ is rational, and any rational number to the power of a natural number is indeed rational, and no irrational number can be written as both $\sqrt[2]{a/b}$ and $\sqrt[3]{a/b}$, (so for example $\sqrt{2}$ is rational if squared, but not if cubed), so yes, $x+\frac{1}{x}\in\mathbb{Q}$ if $x^n+\frac{1}{x^n}\in\mathbb{Q}$


Is there a formula to calculate the sum of all proper divisors of a number? I don't need to list all proper divisors, I just want to get its sum. Because for a small number, checking all proper divisors and adding them up is not a big deal. However, for a large number, this would run extremely slow. Any idea?

Thanks,
Chan Nguyen

If the prime factorization of $n$ is $$n=\prod_k p_k^{a_k},$$ where the $p_k$ are the distinct prime factors and the $a_k$ are the positive integer exponents, the sum of all the positive integer factors is $$\prod_k\left(\sum_{i=0}^{a_k}p_k^i\right).$$

For example, the sum of all of the factors of $120=2^3\cdot3\cdot5$ is $$(1+2+2^2+2^3)(1+3)(1+5)=15\cdot4\cdot6=360.$$

For proper factors, subtract $n$ from this sum.  This may or may not be faster, depending on the number and how you'd get the prime factorization, but this is the typical technique for high school contest problems of this sort.

The typical brute foce approach in, say, C language:

 public int divisorSum(int n){
    int sum=0;
    for(int i=1; i<= n; i++){
        if(n % i == 0){
            sum +=i;
        }
    }
    return sum;
}


Limit of Rational Expression with Repeating Integer I'm wondering if someone could point me in the right direction for the following problem:

I'm trying to express $(a + aa + \cdots + a\cdots a)/10^n$ as a summation that I can take the limit of as $n \to \infty$ and then use geometric series tools to get that down to something I can work with. However, I'm having quite a bit of trouble posing this as a summation. Should I be trying to express it recursively? It seems like that might get me somewhere. All of the things I've done so far have given me sums that diverge, which is clearly incorrect.

If I did not do any mistake, continuing Alex Becker's answer, the sum can be written as $$S(n)=\frac{a}{9}  \left(\frac{10}{9}- \left(\frac{10}{9}+n\right)10^{-n}\right)$$


Proofs without words of some well-known historical values of $\pi$? Two of the earliest known documented approximations of the value of $\pi$ are $\pi_B=\frac{25}{8}=3.125$ and $\pi_E=\left(\frac{16}{9}\right)^2$, from Babylonian and Egyptian sources respectively. I've read that the Egyptian figure at least could be justified through some geometrical diagram which made the approximation a visually obvious statement about the areas of circles and squares. As far as I know, the Babylonian value on the other hand could have simply been obtained empirically through direct measurement of circle diameters and circumferences; I really have no idea. My question is simply can anyone provide simple visual proofs of these approximations? It doesn't matter to me if the proofs happen to be the historically used ones or not, as long as they get the job done.

Side-note: The Egyptian value pertains to the area-$\pi$, whereas the Babylonian one is about the circumference-$\pi$. As far as anyone knew back in the day, the two constants were not necessarily equal a priori. Bonus points go to answers that can demonstrate both approximations for both pi's.

Here we have a visual proof of the egyptian $\pi$:

And here one of the babylonian $\pi$:

Straight edge and compass construction of a quadrature of a circle is possible only with Babylonian value of pi. Try this link : https://www.academia.edu/8084209/Ancient_Values_of_Pi. Though, Egyptian value of Pi (22/7 or 256/81) is rational, compass and straight edge construction of a quadrature of a circle is not possible.


Why is the one quadratic polynomial a perfect square more often than the other? I was solving problem 137 of Project Euler, which led me to find $n$ such that $5n^2+2n+1$ is a perfect square. But such numbers are very rare (the 13th is around 3 billions) so after decomposing into $(n+1)^2 + (2n)^2 = m^2$ and looking for Pythagorean triples and their $(a^2-b^2,2ab,a^2+b^2)$ generation, I ended up having to look for $k$ such that $5k^2+4$ is a perfect square.

This is a much easier task, which retrospectively makes sense since every $k$ will lead to some $n=O(k^2)$, so you only need to iterate to $10^5$ to find the $13$th number.

Question
So we proved that there are more squares in $5n^2+4$ than in $5n^2+2n+1$.
Was there an easier way to spot this without pulling out the Pythagorean triple trick? Is there an intuitive reason or more generic underlying principle solely by looking at the equations?

I should start by clarifying that both equations yield the same number of squares; both yield countably infinitely many perfect squares. Up to any given upper bound, however, the former equation produces roughly twice as many perfect squares as the latter.

Solving $5n^2+4=m^2$ over the integers is equivalent to solving the equation
$$m^2-5n^2=4,$$
Similarly, solving $5n^2+2n+1=m^2$ over the integers is equivalent to solving the equation
$$(5n+1)^2-5m^2=-4.$$
Intuitively the two equations
$$x^2-5y^2=4\qquad\text{ and }\qquad x^2-5y^2=-4,$$
should have roughly the same number of solutions (up to any given upper bound), and indeed their solution sets are in bijective correspondence through the maps
$$(x,y)\ \rightarrow\ \big(\tfrac{x+5y}{2},\tfrac{x+y}{2}\big)
\qquad\text{ and }\qquad
\big(\tfrac{-x+5y}{2},\tfrac{x-y}{2}\big)\ \leftarrow\ (x,y).$$
But for the second equation, you only get valid solutions to the original equation when $x\equiv1\pmod{5}$. So you might expect the first equation to have about five times as many solutions up to any given upper bound. A more careful analysis shows that you only get about twice as many solutions, though.

