Formula for occurrence of leap years in the Jewish calendar Over at Judaism.SE, there was a discussion about a formula to determine leap years in the Jewish calendar. Basically, the calendar follows a 19-year cycle, and seven of those years -- 3, 6, 8, 11, 14, 17, and 19 -- are leap years. Someone reduced this to the simple equation: (7y+1) mod 19 < 7.

My question is whether there is any rhyme or reason as to why this works, and is there a method to go about determining such formulas -- or is it all simply trial and error?

Here's "why this works", and also how to find a formula (with exhaustive search, essentially) when one exists. But really, the answer is that no such formula may exist in general; we're just lucky it exists for this particular set of 7 values mod 19.

The basic idea is that for any integer $k$ relatively prime to $19$, the function $y \mapsto ky$ modulo 19 is invertible. In other words, instead of looking at a set of integers $y$ modulo 19, you can look at the corresponding set of integers $ky$ mod 19, and if that set happens to have a simple characterisation for some $k$, it amounts to a characterisation of your original set of values.

So consider a table of $ky \bmod 19$, for $y$ in your list $[0, 3, 6, 8, 11, 14, 17]$, and integers $k$. Only integers $k$ from $1$ to $18$ need be considered, as $(k+19)y \equiv ky \mod 19$. No need to consider $k=0$ as $0$ is not relatively prime to $19$: $0y \bmod 19$ is always $0$, so it doesn't tell you anything about $y$. Now it so happens that for $k=7$, the values modulo 19 that are attained happen to be consecutive modulo 19. For instance, this Python code:

for k in range(1,19):
    values = [(k*y)%19 for y in [0, 3, 6, 8, 11, 14, 17]]
    attained = ['*' if x in values else '.' for x in range(19)]
    print "%2d: %-30s %s" % (k, values, ''.join(attained))

prints

 1: [0, 3, 6, 8, 11, 14, 17]       *..*..*.*..*..*..*.
 2: [0, 6, 12, 16, 3, 9, 15]       *..*..*..*..*..**..
 3: [0, 9, 18, 5, 14, 4, 13]       *...**...*...**...*
 4: [0, 12, 5, 13, 6, 18, 11]      *....**....***....*
 5: [0, 15, 11, 2, 17, 13, 9]      *.*......*.*.*.*.*.
 6: [0, 18, 17, 10, 9, 8, 7]       *......****......**
 7: [0, 2, 4, 18, 1, 3, 5]         ******............*
 8: [0, 5, 10, 7, 12, 17, 3]       *..*.*.*..*.*....*.
 9: [0, 8, 16, 15, 4, 12, 1]       **..*...*...*..**..
10: [0, 11, 3, 4, 15, 7, 18]       *..**..*...*...*..*
11: [0, 14, 9, 12, 7, 2, 16]       *.*....*.*..*.*.*..
12: [0, 17, 15, 1, 18, 16, 14]     **............*****
13: [0, 1, 2, 9, 10, 11, 12]       ***......****......
14: [0, 4, 8, 17, 2, 6, 10]        *.*.*.*.*.*......*.
15: [0, 7, 14, 6, 13, 1, 8]        **....***....**....
16: [0, 10, 1, 14, 5, 15, 6]       **...**...*...**...
17: [0, 13, 7, 3, 16, 10, 4]       *..**..*..*..*..*..
18: [0, 16, 13, 11, 8, 5, 2]       *.*..*..*..*.*..*..

Now look at each row in the picture on the right, until you notice that for $k=7$, all the asterisks are consecutive (cyclically) — the values taken are $[18, 0, 1, 2, 3, 4, 5]$. You can add 1 to shift it cyclically, then the values will be $[0, 1, 2, 3, 4, 5, 6]$. Thus the set of values $(7y+1) \bmod 19$ for $y$ in your original list is the above, which has a nice simple characterisation as the set of nonnegative integers less than 7. Using this characterisation gives the formula $(7y + 1) \bmod 19 < 7$. (If you look at the table you see that $k=12$ also has the values consecutive; this is not surprising because $12 \equiv -7 \mod 19$.)

We just happened to get lucky. Starting with a different set of 7 integers may not give any such formula. For instance if we change "11" to "12", we get the picture:

 1: [0, 3, 6, 8, 12, 14, 17]       *..*..*.*...*.*..*.
 2: [0, 6, 12, 16, 5, 9, 15]       *....**..*..*..**..
 3: [0, 9, 18, 5, 17, 4, 13]       *...**...*...*...**
 4: [0, 12, 5, 13, 10, 18, 11]     *....*....****....*
 5: [0, 15, 11, 2, 3, 13, 9]       *.**.....*.*.*.*...
 6: [0, 18, 17, 10, 15, 8, 7]      *......**.*....*.**
 7: [0, 2, 4, 18, 8, 3, 5]         *.****..*.........*
 8: [0, 5, 10, 7, 1, 17, 3]        **.*.*.*..*......*.
 9: [0, 8, 16, 15, 13, 12, 1]      **......*...**.**..

(I've omitted $k=10$ to $k=18$) in which the asterisks are not consecutive for any $k$. This doesn't mean that no clever formula of any sort can exist, but it does mean that no formula of the form $ky + l < m$ will work for any triple $(k, l, m)$.

It may be an interesting probability exercise to calculate how lucky we got. :-) (Given a modulus like $19$, what is the probability that for a set of integers of a certain size, for some $k$ the multiples are consecutive?)

It follows the steps in an octave of music.  Full step, full step, half step, full step, full step, full step, half step.  The full steps the third year is leap year.  The half steps the second year is leap year.  Thus the leap years are in a 19 year cycle are as follows: 3, 6, 8, 11, 14, 17, 19.  No trial and error, just know the scale.


Proving $\sum_{i=0}^n \binom{n}{i} = 2^n$ by math induction I am having some trouble using math induction to prove the following problem:

$$\sum_{i=0}^n \binom{n}{i} = 2^n$$ Where n $\geq$ 0

I know the first thing with math induction is substitute the base value for n.

Which we get $$\sum_{i=0}^0 \binom{0}{0} = 2^0$$ which equals to 1 on the left hand side and 1 on the right hand side. The base case is now verified. We move on to substituting n for k (assuming k is true) like so:

$$\sum_{i=0}^k \binom{k}{i} = 2^k$$
If k is true, k + 1 should be true as well. So we must prove k + 1 now:
$$\sum_{i=0}^{k+1} \binom{k+1}{i} = 2^{k+1}$$

This is the part I am having trouble on. Are all my steps correct so far? If so, how do I finish proving this problem?

You can take advantage of the recurrence relation:

$${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k},1 \leq k \leq n-1$$

$${n \choose n} = {n \choose 0} = 1, n \geq 0.$$

Then split apart the $k+1$ sum and watch the indices to make sure you include all of the terms:

$$\sum_{j=0}^{k+1} {k+1 \choose j} \\ = \sum_{j=0}^{k} {k \choose j-1} + \sum_{j=0}^{k} {k \choose j} + {k+1 \choose k+1}  \\ = \left[\left(\sum_{j=0}^{k} {k \choose j}\right) - {k \choose k} \right] + \sum_{j=0}^{k} {k \choose j} + {k+1 \choose k+1} \\ =2^k - 1 + 2^k + 1 \\ = 2 \cdot 2^k \\ = 2^{k+1}.$$

If

$$1+4+6+4+1=2^4,$$

then by the well-known additive property of the binomial coefficients,

$$\begin{align}2^4+2^4=\\1+&4\ +6\ +4\ +1\ +\\
&1\ +4\ +6\ +4\ +1=\\
1+&5+10+10+5+1=2^5.\end{align}$$
This should be accepted as a proof by induction, as it unambiguously shows the pattern, which can be generalized for other $n$. (The first identity can also be used for the base case, but $1+1=2^1$ or $1=2^0$ are even better.)


How many three digit number can be formed? Question:

(a) How many three-digit numbers can be
formed from the digits 0, 1, 2, 3, 4, 5, and 6 if
each digit can be used only once?

(b) How many of these are odd numbers?

(c) How many are greater than 330?

My Solution:

a)(7)(6)(5).

Don't know the answers of other 2.

Correct me if I am wrong. Thanks!!!

There are two possible answers to this depending of if the number can start with a 0 or not.

If it can then the number is $7 \times 6 \times 5 = 210$

However we don't usually write numbers with leading zeros. So assuming we don't allow a leading zero the answer is $6 \times 6 \times 5 = 180$

How many of these are odd?

Lets consider picking the numbers in a special order

First we need to pick a 1, 3 or 5 so in our first draw for the final digit so we have a choice of 3. Next we pick our first digit which can be any thing apart from 0 or the number we have just picked making a choice of 5 and for our final draw the middle digit we can pick any of the 5 remaining digits making $3 \times 5 \times 5 = 75$

Finally For the third one greater than 330 we have two ways to achieve this either draw a 4, 5 or 6 for the first digit then we don't care about the others making $3 \times 6 \times 5 = 90$

Alternatively we must draw a 3 for our first digit then 4, 5 or 6 for our second and finally any digit making $1 \times 3 \times 5 = 15$

Now since both of these groups have no overlap we can simply add them together to get $90 + 15 = 105$.

Any of the 6 nonzero digits can be chosen for the hundreds position, and of the
remaining 6 digits for the tens position, leaving 5 digits for the units position. So,
there are (6)(5)(5) = 150 three digit numbers.


Albanese map for complex compact manifold Let $X$ be a compact Kahler manifold. To it, it is possible to associate a complex torus $Alb(X)$ with a map $$alb: X \to Alb(X) .$$

In class, our teacher claimed the image of this map generates the whole torus, but I'm not able neither to prove this result nor to find it on our textbooks.

As I mentioned in the comment, it is proved in Voisin's Hodge Theory and Complex Algebraic Geometry, Volumn I, Lemma 12.11. I will just paraphrase the proof here.

Let $k\ge 1$ and denote $X^k=X\times\cdots \times X$, the $k$-the Cartesian product. Let $alb^k$ be the map $$alb^k:X^k\to Alb(X):=H^0(X,\Omega^1)^*/H_1(X,\mathbb Z)$$ $$(x_1,...,x_k)\mapsto  (\omega\mapsto \sum_{i=1}^k\int_{x_0}^{x_i}\omega).$$

  Theorem: The map $alb^k:X^k\to Alb(X)$ is surjective for $k>>0$.

Proof: By proper mapping theorem, it suffices to show that $alb^k$ is submersive at some point $\vec{x}=(x_1,...,x_k)$. Dualize the tangent map, it reduces to show injectivity between cotangent spaces

$$T^*_{alb^k(\vec{x})}Alb(X)\to T^*_{x_1}X\times\cdots \times T^*_{x_k}X.\tag{1}\label{1}$$
 at some point $alb^k(\vec{x})$. Since a torus $V/\Lambda$ has trivial (co)tagent bundle, the cotangent space at any of its point is canonically identified with $V^*$. Therefore, the LHS of $(\ref{1})$ is identified with $H^0(X,\Omega^1)$ and the map $(\ref{1})$ is just restriction. Now, it is clear that one can choose integer $k$ and points $x_1,...,x_k$ realizing the injectivity. $\square$

Remark 1: When $\dim X=1$, this is just Jacobi inversion theorem for compact Riemann surfaces. Moreover, $Alb(X)$ coincides with Jacobi variety $Jac(X)$ and one can choose $k=\text{genus}(X)$ to generate the torus. (See Griffiths & Harris p.237 for an elementary proof, without using proper mapping theorem.)

Remark 2: An alternative way to prove this theorem is to apply universality property of Albanese map $alb:X\to Alb(X)$, i.e., for any holomorphic map $f:X\to T$ to some complex torus with $f(x_0)=0$, there exists a unique morphism of complex tori $$g:Alb(X)\to T$$
such that $f=g\circ alb$.

On the other hand, one can show that the analytic subvariety generated by $alb(X)$ in $Alb(X)$ is a subtorus, which still satisfies the universal property. Now, uniqueness guarantees that it has to be the whole $Alb(X)$. (See Birkenhake-Lange's Complex Abelian Varieties, Prop. 11.11.8).

Let $A$ be the set of analytic maps from $X$ to a complex torus, for $h\in A,h:X\to \Bbb{C}^m/L$ then $dh(x)_i$ is an holomorphic one-form on $X$, let $V$ be the complex subspace of $\Omega_X$ generated by all those $dh(x)_i,h\in A$.

$\Omega_X$ thus $V$ is finite dimensional because $\pi_1(X,x_0)$ is finitely generated and if $\omega$ integrates to $0$ on every closed-loop then $\int_{x_0}^x \omega$ is holomorphic $X\to \Bbb{C}$ thus it attains its maximum modulus at some $x\in X$, thus it is constant and $\omega=0$.

Take a basis $V=\sum_{j=1}^d \omega_j \Bbb{C}$, fix some $x_0\in X$, let $\Lambda = \{ (\int_\gamma\omega_1,\ldots,\int_\gamma \omega_d)\in \Bbb{C}^d,\gamma \in \pi_1(X,x_0)\}$ then $$Alb(X)=\Bbb{C}^d/\Lambda, \qquad F: X\to Alb(X),\qquad F(x)=(\int_{x_0}^x\omega_1,\ldots,\int_{x_0}^x\omega_d)$$
$\Lambda$ is a discrete subgroup of $\Bbb{C}^d$ because any $h\in A,h:X\to \Bbb{C}^m/L$ factors as $h=h(x_0)+H\circ F+L$ with $H$ linear $\Bbb{C}^d\to \Bbb{C}^m$ (where $H(y)_i=\sum_j c_{ij} y_j, dh(x)_i = \sum_j c_{ij}\omega_j$) and with enough different $h$ the obtained $H$ is injective. $\Bbb{R}\Lambda=\Bbb{C}^d$ for the same reason.

