How to prove the expressiveness of first-order logic formulas with equality over the empty signature? How can one prove that every first-order-logic formula with equality over the empty signature is equivalent to either False, or "there are exactly n elements in the domain", or "there are at least n elements in the domain"?

I assume you mean a formula with no parameters. In this case take any structure in this language = a set with no additional structure. Now look at all permutations of this structure = all permutations of the set. In this way any subset of size $n$ can be mapped to any other subset of order $n$. That is all finite sets of the same size are conjugate under automorphisms, so no formula can distinguish between one set of size $n$ and another set of size $n$. Thus the formulas you mention are the only possible ones.


Is this second order differential operator Hermitian (2 variables)? Let

$ T= \frac{ \partial^{2}}{\partial _{x} \partial _{y}}+iay\frac{ \partial}{\partial _{y}}+i(1-a)x \frac{ \partial}{\partial _{x}}+ \frac{i}{2}$

be the second order differential operator, where $ i =\sqrt{-1} $ and $ a $ is a real parameter.

Can we prove that this Hamiltonian is Hermitian (so $ T = T^{+} $)?

It is symmetric on $L^2({\mathbb R}^2)$, because $D_x = i \partial/\partial x$ and $D_y = i \partial/\partial y$ are symmetric, and this is $(S + S^+)/2$ where $S = - D_x D_y + a y D_y + (1-a) x D_x$. Self-adjointness might be trickier.


How to prove continuity of the retraction map in proof of Brouwer's fixed point theorem? I was reading Allen Hatcher's proof of the Brouwer's fixed point theorem using the no retraction theorem. That is, we want to prove that every continuous $f: \mathbb{D}^n \to \mathbb{D}^n$ has a fixed point $f(x) = x$, by contradiction.

The no retraction theorem says there is no continuous $r: \mathbb{S}^{n-1} \to \mathbb{D}^n$ such that $r \mid_{\mathbb{S}^{n-1}}(y) = y$. i.e. $r$ restricted to $\mathbb{S}^{n-1}$ is the identity functon.

Suppose a continuous $f$ exists with no fixed point. Then, define $r(x)$ by the point on the line $tx + (1-t)f(x)$ intersecting $\mathbb{S}^{n-1}$. The book, and lecture notes that I looked up, say this is clearly continuous.

Question: How do we proof that $r(x)$ is continuous?

You could try the following: Take the function
$$g(t,x) = |tx+(1-t)f(x)|$$
and apply the implicit function theorem to get a continuous $t(x)$ with
$$g(t(x),x) = 1.$$
Then
$$r(x) = t(x) x + (1-t(x))f(x)$$
clearly is continuous since it is constructed from continuous functions.

One way is by showing that : i ) Multiplication is a continuous function  ii)so is sum, i.e., they are both continuous functions $\mathbb R \times \mathbb R \rightarrow \mathbb R$. And iii) The sum of continuous functions is continuous. Then $r(t)$ is a sum of the continuous functions $tx$ and $(1-t)f(x)$. If you want to make sure that the restriction to $S^{n-1}$ is continuous, you can either just say that a continuous function in a space restricts to a continuous function in a subspace, or you can divide $r(t)$ by the (continuous )norm function.


prove that $u$ is equal a.e. to an absolutely continuous function Prove that if $n=1$ and $u\in W^{1,p}(0,1) $ for some $1\leq p<\infty$, then $u$ is equal a.e. to an absolutely continuous function,and $u'$ (which exists a.e.) belongs to $L^{p}(0,1)$.

My idea:due to $(0,1)$ is a bounded and we can think that $u$ has a compact support,if $u\in W^{1,p}((0,1))$,then the $u^{\varepsilon}=\eta_{\varepsilon}*u$ is a smooth function with
$ u^{\varepsilon}\rightrightarrows u $ when $\varepsilon\to 0^{+}$,and $\|Du^{\varepsilon}\|_{p}\leq \|Du\|_{p} $,then for $x,y\in (0,1),x\neq y$, we have
\begin{align*}
 u^{\varepsilon}(x)-u^{\varepsilon}(y)&= \int_{0}^{1} (x-y)\cdot (Du^{\varepsilon}(tx+(1-t)y))dt
\end{align*}
then we have
$$ |u^{\varepsilon}(x)-u^{\varepsilon}(y)|\leq \|Du^{\varepsilon}\|_{p}|x-y|\leq \|Du\|_{p}|x-y| $$
Let $\varepsilon\to 0$,we get
$$ |u(x)-u(y)|\leq \|Du\|_{p}|x-y| $$

Another approach without convolutions:

Since $Du$ is an $L^p(0,1)$ function, hence $L^1(0,1)$ in particular, for each $x \in (0,1)$ the integral $w(x) = \int_0^x Du(t)\,dt$ makes sense.  By the fundamental theorem of calculus for Lebesgue integrals, $w$ is absolutely continuous and $w' = Du$ almost everywhere.  Intuitively, we should expect that $w$ differs from $u$ by a constant.  To find the constant, set $c = \int_0^1 (u(t) - w(t))\,dt$.  Then set $v = w+c$, which is again absolutely continuous and has $v' = Du$ almost everywhere.  Now for any test function $\varphi \in C^\infty_c((0,1))$, use the definition of weak derivative and integration by parts to show that $\int_0^1 (u(t)-v(t)) \varphi(t)\,dt = 0$.  Conclude that $u-v = 0$ almost everywhere.

Yes, this is one way to do it. I would suggest you to read this book, chapter 8 for a complete introduction on 1 dimensional Sobolev space.

Also, this note collects more results, but without proof, of Sobolev function properties in one dimensions.


Evaluate the integral $\int \frac {1}{x^4 +1} dx$( finding the constants in the partial fraction method). Evaluate the integral $\int \frac {1}{x^4 +1} dx$.

This question is answered here :

Evaluate the following indefinite integral. $\int\frac{1}{x^4+1}\, dx$

But I do not know how they found the constants in the partial fraction method, could anyone explain this for me please?

If you are doing the integral from $0$ to $\infty$ it's worth noting the following. Consider the integral $$\frac{1}{2\pi i} \int \frac{\log z}{1+z^n} dz, \qquad n=4$$ around the classic keyhole contour, that is to say, start at 0 just above the real axis, go along the real axis to $\infty$, then anticlockwise in a circle until you get to $\infty$ just below the real axis, then back to zero. The log funciton is different on the two sides of the branch cut. The rest is residue calculus. Note that this works for any $n\ge 2$, and emerges as $(\pi/n) \mathrm{cosec} (\pi/n)$.

A related trick is $$\frac{1}{2\pi i} \int \frac{z^a}{1+z^n} dz$$ around the same contour, and let $a\to0$ whereupon the branch cut disappears(!) but the answer remains intact.

You can also do this:
When you have factorised the term $x^4+1$ into factors $p(x) & q(x)$ then
$$ \frac{1}{x^4+1} = \frac {Ax+M}{p(x)}+\frac{Bx+N}{q(x)}$$ and then to find  A, B, M and N just take values of x for both sides and equate to find them just like linear equations in two variables.


Pontryagin Duality and Injective Direct Limits Let $(G_n)_{n \in \Bbb{N}}$ be an injective direct limit of discrete groups. How does pontryagin duality play with injective direct limits? More precisely, what is $\widehat{\lim G_n}$ isomorphic to? Is it $\lim \widehat{G_n}$? If it helps, we may assume the $G_n$ are abelian groups.
EDIT: I suspect maybe they aren't isomorphic...I don't think that the direct limit of compact spaces is necessarily compact. Isn't $\Bbb{R}$ isomorphic to the injective direct limit of the compact intervals $[-n,n]$? If so, $\lim \widehat{G_n}$ probably won't always be compact, but $\widehat{\lim G_n}$ is always compact.

Since Pontryagin duality is a contravariant equivalence of categories, it turns colimits into limits.  So, if $G$ is the direct limit of a system $(G_n)$, the Pontryagin dual $\widehat{G}$ is the inverse limit of the dual system $(\widehat{G_n})$ (which is an inverse system, not a direct system, since the arrows got reversed) in the category of compact abelian groups.  Inverse limits in the category of compact abelian groups are computed by just taking the inverse limit as sets and giving it the inverse limit topology (compact Hausdorff spaces are closed under inverse limits of topological spaces) and group structure.

Here is another example. Consider a collection of cyclic
groups $G_n=\langle a\rangle$, where order of $a$ is $p^n$ and
a collection homomorphisms $a_n\rightarrow a_{n+1}^p$.
Then $\lim G_n=\mathbb{Z}(p^\infty)$ is the Pr$\ddot{\rm u}$fer $p$-group.
Since $\widehat{G_n}=G_n$,
we have also $\lim \widehat{G_n}=\mathbb{Z}(p^\infty)$.
However, it is easy to check that $\widehat{\mathbb{Z}(p^\infty)}=\widehat{\mathbb{Z}}_p$
is the group of $p$-adic integers.


What is your expected waiting time if limousine inter-departure times follow an exponential distribution? Limousines depart from the railway station to the airport from the early morning till late at night. The limousines leave from the railway station with independent inter-departure times that are exponentially distributed with an expected value of $20$ minutes. Suppose you plan to arrive at the railway station at three o’clock in the afternoon. What are the expected value and the standard deviation of your waiting time at the railway station until a limousine leaves for the airport?

Shouldn't the expected waiting time be $20$ minutes? Since it's an exponential distribution then for $t\to$  time to the departure, $f(t) = \frac{1}{20}e^{-\frac{1}{20}t}$. The expected value for this distribution would be $20$.

Is this correct?

This is a special case of the so called "stationary excess distribution".

The CDF of the time you have to wait is given by $F(t) = \frac{1}{20} \int_0^t [1-F(s)] ds$ where $F(s)$ is the CDF of an Exponential RV with mean 20.

An alternative way to do this is to use the fact that you're dealing with a poisson process, and if you start a poisson process at any time, it is still a poisson process with the same parameters, so the distribution to the next arrival is still Exponential with mean 20.

This is not exactly correct. If the number of limousines run every day is a constant $n$ = 60*(LateAtNight-YearlyMorning)/20,  and then the answer will be

$$20 * (n/(n+1)*(1+1/(n+1)/(n+2)) + 60*(24-(LateatNight-YearlyMorning))*pow((ArrivalTime-YearlyMorning)/(LateatNight-YearlyMorning),n)$$

As such, for a station opened from 5 AM till 1 AM and you arrived at 3PM, you have to wait ~ 19' 40''.


Given two events, A and B, what is the probability that at most one will occur? I know this is a very basic question, but my mathematics is coming out a bit wonky. Assume Events A, B are independent.

Define:

$Pr(A) = 1/6$

$Pr(B) = 1/4$

Let Event C = "at most one event out of A, B will occur".

So either A, B, or neither A nor B are all the possible outcomes for Event C.

$Pr(\neg A) = 5/6$

$Pr(\neg B) = 3/4$

$\neg A \cap \neg B = \neg A * \neg B = 5/8$

So, Event $C = Pr(A) + Pr(B) + Pr(\neg A \cap \neg B) = 1/6 + 1/4 + 5/8 = 25/24$

This probability is greater than 100%! I'm sure I've messed up somewhere. If it were 100%, that would exclude the possibility that both A and B could ever occur, which is definitely not the case.

UPDATE: exclude $Pr(A \cap B)$

Event $C = Pr(A) + Pr(B) - Pr( A \cap  B) = 1/6 + 1/4 + 1/24 = 3/8$

Hint.  The easy way to do it: the probability that at most one event will occur is the same as the probability that not both will occur, that is,
$$1-P(A\cap B)\ .$$
Using the given information, you should easily be able to work this out.

$C$ is the event that exactly one of the following events:

$A$ but not $B$
$B$ but not $A$
Not B and Not A (Neither A nor B)

1 is the event $A \setminus B$ or $A \cap B^C$

2 is the event $B \setminus A$ or $B \cap A^C$

3 is the event $B^C \cap A^C$ (not B and not A) or $(A \cup B)^C$ (not (A or B))

Observe that the events are mutually exclusive.

Hence,

$$P(C) = P(A \cap B^C) + P(B \cap A^C) + P(B^C \cap A^C)$$

Solution 1:

Recall that if $A$ and $B$ are independent, then so are the following pairs:

$$A, B^C$$
$$A^C, B$$
$$A^C, B^C$$

$$P(C) = P(A \cap B^C) + P(B \cap A^C) + P(B^C \cap A^C)$$

$$P(C) = P(A)P(B^C) + P(B)P(A^C) + P(B^C)P(A^C)$$

$$P(C) = P(A)(1-P(B)) + P(B)(1-P(A)) + (1-P(B))(1-P(A))$$

Solution 2:

$$P(C) = P(A \cap B^C) + P(B \cap A^C) + P(B^C \cap A^C)$$

$$P(C) = 1 - P(A \cap B)$$

$$P(C) = 1 - P(A)P(B)$$


Prove that polynomial $x^{2n}-nx^{n+1} +nx^{n-1}-1$ has root $1$ with multiplicity $3$. Prove that polynomial $x^{2n}-nx^{n+1} +nx^{n-1}-1$ has root $1$ with multiplicity $3$.
I am thinking about using $Horner's$ $method$ but don't know how.

