Square root of a $3\times3$ matrix Here is  $3\times3$ matrix$$\begin{pmatrix}   0&    0&    1\\
    0  & -1 &   0\\
    1&    0  &  0\end{pmatrix}$$
How can I solve this  by using Cayley-Hamilton?
I know how to use Cayley-Hamilton for a $2$-dimensional matrix.
How can it help in finding the square root of a $3\times3$ matrix?
for 2 dimensional matrix we can solve this equation A^2−(trA)A+(detA)I=0
we have A and I,
we can compute det(A^2) so we have det A,
and  we can find A.

for 2 dimensional matrix using above equation we can compute square root.
for example we have this matrix:

A^2= $$\begin{pmatrix}   4&    2\\
        2&    2\end{pmatrix}$$

det A^2= 4

det A=2

4 2             1 0
2 2  +6 *A+ 2*  0 1 =0

by solving above equation we can find A.

Since your matrix $M$ is symmetric you can diagonalize it. You'll get $M=U \cdot D \cdot U^\dagger$, where $D$ contains the eigenvalues and $U$ contains the eigenvectors. Take the square roots of the eigenvalues  to get $D_{\sqrt \cdot}$ and transform the resulting diagonal matrix $U^\dagger \cdot D_{\sqrt{.}}\cdot U=S$. It's obvious to see that $S^2=M$.

EDIT

And even that I don't understand your original approach, you can use it when you rewrite your matrix to
$$
\pmatrix{0 & 1 & 0\\1&0&0\\0&0&-1}.
$$
Now you have one $2\times 2$ block suitable for your approach.

I solved it by Euler's method of solving for the (later named) unit-length eigenvector (a tilted 3D rotational axis), and the Euler 2D angle theta of rotation about the 3D Euler axis.

I did it computationally.

The square root matrix is:
{ { 0.5  , 0.7071, 0.5 }, {-0.7071,0,0.7071}, {0.5,-.7071,0.5}}


Linear Algebra - Homogenous Equations Find a homogenous system of linear equations in five unknowns whose solution space consists of all vectors $\Re^5$ that are orthogonal to the vectors:

$\mathbf v_1$$=\langle3,0,1,2,3\rangle$;

$\mathbf v_2$$=\langle-3,1,0,-1/2,-1\rangle$;

$\mathbf v_3$$=\langle6,0,1,2,-3\rangle$.

What kind of geometric object is the solution set? Find a general solution of the system and confirm that the solution space has the orthogonality and the geometric properties that were stated.

I was told this problem would be really good practice for linear algebra however I have no idea how to approach this problem and would appreciate any guidance.

We are looking for all $\mathbf x=\langle x,y,z,u,v\rangle \in \Re^5$ such that

$\mathbf x \cdot \mathbf v_j=0$ for $j=1,2,3$, where $ \mathbf x \cdot \mathbf v_j$ denotes the usual inner product on $\Re^5$.

A vector $(x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5$ is orthogonal to a vector $(a_1, a_2, a_3, a_4, a_5) \in \mathbb{R}^5$ if and only if $a_1 x_1 + a_2 x_2 + a_3 x_3 + a_4 x_4 + a_5 x_5 = 0$. Here, the set of vectors you are looking for is the set of solutions of the system :
\begin{equation}
  \left\{
      \begin{aligned}
        3x_1& & +x_3& +2x_4&+3x_5&=&0\\
        -3x_1&+x_2& &-\frac{1}{2}x_4&-x_5&=&0\\
        6x_1& &+x_3&+2x_4&-3x_5&=&0\\
      \end{aligned}
    \right.
\end{equation}

Now you get a linear system and you can resolve it with your favorite method.


First write
$$\left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] = \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \dots \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] $$
Also,
$$\det \left( \left[ \begin{array}{ccccc} \text{I}_{n_1} \hspace{-5pt} &&&& \\[-3pt] & \ddots \hspace{-5pt} &&& \\ && A_j \hspace{-5pt} && \\[-3pt] &&& \ddots \hspace{-5pt} & \\ &&&& \text{I}_{n_k} \end{array} \right] \right) = \det (A_j)$$
which can be seen by using the cofactor formula and repeatedly expanding along a row or column with all 0's and one 1

$$ \implies \det \left( \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] \right) = \det (A_1) \det (A_2) \cdots \det (A_k)$$

A "functorial approach" using the exterior product: If $\phi: V \rightarrow V$ is an endomorphism of a vector space, you may calculate the determinant of the endomorphism $\phi$ as the induced map
$$\wedge^n (\phi): \wedge^n V \rightarrow \wedge^n V$$
where $n:=\dim(V)$. Since $\wedge^n$ is a functor you get a canonical map $\wedge^l (\phi)$ for any integer $l \geq 1$.
It follows $\wedge^n (\phi)$ is an endomorphism of a one dimensional vector space $\wedge^n V$ and hence it is given as multiplication with a number $a$. The number $a$ is the determinant: $a=\det(\phi)$ of the map $\phi$. If you choose a basis $B$ of $V$ and the matrix of $\phi$ in this basis is a matrix $A$, it follows $a=det(A)$ is the determinant of the matrix $A$.
There is a formula:
$$\wedge^{n_1+n_2}(V_1\oplus V_2) \cong \wedge^{n_1}V_1 \otimes \wedge^{n_2}V_2,$$
where $n_i:=\dim(V_i)$. Let
\begin{align*} \phi= \begin{pmatrix} \phi_1 & 0 \\ 0 & \phi_2 \end{pmatrix} \end{align*}
where $\phi_i$ is an endomorphism of $V_i$. It "follows"
$$\det(\phi)=\wedge^{n_1+n_2}(\phi) \cong \wedge^{n_1}(\phi_1) \otimes \wedge^{n_2}(\phi_2)$$
But the tensor product $\wedge^{n_1}V_1 \otimes \wedge^{n_2}V_2$ is a one dimensional vector space and any linear endomorphism of such a space is given (in a basis) as multiplication with a number. Choosing a basis it follows the endomorphism $\wedge^{n_1}(\phi_1) \otimes \wedge^{n_2}(\phi_2)$ is multiplication with the number
$$\det(A_1)\det(A_2),$$
where $A_i$ is a matrix of $\phi_i$ in a basis $B_i$ for $V_i$.
Question: "But what about the determinant???"
By induction it follows that if $M$ is a matrix with square matrices $A_i$ along the diagonal, you get the formula
$$\det(M)=\det(A_1) \cdots \det(A_n).$$


Minimizing over partitions $f(\lambda) = \sum \limits_{i = 1}^N |\lambda_i|^4/(\sum \limits_{i = 1}^N |\lambda_i|^2)^2$ I'm trying to characterize the behavior of the the quantity:
$$A = \frac{\sum \limits_{i = 1}^N x_i^4}{(\sum \limits_{i = 1}^N x_i^2)^2},$$
subject to the constraints that
$$ \sum \limits_{i = 1}^N x_i = N, \ x_i \ge 0 \ \forall \ i$$

(I.e., we are working with all integer partitions of size $N$).
My hypothesis is that if there exists $x_i \ge N/2$, then $A \ge \frac{x_i^4 + (N-x_i)^4}{(x_i^2 + (N-x_i)^2)^2}$ (which I've verified experimentally for N up to 40). But I'm having trouble proving this hypothesis.

The closest I've come is to make use of the fact that $\sum \limits_{i = 1}^N x_i^2 \le N^2$, which gives an upper bound for the denominator, but the corresponding inequality for fourth powers in the numerator doesn't seem helpful, since we're trying to find a lower bound.

I'd greatly appreciate any advice. Please ask for any clarifications if anything is unclear.

Edit:
I've decided this question is better phrased explicitly in terms of partitions. So let $$f(\lambda) = \frac{\sum \limits_{i = 1}^N |\lambda_i|^4}{(\sum \limits_{i = 1}^N |\lambda_i|^2)^2},$$
be a function over partitions $\lambda$ of size $N$ and $|\lambda_i|$ is the size of the $i$-th part. The hypothesis now is that in the domain of all partitions $\lambda'$ in which there exists $\lambda_i$ such that $|\lambda_i| \ge N/2$, $f(\lambda')$ is minimized by the 2-element partition $<\lambda_i, \lambda_j>$ (in which $|\lambda_j| = N - |\lambda_i|$).

You can assume that $x_N\ge N/2$ and so you want to prove that
$$
 \frac{\sum \limits_{i = 1}^N x_i^4}{\left(\sum \limits_{i = 1}^N x_i^2\right)^2}\ge \frac{x_N^4 + (N-x_N)^4}{(x_N^2 + (N-x_N)^2)^2}
$$
Set $\lambda:=N-x_N$, and assume $\lambda>0$ (else the inequality is trivially true), and set also
$$
x:= \frac{x_N}{\lambda},\quad\text{and}\quad y_i:=\frac{x_i}{\lambda},\quad\text{for $i=1,\dots, N-1$}.
$$
The inequality now reads:
$$
\frac{x^4+\sum \limits_{i = 1}^{N-1} y_i^4}{\left(x^2+\sum \limits_{i = 1}^{N-1} y_i^2\right)^2}
\ge \frac{x^4 + 1}{(x^2 + 1)^2},
$$
and you want to prove this inequality under the conditions $x\ge 1$ (corresponding to $x_N\ge N/2$) and $S_1:=\sum_{i=1}^{N-1}y_i=1$,
(corresponding to $\sum_{i=1}^N x_i=N$). Set $S_2:=\sum_{i=1}^{N-1}y_i^2$ and $S_4:=\sum_{i=1}^{N-1}y_i^4$. Then you want to prove that
the inequality
$$
(S_4+x^4)(x^2+1)^2\ge (x^4+1)(S_2+x^2)^2
$$
holds, or equivalently, that for the function
$$
f(x):=S_4 x^4+2 S_4 x^2+S_4+2x^6-2S_2 x^6-2x^2S_2-S_2^2-S_2^2x^4
$$
we have $f(x)\ge 0$ for $x\ge 1$.
Below we will prove the following inequality:
$$
(MI) \qquad\qquad\qquad 1+2S_4\ge 2 S_2+S_2^2.
$$
From this inequality it follows that $f(1)=4S_4+2-4S_2-2S_2^2\ge 0$.
Hence it suffices to prove that $f'(x)\ge 0$ for all $x\ge 1$.
But
$$
f'(x)=4S_4 x^3+4 S_4 x+12 x^5-12 S_2 x^5 -4 S_2 x -4 S_2^2 x^3,
$$
and so $f'(x)=2x g(x)$ with
$$
g(x):=2 S_4 x^2+2S_4+6 x^4-6 S_2 x^4-2S_2-2S_2^2 x^2
$$
and we have to prove that $g(x)\ge 0$ for $x\ge 1$.
From $(MI)$ we obtain
$$
2S_4 x^2-S_2^2 x^2\ge 2x^2 S_2-x^2\quad\text{and}\quad
2S_4-2S_2\ge S_2^2-1,
$$
which implies
$$
g(x)\ge 2 S_2 x^2-x^2+S_2^2-1+6x^4(1-S_2).
$$
Now clearly $S_2\le 1$ and so $6x^4(1-S_2)\ge 6x^2(1-S_2)$. We arrive at
$$
g(x)\ge S_2^2+(x^2-1)+4x^2(1-S_2)\ge 0,
$$
which concludes the proof.

Finally we prove
$$
(MI) \qquad\qquad\qquad 1+2S_4\ge 2 S_2+S_2^2.
$$
We will prove a slightly more general result:
Take $x_1,\dots,x_n$ with $x_i\ge 0$ and set
$$
S_1:=\sum_{i=1}^n x_i,\quad S_2:=\sum_{i=1}^n x_i^2\quad\text{and}\quad S_4:=\sum_{i=1}^n x_i^4.
$$
Then we have
$$
S_1^4+2S_4\ge 2S_2 S_1^2+S_2^2.
$$
This inequality, in the case $S_1=1$, gives $(MI)$.

Let's prove the inequality $
S_1^4+2S_4\ge 2S_2 S_1^2+S_2^2
$:

Adding $S_2^2$ and subtracting $2S_4$ and $2 S_2 S_1^2$ it is equivalent to
$$
S_1^4-2S_2 S_1^2+S_2^2\ge S_2^2-2S_4+S_2^2,
$$
hence we have to prove
$$
(S_1^2-S_2)^2\ge 2(S_2^2-S_4).
$$
Now we compute
$$
S_1^2-S_2=\left(\sum x_i\right)\left(\sum x_j\right)-\sum x_i^2=\sum_{i\ne j}x_i x_j
$$
and
$$
S_2^2-S_4=\left(\sum x_i^2\right)\left(\sum x_j^2\right)-\sum x_i^4=\sum_{i\ne j}x_i^2 x_j^2.
$$
Subtracting $\sum_{i\ne j}x_i^2 x_j^2$ from both sides of the inequality
$$
\left(\sum_{i\ne j}x_i x_j\right)^2\ge 2\sum_{i\ne j}x_i^2 x_j^2,
$$
we see that the inequality is equivalent to
$$
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l\end{array}}x_i x_j x_k x_l\ge \sum_{i\ne j}x_i^2 x_j^2.
$$
But we have
$$
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l\end{array}}x_i x_j x_k x_l=
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l \\ (i,j)=(l,k)\end{array}}x_i x_j x_k x_l+
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l \\ (i,j)\ne (l,k)\end{array}}x_i x_j x_k x_l
$$
hence
$$
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l\end{array}}x_i x_j x_k x_l\ge
\sum_{\begin{array}{c} (i,j)\ne (k,l)\\ i\ne j\\ k\ne l \\ (i,j)=(l,k)\end{array}}x_i x_j x_k x_l=\sum_{i\ne j}x_i^2 x_j^2,
$$
which concludes the proof of $(MI)$.