Thus $\Lambda$ is a lattice and $Alb(X)$ is a complex torus.

Every $h\in A$ factors uniquely through $F$. By a dimensionality argument (the image of $\mathrm{x}\in X^d \to \sum_{k=1}^d F(\mathrm{x}_k)\in Alb(X)$ contains  a $d$-dimensional complex ball) any point of $Alb(X)$ is a $\Bbb{Z}$-linear combination of some $F(\mathrm{x}_k)$.


Non-trivial values of error function $\operatorname{erf}(x)$? The so called error function $\operatorname{erf}(x)$ is defined as
$$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt,$$
and it is well known that $\operatorname{erf}(\infty)=1$.

Are there any other known closed-form values of $\operatorname{erf}(x)$, except for $\operatorname{erf}(0)$ and $\operatorname{erf}(\pm\infty)$?

If the values aren't listed on the Wolfram function page, I would be surprised if you found them anywhere else. The only listed closed form values are for $0$, $\pm\infty$, and $\pm i\infty$. However, you can find various equivalent formulations, continued fractions, and the like on that page. A good reference, generally, for most well-known functions.

There are other exact values of erf(x) besides 0 and plus or minus infinity
this graph there are horizontal asymptotes at y=1 and y=-1 because

The Z-score is a probability that certain data fall within a certain interval. For example a Z-score of 1 is represented by erf(1) = 0.682 which means that plus or minus 1 standard deviation from the mean will encompass approximately 68.2% of the population based upon a sample population. A Z-score of 2 erf(2) = 0.955, erf(3)= 0.9995. In order to be sure that you have covered the entire population or be 100% certain about your data you have to cover an infinite number of standard deviations, which is why it would be impossible to do such a thing.

Example mean IQ is 100 and SD is 15, since erf(3) = 0.9995 then 99.95% of the population will have an IQ within 3 standard deviations from the mean.

STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score. https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable


Radius, diameter and center of graph The eccentricity $ecc(v)$ of $v$ in $G$ is the greatest distance from $v$ to any other node.
The radius $rad(G)$ of $G$ is the value of the smallest eccentricity.
The diameter $diam(G)$ of $G$ is the value of the greatest eccentricity.
The center of $G$ is the set of nodes $v$ such that $ecc(v) = rad(G)$

Find the radius, diameter and center of the graph

Appreciate as much help as possible.
I tried following an example and still didn't get it, when you count the distance from a node to another, do you count the starting node too or you count the ending node instead. And when you count, do you count the ones above and below or how do you count? :)

The distance is defined as the number of edges on the shortest path between the vertices. For example, adjacent vertices have a distance of $1$.
In your graph, it might be helpful to explicitly enumerate the eccentricity of each vertex. It is not too difficult to eye-ball the eccentricity for each vertex. I have labelled your graph below with the vertex eccentricities

You can see that in this graph, the larger eccentricities occur at the sides of the graph, with the largest (the diameter) being $6$. The smallest eccentricity occurs at the central vertex with an eccentricity of $3$. This is your radius. Your center consists of all the vertices which have eccentricity equal to the radius, in this case $3$. For this graph, there is only a single such vertex, so your center in the single vertex labelled $3$ in the graph.
You can see that the name center really is quite aptly named. The vertices of the center minimizes the maximum distance to any vertex of the graph and in this sense, really are the most central vertices in the graph.

I think for the path graph  Pn, the diameter is n−1.But the radius is (n-1/2)rounded up to the nearest integer.
For example,P3 it has radius of 1 but not 2.


A counting problem related to Parallelogram There are $n$ distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of $4$-element subsets of these points that can be written as $\{A,B,C,D\}$ such that $AB∥CD$, and $BC∥AD$. No four points are in a straight line.

For example, $n=6$ and the $6$ points are $(0,0),(2,0),(4,0),(1,1),(3,1),(5,1)$ the number of parallelograms is $5$. I can find out the number of parallelograms manually if $n$ is small but how can I find out the number of parallelograms if $n$ is large? Is there any efficient method? If so,what is the method? Please explain it in details.

The following code maybe help you

//C++ code for counting the number of Parallelograms
#include <map>
#include <iostream>
#include <utility> //pair
using namespace std;

int N, x[2002], y[2002];
map <pair<int, int>, int> cnt;
long long ans = 0;

int main()
{
    cin >> N;
    for (int i = 1; i <= N; i++)
        cin >> x[i] >> y[i];
    for (int i = 1; i < N; i++)
    for (int j = i + 1; j <= N; j++)
    {
        ans += cnt[make_pair(x[i] + x[j], y[i] + y[j])];
        cnt[make_pair(x[i] + x[j], y[i] + y[j])]++;
    }
    cout << ans;
    return 0;
}


$a+b=0$ implies $a=b=0$ Let $N$ be a set of non-negative integers. Of course we know that $a+b=0$ implies that $a=b=0$ for $a, b \in N$.

How do (or can) we prove this fact if we don't know the subtraction or order?

In other words, we can only use the addition and multiplication.

Please give me advise.

EDIT

The addition law mean that for $a, b \in N$, there is an element $a+b$ in $N$ and this operation is associative.
The multiplication law means that for $a, b \in N$, there is an element $ab$ in $N$ and this operation is associative.
Also the distribution laws hold.

EDIT2

Let me rephrase the question since I don't want arguments on orders.

Let $N$ be a set with operation $+$ and $\times$.

$N$ is a monoid with the operation $+$ and $\times$ respectively. There is an unit element $0\in N$.

The distribution laws hold as in the case of the set of integers.

Can we prove the fact above with this assumption?

Please set out what you mean by the addition law.  You need the axiom that there is no number whose successor is $0$ or this fails.  That is what distinguishes the integers from the naturals.  It allows you to define order as $x \le y \leftrightarrow \exists (z) x+z=y$

You don't need $+$ or $\times$ as forming monoids on the natural numbers, nor the distributive property.  You can get by with less as follows:

Addition has the property that $(0+b)\leq(a+b)$ and $(a+0)\leq(a+b)$.  This is NOT presupposing the order of the natural numbers, or that no number of the naturals has 0 as a successor, but only a monotonicity property of addition.

We'll also assume the existence of an additive identity.

Consequently, if $(a+b)=0$, then $b=(0+b)\leq(a+b)=0.$  So, $b\leq 0.$  Also, $0=(0+0)\leq (0+b)=b$ by substitution of 0 for a in $(a+0)\leq (a+b)$ and the identity rule.  Thus, $0\leq b.$  So, $0 \leq b\leq 0.$  Consequently, $b=0.$

Similarly, $a=(a+0)\leq(a+b)=0.$  Thus, $a\leq 0$.  Also, $0=(0+0)\leq(a+0)=a.$  So, $0\leq a.$  Thus, $0 \leq a\leq 0.$  Consequently, $a=0.$

Therefore, for an algebraic structure with an identity element "$0$", binary operation "$+$" and where $(0+b)\leq (a+b),$ and $(a+0)\leq (a+b),$ and $a \leq x \leq a \implies x=a,$ it holds that "if $(a+b)=0,$ then $a=b=0$".

As an example of an algebriac structure where this holds, and "$+$" is not natural number addition, let "$+$" denote the maximum of two numbers, and consider $(\{0, 1\}, +)$.  Both suppositions used in the proof above can get verified.


Image of a null set by a Hölder continuous function Consider a continuous function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ and a null set $A \subset \mathbb{R}^2$. We assume that $$\forall (x,y) \in A^2, \ \ |f(y)-f(x)| \le ||y-x||^2$$
In addition, we suppose that $f(A)$ is measurable. Is $f(A)$ also a null set (in $\mathbb{R}$) ?

It is well known that not every continuous function maps null sets to null sets : the devil staircase does not have Lusin N property. But what happens when we assume a Hölder condition of order 2 ?

If we also assume that the Hausdorff dimension of A is $\dim _H (A) < 2$, then classically, as $f$ is Hölder continuous of order 2 on $A$, we have $\dim_H \big( f(A) \big) \le \frac{1}{2} \dim_H(A)$. So $\dim_H \big( f(A) \big) < 1$. All measurable subsets of the line with positive measure have Hausdorff dimension 1, and we assumed that $f(A)$ is measurable, thus $f(A)$ is a null set.

However, (see for instance Measure 0 sets on the line with Hausdorff dimension 1, on mathoverflow, or Are there sets of zero measure and full...),

  There exists null sets $A \subset \mathbb{R}^2$ with Hausdorff dimension $\dim_H(A) = 2$.

Is it possible to give an answer in the general case $\mu (A) = 0$ ?

Subsidiary question : is it really necessary to assume that $f(A)$ is measurable ? Is it not implied by the previous hypothesis ?

Edit : Note that $f$ is supposed Holder continuous of order $2$ only on $A$, so this condition does not a priori prevent $f$ from being constant on no open set.

Yes, $f(A)$ is null. By definition of Hausdorff measure and the coincidence of Lebesgue and Hausdorff measure on measurable sets of $\mathbb{R}^n$, since $A$ is null then for every $\delta>0$ we have
$$H^2_\delta(A):=\inf\left\{\sum_i diam(E_i)^2:A \subseteq \bigcup E_i,\, diam(E_i)\leq\delta\right\}=0.$$
Now if $\{E_i\}$ covers $A$, then $\{f(E_i)\}$ covers $f(A)$, and moreover $diam(f(E_i))\leq diam(E_i)^2$ by the Hölder condition. This implies that
$$H^1_{\delta^2}(f(A))\leq \inf\left\{\sum_i diam(f(E_i)):A\subseteq \bigcup E_i,\,diam(E_i)\leq \delta\right\}\leq H^2_\delta (A)=0.$$
Now taking the supremum as $\delta\to 0$ we obtain $H^1(f(A))=0$.

(More in general $H^s(f(A))\leq C(|f|_{C^{0,\alpha}},\alpha)H^{s\alpha}(A)$ for a $\alpha$-Hölder continuous function $f$, which is slightly stronger than saying that the dimension can increase at most by a factor $\frac{1}{\alpha}$.)

Measurability follows from the fact that $f(A)$ is Hausdorff null, and Hausdorff measure coincides with Lebesgue outer measure. You can have a look at this other question for a related result: Image of $A\subset \mathbb R^d$ under a Lipschitz function is $H^d$ measurable.

If $c>1$ and $f: \mathbb{R}^k \to \mathbb{R}$ satisfy condition $|f(u) -f(v) |\leq |u-v|^c $ then we have $$\left|\frac{f(x+h) -f(x) -0h }{|h|}\right| \leq |h|^{c-1}$$ letting $h\to 0$ we obtain that $f$ is differentiable on whole $\mathbb{R}^k$ and its derivative is equal to $0.$ Thus $f$ is constant on whole $\mathbb{R}^k$. So image of any set by $f $ is one point set and hence measure zero.


Altering an Infinite Set does not change cardinality Let X be an infinite set. Show that adding or subtracting a single point does not change its cardinality.

I have a plan but need help writing the actual proof.
I need to show that it doesn't matter which point is removed, and then I can use the fact that X is in one-to-one correspondence with a proper subset to prove this.

Hint: show that there exists an injective map $\omega\to X$. Then it is possible to "hide" a single point by shifting $\omega$ up/down.

Let $X$ be the infinite set, $Y \subset X$ the set for which there's a bijection $Y \rightarrow X$ (which means $|Y|=|X|$), and $x$ some element in $X$.
Since there's at least one element "missing" in $Y$:
$$|X|=|Y| \leq |X- \{x\}|$$
Using the same reasoning:
$$|X- \{x\}| \leq |X|$$
Conclusion:
$$|X- \{x\}| = |X|$$


Equivalent norms imply equivalent inner products? Let $H$ be Hilbert and let it have two innner products $(\cdot,\cdot)_1$ and $(\cdot,\cdot)_2$.

If the norms $|\cdot|_1$ and $|\cdot|_2$ are equivalent, does this ever imply: there exist constants $C_i$ such that
$$C_1(a,b)_2 \leq (a,b)_1 \leq C_2(a,b)_2$$
(i.e. the inner products are equivalent in some sense)?

under what conditions on $H$ does this happen?

You have such constants that

$$C_1\langle a\mid b\rangle_2 \leqslant \langle a \mid b\rangle_1 \leqslant C_2\langle a\mid b\rangle_2$$

for all $a,\, b \in H$ if and only if the two norms are constant multiples of each other.

Firstly, such an inequality can only hold if $\langle\mid\rangle_2$-orthogonality implies $\langle\mid\rangle_1$-orthogonality.

Assume that to be the case.

Then let $(e_\nu)$ be a $\langle\mid\rangle_2$-ONB.

For $\nu \neq \mu$ (if $\dim H = 1$, there is nothing to show), we have $\langle e_\nu + e_\mu \mid e_\nu - e_\mu\rangle_2 = \lVert e_\nu\rVert_2^2 - \lVert e_\mu\rVert_2^2 = 0$, and therefore

$$\langle e_\nu + e_\mu \mid e_\nu - e_\mu\rangle_1 = \lVert e_\nu\rVert_1^2 - \lVert e_\mu\rVert_1^2 = 0,$$

i.e. $\lVert e_\nu \rVert_1$ is constant, and thus, up to scaling by a factor independent of $\nu$, $(e_\nu)$ is also an ONB for $\langle\mid\rangle_1$.

Your inequality is the definition of equivalent norms so if you have two equivalent norms you will always be able to find $C_1$ and $C_2$ such that your inequality holds. Remember that if $H$ is a finite dimensional Hilbert space then all norms are equivalent to each other. If $H$ is infinite dimensional this is not true.