Let's $P(x) = x^{2n}-nx^{n+1} +nx^{n-1}-1$, we have
$$P'(x) = 2nx^{2n-1}-n(n+1)x^{n}+(n+1)nx^{n-2}$$
$$P''(x) = 2nx^{2n-1}x^{2n-2}-n^2(n+1)x^{n-1}+(n-2)(n-1)nx^{n-3}$$
As $$P(1) = 0$$
$$P'(1) = 0$$
$$P''(1) = 0$$
Then $P(x)$ has root 1 with multiplicity 3
PS: If the OP wants to prove that $P(x)$ has root 1 with multiplicity exactly 3, it suffices to calculate $P'''(x)$ and prove that $P'''(1)  \ne 0$


How to prove that $sign(\alpha \beta) = sign(\alpha)sign(\beta)$? Claim : How to prove that $\text{sign}(\alpha \beta) = \text{sign}(\alpha)\text{sign}(\beta)$, where $\alpha,\beta \in S_n$?

Where $\text{sign}(\alpha) = 1$ if number of transpositions in the representation in $\alpha$ is even.

and it will be -1 otherwise.

Case 1: $\alpha$ and $\beta$ both have even number of transpositions then it is easy to verify that above claim hold

Case 2 : if $\alpha$ has odd number of transpositions and $\beta$ has even number then $\text{sign}(\alpha\beta)$ will be -1 and
$\text{sign}(\alpha) \text{sign}(\beta) $ will also be negative.

Then case 3 and case 4 and we are done. Is there any better proof for this problem?

Define $f:S_n\to GL_n(\mathbb{C})$ by $f(g)e_i = e_{g(i)}$ for the standard basis $e_1, \dots, e_n$ of $\mathbb{C}^n$. This map is clearly a homomorphism. Then $\det f:S_n\to \{\pm 1\}$ has $f(g) = -1$ if $g$ is a transposition (for example, use the invariance of $\det$ under a change of basis to reduce the case of $g = (12)$ and compute it directly), so $f(g) = \operatorname{sign}(g)$ for any $g$.

Given the definition and the observation that the number of transpositions in (some representative of) $(\alpha + \beta)$ is the sum of that in $\alpha$ and in $\beta$, this is equivalent to proving that odd + odd is even, etc. You can phrase that equivalently as the addition being well-defined $\mod 2$ and so on, if that counts as a 'better' proof.


Solve this equation $3^{(x-3)^{(x+3)}} = (3x+2)^{(x-3)}$ Please can someone help with step by step solving of this equation?

Solve

$$3^{(x-3)^{(x+3)}} = (3x+2)^{(x-3)},$$
for x

Thanks
A

As Caife answered, $x=3$ is a trivial solution.

However, taking logarithms and graphing the functions $$f(x)=\log \left(3^{(x-3)^{(x+3)}}\right)\qquad \qquad g(x)=\log \left((3 x+2)^{(x-3)}\right)$$ there is another intersection just above $x=4$ but I suspect that only numerical methods will allow you to find its exact value. Using Newton methods (have fun with the derivatives), the solution is almost $x=4.15532$.

Taking into acount the algebra-precalculus tag, I do not suppose that they expect you to find this one. If you are interested in the manner we can get it, just let me know.

Try substituting $x=3$. Certainly solves it!


Restriction of a diagonalizable linear operator $T$ to any nontrivial $T$-invariant subspace is also diagonalizable using a given lemma. Lemma

Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $W$ be a $T$-invariant subspace of $V$. Suppose that $v_1, \dots , v_k$ are eigenvectors of $T$ corresponding to distinct eigenvalues. Prove that if $v_1+\dots+v_k$ is in $W$, then $v_i ∈ W$ for all $i$.

Use the above lemma and prove the given result.
Question

Prove that the restriction of a diagonalizable linear operator $T$ to any nontrivial $T$-invariant subspace is also diagonalizable.

I could prove the above lemma using induction.
My attempt:
A linear operator $T$ on a ﬁnite-dimensional vector space $V$ is diagonalizable if and only if $V$ is the direct sum of the eigenspaces of $T$. Let $\lambda_1, \lambda_2,..., \lambda_k$ are the eigenvalues of $T$, $V=E_{\lambda_1} \bigoplus E_{\lambda_1}\bigoplus ...\bigoplus E_{\lambda_k}.$ Let $W_{\lambda_j}=E_{\lambda_j}\cap W(1\leq j \leq k)$. Will we get $W=W_{\lambda_1} \bigoplus W_{\lambda_1}\bigoplus ...\bigoplus W_{\lambda_k}?$
Claim 1:

$W=W_{\lambda_1} \bigoplus W_{\lambda_1}\bigoplus ...\bigoplus
 W_{\lambda_k}.$

$$V=E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_1}.$$
$$V\cap W=W \cap (E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_k}).$$
Claim 2:

$W \cap (E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_k})=(W \cap
 E_{\lambda_1}) + (W\cap E_{\lambda_1})+ ...+ (W\cap E_{\lambda_k})$

$x\in (W \cap
 E_{\lambda_1}) + (W\cap E_{\lambda_1})+ ...+ (W\cap E_{\lambda_k})\implies$
$x=x_1+x_2+...x_k \implies x \in W \cap (E_{\lambda_1} + E_{\lambda_1}+ ...+ E_{\lambda_k})$.
How do I prove the converse? Does there exist any counter example to prove that result is false?
How do I use the above lemma and prove the result?

Let $w \in W$. Then $w \in V=E_{\lambda_{1}} \bigoplus E_{\lambda_{2}}\bigoplus\cdots\bigoplus E_{\lambda_{k}}$. So we can write $w=w_1+w_2+\cdots+w_k$ where $w_i \in  E_{\lambda_{i}}$ for each $i$. The lemma now tells you that $w_i \in W$ for each $i$. Claim 2 follows immediately from this.


Set-builder notation proof $S=\{x^2:x \in \mathbb R\}$, $T=\{y^3:y \in \mathbb Z\}$, $V=\{z^{12}:z \in \mathbb Z\}$

  Prove that $V \subsetneq (S \cap T)$.

Having some difficulties with figuring out how to start with these kind of questions.

$S$ is the set of squared reals, $T$ the set of cubed integers, and $V$ the set of twelve-powered integers.

Now,
$V\subsetneq S\cap T$ iff $(\exists s\in S\cap T.s\notin V)~\wedge~(\forall v\in V.v\in S\cap T)$.

Which is to say $(\exists s\,. s\in S\wedge s\in T\wedge s\notin V)\wedge (\forall v\,.v\in V\to v\in S\wedge v\in T)$

Notice that $s\in S$ iff $s^{1/2}\in \Bbb R$, and such.

So is it so that?: There is some number which is in both a squared real and a cubed integer yet not a twelve-powered integer, but all twelve-powered integers are both a squared real and a cubed integer .

Let $a \in V$, then $a = z^{12}$ for some $z \in \mathbb{z}$. But then $a^{1/3} = z^{4}$ is an integer such that cubing gives $a$, thus $a \in T$. Likewise $a^{1/2} = z^6$ is a real number such that squaring gives $a$ so that $a \in S$. This means $a \in V$ implies $a \in S$ and $a \in T$ thus $a \in S \cap T$ which means $V \subseteq S \cap T$. Now we need to show that the sets are not equal. $(1/2)^2 = .25$ is in $S$ but clearly not in $V$ for the integers are multiplicatively closed.


How many ways to make a match if there are 4 guards, 4 forwards, and 2 centers in a gym? We know that one team is made of 2 guards, 2 forwards, and a center. There are 4 guards, 4 forwards, and 2 centers in the gym. How many different matches can they play if everyone plays their own position?
My answer is 4 choose 2 times 4 choose 2 times two choose 1 to equal $72$. Out of 4 guards and forwards, you pick two, and the other two are chosen. Same logic goes for center. Is my answer correct?

Your answer, ${4\choose2}{4\choose2}{2\choose1}=72$, would be correct if you were designating one team as the home team and the other as the visitors. But it would be the same two teams playing.  In other words, you are counting each match twice.

A correct way to count the number of matches is to realize that the two centers must be on opposite teams, so you can think of one center picking his or her teammates, which can be done in ${4\choose2}{4\choose2}=36$ ways.

Yes it's correct.
In the question, choosing the members of one team automatically determines the members of the other team.
4C2 * 4C2 * 2C1 = 72 is the right answer.


$(a\mod m)/(b\mod m) = (a/b)\mod m$? b and m are relatively prime (m is prime and $b \in \mathbb Z_m^* $). In truth, I would like to be able to get to the following point (it is a simplified example):

$\frac{ab \mod m}{b \mod m} = a \mod m$

Is it possible?

EDIT

My apologies. Thanks to Henning Malkholm I noticed I had used the wrong notation. I actually meant this:

$(ab \mod m)(b^{-1} \mod m) \equiv abb^{-1} \mod m$

$abb^{-1} \mod m \equiv (a \mod m)(bb^{-1} \mod m) \equiv (a \mod m)$

Is this valid? Again, I apologise if this seems too silly. I just don't feel secure enough to trust what I've found.

Let $a = 3, b = 4, m = 5$. Then
$$ab \mod m = 12 \mod 5 = 2$$
$$b \mod m = 4$$
$$a \mod m = 3$$
Thus, you can see that your assumption is not true.

It is possible, but it is not necessarily true.

For example, take $m=11$, $a=4$, $b=3$. Then $ab\bmod m=1$, $a\bmod m=4$ and $b\bmod m=3$, yet
$$ \frac{1}{3} \ne 4 $$

In order to extend division to modular arithmetic, what you need to do is find the modular inverse $b^{-1}$ of $b$, defined as the solution to
$$ xb \equiv 1 \pmod m$$
which is $3$.

Then you can let "$a$ divided by $b$, modulo $m$" mean $ab^{-1}\pmod m$.


Prove that if $a \mid b$ then $\gcd(a,b) = |a|$ I am having trouble completing the proof if $a \mid b$ then $\gcd(a,b) = |a|$.
My attempt:
If $a\mid b$ there is a $k \in \mathbb{Z}$ such that $b = ak$. Let $d = \gcd(a,b)$. Then $d \mid a$ and $d\mid b$ so by hypothesis we have $d\mid a$ and $d\mid ak$. Hence, $d \leq \gcd(a,ak)$. We can easily see that the largest common denominator of $a$ and $ak$ is $|a|$, thus $d \leq\gcd(a,ak) = |a|$ so $\gcd(a,b) \leq |a|$.
I am stuck and not sure how to show the reverse inequality $|a| \leq \gcd(a,b)$ to complete the proof.

Clearly $|a|$ is a divisor of $a$.
Also, since $b=ak$. We have $b=\operatorname{sign}(b)|a||k|$, that is $|a|$ is a divisor of $b$.
Hence as a common divisor of $a$ and $b$., it is at most the greatest common divisor of $\gcd(a,b)$.
$$|a| \le \gcd(a,b)$$

"such that b=ak.... d≤gcd(a,ak)

You seem to be overlooking that if $b = ak$ then ..... $b = ak$!
So if $b=ak$ then $\gcd(a,b) = \gcd(a,ak)$.   There is not need to conclude that if $d = \gcd(a,b)$ then $d \le \gcd(a,ak)$.  If $d=\gcd(a,b)$ and $b  = ak$ then $d$ EQUALS $\gcd(a,ak)$.
But I can not accept your claim "We can easily see that the largest common denominator of $a$ and $ak$ is $|a|$".  I think that is the entire point of the proof and must be spelled out.  (I have an explanation below)
====== Complete Answer======
Dont let the absolute value throw you.  The integer factors of a negative number are exactly the same as the factors of the equivalent positive number (i.e. the absolute value), and if $d$ is a factor of an integer then $-d$ is as well, and positive numbers are larger than even numbers so a $\gcd$ is always positive. We can assume without loss of generality all numbers are non-negative.
If $|a| = a'>0$ and $|b| = b' \ge 0$ then $a|b \iff a'|b'$ and $\gcd(a,b) = \gcd(a',b')$ by the argument in the paragraph above.  So we just need to prove the following.

if $a$ and $b$ are non-negative integers and $a|b$ then prove $\gcd(a,b) = a$.