The trick here is to consider the following "switching operation":
$$\sigma_{ij}(\lambda_1,\ldots,\lambda_N) = (\lambda_1,\ldots,\lambda_i+1,\ldots,\lambda_j-1,\ldots,\lambda_N).$$
Now the question becomes: what can we say about $f(\sigma_{ij}(\lambda))$?

Assume without loss of generality that $\lambda_j\ge1$, and that all $\lambda_k\ge0$. A short computation shows:
$$f(\sigma_{ij}(\lambda)) = \frac{\sum_{k=1}^N\lambda_k^4 + 4(\lambda_i^3-\lambda_j^3) + 6(\lambda_i^2+\lambda_j^2) + 4(\lambda_i-\lambda_j) + 2}{\left(\sum_{\ell=1}^N\lambda_\ell^2+2(\lambda_i - \lambda_j) + 2\right)^2}$$
Now write $\lambda_j = \lambda_i+k$ with $k$ an integer. Then the above expression becomes
$$f(\sigma_{ij}(\lambda)) = \frac{\sum_{k=1}^N\lambda_k^4 + 2\lambda_i^2+k(4+2\lambda_i-3\lambda_i^2)+k^2(1-3\lambda_i^2)-k^3}{\left(\sum_{\ell=1}^N\lambda_\ell^2+2(k+1)\right)^2}$$
Now we are left with the tedious work of checking by cases ($k>0$, $k<0$ etc.) when this decreases with respect to $f(\lambda)$. This allows you to minimize the quantity.


Determine price elasticity of demand and marginal revenue Determine price elasticity of demand and marginal revenue if $q = 30-4p-p^2$, where q is quantity demanded and p is price and p=3.

I solved it for first part-

Price elasticity of demand = $-\frac{p}{q} \frac{dq}{dp}$

on solving above i got answer as $\frac{10}{3}$

But on solving for Marginal revenue i am getting -10. But the correct answer given is $\frac{21}{10}$

Any hint is appreciable please help.

Try this. Total Revenue TR is $pq$. Marginal revenue is the change in TR with change in $quantity$ (not price, as I incorrectly stated in my comment) so marginal revenue is $\frac{\partial TR}{\partial q}$ or
$$\frac{\partial (pq)}{\partial p}\frac{\partial p}{\partial q}$$ Revenue is $pq$, or $30p−4p^2−p^3$ so marginal revenue is $30−8p−3p^2$ = 30−24−27=−21. But then, as OP correctly calcualted, $\frac{\partial q}{\partial p}=-10$ so $$\frac{\partial (pq)}{\partial p}\frac{\partial p}{\partial q}= -21/-10.$$

marginal revenue $= p(1+1/elasticity) = 3(1-3/10)= 21/10$.


Find the $E[X^3]$ of the normal distribution Find the $E[X^3]$ of the normal distribution with mean μ and variance $σ^2$ (in terms of $μ$ and $σ$).

So far, I have that it is the integral of $x^3$ multiplied with the pdf of the normal distribution, but when I try to integrate it by parts, it becomes super convulated especially with the e term. I know there's tricks with odd and even functions that may apply and I have a sinking suspicion it might just end up being μ, but I've kind of hit a wall. Any help would be appreciated :)

EDIT:

So I think the answer is either $0$ (because of how the graph looks when I plotted it) or, by simple integration it is $3μσ^2 + μ^3$.

You can observe that it is a integral of a impar function over the all domain $\mathbb R$. Therefore it is equal to zero

Try replacing $x^2$ by $t$ and do the integration by parts.


Show that for a continuous Gaussian martingale process $M$ that $\langle M, M \rangle_t = f(t)$ is continuous, monotone, and nondecreasing Let $M$ be a (true) martingale with continuous sample paths, such that $M_0 = 0$. We assume that $(M_t)_{t \geq 0}$ is also a Gaussian process. Show that there exists a continuous monotone nondecreasing function $f: \mathbb{R}_+ \to \mathbb{R}_+$ such that $\langle M, M \rangle_t = f(t)$ for every $t \geq 0$.

Recall that $\langle M, M \rangle_t$ is the quadratic variation of $M$ and that $\langle M, M \rangle_t = \lim_{n \to \infty} \sum_{i = 1}^{p_n} (M_{t_i^n} - M_{t_{i-1}^n})^2$ for a sequence of partitions with mesh going to zero. We know this is an increasing process.

In a previous part of this question I showed that a consequence of the martingale property is that the increments are independent of the past. I don't know how to show that $\langle M, M \rangle_t$ is continuous.

Define $f(t) = \mathbb{E}[M_t^2]$. It remains to check that $M_t^2 - f(t)$ is a (local) martingale since then $f$ will be the quadratic variation of a continuous martingale and hence will be continuous and non-decreasing.

Recall that a continuous martingale that is also a Gaussian process has independent increments. Hence
\begin{align*}
\mathbb{E}[M_t^2 - M_s^2 \mid \mathcal{F}_s] = \mathbb{E}[(M_t - M_s)^2 \mid \mathcal{F}_s] = \mathbb{E}[(M_t - M_s)^2] = \mathbb{E}[M_t^2 - M_s^2]
\end{align*}
where we used independence of increments in the third equality. In particular, it follows that $M_t^2 - \mathbb{E}[M_t^2]$ is a martingale.

I figured it out on my own.

Continuous martingales are continuous local martingales, and $\langle M, M\rangle_t$ is unique up to indistinguishability, an increasing (thus monotone and nondecreasing) process, and $N_t = M_t^2 - \langle M, M \rangle_t$ is a continuous local martingale. $M_t^2$ is also continuous, so

$$f(t) = \langle M, M\rangle_t = M_t^2 - N_t$$

must also be continuous.


Evaluating $\sum_{k=1}^{n} 2^{k}k$ How can I evaulate the following series?

$$\sum_{k=1}^{n} 2^{k}k$$

I don't know where to begin. $2^k$ alone would be straight forward.

Notice that $\sum_{k=1}^nk2^k=\sum_{m=1}^n\sum_{k=m}^n2^k$. Can you conclude from here?

$$ \sum_{k=1}^{n} 2^{k} = infinity$$

then your sum is greater so it to goes to infinity

The comparison test shows why.

https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-6/v/comparison-test-convergence


What are the complex solutions of a linear homogenous ODE of order $n$ with constant coefficients? What are the complex solutions of a linear homogenous ODE of order $n$ with constant coefficients? Where can I read a proof?

p.s. I don't even see the answer to the first question with a google search.

Also you can consult the following textbook:
L. Perko, "Differential Equations and Dynamical Systems," chapter 1.

The wikipedia page entitled Homogenous Differential Equations gives the method right at the start; a roof (depending on the degree of rigor required) is straightforward because the method is fairly simple.


Finding a matrix of an odd determinant I am trying to find a square matrix of size $n$ consisting of zeros and ones whose determinant is an odd number. Any ideas?

All identity matrices have determinant $1$.
More generally, all permutation matrices have determinant $\pm1$, and remember that adding a multiple of any row to any other row does not change the determinant. This should already give you many, many examples.
Another example (not in the instances described above) of a zero-one matrix with odd determinant is a $2n×2n$ matrix with zeros on the diagonal and ones elswehere – this has determinant $1-2n$.
Finally, the most general way of constructing a zero-one matrix with odd determinant is to take any integral matrix with odd determinant and reduce modulo $2$ (odds to $1$, evens to $0$), since $\mathbb Z_2$ is a field.

Also any diagonal matrix with odd elements on the diagonal, in fact:
$$\det(A)=a_{1,1}  \cdots a_{n,n}$$


Number of partitions of an $n$-element set into $k$ classes A partition of a set $S$ is formed by disjoint, nonempty subsets of $S$ whose union is $S$. For example, $\{\{1,3,5\},\{2\},\{4,6\}\}$ is a partition of the set $T=\{1,2,3,4,5,6\}$ consisting of subsets $\{1,3,5\},\{2\}$ and $\{4,6\}$. However, $\{\{1,2,3,5\},\{3,4,6\}\}$ is not partition of $T$.
If there are $k$ nonempty subsets in a partition, then it is called a partition into $k$ classes. Let $S_k^n$ stand for the number of different partitions of a set with $n$ elements into $k$ classes.

Find $S_2^n$.
Show that $S_k^{n+1}=S_{k-1}^n+kS_k^n$.

--

My work:
From the definition of $S$, $S_2^n=2^n$. I think I am wrong somewhere, because when I put this formula into the second part to prove, I get,
$$k^{n+1}=(k-1)^n+k \cdot k^n.$$
Please tell me where I am wrong. I think this problem cannot be solved by star-and-bar method as that method finds value for $k$ but does not prove it. Please help!

on(i):

$2^n$ gives you the number of all subsets of $S$, but you are looking for the number of subsets that are not empty and have no empty complement. Their total number is $2^n-2$. Note that a nonempty subset $A$ having a non-empy complement $A^c$ corresponds with partion $P=\{A,A^c\}$. However, $A^c$ corresponds with that partition too. So counting these sets gives twice the number of partitions. This amounts in: $$S_n^2=2^{n-1}-1$$
addendum on (ii)

Start with a set $S$ having $n$ elements. Now form $S'=S\cup\left\{ x\right\} $
where $x\notin S$. Partitions of $S'$ in $k$ classes can be made in two
ways:

1) Let $\left\{ x\right\} $ be one of the classes. If $P$ is a partition
of $S$ in $k-1$ classes then $P'=P\cup\left\{ \left\{ x\right\} \right\} $
is a partition of $S'$ in $k$ classes. Here there are $S_{k-1}^{n}$ possibilities.

2) Let $\left\{ x\right\} $ be not one of the classes. For every partition of $S$ in $k$ classes we can put $x$ in
one of these classes wich will induce a partition of $S'$ in $k$
classes. There are $S_{k}^{n}$ such partitions and for $x$ there
are $k$ candidates so there are $kS_{k}^{n}$ possibilities.


Uniformly distributed rationals Is there any algorithm, function or formula $f(n)$, which is a bijection between the positive integers and the rationals in $(0,1)$, with the condition, that for all real numbers $a,b,x$ with $0<a<b<1<x$, if we let $i(x)$ be the number of distinct integers $0<n_j<x$ which satisfy $a<f(n_j)<b$, then we have $\lim_{x\rightarrow\infty}i(x)/x=b-a$?

You are looking for an equidistributed sequence of rationals that uses every rational exactly once.  The Van der Corput sequence mentioned by GEdgar comes close, except that it misses some of the rationals.  But you can just throw in the missing numbers, as long as you spread them out.

For instance, the binary Van der Corput sequence uses all the dyadic rationals.  If you enumerate the non-dyadic rationals and insert the $k$th one into the sequence at position $2^k$, say, you can check that the new sequence is still equidistributed.  The key is that of the first $N$ values in the sequence, only $o(N)$ have been changed.

(I may have misinterpreted this question. I came across this question while wondering if it is possible to construct a "uniform" random variable that generates rationals. I think it's impossible.)

Consider a random variable $X$ over the rationals in (0,1). One way to define a "uniform" distribution is the following rule:

$$ \mathrm{P}(a < X < a+\epsilon) = \epsilon $$

for any $0 < a < a+\epsilon <1$. This says that the probability that a number drawn from $X$ is in the interval $(a,a+\epsilon)$ equals $\epsilon$ and, in particular, the probability is independent of $a$.

The probability that $X$ equals any particular rational, $r$, must be zero. A proof by contradiction: If there existed an $r$ such that

$$ \mathrm{P}(X = r) = p_r > 0 $$

then one could construct an interval around $r$ which is narrower than $p_r$, such as $(r-\frac13 p_r, r+\frac13 p_r)$. By our earlier rule, it should be the case that

$$ \mathrm{P}(r-\frac13 p_r < X < r+\frac13 p_r) = \frac23 p_r $$

but this contradicts the assumption that there already exists a single element in that interval which has probability $p_r$. There is not enough probability mass in the arbitrarily-small interval to allow the individual rationals to have non-zero probabilities.

Therefore, the probability mass associated with any rational must be zero. (Or perhaps hyperreal.)

Now, it is trivial to construct a (possibly non-uniform) distribution over the rationals, which I'll call $Y$. Draw the denominator, $d$, from a suitable integer distribution (such as Geometric) and draw the numerator, $n$, uniformly from the range $0<n<d$. For each rational $r$, there will be a non-zero probability of it being generated from that simple scheme:

$$ \mathrm{P}(Y=r) > 0 $$

and of course, these must sum to 1 to be a proper probability distribution:

$$ \sum_{r \in \mathcal{Q}} \mathrm{P}(Y=r) = 1 $$

Finally (and I'm not sure I've got this right...), the probabilities for every element in X (zero) are smaller than for every element in Y (non-zero).

$$ \forall r \qquad \mathrm{P}(Y=r) > \mathrm{P}(X=r) $$

and therefore the sum for X, $ \sum_{r \in \mathcal{Q}} \mathrm{P}(X=r) $, can't possibly sum to 1. Every term in the sum for X is smaller than the corresponding term for Y. So such a distribution over the rationals cannot exist.

(This answer probably isn't novel, I'm not a specialist. But I am interested in this question. Any feedback appreciated.)


"Negative" versus "Minus" As a math educator, do you think it is appropriate to insist that students say "negative $0.8$" and
not "minus $0.8$" to denote $-0.8$?

The so called "textbook answer" regarding this question reads:

A number and its opposite are called additive inverses of each other because their sum is zero, the identity element for addition. Thus, the numeral $-5$ can be read "negative five," "the opposite of five," or "the additive inverse of five."