Prove by induction, for all positive integers $n$, that $n!\ge2^{n-1}$ So far I have:
Base case: $n=1$: LHS: $1!=1$, RHS: $2^{1-1}=1$, $1\ge1$ so $P(1)$ is true.
Inductive step: Assume $n=k$ is true, that is $k!\ge2^{k-1}$, we must show that $n=k+1$ is true, that is $(k+1)!\ge2^k$.
$(k+1)!=k!(k+1)\ge2^{k-1}(k+1)$ (since we assume $k!\ge2^{k-1}$)

Not really sure what to do after this, I know that $2^{k-1}=$$2^k\over2$ but I don't know if that helps at all.

Thanks in advance for any help!


On the 11 Sylow subgroup of a group of order 792 It seems that the 11-Sylow subgroup of a group of order $792=11×2^3×3^2$ is normal. Could you help me why it is true

If there are $12$ Sylow $11$-subgroups, then the normalizer of a Sylow $11$-subgroup has order $66$. Since $3$ does not divide $11-1=10$, an element of order $3$ centralizes an element of order $11$, giving an element of order $33$, but there is no such element in $S_{12}$.

Sorry, for some reason I thought we were just trying to prove that there was no simple group of order $792$.

OK, carrying on, the element of order $3$ must be in the kernel of the map $\phi:G \to S_{12}$. Now $\phi$ is the action by conjugation on the set of $12$ Sylow $11$-subgroups, which is transitive. Also a Sylow $11$-subgroup does not normalize any other Sylow $11$-subgroup, so its image consists of an $11$-cycle. So the image of $\phi$ is a doubly transitive group. Since $3$ divides $|\ker(\phi)|$, we have $|{\rm im}(\phi)|=132$ or $264$.

If $|{\rm im}(\phi)|=264$, then the element of order $2$ that normalizes the Sylow $11$-subgroup has $5$ $2$-cycles, so it is an odd permutation. So intersecting with $A_{12}$ gives us a group of order $132$.

Either way, we have a group of order $132$ with $12$ Sylow $11$-subgroups. It is actually a $2$-transitive Frobenius group, which is impossible, because it does not have prime power degree. But arguing directly, the counting argument suggested by Ronno starts to kick in. We have $12$ elements of order $11$, leaving only $12$ other elements. So either there must be a unique (and hence normal) Sylow $2$-subgroup or a unique Sylow $3$-subgroup. Either way, it would be centralized by the Sylow $11$-subgroup, and this time we really do get a contradiction!!!

Breaking 792 into prime divisors, we get $792$ = $11*3^2*2^3$.
That means we have 3 Sylow p-groups of order $11$, $3^2$ and $2^3$ by the first Sylow theorem.

Let $n_{11}$ represent the Sylow p-group of order $11$

Let $n_2$ represent the Sylow p-group of order $2^3$

Let $n_3$ represent the Sylow p-group of order $3^2$

Using $n_{11}$ would be the easiest because it's the only order that is prime. So by the third Sylow theorem, $n_{11}$ $\equiv 1$ mod $11$. We know that $n_{11}$ | $11*3^2*2^3$

This would imply that $n_{11}$ = $1$. Thus the Sylow p-subgroup of order $11$ is a normal subgroup.


Finding an example of an analytic function, 0 on set of points (1/n) As per the title, I want a function $f(z)$ which is analytic on $\mathbb{C}$ on the set of points $\{1/n\}$ for $n \in \mathbb{Z}^+$ and with $f(z) \neq 0$.  What would this look like if the function just had to be analytic on $\mathbb{C} - {0}$?


Limit of $\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$ I want to solve a limit without l'Hospital, just with algebraic manipulation:

$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}$$

I started with:

$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{(3-\sqrt{6+x})(3+\sqrt{x+6})}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}=\lim_{x\to 3} \frac{3-x}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}$$

$3+\sqrt{x+6} \to 6$ is determinate, so I only need to calculate:

$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$$

but I could only do it with l'Hospital:

$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{-1}{6\cdot \frac{\pi}{18}\cos \frac{\pi x}{18}-1}=\frac{-1}{\frac{\pi\sqrt{3}}{6}-1}$$

Can I get some help without l'Hospital?

I assume you meant elementary limits are allowed. I'll use:

$$\lim_{x\to 0} \frac{\sin x}{x}=1$$

Instead of the fraction, it's easier to look at its inverse:

$$
\begin{aligned}
\lim_{x\to 3} \frac{6\sin \frac{\pi x}{18}-x}{3-x}&=1+\lim_{x\to 3}\frac{6\sin \frac{\pi x}{18}-3}{3-x}\\
&= 1+ 6\lim_{x\to 3}\frac{\sin \frac{\pi x}{18}-\sin \frac{\pi}{6}}{3-x}\\
&= 1+ 12\lim_{x\to 3}\frac{\sin \left(\frac{\pi x}{36}-\frac{\pi}{12}\right)\cos \left(\frac{\pi x}{36}+\frac{\pi}{12}\right)}{3-x}\\
&= 1+6\sqrt{3}\lim_{x\to 3}\frac{\sin \left(\frac{x-3}{36}\pi\right)}{3-x}\\
&= 1-6\sqrt{3}\lim_{x\to 3}\frac{\sin \left(\frac{\pi}{36}(x-3)\right)}{\frac{\pi}{36}(x-3)} \cdot \frac{\pi}{36} \\
&=1-\frac{\pi\sqrt{3}}{6}
\end{aligned}
$$

Therefore your limit equals $\dfrac{6}{6-\pi\sqrt{3}}$.

Clearly, it is enough to compute$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac x6}{x-3}\tag1$$and, in order to compute that, it will be enough to compute$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}\text{ and }\lim_{x\to3}\frac{\frac x6-\frac12}{x-3},\tag2$$since $(1)$ is equal to$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}-\lim_{x\to3}\frac{\frac x6-\frac12}{x-3}$$(assuming that both limits from $(2)$ exist). But, clearly,$$\lim_{x\to3}\frac{\frac x6-\frac12}{x-3}=\lim_{x\to3}\frac16\frac{x-3}{x-3}=\frac16,$$whereas$$\lim_{x\to3}\frac{\sin\left(\frac{\pi x}{18}\right)-\frac 12}{x-3}$$is the derivative at $3$ of $x\mapsto\sin\left(\frac{\pi x}{18}\right)$, which is equal to $\frac{\pi\sqrt3}{36}$.


Distributing $n$ different things among $r$ distinct groups such that all of them must get atleast $1$ In how many ways can we arrange $7$ different things to $3$ people, such that all of them must get at least one?

We know that if we have $n$ identical items which will be distributed in $r$ distinct groups where each must get at least one then the number of way  is $\binom {n-1}{r-1}$ (i.e in this case $\binom62$. But in the question it is said that $7$ DIFFERENT things then what will be the approach?

MY TRY ::

At first selecting $1$ by $1$ for each $3$ people so that each get at least one : $\binom71 \times \binom 61 \times \binom 51$
Now each of $4$ can go to any of $3$ so :$3^4$ ways, so $\binom71\times\binom61\times\binom51\times (3^4)$

MY TRY #2 ::

total possibilities - {any one get $0$  $(0,6,1)(0,5,2)(0,4,3)$} - {any two get $0$ $ (0,0,7)(0,7,0)(7,0,0)$}
so

$3^7$ - {($\binom76 + \binom 75 + \binom 74$  )$\times$ 3!} - 3

It's actually helpful here to generalize the problem to distributing $n$ different objects to $3$ people (with $n\ge3$ so that each person can be guaranteed getting at least one item).  If you disregard the requirement that each person get at least one item, then you get the overestimate $3^n$.  From this, let's subtract the number of ways you can pick one person to not get any items, distributing the rest to the other two, giving

$$3^n-3\cdot2^n$$

But this now under-estimates because you're now doublecounting occasions when two people fail to receive any items.  So we need to add back in the number of ways this can happen, producing

$$3^n-3\cdot2^n+3\cdot1^n$$

As a sanity check, consider the case $n=3$, where there are clearly $6$ ways to assign an item to each person:

$$3^3-3\cdot2^3+3=27-24+3=6$$

For $n=7$, the formula gives

$$3^7-3\cdot2^7+3 = 2187-3\cdot128+3=1806$$

The general principle here is known as inclusion-exclusion.  Note that it gives the correct answer, $0$, even for $n=1$ and $2$.

Check stars and bars. Say you want to distribute $k$ items among $n$ people, no other restrictions. This means you have to chop a string of $k$ items (stars, $*$) into $n$ stretches, one to each person (with bars, $|$). This requires $n - 1$ bars. So you have a string of $n + k - 1$ stars-and-bars, of which $n - 1$ are bars, so the number of ways to do this is:

$$\binom{k + n - 1}{n - 1}$$

If you restrict each one to get at least one item, it means you first distribute $n$ items, one to each, and then you get to distribute $k - n$ among the $n$ people with no restrictions, for a number of ways (by the above):

$$\binom{k - n + n - 1}{n - 1} = \binom{k - 1}{n - 1}$$


$\frac{\sin(z)}{z}$ Bounded on $\mathbb{C}-\{0\}$? Is $\frac{\sin(z)}{z}$ bounded on $\mathbb{C}-\{0\}$?

Based on real analysis intuition it is, and even though complex $\sin$ is not bounded the $\frac{1}{z}$ makes me think it is.

Notice that it does have a limit at $z=0$. This means it has a removable singularity there.

If it were bounded on $\mathbb{C}\setminus\{0\}$ it would be bounded everywhere. This is because it is also bounded near $z=0$.

Since it is holomorphic everywhere (notice at $z=0$, since it has a limit, it has a removable singularity), then by Liouville's theorem it would be constant.

But it is not, because if it were then $\sin(z)=Cz$, for some constant $C$. But you know that $\sin$ vanishes in ($z=\pi k, k\in\mathbb{Z}$) more places than $Cz$ so they can't be equal.

No, it is not bounded. Since this function is analytic in its domain. if it is bounded then it would be constant by Liouvill's Theorem but function is no constant.


Use implicit differentiation to find an equation of the tangent line to the curve $$x^2+xy+y^2=3, (1,1)$$
I got the derivative as..
$$\frac{2x-2}{x+4}$$
But when I plug in the points I get the equation $y=x/2+2$ which is wrong. Is my derivative wrong? Or am I making a mistake plugging my numbers in. If you could show me where I'm going wrong it'd be much appreciated.

I'm getting a very different derivative. Here's how I did it

$$3=x^2+xy+y^2$$

Now we differentiate

$$0=2x+x \frac{dy}{dx}+y+2y \frac{dy}{dx} $$

Separate terms

$$-(2x+y)=\frac{dy}{dx}(x+2y)$$

so

$$\frac{dy}{dx}=\frac{-(2x+y)}{x+2y}$$

Plugging in $(1,1)$, we get $\frac{dy}{dx}=\frac{-(2+1)}{1+2}=\frac{-3}{3}=-1$, so the tangent line at $(1,1)$, in point-slope form, is $y-1=-(x-1)$, or $y=-x+2$

Ok what about after finding the first derivative I make y the subject in the main function and then substitute the x(1) to find the gradient of the tangent
Please i want a clarification

solution


Suppose $3_1+2_2−_3=_4$. Which of the following statements are true? $\newcommand{\span}{\text{span}}$
Question: Suppose $3_1+2_2−_3=_4$. Which of the following statements are true?

$\span( {_1,_2,_3} ) = \span( {_1,_2,_3,_4} )$

$\span( {_1,_2,_4} ) = \span( {_1,_2,_3,_4} )$

$\span( {_1,_2,_3} )$ is contained in but not equal to $\span( {_1,_2,_3,_4} ).$

$\span( {_1,_2,_4} )$ is contained in but not equal to $\span( {_1,_2,_3,_4} ).$

I chose answers 1 and 2, but I am new to this topic so I am not entirely sure if it's correct.
Does $_3$ qualify as a valid linear combination of all vectors: $$ _3 = 3_1 + 2_2 - _4 + 0_3$$ AND/OR  $$ _3 = 6_1 + 4_2 - 2_4 - _3$$
If yes, then 2nd statment is true right?
Thank you!

The correct answer is the first one since the non trivial containment is due to the fact that you find a $\textbf{non trivial}$ combination of $u_{4}$, expressed as linearly dependant from $u_{1},u_{2},u_{3}$.


Existence of a holomorphic function in the open disc $D=\{z\in \mathbb{C}: |z|<1\}$. Let $D=\{z\in \mathbb{C}: |z|<1\}$. Then there exists a holomorphic function $$f:D\to \bar{D}$$ with $f(0)=0$ with the property

$f'(0)=1/2$
$|f(1/3)|=1/4$
$f(1/3)=1/2$
$|f'(0)|=\sec{\pi/6}$

I have no idea how to come up with such a fuction, which satisfies atmost two of the above conditions as there are contradictory statements.  I tried to guess some but it does not work. Is there any general rule to solve this kind of problems? Which options are true? Please help me out. Any hint is sufficient.

edit - Answer says $1$ and $2$ are correct.

By Schwarz lemma, $1$, $2$ and $3$ are true trivially.

I think 3rd is not true. As by Schwarz lemma $|f(z)|\leq|z|$ which is not true in 3rd option


Invertibility of the matrix whose elements are the cube of the distance of the indices. I would like to prove, for any integer $n>1$, the invertibility of the $n\times n$ matrix $A$ whose elements are given by
$A_{ij}=|i-j|^3$, where $i$ and $j$ are the indices.

To be clearer, for instance if $n=5$ I'm referring to the matrix
$$A=\begin{bmatrix}&0&1&8&27&64\\&1&0&1&8&27\\&8&1&0&1&8\\&27&8&1&0&1\\&64&27&8&1&0\end{bmatrix}\,.$$

Such matrices should indeed be invertible (or at least it seems so looking at the determinant with Mathematica) for any $n>1$ and I do believe that there's should be some easy way to show it. I had some look on invertibility for Toeplitz or Hankel matrices, but I couldn't find any helps there for now.

