Difference equation with the annihilator method $y(n+2)-4y(n)=2^{n+3}-1$, $n\geq 0$ I want to find the general solution of the difference equation $y(n+2)-4y(n)=2^{n+3}-1$, $n\geq 0$ with the annihilator method. I can rewrite the left-hand side as $L(y)(n)$ where $L=\tau^2-4$ and $\tau$ is the backshift operator. But I do not know how to rewrite the right-hand side as a solution of some difference equation. Could someone help me out? Thanks in advance!

The right-hand side (RHS) is the linear superposition of $c_1 2^n$ and $c_2$ for constants $c_1$ and $c_2$. The annihilator for the RHS is $(\tau -2)\tau$, where $\tau$ is the timeshift operator. Like you said, the annihilator for the LHS (if the difference equation was homogenous) is $(\tau^2 - 4)$. Knowing that, we can make the original equation homogenous by multiplying the difference equation by the annihilator of the RHS: $(\tau-2)^2(\tau+2)\tau = 0$. The complete solution is $y(n) = a2^n + bn2^n + c(-2)^n + d$. The annihilator for the LHS corresponds to the homogenous solution $y_h(n) = a2^n + c(-2)^n$, leaving the particular solution $y_p(n) = bn2^n + d$. The coefficients $b$ and $d$ can be solved for, while $a$ and $c$ depend on initial conditions.


Equivalent definition of normalizer of a group. Need help in proof. The definition of normalizer of a non-empty set $A$ in group $G$ is given by $$N_G(A) = \{g\in G\:|\:gAg^{-1} = A\}$$

This definition is from Abstract Algebra by Dummit and Foote.

Another definition I have seen around the internet (here and in this video) is $$N_G(A)=\{g\in G\:|\:gag^{-1}\in A\:\forall\: a\in A\}$$

I try to prove these equivalent. So we need $$gAg^{-1}=A \iff gag^{-1}\in A \:\forall\: a \in A$$

Here is my partial proof:

Assume $gAg^{-1}=A$. Then $gag^{-1}\in gAg^{-1}=A \:\forall \:a \in A$.

Conversely, assume $gag^{-1}\in A \:\forall\: a \in A$. Let $x\in gAg^{-1}$. Then, $\exists\:a_x \in A$ such that $x=ga_xg^{-1}\in A$. It follows that $gAg^{-1}\subseteq A$.

Now , let $y \in A$. This implies $gyg^{-1} \in A$. (This is where I'm stuck).

If I could somehow prove $y \in A \implies g^{-1}yg\in A$, then the result would from the hypothesis. Please let me know if I'm headed in the wrong direction.

I have seen this MathSE post. The definition of the normalizer is a slightly modified version of the first one stated here. However, I would like to know how I can proceed from where I'm stuck.

$gAg^{-1}$ = {$gag^{-1} : a\in A$}

Therefore, $gAg^{-1} = A$ occurs iff $g^{-1}Ag=A$ which both imply $\forall a\in A\; gag^{-1},g^{-1}ag\in A$

Now, assume $\forall a\in A\; gag^{-1}\in A$. From this we have $gAg^{-1}\subset A$. Choose $b\in A$ and note that $y=g^{-1}bg\in A$ and therefore $$b=gyg^{-1}\in gAg^{-1}$$

implying $A\subset gAg^{-1}$. Therefore, $gAg^{-1} = A$ iff $\forall a\in A\; gag^{-1}\in A$


Pigeonhole principle: Asking the minimum number of students The question

What's the minimum number of students, each of whom comes from one of the 50 states must be enrolled in a university to guarantee that there are at least 100 who come from the same state?

The solution

50*99 + 1 = 4951.
4951/50 = 100

My question

I don't understand where 99+1 comes from? 100? and why do we have to formulate 100 to 99+1? Thank you.

.

If you take $50\times 99$ students ($99$ by state), you are sure to have at least $99$ students coming from each state. If you add $1$, whatever the state he comes from, you are sure there are $100$ students coming from the same state (and there is only one state from which $100$ students come).

