Why is the Chevalley-Eilenberg differential a coderivation? For every Lie algebra $\mathfrak{g}$ we can consider the Chevalley-Eilenberg complex given by the exterior powers $\bigwedge^n \mathfrak{g}$ together with the differential $d_{\mathrm{CE}} \colon \bigwedge^n \mathfrak{g} \to \bigwedge^{n-1} \mathfrak{g}$ that is given by
$$
  d_{\mathrm{CE}}(x_1 \wedge \dotsb \wedge x_n)
  =
  \sum_{1 \leq i < j \leq n}
  (-1)^{i+j-1}
  [x_i, x_j]
  \wedge x_1
  \wedge \dotsb
  \wedge \widehat{x_i}
  \wedge \dotsb
  \wedge \widehat{x_j} \wedge \dotsb \wedge x_n \,.
$$
We may regard the Lie algebra $\mathfrak{g}$ as a graded vector space concentrated in degree $1$.
Then the tensor algebra $\operatorname{T}(\mathfrak{g})$ becomes a graded Hopf algebra (such that $\mathfrak{g}$ consists of primitive elements) and its quotient $\bigwedge \mathfrak{g}$ inherits the structure of a graded Hopf algebra.
If I understand the nlab correctly then the Chevalley-Eilenberg differential $d_{\mathrm{CE}}$ is the unique extension of the Lie bracket $[-,-] \colon \bigwedge^2 \mathfrak{g} \to \mathfrak{g}$ to a graded coderivation for the underlying graded coalgebra structure of $\bigwedge \mathfrak{g}$.

  How can we show that $d_{\mathrm{CE}}$ is a graded coderivation of $\bigwedge \mathfrak{g}$, and that is it the unique one that extends the Lie bracket?

I have put my (so far unsuccessful) attempt below.

Coderivation

The differential $d_{\mathrm{CE}}$ is of degree $-1$ so we need to show that
\begin{equation}
  \Delta( d_{\mathrm{CE}}(c) )
  =
  \sum_{(c)}
  d_{\mathrm{CE}}( c_{(1)} ) \otimes c_{(2)}
  +
  (-1)^{|c_{(1)}|} c_{(1)} \otimes d_{\mathrm{CE}}(c_{(2)})
  \tag{1}
\end{equation}
for all $c = x_1 \wedge \dotsb \wedge x_n$ with $x_i \in \mathfrak{g}$ (i.e. that $\Delta$ is a homomorphism of chain complexes with respect to $d_{\mathrm{CE}}$).
I started explicitely calculating the right hand side:
By writing $x_1 \dotsm x_n$ instead of $x_1 \wedge \dotsb \wedge x_n$ we have
\begin{align*}
  {}&
  \Delta(x_1 \dotsm x_n)
  \\
  ={}&
  (x_1 \otimes 1 + 1 \otimes x_1)
  \dotsm
  (x_n \otimes 1 + 1 \otimes x_n)
  \\
  ={}&
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  x_{\sigma(1)} \dotsm x_{\sigma(k)}
  \otimes
  x_{\sigma(k+1)} \dotsm x_{\sigma(n)}
\end{align*}
where $S(k,n-k) \subseteq S_n$ denotes the set of $k$-$(n-k)$-shuffles and
$$
  n_k(\sigma)
  =
  \#
  \{
    (i,j)
  \mid
    1 \leq i \leq k, \,
    k+1 \leq j \leq n, \,
    \sigma(i) > \sigma(j)
  \} \,.
$$
For the right hand side of $(1)$ we hence get
\begin{align*}
  {}&
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  \biggl[
    d_{\mathrm{CE}}(x_{\sigma(1)} \dotsm x_{\sigma(k)})
    \otimes
    x_{\sigma(k+1)} \dotsm x_{\sigma(n)}
  \\
  {}&
  \phantom{
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  \biggl[
  }
    + (-1)^k
    x_{\sigma(1)} \dotsm x_{\sigma(k)}
    \otimes
    d_{\mathrm{CE}}(x_{\sigma(k+1)} \dotsm x_{\sigma(n)})
  \biggr]
  \\
  ={}&
  \sum_{k=0}^n
  \;
  \sum_{\sigma \in S(k,n-k)}
  (-1)^{n_k(\sigma)}
  \\
  {}&
  \biggl[
    \biggl(
      \sum_{1 \leq i < j \leq k}
      (-1)^{i+j-1}
      [x_{\sigma(i)}, x_{\sigma(j)}]
      x_{\sigma(1)}
      \dotsb
      \widehat{x_{\sigma(i)}}
      \dotsb
      \widehat{x_{\sigma(j)}}
      \dotsb
      x_{\sigma(k)}
    \biggr)
    \otimes
    x_{\sigma(k+1)} \dotsm x_{\sigma(n)}
  \\
  {}&
    +
    (-1)^k
    x_{\sigma(1)} \dotsm x_{\sigma(k)}
  \\
  {}&
    \otimes
    \biggl(
      \sum_{1 \leq i < j \leq n-k}
      (-1)^{i+j-1}
      [x_{\sigma(k+i)}, x_{\sigma(k+j)}]
      x_{\sigma(k+1)}
      \dotsb
      \widehat{x_{\sigma(k+i)}}
      \dotsb
      \widehat{x_{\sigma(k+j)}}
      \dotsb
      x_{\sigma(n)}
    \biggr)
  \biggr] \,.
\end{align*}
For the left side of $(1)$ one similarly find that
\begin{align*}
  {}&
  \Delta( d( x_1 \dotsm x_n) )
  \\
  ={}&
  \sum_{1 \leq i < j \leq n}
  (-1)^{i+j-1}
  \Delta( [x_i, x_j] x_1 \dotsm \widehat{x_i} \dotsm \widehat{x_j} \dotsm x_n )
  \\
  ={}&
  \sum_{1 \leq i < j \leq n}
  (-1)^{i+j-1}
  ([x_i, x_j] \otimes 1 + 1 \otimes [x_i, x_j])
  (x_1 \otimes 1 + 1 \otimes x_1)
  \\
  {}&
  \phantom{
    \sum_{1 \leq i < j \leq n}
    (-1)^{i+j-1}
  }
  \dotsm
  (x_i \otimes 1 + 1 \otimes x_i)^{\wedge}
  \dotsm
  (x_j \otimes 1 + 1 \otimes x_j)^{\wedge}
  \dotsm
  (x_n \otimes 1 + 1 \otimes x_n)
\end{align*}
I guess that it could make sense that after multiplying out these terms and rearranging the resulting summands both sides of (1) give the same result
But I don’t really see this (and even less so for the signs).

Uniqueness

I think I have checked that if $C$ is any graded coalgebra and $d_1, d_2 \colon C \to C$ are graded derivations of the same degree $k$ that their equalizer
$$
  E
  :=
  \{
    x \in C
  \mid
    d_1(x) = d_2(x)
  \}
$$
is then a graded subcoalgebra of $C$.
We would like that any coderivation of degree $-1$ of $\bigwedge \mathfrak{g}$ is uniquely determined by the restriction $\bigwedge^2 \mathfrak{g} \to \mathfrak{g}$, so I suspect that $\bigwedge \mathfrak{g}$ is generated as a (graded) coalgebra by its homogeneous parts $\bigwedge^2 \mathfrak{g}$.
But I don’t know how to show this, or if this is correct.

I guess that for finite dimensional $\mathfrak{g}$ one could also dualize the problem and consider instead the cochain complex $\bigwedge \mathfrak{g}^*$ together with the (co)differential $d_{\mathrm{CE}}^*$ induced by $d_{\mathrm{CE}}$.
Then $\bigwedge \mathfrak{g}^*$ should be a graded algebra with derivation $d_{\mathrm{CE}}^*$, so $d_{\mathrm{CE}}^*$ should be uniquely determined by the restriction $\mathfrak{g}^* \to \bigwedge^2 \mathfrak{g}^*$ because $\bigwedge \mathfrak{g}^*$ is generated as an algebra by $\mathfrak{g}$.
But I’m not sure if this actually works.


For which $p>0$ does $\sum_{n=3}^{\infty}\frac{\log(n)}{n^p}$ converge? For which $p>0$ does $\sum_{n=3}^{\infty}\frac{\log(n)}{n^p}$ converge? I tried all the criteria for series convergence I know, but I'm not getting any further with this exercise. I'm not asking someone to do my homework for me, but could somebody tell me what criteria I should try please or how to proceed?

Consider $3$ cases:

Case #1: $p=1$. From this: $log(n)>1 \implies \dfrac{log(n)}{n} > \dfrac{1}{n} $ you can see by comparison test (with harmonic series) that series diverges.

Case #2: $p>1$. You just use integration test and integrate this $\int_{1}^{+\infty} x^{-p} log(x)dx$ via integration by parts (set $u=log(x)$) and you will see that series converges (because the integral converges).

Case #3: $p<1$. You can easily deduce from previous cases that in this case sum diverges.

As already mentioned elsewhere, this is (apart from the sign and the missing contribution from n=2) the first derivative of the Riemann Zeta function at p.


Why can an integer written $2$ times in a row never be a perfect square? It seems to be true for all natural numbers below $1,000,000$.
I am really stuck any kind of help will be appreciated!

This is not true in general.

Here is a counterexample: 4049586776940495867769 is the square of 63636363637

Why did I have to come up with such a huge counterexample? The reason is that the "conjecture" indeed works for small numbers. Here is why : Consider $\overline{NN}$ where $N$ is a $d$-digit number so that $\overline{NN}$ has $2d$ digits. $\overline{NN} = N \times (10^d+1)$. If $10^d+1$ is squarefree, $\overline{NN}$ has to be at least $(10^d+1)^2$ which has more than $2d$ digits, a contradiction. Thus we need $10^d+1$ to be the multiple of the square of some prime for $\overline{NN}$ to have the possibility of being a perfect square. $d=11$ is the smallest value for which this works.

This just might be useful.  If $n$ is even, $10^n$ is a perfect square, so $10^n + 1$ cannot be a perfect square.


Prove that if $A$ is normal, then eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal (alternative proof) The problem statement is as follows:

  Prove that for a normal matrix $A$, eigenvectors corresponding to different eigenvalues are necessarily orthogonal.

I can certainly prove that this is the case, using the spectral theorem.  The gist of my proof is presented below.

If possible, I would like to find a simpler proof.  I was hoping that there might be some sort of manipulation along these lines, noting that
$$
\langle Av_1,A v_2\rangle
= \langle v_1,A^*Av_2\rangle
= \langle v_1,AA^*v_2\rangle
= \langle A^* v_1,A^* v_2 \rangle
$$

Any ideas here would be appreciated.

My proof:

Let $\{v_{\lambda,i}\}$ be an orthonormal basis of eigenvectors (as guaranteed by the spectral theorem) such that
$$
A v_{\lambda,i} = \lambda v_{\lambda,i}
$$
Let $v_1,\lambda_1$ and $v_2,\lambda_2$ be eigenpairs with $\lambda_1 \neq \lambda_2$.  We may write
$
v_1 = \sum_{i,\lambda}a_{i,\lambda}v_{i,\lambda}
.$
We then have
$$
0 = Av_1 - \lambda_1 v_1 = \sum_{i,\lambda}(\lambda - \lambda_1)a_{i,\lambda}v_{i,\lambda}
$$
So that $a_{i,\lambda} = 0$ when $\lambda \neq \lambda_1$.  Similarly, we may write $v_2 = \sum_{i,\lambda}b_{i,\lambda}v_{i,\lambda}$, and note that $b_{i,\lambda} = 0$ when $\lambda \neq \lambda_2$.  From there, we have
$$
\langle v_1,v_2 \rangle = \sum_{i,\lambda}a_{i,\lambda}b_{i,\lambda}
$$
the above must be zero since for each pair $i,\lambda$, either $a_{i,\lambda}=0$ or $b_{i,\lambda} = 0$.

Assume $\;\lambda\neq \mu\;$ and

$$\begin{cases}Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v\\{}\\Aw=\mu w\implies A^*w=\overline\mu w\end{cases}$$

From this we get:

$$\begin{cases}\langle v,Aw\rangle=\langle v,\mu w\rangle=\overline\mu\langle v,w\rangle\\{}\\
\langle v,Aw\rangle=\langle A^*v,w\rangle=\langle\overline\lambda v,w\rangle=\overline\lambda\langle v,w\rangle \end{cases}$$

and since $\;\overline\mu\neq\overline\lambda\;$ , we get $\;\langle v,w\rangle =0\;$

Question: Where did we use normality in the above?

A normal matrix is unitarily similar to diagonal matrix.

$$A = UDU^{-1}$$ where $U$ is Unitary matrix.

Eigen decompositions tells that $U$ is a matrix composed of columns which are eigenvectors of $A$. And matrix $D$ is Diagonal matrix with eigenvalues on diagonal.

  Property: Columns of Unitary matrix are orthogonal.

So, columns of $U$ (which are eigenvectors of $A$) are orthogonal.


Finding the Normal & Tangent Vector and write an Equation. Find a normal vector and a tangent vector at the
point P. Write an equation for the tangent line and an equation
for the normal line.

$x^2 + xy + y^2 = 3; P(−1,−1).$

So what I did first was find the gradient by taking the partial derivatives.

$▽f$ = $(2x+y)i + (x+2y)j$

$▽f$  $(-1,-1)$ = $3i-3j = 3$

Normal Vector $i+j$

Tangent Vector $i-j$

How does one find the tangent line and normal line from this?