Actually I guess I've found a way to tackle the problem. Defining $q$ the matrix with elements $q_{ij}=|i-j|$, which is not hard to prove to be invertible, one has
$$q^{-1}\,A\,q^{-1}=M\,,$$
where $M$ is a sparse matrix (nearly tridiagonal), which can be easily shown to be invertible.
Now I'm looking for a smart way to prove that $M$ has such a simple structure. It's not hard to do it by direct calculations but it's a bit tedious and I'm sure there must be some smarter way to do it.

You can think of rank of matrix.

If [mat A ] is to be invertible A has to to have full rank.


Confusion with sequences I am having some problem solving sequences in the form $x_k$, I have some notes about it however they skip over a lot of steps and I was hoping someone could help me clarify them!

Basically I start with the sequence:

$$x_{k+1}-4x_k+3x_{k-1}=0$$

Then what they say is to look for solutions of the form $x_k =l^k$ ,where $l$ is a constant to be determined.

They then say: Substituting in to the equation and cancelling $l^{k-1}$ we discover $l$ must satisfy the quadratic equation:

$l^2-4l+3=0$

But how on earth do they get this with that substitution? Could anyone shed any light on this? Thank you :)

If $x_k=l^k$ and $x_{k-1}=l^{k-1}$ for some particular $k$ then you can say   $x_{k+1}-4l^k+3l^{k-1}=0$ ,  i.e. you can say $x_{k+1}=4l^k-3l^{k-1}=l^{k-1}(4l-3)$

If you also know  $l^2-4l+3=0$, i.e. $l^2=4l-3$, then you can then say $x_{k+1}=l^{k+1}$

It is then a simple induction to say that any $l$ satisfying $l^2-4l+3=0$ gives $x_k=l^k$ as a solution to  $x_{k+1}-4x_k+3x_{k-1}=0$

You can go further:

with any $l$ satisfying $l^2-4l+3=0$ you have $x_k=al^k$ as a solution to  $x_{k+1}-4x_k+3x_{k-1}=0$ for any constant $a$
with distinct $l_1, l_2$ satisfying $l^2-4l+3=0$ you have $x_k=a_1^{\,}l_1^k+a_2^{\,}l_2^k$ as a solution to  $x_{k+1}-4x_k+3x_{k-1}=0$ for any constants $a_1,a_2$
$x_{k+1}-4x_k+3x_{k-1}=0$, i.e. $x_{k+1}=4x_k+3x_{k-1}$ has two degrees of freedom as knowing $x_0$ and $x_1$ determines $x_k$ for all positive integer $k$.  The same is true for $x_k=a_1^{\,}l_1^k+a_2^{\,}l_2^k$ when $l_1$ and $l_2$ are distinct.  So there can be no other solutions

Here you have the solutions to $l^2-4l+3=0$ being $1$ and $3$ so you have the particular solutions $x_k=1$ and $x_k=3^k$, giving $x_k=a_1^{\,}+a_2^{\,}3^k$ as the general form satisfying $x_{k+1}-4x_k+3x_{k-1}=0$

If $x_k=l^k$, then the equality $x_{k+1}-4x_k+3x_{k-1}$ becomes $l^{k+1}-4l^k+3l^{k-1}=0$. Dividing this by $l^{k-1}$, you get that $l^2-4l+3=0$.


A morphism of sheaves is an epimorphism if and only if it is surjective on the level of stalks Let $\phi: \mathscr{F} \to \mathscr{G}$ be a morphism of sheaves. Recall that $\phi$ is an epimorphism if given $\psi_1, \psi_2: \mathscr{G} \to \mathscr{H}$, then $\psi_1 \circ \phi = \psi_2 \circ \phi$ implies $\psi_1 = \psi_2$.

One direction of this is clear: let $p \in X$ (assuming $\mathscr{F}, \mathscr{G},$ and $\mathscr{H}$ are sheaves on $X$); if $\phi_p: \mathscr{F}_p \to \mathscr{G}$ is surjective for all $p$, then if $\psi_1\circ\phi = \psi_2\circ\phi$, they induce the same map on stalks. $\phi_p$ is surjective on stalks, so is an epimorphism in $\mathsf{Set}$ or $\mathsf{Grp}$ or whatever category. Thus $\psi_{1,p} = \psi_{2,p}$. Maps that are equal on stalks are equal, so $\psi_1 = \psi_2$.

It is the other direction that I'm failing to see clearly. I understand that we must exhibit a sheaf $\mathscr{H}$, along with maps $\psi_1, \psi_2: \mathscr{G} \to \mathscr{H}$ such that the following conditions hold:

$\psi_1 \circ \phi = \psi_2 \circ \phi$, and
$\psi_1 = \psi_2$ implies $\phi_p$ is surjective.

Any hints into what the right sheaf or sheaf morphisms to consider are would be greatly appreciated. For reference, this comes from Exercise 2.4.O in Vakil's Algebraic Geometry notes.

One way to get out of this may be as follows: Let $\phi_1, \phi_2$ be such that $\phi_1\not= \phi_2, \phi_1\circ \phi=\phi_2\circ \phi$. Since $\phi_1\not=\phi_2$, there exist some point $x\in X$ such that the morphisms on the stalks are different. But by definition after compose it with $\phi_{x}$ we get the same map on the stalks. So we reached a contradiction.

The forward proof you claimed seems to be fishy. For a concise proof, see:

An 'easy' way to prove that epimorphism of sheaves implies surjectivity on stalks


How can I show that $(1+2\cos 2\theta)^3=7+2(6\cos 2\theta+3\cos4\theta+\cos6\theta)$ using the tensor product and the Clebsch Gordan Theorem? How can I show that $$(1+2\cos2\theta)^3=7+2(6\cos2\theta+3\cos4\theta+\cos6\theta)$$
using the tensor product and the Clebsch Gordan Theorem?

The Clebsch Gordan formula for characters of $SU(2)$ is the equality
$$\chi_m  \cdot \chi_n = \chi_{m+n} + \chi_{m+n-2} + \cdots + \chi_{|m-n|}$$
where
$$\chi_m(\theta)= e^{im\theta} + e^{i(m-2)\theta} + \cdots e^{-i m \theta}$$

So you want to express $\chi_2^3$ as a combination of $\chi_n$'s. We have
$$\chi_2^2 = \chi_4 +\chi_2+ \chi_0$$
so
$$\chi_2^3 = \chi_2(\chi_4 +\chi_2+ \chi_0)= (\chi_6 + \chi_4 + \chi_2)+ (\chi_4 + \chi_2 + \chi_0)+\chi_2 = \chi_6 + 2 \chi_4+ 3\chi_2+ \chi_0$$

Now we  plug in the expressions for $\chi_0$, $\chi_2$, $\chi_4$, $\chi_6$ and get the result.

We have $$(1+2\cos(2\theta))^3=1+3\cdot 2\cos(2\theta)+3(2\cos(2\theta))^2+(2\cos(2\theta))^3$$ and this is (simplified) $$-1+12\, \left( \cos \left( \theta \right)  \right) ^{2}-48\, \left(
\cos \left( \theta \right)  \right) ^{4}+64\, \left( \cos \left(
\theta \right)  \right) ^{6}
$$
expanding the right-Hand side we get
$$-1+12\, \left( \cos \left( \theta \right)  \right) ^{2}-48\, \left(
\cos \left( \theta \right)  \right) ^{4}+64\, \left( \cos \left(
\theta \right)  \right) ^{6}
$$
this is the same!
$$\cos(2x)=2\cos(x)^2-1$$
$$\cos(4x)=8\cos(x)^4-8\cos(x)^2+1$$
$$\cos(6x)=32\cos(x)^6-48\cos(x)^4+18\cos(x)^2-1$$


How to differentiate $x^TWLW^Tx$ w.r.t $W$? Suppose $x \in \mathbb R^d$, $W \in \mathbb R^{d\times c}$, and $L \in \mathbb R^{c \times c}$ is diagonal.

How do I differentiate $x^TWLW^Tx$ with respect to $W$?

Let $f : \mathbb R^{m \times n} \to \mathbb R$ be defined by
$$f (\mathrm X) := \mathrm a^\top \mathrm X \mathrm B \mathrm X^\top \mathrm a$$
where $\mathrm a \in \mathbb R^m$ and $n \times n$ symmetric matrix $\mathrm B$ are given. Taking the differential,
$$\begin{array}{rl} \mathrm d f &= \mathrm a^\top (\mathrm d \mathrm X) \mathrm B \mathrm X^\top \mathrm a + \mathrm a^\top \mathrm X \mathrm B (\mathrm d \mathrm X)^\top \mathrm a\\ &= \mbox{tr} ( \mathrm a^\top (\mathrm d \mathrm X) \mathrm B \mathrm X^\top \mathrm a) + \mbox{tr} ( \mathrm a^\top \mathrm X \mathrm B (\mathrm d \mathrm X)^\top \mathrm a)\\ &= \mbox{tr} ( \mathrm B \mathrm X^\top \mathrm a\mathrm a^\top (\mathrm d \mathrm X) ) + \mbox{tr} ( (\mathrm d \mathrm X)^\top \mathrm a \mathrm a^\top \mathrm X \mathrm B)\\ &= \langle \mathrm a \mathrm a^\top \mathrm X \mathrm B^\top, \mathrm d \mathrm X \rangle + \langle \mathrm d \mathrm X, \mathrm a \mathrm a^\top \mathrm X \mathrm B \rangle\\ &= \langle \mathrm d \mathrm X, \mathrm a \mathrm a^\top \mathrm X (\mathrm B + \mathrm B^\top) \rangle\\ &= \langle \mathrm d \mathrm X, \color{blue}{2 \mathrm a \mathrm a^\top \mathrm X \mathrm B} \rangle\end{array}$$
and, thus, the gradient of $f$ is
$$\nabla f (\mathrm X) = 2 \mathrm a \mathrm a^\top \mathrm X \mathrm B$$

matrix-calculus scalar-fields gradient

using $\frac{\partial }{\partial X} trace(A X B X^T C)= BX^T CA + B^TX^TA^TC^T$ then $$\frac{\partial }{\partial W} x^T WLW^T x = \frac{\partial }{\partial W} trace(x^T WLW^T x) = LW^Txx^T + L^TW^Txx^T$$


Does every normal number have irrationality measure $2$? A normal number is a number whose digit expansion in any base is "uniform" in the sense that all finite digit strings occur with the "statistically expected" frequency.

I read a sentence somewhere which could be understood as implying that every normal number had irrationality measure $2$.

Is it known if all normal numbers have irrationality measure (approximation exponent) $2$?

Also, are there some non-normal irrational numbers whose irrationality measures are known?

Also, are there some non-normal numbers with irrationality measure $2$ known?

normal $\; \not\Rightarrow \;$ irrationality measure $2$

There exist numbers that are normal with irrationality measure $>2$. In fact, there exist normal numbers (meaning normal with respect to every base) that have irrationality measure $\infty.$ This is Theorem 2 in Bugeaud [1] (2002). For related results, see [2] and [3].

irrationality measure $2$ $\; \not\Rightarrow \;$ normal

There exist numbers with irrationality measure $2$ that are not normal. In fact, there exist numbers with irrationality measure $2$ that fail to be simply normal in every base. Note that this is stronger than failing to be simply normal in base $10,$ which in turn is stronger than failing to be normal in base $10,$ which in turn is stronger than "normal" in the sense that you are asking about. Shallit [4] (1979) showed that the continued fraction expansion of

$$\sum_{n=0}^\infty 2^{-2^{n}}$$

has bounded partial quotients (see Theorem 3 on p. 213 for meaning of $B(u,\infty)$ and Theorem 9 on p. 216 for the result), and thus has irrationality measure $2.$ [Indeed, for this we only need the $n$th partial quotient to be bounded by a linear function of $n.$ See Robert Israel's answer to the math overflow question Numbers with known irrationality measures.] However, it is clear that this number is not simply normal in any base. Indeed, in this number's expansion in any base, the proportion of $0$'s approaches $1,$ and hence the proportions for each of the other digits approaches $0.$

[1] Yann Bugeaud, Nombres de Liouville et nombres normaux [Liouville numbers and normal numbers], Comptes Rendus Hebdomadaires des Séances de l'Académie des Sciences 335 #2 (2002), 117-120.

[2] Verónica Becher, Pablo Heiber and Theodore A. Slaman, A computable absolutely normal Liouville number, preprint, 30 January 2014, 14 pages.

[3] Satyadev Nandakumar and Santhosh Kumar Vangapelli, Normality and finite-state dimension of Liouville numbers, arXiv:1204.4104v2, 21 January 2014.

[4] Jeffrey Outlaw Shallit, Simple continued fractions for some irrational numbers, Journal of Number Theory 11 #2 (April 1979), 209-217.

(UPDATE, 41 MONTHS LATER) A few days ago I happened to come across [5] below, which might be of interest to those finding this web page from an internet search for some known relations between normal numbers and Liouville numbers.

[5] Richard George Stoneham, A general arithmetic construction of transcendental non-Liouville normal numbers from rational fractions, Acta Arithmetica 16 #3 (1970), 239-253.

The Dirichlet transcedentals, which are of the form
$$
a=\sum_{n\in\mathbb N} a_n 10^{-n!}, \quad a_n\in\{1,2\},
$$
are as many the real numbers.

The irrationality measure of these numbers is infinite, and hence they are not normal!


Line Of Regression given x? You have found the regression line for a set of data points to be: ŷ = 30.23x + 173.52. Use the line to predict the value of y when x = 48.

Here's a quick overview of what you mean, by regressing $y$ on $x$.