Think of it this way: if you have $4950$ students total, then that could mean you have exactly $99$ from each state. To guarantee that one state has $100$ students, add one more.


Group homomorphism from $\mathbb{Z}_p$ to $\mathbb{Z}_q$ By $\mathbb{Z}_p$ of course I mean $p$-adic integers.

I am fairly certain that there is no non-trivial group homomorphism between $\mathbb{Z}_p$ and $\mathbb{Z}_q$ for $p\neq q$ as additive groups but I can't come up with a simple proof.

Note that $q$ is invertible in $\mathbb{Z}_p$.  So in particular for any $a$ in $\mathbb{Z}_p$ there is a unique $a'$ such that $qa'=a,$ a unique $a''$ such that $q^2a''=a$, and so on.

However in $\mathbb{Z}_q$, $0$ is the only element infinitely divisible by $q$ in this sense. Therefore everything must get sent to $0$.

For prime $p$ and $q$ you already have the answer.

I would however adjust Nate's reasoning as follows: there is no need for doing "infinite" division by $q$. Simply, since in $\mathbb Z_p$ you can divide by $q$ then for some $a$ we have $qa = 1$. In case homomorphism $h: \mathbb Z_p\rightarrow \mathbb Z_q$ exists we have $h(a+a+...+a) = h(a)+...+h(a)$  ($q$ summands in each) and therefore $h(1) = qh(a) = 0$, so homomrphism is trivial.

Same proof if $p$ and $q$ are just co-prime.


Mathematical Induction problem Can somebody help me with these questions? I can't seem to get started...

Having $P(n) : n^2 + 5n + 1\text{ is even}$.

a) Demonstrate that if $P(k)$ is True to some $k$ natural, then $P(k + 1)$ is also true.

b) Considering the Principle of Complete Induction, can you say that $P(n)$ is true for all natural numbers?

c) Demonstrate that $P(n)$ is False for all n natural.

Thank you!

a) Let $P(n)$ be the assertion that $n^2+5n+1$ is even. We show that if $P(k)$ is true for a particular $k$, then $P(k+1)$ is true.

Suppose that $k^2+5k+1$ is even. Note that
$$(k+1)^2+5(k+1)+1=k^2+2k+1+5k+5+1=(k^2+5k+1)+(2k+6).\tag{1}$$
By the induction assumption, $k^2+5k+1$ is even. But $2k+6$ is even, and if we add an even number to an even number, the result is even. This completes the proof of the induction step.

b) The Principle of Mathematical Induction says that if (i) $P(1)$ is true and (ii) whenever $P(k)$ is true, $P(k+1)$ is true, then $P(n)$ is true for all $n\ge 1$.

We have proved (ii). But we must also verify (i). And notice that when $n=1$, we have $n^2+5n+1=7$, and $7$ is odd. So (i) fails, and therefore the induction "proof" fails.

c) We will show that in fact $n^2+5n+1$ is odd for all $n\ge 1$. Let $P^\ast(n)$ be the assertion that $n^2+5n+1$ is odd.

It is easy to verify that $P^\ast(1)$ is true, we just verified it above.

Now let us show that if $P^\ast(k)$ holds, then $P^\ast(k+1)$ holds. We use the calculation in (1). There we concluded that
$$(k+1)^2+5(k+1)+1=(k^2+5k+1)+(2k+6).$$
By the induction assumption, $k^2+5k+1$ is odd. But $2k+6$ is even, and an odd plus an even is odd, so $(k+1)^2+5(k+1)+1$ is odd. This completes the proof of the induction step.

Remark: Basically the only point of this exercise is to remind you that in an induction argument, you must show two things: that $P(1)$ holds, and that whenever $P(k)$ holds, $P(k+1)$ must hold. I actally don't quite see the point, since students almost never forget to verify that $P(1)$ holds.

A) If $k^2+5k+1$ is even, $$(k+1)^2+5(k+1)+1=(k^2+2k+1)+(5k+5)+1=(k^2+5k+1)+2k+6$$
is even as well.

B) No, you need a base hypothesis.