Pf:  Well, $a = 1\cdot a$ so $a|a$ and $a|b$ so $a$ is a common divisor.  We know that $a\ne 0$ because $0$ does not divide anything.  If $d > a$ then $0 < \frac ad < 1$.  But there is no such integer so $d\not \mid a$.  So $d$ is not a common divisor of $a$ and $b$.  So the greatest common divisor of $a$ and $b$ is $a$.
This assumes your definition of $\gcd(a,b)$ is the largest common divisors of $a$ and $b$.  Many texts uses $\gcd(a,b)$ is the positive common divisor so that every possible common divisor is a divisor of $\gcd(a,b)$.
Proof using that definition:   Now $a$ is a common divisor of $a$ and $b$ so by this definition of $\gcd$ we must have $a|\gcd(a,b)$.  If $a > \gcd(a,b)$ then $a\not \mid \gcd(a,b)$ so $a \le \gcd(a,b)$.  And if $a < \gcd(a,b)$ then $\gcd(a,b) \not \mid a$ and $\gcd(a,b)$ is not a common divisor of $a$ and $b$.  So $a \ge \gcd(a,b)$.  So $a \ge \gcd(a,b)$ and $a \le \gcd(a,b)$.  So $a = \gcd(a,b)$.


Exterior Derivative question In http://mathworld.wolfram.com/ExteriorDerivative.html,

there is a section that starts from:

  Define the exterior derivative by $Dt ≡  \frac{\partial}{\partial x} \wedge t$

First of all, what does this first part quoted mean, in relation to the previous parts?

I then want to know what the rest of the section is talking about.

The notation between the dollar sign is LaTeX (http://www.latex-project.org/) mathematics
notation. It is meant to be very descriptive of mathematical expressions; it is not intended to "look" like standard mathematical notation.

The closest equivalent in Mathematica itself is:

Dt [Congruent] ([PartialD]/[PartialD]x)[Wedge]t


How to get the size of the attempts in this exercise? How many children should a couple have so that, with probability $0.95$, she has at least one son and one daughter?

I've thinked the problem like a Binomial Distribution, Assuming that the probability of having a son is equal to having a daughter then $p=0.5$, if $X:= '\text{Number of sons obtained in n attempts}'$ then $X=B(n,0.5)$.

So, for find the $n$ i need:
$P(1\leq X \leq n-1)=0.95$

Because, i need that at least she have a daughter.

Am I right ?, or how should this exercise be solved?

Yes, you are right. Now calculate the sum and solve the inequality:
$$P(1\leq X \leq n-1)\ge 0.95 \iff \\
\sum_{k=1}^{n-1} {n\choose k}\left(\frac12\right)^k\left(\frac12\right)^{n-k}\ge 0.95 \iff \\
\frac1{2^n}\sum_{k=1}^{n-1} {n\choose k}\ge 0.95 \iff \\
\frac1{2^n}\left(2^n-2\right)\ge 0.95 \iff \\
0.05\cdot 2^n\ge 2 \iff \\
2^n\ge 40 \iff \\
n\ge 6. $$

Not sure this is right but with 50% chance boy or girl is 0.5
If you say the following ( Boy / Girl there is 50% chance each time )  then....
0.5 X 0.5 X 0.5 X 0.5 X 0.5 = 0.03125

the 3.125% here falls under the 5% margin between 1.00 and 0.95
so the answer could be 5 children.

However we all know nature can be unpredicatable and couples have been known to have lots of just boys or just girls.

Regards,

John O.


Solutions to $x+y+z=31$ and $x+2y+3z=41$ For the equations
$$x+y+z=31$$ $$x+2y+3z=41$$
is there a elegant way or method to find all the positive solutions in integers? Thus far, I have been using trial and error (which is time consuming). Another approach I tried was subtracting the first equation from the second to get $y+2z=10$. But I feel like there must be some faster way of doing this that I am not thinking of right now. Any input is appreciated.

No, I think the second approach you had is as quickly as this can be solved. Subtracting the first equation from the second gives $y + 2z = 10$, and the only possible ways that both $y$ and $z$ are positive integers is when $(y,z) = (8,1), (6,2), (4,3), (2,4)$. Plugging each set back into the first equation, we get that the four possible combinations for $(x,y,z)$ are $(22,8,1), (23,6,2), (24,4,3), (25,2,4)$.

Subtract the first equation from the second.

Express $y$ in terms of $z$

Now substitute into the first equation and you'll get and relation for $x$ with respect for $z$.

Because $z \in \mathbb{R}$ there are infinite amount of solutions and all of them are of a closed form, you'll obtain. And this is the fastest way to compute solution, maybe you were confused, because there are infinite amount of them.


Is $f$ uniformly continuous Prove/Disprove : Let $f:(0,1)→\mathbb{R}$ be Continuous. The condition $f(1/n)$$\rightarrow$$1/2$ and $f(1/n^2)$$\rightarrow$$1/4$  imply that $f $ is uniformly continuous.

The condition implies everything! This because it is not true.

($f\left(\frac{1}{n}\right)\rightarrow2$ implies: $f\left(\frac{1}{n^{2}}\right)\rightarrow2$
wich contradicts $f\left(\frac{1}{n^{2}}\right)\rightarrow4$)

$p\rightarrow q$ is true if $p$ is not true.

So indeed the condition implies that $f$ is uniformly continuous. There is no function that satisfies this condition and is not uniformly continuous. You cannot even find a function that satisfies this condition.

As pointed out in other answers, the conditions are self-contradictory; the sequence $f(1/n^2)$ is a subsequence of $f(1/n)$. If they have limits at all, they cannot have different ones.

It seems to me that the spirit of the question is to ask,

  if $a_n,\ b_n\to 0$ but $\lim f(a_n)\neq \lim f(b_n)$, then can $f$ be uniformly continuous?

Now my hint is: $f$ cannot extend to a continuous function $\hat{f}:[0,1]\to\mathbb{R}$. (Can you see why?) Why does this imply $f$ must not be uniformly continuous?


Linear Algebra Polynomial Subspace Let $P_n$ be the set of polynomials of degree at most $n$ with real coefficients. Is the set of polynomials of the form $p(t) = a + t^2,$ such that '$a$' is an element of $\mathbb{R},$ a subspace of $P_n?$

I know I need to check if it contains the zero vector first. How would I do this? I have two ideas:

$1.$ plug $0$ into '$a$' and have a function $g(t) = t^2$ then add it to $p(t)$ to get $p(t) + g(t) = a + 2t^2$ which is not in the form, or

$2.$ plug $0$ into '$a$' and also for the coefficient of $t^2?$

Thanks!

Let $V$ be a vector space and $W\subset V$. We want $W$ to satisfy three key axioms for it to fit the definition of subspace.
$1$. $\{0\}\in W$
$2$. $w_1+w_2=w_3\in W$ (closure under vector addition)
$3$. $cw\in W$ (closure under scalar multiplication)
For the subset of polynomials $W$ defined by $p(t)=a+t^2$, we don't have closure under addition, because we have $p(t)+q(t)=(a+b)+2t^2$, which is not of the desired form.
also, the set fails closure under scalar multiplication as well, since $cp(t)=c(a+t^2)=ca+ct^2$. The only exception is $c=1$, but $W$ still fails the vector addition axiom so it is not a subspace.
Finally, the zero vector (and for polynomials, the zero polynomial -that whose all coefficients $a_0, a_1,..., a_n = 0$, and in this case, only $a = 0$) is also not in the subset except for the single case where $t=0$.

The zero element $P_n$ is not in the set, but I feel some clarification to the above statements is necessary:
The zero of this space is the polynomial with all zero coefficients - not the polynomial that equals zero when you set $t=0$. Thus, multiplication by the zero polynomial will result in zero for all vectors (i.e., other polynomials) in this space. You never really plug in a number for $t$ when dealing with questions on this space, the vectors are the polynomial equations themselves.
This is enough to conclude it is not a vector space, but it's also true that the set is not closed under vector addition or scalar multiplication as well, as outlined above.


Restrictions on universal generalization Wikipedia's article on universal generalization doesn't seem to give a satisfactory explanation of the restrictions on when it can be used:

Assume $\Gamma$ is a set of formulas, $\varphi$ a formula, and $\Gamma \vdash \varphi(y)$ has been derived.  The generalization rule states that $\Gamma \vdash \forall x \varphi(x)$ can be derived if $y$ is not mentioned in $\Gamma$ and $x$ does not occur in $\varphi$.

The article then gives an example of UG being used incorrectly to derive $\exists z\exists w(z\neq w) \vdash \forall x(x\neq x)$, with the restrictions given certainly being violated.  However, wouldn't the following modification of the "proof" be in accordance with the restrictions?

$\exists z\exists w(z\neq w)$
$\exists w(y\neq w)$
$y\neq x$
$\forall\alpha(\alpha\neq x)$
$x\neq x$

Here, the generalization in step 4 has been amended to use $\alpha$ as the quantifier's bound variable, which should be permitted, as $\alpha$ does not occur in $y\neq x$ and $y$ does not occur in the assumption $\exists z\exists w(z\neq w)$, yet this leads to a universal instantiation that is clearly false!  Is the $\Gamma$ of the restrictions meant to be all previous steps of the proof, not just the assumptions?  If so, wouldn't this invalidate the proof given further in the article that includes the steps $P(y)\to Q(y)$ and $P(y)$ before generalizing $Q(y)$?  Does the presence of existential instantiation place further restrictions on universal generalization somehow, with the existential instantiated counting as a mention of the variable introduced?

This sort of thing can only be answered by looking at a particular set of inference rules, rather than looking at one inference rule at a time.  This is an inherent flaw in the way that Wikipedia covers inference rules, because rules that are sound individually may be unsound when combined, as the deduction in the question shows.

Here is how the problem in the question is resolved in Mendelson's logic textbook, which does use a Hilbert-style deductive system. Recall that the deduction of $\phi(c)$ from $(\exists x)\phi(x)$, as in the first two steps of the deduction in the question, is called existential instantiation. In Mendelson's system, this is not formalized as an inference rule, it is treated as a definitional extension of the original theory, in which a new constant symbol $c$ is added along with a new axiom $\phi(c)$. Now Mendelson's version of universal generalization is just that from $\phi$ we can deduce $(\forall x)\phi$ for any variable $x$. Thus there is no way to go from step 3 to step 4 of the deduction above, because Mendelson's universal generalization rule does not have the ability to replace the constant symbol $y$ with a variable $\alpha$ in the formula that is having a quantifier adjoined.  In this way Mendelson is able to avoid any restrictions on the variable in the universal generalization rule.

In a different deductive system where the universal generalization rule does have the ability to replace constant symbols with variables, you're right that extra restrictions will have to be added if the constant symbols can be added by existential instantiation. For example, if we take as a convention that any constant symbol introduced by existential instantiation has been mentioned in $\Gamma$, that would also prevent this sort of problem.

Of course the real test is not whether it appears that problems have been prevented - the test is whether the soundness and completeness theorems can be proved for a particular deductive system. It's easiest to pick one book that has a system that matches your taste, and then stick scrupulously to the system in that book. That avoids all these subtle problems about mismatched inference rules.

I think that this is an excellent question that can be answered in two ways.

The first way, as in the great answer above, is to ensure that we can understand the restrictions for universal instantiation and generalization so that the first-order logic (FOL) can be employed usefully and appropriately.

The second way is to realize that our intuitions about "for all" and "some" as quantifiers do not match the dictates of the FOL for $\forall$ and $\exists$.

A full discussion here would take too much time and space. But, suffice it to say that the systemic goals of consistency ($\vdash A$ if and only if $\models A$) and completeness ($\models A$ if and only if $\vdash A$) are all we need for a FOL, unless we have goals other than those essential to a FOL. For example, soundness, where the requirement is $A \vdash B$ if and only if $A \models B$. Soundness is an idiosyncratic but usual requirement that is the "bugaboo," in my opinion, and we have to distinguish it from consistency. The requirement for universal generalization and consistency is that we preserve logical truth. The requirement for soundness is that we preserve truth. The two are vastly different and we do not yet understand what is going on in this regard, I think.

There are holes in the fundamentals of FOL which we have to discuss. This question is a beauty in that respect.


If $n$ is a positive integer, Prove that $\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac{2329}{3600}.$ If $n$ is a positive integer, Prove that

$$\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac{2329}{3600}.$$

please don't refer to the famous  $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.

I am looking a method that doesn't use $\text{“}\pi\text{''}.$

Unfortunately, I know and tried only $\text{“}\pi\text{''}$ method.