Linear regression essentially tries to establish a straight-line relationship between a variable of interest $y$ and a factor $x$ (sometimes multiple factors $x_{1},x_{2},\cdots,x_{k}$) that is(are) believed to explain(cause) $y$.

Linear regression assumes that $y$ is a linear function of $x$. Any $y_{i}$ can be computed, if $x_{i}$ is known.

$y_{i}=a+bx_{i}$

As an example, you believe that the average returns on Google's stock trading on the stock exchange has a significant relationship to the average returns on the S&P 500 stock index.

Let's denote,

The expected annual returns on Google's stock by $E(R_{google})$.

The expected annual returns on the S&P500 index by $E(R_{market})$.

You would then study the historical data of Google's stock returns, say over the last 25 years and likewise the S&P 500 index returns for the past 25 years. In reality, you are collecting a sample.

Since we are performing linear regression, we assume that $E(R_{google})$ is a linear function of $E(R_{market})$.

$E(R_{google})=a+\beta\cdot{E(R_{market})}$

Using the sample data, mathematicians then use a method like Ordinary Least Squares(OLS) to estimate the coefficients $a$ and $\beta$. You would then arrive at an equation like (I just mocked it up),

$E(R_{google})=4+(1.2){E(R_{market})}$

or like the one you gave us,

$\hat{y}=30.23x+173.52$

Graphically, this is equivalent of plotting the sample data - the set of $(x_{i},y_{i})$'s with $E(R_{market})$ on the X-axis, and $E(R_{google})$ on Y-axis and passing a straight line through these points which is the best fit.

Once you've built a regression model, you could predict the dependent variable $y$ for any value of the independent variable $x$. You forecast $y$ by simply plugging in the x-value in the regression equation.

Regression is used in various disciplines to study different phenomena e.g. man-power planning a company, predicting inflation in an economy, predicting the long-run exchange rates, predicting the volatility of option prices, predicting the energy demands given the temperature and day of the year etc.


Markov Chain Exercise. A Markov chain has the transition probability matrix

$$P=
        \begin{bmatrix}
        0.7 & 0.2 & 0.1 \\
        0 & 0.6 & 0.4 \\
        0.5 & 0 & 0.5 \\
        \end{bmatrix}.
$$

Let us number the states $0,1,2$.

Now I have to determine (a) the conditional probability $Pr{\{X_2=1,X_3=1|X_1=0\}}$

(b) If it is known that the process starts in state $X_0=1$, what is the probability $Pr{\{X_0=1,X_1=0,X_2=2\}}$ ?

My Attempt:

(a) $$Pr{\{X_2=1,X_3=1|X_1=0\}}=Pr{\{X_2=1|X_1=0\}}Pr{\{X_3=1|X_1=0\}}=(0.2)Pr{\{X_3=1|X_1=0\}},$$

for calculating $Pr{\{X_3=1|X_1=0\}}$, I need two step transition probability matrix :

$$P^2=
        \begin{bmatrix}
        0.54 & 0.26 & 0.2 \\
        0.2 & 0.36 & 0.44 \\
        0.6 & 0.1 & 0.3 \\
        \end{bmatrix}.
$$

Hence  $$Pr{\{X_2=1,X_3=1|X_1=0\}}=(0.2)Pr{\{X_3=1|X_1=0\}}=(0.2)(0.26)=0.052,$$ which doesn't match with the result. Where am I doing mistake ?

(b)$$Pr{\{X_0=1,X_1=0,X_2=2\}}=Pr{\{X_0=1\}}P_{1,0}P_{0,2},$$
where $P_{0,2}$ denotes transition from state $0$ to state $2$ in one step.

But my question is  the problem hasn't specified the initial distribution, so I am not getting the value of $Pr{\{X_0=1\}}$, consequently not getting $Pr{\{X_0=1,X_1=0,X_2=2\}}$. But the problem has an answer. How can I proceed ?

(a) Since the process if Markov, we can write

$$
P(X_2 = 1, X_3 = 1 \mid X_1 = 0)
    = P(X_2 = 1 \mid X_1 = 0) \cdot P(X_3 = 1 \mid X_2 = 1)
$$

and you only need the transition matrix itself to answer.

(b) If you're given that $X_0 = 1$, then (as Antitheos also points out) $P(X_0 = 1) = 1$, and you can write

\begin{align}
P(X_0 = 1, X_1 = 0, X_2 = 2)
    & = P(X_1 = 0, X_2 = 2 \mid X_0 = 1) \\
    & = P(X_1 = 0 \mid X_0 = 1) \cdot P(X_2 = 2 \mid X_1 = 0)
\end{align}

and you can proceed as before.

a)
First, by the definition of the term
$$P(X_2=1,X_3=1 |X_1=0) = P(X_3=1|X_2=1,X_1=0)\cdot P( X_2=1|X_1=0)$$
and by Markoc chain definition
$$P(X_3=1|X_2=1,X_1=0)\cdot P( X_2=1|X_1=0) =P(X_3=1|X_2=1)\cdot P( X_2=1|X_1=0) =p_{1,1}\cdot p_{1,0} $$

b)
Read the problem again:

  it is known that the process starts in state $X_0=1$

This means
$$P(X_0=1) =1.$$


the coefficient of $t^{18}$ in $t^6(1+t+\cdots+t^5)^6 $ equals the coefficient of $t^{12}$ in $(1+t+\cdots+t^5)^6 $. Let $X$ be the total from rolling 6 fair dice, and let $X_1,\ldots,X_6$ be the individual rolls. What is $P(X=18)$? Then the PGF of $X$ is $$E(t^X) = E(t^{X_1} \cdots t^{X_6}) = \frac{t^6}{6^6}(1+t+ \cdots+t^5)^6$$
So the number of ways to get a sum of $18$ is the coefficient of $t^{18}$ in $t^6(1+t+\cdot\cdot\cdot+t^5)^6$, which is the coefficient of $t^{12}$ in $(1+t+\cdot\cdot\cdot+t^5)^6 $.

I don't understand why the coefficient of $t^{18}$ in $t^6(1+t+\cdot\cdot\cdot+t^5)^6$ is the same as the coefficient of $t^{12}$ in $(1+t+\cdot\cdot\cdot+t^5)^6 $?

the first one, you'll note, has $t^6$ glommed onto the beginning of its generating function; remove that, and you go from the coefficient of $t^{18}$ to that of $t^{18-6}=t^{12}$.

The number of ways to get $n$ in one throw of the dice is the coefficient of $n$ in $t+t^2+t^3+t^4+t^5+t^6$, in other words it is one for $n=1,2,3,4,5,6$. The probability to get, say, seven, in two throws of the dice is the value of the $t^7$ term in $(t+t^2+t^3+t^4+t^5+t^6)^2$. This is just a "coincidence" in that the number of ways to combine works out to be the same as the coefficient $t^7$. This is quite standard and explained in any textbook.


square of a real number is always positive I was working on a problem and had to prove firsthand that $a^2>0$ if $a$ is not $0$ and I did not find anything on the site that gave a proof using the properties I'm using so I was wondering if mine was good:

$P$ is the set of all positive numbers

Case 1

If $a>0$ then $(a-0) = a\in P$.

Then $(a-0)a\in P$ (closure under multiplication)

$\Rightarrow (a^2 -0)\in P \Rightarrow a^2>0$

Case 2

If $a<0$ then $(0-a) = -a\in P$.

Then $-a(0-a)\in P$ (closure under multiplication)

$\Rightarrow (0 + a^2)\in P$ considering $((-a)(-b) = ab)$

$\Rightarrow (a^2 -(-0)) = (a^2 - 0)\in P \Rightarrow a^2>0$

I feel like $(-a)(-b) = ab$ kind of implies what I'm trying to prove...

Thank you for your help!!

Short answer:

Let $a$ be positive and note that $-a=(-1)a$. Then by associativity of the multiplication,

$$(-a)^2=(-1)^2a^2=a^2.$$

Proposition 10 is the requested result.

Introduction

$+$ is a binary operations from $\mathbb{R}\times \mathbb{R} \to \mathbb{R}$,$-$
is a function from $\mathbb{R} \to \mathbb{R}$, $\frac 1 .$ is a function from $\mathbb{R}\setminus \{0\} \to \mathbb{R}$ and $0$ and $1$ are from $\mathbb{R}$
, $0 \ne 1$

In tthe Axiom section the tags on the right side are the names of the axioms. So the associative law of the $+$ operator is called $(\text{AddAssoc})$

The propositions have names $(\text{P1})$ to $(\text{P10})$.
In the proofs the tag on the right side is the name of the axiom that is used to get from the current line to the next line. So in the proof of proposal 1 to get from the first line
$$a\cdot 0$$
to the second line
$$0\cdot a+0 $$
one uses axiom $(\text{A2})$
$$a=a+0$$
where we substitue $a$ by $0\cdot a$
to get
$$0\cdot a= 0\cdot a +0$$

In the proofs propositions 1 to 6 and 8 the expression in a line is a real number that is equal to the real number  of the next line or an inequality holds.
Proposition 7 and  9 are more complex. These are equivalent logical expressions. Proposition 10 is the most complex. It is an indirect proof where we deduce a contradiction from some assumptions.

$(\text{def})$ means that is a definition or an abbreviation that is used here, e.g. $a^2$ is an abbreviation for $a\cdot a$. We do not explicitly state this definition in this text but assume they are known to the reader.

Propositions 1 to 4 are valid for all fields. Propositions 5 to 10 need the order axioms for their proofs.

Maybe it is also useful to read the post from the bottom to the end, so starting with propostion 10 that is the statement asked by the OP.

Axioms

$$a+(b+c)=(a+b)+c \tag{AddAssoc}$$
$$0+a=a \tag{AddIdent}$$
$$a+0=a \tag{AddIdent}$$
$$a+(-a)=0 \tag{AddInv}$$
$$(-a)+a=0 \tag{AddInv}$$
$$a+b=b+a \tag{AddComm}$$

$$a \cdot (b \cdot c)=(a \cdot b) \cdot c \tag{MulAssoc}$$
$$1 \cdot a=a \tag{MulIdent}$$
$$a \cdot 1=a \tag{MulIdent}$$
$$a \cdot \frac 1 a =1 \tag{MulInv}$$
$$\frac 1 a \cdot a   =1 \tag{MulInv}$$
$$a \cdot b=b \cdot a \tag{MulComm}$$

$$a\cdot(b+c)=a\cdot b + a \cdot c \tag{Distrib}$$
$$(a+b) \cdot c =a\cdot c + b \cdot c \tag{Distrib}$$

These axioms define a field.

$$a<0 \; \lor a=0  \; \lor a>0 \tag{OrdTrich}$$
$$a>0 \;  \land  \; b>0 \Rightarrow a\cdot b >0 \tag{OrdMulClose}$$
$$ a<b \Rightarrow a+c \; < \; b+c\tag{OrdTranInv}$$

This is an ordered field now. For our investigations we do not need the completeness axiom that holds in a complete ordered field.

Proposition 1
$$a\cdot 0=0 \tag{P1}$$
Proof
$$ 0\cdot a \tag{AddIdent} $$
$$ =0\cdot a+0  \tag{AddInv} $$
$$=0\cdot a+(0\cdot a+(-(0\cdot a)))  \tag{AddAssoc}$$
$$=(0\cdot a+0\cdot a)+(-(0\cdot a))  \tag{Distrib}$$
$$=(0+0)\cdot a+(-(0\cdot a )) \tag{AddIdent}$$
$$=0\cdot a+(-(0\cdot a))  \tag{AddInv}$$
$$=0$$

Proposition 2:
$$(-1)^2=1 \tag{P2}$$
Proof:
$$(-1)^2 \tag{AddIdent}$$
$$=(-1)^2+0 \tag{AddInv}$$
$$=(-1)^2+((-1)+1) \tag{AddAssoc}$$
$$=((-1)^2+(-1))+1 \tag{MulIdent}$$
$$=((-1)^2+(-1)\cdot 1)+1 \tag{Distrib}$$
$$=(-1)((-1)+1)+1 \tag{AddInv}$$
$$=(-1)\cdot 0+1 \tag{P1}$$
$$=0+1 \tag{AddIdent}$$
$$=1$$

Proposition 3
$$-a=(-1)a \tag{P3}$$
Proof
$$-a \tag{AddIdent}$$
$$=-a+0 \tag{P1}$$
$$=-a+0\cdot a \tag{AddInv}$$
$$=-a+(1+(-1))\cdot a \tag{Distrib}$$
$$=-a+(1\cdot a+(-1)\cdot a) \tag{MulIdent}$$
$$=-a+(a+(-1)\cdot a) \tag{AddAssoc}$$
$$=(-a+a)+(-1)\cdot a \tag{AddInv}$$
$$=0+(-1)\cdot a \tag{AddIdent}$$
$$=(-1)\cdot a$$

Proposition 4
$$(-a)^2=a^2   \tag{P4}$$

Proof
$$ (-a)^2 \tag{P3}$$
$$= ((-1)a)^2 \tag{def}$$
$$=((-1)a)((-1)a) \tag{MulAssoc}$$
$$=(((-1)a)(-1))a \tag{MulAssoc}$$
$$=((-1)(a(-1)))a \tag{MulComm}$$
$$=((-1)((-1)a))a \tag{MulAssoc}$$
$$=(((-1)(-1))a)a \tag{MulAssoc}$$
$$=((-1)(-1))(a\cdot a) \tag{def}$$
$$=(-1)^2a^2 \tag{P2}$$
$$=1\cdot a^2  \tag{MulIdent}$$
$$=a^2$$

Proposition 5
$$1>0 \tag{P5}$$
Proof:
$$1 \tag{MulIdent}$$
$$=1\cdot 1 \tag{OrdMulClose}$$
$$\ge 0 $$

Proposition 6
$$(-1)^2>0 \tag{P6}$$
Proof:
$$(-1)^2 \tag{P2}$$
$$=1\tag{P5}$$
$$\ge 0 $$

Proposition 7
$$a \ge 0 \Rightarrow a^2\ge0\tag{P7}$$
Proof

This follows from $(\text{OrdMulClose})$
if we replace $b$ by $a$.