C) Obviously $0^2+5\cdot0+1$ is not even (base hypothesis), and by adapting A), if $k^2+5k+1$ is not even, neither is $(k+1)^2+5(k+1)+1$ (induction).


4-dim. generalization of $ab+ac+bc=0$ The equation $ab+ac+bc=0$ can be parameterized by $(a,b,c)=\lambda(-pq, p(p+q), q(p+q))$.

Is there a (similar) parameterization for $ab+ac+ad+bc+bd+cd=0$?

What about the 5-dimensional case?

Edit:
Proof of the parametrization in the 3-dim. case:

Write $x=a+b, y=a-b$.
Then $ab+ac+bc=0 <=> (x-y)(x+y) +2(x+y)c+2(x-y)c=0 <=> (x+2c)^2-y^2-4c^2=0$. Then use the pythogorean triple parametrization to get $x+2c=\lambda(m^2+n^2), y=\lambda(m^2-n^2), c=\lambda mn$.
Now $a=(x+y)/2= \lambda(m^2+n^2-2mn+m^2-n^2)/2=m(m-n), b=(x-y)/2=\lambda n(n-m), c=\lambda mn$. Now set $m=p$ and $n=q+p$ and the result follows (with $b$ and $c$ switched)

Edit2: yes, I am looking for a parameterization in the integers. Sorry, that I didn't make myself clear.

This kind of equations are solved everything. And even what some coefficients.
equation

$ab+ac+ad+bc+bd+cd=0$

Has the following solutions:

$a=(t+k)ps-(t^2+tk+k^2)s^2$

$b=(t-2k)ps+(k^2+tk-t^2)s^2$

$c=-3p^2+4(t+k)ps-(t^2+3tk+k^2)s^2$

$d=(k-2t)ps+(t^2+tk-k^2)s^2$

And more.

$a=3p^2-(t+k)ps$

$b=3p^2-(t+4k)ps+(t+k)ks^2$

$c=-3p^2+2(t+k)ps-tks^2$

$d=3p^2-(4t+k)ps+(t+k)ts^2$

Although I like this equation:

$a^2+b^2+c^2+d^2=ab+ac+ad+bc+bd+cd$

And solutions:

$a=p^2-2tps+(3k^2+t^2)s^2$

$b=p^2+3(k+t)^2s^2$

$c=4p^2-4tps+4t^2s^2$

$d=3p^2+6kps+(3k^2+t^2)s^2$

And more:

$a=7p^2+2(3k-t)ps+(3k^2+6kt+7t^2)s^2$

$b=7p^2+6(k-t)ps+3(k-t)^2s^2$

$c=p^2+2(3k+t)ps+(3k+t)^2s^2$

$d=3p^2-6(k+t)ps+(3k^2+6kt+7t^2)s^2$

number $t,k,p,s$  integers and set us.


fraction division understanding Want to visualize rule division of fraction.

For example

1)

2   2     4
_ * _   = _
2   3     6

in this case we "split" each piece of cake in numerator to the 2 "first fraction" and apply the same operation for the denominator.

Could you please clarify rule with devision, why we should turn fractions and multiply it. How can I visualize this rule?

Thanks.

Here's a good way to think about it. When you divide $6$ by $2$, you're asking how many groups of $2$ apples make up $6$ apples. The answer, of course, is that there are $3$ groups.

Now, when you want to divide $1$ by $2/3$, you're asking how many groups of $2/3$ of a cup there are in $1$ cup. As the second picture clearly shows, there are $1 1/2$ or $3/2$ groups. This same principle generalizes for any fraction division.

Think of it this way:$$a\div b=\frac{a}{b}=\frac{a}{1}\times\frac{1}{b}=a\times\frac{1}{b}$$Now lets set:$$a=\frac{p}{q}, b=\frac{r}{s}$$We then get:$$\frac{p}{q}\div\frac{r}{s}=\frac{p}{q}\times\frac{1}{\frac{r}{s}}=\frac{p}{q}\times\frac{s}{r}$$