Hint:

$\vec{r}$ represents the point (could be any points) on the line , and $\vec{x}$ is a known point on the line

$\vec{r}-\vec{x}$ is the direction of the line.

P.S : I don't know why I earn a downvote . This is very basic knowledge in elementary geometry ( someone call it analytical geometry) . Therefore , I just provide some hints without details. Thanks to your imprudent judgement , I will keep silent.


If $A^2B^2 = ABAB$, for square matrices A,B having same size, then prove that $AB = BA$. I actually know it's false if A and B are not invertible.
But how can i prove that not by using counterexamples

You can prove that by considering the null-space and value space of matrices. Recall that we know that if a matrix is non-invertible then the dimension of the null-space is greater than $0$ and the dimension of the value space is less than the row count. Also we know that $\dim N(M) + \dim V(M)$ is the actual row count of $M$.

What we use that given two subspaces $U$ and $W$ with $\dim U + \dim W$ being the row count then there exist a matrix $M$ that has $U = N(M)$ and $W = V(M)$. This can be seen as the ortogonal complement to $U$ has the same dimension as $W$ and therefore it exists a linear bijection between $U^\perp\to W$ and then we can define a mapping of the whole space by projecting the argument on $U^\perp$.

Now we know that if there's at least two dimensions there therefore exists $A$ and $B$ such that $N(A) = V(B)$ and $N(B) \perp V(A)$, and $\dim V(A), \dim V(B)> 0$. This means that $B$ would map anything to something that $A$ maps to $0$ which means that $AB$ always maps to zero which means that $AB=0$. On the other hand $V(A)$ contains vectors not in $N(B)$ which means that $A$ will map some vectors to vectors not in $N(B)$ which means that $B$ will not map those vectors to $0$, that is there exists vectors $x$ such that $BAx\ne 0$ which means $BA\ne 0$. This clearly means that $AB-BA\ne 0$. But sice $AB=0$ then so is $A^2B^2 = A (AB) B$ and also $ABAB=(AB)(AB)$.

We have concluded that there exists $A$ and $B$ such that $AB\ne BA$ yet $A^2B^2 = ABAB$.

Note that although this may seem to be a proof without using counter-example it's probably not that free of counter-example. For example showing that there exists linear bijection between spaces of the same dimension is probably shown by actually constructing such a mapping. Also see that we construct the mapping with prescribed zero and value space from this through direct construction from the linear bijection to the value space. Also that there are subspaces $N(A), V(A), N(B), V(B)$ with the desired properties is also probably done by direct construction.

Actually normally when we want to show the existence of something one often relies on concrete examples, otherwise one constructs an example abstractly starting from something that is known to exist. Only occationally one proves the existence non-constructively. And disproving a general assertion means that we're to prove that there exists an object that falsifies the assertion.

It is indeed false. For a counterexample, consider:
$$
A = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
\qquad\text{and}\qquad
B = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}
$$
Note that $BA = A$, and:
$$
A^2 = AB = \textbf{0} = \begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}
$$
so we have that $A^2B^2 = \textbf{0}(B^2) = \textbf{0} = \textbf{0}(AB) = ABAB$, yet $AB = \textbf{0} \neq A = BA$.


Applications of coupled systems of $\;2\times 2\;$ linear differential equations I am providing maths help to some students studying just before University level in mathematics.

I am writing some practice questions for them on solving coupled first order linear equations and I was wondering if there are any 'real-life' applications of this.

I am having difficulty finding examples because the systems they need to solve are restricted to two dependent variables and one independent variable and need to be linear. (Most of the easy go-to options, e.g. SIR models, are not linear).

I have already used:

predator-prey problems
two-tanks mixing problems

Are there any others?

[For information, the equations they are required to solve should be of the form

$\frac{dx}{dt} = ax + by + f(t), \hspace{3mm} \frac{dy}{dt} = cx + dy + g(t)$.]

Coupled harmonic oscillators are a good example. The outer rectangles are walls, the inner ones are masses and the lines are springs.  The heavy lines are strong springs.  If you write it in terms of the positions of each mass, the coupling spring shifts the frequencies up and down a bit.  You can do the linear algebra to show that the modes are symmetric and antisymmetric and that the correct basis is in those terms.


Average of sum of unit roots is an algebraic integer Let $\alpha_1,\ldots, \alpha_n$ be roots of unity, and let $a=\frac{1}{n}\sum\alpha_i$. Then if $a$ is an algebraic integer, we have either $a=0$ or $a=\alpha_1=\dots=\alpha_n$. Why?

Suppose that $\alpha_1, \dots, \alpha_n$ are $k$th roots of unity. Since all conjugates of the $\alpha_i$ are again $k$th roots of unity, for any conjugate $a'$ of $a$, we have the (complex) norm of $a'$ is bounded:
$$
|a'| = \frac{1}{n} \sum \left |\zeta_k^{t_i} \right| \leq \frac{1}{n} \sum 1 = 1.
$$
But any nonzero algebraic integer whose conjugates are all bounded by 1 must be a root of unity. So $a = 0$ or $a$ is also a ($k$th) root of unity.

If $a \neq 0$, then the triangle inequality
$$
|a| = \frac{1}{n} | \alpha_i | \leq \frac{1}{n} \sum |\alpha_i| = 1
$$
is an equality. By a simple induction argument, this implies $\alpha_i = \pm \alpha_j$ for $1 \leq i,j \leq n$.

Write $\alpha_1 = \zeta_k^\ell$, so each $\alpha_i = \pm \zeta_k^\ell$. Reordering (and potentially swapping signs) if necessary, we may assume $\alpha_1 = \dots = \alpha_m = \zeta_k^\ell$ and $\alpha_{m+1} = \dots = \alpha_n = - \zeta_k^\ell$ with $\frac{n}{2} \leq m \leq n$.

Then
$$\begin{align*}
a &= \frac{1}{n} \left(\sum_{i=1}^m  \zeta_k^\ell - \sum_{i = m+1}^n \zeta_k^\ell \right)\\
&= \frac{2m - n}{n} \zeta_k^\ell
\end{align*}$$

But $a$ is an algebraic integer, hence $\frac{2m-n}{n} \in \mathbf{Z}$. Thus $m = \frac{n}{2}$ or $m = n$, giving $a = 0$ or $a = \alpha_1 = \dots = \alpha_n$, respectively.

I don't think we need Kronecker? First notice $|a|\le 1$, with equality iff all the $\alpha_i$s coincide, so WLOG assume $|a|<1$. Let $a'$ be any conjugate of $a$. WLOG $a$ is $1/n$ times the sum of $n$ $k$-th roots of unity; and so $a'$ must also be $1/n$ times the sum of $n$ $k$-th roots of unity, since the conjugates of $k$-th roots of unity are still $k$-th roots of unity. Immediately we get $|a'|\le 1$. Now let

$a^m + r_{m-1}a^{m-1}+\cdots + r_1a^1+r_0=0$

be a minimal polynomial for $a$. WLOG (if $a\neq 0$) we may assume $r_0\neq 0$. By Vieta, we have $\pm r_0$ equals the product of $a$ with all its conjugates, counting multiplicities. Thus $|r_0|<1$ and so $r_0=0$ (because it's an integer), our desired contradiction.


Fast way to compare angles w/o their measures I need an efficient way to compare the measures of two angles in $\mathbb{R^2}$ that ideally relies on the smallest number of arithmetic operations and no trigonometric operations (no $\arccos$) or square roots. Given two angles $\angle{p_1p_0p_2}$ and $\angle{p_3p_0p_4}$ whose measures are in the range $\big(0,{\pi}\big)$ radians with vertex $p_0$ at the origin, what is the fastest way to compare the measures of the angles (i.e. find which angle has the largest measure/which one has the smallest measure)?

I understand that the $z$ component of the cross product of two or more pairs of vectors can be used to compare angles where $\theta \in \big(0,\frac{\pi}{2}\big)$ radians. As stated above, however, I need to compare angles with  $\theta \in \big(0,{\pi}\big)$.

Any ideas?

Let $\theta$ be the angle between $p_1$ and $p_2$, and $\phi$ the one between $p_3$ and $p_4$. Since $p_1 \cdot p_2 = |p_1| |p_2| \cos \theta$, the sign (positive or negative) of $p_1 \cdot p_2$ tells you whether $\theta \in [0, \pi/2)$ or $(\pi/2, \pi)$. Similarly for $\phi$, and so you may already be able to tell which is larger.

If both dot products have the same sign, then define
$$ r_\theta := (p_1 \cdot p_2)^2 |p_3|^2 |p_4|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 \cos^2 \theta$$
and
$$ r_\phi := (p_3 \cdot p_4)^2 |p_1|^2 |p_2|^2 = |p_1|^2 |p_2|^2 |p_3|^2 |p_4|^2 \cos^2 \phi$$
and so comparing $r_\theta$ and $r_\phi$ tells you which has the largest cosine squared, from which you can figure out what is the larger angle. At worst, this takes 6 dot products, 4 multiplications, and 4 comparisons.

Try the dot product. You will like the results.

Because, dot product of two vectors will be equal to the cosine of the angle between the vectors, times the lengths of each of the vectors.

You do the rest.


How to calculate the average x-coordinate of a region? I have a region in the x-y plane. Now I randomly select a point inside or on the boundary of the region. What is the expected value of the x-coordinate of my point? I.e. If I perform this operation infinitely many times, what is the average of the x-coordinate?

Thanks for your help!

$E[x] = \frac{\int\limits_{\cal R} x\ dA}{\int\limits_{\cal R} dA}$ where ${\cal R}$ is the two-dimensional region (assumed uniform density) and $dA$ is a differential area, such as $dx\ dy$ or $2 \pi r\ dr\ d\theta$.


multiplying two normal subgroups is still normal? Let $G_i \triangleleft G_{i+1}$ both subgroups of $G$. Let $N$ be a normal group.

Does $G_iN \triangleleft G_{i+1}N$?

Does $(G_iN/N) \triangleleft (G_{i+1}N/N)$?

I know that $q:G\longrightarrow G/N$ preserves normality. Hence if $G_iN \triangleleft G_{i+1}N$ then their quotients would be normal.

I have tried by considering an element in $g\in G_iN$ and $h\in G_{i+1}N$ and trying to find out if $ghg^{-1}\in G_iN$. But I am a bit confused of those multilpication groups. And that is why I can not play with this expression.

I want to change the notation to make stuff clearer: Suppose $H\lhd K< G$, and let $N\lhd G$ (so really $H=G_i$ and $K=G_{i+1}$). Then a proof that $HN\lhd KN$ by considering a conjugator is:

Suppose $k\in K$, $h\in H$, and $n, n'\in N$. Firstly, note that
$$n^{-1}k^{-1}(hn')kn=(n^{-1}k^{-1}hkn)\cdot (n^{-1}k^{-1}n'kn)$$
Clearly $(n^{-1}k^{-1}n'kn)\in N$, while $(n^{-1}k^{-1}hkn)\in HN$ by the following:
$$\begin{align*}
k^{-1}hk&\in H\\
\Rightarrow k^{-1}hk\cdot (k^{-1}hk)^{-1}n^{-1}(k^{-1}hk)&\in HN\\
n^{-1}(k^{-1}hk)&\in HN\\
n^{-1}(k^{-1}hk)n&\in HN
\end{align*}
$$
The result follows.

Hint:

The first one should be almost immediate, and for the second one:

$$G_iN/N\cong G_i/(G_i\cap N)\;,\;\;\; G_{i+1}/(G_{i+1}\cap N)\cong G_{i+1}N/N$$


Given a simple graph and its complement, prove that either of them is always connected. I was tasked to prove that when given 2 graphs $G$ and $\bar{G}$ (complement), at least one of them is a always a connected graph.

Well, I always post my attempt at solution, but here I'm totally stuck. I tried to do raw algebraic manipulations with # of components, circuit ranks, etc, but to no avail.
So I really hope someone could give me a hint on how to approach this problem.

Suppose $G$ is disconnected. We want to show that $\bar{G}$ is connected. So suppose $v$ and $w$ are vertices. If $vw$ is not an edge in $G$, then it is an edge in $\bar{G}$, and so we have a path from $v$ to $w$ in $\bar{G}$. On the other hand, if $vw$ is an edge in $G$, then this means $v$ and $w$ are in the same component of $G$. Since $G$ is disconnected, we can find a vertex $u$ in a different component, so that neither $uv$ nor $uw$ are edges of $G$. Then $vuw$ is a parth from $v$ to $w$ in $\bar{G}$.

This shows that any two vertices in $\bar{G}$ have a path (in fact a path of length one or two) between them in $\bar{G}$, so $\bar{G}$ is connected.

This is another answer, showing G¯ is connected in 3 cases by showing there exists an {a,b} path in each case. Let x,y be vertices in G. Let H be the component that holds the vertex x but not y.

C1: Suppose a and b are arbitrary vertices not in H. Then x~a and x~b in G¯ so there is a unique {a,b} path in G¯, namely {a,x,b}. (~ means adjacent)

C2: Suppose H holds one of the two vertices, a and b. Without loss of generality suppose H holds a. Then x~b and a~b in G¯. So there is a unique {a,b} path in G¯, namely {a,b}.

C3: Suppose H holds both the a and b vertices. Then y~a, y~b, y~x in G¯. So there is a unique {a,b} path in G¯, namely {a,y,b}.

Therefore there is a unique {a,b} path in G¯ for arbitrary a and b vertices in all 3 cases. Therefore G¯ is connected.


For all $a, b \in R $ with $a \neq 0$, the equation $ax = b$ has a solution in$R$ Suppose $R $  is a non zero ring.