Proposition 8
$$a \le 0 \Rightarrow a^2\ge0\tag{P8}$$
Proof:
We have
$$a \le 0 \tag{OrdTranInv}$$
$$a+(-a) \le 0+(-a)$$
applying $(\text{AddInv})$ to the LHS and $(\text{AddIdent})$ to the RHS gives
$$0\le -a$$
which means
$$-a \ge 0$$
So by $(\text{P7})$ we have
$$(-a)^2 \ge 0\tag{1}$$
From $(\text{P4})$
So we have
$$a^2 =(-a)^2 \tag{2}$$
From $(1)$ and $(2)$ follows
$$a^2\ge 0$$

Proposition 9
$$a^2\ge0\tag{P9}$$
Proof:
From $(\text{OrdTrich})$ we know that for each $a \in \mathbb{R}$ holds
$$a<0 \; \lor a=0  \; \lor a>0 $$
So from $(\text{P7})$  and $(\text{P8})$ follows $(\text{P9})$

Proposition 10
$$a\ne0 \Rightarrow a^2\gt0\tag{P10}$$
Proof:
From $(\text{P9})$  we have
$$a^2\ge0$$
if
$$a^2=0\tag{3}$$
and
$$a\ne 0\tag{4}$$
we can assume that there is $\frac 1 a$
and so from $(3)$ we get
$$\frac 1 a a^2=0\tag{MulAssoc}$$
$$\Rightarrow (\frac 1 a \cdot a) \cdot a=0\tag{MulInv}$$
$$\Rightarrow 1 \cdot a=0\tag{MulIdent}$$
$$\Rightarrow a=0$$
which contradicts $(\text{4})$


I need an equation for some data points. My data points are (97.57,6.14), (90.54,7.03), (81.99,8.55), (71.47,10.52), (56.5,14.97)
and (31.88,24.62).  I'm trying to find the nonlinear equation that describes these points, but I'm having trouble.  Can anyone come up with it.

It's pretty much impossible to come up with "the" nonlinear equation that describes the data without more information on the problem, such as the source of the data and the desired fitted form, or at the very least the units.  What's the independent and dependent variables?  Time?  Position?  Elevation? Force? Voltage? Money?  You can't just list random data points and expect to automatically fit the "correct" function to it; we need more information to solve this problem.


Range of entire function. Picard's theorem $f(z)$$=$ 2$z^{14}$$cos^2z$

$f(z)$ is entire function. I want to figure out if $f$($\Bbb C$)$=$$\Bbb C$ or there's a point $a$ such that $f$($\Bbb C$)$=$$\Bbb C$\ {$a$}. So essentially I want to use Little Picard Theorem here.

I thought about using the reflection principle. Let $f(z)$$=$$w$.  So here $f$($\bar{z}$)=$\bar{f(z)}$. This means that if $w$ if this exceptional point then $\bar{w}$ is too. I guess this implies that exceptional points can only be located on the real line?

Can someone please correct me if I am wrong. Any help with solving this problem would be appreciated.

Along the real axis, we have $f(z) = f(x) = 2x^{14}\cos(x)^2 \ge 0$.
Notice $f(0) = 0$ and $\lim\limits_{n\to\infty} f(n\pi) = 2(n\pi)^{14} \to \infty$. Apply IVT along positive real axis, we have

$$f([0,\infty)) \supset [0,\infty)\quad\implies\quad f(\mathbb{R}) = [0,\infty)$$

Along the imaginary axis, we have $f(z) = f(iy) = -2y^{14}\cosh(y)^2 \le 0$.
Notice $f(i0) = 0$ and $\lim\limits_{y\to+\infty} f(iy) = -\infty$. Apply IVT along positive imaginary axis, we have
$$f(i[0,\infty)) \supset (-\infty,0]\quad\implies\quad f(i\mathbb{R}) = (-\infty,0]$$

Combine them, we have $f(\mathbb{R} \cup i\mathbb{R}) = \mathbb{R}$.
As you have already known, by reflection principle, if $f(z)$ avoids any number $\alpha$, $\alpha$ need to be real. Since this have been ruled out by above argument, $f(z)$ does not avoid any number and $f(\mathbb{C}) = \mathbb{C}$.

Notice that $f(z)$ is the product of two entire functions $g(z)=2z^{14}$ and $h(z)=cos^2z$ where $g$ is onto by fundamental theorem of algebra, so $f(\mathbb C)=\mathbb C$.


Definition of Existence of Identity in a Group For a binary operation structure to be a group in a Set, i.e., $G = \langle*,S\rangle$

One must have an identity $e$ in $G$ s.t. $a*e = e*a =a \ \forall a \in G$.

But does it mean that e is one element which is applied to all element in $G$ or just it's okay for $e$ to specifically exist for each $a$ in $G$, e.g., if $(e_0, a_0), (e_1, a_1)\cdots (e_n, a_n)$ where $e_i$'s are different and meeting the criterion $e_i * a_i = a_i * e_i = a_i$ ?

It's important to write this out in quantifiers:

$$\exists e \in G: \forall g \in G: ge =g = eg$$

In other words, the existential quantifier comes for the universal quantifier, and hence it means that the element $e$ has to work for every group element.

In particular, it follows that this element $e$ is unique so the answer to your question is no: we can't have different identities for different elements. Every element has the same identity.

It is just one $e$ that satisfies this criterion. Identity doesn't have to specifically exist for each and every element.
Try proving the theorem:the identity element in a group is unique
Hint: assume it there are two identities and go on to show that they both are the same.


Calculate $\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}$ Question:

Calculate

$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}$$

using substitution, cancellation, factoring etc. and common standard limits (i.e. not by L'Hôpital's rule).

Attempted solution:

It is not possible to solve by evaluating it directly, since it leads to a "$\frac{0}{0}$" situation. These kinds of problems are typically solved by cancellation and/or direct application of a after some artful substitution in a precalculus context.

It was a bit hard to find a good substitution, so I tried several:

$t = 3x$

$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1} = \{t=3x\} = \lim_{x \to 0} \frac{e^{t} - 1}{e^{4\frac{t}{3}} - 1}$$

Although this gets me closer to the standard limit

$$\lim_{x to 0} \frac{e^x - 1}{x} = 1$$

...it is not good enough.

$t = e^{4x} - 1$

$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1} = \{t=e^{4x} - 1\} = \lim_{t \to 0} \frac{e^{\frac{3 \ln (t+ 1)}{4}} - 1}{t}$$

Not sure where to move on from here, if it is at all possible.

Looks like the denominator can be factored by the conjugate rule:

$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1} = \lim_{x \to 0} \frac{e^{3x} - 1}{(e^{2x} - 1)(e^{2x} + 1)}$$

Unclear where this trail can lead.

What are some productive substitutions or approaches to calculating this limit (without L'Hôpital's rule)?

hint: do a simple trick: $\dfrac{e^{3x}-1}{e^{4x}-1} = \dfrac{e^{3x}-1}{3x}\cdot \dfrac{4x}{e^{4x}-1}\cdot \dfrac{3}{4}$. The first two factors tend to $1$ each, leaving you the last factor being the answer.

You can use  l'Hopital's Rule to get
$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}= \lim_{x \to 0} \frac{3e^{3x}}{4e^{4x} }= \frac{3}{4}.$


Does $2\int_{0}^{y^{2}}c(t)\,dt<3\frac{1}{y}\int_{0}^{y^{2}}\sqrt{t}c(t)\,dt$ always hold for positive and increasing $c(t)$? I come across one problem. Does $2\int_{0}^{y^{2}}c(t)\,dt<3\frac{1}{y}\int_{0}^{y^{2}}\sqrt{t}c(t)\,dt$ always hold ? The $y$ is any positive number. The domain of $c(t)$ is $[0,\infty)$ and it is smooth, strictly increasing the positive. I think it is not true. But I cannot find out one counter-example of $c(t)$. Can you help ? Thank you.

The inequality is equivalent to

$$  2y \int_0^{y^2} c(t) dt \le 3 \int_0^{y^2}\sqrt{t}  c(t) dt.$$

Let's prove this by differentiation.

$$ \frac{d}{dy} \left (  2y \int_0^{y^2} c(t) dt\right)  = 2 \int_0^{y^2} c(t) dt + 4y^2 c(y^2) < 6 y^2 c(y^2),$$

because $c$ is strictly increasing. Also,

$$ \frac{d}{dy}\left (  3\int_0^{y^2} \sqrt{t} c(t) dt\right) = 6 y^2 c(y^2),$$

and result follows, because both are equal to zero at $y=0$.

If $c(t)=1$ we have

$2\int_{0}^{y^{2}}c(t)\,dt=3\frac{1}{y}\int_{0}^{y^{2}}\sqrt{t}c(t)\,dt$

Edit: this is no counterexample since $c$ shoul be increasing

FRED


Inequality from Chapter 5 of the book *How to Think Like a Mathematician* This is from the book How to think like a Mathematician,

How can I prove the inequality
$$\sqrt[\large 7]{7!} < \sqrt[\large 8]{8!}$$

without complicated calculus? I tried and finally obtained just $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$

Your inequality is equivalent to
$$(7!)^8 < (8!)^7$$
divide it by $(7!)^7$, and get
$$7! < 8^7$$
and this is clear, since $$1 \cdots 7 < 8 \cdots 8$$

You have already turned the comparison of two geometric means into the comparison of two arithmetic means.  So consider a more general comparison: show that appending a larger number always raises the geometric mean of a list of positive numbers by showing the effect on the arithmetic mean. Suppose the $x_i$ are real and $x_{n+1}$ is strictly largest.
\begin{equation*}
\begin{split}
(1/(n+1)) \sum_{i=1}^{n+1} x_i &= (1/(n+1)) (x_{n+1} + \sum_{i=1}^{n} x_i)
\\
&=(1/(n+1) (n x_{n+1}/n  + n \sum_{i=1}^{n} x_i / n)
\\
&> (1/(n+1) (\sum_{i=1}^{n} x_i/n  + n \sum_{i=1}^{n} x_i / n)
\\
&= (1/(n+1) ((n+1) \sum_{i=1}^{n} x_i / n)
\\
&= \sum_{i=1}^{n} x_i / n
\end{split}
\end{equation*}

Note that we really only needed $x_{n+1}$ to be larger than the previous mean.


Cauchy-Riemann Equations + Locally Invertible implies non-zero derivative Suppose $C^1$ $f:\mathbb{R}^2 \to \mathbb{R}^2$ satisfies the Cuachy-Riemann equations $\frac{\partial f_1}{\partial x} = \frac{\partial f_2}{\partial y}, \frac{\partial f_1}{\partial y} = -\frac{\partial f_2}{\partial x}$ at some point $a \in \mathbb{R}^2$. Additionally, let $f$ be locally invertible. Is it true that $Df(a) \neq 0$? I have been asked to prove that in a homework, but I cannot figure out how to do it without further conditions (e.g. $f^{-1}$ is differentiable). Any tips would be greatly appreciated.

It seems like there are obvious counter examples as given, like $f(x,y) = (x^3, y^3)$.

The function $$f:\quad{\mathbb R}^2\to{\mathbb R}^2,\qquad(x,y)\mapsto(x^3,y^3)$$
is $C^1$, and even globally invertible, if we define (as usual) $$\root 3\of t:={\rm sgn}(t)\root3\of{|t|}\qquad(t\in{\mathbb R})\ .$$
Furthermore we have $$f_{1.1}=f_{2.2}=0,\qquad f_{1.2}=-f_{2.1}=0$$
at $(0,0)$, and therefore $Df(0,0)=0$.

The claim in your homework (as it appears in the question) can therefore not be proven.

I think the professor in our Honors Analysis class stated that local invertibility assumes a smooth inverse which (x^3,y^3) does not have at (0,0).


Suppose $A$ is nilpotent, and $B = c_0I_n + c_1A + \cdots + c_{m-1}A^{m-1}$, show $\det(B) = 0$ iff $c_0 = 0$ Let $A$ be an $n\times n$ matrix, with real entries. If $A$ is nilpotent, so $A^m = 0$ for some $m \ge 1$, $B = c_0I_n + c_1A + \cdots + c_{m-1}A^{m-1}$, where $c_i \in R$ for each $i$, show that $\det(B) = 0$ if and only if $c_0 = 0$.

My attempt:

So, $c_0 = 0$ implies $\text{det}(B) = 0$ is obvious: if $c_0 = 0$, then $\det(B) = \det((c_1 + c_2A + \cdots + c_{m-1}A^{m-2})A) = \det(c_1 + c_2A + \cdots + c_{m-1}A^{m-2})\cdot \det(A) = 0,$ since $\det(A) = 0$ if $A$ is nilpotent.

I'm not sure how to tackle the reverse direction. If $\text{det}(B) = 0$, we know: $B$ is not invertible, we know that $\det(c_0I_n + c_1A + \cdots + c_{m-1}A^{m-1}) = 0$. But I don't know where to go from here! A hint in the right direction would be appreciated.

If $N$ is nilpotent of order $k$, then $I-N$ is invertible because
$$
    (I-N)(I+N+N^{2}+\cdots+N^{k-1})=I-N^{k}=I.
$$
Now suppose $A$ is nilpotent and further suppose $c_0 \ne 0$. Then
$$
         c_0 I + c_1 A + c_2 A^{2}+\cdots + c_{m-1}A^{m-1} = c_0(I-N),
$$
where $N$ is nilpotent. Hence, the above is invertible if $c_0 \ne 0$. However, the above is definitely not invertible if $c_0=0$ because, in that case, the above is nilpotent.