This answer comes from the standard method of solving the Pell equation
$$x^2+Dy^2=C,$$
with parameters $D$ and $C$, where $D$ is a squarefree integer. It shows that all solutions to
$$5n^2+4=m^2,$$
are parametrized by
$$m_k+n_k\sqrt{5}=\pm2\left(\frac{3+\sqrt{5}}{2}\right)^k,$$
and similarly that all solutions to
$$5n^2+2n+1=m^2,$$
are parametrized by
$$n_k+m_k\sqrt{5}=\pm(1+\sqrt{5})\left(\frac{7+3\sqrt{5}}{2}\right)^k.$$
In particular the solution sets to both equations are exponential families, with growth factors
$$\frac{3+\sqrt{5}}{2}\qquad\text{ and }\qquad \frac{7+3\sqrt{5}}{2}=\left(\frac{3+\sqrt{5}}{2}\right)^2,$$
so the former has about twice as many solutions as the latter, up to any given upper bound.

It proves that the problem are related, I won't conclude that the latter has more squares

$5*n^2+2*n+1$ compare with $5*n^2+4$

By inspection if $n = 1$, $3$ it makes $5*n^2+4$ a perfect square, but this doesn't mean it has more square...... What about other greater number

Since $5*n^2+4$ came up will trying to tackle $5*n^2+2*n+1$, I conclude it has lesser squares instead

Check the number $5*n^2+4$ well

also trying do this $5*n^2+2*n+1 = (n+m)^2$

$5*n^2+2*n+1 = n^2+2*n*m+m^2$

$4*n^2+2*n*(1-m) = m^2-1$

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

Therefore $m > 2*n$ and $m > w$

Say $2*n*(1-m)+1 = w^2$

$2*n-2*n*m+1 = w^2$

Now say $m = w + 2*n +x $ where $x \in Z$

I hope you understand what am trying to do, it's going to be messy


Diagonalize a symmetric matrix let $$A = \left(\begin{array}{cccc} 1&2&3\\2&3&4\\3&4&5 \end{array}\right)$$

I need to find an invertible matrix $P$ such that $P^tAP$ is a diagonal matrix and it's main diagonal may have only the terms from the set $\{ 1,-1,0 \}$

I'd be glad if you could explain to me how to solve this. I haven't found the right theorem/algorithm.

Thanks.

This trick is due to Hermite. It is especially useful when you have a symmetric matrix of integers. First, we write a certain function in three variables,
$$ f(x,y,z) = x^2 + 3 y^2 + 5 z^2 + 8 yz+ 6 zx +4xy,  $$ because this is exactly the result of calculating $v^t A v,$ with
$$
v =
\left(
\begin{array}{c}
x \\
y \\
z
\end{array}
\right)
$$

In order to clear the terms with $x,$ we write
$$ (x + 2 y + 3 z)^2 =   x^2 + 4 y^2 + 9 z^2 + 12 yz+ 6 zx +4xy.  $$
So far,
$$ f(x,y,z) -  (x + 2 y + 3 z)^2 = -y^2 - 4 z^2 - 4 y z.  $$
Next we clear all $y,$
$$ (y + 2 z)^2 = y^2 + 4 z^2 + 4 y z,  $$ and
$$  f(x,y,z) -  (x + 2 y + 3 z)^2 +  (y + 2 z)^2 = 0,  $$
$$ \color{red}{ f(x,y,z) =  (x + 2 y + 3 z)^2 -  (y + 2 z)^2 }.  $$

The matrix multiplication that this shows is
$$
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 2 & 0
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 0
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & 0 & 0
\end{array}
\right) =
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{array}
\right)
$$
That is actually the right way to do it.

However, I see that someone asked for invertible $P,$ despite non-full rank. Also can be done, and easily:
$$
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 0
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & 0 & 1
\end{array}
\right) =
\left(
\begin{array}{ccc}
1 & 2 & 3 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{array}
\right)
$$
The effect of this is to add $0$ to the function in the shape of $0z^2.$

ADDED: looking at the question again, we do need the extra $1$ in the lower right, because the matrix $P$ requested is actually the inverse of the on I display above. Life goes on.

Hint

Please follow the procedure, outlined here with an example: http://www.sosmath.com/matrix/diagonal/diagonal.html

Update your question with the results of your work and post a comment to this response. I will be happy to check your work for you.


What's the goal of mathematics? Are we just trying to prove every theorem or find theories which lead to a lot of creativity or what?

I've already read G. H. Hardy Apology but I didn't get an answer from it.

There was a book I read at one point, can't remember which, that had the following parable in the introduction (paraphrasing from memory):

"There was once a group of people who wanted to get to a mountain far in the distance and climb it. Between them and the mountain, was a great desert they would have to cross. It was clear that the people had neither the knowledge to cross the desert nor to climb the mountain once they got there. So they began to study and work. At first, everyone worked on a bit of everything: how to carry water; how to avoid sand traps; how to climb safely; etc. Slowly, some people began devoting more of their time to studying the desert, and some to studying mountain climbing. And so the people studying the desert began to develop theories of sand dunes, wind, erosion, and so on; these theories were not really necessary in order to cross the desert, but the people studying the desert found them interesting nonetheless. Those studying mountain climbing were only interested in the desert-studies in so far as it allowed them to get closer to the mountain, and often complained that the desert-study people were wasting time considering theoretical questions that were unnecessary for the purposes of crossing the desert. The desert study people didn't care: they had fallen in love with the desert and were intent on studying it, regardless of whether it helped them all get to the mountain or not."