I like this kind of numerical challenges. Let us exploit creative telescoping. For any $n\geq 2$ we have
$$ \frac{1}{n^2}-\left[\frac{1}{n}-\frac{1}{n+1}\right]=\frac{1}{n^2(n+1)}\tag{1} $$
$$ \frac{1}{n^2(n+1)}-\left[\frac{1}{2n^2}-\frac{1}{2(n+1)^2}\right]=\frac{1}{2n^2(n+1)^2}\tag{2} $$
$$ \frac{1}{2n^2(n+1)^2}-\left[\frac{1}{6n^3}-\frac{1}{6(n+1)^3}\right]=-\frac{1}{6n^3(n+1)^3}\tag{3} $$
$$ \frac{1}{6n^3(n+1)^3}-\left[\frac{1}{30n^5}-\frac{1}{30(n+1)^5}\right]=\frac{5n^2+5n+1}{30n^5(n+1)^5}\tag{4}$$
$$ \frac{5n^2+5n+1}{30n^5(n+1)^5}< \frac{1}{42 n^7}-\frac{1}{42(n+1)^7}\tag{5} $$
hence by setting $g(n)=\frac{1}{n}+\frac{1}{2n^2}+\frac{1}{6n^3}-\frac{1}{30n^5}$ we have
$$\begin{eqnarray*} \zeta(2)-1=\sum_{n\geq 2}\frac{1}{n^2} &=& \sum_{n\geq 2}\left[g(n)-g(n+1)\right]+\sum_{n\geq 2}\frac{5n^2+5n+1}{30n^5(n+1)^5}\\ &<& g(2)+\frac{1}{42\cdot 2^7}=\color{red}{\frac{5779}{8960}}<\frac{2322}{3600}.\tag{6}\end{eqnarray*} $$

The same technique proves Stirling's inequality and the "acceleration formula"
$$ \sum_{n\geq 1}\frac{1}{n^2} = \sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}\tag{7} $$
leading to the more accurate approximation $\zeta(2)-1\approx \frac{51077}{79200}$.

There also is a simple and accurate lower bound. By choosing $h(x)=\frac{6x-3}{6x^2-6x+2}$ we have $\frac{1}{x^2}\geq h(x)-h(x+1)$ for any $x\geq 2$, with the difference behaving like $\frac{1}{9x^6}$. It follows that
$$ \zeta(2)-1\geq \color{red}{\frac{9}{14}} = h(2).\tag{8}$$
By selecting $k(x)=\frac{60x^2-60x+31}{60x^3-90x^2+66x-18}$ we get $\zeta(2)-1\leq \frac{151}{234}$, simpler but less accurate than $(6)$.

The simplest approach is to approximate the sum by the integral. The crude way of doing it just uses the fact that $1/x^2$ is a decreasing function, so the integral from $k$ to $k+1$ is bigger than $1/(k+1)^2.$ to make it sharper, one notes that since $1/x^2$ is convex, the integral is actually bigger than $\frac12(1/n^2 + 1/(n+1)^2).$


Which methods are used by actuaries in practice? Recently I read a comment from an actuary that a lot of the math they studied as part of the program they never actually used.

I'm not interested in becoming an actuary, but I'm interested in learning more about some of the methods that are commonly used (mainly in non life insurance). Can anyone clarify which statistical / mathematical methods are commonly used by actuaries or insurance analysts in their day to day work?

Is there a big emphasis on bayesian methods such as bayesian credibility? What about more traditional methods such as the chain ladder method. Or perhaps Markov chain methods for no claims discounts?

I've looked at a couple of books such as
Hossack, Pollard. Introductory statistics with applications in general insurance
(seems like a good basic book)
and

Boland. Statistical and probabilistic methods in actuarial science
(claims to be a practical books with useful R code, but is actually full of theorems, proofs, and lemmas).

but I've no idea whether it is worth expending the effort in wading through a book like Boland's in terms of practical knowledge acquired. I tried to find a introductory book that would take an applied approach that mixed theory with analyzing actual datasets using something like SAS or R, but I was unsuccessful.

So can anyone clarify which methods are used on a regular basis by working actuaries. Any book recommendations would also be appreciated. Thanks.

I worked for a few years as life actuary. They study syllabus was a bit different then vs. as I vaguely understand it now. In that era, one studied math courses to become an associate of the Society of Actuaries. Then you studied the actual machinations of the insurance business to become a Fellow.

Passing exams was a key factor in your pay scale which was a plus or a minus depending on results. But for sure, one's career track is heavily dependent on these results.

But, in short, I can tell you that I never used one iota of the math covered on the exams. They are more of a training ground to come up with fast ways to solve problems. The higher level exams require a good deal or practice and are high-pressure considering what is at stake. As an example of the intensity, I got a 10 (out of 10) on the risk theory exam (considered a tough one) answering 7 out of 15 questions - and I yet may have gotten some wrong.

The major function of actuaries was to calculate and substantiate the adequacy of reserves. This relied heavily on mortality tables - an easy math problem. A significant aspect of the endeavor entailed accounting regulations both for GAAP and state of domicile statutory requirements.

Most of the actual work in this regard was modeling of various policy lines. And each company has it's own set of well- worn models.

A latter consideration was the requirement of cash flow testing. This again was a modeling problem as the investment assets had to be tested for their performance under various, usually extreme, interest rate scenarios. This was intended to make sure there is money available to fund the reserves.

So while passing exams entailed cubic splines, the Poisson distribution, the Black-Scholes model, Ito's Lemma, the real work required familiarity with insurance regulations and their related accounting treatment.

For more information on the study materials, esp. books, you can go to the websites of the Society of Actuaries and the Casualty Actuarial Society.

Read this Abstract: Actuarial Methods in Health Insurance Provisioning, Pricing and Forecasting
By Peter Lurie

OSC Valuations/ChainLadder
PPCI/IBNR
Loss Ratio/BF
PCE
(Classical Statistical Regression Methods:
Pricing & 6. Forecasting are in Methods 1-4, Hence determination of Reserves is to calculate the future claims/prices or forward pricing (as an insurer you want to cover future pricing costs).
Simply they calculate the Estimates:
(1) Projects Costs,
(2) Losses/Loss Ratios,
(3) Claims (Outstanding Claims, IBNR, PPCI and Development Claims).


Please only give hints, not answer: if $g(x)=e^{-x}\sin x-\cos x$ what is $g^{-1}(1)$ question
$g(x)=e^{-x}*\sin x-\cos x$

what is $ g^{-1}(1)$

my steps
i know that
$1=e^{-x}\sin x-\cos x$
thus
$\ln 1=-x\sin x-\cos x$

here is a possibility:
$\ln^21=(-x-2x\sin x\cos x)^2 $
and that is where i got stuck

Hint: $e^{−x}$ is difficult to evaluate for even simple values, so if you can find the root by hand, there must be some way to offset, or negate the effort of calculating $e^{−x}$. $\sin x$ must have a value that makes the value of $e^{−x}$ irrelevant.

See the graph. g (x) is period and hence inverse does not exist. So you are left with numerical analysis..


If a is odd, how do I prove that 3a is also odd? I know that if something is odd then $\exists k \in \mathbb{Z}: a = 2k + 1$. But what I get is:

$n = 2k + 1$

$3n = 3(2k+1)$

$3n = 6k + 3 $

But i can't factor 6k + 3 to give me 2k + 1 !

Any suggestions?

If you factor out a 3, $(6k+3)$ $=$ $3(2k +1) = 2(3k + 1) + 1.$ This is now an odd number. You are here in your post, try thinking about taking an odd number and multiplying it by 3. You get an odd number in any example you try. This is because an odd integer multiplied by an odd integer is always odd, the proof of this is another good exercise.


Are We Teaching Pre-Calc Wrong? It took some 1,250 years to move from the integral of a quadratic to that of a fourth degree polynomial.  When we jump too fast to the magical algorithm, when we fail to acknowledge the effort that went into its creation, we risk dragging our students past that conceptual understanding. Source.

By the time Newton and Leibniz were developing Calculus mathematicians already knew how to solve particular problems of derivatives and integration. There was a good understanding on the way to resolve problems, too. Newton's teacher, Isaac Barrow, already had some understandings of the topics Newton would push to the perfection. Source.

For me, pre-calc was plugging and chugging trigonometric, geometric, and algebraic equations. Math books follow a rather narrow and strait path across some imaginary border from the province of pre-calc into calculus. As was mentioned already, that historic transition was anything but discreet. Should Pre-Calc teach the foundations of the fundamental theorem of calculus that were known Pre-Lebniz/Newton? I think that would energize what is otherwise a boring exercise in tedious calculation.

A beautiful example of a Pre-Leibniz integration: Who realized $\int \frac 1x dx =\ln(x)+c$?

We are teaching math wrong in many, many ways.

For some reason, there has been a historical backlash against abstraction in secondary and pre-secondary mathematics. "I'll never need to use this," say so many students and parents.

As a result, what we learn is a hackneyed attempt to apply the mathematical problems to "real world examples."

So when we learn trig, we start talking about things like leaning ladders against houses, and we give students an impression that in order to do such a thing, they need to compute the inverse sin of the height over the length of the ladder.

Of course, by the time students are 14, 15, or 16 years old, they've probably seen someone lean a ladder against a house and do no such thing.

Pre-calculus mathematics is very interesting because it is the first opportunity one has to apply relationships and definitions to problems and to learn how to transform complicated problems into simpler ones. This technique is fundamental no only in mathematics, but in the real world.

At the same time, real-time discovery is often haphazard and chaotic. We don't teach biology in the order it was discovered, principally because we assumed a whole lot of extremely wrong nonsense for most of human history.

The historical motivating factors are not the same as the present ones. Newton and Leibniz were trying to solve specific problems, and they needed new math. But we've solved those problems now, so those motivating factors have perhaps lost their edge.

Instead, we should look at their process, not their imperative. We should be teaching students to ask whether there exists a meaningful relationship between a function and its slope, and whether this can have an effect on a real-world problem. We should teach students about how we can infer new properties from a handful of well-defined conditions. We should teach students to explore "what if..."

Instead, we teach students about billiard balls, ladders, and two-column proofs, as if this bears any relevance to the real world, the mathematical world, or any world. In short, we waste their time. So yes, we're teaching it wrong.

A side story: As a partial counter-example... When I substitute taught math courses, I often got classes full of "Level 2" students, which was a way of saying "remedial". They hated their class, the lessons, the work, everything. So I used to put the Navier-Stokes equations on the board, and I'd tell them that whoever could solve them would win not just $1M dollars, but eternal fame.

I asked the students to name the most popular people they knew. They'd respond with "Avril Lavigne" or "Brittney Spears" or some such.

I asked them if they knew who Sophia Loren was. Or Dom Delouise. Or Patsy Cline. They never heard of them.

I then asked them if they ever heard of Newton. Of Einstein. Of Riemann. Of Euler. They, in fact, did know their names. I told them that mathematics is one of the few ways that your mark can be left on the world permanently. That if you did something truly great, that the high school students in 300 years would be hearing your name.

Of course, none of them really thought they could do that. But it resonated with them, because it made them think, "wow yeah, I have heard of Newton." It made them realize that maybe there was something important, and that unlike being a famous world leader, it was something that was almost purely product of self-actualization.

This context gave them a glimpse into a world they didn't know existed. None of them wanted to compute how to lean a ladder against a house. They wanted to know why math was important -- not microcosmically, but in the context of great things. These were young kids with great thoughts, not computing machines.

I told those students that "yes, it does not matter if you can compute these numbers, but it does matter if you know how." That concept alone motivated them more than anything else, because it showed them that they controlled their own destiny.

For a teenage kid frustrated with school, that's a wonderful feeling. Math is one of the few fields where we can actually deliver that consistently.

There is a good amount of maths in the 'shilling arithmethic' of 1951, (a school text), that one see the mathematicians struggle with.  For example, a whole chapter is devoted to the idea of various forms of compound arithmetic (multiplication and division of several units without conversion).

The whole idea that you can take a lengths in furlongs and perches, and multiply these by way of a division directly into acres, roods and perches, whthout conversions, seems totally alien in these days.  Yet this whole matter is discussed on a single page in an 1895 arithmetic book.  The method for doing this might be described as 'invoice arithmetic', applied twice.  So, one first finds the area of 1 furlong by $x$, and then finds $x$ by $y$, eg.  Here there is no conversion of units.

   Multiply  5 furlongs 22 perches by 8 furlongs 16 perches.
                                                       4,40

        1 furlong   'costs'   10.0.00     5   costs 50.0.00
       10 chains    'costs'    2.2.00     2   costs  5.0.00   (4
        1 chain     'costs'      1.00     2   costs  0.2.00  (10
                        total                 -------------
                                                    55.2.00

        1 furlong  'costs'    55.2.00     8   costs  444.0.00
       10 chains   'costs'    13.3.20     1   costs   13.3.20  (4
        1 chain    'costs'     1.1.22     6   costs    8.1.12 (10
                                              ===============
                                              total  466.0.12
           466 acres, 12 perches.

This is the normal method with dealing with added fractions, without means of conversions. Multiplication by parts is usually dismissed as a kind of 'peasent multiplication', but you see from this example, it is perfectly clear what is going on, and faster than most means of converting back and fourth.

One often sees on this list, 'continued fractions', often written in the quite ugly long vincula.  There is very little comment (if at all), on 'added fractions' or 'continued numerators', which provide a direct way of dealing with quite large fractions without conversions.  I regularly convert things like sixtyfourths into eights and added eights (eg $\frac {35}{64} = \frac 48 \frac 38 = \frac {4 \frac 38}8$).