How can we prove the following conditions are equivalent?

In Proof 2 to 3, how can we prove  from $ a R = R $  that the ring has identity?

  1:  For all $a, b \in  R $ with $a \neq 0$, the equation $ax = b$ has a solution in$R$ .

  2: $ R^{2} \neq 0$  and $ R$ has no right ideals other than $ 0$ and $R$.

  3: $ R$ is a division ring.

The set of all elements $x$ such that $xR=0$ forms a right ideal of $R$.

By hypothesis it can't be all of $R$, so it is just $\{0\}$.

In other words, $xR=R$ for nonzero $x$.

Let $x$ be a nonzero element. Two observations:

The right annihilator of $x$ is a right ideal, and it can't be $R$, so it is zero. Therefore $x$ is left cancelable. (Notice this is true for any nonzero $x$.)
$xe=x$ for some $e$, and furthermore $xee=xe=x$. Canceling $x$ we find that $e^2=e$.

Using the $e$ above, we see that $ey=e^2y$ implies $y=ey$ for any $y\in R$, so $e$ is a left identity.

Finally, there's probably a shorter trick, but this one is the first that occurred to me:
$(y-ye)^2=y^2-y^2e-yey+yeye=y^2-y^2e-y^2+y^2e=0=(y-ye)0$. If $y-ye\neq0$, we could cancel it on the left, obtaining $y-ye=0$, a contradiction. Therefore $y-ye=0$, and $e$ is a right identity too.

If $aR = R$, $a$ is not zero, then $R$ has identity:
we recall from rschwieb that left cancellation is valid. Now:

$aR = R \implies ac = a$ for some $c \in R$

$cR = R \implies cb = a$ for some $b \in R$

Prove that $a = b$ and then $c$ will be our identity:
$ab = acb = a(cb) = aa \implies b = a$,
$ac = cb = ca = a \implies c$ is the identity.


X ∼ Exp(1) and Y ∼ uniform(0, 1) are independent. Compute $E(e^{-XY^2})$. X ∼ Exp(1) and Y ∼ uniform(0, 1) are independent. Compute $E(e^{-XY^2})$.
I am having trouble answering this question, I am not sure if I should take advantage of moment generating functions or multivariable LOTUS. I think multivariable law of the unconscious statistician would be easier, but I am unsure on how to derive the joint pdf.Any help would be greatly appreciated!

We have
\begin{align*}
\mathbb E[e^{-XY^2}] &=\int_0^1\int_0^\infty e^{-xu^2} e^{-x} dx du\\
& = \int_0^1 \Big(-\frac{1}{1+u^2} e^{-(1+u^2)x} \vert_{x=0}^\infty \Big) du\\
&=\int_0^1\frac{1}{1+u^2} du = {\rm arctan}(1) - {\rm arctan}(0)= \frac{\pi}{4},
\end{align*}
since ${\rm arctan}’(x) = 1/(1+x^2).$ Here in the first line I am using the fact that the joint density of $X$ and $Y$ is:
$$
f_{X,Y}(x,u)=e^{-x}1_{x>0}1_{u\in(0,1)}.
$$

If $X$ and $Y$ are independent with respective densities $f$ and $g$, then for any reasonable function $F$,
$$E[F(X,Y)] = \iint F(x,y)f(x)g(y) dxdy.$$
Do you know the densities of $X$ and $Y$ in this case?


X and Y are independent random variables with the same distribution X and Y are independent random variables with the same distribution. If W=X+Y and Z=X-Y why are W and Z not independent?

We give an example that may make it clear why $X-Y$ and $X+Y$ need not be independent.

Let $X$ and $Y$ be (independent and) uniformly distributed on $[0,1]$.

Suppose that $|X-Y|$ is "large" (bigger than $0.9$). Then it is impossible for $X+Y$ to be small (say less than $\frac{1}{2}$). So information about $X-Y$ ccan yield information about $X+Y$.

Remark: One could rewrite the above as $\Pr\left((|X-Y|\gt 0.9)\cap (X+Y\lt \frac{1}{2})\right)\ne \Pr(|X-Y|\gt 0.9)\Pr(X+Y\lt \frac{1}{2})$, but that would somewhat defeat the point of the example.

If two variables are independent, then they must have a covariance of 0.
$$
\begin{align}
Cov(W,Z) &= Cov\left((X+Y),(X-Y)\right)\\
Cov\left((X+Y),(X-Y)\right) &= E\left[(X+Y)(X-Y)\right] - \left[E(X+Y)\right]\left[E(X-Y)\right]\\
&= E(X^2 - Y^2) - \left[E(X) + E(Y)\right]\left[E(X) - E(Y)\right]\\
&= E(X^2) - E(Y^2) - E(X)^2 + E(Y)^2\\
&= \left[E(X^2) - E(X)^2\right] - \left[E(Y^2) - E(Y)^2\right]\\
&= Var(X) - Var(Y)
\end{align}
$$

Just because $X$ and $Y$ are independent does not mean that they have the same variance. For cases where they do not have the same variance, the covariance term is <> 0, and thus the sum and difference of two independent random variables is not necessarily itself independent.


Simplified method for symmetric matrix determinants Preparing for my final exams I have been doing all the exercises in my algebra book. I've seen that there are lots and lots of exercises about determinants of symmetric matrices. Some are easy and others are a bit more twisted, but the basic problem is almost always the same.

I have been trying to come up with something to do these a bit more quickly, since (at least for me) they invariably end with a very ugly stream of numbers and letters, and they are really tedious and time-consuming. I've been two days with the sensation that (being as the matrices are symmetric) there must be a way/catch/method/trick to calculate the determinants in a more elegant way, but I haven't been able to come up with anything.

I started with a $3\times 3$ matrix like this:

$$A= \begin{pmatrix}
\ a & b & c \\
b & a & b \\
c & b & a \end{pmatrix}$$

which looks fairly simple, but the best I could come up with was:

$2b^2(c-a)+a(a^2-c^2)$

and

$a(a^2-2b^2-c^2)+2b^2c$

These look horrific and absolutely not what anyone in his right mind would use. It goes without saying that I haven't even tried this with matrices bigger than 3.

Granted, my math level is basic (I very rarely use linear algebra and calculus out of college, so I tend to completely forget them from year to year) and I don't have many tools to do this, so I don't really know if I'm trying something that is too much for my level or if it simply can't be done. I've consulted all the books about matrices in the college library, but no one mentions anything special about calculating determinants for symmetric matrices.  Is there something I have been missing, or is there nothing to do about it?

There's not a whole lot you can do to simplify that anymore.

In any case, what you've written down is a special case of a symmetric matrix.  In general a symmetric $3 \times 3$ matrix will have the form:

$$A= \begin{pmatrix}
\ a & b & c \\
b & d & e \\
c & e & f \end{pmatrix}$$

which has a determinant of $a(df-e^2) + b(ce-bf) + c(be-dc)$.  Even worse-looking.

The only time it really gets a lot simpler is if you have zeroes in there.  The simplest way to calculate is not to calculate.

I have been doing some try and error in MATLAB for matrices that follows the matrix $A$ from the original question (symmetric with equal terms on their diagonals) and after looking the obtained sequence of determinants in OEIS.org, I found a nice formula for it! $\det(A) = (-1)^{n-1}(n+1)2^{n-2}$.
Now I'm trying to find out why!


$6$ women and $4$ men wait in line. If their order in line is random, find the probability that all of the women are adjacent to one another. My thoughts on the problem are that the number of ways the women can be adjacent to each other is $5!$ and the total number of arrangements for all the people is $10!$. Is this correct?

Some hints: $10$ people can be arranged in a line in $10!$ ways. Four men and a bench can be arranged in $5!$ ways. Six women can be placed on the bench in $6!$ ways.

((women factorial - men factorial) $\times$ possible permutations)/Total possible arrangements irrespective of order factorial

((6!-4!)$\times$5)/10! = 3600/3628800 = 0.00099206
or a 0.0992% probability


Comparing different relativizations in computability Most, but not all, theorems in computability relativize.  In principle, we should go through the original proof to check that a relativized version of a theorem holds.  In practice, we often just wave our hands and say "relativized XXX theorem holds."

But sometimes there can be multiple ways of relativizing a theorem.  For example, Posner-Robinson theorem reads:

  For $X >_T 0$, there exists $G$ such that $G' \equiv_T X \oplus G$.

If I want to relativize this to a set $B$, I could say

  (1) For $X >_T B$, there exists $G \ge_T B$ such that $G' \equiv_T X \oplus G$.

or alternatively

  (2) For $X >_T B$, there exists $G \ge_T B$ such that $G' \equiv_T^B X \oplus G$.

where $C \le_T^B D$ if there's a Turing functional $\Phi_e$ such that $C = \Phi_e^{D \oplus B}$.

For Posner-Robinson theorem, the difference between the two is not a problem, since both of them are true.  But you can think of two relativized versions for a different theorem, and, at least prima facie, they can have different truth values; and relativization of type (2) is more likely to be true. (I barely remember a time in a computability class when a relativization of type (2) is the "correct" way of doing it, but I don't remember what theorem we were talking about.)  And even if they have the same truth value, the "mathematical content" of the two versions could sometimes be different.

So my questions here are:

What is a concrete and natural example of a theorem such that relativization of type (1) fails but (2) holds?
In practice, which versions of relativization should I claim?  Is the possibility of relativization of type (1) failing higher than type (2) to the extent I should be careful?

(I know metamathematical results about elementary equivalence between cones in the degree structure; my question is more about the practice.)

I would argue that the "right" relativization of a statement to (the degree of) a noncomputable set $B$ is the version where all the quantifiers range over the cone above $B$, $\{C: C\ge_T B\}$. On this cone, $\equiv_T$ and $\equiv_T^B$ coincide, so it doesn't matter which we use.

If we look at the co-lower cone, $\{C: C\not<_TB\}$, instead, things are much weirder - this isn't a good class of degrees! - and using $\equiv_T^B$ in place of $\equiv_T$ has the effect of "smooshing" everything into the cone above $B$, so - if you really want to work in the co-lower cone, for some reason - is the right equivalence to use.

From my understanding, the equivalence in $G' \equiv_T^B X \oplus G$ means that $G' \ge_T^B X \oplus G$ and $G' \le_T^B X \oplus G$, and according to your definition of a Turing functional we therefore have both $C = \Phi_e^{D \oplus B}$ and $D = \Phi_e^{C \oplus B}$. So we must have $B$ as some kind of identity, as for a set $S$, $S \oplus \{e\} = S$. We must also practically have $C=D$.

Effectively, the second type allows no more possibilities than the first type, if not less as it implies $B=I$.


How to calculate opposite direction angle I searched but couldn't find on Google:

My question is, how do I find the opposite direction of an angle, for example 170 degree, how do I calculate the opposite direction in degrees?

Thanks in advance.

If you are given an angle $\alpha$, the oppossite angle would be $\alpha+180$. If you need to remain in $[0,360]$ then, you should take $(\alpha+180)\ mod\ 360$ (what in this case it is simply taking $\alpha-180$ if $\alpha+180\ge360$)

Considering an angle normallized between -180° and 180°, the opposite angle is given by :
if angle == 0 :
  opposite = 180 // or -180 as you wish
else :
  if angle > 0 :
    opposite = angle - 180
  else :
    opposite = angle + 180


Evaluating $\lim\limits_{x\to 0} \frac{\ln(1+\sqrt[3]{x})}{\sin^4(\sqrt[3]{x})}$ I am kind of confused trying to compute  $$ \lim_{x\to 0}  \frac{\ln(1+\sqrt[3]{x})}{\sin^4(\sqrt[3]{x})}. $$

I got that it is of the form $(\frac {0}{0})$, since  $\ln(1+x)=x $ so it is  $ \ln(1+\sqrt[3]{x})= \sqrt[3]{x}  $, and $\sin x=x $ so   $ \sin^4(\sqrt[3]{x}) $ =$\sqrt[3]{x}$, so the limit becomes
$\lim_{x\to 0}  \frac{\sqrt[3]{x}}{(\sqrt[3]{x})^4} $.

Am I right this far?
Still I don't know what to do next because I still get 0 as the answer.

Any suggestions?

I'm going to make a small substitution to make this a bit easier, $u=x^{1/3}$.  Conveniently, we still take the limit toward zero, $ \lim_{u \to 0} \frac{ \ln{ (1 + u) }}{\sin^4 (u)}$.

If we evaluate the expression at $u=0$, it's still indeterminate, but a little less messy.  If you've used $\epsilon - \delta$ definitions for the limit, you know the trick is to go away from the limiting value (zero in this case) just a bit.  Maybe to $0.1$ and $-0.1$.  For $u = 0.1$ the expression evaluates to $959.5$.  You should be thinking, "Hm, maybe this thing tends toward infinity."  For $u = -0.1$, however, the expression evaluates to $-1060.7$.  That's sounding like negative infinity.  Sure, it's possible that this expression just has really big numbers and will eventually converge toward a value at zero, but it's unlikely.  More likely, the limits from the left and the from the right don't converge, i.e., $\lim_{u \to 0 +} \neq \lim_{u \to 0 -}$, in which case the two sided limit $\lim_{u \to 0}$ is undefined.  That is, in fact, the case, but I'll prove it.