In the analogy, the mountain was the Theory of Everything from Physics; the Desert was mathematics. There was a time when mathematics' main concern was really to help study the world (help us cross the desert), but it has moved on from that.

This doesn't really answer your question. Perhaps the following won't either: when I was in my early second year in grad school, and looking for an advisor, Hendrik Lenstra told me: "Ask yourself why you do mathematics. If the answer is not 'Because I have to' then quit grad school now." Meaning: unless you would be doing mathematics in your free time, then don't try it as your profession.

It is quite simple. Consider your question. What's the goal of mathematics?

Why Mathematics? This is a simple as the question can get.

Its about questions. How? what? where? when? why? Think about some of the symbols used for those question words. (t)=when f(x)=what/where... And those are high school math symbols.

Mathematics is just a tool to answer questions

Mathematics is probably the best tool at solving some problems.

It does not really get more complicated then that. A tool and a system to solve problems and answer questions. As long as people ask questions, as long as people seek answers there will be maths.


Improvement on the concept of separating families for the union-closed sets conjecture? The union-closed sets conjecture states the following. Let $F$ be a finite family of finite sets that is union-closed and let $\cup (F)$ be the union of all sets in $F$. Then we can find an element in $\cup (F)$ that appears in at least half of the sets in $F$.

One can show that w.l.o.g. we can assume $F$ to be separating, that is for any two elements in $\cup (F)$ we can find a set in $F$ that contains exactly one of the two elements. So in the literature many statements concerning the conjecture are proven for separating families.

However we can do better. Say a family $F$ is given. We now chose an element in $\cup (F)$ and remove it from every set in $F$ if the number of sets doesn't decrease by doing so. We repeat this process as long as we can. After that we obtain a family $\tilde{F}$ with the following property. For every $x \in \cup(\tilde{F})$ we can find $S \in \tilde{F}$ such that $x \in S$ and $S\backslash \{x\} \in \tilde{F}$. We call families with this property minimal. It is easy to see that if $\tilde{F}$ fulfills the conjecture then so does $F$. Thus w.l.o.g. we can assume $F$ to be minimal.

Now we make the observation that being separating is strictly weaker than being minimal. Say $F$ is a minimal family and let $x, y \in \cup (F)$. Then we can find $S \in \cup (F)$ such that $x \in S$ and $S\backslash \{x\} \in F$. If $y \notin S$ then $S$ contains exactly $x$ and if $y \in S$ then $S\backslash \{x\}$ contains exactly $y$. Thus $F$ is separating. Now consider the family $\{\emptyset, \{1,2\},\{2,3\},\{1,2,3\} \}$. One can easily check that it is separating but not minimal.

This leaves me with the following questions:

Is there any literature about the union-closed sets conjecture
that introduces a concept like minimal families?

If not, is an inductive proof for the union-closed sets conjecture plausible
abusing minimal families (reduce the number of sets in a minimal family by removing an arbitrary element from all sets)?

Edit: Here is a follow up.

If I am understanding your idea correctly, I believe you are saying that you can suppose your family doesn't have sets of the form $S$ and $S\cup \{a\}$ for some $a\in U(\mathcal{F})$

You can check Lo Faro's paper "A note on the union-closed sets conjecture" and see that a counterexample must have some sets of that form. Otherwise you wouldn't have a minimal one.


Show that $g(x)=[\frac1x]\sin x$ has a limit in $x=0$. Show that $g(x)=\left[\frac1x \right]\sin x$ has a limit in $x=0$.

( $[1/x]$ as greatest integer less than or equal to $1/x$)

I tried to use squeeze theory to find the limit of this function, but I can’t find two another functions which have obvious limits.
I know how to evaluate a limit when we have 0*bounded function, but in this case, the function isn’t bounded .

I know :
$$\frac{1}{x}-1\le\left[\frac{1}{x}\right]\le\frac{1}{x},x\not=0.$$

And I know :

$$\lim_{x\rightarrow 0} \frac{1}{x}\sin(x)=1$$

I just don’t know how to use this stuffs to solve the question.

Thanks in advance fo the help .

If you mean $[1/x]$ as greatest integer less than or equal to $1/x$ then for sufficiently small $x>0$ we have $$\left(\frac{1}{x}-1\right)\sin(x)≤[\frac{1}{x}]\sin(x)≤\frac{1}{x} \sin(x)$$ since $$\frac{1}{x}-1≤[\frac{1}{x}]≤\frac{1}{x},x\not=0.$$ Also for sufficiently small $x<0$ we have $$\left(\frac1x-1\right)\sin(x)≥[\frac{1}{x}]\sin(x)≥\frac{1}{x} \sin(x).$$ Now $$\lim_{x\rightarrow 0} \frac{1}{x}\sin(x)=1$$ and $$\lim_{x\rightarrow0}\sin(x)=0.$$ Hence $$\lim_{x\rightarrow 0}[\frac1x]\sin(x)=1.$$

The function $\sin$ is positive for small positive numbers and negative for small negative numbers.