Decimals are simply added fractions, rather than increasingly large exponents.  That is, when we upgrade $1.61$ to $1.618$, we are adding $\frac 8{10}$ to the end of such a fraction.  One sees, especially in money, amounts like $£ 5.15\frac 12$, exactly with this meaning: the fraction is against the unit of the last place.

One talks of the sumerian numbers, but totally ignores the fact, that well into the fourteenth century, that the English used a hundred of six scores, and that the great multitude of weights and measures used in pairs or tripples, is as much to avoid using hundred-numbers (eg a weight of seventeen two = 17-2 = 17 stone 2 lb = 240 lb, avoids the hundred lb issue).

Even the idea of 'english number' is given as absurd.  That is, to write eg $720$ as vi C (ie six C, where the hundred is six score), rather than 'DC', is taken as entirely strange.  Yet there are plenty of examples of this sort of writing in Zupko's "Dictionary of English Weights and Measures".

The sumerian system is not a 'regular base' as we know it, but a 'division base'.  The most significant digit is the units, and subsequent places are divisions or added fractions of sixtieths.  Division bases have leading siginficant zeros, so 00.01 means 1/60, and 00,00.01 means 1/3600.  Yet you often see people translate $44.26.40$ as $160,000$, rather than its correct value of $44 \frac 49$.

One of course, seriously objects to this absurdity of introducing radians into every discussion of angle.  Not only is it possible to do angle calculations without such absurd units, but geometry had gone a long way before anyone decided to invent it.

One can do rather complex non-euclidean geometry (including hyperbolic geometry) without recourse to things like $\tanh$ etc or radians.  The coherent angle for higher geometry is the fraction of all space, something i usually write in fractions against base 120, to simplify calculations.  In such units, i found the solid angles occupied by the vertices of all of the four-dimensional regular polytopes (polychora), well before the cubic radian was known.

In practice, the use of radians and radii simply impede the process: nearly everyone in the real world measure circles by their diameter.  You go an buy screws or plates or pipes or whatever, and it is the diameter that is quoted.  The unit circular inch is the area of an inch of unit diameter.  It is interesting that the Sumerians, according to Sir Thomas Heath, rated the circumference of 'real' circles (ie ones that you could walk around or hold), as if their diameter were 60, and $\pi=3$, ie 180 ells, divided to 24 digits.  It is only circles you stand in the middle of (like the sky), that has $360$ degrees = $2(\pi=3) 60$.  Yet i have been caught out here even suggesting that formulae for volumes by dimaeter be discussed.


How to calculate surface area of a curved plane? could anyone explain how to calculate the surface area of a curved plane?  I am trying to calculate the surface area of a "vaulted" ceiling that is 24' long, 7' wide, and the height of the curve is 4' at the mid-point so that I can figure out how much paint I need to buy.

Here is an illustration:

If it were a simple flat ceiling, I would just multiply length x width, but that's not what I have here and a google search is leaving me empty-handed.  Any ideas?

Thank you very much!
-Neal

use the formulae you derived in part 1 to calculate the curved surface area and the total surface area of the cylinder given below

Just calculate the cross-section length and multiply it with 24".
This is the simplest solution, without integration.

In order to deal with this more precisely, we must know the exact shape of the cross-section i.e. if it's a part of a circle or not

EDIT: I just did some thinking, and this cannot be a part of a circle obviously.
It is illogical to me that what we are looking at are actually two arcs glued together.
This is a parabola - once we find out its equation, we will find the surface - I'm working on that.

Here - the equation of parabola is $$h(1-\frac{x^2}{a^2})=y$$

The length of parabola arc is $$\sqrt{(a^2+4*h)}+\frac{2a^2}{2h}sinh^{-1}(\frac{2h}{a})$$

Here, a and h are the half-width which is 3.5 and height which is 4.

When you enter the numbers, the arc length is 7.418', and the surface is $$7.418*24=178.032$$


Modular arithmetic: How to solve $3^{n+1} \equiv 1 \pmod{11}$? Please, I can't solve this equation:

$$3^{n+1}\equiv 1 \pmod{11}$$

for $n \in \mathbb{N}$.

So what should I do please? Thanks.

$3^5=243=11(22)+1\equiv 1\mod 11$

So if $3^{(n+1)}\equiv 1 \mod 11$ we must have $5|(n+1)$(as $5$ is the smallest integer such that $3^5\equiv 1\mod 5$)


Finding the max/min turning points of $y=\sin x$ Use differentiation to show that

  $$y = \sin x$$ has a maximum turning point at $\left(\frac{\pi}2, 1\right)$ and a minimum turning point of $\left(\frac{3\pi}2, -1\right)$.

I know that the turning points occur when $\frac{dy}{dx} = 0$. So ...

$$\begin{align}\frac{dy}{dx} &= \cos x \\[6pt]
\implies\quad \cos x &= 0 \\[6pt]
\implies\quad \cos^{-1}(0) &= 90^\circ
\end{align}$$

$\cos x $ is positive in the 1st and 4th quadrants, so $x = 90^\circ$ or $x = 270^\circ$

I am not sure how to progress from here or if indeed this is the correct approach.

You're correct. Just change your measures from degrees to radians and you will get $$x=90^\circ = \frac{\pi}{2} \\ x=270^\circ = \frac{3 \pi}{2}$$ If you plug these back in the original equation you will get $$ \sin\left(\frac{\pi}{2}\right)=1 \\ \sin\left(\frac{3\pi}{2}\right)=-1$$ Which gives you your points $\left(\frac{\pi}{2},1\right),\left(\frac{3\pi}{2},-1\right)$ as extrema (maximum and minimum respectively) to the function.

Let $f(x) = \sin x$, then $f'(x)=\cos x = 0 \iff x = 90^{\circ}$ or $x = 270^{\circ}$, and $f''(x) = - \sin x$ has $f''(90^{\circ}) = -1< 0$, and $f''(270^{\circ}) = 1 > 0$. Thus the max occurs at $x = 90^{\circ}$, and min happens at $x = 270^{\circ}$.


Find all $x$ such that $2^x,2^{x^2}$ and $2^{x^3}$ form $3$ terms of an A.P. I know that if $a,b,c$ are in Arithmetic Progression, then $2b=a+c$, but in this case, I am unable to solve for $x$. Hints are appreciated. Thanks.

Since $2b=a+c$ we have
$2. 2^{x^{2}}=2^{x}+2^{x^{3}}$

which implies that $2^{x^{2}+1}=2^{x}+2^{x^{3}}$

Now taking log on both sides we get
$x^{2}+1$ log2=log $(2^{x}+2^{x^{3}})$ solving this we can find $x$


When two functions are equal, but not. I haven't looked into it much, but this is something I've been aware of that I know I need to look into.

When I have a function $f(x)=\frac{x+1}{x+1}$, There is a discontinuity at $x=-1$, yet $\frac{x+1}{x+1}=1$ and has no discontinuity.  It's like they're equal but not.

The qualities of the function are not preserved after the algebraic manipulation, so I can't strictly say that $\frac{x+1}{x+1}=1$.

This is an issue for me when understanding integrals.  For instance, finding the definite integral of the quotient, if the discontinuity is within my limits, doesn't make sense.  But after changing the quotient to a constant, it's possible: but I've found the area under a curve that wasn't complete.  I've found a solution for an unanswerable, insensible question.

I hope I've made this clear.  My question is, is this right?  How do I come to terms with this?

$x=-1$ is what's called a removable singularity of your function $f$: if you redefine your function at that point it becomes continuous.

As for integration, the definition of the Riemann integral does require the function to be defined everywhere in the interval in question.  However, for a function which is undefined at some point you could consider an improper integral,
which would converge to the correct value.  When you get to Lebesgue integration,
having the function undefined at some point is allowed, and won't make a difference.

There is no discontinuity at x=−1, just take the right and left limits and you'll see that. Hence f = 1.


Does the set $A$, of all permutations of $\{1,2,3,4\}$ that fixes $2$ and $4$ form a subgroup of $S_4$? Does the set $A$, of all permutations of $\{1,2,3,4\}$ that fixes $2$ and $4$ form a subgroup of $S_4$?
I’ve got as far,
The set $A$ will be the identity where all $4$ numbers map to themselves (as all fixed) and $(1, 3)$ as $2$ and $4$ are fixed. So the set $A =\{\operatorname{id}, (1 3)\}$
Now I need to show that this is a sub group of $S_4$, by identity closure and inverses.
Identity is in both so ok but I’m stuck on the other $2$.
Any help would be great
Kara x

Hint: Each such permutation has the form $\pi = \begin{pmatrix} 1 & 2 & 3 & 4 \\ x_1 & x_2 & x_3 &x_4 \end{pmatrix}$ with $x_2=2$ and $x_4=4$.
Show that the composition of two such permutations has the same shape.
Moreover show that the inverse permutation of such a permutation has the same format.
You're right about the structure of $A$, consisting of $(1)(2)(3)(4)$ and $(13)(2)(4)$.

You're correct: $A=\{{\rm id}, (13)\}$.
I will use the two-step subgroup test.
Observe that $\varnothing \neq A\subseteq S_4$.
Consider the multiplication table for $A$:
$$\begin{array}{c|cc}
\circ & {\rm id} & (13) \\
\hline
{\rm id} & {\rm id} & (13) \\
(13) & (13) & {\rm id}.
\end{array}$$
From the table, one can see that for any $a,b\in A$, we have $a\circ b\in A.$
Since $(13)^{-1}=(13)\in A$ and ${\rm id}^{-1}={\rm id}\in A$, $A$ is closed under inverses.
Hence $A\le S_4$.


The polynominal P is defined as $P ( x ) = x^3+ 4x^2 + 5x + 3$ The polynomial $P$ is defined as $P ( x ) = x^3+ 4x^2 + 5x + 3$

Express $P ( x )$ in the form $P ( x ) = ( x + 2 ) ( x^2+ ax + b ) + c$, where $a$, $b$ and $c$ are whole numbers.

any one tell me the idea of solution ♥ ♥ ♥

Hint: open up parentheses on the right hands side and compare coefficients:

$$(x+2)(x^2+ax+b)+c=x^3+(a+2)x^2+(b+2a)x+(2b+c)$$

Thus, for example, $\,a+2=4\;,\;\;b+2a=5\,$ and etc.


Prove $\sqrt{n}+\sqrt{n+k}$ is irrational For what $k\in\mathbb N$, $\sqrt{n}+\sqrt{n+k}$ is irrational? ($\forall n\in\mathbb N$)

Well, a possible but perhaps in the long run not exhaustive method, since you must find for what $k \in \mathbb{N}$ this is true; consider
$$(\sqrt{n}+\sqrt{n+k})(-\sqrt{n}+\sqrt{n+k})=-n+(n+k)$$
Which of course gives $k$.

now, since $k$ is rational, it means that $\sqrt{n}$ and $\sqrt{n+k}$ is rational thus
\begin{align}
n&=&p^{2} \\
n+k&=& q^{2}
\end{align}
Perhaps you can see where I am going with this and finish it off?

Suppose that $$\sqrt n +\sqrt {n+k}=r, \quad r\in\mathbb{Q}$$

Now we have: $$\sqrt {n+k} =r-\sqrt n $$
$$n+k=r^2-2r\sqrt n +n$$
$$2r\sqrt n=r^2-k$$
$$\sqrt n=\frac{r^2-k}{2p}$$

This is a contradiction because the left is $\sqrt n\in\mathbb{I=\mathbb{Q^c}}$, and the right is $\frac{r^2-k}{2p}\in\mathbb{Q}$. The contradiction is due to a wrong assumption that $\sqrt n+\sqrt {n+k}$ was a rational number.