This question involves two separate, but related issues; the first is discussed at an elementary level here. While the second, and more advanced, issue is discussed here. I also found this concerning use in elementary education.

I recently found an excellent historical/cultural perspective on What's so baffling about negative numbers? written by a Fields medalist.

I am fully comfortable with "minus $x$," and indeed like it better than "negative $x$," and have seldom used the latter in lectures.

There is no problem with the binary operator and the unary operator having the same name. Speaking and writing mathematics would be more awkward if we did not allow useful abus de langage.

In logic

the negation of a certain element in a set is all the other terms in the set

for the set     $\{1,2,3,4\}$

the negation of the element 2 is

$\neg 2=\{1,3,4\} $


Why parameterize systems of equations that contain "free variables"? As an example, let's say you have the system of equations given by

$$x+y+z = 1$$

$$2x+2y+2z = 2$$

$$3x+3y+3z=3$$

As the first few steps, we recognize that solutions to the first equation are solutions to all of them, so we parameterize the first, for instance by writing

$$x=r$$

$$y=s$$

$$z = 1-r-s$$

Point being to show that we can insert arbitrary values into the parameters $r, s$, and produce a $z$ which gives us a guaranteed solution to the system above.

Question

Why go through the step of parameterizing the "free variables"? It always seemed like a redundant step to me. I've perused at least 7 Linear Algebra books hoping that one would give a reason, but they all just tell us to do it.

Parametrizing a space with coordinates from a know space gives you a more concrete description of the solution space. It can tell you the dimension of the solution space, it can give you explicit descriptions of what the solutions look like, and it can facilitate checking whether a given point is a solution or not.

In this particular case there is indeed no need to introduce new variables $r$ and $s$. You can simply parametrize the space of all solutions $V$ as $(x,y,1-x-y)$ with $x,y\in\Bbb{R}$. But introducing these new variables comes from the view that the parametrization is in fact a bijection
$$f:\ \Bbb{R}^2\ \longrightarrow\ V:\ (r,s)\ \longrightarrow\ (r,s,1-r-s).$$
From this point of view the parameters $r$ and $s$ are coordinates in a new space, the parametrizing space $\Bbb{R}^2$, and hence they get new names.

Now it turns out that for systems of linear equations over a field you can always choose a parametrization in terms of the coordinate functions (here in terms of $x$ and $y$). In this case these new parameter names seem redundant (and perhaps they are?). But outside the scope of linear algebra these are not at all redundant. For example, it is impossible to parametrize the rational points on the circle in terms of any of its coordinates; here it is necessary to introduce a new parameter.

The set $\{(x,y,z)\in \Bbb R^3 \; : x+y+z=1\}$ is a plane whose cartesian equation is $x+y+z=1$ and whose parametric equations are
$x=x, y=y, z=1-x-y$

or

$x=a, y=b, z=1-a-b$.

The parametric equations give the two vectors director of the plane.
in your case, the vectors are
$$u=(1,0,-1) \text{ and } v=(0,1,-1)$$


Prove that $ AA^T=0\implies A = 0$ Let $A$ be an $n \times n$ matrix with real entries, where $n\geq2$.
  Let $AA^T = [b_{ij}] $, where $A^T $ is the transpose of $A$. If
  $b_{11} + b_{22 }+\cdots+ b_{nn} = 0$, show that $A = 0$.

From what I've gleaned so far, $AA^T$ is a symmetric matrix, and the diagonals are zero. I can't figure out how to solve this question. Is there some property that exists that I'm missing for handling this question?

Let

$$A=(a_{ij})\implies A^t=(a_{ji})\implies AA^t=(b_{ij})=\left(\sum_{k=1}^ma_{ik}a_{jk}\right)$$

so that

$$0=\sum_{i=1}^nb_{ii}=\sum_{i=1}^n\sum_{k=1}^na_{ik}a_{ik}$$

Complete the proof now.

I thought about this proof, but I think it is not totally correct:

Supose $A≠0$.

Then, for any $X$, $A+X≠X$.

Multiplying by $A^T$ at the right, we get:

$(A+X)A^T≠XA^T$

$AA^T+XA^T≠XA^T$

$XA^T≠XA^T$

Arriving in a contradiction, we get $A=0$.
I realise it may not be totally correct to multiply by $A^T$ without making some proper assumptions, but I know I can pick any $X$. Is there a way to formalise this proof?


If $\sum_{n=2}^{\infty} a_n$ is convergent, is $\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)$ convergent as well? Suppose $\sum_{n=2}^{\infty} a_n$ is a convergent series. Is the following series convergent as well?
$$\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)$$
The part with $n^{a_n}-1 $ looks quite similar to $\sqrt[n]{n}-1$ which behaves like $\ln{n}$, however, I'm not able to tie this fact to the behavior of the whole expression. Any suggestions?

We can rewrite $n^{a_n}$ as $\exp (a_n \ln n)$, and then

$$\sum_{n = 2}^{\infty} \frac{\sqrt{a_n}}{\ln n}\bigl(n^{a_n} - 1\bigr) = \sum_{n = 2}^{\infty} a_n^{3/2}\cdot \frac{\exp (a_n\ln n) - 1}{a_n\ln n}.\tag{1}$$

Now, assuming $a_n \geqslant 0$, everything is fine and dandy if $a_n\ln n$ remains bounded. Of course, if a series of nonnegative terms converges, the terms must on average decay much faster than $\frac{1}{\ln n}$, so it may look as though we can deduce convergence. But that's only on average, and for sparse enough indices, we can have arbitrarily large $a_n\ln n$. So large that the series in $(1)$ diverges. For $k \in \mathbb{N}$, let $n_k = \lceil \exp (4^k)\rceil$, and

$$a_n = \begin{cases} k^{-2} &, n = n_k \\ 2^{-n} &, n \notin \{ n_k : k \in \mathbb{N}\}.\end{cases}$$

Then it's easily seen that $\sum a_n$ converges, yet we have

$$(a_{n_k})^{3/2}\cdot \frac{\exp (a_{n_k}\ln n_k) - 1}{a_{n_k}\ln n_k} > k^{-3}\cdot \frac{a_{n_k} \ln n_k}{2} = \frac{1}{2} k^{-5} \ln n_k > \frac{4^k}{2k^5}$$

since $\frac{e^x-1}{x} > \frac{x}{2}$ for $x > 0$, and we see that the terms in $(1)$ are unbounded, whence $(1)$ diverges for this choice of $(a_n)$.

For large values of $n$,

$$n^{a_n}-1={1+a_n{\ln\, n}}-1 =a_n{\ln\, n},$$

so the given series becomes

$$\sum_{n=2}^{\infty}\frac{\sqrt{a_{n}}}{\ln\, n}(n^{a_{n}}-1)=\sum_{n=2}^{\infty}\frac{\sqrt{a_{n}}}{\ln\, n}(a_n){\ln\, n} = \sum_{n=2}^{\infty}{a_n}^\frac{3}{2},$$

which converges , if
$$\lim\limits_{n\to\infty} a_{n}{\ln\, n}=0
$$


Need explanation for a combinatorics riddle (full answer provided) 11 people in a certain company have access to a safe. The company
  owner wants any group of six people out of the 11 to open the safe,
  But no five-person group can open the safe itself. To achieve this
  goal he decided to put more than one lock on The safe, and give each
  person keys only to some of the locks.

  How many locks he has to put on the safe and how many keys each person
  will have to achieve his goal (the company owner wants to reduce the
  number as much as possible The locks, and as much as possible reduce
  the number of keys each person receives)?

Answer:

Each subgroup of 5 people will not be able to open the safe, so each subgroup should have a lock so that the members of the group do not have a key for it.

On the other hand, a key for the same lock is shared for all but 5 members of the subgroup. We achieved two goals in this: each sub-group of 5 people could not open the safe and any subset of 6 you can.

so we need $\binom{11}{5}$ Locks and $\binom{10}{5}$ keys

My question:
Can I get more elaboration on the answer?

Let’s rephrase this as a question about sets. Let $K$ be the set of all keys (that unlock one of the locks on the safe) and $K_i$ be the set of keys held by person $i$, $1\le i\le 11$. Then we have that $K_i\subset K$. The other conditions of the problem require that
$$K_u\cup K_v\cup K_w\cup K_x\cup K_y\cup K_z=K$$
for all distinct $u,v,w,x,y,z$, and
$$K_v\cup K_w\cup K_x\cup K_y\cup K_z \ne K$$
for all distinct $v,w,x,y,z$.

Solution: The pigeonhole principle will help us out a lot here.

If, for any particular $k\in K$, at least $6$ of the $11$ people hold the key (that is, $k\in K_i$ for exactly $5$ distinct values of $i$), then it is guaranteed by the pigeonhole principle that any group of $6$ people will contain at least one person holding $k$. Conversely, if there exists a key $k\in K$ such that fewer than $6$ of the $11$ people hold that key, then it will be possible to select $5$ people none of whom hold the key, making them unable to open the safe and violating a necessary condition. Therefore, we may conclude that at least $6$ people hold every key.

We may use similar reasoning to show that at most $6$ people hold every key $k\in K$.

Finally, we reach the conclusion that there is a one-to-one mapping between keys $k\in K$ and five-person groups (i.e. for every five-person group, there is exactly one key $k$ that is not held by any person in that group, and is held by all $6$ people not in that group). This entails that there must be $\binom{11}{5}$ keys, as desired.

However, I believe that $\binom{10}{5}$ is not the correct number of keys. Since each key is held by $6$ people as explained above, there should be $6\cdot \binom{11}{5}$ keys.

Each subgroup of 5 people will not be able to open the safe, so each subgroup should have a lock so that the members of the group do not have a key for it.

$11 \choose 5$ is the number of different five person subgroups in a population of eleven employees. You need this many locks to cover every possible instance of five people being unable to open the door. The same lock can't be used to lock out two different five person subgroups, else it would lock out at least six people.

  On the other hand, a key for the same lock is shared for all but 5 members of the subgroup.

Again, since we can't lock out a six person group, everyone not in a five person subgroup must have a key to the lock that locks out that group. Everyone gets the number of keys equal to the number of five person subgroups they’re not in. $10 \choose 5$ is the number of five person subgroups that don't include a particular person. From the ten people who arent person X, choose five of them.

  We achieved two goals in this: each sub-group of 5 people could not open the safe and any subset of 6 you can.

This solution satisfies the stated goal.

The text hints that this is the minimum solution without laying out the above logic. In any case, the answer is correct and the proposed solution is wildly impractical.


How to define the exponential function without calculus? For fun, I would like to define the complex exponential function from these two properties:

$\exp(0) = 1$
$\exp(z + w) = \exp(z) \exp(w)$

From here, I would like to find a way to compute values of $\exp(z)$, or at least to compute $\exp(1)$.

So far, I found only two ways:

Noting that $\exp'(z) = \exp(z)$ and solving the differential equation, which leads to $\int \frac{\exp'(z)}{\exp(z)} dz = \log(\exp(z)) + C = z$.
Noting that $\exp'(z) = \exp(z)$, computing its Taylor series and checking that what I get is an entire function.

The first approach is simply wrong because it involves logarithms, which I have not defined yet. The second approach looks much better. I haven't tried, but I guess I can find a way to manipulate the Taylor series to obtain the limit definition of $e$ and conclude that $\exp(1) = e$, which is my aim.

However, I'm struggling to find another way that does not involve differentiation or limits in general. I would be happy to find a way to say $\exp(1) = e$ without calculus. I think that the irrational nature of $e$ forces me to use limits -- am I right?

You won't be able to do derive

$exp(1)=e$

from your definition, since your definition works for the exponential function for any base $b$:

$b^0=1$

and

$b^{x+z}=b^x\cdot b^z$

are true for any $b$!

(An echo of previous answers)

If
$f(x+y)
= f(x)f(y)
$
and $f$ is differentiable,
then
$f(0) = 1$
(set $y=0$),
so
$\lim_{x \to 0} f(x)
= 1
$.

Switching to standard calculus variables,

$\begin{array}\\
f(x+h)-f(x)
&=f(x)f(h)-f(x)\\
&=f(x)(f(h)-1)\\
&=f(x)(f(h)-f(0))\\
\text{so}\\
\dfrac{f(x+h)-f(x)}{h}
&=f(x)\dfrac{f(h)-f(0)}{h}\\
\end{array}
$

Letting
$h \to 0$,
$f'(x) = f(x)f'(0)$.

$f(x)=a^x$ satisfies
$f(x+y)=f(x)f(y)$.
$a=e$ is special
because that makes
$f'(0) = 1$.

In general,
$(a^x)' = a^x\ln(a)
$.


What are all the homomorphisms between the rings $\mathbb{Z}_{18}$ and $\mathbb{Z}_{15}$? Any homomorphism $φ$ between the rings $\mathbb{Z}_{18}$ and $\mathbb{Z}_{15}$ is completely defined by $φ(1)$. So from

$$0 = φ(0) = φ(18) = φ(18 \cdot 1) = 18 \cdot φ(1) = 15 \cdot φ(1) + 3 \cdot φ(1) = 3 \cdot φ(1)$$

we get that $φ(1)$ is either $5$ or $10$. But how can I prove or disprove that these two are valid homomorphisms?

If one has a homomorphism of two rings $R, S$, and $R~$ has an identity, then the identity must be mapped to an idempotent element of $S$, because the equation $x^2=x$ is preserved under homomorphisms.  Now $5$ is not an idempotent element in $\Bbb Z_{15}$, so the map generated by $1 \to 5$ is not a homomorphism.

However, $10$ is an idempotent element of $\Bbb Z_{15}$. In particular, the subring $T \subset \Bbb Z_{15}$ generated by $10$ has unit $10$.  Since it is annihilated by $3$, and consequently by $18$, there is a unital homomorphism $\Bbb Z_{18} \to T$ (i.e., mapping $1$ to $10$).  So your second map is a legitimate homomorphism of rings (composing with the injection $T \to \Bbb Z_{15}$).