HINT

You are mostly done, indeed since by floor function definition we have

$$\frac1x-1\le\left[\frac1x\right]\le \frac1x$$

then assuming wlog $0<x<1$

$$ \left(\frac1x-1\right)\cdot\sin x\le\left[\frac1x\right]\cdot\sin x\le \frac1x\cdot\sin x$$

and then use squeeze theorem.


algebra related question i meet  following problem and want to clarify   it's statement.problem is following

to reproduce an old photograph,a photographer   charges x dollars to make a negative.$3x/5$ dollars for each of first $10$ prints and $x/5$ dollars for each print in excess of $10$ prints.if $45$ is the total charge to make a negative  and 20 prints from  an old photograph,what is  the value of $x$?
I don't understand one thing,   if  it takes  $3x/5$ dollar for each of first $10$ prints,then it will take $10*3*x/5$  or $6*x$ dollar right? but what does mean in excess of $10$?if it is additional  printing,   and because we have $10+10=20$ and also total number of   charge to make negative,we may would  have $6*x+2*x=45$  or $8x=45$  so $x=5.625$ but in answers list there is only  value $5$,maybe i misunderstand problem,so please help

You only missed the negative, what costs $x$. So we have
$$x+6x+2x=45$$

The mistake was that you missed the negative one i.e. $8x+x=45 \Rightarrow 9x=45 \Rightarrow x=5$.


What does this free quotient space look like? Let $S^2=\{(x,y,z)\in \mathbb{R}^3|x^2+y^2+z^2=1\}$ and $S^1=\{(s,t)\in \mathbb{R}^2|s^2+t^2=1\}$. Suppose that $\mathbb{Z}/2\mathbb{Z}$ acts on $S^2\times S^1$ in such a way that the generator of $\mathbb{Z}/2\mathbb{Z}$ maps
$$
(x,y,z)\mapsto(-x,-y,-z) \ \ \ and \ \ \ (s,t)\mapsto (-s,t)
$$
(Note that $t$ is invariant). I would like to know what the free quotient space $(S^2\times S^1)/\mathbb{Z}/2\mathbb{Z}$ looks like. Since the action preserves orientation of $S^2\times S^1$, it should be oriented 3-manifold.

$S^2/\mathbb{Z}_2 \simeq \mathbb{R}P^2$ this is the definition of projective space.

$S^1/\mathbb{Z}_2 \simeq [0,1]$.  You fold the circle, it is homeomorphic to a line segment.

Your resulting space is $(S^2 \times S^1)/\mathbb{Z}_2 \simeq \mathbb{R}P^2 \times [0,1]$.

This 3-manifold is not oriented since $\vec{x} \mapsto -\vec{x}$ preserves orientation of $S^2$.


Cut up a cube into pieces that form 3 regular tetrahedra? Everyone knows that a regular tetrahedron fits inside a cube, and that the volume of the tetrahedron is 1/3 that of the cube.

(For a picture, see this question or this Google image search.)

The question is, once you've cut the corners off the cube to make that tetrahedron, what's a simple way to cut up those four corner pieces so that their parts can be rearranged to form two additional regular tetrahedra?

You can't do it.

This is because of
Hilbert's third problem:
https://en.wikipedia.org/wiki/Hilbert%27s_third_problem

(At least, I think this is true.)

How are you defining "simple"?  Anyone who has taken calculus knows that the pieces can be atomized into miniscule microcubes, and these can be reassembled into any shape having the same volume.  I suggest your problem statement be tweaked to exclude Newtonian atomization.  Conceputally, atomization is simple...it is the basis for plastics molding and metal castings.


Dual cone and polar cone Dual cone and polar cone http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone are defined only on $\mathbb R^n$. Has anyone seen the extension to $\mathbb C^n$? Any references for these?

As I see it, the definition in Wikipedia is for any Linear Space. The book by Aliprantis and Border defines polars for any Dual Set, which is even more general. These general definitions are very useful in Convex Optimization, but I've never seen it used on $\mathbb{C}^n$.


Why is "$Ax=b$ is solvable $\iff AA^\mathsf{T}y=b$ is solvable" true? We have a random matrix A ($m\times n$), a vector $x$ from $\mathbb{R^{n}}$ and vectors $y$ and $b$ from $\mathbb{R^{m}}$.

Why is it true that $Ax=b$ has a unique solution $\iff$ $ A A^\mathsf{T} y=b$ has a unique solution?

Is it reasonable to approach this from a column space point of view? If $Ax=b$ has a unique solution then $b$ is in the column space of $A$, $b\in C(A)$. How can it be proved that the column spaces of $A$ and $AA^\top x$ are the same?

Indeed, this is not true as stated. Take $m = 2$, $n=1$, $A = b = \begin{pmatrix}1 \\ 0 \end{pmatrix}$.

Then
$$Ax = b$$
is the same as
$$\begin{pmatrix}1 \\ 0 \end{pmatrix} x = \begin{pmatrix}1 \\ 0 \end{pmatrix}$$
which has the unique solution $x=1$.

On the other hand, $AA^T y = b$ is the same as
$$\begin{pmatrix}1 & 0 \\ 0 & 0 \\ \end{pmatrix} y = \begin{pmatrix}1 \\ 0 \end{pmatrix}$$
which has multiple solutions, e.g. $y = \begin{pmatrix}1 \\ 0\end{pmatrix}$ or $y = \begin{pmatrix}1 \\ 1\end{pmatrix}$.