I will use a first order approximation for the numerator and denominator, $f(x) \approx f(0) + \left. \frac{df}{dx} \right|_{x=0} \Delta x $.  This can be made arbitrarily accurate simply by choosing smaller and smaller $\Delta x$.  With the $\epsilon - \delta$ definition of the limit, that's exactly what we're doing.  So, for our two expressions,
$$
\ln (1 + u ) \approx \ln (1 + 0 ) + \left. \frac{1}{1+u} \right|_{u = 0} \Delta u = \Delta u
\\
\sin^4 (u) \approx \sin^4 (0) + \left. 4 \sin^3 (u) \cos{(u)} \right|_{u=0} \Delta u = 0
$$

From this, we see that if we go slightly to the right of $u=0$, for example, $u=0.1$, the numerator will be a small positive number, and the denominator will be zero.  Well, a positive number divided by zero is $\infty$.  However, if we go slightly left, for example, $u = -0.1$, we see that the numerator of the expression will be a small negative number divided by zero.  That limits to $-\infty$.  The result taken together is
$$
\lim_{u \to 0+} \frac{ \ln{ (1 + u) }}{\sin^4 (u)} = +\infty
\\
\lim_{u \to 0-} \frac{ \ln{ (1 + u) }}{\sin^4 (u)} = -\infty
$$

Because the right and left limits do not converge, the two sided limit is undefined.

the limit does not exist.because$\lim_{x\to0+}$ is $\frac{+ve}{+ve}=+ve$ and $\lim_{x\to0-}$ is $\frac{-ve}{+ve}=-ve$


Is it possible to derive the axiom of induction from a construction of the natural numbers? If I start by constructing the natural numbers formally in a reasonably standard way...

Type: NaturalNumber
Constructors: zero:NaturalNumber;successor(NaturalNumber):NaturalNumber

...rather than axiomatising them, then it is straightforward to derive the first eight Peano axioms as theorems, in the sense that it would be relatively easy to write a verifier to do it. I asked myself then why it isn't possible to derive the ninth axiom, namely the axiom of induction?

My motivation for this is that, if you axiomatise the natural numbers rather than construct them, then as I understand it the axiom of induction is necessary in order to exclude the possibility of non-standard sets of natural numbers. Put simply, it rules out the circle of dominoes and allows only the straight line of dominoes in the following picture:

Wikipedia: Domino effect visualizing exclusion of junk term by induction axiom.jpg

I'm aware of, although can hardly claim to be fully versed in, results such as Dedekind's, namely that with the axiom of induction stated using second order logic the Peano axioms define the natural numbers up to isomorphism. And I'm wondering whether these kinds of meta-logical arguments are needed at least to prove that such a proof as this isn't possible, perhaps?

Yes, it is possible to prove that a "natural numbers type structure" satisfies the induction axiom. The proof will be carried out in the same set theory that is used to form the natural number type structure. The only challenge is to formalize the structure appropriately. There are many ways of specifying a "natural numbers type structure".

However, even though there are several ways to formalize inductively defined sets such as the one mentioned in the question, as long as care is taken to do it appropriately the nature of the inductive definition will ensure that the resulting structure satisfies induction. This is because, by nature, an inductive definition forms the smallest set possible. If the resulting set $S$ failed to satisfy the induction axiom because of some proper inductive subset, this would show that the resulting set $S$ is not actually as small as possible, leading to a contradiction.

For example, in the proof at the end of the question, we would look at the set $\{n : P(n)\}$. This would be a set that contains $0$ and is closed under successor. By the inductive definition of the construction, that set has to contain every number in the structure that was constructed - there is no need to worry about a least element or anything like that.

So, at last, I have what I think is an acceptable answer. I'll write out the vernacular and explain as I go along.

Firstly, the construction of the natural numbers (which would be imported from elsewhere):

Type NaturalNumber

Constructors zero:NaturalNumber;successor(NaturalNumber):NaturalNumber

Next, some abbreviations and declarations:

Abbreviations z for zero,s for successor,Nat for NaturalNumber

Metavariables R,S,T(Nat)

Contexts ρ,σ,τ(Nat)

Next, the rule itself, which you would need if you wanted to prove things by Induction. I'll leave off the proof for a moment:

Rule (Induction)
  Premises
    ρ ⊢ R::P(z)
    σ ⊢ S::P(k)=>P(s(k))
  Conclusion
    τ(n) ⊢ T(n)::P(n)

The first premise ρ ⊢ R::P(z) reads 'given a context ρ, the theorem R proves P(z)'. This is the propositions as types paradigm in action, if you like. The context is the axioms and proven theorems that are referenced by the theorem. Similarly for the second premise. So  we are taking as premises for the rule the base case and the induction step.  The conclusion τ(n) ⊢ T(n)::P(n) states that for each natural number n,  a theorem T(n) can be constructed that proves P(n), given a context τ(n).

So the job is to define these contexts τ(n) and to construct the theorems T(n). I'll give the theorems first:

Theorem (T(z))
  P(z)
Proof
  P(z) from T

Theorem (T(s(k)))
  P(s(k))
Proof
  P(k)=>P(s(k)) from S
  P(k) from T(k)
Therefore
  P(s(k)) by ModusPonens

These are worth a few comments. They would not verify standalone, so to speak, because they both rely on contexts that won't be in the main context. The verifier might return something like the following message (recall that this question is about how to implement a verifier as much as anything):

Theorem T(z) verified but the conclusion P(z) has not been added to the
main context because T is not in the main context.

These are generic theorems that are structurally correct but do not hold standalone. It should be clear that the context needed for the first to verify fully is just going to be ρ. Similarly for the second theorem. The context that it needs is  going to be σ,T(k)::P(k). This leads to the recursive definition of the contexts, which is the crucial bit and comes at the top of the proof that follows:

let τ(z)=T
let τ(s(k))=S,T(k)::P(k)

After the above definitions the proof of the Induction rule itself is broken down into the cases n=z and n=s(k). I'll leave out the derivations which prove the implications for the moment, so as to get across the gist of the proof as a whole:

Proof
  ...
  n=z ∨ n=s(k)
  Suppose
    n=z
  Then
    ...
  Hence
    ρ(n) ⊢ T(n)::P(n)
  n=z => ρ(n) ⊢ T(n)::P(n)

  Suppose
    n=s(k)
  Then
    ...
  Hence
    ρ(n) ⊢ T(n)::P(n)
  n=s(k) => ρ(n) ⊢ T(n)::P(n)

Therefore
  ρ(n) ⊢ T(n)::P(n) by ProofByCases

Now I'll write out the derivations themselves. Firstly for the case n=z:

  Suppose
    n=z
  Then
    ρ ⊢ R::P(z)
    τ(z) ⊢ R::P(z)
    τ(z) ⊢ T(z)::P(z)
  Hence
    τ(n) ⊢ T(n)::P(n)

In the derivation we begin by re-stating the first premise and then, since we have defined τ(z) to be ρ, the next line follows. And now note that the next line verifies. It reads 'theorem T(z) is a proof of P(z) given τ(z). Well, it verifies! Because given ρ we can prove R, whatever R happens to be, and since our own generic theorem T(z) relies on nothing but R, the assertion τ(z) ⊢ T(z)::P(z) holds. And since n is just z, we are done. For the case n=s(k), the reasoning is along similar lines.

Update: The complete proof is now here.


Eigenvalues and eigenspaces with isomorphisms I have the following question about eigenvalues and eigenspaces in the case of an isomorphism. I am not sure how to prove those 2 statements. My intuition tells me it has something to do with diagonalizability.

Suppose that $f(v)=\lambda v$ for some non-zero vector $v$, then $g(f(v))=v=\lambda g(v)$. Hence $g(v)=\frac{1}{\lambda}v$. This is sufficient. It has nothing to do with diagonalizability.

Since $f$ is an isomorphism, you know it's 1-1 and onto, and since you know the inverse $g$ exists, you know none of the elements of $V$ can have zero eigenvalue, otherwise they're non-invertible.  (b) follows from (a) since all $v$ in $V$ are invertible, and a matrix and its inverse have the same eigenvectors.


The derivative of a matrix transpose with respect to the original matrix I am just stepping into matrix calculus and I wonder what the following differential is. Thanks.

$$ \frac{\partial(A^T)}{\partial A} $$

The transposition of a $3 \times 2$ matrix is

$$\begin{bmatrix} x_{11} & x_{12}\\ x_{21} & x_{22}\\ x_{31} & x_{32}\\ \end{bmatrix} \mapsto \begin{bmatrix} x_{11} & x_{21} & x_{31}\\ x_{12} & x_{22} & x_{32}\end{bmatrix}$$

Hence,

$$\dfrac{\partial x_{11}}{\partial \mathrm X} = \begin{bmatrix} 1 & 0\\ 0 & 0\\ 0 & 0\\ \end{bmatrix} \qquad \qquad \dfrac{\partial x_{21}}{\partial \mathrm X} = \begin{bmatrix} 0 & 0\\ 1 & 0\\ 0 & 0\\ \end{bmatrix} \qquad \qquad \dfrac{\partial x_{31}}{\partial \mathrm X} = \begin{bmatrix} 0 & 0\\ 0 & 0\\ 1 & 0\\ \end{bmatrix}$$

$$\dfrac{\partial x_{12}}{\partial \mathrm X} = \begin{bmatrix} 0 & 1\\ 0 & 0\\ 0 & 0\\ \end{bmatrix} \qquad \qquad \dfrac{\partial x_{22}}{\partial \mathrm X} = \begin{bmatrix} 0 & 0\\ 0 & 1\\ 0 & 0\\ \end{bmatrix} \qquad \qquad \dfrac{\partial x_{32}}{\partial \mathrm X} = \begin{bmatrix} 0 & 0\\ 0 & 0\\ 0 & 1\\ \end{bmatrix}$$

The gradient $\,\,{\mathcal H}=\frac{\partial A^T}{\partial A}\,\,$ is a fourth order tensor whose components are given by
$$\eqalign{
{\mathcal H}_{ijkl}
 &= \frac{\partial A^T_{ij}}{\partial A_{kl}} = \frac{\partial A_{ji}}{\partial A_{kl}} \,\,\,= \delta_{jk}\,\delta_{il} \cr
}$$
where the Kronecker delta has components
$$\delta_{jk}= \begin{cases}1 &\text{if}\,\,\,\,j=k \\ 0 &\text{otherwise} \end{cases}$$


How to solve a inequality with fractions and roots in denominator and numerator The inequality is like that:
$$
    \sqrt{\frac{3x+1}{2}}>1
$$
I have no idea how should i begin with it.

Hint: The value $\sqrt a$ is greater than $1$ iff $a>1$.

Square both sides since 1 is positive; then cross multiply; then solve. The question will then become like this: $3x - 3 > 0$.

Now, equate $3(x-1)=0$ giving $x=1$.

Next, consider the real number line as follows:

-           +      ( always take + from right and then - if power is odd;

---------1---------     otherwise the sign remains the same)

And since it gives us that $3x-3>0$, it follows that $x > 0$; Hence, we will take the positive value. And so, the answer $x$ belongs to $(1, \infty).$


Introduction to Real analysis I have a somewhat difficult time understanding this line. Can someone please explain a little bit about this?
A ball of radius $\varepsilon$ around $x$ in $\mathbb{R}$ is a set deﬁned by: $$B_\varepsilon(x) = \{y \in \mathbb{R} : |x−y|< \varepsilon\}$$

Thanks for everyone

... and therefore
$$
B_\epsilon(x) = (x-\epsilon,x+\epsilon),
$$
since $|x-y|<\epsilon$ is equivalent to $x-\epsilon<y<x+\epsilon$.

in this case we talk about two distinct point (x,y) in R  |x−y|<ϵ say, ϵ always greater than distance of x & y. this can be easily understand in geometric representation


$|f(1)^2-f(0)^2|\le 2$ Let H the set of the functions $f:[0,1]\to\mathbb{R}$ differentiable, with $f'(x)$ continuous and $\int^1_0 (f(x)^2)\,dx =\int^1_0 (f'(x)^2)\,dx=1$
Prove that
$|f(1)^2-f(0)^2|\le 2$ for any $f\in H$.
I did not had any idea at it other than using the mean value theorem but I do not think it is helpful.

You can't prove something that is not true. If you use in particular $f(x) = e ^x $ then
$$ \int_0^1 {f(x)^2} dx = \int_0^1 e^{2x} dx = \frac{e^2 - 1}{2} $$
$$ \int_0^1 {f'(x)^2} dx = \int_0^1 e^{2x} dx = \frac{e^2 - 1}{2} $$
So we can say:
$$ \int_0^1 {f(x)^2} dx =  \int_0^1 {f'(x)^2} dx $$
But:
$$ \vert{f(1)^2 - f(0)^2 }\vert = \vert e^2 - 1 \vert = 6.38 > 2 $$


Likelihood Function for the Uniform Density. Let the random variable $X$ have a uniform density given by
$$
f(x;\theta)=I_{[\theta-\frac{1}{2},\theta+\frac{1}{2}]}
$$

where $-\infty\leq\theta\leq\infty $

the likelihood function for a sample of size $n$ is

$$L(\theta;x_1,\ldots,x_n)=\prod_{i=1}^nf(x_i;\theta)=f(x_1;\theta)\ldots f(x_n;\theta)$$

Why the likelihood function for a sample of size $n$ is the "joint density" of the $n$ random variables?

I am asking the question because I knew that we have to select $\hat \theta$ in such a way that the random variables $X_1,\ldots,X_n$ assume a particular value $x\prime_1,\ldots,x\prime_n$ such that $f_{X_1,\ldots,X_n}(x_1,\ldots,x_n;\theta) $ is a maximum.