This depends on you knowing that:$$\frac{1}{\frac{r}{s}}=\frac{s}{r}$$This can be shown by first letting:$$\frac{1}{\frac{r}{s}}=x\tag{1}$$and then multiplying both sides by $\frac{r}{s}$ to get:$$\require{cancel}\frac{1}{\cancel{\frac{r}{s}}}\times\frac{\cancel{\frac{r}{s}}}{1}=x\times\frac{r}{s}$$$$\therefore 1=x\times\frac{r}{s}$$Now just multiply both sides by $\frac{s}{r}$ to get:$$\require{cancel}1\times\frac{s}{r}=x\times\frac{\cancel{r}}{\cancel{s}}\times\frac{\cancel{s}}{\cancel{r}}=x$$$$\therefore x=\frac{s}{r}$$
Now substitute this back into (1) to get:$$\frac{1}{\frac{r}{s}}=x=\frac{s}{r}$$


flat trivial vector bundle Let $E=M\times V$ be a trivial vector bundle over $M$. If $\nabla$ is a flat linear connection on $E$, this implies that $(E,\nabla)$ is isomorphic to $(E,\nabla^{\mathrm{triv}})$??, where $\nabla^{\mathrm{triv}}$ is the trivial connection on $E$. In other words, except by isomorphism, every flat connection on trivial bundles must be trivial?

No. For any choice of base point $p\in M$, a flat connection determines a holonomy representation $\pi_1(M,p) \to \operatorname{GL}(E_p)$ (by parallel translating around closed loops), and equivalent connections determine isomorphic representations. The flat connections associated with global frames have trivial holonomy, any nontrivial representation of $\pi_1$ will correspond to a flat connection that doesn't have a global parallel frame.

A flat connection on $E$ is defined by a foliation ${\cal F}$ of dimension $dim(V)$ whose leaves are transverse to the vertical foliation $x\times V$. Let $p:M\times V\rightarrow M$ be the canonical projection and $x_0\in M$. You can define $\Phi:M\times V\rightarrow M\times V$ such that $\Phi(z)=(p(z),u(z))$ where  the intersection of ${\cal F}_z$ and $x_0\times V$ is $(x_0,u(z))$. $\Phi$ defines an isomorphism between the connection defined by $M\times y$ and the connection defined by ${\cal F}$.


Let $T : \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $T(a, b, c) = (a+b, 2a-c)$. Determine $T^{-1} (1, 11)$. Let $T : R^3 \to R^2$ be defined by $T(a, b, c) = (a+b, 2a-c)$. Determine $T^{-1} (1, 11)$. (Exercise 3.3.6 from Linear Algebra by Friedberg)

The solution manual solve a system of linear equation $ a + b =1 $ and $ 2a - c = 11$, and obtain
$T^{-1} (1, 11) = \{(a, 1-a, 2a - 11) : a \in \mathbb{R}\}$.

But, my question is whether $T$ is invertible or not. I have a theorem that $T$ is invertible iff $[T]_{\beta}^{\gamma}$ is invertible, where $\beta$ and $\gamma$ are standard ordered basis for $\mathbb{R}^3$ and $\mathbb{R}^2$. Here, $[T]_{\beta}^{\gamma}$ is $2$ by $3$ matrix, which implies $T$ is not invertible. Am I missing something or is this question wrong?

Your thinking is correct; only square matrices can have an inverse in the usual sense of one-to-one mapping. The fact that $T^{-1}(1,11)$ is an infinite set also means that $T$ is non-invertible.

You can still write $T^{-1}$, but this is the preimage function that takes a codomain vector and returns all its preimages in the domain. If $T$ is an invertible transformation, $T^{-1}$ will be a proper one-vector-to-one-vector function.

The notation $T^{-1} (1, 11)$ is not completely correct. It should read $T^{-1} \{(1, 11)\}$. This means

$$T^{-1} \{(1, 11)\}=\{(a,b,c) \in \mathbb R^3 : T(a,b,c)=(1,11)\}.$$

Hence $T^{-1} \{(1, 11)\}$ is the set of solutions of the linear system

$a+b=1$;

$2a-c=11.$


How to solve trigonometric equations with a domain involving negative values of $x$? I don't seem to understand the concept of a negative domain when solving trigonometric equations on "another interval" For example:

  Solve $\cos x=-\sqrt{3}/2$  given that the domain is  $-\pi \le x\le \pi$.