Finding coefficients in expansions Show that the coefficient of $x^{−12}$ in the expansion of
$$\left(x^4−\frac{1}{x^2}\right)^5\left(x−\frac{1}{x}\right)^6$$
is $−15$, and calculate the coefficient of $x^2$. Hence, or otherwise, calculate the coefficients of $x^4$ and $x^{38}$ in the expansion of $$(x^2−1)^{11}(x^4+x^2+1)^5.$$
The first part of this exercise is solved easily considering "partitions", i.e. what can be added to get the exponents and turns out to be $-15$ and $215$ respectively. I was able to solve the second part of this exercise easily using multinomial theorem. However, this exercise is given before multinomial theorem is introduced, and I would really like to know what the "Hence" was gunning for.  Once I factor out to get$$x^{21}\left(x-\frac{1}{x}\right)^{11}\left(x^2+\frac{1}{x^2}+1\right)^5,$$I can see similarities with the first question. I can even deduce that since $4-21=-17$, $38-21=17$, and inspecting signs, that the coefficients are equal in magnitude and opposite in sign. Please explain how I can use the first part of the question to proceed, rather than going "otherwise"/multinomial theorem.

\begin{align*}
(x^2 - 1)^{11}(x^4 + x^2 + 1)^5
&=(x^2 - 1)^{5}(x^4 + x^2 + 1)^5(x^2 - 1)^{6} \\
&=(x^6 - 1)^5(x^2 - 1)^6 \\
&=x^{16}\left(x^4 - \frac{1}{x^2}\right)^5\left(x - \frac{1}{x}\right)^6
\end{align*}
Thus, the coefficient of $x^4$ in the new equation is the same as the coefficient of $x^{-12}$ in the original equation, which is $-15$.
And like you mentioned, the coefficient of $x^4$ and $x^{42 - 4} = x^{38}$ should be the same but opposite in sign. So the coefficient of $x^{38}$ is $15$.

In
$$(x^4-\frac{1}{x^2})^5(x-\frac{1}{x})^6=\frac{(x^6-1)^5(x^2-1)^6}{x^{16}}=\frac{(y^3-1)^5(y-1)^6}{y^8}
$$
notice that every $x$ is in power of 2, so let $y=x^2$ then create the polynomial
$$
P(y)=(y^3-1)^5(y-1)^6=(y^2+y+1)^5(y-11)^{11}
$$
to which you try to find coefficients to $y^2=x^4$ and $y^9=x^{18}$. It was too tedious to expand by hand, so using the expansion calculator https://www.dcode.fr/math-expression-expand I got quite large numbers.


Translation from English to Predicate Logic I am studying Predicate Logic now.
Could you please help me how to translate a sentence into predicate logic?

How can I translate the following sentence into predicate logic?

"No student read exactly one book."

My translation is...
∃x(student(x)∧read(x, x1)∧book(x1)∧book(y)∧¬(x1=y))

I've already translated this but I am not sure about the part "exactly one" in the sentence..

Is this correct? If there is something wrong, could you please tell me what the mistakes are?

Thank you for your help.
I really appreciate it :>

Your symbolization

  ∃x(student(x)∧read(x, x1)∧book(x1)∧book(y)∧¬(x1=y))

has a few problems. First of all, there are no quantifiers for $x1$ and $y$. I assume you want those to be existentials, so let's add them:

∃x∃x1∃y(student(x)∧read(x, x1)∧book(x1)∧book(y)∧¬(x1=y))

OK, but that is still not right: this statement says that there is some student who has read two or more books (and thus not exactly one book), but the statement says that there is no student who has read exactly one book, which is the same as saying that it is true for all students that they have not read exactly one book.

OK, so maybe we should change the existential into a universal? Let's see:

$\forall x∃x1∃y(student(x)∧read(x, x1)∧book(x1)∧book(y)∧¬(x1=y))$

No, that is still not right, because a student not reading exactly one book does not mean they have to read 2 or more books ... they could also read no books at all. So ... we should give every student the option of either reading no books, or two or more books:

$\forall x (student(x)\rightarrow (\neg \exists x1 (book(x)∧ read(x,x1)) \lor ∃x1∃y(read(x, x1)∧ book(x1)∧book(y)∧read(x,y)∧¬(x1=y)))$

(oh yeah, you also forgot to include the $read(x,y)$ statement)

OK, here is another way of doing this one.  The statement effectively denies that there is a student that reads exactly one book.

So, we can use as a basic setup:

$$\neg \exists x (Student(x) \land 'x-reads-exactly-one-book')$$

Now, how can we symbolize that $x$ reads exactly one book?  To do that, we can say that there is a book that $x$ reads, and that all the books that $x$ reads are that one book. So, '$x$ reads exactly one book' becomes:

$$\exists y (Book(y) \land Reads(x,y) \land \forall z ((Book(z) \land Reads(x,z)) \rightarrow z=y))$$

OK, so fill this in the larger statement, and we get:

$$\neg \exists x (Student(x) \land \exists y (Book(y) \land Reads(x,y) \land \forall z ((Book(z) \land Reads(x,z)) \rightarrow z=y)))$$

$\forall x \forall y \exists z ((S(x) \wedge B(y) \wedge R(x,y)) \rightarrow (B(z) \wedge R(x, z) \wedge y \neq z))$

Bring the quantifiers to the front. ("If any student has read any book, then something else is also a book read by that student".)


Can someone clearly explain about the lim sup and lim inf? Can some explain the lim sup and lim inf?
In my text book the definition of these two is this.

Let $(s_n)$ be a sequence in $\mathbb{R}$. We define
$$\lim \sup\ s_n = \lim_{N \rightarrow \infty} \sup\{s_n:n>N\}$$
and
$$\lim\inf\ s_n = \lim_{N\rightarrow \infty}\inf\{s_n:n>N\}$$

The right side of these two equality, can I think $\sup\{s_n:n>N\}$ and $\inf\{s_n:n>N\}$ as a sequence after $n>N$?
And how these two behave as $ n$ increases? My professor said that these two get smaller as $n$ increases.