Basically, the point of this answer is to check that one of your maps preserves the relations of the two rings, while the other doesn't.

I will quote Wikipedia on the definition of a ring homomorphism.

  More precisely, if R and S are rings,
  then a ring homomorphism is a function
  f : R → S such that1

  f(a + b) = f(a) + f(b) for all a and b in R
  f(ab) = f(a) f(b) for all a and b in R
  f(1) = 1

The last requirement is being relaxed. We can then sub in our f, in this case f(a1)=5a or f(a1)=10a and see if these relations hold. Given that these are small, finite and well understood groups, the problem is easy to solve from here.


Find the maximum possible value Help me to find the maximum value of $T$ with $x, y, z \in \Bbb{R_+}$
$$T=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$$
Thanks :D


If $x≥0$ and $0≤x0$, then $x=0$? I tried to let ϵ=0.5x, but find ϵ might be zero in this way, which is contradict to for all ϵ>0. Is this statement false?

Prove it by contradiction.

Suppose $x \ge 0$ but $ x \ne 0$. So, $x > 0$

Then for $\epsilon = x/ 2 \gt 0$ we have $ x \not \lt \epsilon$.

You can either take the limit of $\epsilon \to 0$ or suppose that $x \not= 0$ and try to find a contradiction.


References for Algebraic number theory I am doing algebraic number theory first time. I have done all ring theory and field theory. I am interested in algebra , so also pretty much excited about algebraic number theory. I have a month's break before my semester begins. I want to self study some beginner's text covering my syllabus so that in class I can understand it much better, and may be it will turn up a research interest for me. Can somebody please suggest me some wonderful text for beginners in algebraic number theory, I do not want any number theory book using analytical methods. Strictly algebraic. Here are my contents...

Course Outline
Characteristic and minimal polynomial of an element relative to a
finite extension, Equivalent definitions of norm and trace, Algebraic
numbers, algebraic integers and their properties.
Integral bases, discriminant, Stickelberger’s theorem, Brille’s
theorem, description of integral basis of quadratic, cyclotomic and
special cubic fields.
Ideals in the ring of algebraic integers and their norm, factorization
of ideals into prime ideals, generalised Fermat’s theorem and Euler’s
theorem.
Dirichlet’s theorem on units, regulator of an algebraic number fields,
explicit computation of fundamental units in real quadratic fields.
Dedekind’s theorem for decomposion of rational primes in algebraic
number fields and its application, splitting of rational primes in
quadratic and cyclotomic fields.

Any book that covers all these with best conceptual approach. Thanks!
Anyways here are some suggested readings by my institute-
Saban Alaca and Kenneth Williams, Introductory Algebraic Number Theory, Cambridge University Press (2003).
M. Ram Murty and J. Esmonde Problems in Algebraic Numbers Theory, Springer-Verlag (2004).
Erich Hecke, Lectures on the Theory of Algebraic Numbers, Springer-Verlag (1981).
Paula Ribenboim, Algebraic Numbers, John Wiley & Sons (1972).
Harry Pollard and Harold Diamond, The Theory of Algebraic Numbers, Dover Publications (2010).
Which one I should go for among them,  or is there some other than these all, please suggest if any? Thanks!

Serge Lang's. Algebraic Number Theory is a good text also.


How can I find $\gcd(n^a-1,m^a-1)$? From Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ , we have $$\gcd(a^n-1,a^m-1)=a^{\gcd(n,m)}-1$$
for every positive integers $a,n,m$.

I reversed $a$ with $n,m$, and I had this question:

  Find $\gcd(n^a-1,m^a-1)$ for every positive integers $a,n,m$

My attempt is to find the greatest common divisor of $n^p-1$ and $m^p-1$ for every prime $p$ such that $p|a$, but I couldn't get any further.

How can I find the general formula of $\gcd(n^a-1,m^a-1)$ ?

(sorry for my grammar mistake, English is my second language)

Welcome. If n and m are as following forms:

$n=k b+1$ ⇒ $n≡1\mod b$⇒ $n^a≡1\mod b$

$m=t b+1$ ⇒ $m≡1 \mod b$⇒$m^a≡1\mod b$

Then the common divisor will be:

$gcd(n^a-1, m^a-1)=b$

In other words if $n-1$ and $m-1$ have a common divisor like b then $n^a-1$ and $m^a-1$ have a common divisor like b. This can also seen in following relations:

$n^a-1=(n-1)(1+n+n^2+ . . .+n^{a-1})$

$m^a-1=(m-1)(1+m+m^2+ . . .+m^{a-1})$

Finally the possible forms may be as follows:

$gcd(n^a-1, m^a-1)= gcd(n-1, m-1)$

Example: $n=4, m=7$ ⇒ $gcd(4^a-1, 7^a-1)=3$

$gcd(n^a-1, m^a-1)= gcd(n-1, 1+m+m^2+. . .+m^{a-1})$

Example: $n=32$, $m=2$, $a=5$ ⇒ gcd(32^5-1, 2^5-1)= 31

$gcd(n^a-1, m^a-1)= gcd(m-1, 1+n+n^2+. . . + n^{a-1})$

$gcd(n^a-1, m^a-1)= gcd[(1+m+m^2+. . .+m^{a-1}),(1+n+n^2+. . . + n^{a-1})]$

Example: $n=2$, $m=4$, $a=3$ ⇒ $gcd(2^3-1, 4^3-1)=7$


An example of a sequence that has no accumulation points For my homework is say to prove that you are correct and that you may quote theorems.
I am  unsure about what accumulation points are. thank you!!

The sequence $(-1)^n$ has not accumulation point Because the values are between -1 and 1 and it doesn't have limit.


Is there an easy way of finding the taylor series for $1/(1+x^2)$? I was trying to calculate the fourth derivative and then I just gave up.

Recall that a geometric series can be represented as a sum by

$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty}x^n \quad \quad|x| <1$$

Then we can simply manipulate our equation into that familiar format to get

$$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^{\infty}(-x^2)^{n} = \sum_{n=0}^{\infty}(-1)^n x^{2n}$$

Fun Alternative:

Note that $$\frac{d}{dx} \arctan x = \frac{1}{1+x^2}$$
and that the Taylor Series for $\arctan x$ is

$$\arctan x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$$

$$\implies \frac{d}{dx} \arctan x = \frac{d}{dx}\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$$

Interchange the sum and $\frac{d}{dx}$ (differentiate term-by-term) on the RHS to get

\begin{eqnarray*}
\frac{1}{1+x^2} &=& \sum_{n=0}^{\infty}\frac{d}{dx}(-1)^n\frac{x^{2n+1}}{2n+1}\\
&=& (-1)^n \frac{1}{2n+1}\sum_{n=0}^{\infty} \frac{d}{dx}x^{2n+1}\\
&=& (-1)^n \frac{1}{2n+1}\sum_{n=0}^{\infty} (2n+1)x^{2n}\\
&=& \sum_{n=0}^{\infty} (-1)^n\frac{(2n+1)x^{2n}}{2n+1}\\
\end{eqnarray*}

Cancel the $(2n+1)$ on the RHS to arrive at
$$\frac{1}{1+x^2}  = \sum_{n=0}^{\infty}(-1)^n x^{2n}$$


Finding the centralizers of $Q_8$ For $x\in G$, the centraliser in $G$ of $x$ is $C_G(x)=\{g\in G: gx=xg\}$. I'm trying to work out the centralisers in $Q_{8}=\langle a,b:a^4=1,b^2=a^2,ba=a^{-1}b\rangle$.

So to do this I am focusing on one element $x$ at a time comparing $gx$ to $xg$ for all $g$ in the group.

Elements of the group have the form $a^m$ and $a^mb$

So $C_G(1)=\{G\}$

To find $C_G(a):$

$a^ma=a^{m+1}=aa^m$ so hence $a^m$ is in the centraliser

but $a^mba=a^{-1}a^mb$ so $a^mb$ are not in the centraliser, hence $C_G(a)=\{1,a^m\}$

Next to find $C_G(a^2)$:

$a^ma^2=a^{m+2}=a^2a^m$ so hence $a^m$ is in the centraliser

and $a^mba^2=a^{-2}a^mb=a^2a^mb$ so $a^mb$ is in the centraliser = $\{1,a^m,a^mb\}=\{G\}$

For
$C_G(a^3)$ we have the same logic as $C_G(a)$ = $\{1,a^m\}$

Then finally for the elements of the form $a^kb$, nothing commutes with these? so $C_G(a^kb)=\{1\}$

So the center of the group is $\{1,a^2\}$

Is the logic of what I have done correct? Is there a better way of doing it?

First, recall that for a group $G$ and $g \in G$, the centralizer $C_{G}(g)$ is a subgroup of $G$. In the case of finite groups, this is especially helpful, as we can leverage Lagrange's Theorem. Note as well that $\langle g \rangle \leq C_{G}(g)$ (why is this true?).

So a good strategy is to find individual elements that commute with $g$. Once you know $a \in C_{G}(g)$, you immediately have $\langle a \rangle \leq C_{G}(g)$ as well. In this way, you can begin narrowing down candidates based on the lattice of subgroups of $G$.

Similarly, if you can find elements not in $C_{G}(g)$, you can narrow down upper bounds for $|C_{G}(g)|$. Examining the lattice of subgroups of $G$ will also be quite helpful here.

This is basically a question from text of abstract algebra by dummit and foote 4th edition.
Still you have used the definition though langrange's theorem is available to you as discussed by @ ml0150


If $G$ is solvable and $H\trianglelefteq G$ then $G/H$ is solvable? I'd like to show the following result of group theory:

Problem: Let $G$ be a solvable group and $H\trianglelefteq G$. Then $G/H$ is solvable.

Definition: A group $G$ is said to be solvable if there is a normal series of $G$ such that the factors are abelian, that is, there exists a sequence of normal subgroups:$$G=G_0\trianglerighteq G_1\trianglerighteq \ldots \trianglerighteq G_n=\{1\},$$ such that the factors $G_{i}/G_{i+1}$ are abelians.

Sketch: Let $$G=G_0\trianglerighteq G_1\trianglerighteq \ldots \trianglerighteq G_n=\{1\},$$

a normal series of $G$ such that $G_i/G_{i+1}$ are abelians. The first try would be to consider

$$\frac{G}{H}=\frac{G_0}{H}\trianglerighteq \frac{G_1}{H}\trianglerighteq \ldots \trianglerighteq \frac{G_n}{H}.$$

Of course this makes no sense for we don't even have $H\subset G_i$ for some $i$. But we can get subgroups of $G$ containing $H$ considering $HG_i$. Since $H\trianglelefteq G$ we have $HG_i\leq G$ for all $i$. So it is natural to consider the sequence: $$G=HG_0\geq HG_1\geq \ldots \geq HG_n=H\geq \{1\}.$$

It is not hard to verify this is a normal series of $G$. I conjecture the normal series for $G/H$ will arise from this but I still can't see how.

Can anyone help me finishing this proof?

Consider the series
$$
G/H=(HG_0)/H \ge (HG_1)/H \ge \dots\ge (HG_n)/H=\{1\}.
$$

Homomorphic image of a solvable group is solvable.
Consider  natural homomorphism from G to G/H.


Show that some of the root of the polynomial is not real. \begin{equation*}
p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_3x^3+x^2+x+1.
\end{equation*}

All the coefficients are real. Show that some of  the roots are not real.

I don't have any idea how to do this, I only have an intuition that something is related to the coefficients of $x^2$ and $x$ and the constant 1.

Kindly help me!!

Adapting the argument from this answer by Noam Elkies.

If $p(x)$ has only real zeros, then so does its reciprocal
$$
x^np(\frac1x)=x^n+x^{n-1}+x^{n-2}+\cdots+a_{n-1}x+a_n.
$$
But by Vieta relations the sum of the roots of this polynomial is $-1$ and the sum of their pairwise products is $1$. Therefore the sum of the squares of the roots is
$$(-1)^2-2\cdot1=-1<0.
$$
Consequently some of the roots must be non-real.

Counterexample: If $n=0$, then $p$ has no roots.


How make summation for a series which contains arbitrary elements I am studding a research paper in winch author presented a analytical model for set traversal and different cases of time complexity.
I am not understanding the one point in the model that is related to summation formulation equation n# 18.

$\frac{m_1 + 1}{2}t + \frac{l_2 + 1}{2}t + .... + \frac{l_n + 1}{2}t$

$ = \frac{1}{2} \bigr(m_1 + n + \sum_{a=l_2}^{l_n} a\bigr) t$

Why $l_2 + l_3 + l_4 + ..... + l_n = \sum_{a=l_2}^{l_n} a$

why its not $\sum_{a=2}^{n} l_a$

Your doubts are reasonable and your proposal is correct. The formula (18) in the referred paper and some more show all  the same kind of typos (or miscalculations).

  We have
  \begin{align*}
l_2 + l_3 + l_4 + ..... + l_n\color{blue}{=\sum_{a=2}^{n} l_a}
\end{align*}

  and not  $l_2 + l_3 + l_4 + ..... + l_n = \sum_{a=l_2}^{l_n} a$, since $$\sum_{a=l_2}^{l_n} a=\sum_{l_2\leq a\leq l_n}a$$

Sloppy notation.

The author intended $\sum\limits_{a \in \{l_2, l_3, .... , l_n\}} a$ which would be the same thing as $\sum_{a=2}^n l_a$.

However I'd argue that $\sum\limits_{a=l_2}^{l_n} a$ is not actually wrong per se.