I don't know what to do or understand the process at all! Please help, thank you!

$$\cos x=-\frac{\sqrt3}2=-\cos\frac\pi3=\cos\left(\pi-\frac\pi3\right)$$

$$x=2m\pi\pm\left(\pi-\frac\pi3\right)$$ where $m$ is any integer

Now find $m$ such that $-\pi\le x\le\pi$

Simple. $x = -pi/6 $ will do nicely. To see this, note that cos is periodic with period $\pi$. This means that $x+\pi = x$ for all $x$. Note however this does not mean that $x=x-\pi$ for all $x$. Anyway, you can find the answer in the interval $0<x<\pi$, then just translate the interval.


How do you determine the smallest possible perimeter in a whole number not expression after factoring area? The area of a square garden, in square meters, is represented by the expression 169 - 182x + 49x2  What is the smallest possible perimeter of the garder if x is a positive integer?

I factored 169 - 182x + 49x2 to be (7x-13)2 so the perimeter would be be 28x-52.  But the answer to the perimeter is 24m with no variable.

How does the process to this answer work?

See that since $x$ is positive then the minimal value for the perimeter (an increasing function) is obtained by taking $x=0$.


Assume $ p \in$ null(A). Show that if $a$ is any vector such that $a^{T}p \neq 0$, then $a$ must be linearly independent of the rows of A. I am pretty sure that the converse holds, i.e., assume $a$ is linearly indepent of the the rows of matrix $A$.

$a$ can be decomposed as $a_{N}+a_{R}$ where $a_{N} \in$ null(A) and $a_{R} \in$ range($A^{T}$).  Thus $a_{n} \neq 0$ since if it was, $a$ would be in the range($A^{T}$), thus $a$ would be linearly independent of the rows of A.

We can construct a $p$ such that it is in the null(A) and $a^{T}p \neq 0.  For example:

$p=a_{N}$

Then $a^{T}p = |a_{N}|^{2} \neq 0$.   (since $a_{n} \neq 0$).

So if $a$ is linearly independent of the rows of $A$ and $p \in $ null(A), then $a^{T}p \neq 0$.

Can someone help me go the other way with this proof.

Hint: $p$ must be linearly independent, (in fact, orthogonal) of the span of the rows of $A$.


How to find the derivative of a piece-wise function? Find the derivative of the function at $x=0$
$$f(x) = \begin{cases}
e^x + x^3\cos\frac{1}{x}, &x\ne 0,\\
1, &x = 0.
\end{cases}$$

Now isn't this is trivial? Since $f(x)  = 1$ when $x = 0$ so the derivative at this point would be $0$ since it is a constant? Do I have to even compute the derivative for the first expression? But this seems to simple relative to the points offered for this question?

You could only apply that reasoning if the function were constant on an interval around $x=0$, which is not the case here.

The derivative of $e^x+ x^3 cos(1/x)$ is $e^x+ 3x^2 cos(1/x)+ x sin(1/x)$.  This function is NOT differentiable at x= 0.


Prove linear vector space consider a set of vectors $H = \{e^{a_1x}, e^{a_2x}, ..., e^{a_nx}\}$ where $a_i \in \mathbb R$ and $i= 1,2,...,n$ are distinct

how can I prove that $V= \text{span}~ H$  is a linear vector space?

I know I need to  prove closure under addition, scalar multiplication, and the zero vector

Could I do this by say:

$f(x) = m_1* e^{a_1x} + m_2* e^{a_2x} +  ... + m_n * e^{a_nx}$

$g(x) = l_1* e^{a_1x} + l_2* e^{a_2x} +  ... + l_n * e^{a_nx}$

for some $m_i$ and $l_i, i= 1,2,...,n$

and then try $f(x) + g(x)$?

and for scalar try:
$r*f(x) = r (m_1* e^{a_1x} + m_2* e^{a_2x} +  ... + mn * e^{a_nx})$

I am not sure what to do about the $0$ vector

If you're taking $V = \text{span}~ H$, then by definition, $V$ is the closure under addition and scalar multiplication of $H$, including the zero vector, so $V$ is a linear vector space.