But here, for the Uniform density function, there is no hint that the value $x_1$ of the random variable $X_1$ maximizes the function $f_{X_1}(x;\theta),\ldots,$ the value $x_n$ of the random variable $X_n$ maximizes the function $f_{X_n}(x;\theta).$ Then how $x_1,\ldots,x_n$ maximizes $f_{X_1,\ldots,X_n}(x_1,\ldots,x_n;\theta) $ for the given density function , ie, for the Uniform density function?

Again,

$$L(\theta;x_1,\ldots,x_n)=\prod_{i=1}^nf(x_i;\theta)=\prod_{i=1}^n I_{[\theta-\frac{1}{2},\theta+\frac{1}{2}]} (x_i)=I_{[y_n-\frac{1}{2},y_1+\frac{1}{2}]} (\theta)$$

where $Y_1$ is the smallest of the observations & $Y_n$ is the largest.

Why have we changed the range? Is it so as Likelihood-function is a function of $\theta$ and here $x_1,\ldots,x_n$ fixed?
Again i have not understand the point: if we subtract $\frac{1}{2}$ from the largest value $y_n$ then it will be the lower limit of $\theta$ and if we add $\frac{1}{2}$ to the smallest value $y_1$ then it will be the upper limit of $\theta$.
Can you please give me a numerical example of the situation?

1) If n rv's (not one) are independent, then their joint density is the product of their individual densities (this extends the notion of the probability of intersection of n independent events in the case of rv's). If, moreover, they are (assumed) identically distributed, then
$$ L(X_1,\ldots,X_n| \theta)=\prod_{i=1}^nf(X_i|\theta)=f(X_1|\theta)\ldots f(X_n|\theta)$$
is the joint density.
If we view this expression as a function of $\theta$, then it becomes

$$ L(\theta|x_1,\ldots,x_n)=\prod_{i=1}^nf(\theta|x_i)=f(\theta|x_1)\ldots f(\theta|x_n)$$
and now it is the likelihood function of the sample. Is there a difference? Big one: In the joint density, the argument/variable of the function is the X's while the $\theta$'s are treated as constants. In the likelihood function, the arguments/variables are the $\theta$'s while the x's are treated as constants (changing from uppercase to lowercase for the x's is a usual -and good- mnemonic) So the likelihood function is not the joint density - they may have the same functional form, but they are functions of different variables.

2) In view of 1), we do not select the $\theta$'s "so as the rv's assume a particular value...etc" as you write. We select the $\theta$'s given the specific values that the r.v's have already taken (values that constitute our sample), so as the resulting value of the function is maximum. Then we flip our point of view again and say (very loosely speaking), that this value of $\theta$ "maximizes the probability that the sample comes from the assumed distribution".

3) The uniform distribution is special in that the variable itself does not appear in the density function. The specific case we study is even more special, because not even the limits of the interval into which the rv ranges appear in it: the density is equal to unity for the specified range, and zero elsewhere (even though the range of the rv is not [0,1]). Instead of writing a piece-wise density we write compactly $$ f_U (X_i)=I_{[\theta-\frac{1}{2},\;\theta+\frac{1}{2}]} (X_i)$$ Then the joint density for the case of n i.i.d rv's is $$ L(X_1,\ldots,X_n| \theta)=\prod_{i=1}^nf(X_i|\theta)=\prod_{i=1}^n I_{[\theta-\frac{1}{2},\;\theta+\frac{1}{2}]} (X_i) =\min\{I_{[\theta-\frac{1}{2}\le X_1\le\theta+\frac{1}{2}]} ,...,I_{[\theta-\frac{1}{2}\le X_n\le\theta+\frac{1}{2}]}\}$$ where the last equality comes from basic properties of indicator functions.
Consider now the joint density as a likelihood function of $\theta$, with given x's. Then we should maximize this likelihood function? No. We are applying here maximum likelihood estimation, and the ML estimators are M-estimators, which maximize the expected value of the likelihood function (or its sample analogue, whatever is feasible). So our estimation problem here can be stated as
$$  \max_{\theta}E\min\{I_{[\theta-\frac{1}{2}\le x_1\le\theta+\frac{1}{2}]} ,...,I_{[\theta-\frac{1}{2}\le x_n\le\theta+\frac{1}{2}]}\}$$

Let's start the optimization procedure. Note that if we choose some $\hat \theta$ such that some of the realized values in the sample fall outside the specified range, the min{} expression will be equal to zero. But if the chosen $\hat \theta$ has a value that leaves all realized values of the rv's inside the resulting common interval, then the min{} expression will be equal to unity. I adopt the notation $x_1=x_{min}$ and $x_n=x_{max}$. So a first conclusion is that we must have  $$ x_1\ge \theta-\frac 12  \text{and} \ x_n\le \theta+\frac 12 \\ \Rightarrow x_n-\frac 12\le \theta\le   x_1+\frac 12$$
(Parenthesis: For $\hat\theta$ to be able to obtain a value in the reals this interval must exist. So we must have $$x_n-\frac 12\le  x_1+\frac 12 \Rightarrow x_n-x_1 \le 1$$
In other words, if in the sample at hand the difference between the maximum realized value and the minimum realized value is greater than unity, then one of the initial assumptions is wrong: either the n rv's do not have the postulated distribution, or they are not identically distributed). Assume that $x_n-x_1 \le 1$ holds. Then our maximization problem has been transformed (after this first step of optimization) as
$$  \max_{\theta}EI_{[x_n-\frac 12\ \, \le \,  \theta \, \le  \, x_1+\frac 12]}$$

The $\min$ operator has disappeared because all indicator functions have become identical.
Now by the properties of indicator functions,
$$  EI_{[x_n-\frac 12\ \, \le \,  \theta \, \le  \, x_1+\frac 12]} = P\left(\left[x_n-\frac 12\ \, \le \,  \theta \, \le  \, x_1+\frac 12\right]\right)$$
where P() is the probability of the interval. Next we can write the upper bound for $\theta$ as a constraint and solve the constrained maximization problem

$$  \max_{\theta}P\left(\left[x_n-\frac 12\ \, , \,  \theta\right]\right) \qquad \text{s.t.}\ \theta\le   x_1+\frac 12$$

Obviously this probability increases with the length of the interval, and so
$$ \hat \theta : [x_n-\frac 12\ ,  \hat \theta] = \text{maximum}$$
which obviously results in
$$ \hat \theta_{\text{ML}} = x_1+\frac 12$$

Then the maximum likelihood estimate of the common distribution of the n rv's is $$U\left(x_1\, , \, x_1+1\right).$$


A trigonometric integral (guessed from a combinatorics formula) In class, I defined the binomial coefficient using an integral:

  $$\binom{n}{k} = \displaystyle \int_0^{2\pi}\dfrac{dt}{2\pi} e^{-ikt}(1+e^{it})^n.$$

I succeeded in demonstrating many standard properties of the binomial coefficient directly using integration: Pascal's identity, Vandermonde identity, Hockey stick. But I could not show that $$\sum_{k=0}^n \binom{n}{k}=2^n.$$

It turns out I have to show the following:

$$\int_{0}^{2\pi}\dfrac{dt}{2\pi}\dfrac{\sin\left(\dfrac{(n+1)t}{2}\right)}{\sin\dfrac{t}2}\cos^n\dfrac{t}{2}=1$$

I do not know how to perform this integration! I need help. It is better if the solution did not involve contour integration.

Hint

Notice that inside the integral we have $$\frac {\sin\left(\frac {(n+1)t}{2}\right)}{\sin\frac t2}$$ which is equal to $D_{\frac n2}$ where $D_n$ denotes the Dirichlet kernel.

Hence using it's properties we have $$\frac {\sin\left(\frac {(n+1)t}{2}\right)}{\sin\frac t2}=D_{\frac n2}=1+2\sum_{r=1}^{\frac n2} \cos (rt) $$

Hope you can continue further

Using mathematical induction. Base case $n=0$, ok!
Assume that is true for an arbitrary natural number $n$. Since
$$ \sum_{k=0}^{n+1} \binom{n+1}{k} = \sum_{k=0}^n \binom{n+1}{k} + \binom{n+1}{n+1} $$
Using your definition, $\binom{n+1}{n+1}$ equals to $1$, then using Pascal's identity
$$ \binom{n+1}{k} = \binom{n}{k}+\binom{n}{k-1} \text{ for all } 1\leqslant k\leqslant n $$
You get the following :
$$ \sum_{k=0}^{n+1} \binom{n+1}{k} = 2^n + 2^n - 1 +1 = 2^{n+1}$$


orthogonal basis - counter example I need a counter example for this:

Let V be inner product space, B={u1,u2,...,un) a basis of V. Suppose there are scalars a1,a2,...,an.

If

then B is orthonormal basis of V.

I cant find an counter example.
Any hint?

It seems that $\langle u_i,u_j\rangle=\delta_{ij}$, thus $B$ is indeed  an orthonormal base: 1) Putting $\alpha_j=\delta_{ij}$ shows that $\Vert u_i\Vert^2=1$. Putting $\alpha_i=1=\alpha_j$ and $\alpha_k=0$ for $k\not=i,j$ if $i\not=j$ results in $\Vert u_i+u_j\Vert^2=2$. Therefore $2=\langle u_i+u_j,u_i+u_j\rangle=\langle u_i,u_i\rangle+2\langle u_i,u_j\rangle+\langle u_j,u_j\rangle=2+2\langle u_i,u_j\rangle$ and thus $\langle u_i,u_j\rangle=0$.

EDIT: This of course only holds when one may choose the coefficients arbitrarily. Otherwise there are, as shown, counterexamples.


Finding particular integral to $ (D^3 + 4D)y = \sin2x$ The question is to find particular integral of this differential equation

$$ (D^3 + 4D)y = \sin2x.$$

where $$ D = \frac{d}{dx} $$
Please looking for your quick tip guys.

the first step is finding the complementary solution
$$D(D^2+4)=0$$
$$D=0$$
$$D=\pm2i$$
so the complementary solution is
$$y_c=c_1+c_2\cos2x+c_3\sin 2x$$
now assume the particular solution as follow
$$y_p=A\cos 2x+B\sin 2x$$
because the similarity with complementary solution, we should multiply the particular solution by $x$, so
$$y_p=x(A\cos 2x+B\sin 2x)$$

then you can complete the solution

Given:- (D³+4D)y=sin2x
Condition : 1/f(D), if after equal to sinax or cosax given means you have to replace D² into -(a²)
Ans-  (1/D³+4D)*sin2x
[1/D(D²+4)]*sin2x    Here, D² is replaced by -(a²) that is
D²= -(2²)
[1/D(-4+4)]*sin2x  (important first square the number and put (-) symbol.)
[1/D(0)]*sin2x [1/0]*sinx ( condition if denomerator  is zero multiply  numerator by x and differentiate f(D)).
[x/3D²+4]*sin2x ( substitute D² value)
[x/3(-4)+4]*sin2x [x/(-12+4)]*sin2x
Therefore [-x/8]*sin2x.


Efficient way of checking linear independence Suppose I have a $4 \times 4$ matrix $A$ whose columns represent vectors $v_1,v_2,v_3,v_4$ in $\mathbb{R}^4$. Now, given that $\det{A} = 0$ (i.e. the vectors are linearly dependent), I want to make sure that any three vectors out of the given 4, are linearly independent. What is the most efficient way of doing this? I can only think of checking linear independence of each three vectors out of all the possible combinations. But I feel that there must be some easier way to accomplish this.

In order to check whether any three columns are linearly independent, you would unfortunately have to examine all subsets of $3$ columns.

As a sidenote, this question is related to computing the spark of a matrix (see here: http://en.wikipedia.org/wiki/Spark_%28mathematics%29).
If all sets of $3$ columns have full rank (rank equal to $3$), then the spark of $A$ in your case is equal to $4$: $4$ is the smallest number of columns that are linearly dependent.  But, if there exists a subset of $3$ columns that is linearly dependent, then the spark is at most $3$.
Computing the spark is an NP-hard problem.

Best way is as copper.hat suggested. Row-reduce. That will give you the linearly independent vectors. If you are asking which three because (for instance, $\vec v_3=2\vec v_2$), this procedure will tell you. As far as computational complexity, I believe that it is $N^3$, which isn't bad.

The reason that row-reduction works is it gives you the pivots. The columns with leading non-zero entries are the linearly independent set of vectors. If you are trying to figure out which other combinations work, try swapping the order of the columns in the matrix, and compute the row-reduction again.


Injection from natural numbers to union of two aleph null sets. Let $X$ and $Y$ be disjoint sets, each having cardinality of the set of all natural numbers. How do I see that there is an injection $f: \mathbb{N} \to X \cup Y$?

If $f\colon \mathbb{N}\to X$ and $g\colon \mathbb{N}\to Y$ are your bijections, then $f\colon \mathbb{N}\to X\cup Y$ is also an injection. Or you can use $g$.

$$f: (\mathbb{N},\mathbb{N} )\to \mathbb{N}$$

$$(a,b)\to 2^{a}3^{b}$$

Obviously, it is injective


How many 10-digit numbers whose first four digits match either the next four digits or the last four digits? If in a 10-digit number, the first four digits are either exactly the same as the next four digits or the last four digits. How many such 10 digit numbers are there?

My attempt:

First I consider the case with first four and next four are same.
For the first four places, we have $9\times 10\times 10\times 10$ ways of choosing digits as zero cant come in first digit.

The digits for next four places are already choosen as they are the same.
The remaining two places can be filled in $10\times 10$ ways.

Therefore total is $9\times 10^5$ ways.

Since the second case first four and last four digits are same, the answer will again be $9\times 10^5$ ways.

Therefore, total numbers are $2\times 9\times 10^5$. However, correct ans is $9\times 10^5-90$ .