The notation for $\sum$ is $\sum\limits_{\text{some condition}} term = $ "the sum of all the $terms$ for which that condition is true".

(A practical example of this would be $\sum\limits_{a|24;a\in \mathbb N} a   = $ the sum of all the divisors of $24=1 + 2 + 3 + 4+ 6 + 8 +12 + 24$.)

This means the familiar $\sum\limits_{a=2}^n l_a$ actually means $\sum\limits_{a\in \{2,3,4,...n\}} l_a$ and that $_{a=2}^n$ is just shorthand for $a\in \{2.....n\}$.

Except when were we ever actually TAUGHT that $_{a=2}^n$ is just shorthand for $a \in \{2.....n\}$?  Well, maybe you were but I never was.  I had to pick it up on the streets...

Anyway,  if $_{a=2}^n$ is proper shorthand for $a \in \{2,.... n\}$ then isn't it okay that $_{a=l_2}^{l_n}$ proper shorthand for $a \in \{l_2, .... l_n\}$?

Meh... maybe.  I admit it looks weird and we have to think but... is it any more weird than $_{a=2}^n$?

(That was rhetorical.)

(Anyway, welcome to learning mathematics on the street.)

====ADDENDUM====

In a comment in Marcus Scheurer's answer, I realize that $\sum\limits_{a=l_2}^{l_n} a$ could just as legitimately be interpretated as $l_2 + (l_2 + 1) + (l_2 + 3) + ..... +(l_2 + \lfloor l_n - l_2\rfloor)$.

It would take a bit bit of a bean-counter mind to interpret it that way if the individual $l_i$ values weren't specifically spelled out.

But that is a technical and correct interpretation.

So...  was the author wrong or right?  Arguments can be made both ways.

But it's what the author did.  And the author clearly meant it to be interpreted as $\sum\limits_{i=2}^n l_i$.

Those are the existential facts.  What we do with them is up to us.


Do Holder continuous functions preserve rate of uniform convergence sequence? Suppose $g$ is $\alpha$-Hölder continuous and $f_n$ converges uniformly to $f$ at rate $a_n$, so that

$a_n \sup_{x}|f_n(x) - f(x)| \to 0 $ as $n \to \infty$

Then does $g \circ f_n$ converges uniformly to $g\circ f$ at rate $a_n$?

The claim is false in general. To see it,
take $f_n(x) = x + \frac 1n$, $f(x) = x$, and set $a_n = n^{1/2}$ (for instance), let also $g(x)= x^{1/2}$, where $x\geq 0$.

We have that $g$ is Holder-$\frac 12$, but the claim in your question fails for these data.

Indeed, $f_n$ converges to $f$ uniformly on $\mathbb{R}$, in particular
$$
n^{1/2} \sup_{x \in [0,1]} |f_n(x) - f(x)| \to 0.
$$
On the other hand,
$$
n^{1/2} \sup_{x\in [0,1]} |g(f_n(x)) - g(f(x))| = n^{1/2} \sup_{x\in [0,1]} \left| \frac{ f_n(x) - f(x) }{\sqrt{f_n(x)} + \sqrt{f(x)}} \right| = \\
n^{1/2} \frac 1n \sup_{x\in [0,1]} \left| \frac{ 1 }{\sqrt{x+1/n} + \sqrt{x}} \right| \geq  (\text{ take } x=0) \\ n^{-1/2} \frac{1}{\sqrt{1/n}} = 1,
$$
so the convergence fails.

No.

We need $a_n \sup_x|f_n(x) - f(x)|^\alpha \to 0$

From the Hölder continuity of $g$, there are non-negative constant $C$, $\alpha$ such that

$ a_n|g(x) - g(y)| \le a_n  C |x - y |^{\alpha} $

For $\alpha > 1$, the function $g$ is constant and the convergence is automatic. For $\alpha < 1$, we have, from the uniform convergence of $f_n$ at rate $a_n$, that for any $\epsilon > 0$, $\alpha \le 1$, $x$, there exists a $N > N_*$ such that for $n  > N$

$a_nC|f_n(x) - f(x)|^\alpha <  \epsilon $

Which implies that for any $\epsilon > 0$ and $x$, we have a $n$ so that

$ a_n|g(f_n(x)) - g(f(x))| \le a_n  C |f_n(x) - f(x) |^{\alpha} < \epsilon$

which proves the result.


Is this of any real importance to the mathematical scientific community? I'm a 31 year old engineer, and I've recently came up with a way to exactly predict the probability of the number of prime numbers between two different integers.

For example using my way, the number of prime numbers between $0$ and $100$ is between $0$ and $50$. And it turns out that it is correct, since there are $25$ primes between $0$ and $100$.

But is this of any real importance that would lead me to publish a paper? Also my way is purely elementary and so I suspect that mathematicians would even bother to give it a look.

The method in question is noting that no primes other than 2 and 5 are divisible by 2 or 5, and thus any ten numbers $10n,10n+1,\ldots10n+9$ (really, any ten consecutive numbers) contain at most 4 primes, assuming the first term is larger than 5.

This can be improved by removing multiples of 3, so that no 30 consecutive numbers can have more than 8 primes (as long as the smallest is more than 5). It's an improvement since the basic method would give only a bound of 12 for such an interval. Of course this can be increased by throwing in new primes like 7.

Indeed, with more subtlety you can prove things about intervals smaller than the least common multiple of the primes used, and this has been done for intervals of lengths up to perhaps 3000. For larger intervals analytic methods have been used which restrict the number of primes which can appear.

This is not a new result, but the family of results is important! In particular, they were an ingredient in improving Zhang's theorem, a major step toward the twin prime conjecture. So what you have is just a piece of the puzzle, but it's a very large, intricate puzzle. Be glad you have seen part of it!

Yes, probabilistic primality testing and related gadgets are quite common. They assist encryption, for example, by providing candidate primes.

However, it's hard to say whether your particular way of doing it is worthy of being published, especially without seeing it. This is quite an extensively studied area.


If $A^2+A=0$,then $\lambda=1$ cannot be an eigenvalue of A. Prove the following statement:
If $A^2+A=0$,then $\lambda=1$ cannot be an eigenvalue of A.

I've been struggling on this question for a couple of hours and don't know how to approach it.

Let $\lambda$ be an eigenvalue of $A$ with eigenvector $v$.  What is $(A^2+A)v$?

I believe this may be a solution as well...

$A^2+A=0$

$A(A+I_n)=0$

$\therefore$ $A=[0]$  or  $A=-I_n$

Since $A=0$ and $A=-I_n$ are both diagonal matrices, the possible eigenvalues for $A$ are simply the entries of the diagonal.

$\therefore$ the possible eigenvalues of $A$ are:

$\lambda=0$, and $\lambda=-1$

Therefore, it is not possible for $\lambda=1$ to be an eigenvalue of $A$


Two dependent random variables with standard normal distribution and zero covariance I need to find two dependent random variables with standard normal distribution, but with zero covariance. It is easy too find just two dependent random variables with such a distribution (X and -X, for example), but how I can reach zero covariance?

Thanks in advance.

Try $X \sim N(0,1)$ and $Y=X$ when $|X|\lt k$ and $Y=-X$ when $|X|\ge k$ for some non-negative $k$.

Then $X$ and $Y$ have standard normal distributions and $cov(X,Y)$ is a continuous  increasing function of $k$, negative when $k$ is close to $0$ and positive when $k$ is large.  So for some $k$ you will have $cov(X,Y)=0$

Although I haven't worked through all aspects that must be checked, I believe that if you specify

$$[X \mid Y] = Y$$

i.e. that $X$ becomes $Y$ when conditioning on $Y$, you obtain

$$\text {Cov}(X,Y) = E(XY) = E[E(XY\mid Y)] = E[YE(X\mid Y)]= E[YE(Y)] =0$$

Note that
$$E(X) = E[E(X\mid Y)] = E[Y] = 0$$

and by the decomposition-of-variance formula

$$\text {Var}(X) = \text {Var}(E[X\mid Y]) + E[\text {Var}(X\mid Y)]=  \text {Var}(E[Y])+E[\text {Var}(Y)] = 0 + 1 = 1$$

i.e. this conditional specification is consistent with the unconditional moments.

So these variables are uncorrelated but not independent.


Volume of a parallelpiped from its sides and diagonals? If a parallelogram has sides of length $a$ and $b$, and diagonals of length $d$ and $e$, then we can find its area in the following way.  By the polarization identity, we have $a b \cos\theta = \frac{1}{4}|d^2-e^2|$.  Now the area is $a b \sin \theta$, which is equal to $\sqrt{a^2 b^2 - (a b \cos \theta)^2}$, or
$$\sqrt{a^2 b^2 - \frac{1}{16}(d^2-e^2)^2}.$$
This is not an exceptionally pretty formula, but maybe it's the best one can do.

How can such a formula be generalized to higher dimensions?  In particular, what is the volume of a parallelpiped, given its edge and diagonal lengths?

Let the parallelepiped with a vertex $O$ at the origin have vertices $A$, $B$, $C$. Write $D = B+C$; $E= C+A$; $F=A+B$; and $G=A+B+C$. Further, assign these names to the lengths of edges and diagonals:
$$a := |OA|\quad b := |OB| \quad c := |OC| \quad d := |AD| \quad e := |BE| \quad f := |CF| \quad g := |OG|$$

Note that a parallelepiped (and it properties) is completely determined by six quantities: the lengths of three edges meeting at a vertex, and the measures of the three angles between those edges. The total number of edge-lengths and diagonal-lengths is seven, so there's a dependency among these values, namely
$$4 \;(\; a^2 + b^2 + c^2\;) = d^2 + e^2 + f^2 + g^2$$
Although one diagonal (say, $g$) is unnecessary, it helps to simplify the volume formula a bit:

$$\begin{align}
32 V^2 &= 32 a^2 b^2 c^2 + d^2 e^2 f^2 + g^2 ( d^2 e^2 + e^2 f^2 + f^2 d^2 )\\
&-  2 g^2 ( a^2 d^2 + b^2 e^2 + c^2 f^2 ) -  2 ( a^2 e^2 f^2 + d^2 b^2 f^2 + d^2 e^2 c^2 )
\end{align}$$
In the case of a rectangular parallelepiped, for which $d^2=e^2=f^2=g^2=a^2+b^2+c^2$, the formula reduces to $V = a b c$, as expected.

I derived the formulas using coordinates
$$A = (a_x,0,0) \qquad B = (b_x, b_y, 0) \qquad C = (c_x, c_y, c_z)$$
Expressing the various distances in terms of these
$$a^2 = A\cdot A = a_x^2 \qquad b^2 = B\cdot B = b_x^2 + b_y^2 \qquad c^2 = C\cdot C = c_x^2 + c_y^2 + c_z^2$$
$$d^2 = \overrightarrow{AD}\cdot \overrightarrow{AD}=\dots \quad e^2 = \overrightarrow{BE}\cdot \overrightarrow{BE} =\dots \quad f^2 = \overrightarrow{CF}\cdot \overrightarrow{CF}=\dots$$
along with
$$V = a_x b_y c_z$$
I systematically eliminated $a_x$, $b_x$, $b_y$, $c_x$, $c_y$, $c_z$ from the system with the help of Mathematica's Resultant[] function. (Appropriate applications of the Law of Cosines should work just as well, but the Resultant process lets me crank through the equations without having to think too hard. :)

An $n$-dimensional parallelotope is determined by $\frac{1}{2}n(n+1)$ values. (These are the triangular numbers. You can see their relevance by noting the coordinatization done above considers vectors using $1$, $2$, $3$, ..., $n$ non-zero coordinates.) The figure has n characteristic edge-lengths and $2^{n-1}$ diagonals, so expressing volume in terms of these is certainly possible. With an increasing number of "extra" lengths, the resulting formula can take a variety of forms.

Let the  parallelpiped be defined by the vectors a, b, c.

Assume that its height is h wrt the base formed by a, b with θ as the angle between a and b. We further let n be unit normal vector wrt the base plane.

Notation:- * = common product (multiplication); x = cross product; . = scalar product

(a x b) . c

= (a * b * sin θ  n) . c

= (a*b*sin θ)* (n . c)

= (a*b*sin θ)* h

= volume of the parallelpiped


What is the operator "capital D" and how can the chain rule be used in this way I ought to know this but I was somehow always able to avoid mixing partials and $d$s

My book notes the following: (with some intermediate steps which are less important than the time taken to write)

Here $f$ is a function of $x(t),y(t),z(t),t$ and $u=\frac{dx}{dt}$ and "" for others.

$\frac{Df}{Dt}=\frac{\partial f}{\partial t}+(\mathbf{u}\cdot\nabla)f$

I was taught to use $D$ for total derivative but there's no malformed interpretation I could apply here.

What is going on and if you were to look it up in an index of a book what would the book be on, and what would you look up? (E.G: Real analysis, partial differentiation)

Also I'm not sure how to use bold with LaTeX nor get a special dot. If someone could tell me in the comments I'd be very grateful.

This is just the chain rule.  It may help to think of this as follows:

We have
$f = f(x(t),y(t),z(t),w(t))$ where it so happens that $w(t) = t$.  We then have
$$
\newcommand{\pwrt}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\dwrt}[2]{\frac{d #1}{d #2}}
D_tf = \pwrt fw \dwrt wt  +
\pwrt fx \dwrt xt +
\pwrt fy  \dwrt yt +
\pwrt fz  \dwrt zt =\\
\pwrt fw + (\mathbf u \cdot \nabla)f
$$
you may consider rewriting $\pwrt fw$ as $\pwrt ft$ a slight abuse of notation.

It's just the chain rule.