Consider this example:
$$
3-\frac12,\quad 5+\frac13,\quad 3-\frac14,\quad 5+\frac15,\quad 3-\frac16,\quad 5+\frac17,\quad 3-\frac18,\quad 5+\frac19,\quad\ldots\ldots
$$
It alternates between something approaching $3$ from below and something approaching $5$ from above.  The lim inf is $3$ and the lim sup is $5$.

The inf of the whole sequence is $3-\frac12$.

If you throw away the first term or the first two terms, the inf of what's left is $3-\frac14$.

If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac16$.

If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac18$.

If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac1{10}$.

. . . and so on.  You see that these infs are getting bigger.

If you look at the sequence of infs, their sup is $3$.

Thus the lim inf is the sup of the sequence of infs of all tail-ends of the sequence.  In mathematical notation,
$$
\begin{align}
\liminf_{n\to\infty} a_n & = \sup_{n=1,2,3,\ldots} \inf_{m=n,n+1,n+2,\ldots} a_m \\[12pt]
& = \sup_{n=1,2,3,\ldots} \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} : n=1,2,3,\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\{ a_m : m\ge n\} : n=1,2,3,\ldots \right\}.
\end{align}
$$

Just as the lim inf is a sup of infs, so the lim sup is and inf of sups.

One can also say that $L=\liminf\limits_{n\to\infty} a_n$ precisely if for all $\varepsilon>0$, no matter how small, there exists an index $N$ so large that for all $n\ge N$, $a_n>L-\varepsilon$, and $L$ is the largest number for which this holds.

Think of it this way. In the $\limsup$ , you are taking the biggest value past a certain $N$. As $N$ increases, there are "less and less" value to choose from, hence the $\limsup$ can only decrease (or stay constant).

Same thing applies with $\liminf$ except that as you get "less and less value" you can only increase (or keep it the same) the value of your $\liminf$.

As a simple example, take a sequence to be
$$
s_n=(4,-4,3,-3,2,-2,1,-1,0,0,\ldots)
$$
Fix $N=1$ then the largest value past or at $N=1$ is $4$ and the smallest is $-4$. A few steps later, say $N=4$, the largest value past or at $N$ is $2$ and the smallest is now $-3$. Further away, at $N=10$ we have $\sup_{n\ge 10}=\inf_{n\ge 10}=0$.

From this you see that $\limsup$ decreases and $\liminf$ increases.

Exercise: What needs to happen for the sequence to converge?


What is the difference between a sound argument and a valid argument? In my notes, these are the definitions of a valid argument

  An argument form is valid if and only if whenever the premises are all true, then
  conclusion is true. An argument is valid if its argument form is valid.

For a sound argument,

  An argument is sound if and only if it is valid and all its premises are true.

Okay so to me, both definitions pretty much says the same thing to me. On a philosophy forum, I see that they distinguish the two by saying a valid argument is such that the truth value of the premises necessarily imply the truth values of the conclusion.

For example, the "Elimination" method say

$p \vee q$

$\sim q$

$\therefore p$

So the premises are $p \vee q$ and $\sim q$

Now if I were to substitute $p$ and $q$ for $p$ := "Jesse is my husband" and q:= "I am Jesse's wife" (assume p is true and q is true)

Then we have

Either "Jesse is my husband" or "I am Jesse's wife"

"I am not Jesse's wife"

Therefore, "Jesse is my husband"

So is this technically still valid or sound? (can't tell the difference) Both premises are true, but the conclusion is false? It should invalid right? Yet the method of elimination is said to be valid?

A sound argument is necessarily valid, but a valid argument need not be sound. The argument form that derives every $A$ is a $C$ from the premises every $A$ is a $B$ and every $B$ is a $C$, is valid, so every instance of it is a valid argument. Now take $A$ to be prime number, $B$ to be multiple of $4$, and $C$ to be even number. The argument is:

  If every prime number is a multiple of $4$, and every multiple of $4$ is an even number, then every prime number is even.

This argument is valid: it’s an instance of the valid argument form given above. It is not sound, however, because the first premise is false.

Your example is not a sound argument: $q$ is true, so the premise $\sim q$ is false. It is a valid argument, however, because for any $p$ and $q$, if $p\lor q$ and $\sim q$ are both true, then $p$ must indeed be true.

Note that an unsound argument may have a true or a false conclusion. Your unsound argument has a true conclusion, $p$ (Jesse is my husband); mine above has a false conclusion (every prime number is even).

Perhaps a valid argument is one that causes the listener to agree to some form of truth about a subject. A sound argument causes the listener to admit that there is more than one truth about the subject. It has the ability to raise the question of doubt. The difference lies in the listeners interpretation of the information provided. Arguments may be sound, but not necessarily valid. And valid arguments are not always sound. A good example of this is our judicial system....defense attorneys argument, prosecution attorneys argument and the jury or judges decision on the matter.


Zero state/input vs homogenous/particular solutions Solution of a differential equation can be written as a sum of its homogenous and particular part:
$$y = y_h + y_p,$$
or as a sum of zero-state and zero-input solutions:
$$y = y_{ZI} + y_{ZS},$$
where zero-input solution $y_{ZI}$ is the homogenous solution $y_h$ with initial conditions applied, and zero-state solution $y_{ZS}$ is the "full solution" $y_h + y_p$, but with the initial conditions set to zero.