Do the following:

Consider $B^n$ the n-th power of B, with $n\geq m$

$B^n = (c_0 I + c_1 A + \cdots + c_{m-1} A^{m-1}) \cdot \ldots \cdot (c_0 I + c_1 A + \cdots + c_{m-1} A^{m-1})$

$B^n = c_0^n I^n + \sum_{2n \geq i\geq n} \alpha_i A^i  = c_0^n I$

With $\sum_{2n \geq i\geq n} \alpha_i A^i$ I mean that $B^n$ is equal to $c_0^n I$ plus some powers of $A$ multiplied by an scalar but the exponent of those powers of $A$ is grater than $m$ therefore $A^i=0$ and $B^n = c_0^n I$.

Now use what you know $0 = det(B)^n = det(B^n) = det(c_0^n I) = c_0^k det(I) = c_0^k$ therefore $c_0 = 0$.


Is is true that $\lim(\ln x) = \ln(\lim x)$? As the title says, I want to ask everyone if $\lim(\ln x) = \ln(\lim x)$ when x approach to infinite with any function.

The function $\ln x:\mathbb{R^{+}}\to\mathbb{R}$ is a continuous function. So $\forall x_0 \in \mathbb{R^{+}} \ \ \lim_{x\to x_0 }\ln x =\ln x_0=\ln\lim_{x\to x_0}x$.

$$\lim_{x\to\infty} \ln(x)=\infty$$
$$\ln\left(\lim_{x\to\infty}x\right)=\infty$$
If I understood what you were saying properly, it was a bit unclear..


convergence of $\sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$ Does anybody know how to show analytically the convergence (divergence ?) of $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}\enspace$ or some literature for it ? (Thanks!)

Note: Unfortunately I am still missing a useful approach.

EDIT:

I want to thank all here for the nice help.

Result: The series is divergent (but limited) as it is written.

It's possible to change to $\enspace\displaystyle\sum\limits_{n=1}^\infty (-1)^n (-2+\sum\limits_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}})\enspace$ to avoid the oscillation and to get convergence.

You can prove that it diverges by observing that the sum of each "block" is greater than a constant, that is, $$\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}>C$$for some positive $C$.

It's alternating series (with periodically strange alternation, but alternation nonetheless) and thus converges according to alternating series test.


How to count the number of faces or regions of a graph How to count the number of faces or regions of a graph such as the below graph:

The graph has 12 vertices and 17 edges. But I am unable to find the number of faces of the graph. I don't know how to find the faces. Help me to find the number of faces.

The depicted graph has a $K_{3,3}$ (complete bipartite graph on six vertices) as a minor. By Kuratowski's theorem such graph is not a planar graph, hence it makes no sense to talk about its faces, unless you embed such graph in a torus or in a surface with genus $\geq 1$:

Euler's theorem $f+n-m=2$.
$n$: number of vertices.
$m$: number of edges.
$f$: number of faces.


find the extremal of a function i need to find the extremal of the functional $\int I(y,y'') dt$. Could anyone tell me the concepts of finding the extremal so that i can go about solving this one?

Update:

So my function was: G= $\int(ay+ $$\frac{1}{2}$$ b$y''^2$ )dt$
solving it using using the Euler–Lagrange equation as given in the link below, we get:

a+by''''=0

which on solving gives:

y=-(a(t^4))/(24b).

Am i right in solving this one?

Principally this should be related to the space on which the functional is defined. This depends on the exact form of $I$. However, if you forget these things, consider for $J(y)=\int I(y,y'')$, the expression $J(u+\tau v)$, where $\tau$ is a parameter, and $u,v$ are functions. Then differentiate the expression which respect to $\tau$, and put $\tau=0$. Rewrite the expression, using partial integrations (this only works if the space you decide to define the functional on is nice enough, for example compactly supported functions on the domain you are working on). You should then find some expression of the form $\int B(u,u'') v$. Since for a critical point this must be zero for all $v$, we conclude $B$ must be identically zero.

of course, the expression containing the terms that are unknown in the sense we trying a unknown rate of the function for the application with domain as the variable x. hence your answer in the predict of the Euler-Lagrange equation concept must needed the exact differentials in order to meet the bijection principle of a mapping.

ok, Asutosh sang K.M.P


Prove $(A^T)^{-1}$ = $(A^{-1})^T$ whenever $A$ is invertible. Prove $(A^T)^{-1}$ = $(A^{-1})^T$ for any invertible matrix $A.$
I actually don't know where to start. I do not think I can just apply index laws.
Any help is cool! Thanks.

Try taking the transpose of the equation
$$AA^{-1}=I.$$

$A^T(A^T)^{-1} = I$   ;                        $ eq^n 1$
taking   transpose  on  both   side of $ eq^n 1 $
$[(A^T)^{-1}]^T A = I $;    $eq^n 1.1$
[Since $(AB)^T= B^TA^T $ & $I^T=I$ , $(A^T)^T=A$]
Now  to hold above $eq^n 1.1$ it must satisfy the relation below
$[(A^T)^{-1}]^T  =  A^{-1} $   ;           $eq^n 2 $
taking transpose on $eq^n 2 , $ we get
$ (A^T)^{-1} = (A^{-1})^T $


Is Russell's paradox really about sets as such? It seems to me that Russell's paradox rather is a "paradox" concerning relations.

  Suppose we want to construct a graph (with finite or infinite number
  of nodes) and want some node to be adjacent to exactly those nodes
  that are not adjacent to them selves.

It's the same problem, which seems to arise from the fact that it is not possible to define relations with nodes of certain adjacent specifications.

And there are a lot of other examples of impossible constructions of relations.

  Suppose we want to construct a graph and want some node $s$ to be adjacent
  to exactly those nodes $x$ such that:

  all chains $x\to x_1\to x_2\to x_3\to\cdots$ are finite;
  given a surjection $f$ for the construction, $f(x)\nrightarrow x$. $($Set $s=f(x)\dots)$

It seems to be necessary to point out that I don't mean that there is a paradox of Russell (it was just a paradoxical consequence of a construction), and that I don't know if mathematicians really mean that the construction of Russell say something about sets as such.

But I do believe that a lot of people think that Russell's paradox really is just about sets.

The interesting thing about Russell's paradox is not that it involves an object that can't exist, but that that object is embedded in a theory that seemed sound until Russell pointed out the contradictory object.

Certainly one can invent all sorts of false principles about nonexistent objects.  For example, let $V$ be a village in which there live two men, $A$ and $B$, where $A$ is taller than $B$ and $B$ is taller than $A$. Now build a theory of such villages.

Well, nobody cares, because it is obvious that there are no such villages  and that any theory of such villages is a waste of time.  Or similarly, a graph $G$ with a node $n$ that is adjacent to all the nodes that are not adjacent to themselves. But there is obviously no such graph, so why would you do that?

The historical crisis caused by Russell's paradox  was that  mathematicians as a group were taken in by the seductive general comprehension principle that

  for each property $\Phi$ there is a collection $\{x\mid \Phi(x)\}$ of everything with property $\Phi$

and then later it transpired (as shown by Russell) that this principle is false.

If everyone was fooled by the general comprehension principle then  how can you be sure that they are not still fooled by some other plausible-seeming idea? 20th-century mathematicians have done  a lot of work on algebraic geometry.  Bézout's theorem says that, given the right context, two algebraic curves of degree $m$ and $n$ have exactly $mn$ intersections.  How can you be sure that some new Russell won't find an argument tomorrow  that actually, no such curves exist and the whole theoretical edifice of algebraic geometry is complete nonsense? But that's just what happened in set theory.

It's easy to construct objects with contradictory properties.  People appear on this web site every week to ask about the largest real number less than 5. (1 2) These questions do not precipitate theoretical crises.  The crisis of Russell's paradox was not the paradoxical object, but the failure of the theory in which that object was embedded.

Technically, Russell's paradox points to a rather interesting proof-technique to show that a certain  mathematical object does not exists:

  To prove that it doesn't exist a surjection $p:A\to\mathcal P(A)$,
  define the set $B=\{x\in A|x\notin p(x)\}$. Obviously,
  $B\notin\mathrm{Im}\; p$.

Historically, Russell's paradox shows that the traditional trust in the correspondence between classes and predicates was faulty. Even in the sixties I was taught in philosophy lessons that there was a correspondence between a property and the class of subjects having that property, a fallacy that of course had influence on early set theory.


Prove that the range is a subspace In Linear Algebra Done Right, it said

If $T \in \mathcal{L}(V,W)$, then range $T$ is a subspace of $W$.

Proof:

Suppose $T \in \mathcal{L}(V,W)$. Then $T(0) = 0$, which implies that $0 \in \text{range } T$.

If $w_1,w_2 \in \text{range } T$, then there exist $v_1,v_2 \in V$ such that $Tv_1 = w_1$ and $Tv_2 = w_2$. Thus
$$T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2$$
Hence $w_1+w_2 \in \text{range } T$

Why $w_1+w_2 \in \text{range } T$ ? I can follow the steps but don't understand why it is in the range

Similarly, I don't understand why $\lambda w$ is in the range of $T$

Range of $T$ consists all vectors of the form $T(v)$ where $v \in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $\lambda w=T(\lambda v)$ so $\lambda w$ is in the range of $T$.

Note the previous line: $T(\text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $\lambda w$ will be similar: since $T$ is linear, $T(\lambda v) = \lambda T(v) = \lambda w$, so there is something that, when we hit it with $T$, we get $\lambda w$, so $\lambda w$ is in the range of $T$.


Probability of having a disease and testing positive for it In a random test you test positive for a disease.In  $5$%  of  cases,  the  test  shows positive  even  when  the  subject does  not  have  the  disease. In  the  population  at  large,  one person  in  $1000$  has  the  disease. What  is  the  conditional  probability  that you  have  the  disease  given  that  you have  been  tested  positive,  assuming that  if  someone  has  the  disease,  he  will test  positive  with  probability  $1$?
My approach was
$P($Teating positive)=P(test positive).$P($you dont have the disease)+$P($test positive).$P($you have the disease)$$=\frac {999}{1000}.\frac{5}{100}+\frac {1}{1000}=\frac {5095}{100000}$$
P(You have the disease$|$you test positive)=$\frac {P(\text{you have the disease,you test positive})}{P(\text{test positive})}$
$$=\frac {1}{1000}.\frac{100000}{5095}=\frac {20}{1019}$$
Is this correct?

Considering, 1000 people were taken for the test.  So, 50 people were positive by test and 1 person has the disease( as well as positive). So, the total test positives = 51 and true positive= 1. So the probability of having the disease and being positive = 1/51. Shouldn't this be the answer?


How to understand combination? $2$ fair tetrahedral dice each showing the numbers from $1$ to $4$ on each of their faces are rolled and their face values are added together. How can I calculate the discrete probability distribution?
My workings:

I am struggling to understand the use of combinations. So lets say I want to know the probability of sum of $3$ how can I do that? So I can get $1$ on the first die and $2$ on the second die. and I can get $2$ on the second die first and they can be counted as one? Likewise I can get $1$ on the second die and $2$ on the first die. I can also get $1$ on the second die first or second, either way they counted as one?
Can someone please provide an intuitive way to understand combination? What is I want to calculate the probability of getting a sum of 5?

So the sample space for all possible outcomes of rolling those two dice will be $S ={\{2,3,4,5,6,7,8\}}$. There are 16 different outcomes of the rolls. Only 1 of them will be equal to $2$ (when you get $1$ and $1$ on both dice), hence probability of getting a sum of $2$ will be $1/16$. To get the sum equal to $3$ you can get either $2,1$ or $1,2$ which means only two different options, so probability of sum $3$ will be $2/16$ and so on. The intuition is to calculate the possible ways of getting the sum and divide that by total possible outcomes.


Subtracting powers with variable in exponent I am having some troubles with a question that subtracts powers.

Solve for unknown:
$$3^{x+4} - 5(3^x) = 684$$

I have a hunch that I should apply factorization somehow. Do I multiply 5 and 3 to begin or should I change to logarithm form and have the 5 act as an exponent on $3^x$, and then factorize?

Any recommendations are appreciated!

I'm not sure if you mean $3^{x+4}-5(3^x)=684$ or $3^x+4-5(3^x)=684$.  If you mean the first, then rewrite $3^{x+4}$ as $3^x\cdot3^4$ and then factor $3^x$ out.  If you mean the second way, subtract $4$ from each side and then factor out $3^x$.  Hope this helps.

You factor out the $3^x:$
$(3^4-5)(3^x)=680$
$\longrightarrow$
$76(3^x)=680$
Now, you can solve this equation with calculator.


How to find an angle of a non-right angle triangle when given two sides and an area? How would I go about finding an angles of a non-right angled triangle when given the area and two of its sides.

For example:

In the triangle $ABC$, $a = 5$, $b = 6$ and the area is $11~\text{cm}^2$. Find the size of Angle $C$.

Thanks for all of your help! :-)

Note:

What is wrong with the current method I am using? (method below)

\begin{align*}
Area & = \frac{ab\sin C}{2}\\
\Rightarrow 2Area & = ab\sin C\\
\Rightarrow \frac{2Area}{ab} & = \sin C\\
\sin C & = \frac{11^2 \cdot 2}{5 \cdot 6}\\
\Rightarrow \sin C & = 8.06
\end{align*}

You still need to do this ~ C=sin-1(11/15)
Because you are trying to find C alone.
Correct me if I'm wrong.


Find for which values of $k$ the system has none, one or infinitely many solutions Given the following linear system
$$\left[\begin{array}{cccc|c} k & 0 & 1 & 2 & 0\\ -1 & 1 & k & 1 & 0\\ 1 & k & 1 & 3 & 0\end{array}\right]$$
find for which values of $k$ the system has none, one or infinitely many solutions.