To be more precise, write $x=X(t)$, $y=Y(t)$, $z=Z(t)$ in order to distinguish the quantities $x$, $y$, $z$ from the functions $X$, $Y$, $Z$ which say how they depend on $t$. Now take your function $f(x,y,z,t)$ and define
$$
g(t) = f(X(t),Y(t),Z(t),t)
.
$$
Then the notation $Df/Dt$ (by definition) means exactly $g'(t)$, which can be computed using the chain rule:
$$
\begin{align}
g'(t) &= f'_x(X(t),Y(t),Z(t),t) \, X'(t) \\ &+ f'_y(X(t),Y(t),Z(t),t) \, Y'(t) \\ &+ f'_z(X(t),Y(t),Z(t),t) \, Z'(t) \\ &+ f'_t(X(t),Y(t),Z(t),t),
\end{align}
$$
which equals the given expression since $X'(t)=u$, etc.


A problem on interchange of limit and integration Suppose $\lim_{h\to 0}f_n(h) = 0$ such that $g(h)=\sum_{n=1}^{\infty}f_n(h)$ converges for any $h$. Can we tell that $\lim_{h\to0}g(h) = 0$ ? i.e can we change the order of limit and sum here ? If not what is needed to make it happen ?

I don't know how to use DCT/BCT here.

Is it true if $f_n(h) = \int_{E_n} |f(x+h) -f(x)| dx$ where $E_n$ is an interval of length $[nh,(n+1)h]$ where $f$ is bounded and integrable.

No: Consider $f_n(h)=h^{n-1}$ for $h\in(-1,1),n\in\mathbb N$. Then $f_n(h)\to0$ for $h\to0$ and any $n$, but
$$
g(h)=\sum_{n\in\mathbb N}f_n(h)=\sum_{n\in\mathbb N}h^{n-1}=\frac{1}{1-h},
$$
which does not tend to 0 as $h\to0$.
To interchange limits you need that the series converges uniformly

EDIT: This example doesnt work (see comment). But how about this:
Since $\mathbb Q\backslash\{0\}$ is countable, there is a bijective mapping $a:\mathbb N\to\mathbb Q\backslash\{0\}$. Now define $f_n(h)=0$ if $h\neq a(n)$ and $f_n(h)=1$ if $h=a(n)$. Then each $f_n$ is continuous at $0$, since $f_n$ is 0 everywhere except at $a(n)\neq0$. Since $a$ is bijective we have that $g(h)=1$ if $h\in\mathbb Q\backslash\{0\}$ and $g(h)=0$ otherwise. In particular, $g(1/n)=1$ for any $n\in\mathbb N$.


Solve differential equations containing $x''$ Solve the equation $$x'' + 3x' = 20e^{2t}$$ if $x(0) = 0$ and $x'(0) = 1$.

I am pretty sure $x$ must contain $e^{2t}$, but other than that, I'm not really sure how to proceed from here. Any help would really be appreciated!

Hints: Put $y=x'$. The homogeneous  equation becomes $y'+3y=0$ whose solution is $y=ae^{-3t}$. Integrate to get $x$. A particular solution of the type $ce^{2t}$ exists. Add the particular solution to the general solution of the homogeneous  equation and the apply the intial conditions.

Answer: $e^{-3t}-3+2e^{2t}$.

For your work: The solution is given by $$x(t)=-\frac{1}{3} c_1 e^{-3 t}+c_2+2 e^{2 t}$$


Is there an identity to combine a sum of more than two sines; eg $\sin(a)+\sin(b)+\sin(c)+\sin(d)$? I get these trigonometric product to sum formulas like:
$$\sin(a)+\sin(b)=2\sin\frac12(a+b)\cos\frac12(a-b)$$
And that's useful, but I'm not too sure what to do if I need to turn a product into a sum if there's more than two variables.
What would I do with something like this?
$$\sin(a)+\sin(b)+\sin(c)+\sin(d)$$

The following formula works:
$$
\begin{align}
\sin(a)+sin(b)+sin(c)+sin(d) = 4*\sin\left(\frac{a+b+c+d}{4}\right)*\cos\left(\frac{a-b+c-d}{4}\right) \\
*\cos\left(\frac{a+b-c-d}{4}\right)*\cos\left(\frac{a-b-c+d}{4}\right)- \\
4*\cos\left(\frac{a+b+c+d}{4}\right)*\sin\left(\frac{a-b+c-d}{4}\right) \\
*\sin\left(\frac{a+b-c-d}{4}\right)*\sin\left(\frac{a-b-c+d}{4}\right) \\
\end{align}
$$
I wrote this Python script in Google colab to "prove" it:
    import numpy as np

    def C(x):
      return np.cos(180/np.pi*x)

    def S(x):
      return np.sin(180/np.pi*x)

    a = 37
    b = 6
    c = 88
    d = 7

    x = S(a) + S(b) + S(c) + S(d)
    y1 = 4*S((a+b+c+d)/4.0)*C((a-b+c-d)/4.0)*C((a+b-c-d)/4.0)*C((a-b-c+d)/4.0)
    y2 = 4*C((a+b+c+d)/4.0)*S((a-b+c-d)/4.0)*S((a+b-c-d)/4.0)*S((a-b-c+d)/4.0)
    print(x)
    print(y1-y2)

When the input is $(a,b,c,d) = (37,6,88,7)$, with angles measured in degrees, the outputs for $x$ and $(y1-y2)$ agree to $11$ decimal places at $-1.02707706592$.  I also tried $(a,b,c,d) = (10,27,18,68)$, and the two sides of the equation again agreed to $11$ decimal places. That isn't a mathematical proof, but the probability of my proposed identity holding true by accident for randomly selected angles is essentially nil. I tried making one of the angles obtuse, and it still worked.
Steps in the derivation:

Use angle sum formulas to derive an expression for $\sin\left(\frac{A+B+C+D}{4}\right)$ in terms of $\sin\left(\frac{A}{4}\right)$, $\sin\left(\frac{B}{4}\right)$, $\sin\left(\frac{C}{4}\right)$, $\sin\left(\frac{D}{4}\right)$, $\cos\left(\frac{A}{4}\right)$, $\cos\left(\frac{B}{4}\right)$, $\cos\left(\frac{C}{4}\right)$ and $\cos\left(\frac{D}{4}\right)$.

Use the first expression, and the fact that sine and cosine are odd and even functions, respectively, to derive expressions for $\sin\left(\frac{A-B+C-D}{4}\right)$, $\sin\left(\frac{A+B-C-D}{4}\right)$ and $\sin\left(\frac{A-B-C+D}{4}\right)$.

Add the four expressions together.

Make the following substitutions:
$$
A = a+b+c+d \\
B = a-b+c-d \\
C = a+b-c-d \\
D = a-b-c+d
$$

...and you should get the formula.


Solving Matrix question AX=B to find the possible solutions of B Question

  Find $b_1$ and $b_2$ so that the equation $Ax = b$ has solutions where
  $$A =
\begin{bmatrix}
1 & 2 \\
0 & 1\\
-1 & 2
\end{bmatrix};\quad
b =
\begin{bmatrix}
b_1 \\
b_2 \\
0
\end{bmatrix}.$$
  Can this equation have a unique solution? Why or why not?

Work

$$\begin{bmatrix}1&2 \\0&1\\-1&2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}b_1\\b_2\\0\end{bmatrix}$$

$$\begin{bmatrix}
x_1 + 2x_2 \\
x_2 \\
-x_1 + 2x_2
\end{bmatrix} =
\begin{bmatrix}
b_1 \\
b_2 \\
0
\end{bmatrix}$$

$$\left.\begin{array}{ll}
x_1 + 2x_2 = b_1 \\
x_2 = b_2 \\
2x_2 = x_1
\end{array}\right\} \Rightarrow \left.\begin{array}{ll}x_1 = b_1/2 \\
x_2 = b_1/4 \end{array}\right\} \Rightarrow b_1/4 = b_2 \Rightarrow b_1 = 4b_2$$

Hint: the equation $Ax=b$ has solutions $ \iff rank(A|b)=rank(A)$.


What is the initial algebra of the identity functor? I am trying to solve some of the exercises in Bird's Algebra of Programming. He defines an initial algebra as the initial object in the category of some F-algebra.

In the book, he says that

data Nat = Zero | Succ Nat

is the initial algebra of the Maybe Functor, i.e, the functor F, where  $FA = 1 + A$ and $Ff = id_1 + f$.

Then he asks the following question as an exercise:

  What is the initial algebra of the Identity Functor?

I have two guesses for this:

My first guess, going by the pattern indicated for Nat, would be some bizzare object like this: data X = Tag X. But in this case, I cannot figure out what the homomorphism from X to an arbitrary A be.
My second guess is that the initial object is some kind of tagged union of all the types and the homomorphism is an appropriate projection.

Any help is appreciated

NOTE: copy-paste the sections in this answer to play with inside ghci

Note that the initial object can be represented in Haskell as as an inhabitant of the type Mu (as shown below).

{-# LANGUAGE NoImplicitPrelude #-}
{-# LANGUAGE EmptyCase #-}
import Prelude hiding (Maybe, Just, Nothing)

-- | Mu F = initial object with respect to functor F
data Mu f = Mu (f (Mu f))

I now  exhibit the ismorphism  between Nat and Mu Maybe below:

-- | Maybe
data Maybe a = Just a | Nothing deriving(Show)

-- | Naturals
data Nat = Zero | Succ Nat deriving(Show)

-- | Bijection between naturals and initial object of Maybe functor
natfwd :: Nat -> Mu (Maybe)
natfwd Zero = Mu Nothing
natfwd (Succ n) = Mu (Just (natfwd n))

natbwd :: Mu (Maybe) -> Nat
natbwd (Mu Nothing) =  Zero
natbwd (Mu (Just mu)) = Succ (natbwd mu)

Next, I show that the initial object generated by Id is isomorphic to Void --- that is, it is uninhabited.

Intuitively, the initial object contains "arbitrary many nestings" of the original functor F. In the Maybe case, we had the ability to use Nothing to terminate this nesting. However, in the case of Id, this is no longer the case: there is no "branch" in the type Id to break this arbitrary recursion. Hence, it cannot be inhabited by any inductive value.

Void is a way to declare values in haskell that cannot be inhabited, in case you have not seen it before. Void has no data constructors.

data Void

data Id a = Id a

isoFwd :: Mu Id -> Void
isoFwd (Mu (Id x)) = isoFwd x

isoBwd :: Void -> Mu Id
isoBwd v = case v of

Claim: isoFwd : Mu Id -> Void proves that Mu Id is not inhabited

Assume for contradiction that Mu Id is inhabited. Hence, there exists an x : Mu Id. Now, compute y = (isoFwd x) : Void. But we know that Void is uninhabited. Hence, contradiction. Therefore, Mu Id must be uninhabited.

Claim: isoFwd, isoBwd define an isomorphism between Void and Mu Id

To discuss equality of functions, we define:
\begin{align*}
\forall f, g : X \rightarrow Y, \quad
f ~\texttt{==}~g \equiv \forall x \in X, f (x) = g(x)
\end{align*}

Note that if $X$ is uninhabited, then the relation ${f~\texttt{==}~g}$ is vacuously true. Armed with this, we show that isoFwd. isoBwd = id = isoBwd . isoFwd.

isoFwd . isoBwd :: Void -> Void == id :: Void -> Void since Void is uninhabited, and hence the two LHS and RHS are vacuously equal.
isoBwd . isoFwd :: Mu Id -> Mu Id == id :: Mu Id -> Mu Id since Mu Id
is uninhabited, and hence the LHS and RHS are vacuously equal.

Further reading

For more of this kind of computational perspective, I can recommend reading about recursion schemes and cata/anamorphisms, from Bananas, Lenses, and Barbed Wire as well as the recursion-schemes library.

The initial algebra of the identity functor is a set $X$ together with a map $X\to X$ that is initial among all such maps : for any set $Y$ and map $Y\to Y$, there is a unique map $X\to Y$ making

$$\require{AMScd}\begin{CD} X @>>> X \\
@VVV @VVV \\
Y@>>> Y\end{CD}$$commute. In particular, for any $Y$ there is a unique map $X\to Y$ (taking $id_Y :Y\to Y$), so $X$ is initial.

Conversely, if $X$ is initial...


Parametrization of $x^2-y^2=1600$ While trying to compute the line integral along a path K on a function, I need to parametrize my path K in terms of a single variable, let's say this single variable will be $t$.
My path is defined by the following ensemble: $$K=\{(x,y)\in(0,\infty)\times[-42,42]|x^2-y^2=1600\}$$ I know how to calculate the line integral, that is not my issue. My problem is to parametrize $x^2-y^2=1600$. I tried using the identities: $$\sin^2(t)+\cos^2(t)=1$$ $$\sec^2(t)-\tan^2(t)=1$$
But I did not get anywhere with my parametrization (see below for my poor try into parametrizing). I would welcome any help/hints and if you happen to know some good reading to learn more about parametrization, I am also interested. $$r(t)=1600\sec^2(t)-1600\tan^2(t)=1600$$
for $$x=40\sec(t) \land y=40\tan(t)$$

I think that using trigonometric function is overcomplicating it in this case. You can let $y$ correspond to a parameter $t$, then, since $x$ is given to be positive, we can say that $x$ is the following positive root $$x = \sqrt{1600 + t^2}.$$ Your parameterised curve is subsequently given by: $$\left\{\left(\sqrt{1600 + t^2},t\right): t \in [-42,42]\right\}.$$

Letting $y = 40\sinh(t)$ is also an option, in which case the parameterisation is given by $$\left(40 \cosh(t), 40 \sinh(t) \right).$$ This perhaps looks more appealing although finding the correct bounds on $t$ now involves inverse hyperbolic functions, which I will leave up to you if you are willing to do it.

$$x^2-y^2=1600$$
$$(\frac{x}{40})^2 - (\frac{y}{40})^2=1$$

Let $x=40\cdot(e^t+e^-t)/2=40\cdot\cosh{t}$.