Asymptotic vlaue of $ f(n)=\sum_{i=0}^n\lfloor \sqrt{i}\rfloor\binom{n}{i} $ Inspired by this question I tried to find an asymptotic formula for
$$
f(n)=\sum_{i=0}^n\lfloor \sqrt{i}\rfloor\binom{n}{i}
$$
With the observation:
$$
f(n)=\sum_{i=0}^n\frac{\lfloor \sqrt{i}\rfloor+\lfloor \sqrt{n-i}\rfloor}{2}\binom{n}{i}
$$
And the a bit naive approximation
$$
\frac{\lfloor \sqrt{i}\rfloor+\lfloor \sqrt{n-i}\rfloor}{2}=\frac{1}{n}\sum_{i=1}^{n}{\lfloor \sqrt{i}\rfloor}=\frac{n\lfloor \sqrt{n}\rfloor-\frac{1}{6}\lfloor \sqrt{n}\rfloor\left(\lfloor \sqrt{n}\rfloor-1\right)\left(2\lfloor \sqrt{n}\rfloor+5\right)}{n}\approx\frac{2}{3}\sqrt{n}
$$
We obtain $f(n)\approx\frac{2}{3}\sqrt{n}2^n=:g(n)$. The approximation doesn't seem that bad; I computed with wolfram alpha:
$$
$$
\begin{array}{c|c|c|c}
n & f(n) & g(n) & f(n)/g(n) \\
\hline
1 & 1 & 1.333 & 0.75 \\
2 & 3 & 3.771 & 0.795 \\
3 & 7 & 9.238 & 0.758 \\
4 & 16 & 21.333 & 0.75 \\
5 & 37 & 47.703 & 0.776 \\
10 & 1882 & 2158.782 & 0.872 \\
50 & 5.128\cdot10^{15} & 5.308\cdot10^{15} & 0.996 \\
100 & 8.391\cdot10^{30} & 8.451\cdot10^{30} & 0.993 \\
200 & 1.531\cdot10^{61} & 1.515\cdot10^{61} & 1.011 \\
350 & 2.929\cdot10^{106} & 2.86\cdot10^{106} & 1.024 \\
\end{array}
$$
$$
The error term fluctuates a bit so it seems natural to ask:

Is it true that:
$$
\lim_{n\to\infty}{\frac{f(n)}{g(n)}}=1
$$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{\mathrm{f}\pars{n}} & =\sum_{k = 0}^{n}\left\lfloor \root{k}\right\rfloor{n \choose k}\ =\
1\times\overbrace{\bracks{{n \choose 1} + {n \choose 2} + {n \choose 3}}}
^{\ds{\color{#f00}{3}\ terms}}
\\[3mm] & +
2 \times\overbrace{\bracks{{n \choose 4} + {n \choose 5} + {n \choose 6} + {n \choose 7} + {n \choose 8}}}
^{\ds{\color{#f00}{5}\ terms}}
\\[3mm] & +
3 \times\overbrace{\bracks{{n \choose 9} + {n \choose 10} + {n \choose 11} + {n \choose 12} + {n \choose 13}+ {n \choose 14}+ {n \choose 15}}}
^{\ds{\color{#f00}{7}\ terms}}\ +\ \cdots
\\[5mm] &=
1 \times \sum_{k = 1}^{3}{n \choose k} +
2 \times \sum_{k = 4}^{8}{n \choose k} +
3 \times \sum_{k = 9}^{15}{n \choose k} + \cdots
\\[3mm] & =
\sum_{m = 1}^{\infty}m\sum_{k = m^{2}}^{m^{2} + 2m}{n \choose k} =
\sum_{m = 0}^{\infty}m\sum_{k = 0}^{2m}{n \choose k + m^{2}} =
\sum_{k = 0}^{\infty}\sum_{m\ \geq\ k/2}^{\infty}m{n \choose k + m^{2}}
\end{align}

  I stop here. Some of the binomial sum seems to be related to Hypergeometric Functions. So far, I was able to get rid of the $\ds{\left\lfloor\cdots\right\rfloor}$ function and along the way to find a pattern.


Why does $a^{b}·a^{c}=a^{b+c}$? Why does $a^{b}·a^{c}=a^{b+c}$ ?

I want proof for it, I asked professor and he replied: "I don't know, it's a property anyway that is true and that's all what you need to  know."

It is easy to see this using the definition of a power , if $b$ and $c$ are natural numbers.

$$a^b \cdot a ^c = \overbrace{a \cdots a}^{b\ times} \cdot \overbrace{a \cdots a}^{c\ times} = \overbrace{a \cdots a}^{b+c\ times}
=a^{b+c}$$

With the rules

$$a^{-b}=\frac{1}{a^b}$$

and

$$a^{\frac{1}{b}}=\sqrt[b]{a}$$

it should be possible to extend it to the negative and rational numbers.

HaHa, interesting!

Let us define an operation "$*$" on the set of positive integers $\mathbb{Z}_+\times \mathbb{Z}_+\to \mathbb{Z}_+$ as follows:
$a*b=\underbrace{b\times b\times\cdots \times b}_{a\text{
    times}}=b^a(\text{ usual exponentiation here })$. Define $a^2:=
a*a$, inductively define
$a^{n+1}:=a^{n}*a$. Then $a^3=a^2*a\neq a*a^2$ in general. For
example,
$3^2*3=\underbrace{3\times \cdots \times 3}_{27\text{ times}}$
but $3*3^2=27\times 27\times 27$. Therefore $3*3^2\neq 3^3$.

Just for fun!


subscheme where two morphisms agree is points where they agree on residue fields Let $X, Y, Z$ be schemes, where $X, Y$ are $Z$ schemes.  I know the definition of "the locally closed subscheme of $X$ where two $Z$-  morphisms $\pi, \pi': X\rightarrow Y$ agree" from its universal property.  Also I can define it as the fiber product of the diagonal $$\delta :  Y\rightarrow Y\times_Z Y$$ with $$(\pi, \pi'): X\rightarrow Y\times_Z Y.$$
My question:  how to prove that the underlying set of "the locally closed subscheme where the two morphisms agree" is the same as the set of points where the two morphism agree on the residue field.
It is probably clear thatthe former is contained in the latter, but why is it all of them?  That is, why is a point where $\pi, \pi'$ agree on the residue field necessarily contained in "the subscheme where $\pi, \pi'$ agree"?

Let "the set where the morphisms agree" be called $A$ and the fiber product $X\times_{Y\times_Z Y} Y$ be called $V$. Our goal is to show the underlying sets are equal. You state in your post that you are ok with $V\subset A$ set-theoretically (where we fudge things a little and identify $V$ with it's image under the immersion $V\to X$).
To show that $V=A$ set theoretically, it is enough to show that if $a\in A$, the fiber $V \times_X \operatorname{Spec} k(a)$ is nonempty. To start, we define the maps $i:\operatorname{Spec} k(a)\to X$ given by the standard inclusion and $j:\operatorname{Spec} k(a)\to Y$ given by $j((0))=f(a)$ as sets and $j^\sharp:\mathcal{O}_Y \to j_*\mathcal{O}_{\operatorname{Spec} k(a)}$ is given by $\mathcal{O}_Y(U)\to \mathcal{O}_{Y,f(a)} \to k(f(a)) \to k(a)$ if $a\in U$ and $0$ otherwise. As $i$ and $j$ agree when composed with the maps to $Y\times_ZY$, this gives a map $\operatorname{Spec} k(a)\to V$ so that the composite of this map with the natural projections $V\to X$ and $V\to Y$ agree with $i$ and $j$ by the universal property of the fiber product. But this means that there's a point of $V$ which maps to $a$ and we're done.

Disclaimer: I have just finished this section in Vakil myself, so I am happy to be corrected by more experienced AGers.
Claim (see the discussion above Vakil 10.2 B): The set of points where two morphisms of $Z$-schemes $\pi, \pi' : X \to Y$ agree is the set of point on which those morphisms agree (on the level of points) and on induced maps of residue fields.
Let $\pi^{\#}_x$ and $\pi'^{\#}_x$ denote the maps of residue fields at $x \in X$ induced by $\pi$ and $\pi'$, respectively. By the universal property of the locus where two schemes agree, we must show
$$X \times_{Y \times_Z Y} Y$$
is in bijective correspondence as a set to
$$\{x \in X \ | \ \pi(x) = \pi'(x), \pi^{\#}_x  =\pi'^{\#}_x \}$$
Recall that points of $$X \times_{Y \times_Z Y} Y$$ are in bijective correspondence with tuples $(x, y, z, P)$ such that $\delta(y) = (y, y) = (\pi, \pi')(x) = z$ and $P \subset \kappa(x) \otimes_{\kappa(z)} \kappa(y)$ is prime. See Stacks Project, Lemma 26.17.5.
Clearly it is necessary that $\pi(x) = \pi'(x)$ for these conditions to be satisfied.
Now, again by Lemma 26.17.5, primes $P \subset \kappa(x) \otimes_{\kappa(z)} \kappa(y)$ have $\kappa(P)$ corresponding to $\kappa(z) = \kappa(\pi(x), \pi(x)) = \kappa(\pi'(x), \pi'(x))$ in the lower right corner of the tensor product diagram.
But $\kappa(P)$ then corresponds to the diagonal map $\kappa(y) = \kappa(\pi(x)) = \kappa(\pi'(x)) \to \kappa(x)$ found in the tensor product diagram. Hence, given such a prime, since $z$ is both $(\pi(x), \pi(x))$ and $(\pi'(x), \pi'(x))$, this diagonal map is both the map $\pi^{\#}_x$ and $\pi'^{\#}_x$, and so they are the same map.


Chinese Remainder Theorem and polynomial roots Suppose $f$ is a polynomial with integer coefficients and that $f(3) \equiv 0 \mod 7$ and $f(5)\equiv 0\mod 11$. Use the Chinese Remainder Theorem to show that there exists $x \in \mathbb{Z}$ such that $f(x)\equiv 0 \mod 77$.

This is a CRT problem where we are given two equivalences with relatively prime moduli.

All I can see is that the CRT gives us some unique $K \mod 77$
$\\$ such that $K\equiv f(3)\equiv 0 \mod 7 \quad \operatorname{and} \mod 11$, but I have no idea why there must be an $x$, such that $f(x)\equiv K\equiv 0 \mod 77$

More generally,

Assume f(x) is a polynomial with integer coefficients. Assume the polynomial has a root modulo m and a root modulo n, and assume gcd(m, n)=1. Prove that the polynomial has a root
modulo mn.

By the CRT, $\exists x\in\mathbb{Z}$ such that

$x\equiv 3 \mod 7
\\
x\equiv 5 \mod 11$.

Since $7 |f(x)$ and $11|f(x)$ and 7 and 11 are coprime, $\quad 77|f(x)$

If $K \equiv 0 \mod 7$ and $K \equiv 0 \mod 11$ then (since $7$ and $11$ are co-prime... indeed each is prime) $K \equiv 0 \mod 77$.


Maximization of multivariable function $ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$ subject to constraint This is purely out of curiosity. I have this set of a high school entrance Math exam in Vietnam (a special high school for gifted kids, the exam was held 3 days ago, maybe 4, taking into account the time zone). Here's one question that I've been stuck with:

With $a, b, c$ $\in \mathbb{R}^{+}$ such that $ab + bc + ca + abc =2$, find the max of

$ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$

I'm just curious what kind(s) of techniques these junior high students can use, since as far as I remember when I was at that age, calculus and derivatives aren't taught until high school.

The functional form of $M$ suggests us to change variable to $u = a+1, v = b+1, w = c + 1$.
In terms of them, the expression of $M$ simplifies to

$$M = \frac{u}{1+u^2} + \frac{v}{1+v^2} + \frac{w}{1+w^2}$$

The constraint $abc + ab + bc + ca = 2$ can be rewritten as

$$(a+1)(b+1)(c+1) - (a+b+c) - 1 = 2\quad\iff\quad u + v + w - uvw = 0$$

Since $a,b,c > 0$, $u,v,w \in (1,\infty)$. Pick $A, B, C \in (\frac{\pi}{4},\frac{\pi}{2})$ such that $u = \tan A$,$v = \tan B$ and $w = \tan C$.
The constraint tell us
$$\tan(A+B+C) = \frac{u+v+w - uvw}{1 - uv - vw - wu} = 0
\quad\implies\quad A+B+C = k\pi, \quad k \in \mathbb{Z}
$$
Since $A + B + C \in (\frac{3\pi}{4}, \frac{3\pi}{2})$, $k$ equals to $1$
and  $A + B + C = \pi$.

In terms of $A,B,C$, the expression we want to maximize becomes

$$M = \frac{\tan A}{1+\tan^2 A} + \frac{\tan B}{1+\tan^2 B} + \frac{\tan C}{1+\tan^2 C} = \frac12 \left[ \sin(2A) + \sin(2B) + \sin(2C) \right]$$

This is nothing but the area of triangle inscribed inside unit circle with
angles $2A$, $2B$ and $2C$ subtended at center. It is well known for such triangles, the one which maximize the area is an equilateral triangle. This implies $M$ is maximized when

$$A = B = C = \frac{\pi}{3} \implies a = b = c = \tan\frac{\pi}{3} - 1 = \sqrt{3} - 1$$

and the maximum value of $M$ is $\displaystyle\;\frac{3}{2}\sin\frac{2\pi}{3} = \frac{3\sqrt{3}}{4}$.

Here Laplace multipliers, derivatives and gradients do not exist. Only multiplying and checking if something simplifies.

In this case, symmetry for simplifying the problem into:
$$
\mathcal{M}: 3{x+1 \over x^2+2x+2}\\
st. x^3+3x^2=2
$$

This holds because every term on the function is a sum of independent terms. In here symmetry is not a false friend.