  Why am I wrong?

If I'm correct, the first digit would be 1-9. Thus, we start the equation with a 9.
The second digit can have 10 different numbers, so the next step of the equation is 10
The third and fourth digit can also have 10 different numbers, which brings us 10*10 numbers.

The first four numbers will have 9*10*10*10= 9000 possibilities

Now, we calculate the remaining 2 digits for both cases, which will be (10*10)*2 =200

Since there are 2 different types of combinations, we multiply 9000 by two and add 200, which will be 18200 different combinations.

I'm a new member so I don't really know where to put my answer if I'm not sure about it, and if I'm wrong, please tell me why and I'll delete my post immediately. :)

Edit: Based on @dhanesh vijay 's answer, we should minus 90 as it is the possibilities with both conditions being true as they are already counted. Thus, the answer would be 18200-90=18110 different combinations.


Linear Algebra- independence and dependence Let $u$ and $v$ be non-zero vectors in $V$. Prove or disprove the following claim.
$u$ and $v$ are linearly dependent $\implies$  $(u+v)$ and $(u-v)$ are linearly dependent.

Is the following proof correct?
Otherwise can someone answer the given question? Thanks in advance. :)

Here is my answer.

Since $u , v$ are two linearly dependent vectors, $au + bv = 0$ with $a,b \neq 0$

we can write,
$[(a+b)/2]\cdot(u+v) + [(a-b)/2] \cdot(u-v) = 0$

case 1 ($|a|=|b|$) :
if $a=b$ then $(a+b)/2 \neq 0 \implies (u+v)$ and $(u-v)$ are linearly dependent.

if $a= -b$  then $(a-b)/2 \neq 0\implies (u+v)$ and $(u-v)$ are linearly dependent.

case 2 ($|a| \neq|b|$) :
then $(a+b)/2, (a-b)/2 \neq 0\implies(u+v)$ and $(u-v)$ are linearly dependent.

So,
u and v are linearly dependent $\implies(u+v)$ and $(u-v)$ are linearly dependent.

Your proof is correct. However, I don't see why you need to break it up into those 2 cases.

Like you said, $au + bv = 0 \Leftrightarrow (a, b)=(0,0)$.

Now, assume that $a(u+v) + b(u-v) = 0$, so $(a+b) u + (a-b) v = 0$. Hence, we know that $a+b = 0 $ and $a-b=0$, which gives that $2a=0, 2b=0$. (Assuming that your field does not have characteristic 2) hence $a=0, b=0$.

In the event that your field has characteristic 2, then $u+v = u-v$ so these vectors are not linearly independent.

Edit: OP wanted a direct proof along the lines he started, and I pointed out that he doesn't need to split it out.

Assume that $u, v$ are not linearly independent. Hence, there exists $(a, b) \neq (0,0)$ such that $au + bv = 0$.

This implies that $ \frac {a+b}{2} (u+v) + \frac {a-b}{2} (u-v) = au +bv = 0$ (assuming that the field does not have characteristic 2). For this to actually show that $(u+v), (u-v)$ are not linearly independent, we have to show that $(\frac {a+b}{2}, \frac {a-b}{2}) \neq (0,0)$. This is true because $ (\frac {a+b}{2}, \frac {a-b}{2}) = (0,0) \Leftrightarrow (a, b) = (0,0)$.

since {u,v} are linearly independent, the set {u+v,u-v} is LI if and only if the determinant formed by the coefficients that is
        1  1
        1  -1

is non zero


Let $(G,\cdot)$ be a finite abelian group and $m\in\Bbb{N}$. Prove that $S=\{g\in G\mid(o(g),m)=1\}\le G$. I am aware that I need to show that given any $g_1,g_2$ in $S$ if $(o(g_1g_2^{-1}),m)=1$ then $S$ will be a subgroup of $G$.

Now let $o(g_1)=r,o(g_2)=s,o(g_1g_2^{-1})=t$. Since $G$ is abelian we have

$$\begin{align}
(g_1g_2^{-1})^{rs}&=g_1^{rs}(g_2^{-1})^{rs}\\
&=(g_1^r)^s((g_2^s)^{-1})^r\\
&=e^s(e^{-1})^r\\
&=e^se^r.
\end{align}$$

This implies that $t\mid rs$. Now we must show that $(t,m)=1$. I could not find a way to show this.

Any help here will be appreciated. Thanks in advance.

If $g,h\in S$, if $a=o(g)$ and $b=o(h)$, then $o(h^{-1})=b$ and (since $G$ is abelian)$$o(gh^{-1})\mid o(g)o(h^{-1})=ab.$$Since both $a$ and $b$ are coprime with $m$, then so is $ab$ and therefore then so is $o(gh^{-1})$. In fact, if $o(gh^{-1})$ and $m$ are not coprime, then there is a prime number $p$ such that $p\mid o(gh^{-1})$ and $p\mid m$. But$$p\mid o(gh^{-1})\implies p\mid ab\implies p\mid a\text{ or }p\mid b.$$But this is impossible, since $(a,m)=(b,m)=1$.

If $t|rs$, and $(r,m)=(s,m)=1$, then $(t,m)=1$.  For suppose $p|t$, where $p$ is prime.  Then  $p|r\lor p|s$.  This is Euclid's lemma.

If you're having trouble seeing it, perhaps think of the unique prime factorizations guaranteed by the fundamental theorem of arithmetic.


Can there be a magic square with equal diagonal sums different from equal row and column sums? I got a task in programming a program that can detect whether a 4x4 square is a magic square or not. At first, I wrote code that met the requirements for all given examples but I noticed one flaw. I used 2 variables to indicate sums. I used them once to calculate the sums of the rows and columns and compare them, then I reset them back to 0 and used them to calculate the diagonal sums and check if they were equal. The thing was, is that I did not actually compare the diagonal sums to the original row and column sums, and that got me thinking. Can there exist a "magic square" where the diagonal sums are equal and the row and column sums are equal, but the diagonal sums are different from the row and column sums? Is there any actual way to prove this? I tried to come up with examples but nothing came to me. An example would disprove this and make me rewrite my code. For simplicity, I would rather know about a 4x4 square, but if you can I'll be happy to hear a proof for any $n$ x $n$ square. Thanks in advance.

Edit: I already check to see if the integers are all different, so I'd rather know if one exists where all of the integers are different.

I assume you are looking for one with not just 'all different numbers', for that is trivial: just take any $n \times n$ magic square and add the same amount (larger than $n^2$) to the cells on the diagonals. For example, you can go from:

\begin{array}{|c|c|c|c|}
\hline
16&3&2&13\\
\hline
5&10&11&8\\
\hline
9&6&7&12\\
\hline
4&15&14&1\\
\hline
\end{array}

to:

\begin{array}{|c|c|c|c|}
\hline
116&3&2&113\\
\hline
5&110&111&8\\
\hline
9&106&107&12\\
\hline
104&15&14&101\\
\hline
\end{array}

... and that's just too easy!  So, I assume you mean that you have to use all numbers $1$ through $n^2$.

Well, after trying a bunch of things I am fairly convinced that you cannot have a $4 \times 4$ square with numbers $1$ through $16$ where are rows and columns sum p to the same amount (this is actually called a 'semi-magic square') but where the diagonals sum up to the same amount, yet different from the rows and columns. In all $4 \times 4$ semi-magic squares that were not $4 \times 4$ magic squares that I looked at, I found the diagonals still adding up to exactly twice the sum of a row. I don't have a proof though that this is really impossible.

I did, however, find a $6 \times6$ square using numbers $1$ through $36$ with all rows and columns adding up to $111$ but both diagonals adding up to only $97$:

\begin{array}{|c|c|c|c|c|c|}
\hline
6&34&1&28&24&18\\
\hline
14&7&36&4&20&30\\
\hline
33&3&31&2&17&25\\
\hline
27&22&23&16&12&11\\
\hline
10&13&5&35&29&19\\
\hline
21&32&15&26&9&8\\
\hline
\end{array}

And so, the answer to your question is Yes!  (But no, I did not look at the $5 \times 5$ case (although my guess is that you can find an example of what you want for the $5 \times 5$ case and up), and no, I did not create this example in a systematic way: I started with a known semi-magic $6 \times 6$ square and kept swapping rows and columns in a semi-random fashion until I found this one).

Finally, in my research I found that there are many different kinds of magic squares (I had no idea!), such as these 'semi-magic squares' or
'extremely magic squares', or .... but what you are asking about I did not see a name for. Given that they apparently exist you should definitely come up with a name for these!

EDIT

Aha! As I thought, it also works for $n=5$. Here is one:

\begin{array}{|c|c|c|c|c|}
\hline
19&6&15&2&23\\
\hline
9&17&10&13&16\\
\hline
21&24&14&5&1\\
\hline
4&11&8&20&22\\
\hline
12&7&18&25&3\\
\hline
\end{array}

Rows and columns sum to $65$, but columns sum to $73$

EDIT 2:  Aha!  I was wrong about the $4 \times 4$: it is possible to have one!! See Taneli Huuskonen's answer.

there is a geometrical method to systematically construct not even (2n+1) magic squares, quite easy, even by hand with a pen!
I will post an example for 5x5 and 7x7 squares, as soon as I get hold of a scanner. see below a 5x5 one
Greg

3 16 9 22 15
20 8 21 14 2
7 25 13 1 19
24 12 5 18 6
11 4 17 10 23


Is the event $\{\max\{X_1,X_2\}=X_2\}$ measurable with respect to $\sigma(\max\{X_1,X_2\})$? Let $X_1$ and $X_2$ be arbitrary random variables defined one the same probability space $(\Omega,\mathcal{F},P)$. I am trying to determine if the event $\{\max\{X_1,X_2\}=X_2\}$ is measurable with respect to $\sigma(\max\{X_1,X_2\})$? Normally I would try to decompose the event into a countable union or intersection of events which are clearly elements of the $\sigma(\max\{X_1,X_2\})$. However, the only way that I can think to rewrite the original set is as $\{\max\{X_1,X_2\}=X_2\}=\{X_1\leq X_2\}$. I'm not sure where to go from here.

It is not true. Let $X_1$ be a Bernoulli(1/2) random variable and set $X_2=1-X_1$. Then $\max(X_1,X_2)$ is the constant random variable 1. The generated $\sigma$-algebra  $\sigma(\max\{X_1,X_2\})$ is trivial, but the event $\{\max\{X_1,X_2\}=X_2\}$ is not.

Convince yourself that $\exists A \in \mathscr F$ s.t. we can write:

$$X: = \max(X_1, X_2) = X_1 1_A + X_2 1_{A^C}$$

Note that:

$\sigma(X) \subseteq \sigma(\sigma(X_1) \cup \sigma(X_2) \cup \sigma(A))$
Almost surely, $1_{A^C} = 1_{\{\max\{X_1, X_2\} = X_2\}}$
$A^C \in \sigma(\sigma(X_1) \cup \sigma(X_2) \cup \sigma(A))$

Q: $A^C \in \sigma(X)$ ? (Alternatively, $A \in \sigma(X)$ ?)

We want to know if we can determine the maximum of $X_1$ and $X_2$ given that we know the value of $X$. We cannot. Just pick $X_1$ and $X_2$ s.t. the value of $X$ is constant.

Let $B \in \mathscr F$ and $a \in \mathbb R$ s.t.

$$X_1 = a \times 1_B$$

$$X_2 = a \times 1_{B^C}$$

Then $X = a \times 1_{A \cap B} + a \times 1_{(A \cap B)^C} = a \times 1_{\Omega} \to$ (almost?) surely, $X \equiv a$.

$\to \sigma(X) = \{\{X=a\}, \{X \ne a\}, \emptyset, \Omega\} = \{\Omega, \emptyset, \emptyset, \Omega\} = \{\emptyset, \Omega\}$

I don't see $A$ or $A^C$ in $\sigma(X)$.


How to tell if a line segment intersects with a circle? Given a line segment, denoted by it's $2$ endpoints $(X_1, Y_1)$ and $(X_2, Y_2)$, and a circle, denoted by it's center point $(X_c, Y_c)$ and a radius $R$, how can I tell if the line segment is a tangent of or runs through this circle? I don't need to be able to discern between tangent or running through a circle, I just need to be able to discern between the line segment making contact with the circle in any way and no contact. If the line segment enters but does not exit the circle (if the circle contains an endpoint), that meets my specs for it making contact.

In short, I need a function to find if any point of a line segment lies in or on a given circle.

EDIT:

My application is that I'm using the circle as a proximity around a point. I'm basically testing if one point is within R distance of any point in the line segment. And it must be a line segment, not a line.

Find the intersections of the line containing the segment and the circle. This amounts to solving a quadratic equation. If there are no intersections (i.e. the solutions of the corresponding equation are non-real), then your segment does not intersect the circle. Now, if there are intersections, see whether they are inside the segment or not.

To implement this, let $P$ and $Q$ be the endpoints of your segment and $C$ and $r$ be the center and the radius of your circle. Then every point of the line through $P$ and $Q$ is given by the formula $$t P + (1-t) Q$$ for exactly one value of $t\in\mathbb R$, and the points in the segment are precisely those for which the corresponding $t$ is in the interval $[0,1]$.

Now the point corresponding to $t\in\mathbb R$ is on the circle if and only if $$\langle t P + (1-t) Q - C,t P + (1-t) Q - C\rangle = r^2,$$ where $\langle\mathord\cdot,\mathord\cdot\rangle$ is the usual inner (dot) product of vectors. If you solve this equation—this is easy, as it is a quadratic equation for $t$—, you find the $t$'s corresponding to the points of intersection of the circle and the line, and if at least one of those $t$'s belong to the interval $[0,1]$, then the segment intersects the circle.