Let $y=40\cdot (e^t-e^{-t})/2=40 \cdot\sinh{t}$

Use positive values of t if you need positive values of $x$.

$d/dt(\cosh{t})=\sinh{t}$

$d/dt(\sinh{t})=\cosh{t}$


Dedekind cuts for $\pi$ and $e$ I tried to search in the internet about this but did not get any exciting answers. So my question is: How is construction of transcendental numbers like  $\pi$ and $e$ explained via Dedekind cuts?

Here is a simple description for $e$. The left set consists of all rationals $r$ such that
$$r\lt 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac{1}{n!}$$
for some $n$. This description is close in spirit to one of the many definitions of $e$.

One can give a similar description for $\pi$, though there is nothing as natural. We could use the following variant of the "Leibniz" series, using for the left set all rationals $r$ such that
$$r\lt 4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+ \cdots +\frac{4}{4n+1}-\frac{4}{4n+3}$$
for some $n$. Note that we stop with a "$-$" because we want to make sure we are below $\pi$.

The real numbers are really troublesome due to the infinities that enter the whole game. It is possible to show that $\sqrt2$ is a cut, because you can easily go to a rational by squaring. With transcendental numbers this is much more difficult, and actually impossible. You simply have to believe that transcendentals are good numbers. It is a choice so to say.

What is done above actually, is that $e$ and $\pi$ are represented by some finite rational interval that can be chosen arbitrary small. But rational intervals do not obey the field rules, since multiplication is not distributive over addition in an unambiguous way no matter how small you choose them. Also the interval that contains zero causes serious problems in elaborate calculations. Calculations based on intervals always can result in an interval that contains zero. Personally I would rather consider real values instead of real numbers, since there is no clarity whether real values are good numbers in general. It is still debated among mathematicians whether the reals are good numbers.


Difference between $\log n$ and $\log^2 n$ I'm researching the different execution time of various sorting algorithms and I've come across two with similar times, but I'm not sure if they are the same.

Is there a difference between $\log n$ and $\log^2 n$?

EDIT:

Follow up question: in terms of complexity , which would be faster, $O(\log n)$ or $O(\log^2 n)$? My guess would be the first one. (Note, this is not homework, I'm just trying to understand the difference between quicksort and bitonic sort on a hypercube topology.
)

O(log^2 N) is faster than O(log N) because of

O(log^2 N) = O(log N)^2

           = O(log N * log N)

Therefore Complexity of O(log^2 N) > O(log N).

Just take n as 2, 4, 16;

          O(log^2 N)         O(log N)

2  -->     1^2 = 1               1
4  -->     2^2 = 4               2
16 -->     4^2 = 16              4


One-one correspondence between maximal ideals and maximal elements. Let $S$ be a multiplicatively closed set, i.e. if $a\in S$ and $b\in S$, we must have $ab\in S$. Then $R_{S}=\left\{\dfrac{r}{s}|r\in R, s\in S\right\}$, where $R$ is a commutative ring containing $S$. $R_{S}$ can be thought of as a subring of quotient field of $R$. Now we want to show that there is a one-one correspondence between totality of maximal ideals of $R_{S}$ and maximal elements $A=\{\alpha|\alpha$ is an ideal of R and $\alpha\cap S=\emptyset\}$ with respect to set inclusion.

First of all, we can establish a homomorphism $\phi : R\rightarrow R_{S}$ where $\phi(r)=\dfrac{rs}{s}$. Now I for every proper ideal $I$ in $R_{S}$, we have $1\notin I$. So $I$ can't contain any element like $\dfrac{rs}{s}$ where $r\in S$. $\phi^{-1}(I)$ is an element in $A$. Now we have to prove that if $I$ is maximal ideal, then $\phi^{-1}(I)$ is a maximal element. Suppose $\phi^{-1}(I)$ is not a maximal element, we have $\phi^{-1}(I)\subset L$ where $L$ is another element in $A$. Now $<\phi(L)>$ is an ideal in $R_{S}$ generated by $\phi(L)$. I don't know how to show that this is a proper ideal.

Conversely, for every maximal element $L$ in $R$, we have to show $<\phi(L)>$ is a maximal ideal in $R_{S}$. Suppose $<\phi(L)>$ is not maximal ideal, we have $<\phi(L)>\subsetneqq M$ where $M$ is maximal ideal in $R_{S}$. Then if the statement in the first paragraph is true, then we have $\phi^{-1}(M)=L$. How can we accurately prove that this is a contradiction.

Appreciate it in advance.

We know the prime ideals of $R_s$ are 1:1 correspondence to the prime ideal of the prime ideals of $R$ which has no elements of $S$.So what you said before may be trivial.


Showing that $x^2\cos^2(\pi/(x^2))$ is not of bounded variation on $[0,1]$ Problem: Let $f(x) = x^2\cos^2(\pi/(x^2))$.  Prove that $f$ is not of bounded variation on $[0,1]$ even though $f$ is differentiable on
  all of $\mathbb{R}$.

Visual Intuition:

The following is a visual of $f(x)$ that provides intuition for how $f$ could not be of bounded variation on $[0,1]$.  Clearly our attention should be on the behavior of $f(x)$ when $x$ is near $0$.

Attempt:

Recall that $T_0^1(f)$ denotes the total variation from $0$ to $1$ on $f$ and is defined as follows:

$$
T_0^1(f) = \sup \left\{\sum_{k=1}^n \left| f(x_i) - f(x_{i-1}) \right|\right\} \text{ across all subdivisions of $[0,1]$}
$$
My idea is to show that

$$
T_0^1(f) \ge \sum_{n=1}^{\infty} T_{1 \over n+1}^{1 \over n}(f) = \infty
$$

But I'm not sure how to execute on this.

$\fbox{EDIT: New proof using fgp's comments:}$

We have$$
  f(x) = x^2\cos^2\left(\tfrac{\pi}{x^2}\right) \text{.}
$$
So let $x_k = \sqrt{\frac{1}{k}}$. Then $f(x_k) = \frac{1}{k}\cos(\pi k)$, which means  $$
  f(x_k) = \begin{cases}
    \frac{1}{k} &\text{if $k$ is even (because then $\cos \pi k = 1$)} \\
    -\frac{1}{k} &\text{if $k$ is odd  (because then $\cos \pi k = -1$).} \\
  \end{cases}
$$
Thus we have

$$
  T_0^1(f) \geq \sum_{n=1}^\infty |f(x_{2n}) - f(x_{2n+1})| = \sum_{n=1}^\infty \frac{1}{2n} + \frac{1}{2(n+1)} \geq {1 \over 2} \sum_{n=1}^\infty \frac{1}{n} = \infty \text{.}
$$

which completes the proof.

As a first step, let's look at $$
  f(x) = x\cos\left(\tfrac{a}{x}\right) \text{.}
$$
Let $x_k = \frac{a}{k\pi}$. Then $f(x_i) = \frac{a}{k\pi}\cos(\pi k)$, which means $$
  f(x_k) = \begin{cases}
    \frac{a}{k\pi} &\text{if $k$ is even (because then $\cos \pi k = 1$)} \\
    -\frac{a}{k\pi} &\text{if $k$ is odd  (because then $\cos \pi k = -1$).} \\
  \end{cases}
$$
Thus (assuming $a\pi \geq 1$, if that isn't the case, just start the sum at some other index. That won't change anything, it'll still diverge)$$
  T_0^1(f) \geq \sum_{n=1}^\infty |f(x_{2n}) - f(x_{2n+1})| = \sum_{n=1}^\infty \frac{a}{2n\pi} + \frac{a}{2(n+1)\pi} \geq \frac{a}{2\pi} \sum_{n=1}^\infty \frac{1}{n} = \infty \text{.}
$$

For $x^2\cos^2\left(\frac{\pi}{x^2}\right)$, just expand $\cos^2(x)$ (it's something like $\frac{1}{2} + \frac{1}{2}\cos(2x)$ or so), and then substitute $u=x^2$.

Just compute the derivative and integrate its absolute value and you'll be able to proof that it (the function) is unbounded.
The latter is easiest done by substituting z=1/x^2 and using a simple estimated lower bound for the contribution of each interval between two integers.


Sum of series $\sin x + \sin 2x + \sin 3x + \cdots $ Please help me compute the sum of the series:
 $$\sin(x)+\sin(2x)+\sin(3x)+\cdots$$

The series does not converge for all $x$. There are some $x$, for instance $x=0$ or $x=\pi$, for which the series converges to $0$, however if we consider $x=\frac \pi 2$ we find that our series is $1+0+-1+0+1+\cdots$ which does not converge.

If you think of the unit circle, imagine a line whose angle from the positive x-axis is the value of $x$. Then the x-coordinate of this point on the unit circle is the value of $\sin x $. Doubling the angle yields a point whose x-coordinate is $\sin 2x$. Tripling it yields a point whose x-coordinate is $\sin 3x$. Continuing this, you can see that the pattern will continue with varied positive and negative values, not approaching any particular limit unless the line representing an angle of $x$ was aligned with the positive or negative x-axis. (This is not rigorous, but can be made to be so.)

More technically, we have that $\lim_{n\to\infty} \sin nx =0$ iff $x=k\pi$ for some $k\in\Bbb Z$, so the series trivially converges to zero for such $x$ and diverges for all other $x$.

There is much more easier way to prove that the series does not converge when $x \ne k\pi$.

$\liminf\limits_{n\to\infty}\sin nx=-1$

The corresponding subsequence is $n_k=-\frac{\pi -4\pi k}{2x}$

$\limsup\limits_{n\to\infty}\sin nx=1$

The corresponding subsequence is $n_k=\frac{\pi + 4\pi k}{2x}$

Therefore there is no limit of $\sin nx$ and hence a necessary condition of convergence is violated.


If $X,Y\subset\mathbb{R}$ are not measure zero sets, how can I show that $X\times Y\subset \mathbb{R}^2$ is not a measure zero set too? If $X,Y\subset\mathbb{R}$ are not measure zero sets, how can I show  that $X\times Y\subset \mathbb{R}^2$ is not a measure zero set too?

or (the following is an easy case)

If $X\subset\mathbb{R}$ is not a measure zero set, how can I show that $X\times\mathbb{R} \subset \mathbb{R}^2$ is not a measure zero set too?

How can I show the assertion by using the definition of measure zero (as follows)?

(A subset $Z\subset \mathbb{R}$ is a mesuare zero set if $\forall\varepsilon>0$, $\exists$ countable open intervals $I_1, I_2, \cdots$
s. t. $Z\subset \cup_k I_k$ and $\sum_{k}length[I_k]<\varepsilon$).

We want to prove that if $X$ and $Y$ are Borel measurable sets and have nonzero measure, then $X \times Y$ has nonzero measure.

Take any set $X$ of nonzero measure and now ask yourself what sets $Y$ behave nicely.

Of course if $Y$ is an interval your approach gives you the desired result. Suppose that $\lambda(X \times Y) = 0$ and choose intervals $I_k, J_k$ such that $$X \times Y \subset \bigcup_k I_k \times J_k$$ where the latter union is disjoint and it holds that: $\sum \lambda(I_k \times J_k) \le \epsilon.$

Without loss of generality you can take all $J_k = Y$ and you would get $\lambda(X)  \le \sum \lambda(I_k) \le \epsilon / \lambda(Y).$ So since $Y$ has nonzero measure you would get that $X$ has zero measure. This is a contradiction.

Now you can build the class of sets that behave nicely, i.e the class $\mathcal{M}$ of all the sets $Y$ such that if $X \times Y$ has measure zero than $Y$ must have measure zero: $$\mathcal{M} = \{ Y \subset \mathbb{R} \text{ measurable, s.t. either } \lambda(Y) = 0 \text{ or } \lambda(X \times Y) > 0\}$$

This class is closed under countable monotone unions and intersections and contains intervals. Hence it is a monotone class which contains intervals (which in turn form a $\pi-$system). Hence by the monotone class theorem it contains all Borel sets on the real line.

If $E\subset\mathbb{R}^{p+q}\quad$and$\quad mE=0$then $mE_x=0$a.e. in  $\mathbb{R}^p$
where$ E_x=\{y:y\in\mathbb{R}^p (x,y)\in E\}$


Does $\pi$ satisfy the law of the iterated logarithm? It is widely conjectured that $\pi$ is normal in base $2$.

  But what about the law of the iterated logarithm?

Namely, if $x_n$ is the $n$th binary digit of $\pi$, does it seem likely (from computer experiments for example) that the following holds? $$\limsup_{n\rightarrow\infty} \frac{S_n }{\sqrt{n\log\log n}}=\sqrt{2}\quad\text{where}\quad S_n=2(x_1 + \ldots + x_n) - n$$

  What about other (conjectured) normal numbers like $e$ and $\sqrt{2}$?

I am sorry if this is too easy, but I tried to search for it and I could not find in on the Internet.  I suppose I could run an experiment myself, but I assumed this is well known, and I would need to brush up on my programming skills to do so...

Update 8/9/2013:

I found a website with the first 32,000 binary digits of $\pi$ and (using a spreadsheet program) graphed out the average of the bits $S_n/n$, comparing it to $\sqrt{\frac{2 \log \log n}{n}}$.  The results were inconclusive.  The average never got close to $\sqrt{\frac{2 \log \log n}{n}}$ (except at the very beginning when it was way past it). However, I had the same result with a source of randomness (the one built into the spreadsheet program).  My conclusion is that 32,000 bits is not enough to see if the law of the iterated logarithm (experimentally) holds for $\pi$. (The picture in the Wikipedia article uses at least $10^{50}$ bits, and the pattern is clear at about $10^{12}$ bits.  However, I don't know where to get even 1,000,000 binary digits of $\pi$ on the Internet.