Maybe you want to ask what this linear vector space looks like?

Because all $a_i$ are distinct we can order them without loss of generality. So let them be ordered like this: $a_1<a_2<...<a_n$. Now suppose that $e^{a_jx}$ is a linear combination of $e^{a_1x},e^{a_2x},...,e^{a_{j-1}x},e^{a_{j+1}x},...,e^{a_nx}$, so that $e^{a_jx}=\sum_{i=1}^{n}b_ie^{a_ix},(i\neq j)$. (I do not know how to write that the sum excludes the case $i=j$ so I wrote it in the brackets). Now we divide this expression by $e^{a_jx}$ to get $1=\sum_{i=1}^{n}b_ie^{(a_i-a_j)x}$ and now let $x\to\infty$. When you go to the limit the terms $e^{(a_i-a_j)x}$ go to $0$  if $(a_i-a_j)<0$ and to $\infty$ if $(a_i-a_j)>0$ so the right hand side is not equal to $1$ so the vectors in the set are linearly independent and because of that they span a vector space.


Prove that $g(x)=\frac{\ln(S_n (x))}{\ln(S_{n-1}(x))}$ is increasing in $x$, where $S_{n}(x)=\sum_{m=0}^{n}\frac{x^m}{m!}$ I want to prove that the function
$$g(x)=\frac{\ln(S_n (x))}{\ln(S_{n-1}(x))},\,x >0$$ is increasing in $x$ for all $n$, where
$ S_n(x)= \sum_{m=0}^{n}\frac{x^m}{m!}$. Differentiating gives something messy that I have not been able to prove it is non-negative. I have also been trying to find an appropriate $h(x)$ increasing and proving that $  h(g(x))$ is increasing $\ln(\cdot)$ seems a good candidate.

By plotting I am almost convinced the statement is true.

Note: Here is a first step towards a complete answer. The point I like to address is the discrete log-concavity of the function $S_n$ for fixed $x>0$. This means $S_n$ fulfils

  \begin{align*}
S_{n-1}^2(x)\geq S_n(x)S_{n-2}(x)\qquad\qquad x>0, n\geq 2\tag{1}
\end{align*}

  A corresponding example can be found e.g. in the slides about Log-convexity and Log-concavity by Dmitry Karp. You might have a look at the section about Bessel functions where the inequality (1) for these functions is stated and some sharper results are presented afterwards.

  Please note that in order to show $g(x)$ is increasing an even sharper inequality than (1) has to be proved.

We start as @Arashium did. Using the differential operator $D_x$ we obtain
\begin{align*}
D_xS_n(x)=D_x\sum_{m=1}^{n}\frac{x^{m-1}}{(m-1)!}=\sum_{m=0}^{n-1}\frac{x^m}{m!}=S_{n-1}(x)
\end{align*}
and by omitting the argument $x$  we get
\begin{align*}
D_x\ln(S_n)&=\frac{1}{S_n}D_x(S_n)=\frac{S_{n-1}}{S_n}\\
\end{align*}
and
\begin{align*}
D_xg&=D_x\frac{\ln(S_n)}{\ln(S_{n-1})}
=\frac{\ln(S_{n-1})D_x\ln(S_n)-\ln(S_n)D_x\ln(S_{n-1})}{\ln^2(S_{n-1})}\\
&=\frac{1}{\ln^2(S_{n-1})}\left(\frac{S_{n-1}}{S_n}\ln(S_{n-1})-\frac{S_{n-2}}{S_{n-1}}\ln(S_n)\right)\tag{2}
\end{align*}

  In order to show that the function $g$ is increasing we need $D_xg\geq 0$.