So,
$$y = y_h + y_p = y_{ZI} + y_{ZS}.$$

Although it is intuitive, and for $t=0$ we can see the equivalence, I would like to see some real proof that this zero-state/zero-input combination truly holds for every $t$, e.g. is always equal to the "default" solution $y_h + y_p$.

Can someone help me with this?

This idea is really Duhamels principle, let us apply it to the general first order ODE:

$y'(t)+ay(t)=f(t)$, $y(0)=g$

Multiply by the integrating factor $e^{at}$ we get:

$\frac{d}{dt}(y(t)e^{at})=f(t)e^{at}$, integrating with respect to $t$ we get:

$y(t)e^{at}-y(0)=\int_0^tf(\tau)e^{a\tau}d\tau$

$\Rightarrow y(t)=y(0)e^{-at}+e^{-at}\int_0^tf(\tau)e^{a\tau}d\tau$

$=y(0)e^{-at}+\int_0^tf(\tau)e^{-a(t-\tau)}d\tau=y_c+y_p$,

it is now very clear that $y_c(0)=y(0)$, and $y_p(0)=0$

Now let $S(t)=e^{-at}$ and we get:

$y(t)=y(0)S(t)+\int_0^tf(\tau)S(t-\tau)d\tau=y(0)S(t)+f*S$. (Where $*$ denotes the convolution).

Also this still holds for non constant coefficients, I.e. consider:

$y'(t)+a(t)y(t)=f(t)$, $y(0)=g$ here we take $S(t)=e^{-\int_0^ta(\tau)d\tau}$, and we obtain:

$y(t)=y(0)S(t)+f*S$

ok.... the theory goes like this.....
Classical solution to the constant coefficient Differential equations D(y) + y = X
is the sum of both homogenous (complementary) solution and Particular solution.
ie y(t) = y(h) + Y(p)

whereas, when we are considering LTI systems, the solution to the Linear system is defined by the sum of its total responses ie the zero input response and zero-state response....
y(t) = y(ZI) + y(ZR)

Also, now remember that in case of linear systems, there are natural and forced responses that are often mistaken to be zero input or zero state response wherein the natural response is the solution of the linear systems that are connected to the characteristic modes of the system.

generally when we are considering to understand the behaviour of an LTI systems the zero input and zero state response method is best because it gives us distinct information on the internal conditions and  external inputs whereas, in case of the classical solution it does not give us the advantage of defining the system response to an input x(t) as an explicit function of x(t).


Can you provide us a good approximation for $\sum_{n=1}^{\infty} \left| \log \left( 1+\frac{\mu(n)}{n^2} \right) \right|$? Let $a_n=\frac{\mu(n)}{n^2}$, where $\mu(n)$ is the Möbius function. Since $\sum \left| a_n \right| $ is convergent by the comparison test, then a proposition from analysis ensures that $$\mathcal{S}=\sum_{n=1}^{\infty} \left| \log \left( 1+\frac{\mu(n)}{n^2} \right) \right| $$
is also convergent.

  Question. Can you give a good approximation or get in a closed-form $S$? If you use computational methods (this is with a computer), please explain us how you are ensuring a good approximation. My goal is learn if my claims are rights and learn or refresh more easy facts from your answer. Thanks in advance.

One has that $\mathcal{S}$ seems that is around $1.256$. I don't know if the following could be useful facts to compute it:

1) Combining the Prime Number Theorem that claims that $\sum_{n=1}^\infty\frac{\mu(n)}{n}$ converges with sum $0$, with the proposition showing that if one has $b_n=\frac{\mu(n)}{n}$ with $\sum b_n^2$ convergent, by the comparison test it is, then using a proposition of analysis we can claim that $\prod_{n=1}^\infty  \left( 1+ \mu(n)/n\right) $ is also convergent.

2) $S$ can be written as the logarithm of a quotient of products, from $$\log 2-\sum_{n\in A}\log \left( 1-\frac{1}{n^2} \right) +\sum_{m\in B}\log \left( 1+\frac{1}{m^2} \right),$$
where $A$ is the set of square-free integers $n$ with an odd number of distinct prime factors, and $B$ is the set of square-free integers $m$ with an even number of distinct prime factors.

3) And we know the approximation $\log(1+x)  \approx x $, that we can use in each summand, for $n\geq 2,$ of the series defining $\mathcal{S}$, and the value for $1/\zeta(2)$. Is it possible combine these facts in 3)?

A trivial approximation can be the following. Since $\left|\frac{\mu\left(n\right)}{n^{2}}\right|<1
 $ if $n\geq2
 $ we have $$\sum_{n\geq1}\left|\log\left(1+\frac{\mu\left(n\right)}{n^{2}}\right)\right|=\log\left(2\right)+\sum_{n\geq2}\left|\sum_{m\geq1}\frac{\left(-1\right)^{m+1}\mu^{m}\left(n\right)}{n^{2m}}\right|\leq\log\left(2\right)+\sum_{n\geq2}\sum_{m\geq1}\frac{1}{n^{2m}}
 $$ $$=\log\left(2\right)+\sum_{n\geq2}\frac{1}{n^{2}-1}=\log\left(2\right)+\frac{3}{4}\approx1.4431
 $$ since the last series telescopes. I don't know if it is a good approximation for you. If it is not please tell me and I will delete the answer.