Later: I tried to have Mathematica do the computation for me. Assuming $C=(0,0)$ and $r=1$, as we may up to translation and rescaling, and letting $P=(p1,p2)$, $Q=(q1,q2)$, the following computes the $t$'s:

P = {p1, p2};
Q = {q1, q2};
ts = t /. Solve[Norm[t P + (1 - t) Q]^2 == 1, t];

This ends up with the variable ts having the value

$$\left\{\frac{-p_1 q_1-p_2 q_2-\sqrt{p_1^2
   \left(-q_2^2\right)-2 p_1 q_1+2 p_2 p_1 q_1
   q_2-p_2^2 q_1^2-2 p_2
   q_2+p_1^2+p_2^2+q_1^2+q_2^2}+q_1^2+q_2^2}{-2 p_1
   q_1-2 p_2 q_2+p_1^2+p_2^2+q_1^2+q_2^2},\frac{-p_1
   q_1-p_2 q_2+\sqrt{p_1^2 \left(-q_2^2\right)-2 p_1
   q_1+2 p_2 p_1 q_1 q_2-p_2^2 q_1^2-2 p_2
   q_2+p_1^2+p_2^2+q_1^2+q_2^2}+q_1^2+q_2^2}{-2 p_1
   q_1-2 p_2 q_2+p_1^2+p_2^2+q_1^2+q_2^2}\right\}$$

(This was a huge formula, which will likely not fit on your browser window...)
If the expression inside the square roots is negative, there are no real $t$'s, so the line (hence, a fortiori the segment) and the circle are disjoint. If not, now we need to see if at least one of the roots in in $[0,1]$.

If I now tell Mma to tell me when then first of the roots is in $[0,1]$, by telling
her to compute

Reduce[0 <= ts[[1]] <= 1, Reals]

it works for a while, and comes up with a huge answer, presumably equivalent to the original

0 <= ts[[1]] <= 1

Can anyone make sense of the huge answer? (if one asks instead for the more meaningful

Reduce[0 <= ts[[1]] <= 1 || 0 <= ts[[2]] <= 1, t, Reals]

the same thing happens)

PS: Please notice that I have made absolutely no attempt at being fast or particularly smart with this idea: it is just the straightforward was to set up the problem and solve it.

Dear God you guys over complicated things...

Not being a math expert along the lines of you guys, I set the problem down and came back to it later. Upon looking it over again, I realized I could really simplify this by using the given data to look at it as different geometry. Instead of a line segment and a circle and testing for intersection, I used the radius of the circle to convert the line to a rectangle and then checked if the center of the circle (as just a point, no longer as a circle) is inside the new rectangle.

Please excuse my rectangle not having perfect dimensions.

EDIT:

Although my answer meets my needs for my specific application, to better answer my own question, about the semi circles at the end of the rectangle (to test if an endpoint of the line segment is contained in the circle), I can compare the distance between an endpoint and the center of the circle, if that distance is less than the radius of the circle, the circle contains the end point.

My application was for a Tower Defense game. The line represented some sort of laser stretched across the trail that enemies (the circle) would walk past, I needed to know when the laser contacted the enemy. The laser endpoints are attached to towers that are not positioned on the path the enemy walks down, so the enemy's circle could not contain an endpoint. I realize I was unclear during the original question about some of these details.


Prove that $x^2 + y^2 -1$ can't be a product of two linear equations How do I prove that $x^2 + y^2 -1$ can't be a product of two linear equations?
I tried assuming that it is a product of $f(x,y) = ax + by +c$ and $g(x,y) = dx + ey +f$
and I finally found the following set of equations:
$$ad = 1,$$
$$be = 1,$$
$$af + cd = 0,$$
$$bf + ce = 0,$$
$$cf = -1.$$
I should probably find a contradiction somewhere, but I am not sure how to solve this system and find this contradiction.

I assume you multiplied and compared coefficients.

You should have six equations:

Coefficients of $x^2$ gives $ad=1$

Coefficients of $y^2$ gives $be=1$

Coefficients of $xy$ gives $ae+bd=0$

Coefficients of $x$ gives $af+cd=0$

Coefficients of $y$ gives $bf+ce=0$

Constant terms gives $cf=-1$

You missed out the coefficients of $xy$

Use just the top three equations: $ad=1$ means that $a$ and $d$ are both positive or both negative. $be=1$ means that $b$ and $e$ are both positive or both negative.

That means that $ae$ and $bd$ have the same sign as each other.

But $ae+bd=0$ means that they can't have the same sign as each other unless they are both equal to zero, and that is your contradiction.

That means we've to prove that the given equation represents pair of straight lines. So, for a general second degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ to represent pair of straight line
$$\Delta=\begin{vmatrix}
a&h&g\\
h&b&f\\
g&f&c
\end{vmatrix}=0$$

But here,$\Delta=-1\ne 0$. Hence the given expression can't be represented as product of two linear equations. Check this.


Fréchet differentiability from Gâteaux differentiability Let $X$ be a Banach space and $\Omega \subset X$ be open.

The functional $f$ has a Gâteaux derivative $g \in X'$ at $u \in \Omega$ if, $\forall h\in X,$ $$\lim_{t \rightarrow 0}[f(u+th)-f(u)- \langle g,th \rangle]=0$$

How can I prove the following:

If $f$ has a continuous Gâteaux derivative on $\Omega$, then $f \in C^1(\Omega,\mathbb R)$.

It seems worth expanding user31373's answer above for the benefit of those who could come to this page by chance (as I did).

The Mean Value theorem discussed above is the following statement:

Let $f : X \to Y$ be a map of normed linear spaces. Given $x_0 \in X$ and a direction $h \in X$ such that the directional derivative $\delta f(x_0+th)(h)$ exists for all $t \in [0,1],$ we have that
$$
\|f(x_0+h)-f(x_0)\| \leqslant \|\delta f(x_0+th)(h)\|
$$
for a suitable $t \in [0,1]$

(P.Drabek, J.Milota, Methods of Nonlinear Analysis, 2007; Th. 3.2.7).  It follows that
\begin{equation} \tag{$*$}
\|f(x_0+h)-f(x_0)-\delta f(x_0)h\| \leqslant \|\delta f(x_0+th)(h)-\delta f(x_0)(h)\|
\end{equation}
(the latter statement is left without the proof in Drabek--Milota, and, unfortunately, the hint that is given there is misleading; however the book is one of the best sources on Fréchet and Gâteaux differentiation). In fact, one can use the idea from user31373's answer. We consider the map $g : X \to Y$ such that
$$
g(x)=f(x)-\delta f(x_0)(x-x_0)
$$
for all $x \in X.$ It is easy to see that the Mean Value theorem applied to $g$ gives us $(*).$ Let us verify, for instance, the key fact of existense of the directional derivatives $\delta g(x_0+th)(h).$ First, recall that the Gâteaux differential preserves the scalar multiplication: $\delta f(x)(r h)=r \delta f(x)(h)$ for all $x,h \in X$ and for all $r \in \mathbf R$ (the homogeneity of $\delta f(x)$). We have that
\begin{eqnarray}
\delta g(x_0+th)(h) &=& \lim_{u \to 0} \frac{g(x_0+th+uh)-g(x_0+th)}u  \\
&=&\lim_{u \to 0} \frac{f(x_0+th+uh)-\delta f(x_0)( (t+u)h) - f(x_0+th)+\delta f(x_0)(th)}u  \\
&=&\delta f(x_0+th)(h)-\delta f(x_0)(h)
\end{eqnarray}
where we have used the homogeneity to justify the last equation.

Now to address the question under what conditions existense of Gâteaux differentials implies existense of Fréchet differential. The standard result is the following one (quoting again Drabek--Milota):

if $f : X \to Y$ is a map of linear normed spaces such that in a certain open neighbourhood $O_{x_0}$ of $x_0 \in X$ the Gâteaux differential $\delta f(x)$ exists and is linear and continuous for all $x \in O_{x_0}$ and the map
$$
x \mapsto \delta f(x)
$$
from $X$ into $L(X,Y) $ is continuous at $x_0,$ then $f$ is Fréchet differentiable at $x_0$

(Prop. 3.2.15 in Drabek--Milota); $L(X,Y)$ is the normed space of continuous linear maps from $X$ to $Y$ equipped the uniform norm
$$
\|\varphi\| =\mathrm{sup}_{x \ne 0} \frac{\|\varphi(x)\|}{\|x\|} \qquad (\varphi \in L(X,Y)).
$$
The proof is based on the equality $(*)$ above.

If $\phi$ is continuous at $x_0$ and ∂ϕ(x0) contains a singleton element,
then ϕ is Gâteaux differentiable at x0 and ϕ(x0) = ∂ϕ(x0).


$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$, $\frac{P(x)}{x^3+x}$ remainder Given the polynomial:

$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$

What is the remainder of $\frac{P(X)}{x^3+x}$?

I don't think the long division is efficient the way to go, and the remainder theorem doesn't seem to be applicable here as $x^3+x$ is not linear. Could I have some hints on how to approach this? Thank you.

We have
$$
P(x) = Q(x)(x^3 + x) + r(x)
$$
where $Q(x)$ and $r(x)$ are polynomials, and $r(x)$ has degree at most $2$. Now, considering that $Q(x)(x^3 + x)$ has roots at $x = -i, 0$ and $i$, we can use this to get information about $r(x)$, namely that $r(\pm i) = P(\pm i)$ and $r(0) = P(0)$. These three points together with what we know about the degree of $r$ is enough to find $r$ exactly.

The remainder is
$$\frac{x+2}{x^2+1}\:+\:\frac 1x\qquad\text{or}\quad2x^2+2x+1\;\;\text{respectively}\\[3ex]$$
obtained by taking a lazy approach: Feeding the command

apart((x**100+x**50-2*x**4-x**3+x+1)/(x**3+x))

into http://live.sympy.org/ ,
thus asking Python to go through the long division in short time,
yields
$$\frac{P(x)}{x(x^2+1)}\;=\;\sum_{m=48}^{24}(-1)^mx^{2m+1}-2x-1+\frac 1x+\frac{x+2}{x^2+1}\\[3ex]$$

Note that the $\,\sum\,$ equals $(x^{100}+x^{50})/(x^3+x)=x^{49}(x^{50}+1)/(x^2+1)\,$, whence no contribution to the remainder.

... did I fail to address the issue in question?


Rational + irrational = always irrational? I had a little back and forth with my logic professor earlier today about proving a number is irrational. I proposed that 1 + an irrational number is always irrational, thus if I could prove that 1 + irrational number is irrational, then it stood to reason that was also proving that the number in question was irrational.

Eg. $\sqrt2 + 1$ can be expressed as a continuous fraction, and through looking at the fraction, it can be assumed $\sqrt2 + 1$ is irrational. I suggested that because of this, $\sqrt2$ is also irrational.

My professor said this is not always true, but I can't think of an example that suggests this.

If $x+1$ is irrational, is $x$ always irrational?

Actually, a better question is: if $x$ is irrational, is $x+n$ irrational, provided $n$ is a rational number?

Suppose $x$ is irrational and $x+\dfrac pq=\dfrac mn$ then, $x=\dfrac mn-\dfrac pq=\dfrac{mq-np}{nq}$ so, $x$ would then be rational. :)

This follows essentially from the fact that as additive abelian groups $\mathbb Q$ is a normal subgroup of $\mathbb R$, which is true since both groups are abelian and the latter contains the former. Hence we have a quotient homomorphism $\varphi: \mathbb R \rightarrow \mathbb R / \mathbb Q$. $r \in \mathbb R$ is rational iff $\varphi (r) = 0_{\mathbb R / \mathbb Q}$, where $0_{\mathbb R / \mathbb Q}$ is the identity element of $\mathbb R / \mathbb Q$.

Consider then for $x \in \mathbb R - \mathbb Q$, $r \in \mathbb Q$, that
\begin{align*}
 \varphi (x+r)&= \varphi(x) + \varphi(r) \\
&= \varphi(x) + 0_{\mathbb R / \mathbb Q} \\
&= \varphi(x) \\
& \ne 0_{\mathbb R / \mathbb Q} \text{ by assumption that }x \not\in \mathbb Q,
\end{align*}

so $x+r \not \in \mathbb Q$.

Of course, the above is just reinterpreting the above elementary proofs in a more general context, but this lets us apply the same line of reasoning to a wide variety of things including modular arithmetic, rotations/reflections of geometric objects, Rubik's cube moves, matrix multiplication, permutations of sets of numbers, exotic differentiable structures on spheres, ...


Convergence of $\int_0^1 \frac{\ln(1-x)\sqrt{x-x^2}}{\sin(\pi x)} \, dx$ I have difficulties with convergence of this integral:
$$\int_0^1 \frac{\ln(1-x) \sqrt{x-x^2}}{\sin(\pi x)} \, dx$$
I found similar problem here Covergence of integral but I don't get the solution from there.

I split the interval on two intervals: $(0,\alpha]$ and $[\alpha,1)$. The integral on the first interval convergs following a limit comparison test with $\frac{x^{3/2}}{\pi x}$.

But I am struggling with the second interval. Can somebody show me what to do on the second interval $[\alpha,1)$?