  Therefore the following is to prove according to (2)
  \begin{align*}
S_{n-1}^2\geq S_{n-2}S_n\frac{\ln(S_n)}{\ln(S_{n-1})}\tag{3}
\end{align*}

Here's the proof of the weaker inequality (1) showing that $S_n(x)$ is discrete log-concave for fixed $x>0$.

\begin{align*}
S_{n-1}(x)^2&-S_{n-2}(x)S_n(x)=\\
&=\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}\right)^2-S_{n-2}(x)\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}+\frac{x^{n}}{n!}\right)\\
&=S_{n-2}(x)\left(\frac{x^{n-1}}{(n-1)!}-\frac{x^{n}}{n!}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}S_{n-2}(x)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}\right)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=0}^{n-2}\frac{x^{m+1}}{m!}\right)
+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=1}^{n-1}\frac{x^{m}}{(m-1)!}\right)
+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}-\frac{x^{n-1}}{n(n-2)!}\right)
+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}\right)\\
&\qquad\qquad+\frac{x^{2n-2}}{(n-1)!(n-2)!}\left(\frac{1}{n-1}-\frac{1}{n}\right)\tag{4}\\
&>0\\
\end{align*}

  From the line (4) it's obvious that the inequality is valid and we may therefore conclude that $S_n$ is discrete log-concave.

  The challenge is of course to sharpen the inequality (4) in order to obtain (3).

Note: Since the function $\ln$ is monotonically increasing we obtain due to the log-concavity of $S_n$

\begin{align*}
\ln\left(S_{n-1}^2(x)\right)&\geq\ln\left(S_n(x)S_{n-2}(x)\right)\\
2\ln\left(S_{n-1}(x)\right)&\geq\ln\left(S_n(x)\right)+\ln\left(S_{n-2}(x)\right)\\
\end{align*}

which may of some use for further calculations.

It's not hard to prove that  $ dS_n/dx = S_{n-1} $, and since $ Sn(x) > 0 $ for all x > 0 and any natural n, the derivative of g(x) is positive for all x.


Why is $1 - \frac{1}{1 - \frac{1}{1 - \ldots}}$ not real? So we all know that the continued fraction containing all $1$s...

$$
x = 1 + \frac{1}{1 + \frac{1}{1 + \ldots}}
$$

yields the golden ratio $x = \phi$, which can easily be proven by rewriting it as $x = 1 + \dfrac{1}{x}$, solving the resulting quadratic equation and assuming that a continued fraction that only contains additions will give a positive number.

Now, a friend asked me what would happen if we replaced all additions with subtractions:

$$
x = 1 - \frac{1}{1 - \frac{1}{1 - \ldots}}
$$

I thought "oh cool, I know how to solve this...":

\begin{align}
x &= 1 - \frac{1}{x} \\
x^2 - x + 1 &= 0
\end{align}

And voila, I get...

$$ x \in \{e^{i\pi/3}, e^{-i\pi/3} \} $$

Ummm... why does a continued fraction containing only $1$s, subtraction and division result in one of two complex (as opposed to real) numbers?

(I have a feeling this is something like the $\sum_i (-1)^i$ thing, that the infinite continued fraction isn't well-defined unless we can express it as the limit of a converging series, because the truncated fractions $1 - \frac{1}{1-1}$ etc. aren't well-defined, but I thought I'd ask for a well-founded answer. Even if this is the case, do the two complex numbers have any "meaning"?)

You're attempting to take a limit.

$$x_{n+1} = 1-\frac{1}{x_n}$$

This recurrence actually never converges, from any real starting point.
Indeed, $$x_2 = 1-\frac{1}{x_1}; \\ x_3 = 1-\frac{1}{1-1/x_1} = 1-\frac{x_1}{x_1-1} = \frac{1}{1-x_1}; \\ x_4 = x_1$$

So the sequence is periodic with period 3.
Therefore it converges if and only if it is constant; but the only way it could be constant is, as you say, if $x_1$ is one of the two complex numbers you found.

Therefore, what you have is actually basically a proof by contradiction that the sequence doesn't converge when you consider it over the reals.

However, you have found exactly the two values for which the iteration does converge; that is their significance.

Alternatively viewed, the map $$z \mapsto 1-\frac{1}{z}$$ is a certain transformation of the complex plane, which has precisely two fixed points. You might find it an interesting exercise to work out what that map does to the complex plane, and examine in particular what it does to points on the real line.