So I can not add anything using any sort of logical approximation, but I happen to have a list of the first 10000 terms of the möbius function, which I calculated some time ago, so I used said list and came up with these
 results:
$$$$1) The Series S(n) = $sum_{k=1}^n |ln(1+µ(k)/k^2)|$ converges logarithmically
$$$$3) S(10000) ~ 1.25789869841338


How to find a general sum formula for the series: 5+55+555+5555+.....? I have a question about finding the sum formula of n-th terms.

Here's the series:

$5+55+555+5555$+......

What is the general formula to find the sum of n-th terms?

My attempts:

I think I need to separate 5 from this series such that:

$5(1+11+111+1111+....)$

Then, I think I need to make the statement in the parentheses into a easier sum:

$5(1+(10+1)+(100+10+1)+(1000+100+10+1)+.....)$

= $5(1*n+10*(n-1)+100*(n-2)+1000*(n-3)+....)$

Until the last statement, I don't know how to go further. Is there any ideas to find the general solution from this series?

Thanks

The first order difference of the series is $\Delta_n^1=55,555,5555,55555...$.

The second order difference is $\Delta_n^2=500,5000,50000...=50.10^n$.

As the finite difference of a power is the same power, we have (adding an affine term that vanishes with the second order difference)

$$S_n=a.10^n+bn+c,$$
$$\Delta_n^1=a.9.10^n+b,$$
$$\Delta_n^2=a.81.10^n.$$

Compute the unknowns from $\Delta_1^2=500, \Delta_1^1=55, S_1=5$ to get
$$S_n=\frac{50}{81}.10^n-\frac59n-\frac{50}{81}.$$


If every infinte subset of E has a limit point in E then E is compact in X Let X be a metric space. Show that a subset E of X is compact if and
only if every infinite subset of E has a limit point in E.
How to prove the converse statement ?
i.e. if every infinite subset of E has a limit point in E then E is compact in X.
I want to prove from the definition of compact sets.

First note that by this answer we know that if $X$ limit point compact (i.e. every infinite set has a limit point), then $X$ is Lindelöf (i.e. every open cover has a countable subcover), otherwise $X$ would have a closed and discrete uncountable subspace, contradicting the limit point compactness.

Then as metric spaces are $T_1$, this answer shows that a limit point compact metric $X$ is also countably compact (every countable cover has a finite subcover).

And clearly then $X$ is compact (we reduce an open cover to a countable one via Lindelöfness, and then to a finite one using countable compactness). Also see this answer for an overview of similar ideas.


Finding the pattern to a general matrix Consider the following $(n+1) \times (n+1)$ matrix:

$$\begin{bmatrix} n & 1 & 0 & 0&\cdots &0& 0 & 0 & 0 \\ -n & n-2 & 2 &0& \cdots &0& 0 & 0 & 0 \\ 0 & 1-n & n-4 &3& \cdots &0& 0 & 0 & 0 \\ 0 & 0 & 2-n &n-6& \cdots &0& 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots & 6-n & n-2 & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots &-3& 4-n & n-1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & -2 & 2-n & n \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 & -1 & -n \\\end{bmatrix}$$

Show the eigenvalue is $0$ with multiplicity $n+1$.

I have tried row reducing to no avail, it is doable for any particular instance but surely there is a simpler way to find the determinant with a more connected pattern across all integers $n$.

For clarity, the problem is equivalent to showing:

$\begin{bmatrix}x+ n & 1 & 0 & 0&\cdots &0& 0 & 0 & 0 \\ -n &x+ n-2 & 2 &0& \cdots &0& 0 & 0 & 0 \\ 0 & 1-n &x+ n-4 &3& \cdots &0& 0 & 0 & 0 \\ 0 & 0 & 2-n &x+n-6& \cdots &0& 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots &x+ 6-n & n-2 & 0 & 0 \\ 0 & 0 & 0 & 0 & \cdots &-3&x+ 4-n & n-1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & -2 &x+ 2-n & n \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0 & -1 &x -n \\\end{bmatrix} = x^{n+1}$

Add row 1 to row 2
and row 2 is
[0, n-1, 2, 0, ...].

Then add row 2 to row 3
and row 3 is
[0, 0, n-2, 3, 0, ...].

Then add row 3 to row 4
and row 4 is
[0, 0, 0, n-3, 4, 0, ...].

I think that,
with a proper specification
of each row,
this can reduce the matrix
to upper triangular
with the main diagonal being
[n, n-1, n-2, ...]
and the diagonal above it
being
[1, 2, 3, ...],
and the rest of the matrix being zero.

This might be enough
to show what is wanted.


right continuous implies countably additive Let $F:\mathbb{R} \rightarrow \mathbb{R} $ be a monotonically increasing function. Let $\mu_F$ be defined on the half open real intervals such that $\mu_F((a,b]) = F(b) - F(a)$. Then $\mu_F$ is countably additive if $F$ is right-continuous. (defined here (screenshot), definition has a bigger domain but I think restricting it here still works).

But why is this? I've proved finite additivity. Also tried to use the continuity of probability measures, but I couldn't get that proof to work here since the half open intervals isn't a $\sigma$-algebra.

Here's a proof idea which I got working in the end, use the proof for countable additivity of the length function for intervals (http://www.maths.tcd.ie/~richardt/MA2224/MA2224-ch2.pdf Proposition 2.1.12), and where it uses $\epsilon/2^n$, we take a $\delta_i$ s.t. $0 < t - d_i <= \delta_i$ we have that $F(t) - F(d_i) < \epsilon/2^i$. Nothing else changes much.

The key is the Heine Borel theorem, it reduces the countable case to a finite case, which is easy to handle.