For $x$ close to $1$ the integrand

$$ \frac{\ln(1-x)\sqrt{x-x^3}}{\sin \pi x} \sim c\frac{\ln(1-x)}{\sqrt{x-1}} $$

which is an integrable function


Show that $\lim\limits_{n\to\infty}\sqrt [n] {n^{2}+n+1}=1$ Proof. Note that $1\leq \sqrt [n] {n^2+n+1}\leq \sqrt [n] {3n^{2}}$. But,
$\lim\limits_{n\to\infty}\sqrt [n] {3n^{2}} =1$. Therefore, by the squeeze lemma, we have $\lim\limits_{n\to\infty} \sqrt [n]{n^2+n+1}  =1$

My question is: How did $\lim\limits_{n\to\infty} \sqrt [n] {3n^2} =1$ be? So, how can it prove as primitive methods? We cannot use L'hopital rule, rules of derivative, exp functions.

Write $n^{1/n} = 1 + a_n.$ We have $0<a_n$ for all $n.$ The binomial formula shows

$$n = (1 + a_n)^n = 1 + na_n + [n(n-1)/2]a_n^2 + \cdots$$

for $n>1.$ Thus for $n>1$ we have $n>[n(n-1)/2]a_n^2,$ which implies $0 < a_n < \sqrt {2/(n-1)}.$ It follows that $a_n \to 0,$ giving $n^{1/n} \to 1.$

see my nice answer:
$$\lim_{n\to \infty}n^{1/n}=e^{\lim_{n\to \infty}\large \frac1n\ln n}$$
apply l'hopital's rule,
$$=e^{\lim_{n\to \infty}\large \frac{1/n}{1}}=e^{\lim_{n\to \infty}\large \frac1n}=1$$


Why is 1+1=0 in a finite field F={0,1}? This table:
$$\begin{array}{|c|cc|}
\hline
+& 0& 1\\
\hline
0& 0& 1\\
1& 1& 0\\
\hline
\end{array}$$
"feels" right, but how can you prove that $1+1=0$? What is the reason? I assume that due to $F \times F \rightarrow F$, the result of $1+1$ must be within the field F after all.

I'm looking for a logical explanation.

Recall the two axioms about addition:

There exists a unique element $0$ such that for every $x$, $x+0=x$.
For each element $x$ there exists a unique element $y$ such that $x+y=0$.

If $1+1=1$ it has to be the case that $1+0=0$. But in that case you just reversed the roles of $0$ and $1$, as dictated by the axioms. That's fine, and it still defines a group structure, but it's easier to just relabel them to the traditional roles, where $0$ is the additive unit.

So if $1+1\neq 1$, it has to be the case that $1+1=0$.

As a Finite Field, the distribute law must hold:
The distributive law holds: (a+b) · c = a · c+b · c for all a,b, c ∈ F.
You can chose 1 or 0.

Now, consider Boolean Logic.
Using the Or operator, then 1 is the correct answer. 0,1,1,1
If Exclusive Or, then 0,1,1,0 (as you have in your table)


How do I prove whether this formula is valid or satisfiable? $\exists xy[P(x,y) \rightarrow \forall xy\ P(x,y)]$ By valid/satisfiable I mean in the context of propositional logic. The following formula is:

$$\exists xy[P(x,y) \rightarrow \forall xy\ P(x,y)]$$

I understand that P is considered to be a predicate symbol, by P(x,y) is a formula that should evaluate to true or false. However, how do I go about solving this kind of problems if I don't have information on what x and y are?

By the rules I learned the sentence is not well formed, so it is neither valid nor satisfiable.  The square brackets mean the existential quantifier applies to everything inside, so you can't requantify over them with the universal quantifier.  It would be legal to write $\exists xy P(x,y) \implies \forall xy P(x,y)$ but this still creates an opportunity for confusion because of the reuse of $x,y$.  I find it better to write
$$\exists xy P(x,y) \implies \forall wz P(w,z)$$
where the scope of the quantifiers is obvious.  I have seen authors who specifically allow reuse of variables, but they have to be very careful about the rules for substitution.  Some comments have said the rules they are used to accept your sentence, but it is equivalent to the one I set in display mode.

This makes it easier to see that the sentence is satisfiable but not valid.  If $\forall wz P(w,z)$ is true the implication is true as well.  If $P(x,y)$ is true sometimes and false sometimes the implication is false.  We could easily construct a model for each of the previous sentences to prove our claim.


Order of general linear group of $2 \times 2$ matrices over $\mathbb{Z}_3$ From problem 2.3.25 in Topics in Algebra, 2$\varepsilon$ by I. N. Herstein:

  Let $G$ be the group of all $2 \times 2$ matrices $\left(\begin{array}{c c}a & b \\ c & d\end{array}\right)$ where $ad-bc \ne 0$ and $a,b,c,d$ are integers modulo 3, relative to matrix multiplication. Show that $o(G) = 48$.

I know that $o(G) \le 3^4 = 81$, since $a,b,c,d$ can each take one of 3 values (mod 3). I attempted to tighten this bound by finding the number of matrices such that $ad=bc$ (mod 3):

Suppose $ad=bc=0$ (mod 3). Then ($a = 0$ or $d = 0$) and ($b = 0$ or $c = 0$), leading to 36 possible values for $(a,b,c,d)$.
Suppose $ad=bc=1$ (mod 3). Then ($a=d=1$ or $a=d=2$) and ($b=c=1$ or $b=c=2$), leading to 4 possible values for $(a,b,c,d)$.
Suppose $ad=bc=2$ (mod 3). Then ($(a,d)=(1,2)$ or $(a,d)=(2,1)$) and ($(b,c)=(1,2)$ or $(b,c)=(2,1)$), leading to 4 possible values for $(a,b,c,d)$.

So, there are in total $36+4+4 = 44$ such $\left(\begin{array}{c c}a & b \\ c & d\end{array}\right)$ where $ad-bc=0$ (mod 3). That means there are at most $81-44 = 37$ such $\left(\begin{array}{c c}a & b \\ c & d\end{array}\right)$ where $ad-bc\ne 0$, i.e., $o(G) \le 37$. However, this contradicts the problem. Where did I go wrong? Can someone set me on the right path?

Vector $a\choose c$ must be non-zero, this allows $p^2-1$ choices.
Vector $b\choose d$ must not be a multiple of $a\choose c$, this allows $p^2-p$ choices. Thus, $|G|=(p^2-1)(p^2-p)$.

Let G be group of all 2 cross 2 matrices(a b c d){where in 1st row has a and b,2nd row has c and d}and a,b,c,d are integers modulo 3,relative to matrix multiplication.show ghat o(G)=48.(topics in algebra by i n hetstein with 2nd edition.capter-2 problem number 25)
Solution:-According to the question this is a general linear group of integers modulo n.for this in general we have a formula to find the order of group.
o(GL(n z_p))=(p^n-1)(p^n-p)...(p^n-p^n-1)
So according to this formula here n=2 and p=3.Now put these value in the above formula we will get order of general linear equal to 48.


Finding the number of permutations in$S_9$ of the form $(a_1a_2)(a_3a_4)(a_5a_6)(a_7a_8a_9)$ How many permutations, ρ, are there in $S_9$(the group of permutations of order 9!) whose decomposition into disjoint
cycles consists of three 2-cycles (transpositions) and one 3-cycle? In other words, how
many permutations are there in $S_9$ whose decomposition into disjoint cycles is of the form
$(a_1a_2)(a_3a_4)(a_5a_6)(a_7a_8a_9)$?
would I be able to assume that,  $a_1 = 1$, so $a_2$ would have 8 possibile values, $a_3$ would have 7 and so on, for a total of $8!$ such permutations?

You need to choose the support of the first transposition : $\begin{pmatrix}9\\2\end{pmatrix}$ choices.
You need to choose the support of the second transposition in the remaining set : $\begin{pmatrix}7\\2\end{pmatrix}$ choices.
You need to choose the support of the third transposition in the remaining set : $\begin{pmatrix}5\\2\end{pmatrix}$ choices.

You need to divide this number by the number of ways to order the three transpositions (since they all give the same permutation) e.g. : $(1,2)(3,4)(5,6)=(3,4)(1,2)(5,6)$. Hence you need to divide by $3!$.
Choose the support of the $3$-cycle in the remaining set $\begin{pmatrix}3\\3\end{pmatrix}=1$ and on a support of size $n$ you know that you have exactly $(n-1)!$ different $n$-cycles, here this gives $(3-1)!=2$ choices.
Gather everything :

$$\frac{1}{3!} \begin{pmatrix}9\\2\end{pmatrix}\begin{pmatrix}7\\2\end{pmatrix}\begin{pmatrix}5\\2\end{pmatrix}2=2\frac{9\times 8\times 7\times 6\times 5 \times 4}{8\times 6}=9\times 7\times 5\times 8$$

Step 1: Choose, out of 9 possible numbers, the first two for the first 2-cycle. (I believe the number of ways you can do this is $\binom{9}{2}$)

Step 2: Choose, out of the remaining 7 possible numbers, the next two for the second 2-cycle. (I believe the number of ways you can do this is $\binom{7}{2}$)

Step 3: Choose, out of the remaining 5 possible numbers, the next two for the third 2-cycle. (I believe the number of ways you can do this is $\binom{5}{2}$)

Step 4: Arrange the last 3 remaining numbers for the 3-cycle. (I believe the number of ways you can do this is $3!$)

[EDIT] Last Step: Take the product of the number of possible combinations. $$\binom{9}{2}\binom{7}{2}\binom{5}{2}(3!)$$

That is the long way to do it. However, if you simplify that, that would just be $$\frac{9!}{(2!)(2!)(2!)}$$.

By the way, $8!$ counts the number of 8-cycles in $S_9$.


How to predict symmerty of parametric curve Suppose we are given a curve as $$x^{2/3} + y^{2/3} = 1 $$
In parametric form it can be written as $$x=\cos^{3}(\theta)$$
and
$$y=\sin^{3}(\theta)$$ now how can we predict if curve will be symmetric about $x$ axis or $y$ axis? In Cartesian curve we just replace $$x=-x$$  and $$y=-y$$ to get an idea about symmetry. Someone told me to replace $$\theta=\pi-\theta$$ but why? How will that get me information on symmetry ?

Try and take a look at the parametric equation in this perspective:

The most $x^{2/3}+y^{2/3}$ will ever be equal to is 1. This may sound trivial but you can glean a bit of information out of this. If the $x^{2/3}$ term is 1, the $y$ term must be zero. Conversely, when the $y^{2/3}$ term is 1, the $x$ term must be zero. We can imagine these points on the Cartesian plane. Continuing to work from intuition alone, when the $x^{2/3}$ term increases from zero, the $y^{2/3}$ must decrease from one. Because $f(a)=a^{2/3}$ is strictly increasing when examined over the positive reals, and because the exponents for the parameters are the same, it may help to imagine $x+y=1$, and note that when you modify the equation to $x^{2/3}+y^{2/3}=1$, the curve is no longer linear - it will pucker towards the origin. (As for some intuition behind that, why is $({1/2},{1/2})$ not part of the solution set of your parametric equation?)

If there is no change in x by changing sign of x it has x axis symmetry.If there is no change in y by changing sign of y it has y axis symmetry. If either, symmetry about both axes. No change in sign .. this is brought about by even functions like cos, square of arguments etc.


Simplifying fractional exponents I am very confused about the following:

whenever I put in into wolfram alpha the answer it gives me is "indeterminate", is it not possible to simplify fractional exponents or something? if the exponent is 2/6 can i simplify it to 1/3 and get -1 as the answer? Provided I'm working with real numbers.

What I put in wolfram alpha was "(-1)^(1/3) real numbers" to which it answered "-1", then when I tried "(-1)^(2/6) real numbers" it answered "indeterminate".

I figured it out, apparently wolfram first takes the 6th root of -1 and since (-1)^(1/6) is indeterminate since we're dealing with real numbers and can't take an even root of a negative number, it says "indeterminate".

Also, Rory's first solution is absolutely wrong (unless he corrects if when you see this), because we can't take a root of a real negative number raised to a power greater than 1, if what he wrote were true I can then prove that -1 = 1 like so: -1 = (-1)^(1/3) = (-1)^(2/6) = (1)^(1/6) = 1.

The usual definition (used in high schools) of $x^{a/b}$ is $\sqrt[b]{x^a}$. Therefore, your expression $(-1)^{2/6}$ could mean either:

$$(-1)^{2/6}=\sqrt[6]{(-1)^2}=\sqrt[6]{1}=1$$ or
$$(-1)^{2/6}=(-1)^{1/3}=\sqrt[3]{(-1)^1}=\sqrt[3]{-1}=-1$$

That seems pretty indeterminate to me!

You could complain that it is obvious that the fraction $2/6$ should be simplified before applying the definition of a rational power, but it apparently is not obvious to Wolfram Alpha. Wolfram does not know you or your particular definition, it could be argued that Wolfram took the correct approach.


Compute the Euler function $\phi(n)$ for $n = 360$ as well as the number of divisors $d(n)$. Compute the Euler function $\phi(n)$ for $n = 360 $ as well as the number of
divisors $d(n)$.

Is this correct?

$360 = 2^3\cdot 3^2\cdot 5$ thus $\phi(n) = 2^2\cdot 2\cdot 3\cdot 4 = 96$.

$d(n) = 4 \times 3 \times 2 = 24$.

$360 = 2^3\times 3^2\times5$. Since Euler Phi function is multiplicative, $\phi(360)=\phi(8)\times\phi(9)\times\phi(5)=(8-4)\times(9-3)\times4=96.$

Which means you are correct.

You are correct, although note that you're better using \times or \cdot rather than just a space, because in LaTeX (and MathJax) a space will be omitted.