Which reduces to finding the positive roots of $x^3+3x^2-2=0$. We do not know to solve a cubic equation, so we only can guess to break the polynomial. Our first guess is $x=-1$. Broken:
$$
(x^3+x^2+2x^2+2x-2x-2)/(x+1)=x^2+2x-2
$$

And the others two are given by, yes, we know how to solve that already:
$$
x=\frac 12 (-2\pm\sqrt{4+8})=-1\pm \sqrt{3}
$$

Hence the only positive solution is $x=\sqrt{3}-1=a=b=c$ and $M=3{\sqrt{3}\over 3+1-2\sqrt{3}+2\sqrt{3}-2+2}=3{\sqrt{3}\over 4}\sim 1.3$.

What if they are not equal?. Suppose $a=b=1,c=x$:
$$
\mathcal{M}: \frac45+{x+1 \over x^2+2x+2}\\
st. 3x=2
$$
then$x=2/3$ and $M=\frac45+{2/3+1 \over 4/9+4/3+2}=\frac45+{15 \over 34}\sim 1.2<1.3$

Yes, symmetry here helps, and the solution is a maximum.


Orthonormal Matrices-Intuition Why is it, geometrically, that the row space is ALSO orthonormal? What exactly does the transpose LOOK like? Normally the row and column space are two separate things, but in the case of an orthonormal matrix, you have that the row space is actually the inverse of the column space, so also, what does an inverse space look like? I feel there should be a very good geometric interpretation, but I don't see it.

I think this is actually a good question. We can talk about two things: the geometric intuition of an orthogonal matrix and the relationship between row and column space. The latter is slightly more general.

In the first place, let $O(n) = \{A \in M_n(\mathbb{R}) : A^T A = 1\}$, and let $SO(n)$ be the restriction of $O(n)$ to matrices with determinant 1. Note that this basically gets rid of the $A$ with negative determinant, since the condition $A^TA = I$ implies
\begin{gather*}
1 = \det(I_n) = \det(A^T A) = \det(A^T)\det(A) = \det(A)^2,
\end{gather*}
since the transpose does not affect the determinant (prove it or trust me, it's probably not the key idea on a first reading, though it is a consequence of and implies non-trivial things). Therefore $\det(A) = \pm 1$.

Always give yourself simple examples (and counter-examples!). For instance, take
$$A = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}.$$
This matrix is not in $SO(n)$, since $\det(A) \neq 1$. Geometrically, this matrix is a reflection across the $x$-axis. To see this, see what happens to the basis vectors $(1,0)$ and $(0,1)$. The $x$-coordinate (1,0) stays fixed, while $y \rightarrow -y$. Note that this does not preserve orientation. That is,

Flips aren't all bad though. They still preserve volume. Matrices in $SO(n)$, however, preserve both orientation and volume and can be viewed as rotations. Consider for instance the matrix
$$R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$
This is a rotation matrix in $\mathbb{R}^2$. Its characteristic polynomial is $\lambda^2 -2\lambda \cos\theta + 1$, and the discriminant of this polynomial is $b^2 - 4ac = -4\sin^2(\theta).$ Therefore, whenever $\theta \neq 0 (mod 2\pi)$, the eigenvalues are imaginary. The eigenvalues of an orthogonal matrix always have norm $1$, so they are of the form $e^{i\theta_k}$ for various $k$.

If you want to talk about row and column space, you're going to have to think about the adjoint map a bit, i.e. $A \rightarrow A^*$, where $A^*$ is the matrix which satisfies $(Au,v) = (u,A^*v)$ for every $u, v$ in the vector space. Here $( \cdot,\cdot)$ denotes the inner product on the space (say the Euclidean inner product $x \cdot y$).

It turns out (one can prove by straight linear algebra by expanding sums and re-arranging terms) that the matrix which satisfies precisely that condition is always the transpose of the matrix. One can then ask what the relationship between these matrices is, or how one can think of them.

One interpretation is that if $A$ is a map from the vector space $V$ to v.s. $W$, then $A^* : W^* \rightarrow V^*$. Here, $V^*$ denotes the dual space of $V$ (ditto $W$). It can be seen as the map on row vectors ('dual vectors') induced by the map on column vectors.

Note: the reason I call row vectors 'dual vectors' is because one can associate to any row a linear functional on the vector space: write the row as $v^T$ and act on vectors $u$ by $u \rightarrow v^Tu \in \mathbb{R}$.

If you want I can get into this, depends on how much of this is below/above your level.

A real orthogonal matrix is geometrically a rotation, a reflection or a permutation of the axes.  For example, the matrix M =
$\begin{pmatrix}
cos(t) & -sin(t)\\
sin(t) & cos(t)\\
\end{pmatrix}$ rotates any vector through an angle of t.  $M^{-1}$ undoes what M did, which is to say, it rotates every vector back through the angle (-t) and leaves it where it started.  The matrix
$\begin{pmatrix}
1 & 0\\
0& -1\\
\end{pmatrix}$
reflects a vector across the x axis and its inverse reflects it back. This matrix can also be expressed in terms of sines and cosines if you wish.

The matrix
$\begin{pmatrix}
0 & 1\\
1& 0\\
\end{pmatrix}$ maps the x-axis into the y-axis and the y-axis into the x-axis.  It inverse does the opposite.  It also can be expressed in terms of sines and cosines.

To be sure this is right, just compute what each matrix does to a sample of vectors; figure out what its inverse is, and see that it undoes what M did to your sample.


show $\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^{x}=1+\frac{x(x-1)}{2j^{2}}+O(\frac{1}{j^{3}})$ I should show the following equality provided $x\in\mathbb{C}$ without negative integers:
$$\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^{x}=1+\frac{x(x-1)}{2j^{2}}+O(\frac{1}{j^{3}})$$
I tried using binomial theorem, then I get for $(1+\frac{1}{j})^{x} = 1+ \frac{x}{j} +\frac{x(x-1)}{2j^{2}} +O(\frac{1}{j^{3}}) $. So it seems to go in the right direction but I don't know how multiplying with $\left(1+\frac{x}{j}\right)^{-1}$ brings me to the right result. Any help is greatly appreciated.

Geometrically & binomially expand ...
\begin{eqnarray*}
\left(1+\frac{x}{j}\right)^{-1}\left(1+\frac{1}{j}\right)^{x}= \left(1-\frac{x}{j}+\frac{x^2}{j^2}+ O(\frac{1}{j^{3}}) \right)\left(1+\frac{x}{j}+\frac{x(x-1)}{2} \frac{1}{j^2} +O(\frac{1}{j^{3}}) \right)
\end{eqnarray*}

okay so if we have some $x\in\mathbb{C}$ then it is fair to write it as $x=a+bj$
now:
$$\frac1j=\frac1j \frac jj=-j$$ and so:
$$\frac xj=-j(a+bj)=-aj+b$$
see if you can do anything with this

looking at the other part:
$$\left(1+\frac1j\right)^x=\left(1-j\right)^x=\left(\sqrt{2}e^{7\pi j/4}\right)^x$$
Now i suggest multiplying this term by the expansion of $(1-xj)^{-1}$ and see what you get


Prove $\,\gcd(a,b)\mid {\rm lcm}(a,b)\ $ I have a hard time with this question, can someone explain?
"Prove that for any number $a$ and $b$, $(a,b)$ goes into $\text{lcm}(a;b)$"

Hint: by gcd & lcm universal property $\,\gcd(a,b)\mid a\mid {\rm lcm}(a,b)\,$ so by transitivity of divisibility ...
Remark $ $  The hint boils down to $\,\inf S \le \sup S\,$ in poset / lattice language. More generally, the hint shows that every common divisor of a set of integers divides every common multiple.

I think the question is asking you to prove that $a\times b$ divides $\mathrm{lcm}(a,b)$. That is, prove that $\mathrm{lcm}(a,b)\times n=a\times b$ where $n\in\mathbb{Z}$.


Subspace criterion (zero vector?) This may be a poor question but it is one that I cannot wrap my head around.

The theorem I am presented with is as follows:

  (Subspace criterion) Suppose $V$ is vector space over $F$.
  A nonempty subset $U ⊆ V$ is a subspace if and only if $U$ is closed for addition
  and multiplication with scalars.

When proceeding to worked examples, I noticed that the first property that is checked when proving a set is a subspace using the above criterion is whether the zero vector exists. However, this property is not mentioned in the criterion?

I assume that there is some connection between a set being closed under addition and it has a zero vector but I am failing to deduce it.

Any help is greatly appreciated.

Consider a subspace of $V$ of dimension $1$, $U$, where the entire subspace is spanned by scalar multiples of some element $\vec{u}$.

No matter what $\vec{u}$ is, we can always construct a sum of the form:

$$\vec{u} + -\vec{u}$$

This is equal to $\vec{0}$ by definition, so the zero vector needs to be in the subspace in order for the subspace to be closed with respect to vector addition.

Let 0'=additive identity of field F,now since vector subspace S is closed with respect to scalar multiplication,hence 0'(a)€S
0'(a)=(0'+0')a  {a€S}
0'(a)=0'a+0'a
0'(a)=0''€S{where 0''=zero of vector subspace},,Hence there is a connection between closed wrt scalar multiplication and existence of Zero of subspace


How do you find the closest square number to another number without using a calculator Say we try to find the closest square number to 26. we already know the closest square number is $25$.
However, how do I calculate out 25?

  Because, if I try to prime factorize it like so:
  $\sqrt{26}$ = $\sqrt{2*13}$

  And I try to round 13 to the closest even number, I would get this equation here from prime factorization:
  $\sqrt{26}$ = $\sqrt{2*13}$
  -> $\sqrt{2*2*2*3}$

Then, if I write out the equation, it wouls turn up as:
$\sqrt{24}$ ,
Which is not 25.

I there a way to find out the closest square number to a certain number without using a calculator?

You have to subtract all the digits but the first one from the number then add this to the original number. e.g  with 745, we would subtract the 45 as those are the digits after the first one. We are left with 700, we add 700 to 745 therefore are answer will be 1445. Weeeee did it!!


Pigeonhole Principle - Pairs of Sets with distinct sums (a) Each of 15 men and 15 woman is given an integer between 1 and 100 inclusive with no integer appearing more than once. The value of a man-woman pair is the sum of their numbers. Prove that there are two man-woman pairs with the same value.

(b) Suppose there are 14 men and 14 women as above. Prove that there are still two man-woman pairs with the same value.

(c) Give an example with 13 men and 13 women where the value of every man-woman pair is unique.

I can prove (a), namely that the set of possible sums is $S$ = {$3,4,5,...,197,198,199$} with |$S$| = $197$. With 15 men and women there are $15\cdot 15 = 225$ pairs so by the pigeonhole principle there will be at least two pairs with the same sums because $225>197$.

With 14 men and women there are $14\cdot 14 = 196$ possible pairs but this is not larger than |$S$| = $197$. I assume that due to the way this is set up it is impossible to have certain pairs together so the actual amount of sums you could have is less than 197 but I don't know how to prove this.

For (c) I really have no clue how to come up with the lists without brute-force which isn't really working for me.

Edit: A basically identical problem has already been asked and solved here: Pigeon-Hole Principle Common Sum

You cannot get all of {3,...,17}.  If you have both 3 and 4, they must be 1+2 and 1+3, so 2 and 3 are on the same side. Then to also get 5, 4 must be on the same side as 2 and 3. Repeat this argument up to 16.  But then you cannot get 17.  A similar argument shows that you cannot get all 15 of {185,....,199}.


Why does sample standard deviation underestimate population standard deviation? Refering to this wikipedia page Unbiased estimation of standard deviation, it says that "it follows from Jensen's inequality that the square root of the sample variance is an underestimate".

I do know that for the concave square root function, Jensen's inequality says that the square root of the mean > mean of the square root.

So, how do we conclude that the square root of the sample variance underestimates population standard deviation?

Since we know from Jensen's inequality that square root of the mean > mean of the square root, does "square root of sample variance" somehow relate to "mean of the square root" while "population standard deviation" somehow relates to "square root of the mean"?

Added after joriki's response:

Given joriki's response about using a single sampling of data, I am now left with why $s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N{(x_i-\overline{x})^2}}$ will underestimate pop std dev.  In order to use Jensen's inequality (mean of the square root < square root of the mean).  I need to somehow relate the expression for $s$ to "mean of square root".  I do see the square root sign in the expression for $s$ but where is the "mean" of this square root quantity?

The mean is part of what it means for an estimator to be biased. You can't make the estimator unbiased by averaging over several estimates; to the contrary, you can show that it's biased by averaging over estimates and showing that the expected average isn't the value to be estimated. (You can reduce the bias and the variance of the estimator by averaging several estimates, but as discussed above you can do that even better by using all the data for one estimate.)