I got to
$$\left[\begin{array}{cccc|c}1&k&1&3&0\\0&1+k&1+k&4&0\\0&-k^2&1-k&2-3k&0\end{array}\right]$$
but now I'm stuck. How do I eliminate $-k^2$ using $1+k$?
This is not the first time I got a question like this, how do I solve it?

You’ll have to do a case analysis to proceed with the row-reduction. Dividing the second row by $1+k$ is legal as long as $k\ne1$, which you’ll have to deal with separately. You could instead proceed by swapping the last two rows and dividing the second row by $-k^2$, but that gives you a different special case, $k=0$ to deal with.

However, the highlighted question can be answered without doing any calculations whatsoever. The system is homogeneous, so it has at least one solution: the zero vector. Moreover, the system is underdetermined, so it has an infinite number of solutions for all $k$.

Now, if you were asked for the dimension of the solution space for different values of $k$, that would be a different matter. You’d have to do some calculations to find values of $k$ for which the null space of the coefficient matrix is more than one-dimensional. As an alternative to row-reduction, you could examine the $3\times3$ minors of the coefficient matrix. It turns out that there’s no value of $k$ for which they all vanish, so the matrix has full rank for all $k$.

To perform the second step in row reduction, you need to find the first non-vanishing element in the second row.  That would be $k+1$ iff $k\neq-1$.  In that case multiply the second row by $k^2$ and the third one with $k+1$ and add both.

Now if $k=-1$ you get
$$\left[\begin{array}{cccc|c}1&-1&1&3&0\\0&0&0&4&0\\0&-1&2&5&0\end{array}\right],$$
which is easily solved.


Show that : $ \frac{f'(x)}{g'(x)}\xrightarrow[x \to a]{} l \implies \frac{f(x)}{g(x)}\xrightarrow[x \to a]{} l\quad (l\in \mathbb{\bar{R}}) $ Let $f$,$g\in \mathcal{C}((a,b),\mathbb{R})$ are differentiable on $(a,b)$

we assume that $f(a)=g(a)=0$ and $g'\neq  0  $ on $(a,b)$

Show that $$ \frac{f'(x)}{g'(x)}\xrightarrow[x \to a]{} l \implies \frac{f(x)}{g(x)}\xrightarrow[x \to a]{} l\quad  (l\in \mathbb{\bar{R}}) $$

My thoughts

by Cauchy's mean value theorem : $\exists\ C\in (a,b)$ such that $\left(f(b)-f(a)\right)g'(c)=\left(g(a)-g(b)\right)f'(c)$

since $g'\neq 0$ then $\dfrac{f(b) - f(a)}{g(b) - g(a)} = \dfrac{f'(c)}{g'(c)}.$

or $f(a)=g(a)=0$ then $\dfrac{f(b)}{g(b)} = \dfrac{f'(c)}{g'(c)}.$

i want to use L'Hôpitals rule but i'm stuck

Clearly
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f(x) - f(a)}{g(x) -g(a)}.$$

For each $x > a$, use Cauchy's Mean Value Theorem on $(a,x)$ to say that there exists $c = c(x) \in (a,x)$ such that $\frac{f'(c)}{g'(c)} = \frac{f(x)-f(a)}{g(x) - g(a)}$. Notice that $c(x) \to a$ as $x \to a$, so that

$$\lim_{x \to a}\frac{f(x) - f(a)}{g(x) -g(a)} = \lim_{x\to a} \frac{f'(c(x))}{g'(c(x))} = \lim_{c \to a} \frac{f'(c)}{g'(c)} = l$$


Volume of region in the first octant bounded by coordinate planes and a parabolic cylinder? Find the volume of the solid region in the first octant bounded by the coordinate planes, the plane $y + z = 2$ and the parabolic cylinder $x = 4 - y^2$.

I have a final answer, I would just like to make sure I am correct.

Because it is the first octant bounded by the coordinate planes: $x \geq 0, y \geq 0, z \geq 0$. This means that we can define the boundaries of the region as $0 \leq z \leq 2 - y, 0 \leq y \leq \sqrt{4 - x}, 0 \leq x \leq 4$ (from substituting $y = 0$ into $x = 4 - y^2$).

So E = { $(x,y,z) | 0 \leq x \leq 4, 0 \leq y \leq \sqrt{4 - x}, 0 \leq z \leq 2-y$} and the projected xy plane would be the function $y = \sqrt{4 - x}$.

V = $\int_0^4 \int_0^{\sqrt{4 - x}} \int_o^{2-y}dzdydx$  (I wanted to make sure this part was accurate in terms of order and logic).

V = $\int_0^4 \int_0^{\sqrt{4 - x}} (2-y)dydx$

V = $\int_0^4 [2 \sqrt{4-x} - \frac{4 - x}{2}]dx$

V = $2\int_0^4 \sqrt{4-x}dx - \frac{1}{2}\int_0^4(4 - x)dx$

V = $6\frac{2}{3}$ $units^3$

The solution appears to be correct. Personally, I used different construction of the integral, which is

$\int_0^2dy\int_0^{2-y}dz\int_0^{4-y^2}dx = \frac{20}{3}$.

Hope this helps.

Find the volume of the region in the first octant
bounded by the coordinate planes, the plane $ +
 = 2$, and the cylinder $
^2 + 4^
2 = 16$.


How to find all solutions of $y'=c-y/x$? How to show that every solution of the ODE $f'(x)=c-f(x)/x$ is of the form

$f(x) = cx/2 + k_1/x$ for some $k_1 \in \mathbb{R}$?

(Let's say that we limit the domain of $f$ to $x>0$ so $1/x$ is a valid solution).

$$y' + \frac{1}{x}y = c$$

Multiplying both sides by $x$

$$xy' + y = cx$$
$$(xy)' = cx$$
$$xy = c\frac{x^2}{2} + d$$
$$y = \frac{cx}{2} + \frac{d}{x}$$

All assuming $x\gt 0$

Let $z(x)=\frac{y(x)}{x}, $ then $c-z=y'=z+xz'.$

Can you proceed ?


What's the difference/connection between PCA and inverse Fourier transform? Principle Component Analysis (PCA) finds the component with the highest contribution, which is very similar to the idea of inverse Fourier transform, which finds the frequency with the highest weight. Could someone help clarify their difference/connection. It seems that they are connected in some mathematical forms.

There is indeed a connection but with the Fourier transform, and not it's inverse.
PCA is also known by Karhunen-Loève Expansion, this finds the optimal basis for your data (optimal in the sense that it minimizes the euclidean distance between the data points and their projections onto the new basis). If your data is translationally invariant (i.e. permutation of one data point generates another data point), then the optimal basis consists of Fourier vectors.
In other words, PCA on translationally invariant data will yield the DFT.
See https://en.wikipedia.org/wiki/Circulant_matrix but I will give a brief, handwaving explanation:
If you have translationally invariant data matrix X:
$X =
    \begin{pmatrix}
    1 & 2 & 3 \\
    2 & 3 & 1 \\
    3 & 1 & 2 \\
    \end{pmatrix} $
A circulant matrix $C_{ij} = c_{i-j}$ will circularly permute matrix X. Associated with $C$ is a permutation matrix $P$ which will rearrange the rows of $X$ into their original positions:
$PXC = X$
We can get the right singular vectors from:
$X^T X = (PXC)^ T PXC = C^T X^T P^{-1} PXC = C^T X^T XC$
and
$CX^T X = X^T XC$
(here I am assuming $C$ is also a permutation matrix, because I am both too lazy to prove the general case).
Since $C$ commutes with $X^T X$ they can be simultaneously diagonalized. The eigenvectors of a circulant matrix are the Fourier modes and thus the optimal basis for translationally invariant data are Fourier vectors with the singular values being the discrete Fourier coefficients.
Please also see https://en.wikipedia.org/wiki/Karhunen%E2%80%93Lo%C3%A8ve_theorem

http://shawnleezx.github.io/blog/2015/06/05/connection-between-pca/

"We could unify PCA and Fourier Transform from filter bank or linear operator point of view. Actually those two point of views are just different way to say the same thing in signal processing and mathematics.

From filter bank point of view, the process of getting the PCA coefficients and Fourier coefficients could be regarded as passing the original signal to a Linear Time Invariant system. Each Fourier basis or eigenvector is a filter. All of them make up the filter bank."


Appropriate symbol to denote feasible region I have an optimization problem such as
$$
\min_{x\in\mathop{Range}(\mathbf A) \text{ s.t. } \mathbf B x = b} \|x\|
$$
which I would like to write more concisely as
$$
\min_{x\in\mathcal{X}} \|x\|
$$
where $\mathcal{X} = \{x \in \mathrm{Range}(A) : \mathbf{B}x = b\}$.

What symbol is traditionally used in place of $\mathcal{X}$ to denote the set of feasible $x$?

If the answer depends on whether the constraint set is linear or nonlinear, I am interested in both cases.

Step two1 in "how to do a math problem": label the parts of the problem.  So give the set you want a name.

"Let $S = \{x \in \mathrm{Range}(A) : \mathbf{B}x = b\}$.  ...
$$  \min_{x \in S} ||x||  $$
... "

1 : The number of this step in any particular "how to do a math problem" may vary.


Pointwise vs. Uniform Convergence This is a pretty basic question. I just don't understand the definition of uniform convergence.

Here are my given definitions for pointwise and uniform convergence:

Pointwise convergence: Let $X$ be a set, and let $F$ be the real or complex numbers. Consider a sequence of functions $f_n$ where $f_n:X\to F$ is a bounded function for each $n\in \mathbb N$. $f:X\to F$ is the pointwise limit of $f_n$ if for every $x \in X$, $$\lim_{n\to \infty}f_n(x)=f(x).$$

Uniform convergence: Let $f_n$ be a sequence of functions in the set of all bounded functions from $X$ to $F$ where $F$ is the real or complex numbers. The sequence is said to converge uniformly to a bounded function $f:X \to F$ if, given $\epsilon>0$, there exists an $N\in \mathbb N$ s.t. $\sup\{|f_n(x)-f(x)| : x \in X \}<\epsilon$ for $n\ge N$

I'm sorry I don't have a more specific question. I just don't see the exact relation/difference between the two definitions. I've asked two different professors to explain this to me but neither of their explanations helped.

Edit: Attempting to show that uniform convergence implies pointwise convergence
if $f_n$ converges uniformly to f, then $\sup\{|f_n(x)-f(x)| : x\in X \}$ for $n\ge N$. Thus, $|f_n(x)-f(x)|<\epsilon$ for $n\ge N$, which is the definition of pointwise convergence.

Comparison

Pointwise convergence means at every point the sequence of functions has its own speed of convergence (that can be very fast at some points and very very very very slow at others).

Imagine how slow that sequence tends to zero at more and more outer points:
$$\frac{1}{n}x^2\to 0$$

Uniform convergence means there is an overall speed of convergence.

In the above example no matter which speed you consider there will be always a point far outside at which your sequence has slower speed of convergence, that is it doesn't converge uniformly.

Another Approach

One can check uniform convergence by considering the "infimum of speeds over all points". If it doesn't vanish then it is uniformly convergent. And that gives another characterization as the ones with nonvanishing overall speed of convergence.

The pointwise convergence depends on each $x$, that is you need to count and fix every $x$ before passing to the limit. On the other hand, uniform convergence is independent of $x$.
Hence, the pointwise convergence is enough when your space $X$ is a countable set and not enough when you are working with an uncoutable set like $[0,1]$, in such case you need a uniform convergence to generalise your convergence on $[0,1]$.


Is there a function over which multiplication is distributive which isn't addition? Does there exist a binary operation on $\mathbb{Z}$ over which multiplication is distributive which isn't addition? Formally does there exist $f:\mathbb{Z^2}\to\mathbb{Z}$ s.t. $af(b,c)=f(ab,ac) \forall a, b, c \in \mathbb{Z}$ and $f$ isn't just the usual addition? It would also be nice if this made $(\mathbb{Z},f)$ into an Abelian group.

Edit 1. Ok, there are lots of examples, but are there any which make an Abelian group?

Edit 2. Ok, Andreas Blass has given a rather good example, but I believe his example is a group which is isomorphic to the usual additon. Are there any which make an Abelian group which is not isomorphic to $(\mathbb{Z},+)$?

Let $\pi$ be any permutation of the set of prime natural numbers, and extend it to map $\mathbb Z$ to itself by sending any integer, written as a signed product of primes $\pm p_1^{e_1}\cdots p_k^{e_k}$, to $\pm \pi(p_1)^{e_1}\cdots \pi(p_k)^{e_k}$ and sending $0$ to $0$.  This extended $\pi$ commutes with multiplication, but it transforms addition into a new function $f(a,b)=\pi(\pi^{-1}(a)+\pi^{-1}(b))$, and multiplication distributes over $f$.  Indeed, $\pi$ is an isomorphism from $\mathbb Z$ with ordinary addition and multiplication to $\mathbb Z$ with $f$ and ordinary multiplication.

I thought of $\max(x,y)$, but this fails for negative $a$.
Maybe only linear functions work.
I'll play around a bit and see what happens.

Set $c = 0$ and $g(x) = f(x, 0)$.
Then $ag(b) = g(ab)$.
Let $b = 1$, so $a g(1) = g(a)$,
so $g$ is multiplication by $g(1)$.

Similarly, if $h(x) = f(0, x)$,
$a h(1) = h(a)$, so $h$ is also linear.

Setting $a = 0$, $f(0, 0) = 0$.

Sure looks like only linear might work.
The problem seems to me to be that
the two arguments to $f$
are completely independent,
so there is no way to get them to interact.

I'll leave it at this for now.