[End Update]

Also, I am sorry that I really don't know how to properly tag this.

I favourite'd this two years ago and recently stumbled upon it again. Here's some code I wrote to answer this numerically.

The dotted line is $\sqrt{2 \log\log n/n}$
The solid line is $S_n/n$
$S_n = 2 \sum_{k=1}^n x_k - n$ where $x_k$ is the k-th decimal digit of Pi in binary.

You should be able to zoom in by running the code. You'll need matplotlib and numpy. I used y-cruncher to generate the digits.

I am not sure if this is the answer you are looking for (or if you want something more computational specifically for $\pi$), but if a number is normal in base 2, then the sum $S_n$ that you defined above will satisfy the law of the iterated logarithm.

First, let $X_i$ be the $i$th binary digit.  Assuming normality, the digits $X_i$ are iid and $X_i\sim\text{Bernoulli}(1/2)$.  Then, let $Y_i = 2X_i-1$.  Consequently, $\mathrm{EY_i=0}$, $\text{Var}(Y_i)=1$, and $S_n = Y_1+\ldots+Y_n$.  Consequently,
$$
  \mathrm{P}\left(
    \limsup_{n\rightarrow\infty} \frac{S_n}{\sqrt{2n\log\log n}}=1
  \right)=1.
$$
This result and an accompanying proof, which relies on the result for Brownian motion and Skorokhod embedding, can be found in Section I.16 of Rogers and Williams' Diffusions, Markov Processes and Martingales: Volume 1.

In general, if a number is normal in base $b$, then the digits $X_i$ have a discrete uniform distribution on $0,1,2,\ldots,b-1$.  Hence, $\mathrm{E}X_i=\frac{1}{2}(b-1)$ and $\text{Var}(X_i) = \frac{1}{12}(b^2-1)$.  Therefore, setting
$$
  Y_i = \frac{2\sqrt{3}\left(X_i-\frac{1}{2}(b-1)\right)}{\sqrt{b^2-1}},
$$
gives $Y_i$ zero mean and unit variance.  The sum of the $Y_i$
$$
  S_n = \sum_{i=1}^n Y_i = \frac{2\sqrt{3}}{\sqrt{b^2-1}}\sum_{i=1}^n X_i -n\sqrt{3}\sqrt{\frac{b-1}{b+1}},
$$
once again satisfies the law of the iterated logarithm.


Find the homogeneous linear DE with constant coefficients of least order that has $y = 1 + 2e^{-2x}\cos x$ as a solution. Find the homogeneous linear DE with constant coefficients of least order that
has
$$y = 1 + 2e^{-2x}\cos x$$

as a solution.

Your solution function has terms for the eigenvalues $0,-2\pm i$ without polynomial factors for multiplicities, so that your DE can be read off as
$$
(D-0)(D+2+i)(D+2-i)y=0
$$

The general solution of a first order homogeneous DE of the form
$$ y'(x) + p(x)y = q(x)\tag{1}$$
is given by
$$ y(x) = \frac{1}{\mu (x)}\int \mu(x) q(x)\tag{2},$$
where $\mu(x) = e^{\int p(x) \ dx}$ is the integrating factor.
\begin{align}
1+2e^{-2x} \cos(x) &= \mu^{-1} (x) \int \mu(x) (0) \ dx = \frac{C}{\mu(x)} \\ \implies& \mu(x) = \frac{C}{1+2e^{-2x}\cos(x)}.
\end{align}
Therefore,
\begin{align}
 \mu(x) = e^{\int p(x) \ dx} \implies \frac{C}{1+2e^{-2x}\cos(x)} &= e^{\int p(x) \ dx}\\
 \ln\biggl |\frac{C}{1+2e^{-2x}\cos(x)} \biggr| &= \int p(x) \ dx\\
\frac{d}{dt}\ln\biggl |\frac{C}{1+2e^{-2x}\cos(x)} \biggr| &= p(x) \\
\therefore \frac{2\sin(x)+4\cos(x)}{e^{2x}+2\cos(x)} &= p(x).
\end{align}
Finally , using $(1)$, the initial differential equation is
$$y' + \left(\frac{2\sin(x)+4\cos(x)}{e^{2x}+2\cos(x)} \right)y = 0$$


Classify the critical point $x=2$ of $f(x)=(x-2)^{17}\left(x+5\right)^{24}$ Question

  Let $f(x)=(x-2)^{17}(x+5)^{24}$. Then is $x=2$ a maximum, minimum, or neither?

Book's Approach

  $f'''\left(2\right)\ne 0$. S
  Since the odd integral derivative of the function is non-zero, $x=2$ is neither a minimum nor maximum

Please Explain

How did they conclude odd integral derivative
of the function is non-zero?

$f'''(2)\ne 0$ it is $2^{nd}$derivative. I don't how they did it.

The hint:

For $x>2$ we have $f(x)>0$ and for $x<2$ around $2$ we have $f(x)<0$.

Just a typo in the book. Should say: $f''(2)=0$, therefore neither maximum, nor minimum.

   f"(x) < 0 - local maximum
   f"(x) > 0 - local minimum
   f"(x) = 0 (your case)- neither of those


Usage of $\Rightarrow$ – Implication or type declaration? I have another notational question about the usage of $\Rightarrow$. This symbol is usually understood to indicate "material implication" (that is, $A\Rightarrow B$ means that the truth value of $B$ is at least as big as $A$, there hasn't to be a causal link between the two statements $A, B$). $A\Rightarrow B$ is often prononounced "If $A$, then $B$", and if there are free variables involved, then these are implicitly understood to be universally quantified.

Now, I noticed that "If $A$, then $B$" is not always a material implication in mathematics. It could as well mean a type declaration (or universal quantification): The statement "If $f$ is a function $A\to B$ and $A$ a nonempty set, then: blablabla" shouldn't be understood to mean "For any $f$, if $f$ happens to be a function with nonempty domain, then blablabla". Also, "For all real numbers x, ..." shouldn't mean "for any object x, if x is a real number, then ..."

Question: When "If $A$, then $B$" is used in such a context where it shouldn't be understood as a material implication, but as a type declaration (or universal quantification), can this also be abbreviated by the $\Rightarrow$ sign?

For example, I think I've seen in the definition of the term "first-order formula"
$$\text{$\phi$, $\psi$ formulae $\Rightarrow$ $\phi\land\psi$ formula}$$

But on the other hand, it would seem strange to abbreviate the statement "For all natural numbers $n$, we have $n^0 = 1$" by

$$n\in\mathbb N\Rightarrow n^0 = 1.$$

I am going to give two technically grounded arguments.  One contradicting what you state, and one agreeing with what you state.  They come to roughly the same conclusion (for different reasons).

First, the set theory view.  In (material) set theory, e.g. ZFC, all the things you say "shouldn't be understood to mean" or "seem strange" are completely legitimate.  While $f : A \to B$ is not usually a primitive of set theory (but it is in some), it is completely reasonable to define it as $$f\text{ is a function }\land \pi_1(f) = A \land \pi_2(f) \subset B$$ where "is a function" is an abbreviation for the set theoretic definition of a function as a binary relation satisfying certain properties and $\pi_1$, $\pi_2$ are projections lifted to sets.  The upshot is: it very much can be viewed as stating "filter $f$ to the things that satisfy this property".  The next example is even simpler. $\forall x \in \mathbb{R}.P(x)$ is often (in set theory) defined to mean $\forall x.x\in \mathbb{R} \Rightarrow P(x)$.  This also addresses your last example: $\forall n.n \in \mathbb{N} \Rightarrow n^0 = 1$ is a completely legitimate formal term of ZFC (given suitable standard definitions).  So this is not strange, this is what set theorists do all the time.

Now, a type theorist says all the above was drivel.  In type theory a statement like $$n :\mathbb{N} \Rightarrow n^0 = 1$$ is simply syntactically meaningless.  That is, it's not even a well-formed formula even in a context where $n$ is bound.  A type declaration like $n : \mathbb{N}$ is not a proposition that can be true or false.  In type theory, there is no alternative to $\forall x:\mathbb{R}.P(x)$.  To a type theorist $(x \mapsto x^2) : \mathbb{R} \to \mathbb{R}$ and $(x \mapsto x^2) : \mathbb{R} \to \mathbb{R}^+$ are just totally different things even though they are literally identical as set theoretic functions. But in (dependent) type theory there is the dependent product type often written $\Pi x:T.E(x)$ for some type $T$ and term $E$.  In (particularly constructive) type theory it is common for function types, universal quantification, and implication all to be special cases of $\Pi$-types.  In most implementations of type theory, $A \to B$ is defined to be $\Pi x:A.B$ (where $x$ is not free in $B$, i.e. we just ignore $x$).  In NuPRL, $\forall$ and $\Rightarrow$ are simply $\Pi$ and $\to$ at the type Prop. (NuPRL uses the syntax $x:A \to B(x)$ for $\Pi x:A.B(x)$.)

To summarize, in (material) set theory, the cases you are worried about are fine; a lot of high level syntax expands to the very constructs you are worried about.  That said, implication and universal quantification and functions, while having interrelations are quite distinct concepts.  In type theory, all of these concepts are literally special cases of the same concept.

$A\Rightarrow B$ is often prononounced "If $A$, then $B$", and if there are free variables involved, then these are implicitly understood to be universally quantified.

Free variables are universally quantified over their entire deduction, not the formula they appear in.

  Now, I noticed that "If $A$, then $B$" is not always a material implication in mathematics.

Yes it is.

  It could as well mean a type declaration (or universal quantification):

No it can't.

  The statement "If $f$ is a function $A \to B$ and $A$ a nonempty set, then: blablabla" shouldn't be understood to mean "For any $f$, if $f$ happens to be a function with nonempty domain, then blablabla".

Of course they aren't the same, you left off the restriction about the domain and codomain of $f$.

  Also, "For all real numbers x, ..." shouldn't mean "for any object x, if x is a real number, then ..."

Yes it does mean that.  Who told you it doesn't?  Don't listen to that person.

  Question: *When "If $A$, then $B$" is used in such a context where it shouldn't be understood as a material implication, but as a type declaration (or universal quantification),

There is no such situation.

  But on the other hand, it would seem strange to abbreviate the statement "For all natural numbers $n$, we have $n^0 = 1$" by
  $$n\in\mathbb N\Rightarrow n^0 = 1.$$

Those are equivalent statements as long as no where else is anything assumed about $n$.

There is a very surprising isomorphism between $\Rightarrow$ in logic and $\rightarrow$ in set theory.  But they are not the same thing.  $A \rightarrow B$ is defined in set theory to be the set of all function with a domain of $A$ and a codomain of $B$, and is sometimes written $B^A$.  For example:

$$\{x, y\} \rightarrow \{1, 2\} = \{ \{(x, 1), (y, 1)\}, \{(x, 1), (y, 2)\}, \{(x, 2), (y, 1)\}, \{(x, 2), (y, 2)\}\}$$


distributive law in polish notation On page 18 "Logic as Algebra" Halmos&Givant wrote the distributive law in Polish notation as
$$ = \times a + bc + \times ab \times ac  $$
I fail to see anything remarkable here, is there a combinatorial pattern that I'm missing?

I'm not quite sure what you're looking for, but here's a bit about the distributive laws.

The polish notation: $=\times a+bc+\times ab \times ac$ in standard infix notation is: $a \times (b+c) = a \times b + a \times c$.

What this means is one gets the same result if one multiplies then adds or adds then multiplies.

Using abstract algebra terminology, the distributive laws say that multiplication operators are group homomorphisms addition preserving maps. An group homomorphism (using additive notation) addition preserving map is a function such that $\varphi(b+c)=\varphi(b)+\varphi(c)$ (you can add then map or map then add). So in polish notation this is: $= \varphi + b \; c + \varphi \; b \; \varphi \; c$. This is the general pattern you're looking for (I guess).

This property is capturing a kind of commutativity: It doesn't matter which is done first: map or add. Or in your context, it doesn't matter which is done first: add or multiply.

I'd say that Polish notation (or reverse Polish notation) makes it clearer that the "the outer operation moves to the inside and doubles, and the inner operation moves to the outside" when applying distributivity.  Syntactically speaking, expressions in Polish notation (and reverse Polish notation) come as almost always clearer than in ubiquitous infix notation.

If you consider ax(b+c)=ab+ac, or a(b+c)=(axb)+(axc) and most other infix notation expressions you have to consider more than just the syntax of the expressions involved (such as what the author intends to say, which he may have stated earlier in the text, however, this is not part of the syntax of expressions).  In other words, if you interpret either ax(b+c)=ab+ac or a(b+c)=(axb)+(axc) purely in terms of what your formation rules (rules for what well-formed formulas, or formulas) say exactly, they'll come out as nonsensical, or at best as ambiguous.  On the other hand in Polish notation, so long as you know the arity of the operations, predicates/relations (for "=") something like =×a+bc+×ab×ac comes as clear from basically the formation rules.

For instance, with this condensed formation rule

If "p" and "q" all represent well-defined values, then +pq, ×pq, and =pq each represent well-defined values.

and this axiom:

"a", "b", and "c" represent well-defined values,

then you can formally prove =×a+bc+×ab×ac as well-defined also as follows:

1 +bc by rule 1, and axiom 2
2 ×ab by rule 1, and axiom 2
3 ×ac by rule 1, and axiom 2
4 +×ab×ac by steps 2 and 3, and rule 1
5 ×a+bc by axiom 2, step 1, and rule 1
6 =×a+bc+×ab×ac by steps 3 and 4 and rule 1

You have to fully parenthesize an infix expression in order to do something like this.