For example, if your population has equidistributed values $-1,0,1$, with variance $\frac23$, and you take a sample of $2$, you'll get variance estimates of $0$, $\frac12$ and $2$ with probabilities $\frac13$, $\frac49$ and $\frac29$, respectively, yielding the correct mean $\frac13\cdot0+\frac49\cdot\frac12+\frac29\cdot2=\frac23$, whereas the estimates for the standard deviation, $0$, $\sqrt{\frac12}$ and $\sqrt2$ average to $\frac13\cdot0+\frac49\cdot\sqrt{\frac12}+\frac29\cdot\sqrt2=\frac49\sqrt2\neq\sqrt{\frac23}$, with $\frac49\sqrt2\approx0.6285\lt0.8165\approx\sqrt{\frac23}$, an underestimate as expected. If you take a sample of $3$ instead, the mean improves to $\frac19\cdot0+\frac49\cdot\sqrt{\frac13}+\frac29\cdot\sqrt{\frac43}+\frac29\cdot1=\frac19(8\sqrt{\frac13}+2)\approx0.7354$.

Reality is that the sampling distribution of $S$ is skew to the right, more so for s from small samples, less so for $S$ from large samples.

Fact 1: the median of a distribution that is skew to the right is less than the mean of the distribution.

Fact 2: The median $S$ is the $50^{th} \% $ile.

Fact 3: the mean $S$ is equal to the population standard deviation.

Therefore, the mean has a percentile larger than $50\%$, and a single randomly drawn $S$ is more likely to underestimate sigma than to overestimate sigma, even though the mean $S$ is sigma.
The discrepancy  between the median s and the mean s vanishes as sample size approaches infinity, which is why $t$ converges to $Z$ for progressively larger sample sizes.


Why can't you have more turning points than the degree? I get that each degree can correspond to a factor.   $x^5=(x+a)(x+b)(x+c)(x+d)(x+e)$ and that results in 4 turning points, so the graph can "turn around" and hit the next zero.    Why can't a curve have more turning points than zeros?

In the graph below, there are 4 roots, so degree is 4, but way more turning points than 4.    What gives?   Are those additional turning points represented by imaginary roots?

The problem is that you are confusing real zeros of a polynomial with the degree.  These are not the same.  The degree of a single variable polynomial is the highest power the polynomial has.

Your hand drawn graph has only 4 real roots, but if it was a polynomial it must have more complex roots.  You could not make all those turning points without this been true.  You may not be aware of complex numbers.

Although you mention this as precalculus, this does become clearer with calculus, where you find the turning points ( "local maxima and minima" ) by equating the derivative of the polynomial to zero.  The derivative of an n-th degree polynomial is an (n-1)th degree polynomial, so their can be as many as (n-1) turning points.  However, the derivative's roots need not all be real, and in that case the original polynomial would have fewer real local maxima and minima than n-1.

So the problem is equating the number of real roots with the degree.  You can really only know the degree by knowing the highest power the polynomial has.  This is not always immediately obvious from the shape of a graph.  You also need to be aware of the possibility of complex roots.

If there is a fifth degree polynomial there are five roots, some real some complex conjugates, adding to total five roots. And four (one less) turning points.

EDIT1;

It is from the IVT and mean value theorems that are operating.Between two roots there is a maximum turning point,at its tangent is parallel to line joining roots. With two roots, one turning ... five roots four turnings. The maxima / minima tally (one less)  occurs due to  number of roots irrespective of real or complex .

The graph you drew has a degree much higher.For example in $ y= 1 + a \sin x + \sin b x $ it has infinitely many turning points and infinitely many roots. They cannot be fully indicated  as in the domain that you have pictured. Here is a graph for $ (a = 1.1, b = 0.7 ). $


how do you ban things from subsets How can you get the total number of subsets, if you generate all subsets from the set $\{A,B,C,D,E,F\}$ but BAN all subsets that either include $A$ and $B$, $B$ and $C$, or $C$ and $D$.

Is there a way to do this non-manually?

I would also appreciate links and general information about the topic!

There are $2^6$ total subsets.  Now let us subtract the banned ones-there are $2^4$ subsets containing $A$ and $B$ and also $2^4$ with each other pair.  We have subtracted the $2^3$ that have $A,B,C$ twice and the $2^3$ that have $B,C,D$ twice, so we have to add them back in once.  Finally the $2^2$ that have $A,B,C,D$ have been counted, subtracted three times, and added twice, so no more correction is necessary.  The final answer is
$$2^6-3\cdot 2^4+2\cdot 2^3=32$$
This is an example of the inclusion-exclusion principle.

To "ban" $A$ and $B$, take all the subsets of:
$$
\{C,D,E,F\}.
$$
In more detail, first carry out the calculation
$$
\{A,B,C,D,E,F\} \setminus \{A, B\},
$$
which results in the set $\{C,D,E,F\}$.  Take all subsets of this set.


$\require{enclose}\enclose{horizontalstrike}{\rm Greatest}\!$ Least prime factor of $n$ is less than square root of $n$, proof I remember reading this somewhere but I cannot locate the proof.

It is the smallest prime factor that is less than or equal to $\sqrt{n}$, unless $n$ is prime.  One proof is as follows: Suppose $n=ab$ and $a$ is the smallest prime factor of $n$, and $n$ is not prime.  Since $n$ is not prime, we have $b\ne1$.  Since $a$ is the smallest prime factor of $n$, we have $a\le b$.  If $a$ were bigger than $\sqrt{n}$, then $b$ would also be bigger than $\sqrt{n}$, so $ab$ would be bigger than $\sqrt{n}\cdot\sqrt{n}$.  But $ab=n$.

The answer is quite simple.

Given that n is composite so $n=p\cdot q\cdot r\cdot s\cdot t\cdots$ where $p,q\cdots$ are primes by fundamental theorem of arithmetic.

Now assume without loss of generality that $p<q<r<\cdots$. So $p<q\cdot r\cdot t\cdots$.

Now take square root on both sides then $\sqrt{p}<\sqrt{q\cdot r\cdot s\cdot t\cdots}$. Now multiply both sides by $\sqrt{p}$. We get $p<\sqrt{p\cdot q\cdot r \cdot s \cdot t\cdots}$ and $\sqrt{p\cdot q\cdot r\cdot s\cdot t\cdots}$ is equal to $\sqrt{n}$.

Hence established that $p \leq \sqrt{n}$.


How to check if a matrix is positive definite I want to know how to check if a matrix M is positive definite
,assume that M is 3x3 real numbers matrix

I think one way is to put the matrix in a quadratic form  $X^TMX$
 , where X is a vector $X^T=[x_1   x_2   x_3]$
 , my question is if I found that $X^TMX = ax_1^2 + bx_1*x_2+ ........$ can I say that the matrix M is not positive definite because the term $bx_1*x_2$ can be negative or I have to try to put the value of $X^TMX$ in the form of sum of squares e.g.,$()^2+()^2+.....$ and then decide?

and what is the relation between the positive definiteness of a matrix and its determinant?

I don't think there is a nice answer for matrices in general. Most often we care about positive definite matrices for Hermitian matrices, so a lot is known in this case.

The one I always have in mind is that a Hermitian matrix is positive definite iff its eigenvalues are all positive.

Glancing at the wiki article on this alerted me to something I had not known, Sylvester's criterion which says that you can use determinants to test (a Hermitian matrix) for positive definiteness by checking to see if all the square submatrices whose upper left corner is the $(1,1)$ entry have positive determinant.

Sorry if this is repeating things you already know, but it's the most useful information I can provide. Good luck!

Just do calculation of the term X^TAX and then check whether the answer can be negative or not. if it can be negative then it is not positive definite or vice versa for example if answer comes to be x1^2+x2^2+x3^2 then it can never be negative as there are squared terms so in this case matrix A will be positive definite.


What are the two 'sides' of a decimal number called? Is there a fancy name for the "left side" and "right side" of a decimal number?

(That is, the pre-decimal part and the post-decimal part.)

We really do use "integer part" and "fractional part" respectively: see Wikipedia, e.g., on decimal fractions.

  The integer part, or integral part of a decimal number is the part to the left of the decimal separator. (See also truncation.) The part from the decimal separator to the right is the fractional part.

The left side of the decimal is called whole number and the right side of the decimal is known as decimal.


Is the product of two Gaussian random variables also a Gaussian? Say I have $X \sim \mathcal N(a, b)$ and $Y\sim \mathcal N(c, d)$.  Is $XY$ also normally distributed?

Is the answer any different if we know that $X$ and $Y$ are independent?

The product of two Gaussian random variables is distributed, in general, as a linear combination of two Chi-square random variables:

$$ XY \,=\, \frac{1}{4} (X+Y)^2 - \frac{1}{4}(X-Y)^2$$

Now, $X+Y$ and $X-Y$ are Gaussian random variables, so that $(X+Y)^2$ and $(X-Y)^2$ are Chi-square distributed with 1 degree of freedom.

If $X$ and $Y$ are both zero-mean, then

$$ XY \sim c_1 Q - c_2 R$$

where $c_1=\frac{Var(X+Y)}{4}$, $c_2 = \frac{Var(X-Y)}{4}$ and $Q, R \sim \chi^2_1$ are central.

The variables $Q$ and $R$ are independent if and only if $Var(X) = Var(Y)$.

In general, $Q$ and $R$ are noncentral and dependent.

For a general answer to your question please refer to the Wishart distribution.


Can i find a 3D function given some points? is it possible to find a 3D function given a set of data points? i tried plane-fitting it did not work, too chaotic for a plane.

I am trying to find a 3D equation that cover most of points, how can i get to it ?
Is there some terminolgy for this, maybe something i can study?

points are:

$( 3 , 0.48 , 1.25 )$

$( 15 , 3.77 , 5.5 )$

$( 5.5 , 1.3 , 2.75 )$

$( 12.5 , 3.08 , 5.25 )$

$( 2.2 , 0.1 , 0.2 )$

$( 2.2 , 0.2 , 0.3 )$

$( 2.2 , 0.3 , 0.5 )$

You can fit anything. You have 7 points. Assume your function is of the form $f(x,y)=ax+by+cx^2+dy^2+ex^3+fy^3+gx^4$. Now substitute your given points and solve simultaneous linear systems in variables $a,b,c,d,e,f,g,$ in seven variables.  If mine choice of $f$ doesn't work, you can adjust $f$ to make the linear system consistent.

Take a look at something like a "thin-plate spline" (Duchon $D^m$ splines) for a $3D$ case, or at rbf interpolation.


Is the ring $3\mathbb Z$ a ring homomorphic image of the ring $2\mathbb Z$. Is the ring $2\mathbb Z$ isomorphic to the ring $3\mathbb Z$ ?

Solution: Let if possible,$\phi:\mathbb {2Z\to 3Z}$ be a ring isomorphism.

Then $\phi$ is a group isomorphism between the additive groups $2\mathbb Z$ and $3\mathbb Z$.

Now,$\phi(2)$ completely determines $\phi$.Let $\phi(2)=3x$ where $x\in \mathbb Z$.

Then $\phi(2y)=3xy $ for all $y\in \mathbb Z$.

Now,$\phi$ is surjective,so $3\in 3\mathbb Z$ has a preimage.So,$\phi(2y)=3$ for some $y\in \mathbb Z$.

Which implies,$3xy=3\implies xy=1$.So,$x,y\in \{-1,1\}$.But then $\phi(2)=\pm3$ and hence $\phi(4)=\phi(2.2)=\phi(2).\phi(2)=9$.

Again,$\phi(4)=\phi(2+2)=\phi(2)+\phi(2)=\pm 6$,but that is not possible.

So,$2\mathbb Z$ and $3\mathbb Z$ are not ring isomorphic.

Note: $3\mathbb Z$ is not a ring homomorphic image of $2\mathbb Z$,as we did not make use of injectivity of $\phi$ in the above proof.

Is the above conclusion correct?

Your argument is correct.

But if you strip it of all inessential parts, you get a shorter, in my opinion also clearer, argument that actually proves something more.

The key point is, as you did, to look at $\phi(4)$, where $\phi$ is an arbitrary rng-homomorphism from $2\mathbb{Z}$ to some sub-rng of $\mathbb{Z}$. Since $4 = 2\cdot 2 = 2 + 2$ it follows that
$$0 = \phi(4) - \phi(4) = \phi(2\cdot 2) - \phi(2+2) = \phi(2)^2 - \bigl(\phi(2) + \phi(2)\bigr) = \phi(2)\cdot \bigl(\phi(2) - 2\bigr)\,.$$
Thus either $\phi(2) = 0$ or $\phi(2) = 2$. Hence either $\phi = 0$ or $\phi$ is the inclusion of $2\mathbb{Z}$ in $\mathbb{Z}$, or $\phi$ is the identity map on $2\mathbb{Z}$.

In particular, the only rng-homomorphism $2\mathbb{Z} \to 3\mathbb{Z}$ is the zero map.

Your argument seems correct. But there's an easier way: if $2\Bbb Z$ and $3\Bbb Z$ were isomorphic, then the quotient rings $\Bbb Z/3\Bbb Z$ and $\Bbb Z/2\Bbb Z$ would be isomorphic too. But this can't be possible since $\Bbb Z/3\Bbb Z$ has precisely 3 elements and $\Bbb Z/2\Bbb Z$ has precisely 2 elements.