Let consider a sequence $\left\{1,1-1,1-\frac{1}{1},1-\frac{1}{1-1},1-\frac{1}{1-\frac{1}{1}}\cdots\right\}$ that the limit is the answer. However, start from the third term, all the values are undetermined because there are $\dfrac{1}{0}$ appeared in the terms. Therefore, it has no answer of $1-\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{1-\cdots}}}$.


Integer solution for $2^{3p}+2^{4q}=2^{5r}$ If $x,y,z$ are integer powers of $2$, prove that equation $$x^3+y^4=z^5$$
does not have solution, otherwise give a counterexample.
I tried to do this exercise, I firmly believe that the equation is not fulfilled, the detail is in how to show what is requested, can someone give me a hint please?

$$2^{3p}+2^{4q}=2^{5r}$$
We have:
$$2^{3p-4q}+1=2^{5r-4q}$$
$$\begin{align}\begin{cases} 3p-4q=0 \\5r-4q=1\end{cases}
&\implies \begin{cases}3p=4q\\ 5r-4q=1\end{cases} \\
&\implies
\begin{cases} p=4m,q=3n \\ 5r-4q=1  \end{cases} \\
&\implies \begin{cases}5r-12n=1 \\m =n.\end{cases}\end{align}$$
Applying Euclidean algorithm, we can find a counter-example:
$$p=28, q=21, r=17 $$ which is correct.
$$2^{84}+2^{84}=2^{85}$$

The only way to get a power of $2$ by adding two powers of $2$ is when the exponents are equal. Hence
$$3p=4q=5r-1.$$
As $3,4$ are relative primes, $p=4t,q=3t$ and
$$12t=5r-1.$$
The smallest solution is $t=2,r=5.$

$$256^3+64^4=32^5.$$


Prove that this sequence diverges to infinity. $\lim_{n\to \infty} (1 +\frac{1}{n}\tag{displayed})^{n^2} = \infty$

I don't know how to tackle this one. Knowing that it diverges to infinity and thus does not have an upper bound, should I try to prove that it's an unbounded subsequence, if so how? Is that sufficient to show that $\infty$ is the limit?

Any help would be appreciated.

An approach would be using the fact that:

$$
\lim_{x \to \infty} \left(1+\frac{1}{n}\right)^n = e
$$


Show $\frac{du}{dx}=|u| \ (x\in\mathbb{R}$) has solutions $u=Ae^x$ Show the following ODE has solutions of the form $u=Ae^x$: $$\frac{du}{dx}=|u|, \ x\in\mathbb{R}.$$

My attempt:

I first considered the case where $u>0$.
So, \begin{align}
\frac{du}{dx}&=u \\
\int \frac{du}{u}&=\int dx \\
\ln|u|&=x+C, \ \ C\in\mathbb{R} \\
u&=Ae^x, \ \ A=e^c\in\mathbb{R}
\end{align}
Next, for the case where $u<0$ (using separations of variables as above), I get that,
$$u=Ae^{-x}.$$
For the final case, $u=0$,
$$u=0.$$
Only one of these cases admits the required form. Where have I gone wrong?

Actually, $u>0$ and $u<0$ are not both possible. Either $u>0$ everywhere or $u<0$ everywhere. Here is the argument: consider the open set $U=\{x: u(x)>0\}$.  On this set we have $u(x)=Ae^{x}$. Clearly $A >0$. Let us show that the open set $U$ has no boundary point. If there is a boundary point $x$ then there is a sequence $x_n \in U$ converging to $x$. We then have $u(x) =0$ and $u(x)=\lim u(x_n)$ so $Ae^{x_n} \to 0$, hence $e^{x_n} \to 0$, But this is possible only when $x_n \to -\infty$ contradicting the fact that $x_n \to x$. This proves that $U$ is an open set with no boundary points. This implies that it is either empty or the whole line. Thus the cases $u>0$ and $u<0$ cannot both occur.

$u=0$ is part of the first case for $u=Ae^x$ if you were to say $A \ge 0$. Your first case can actually be $A>0$ because you assumed $u>0$. For your second case you should get

$$\ln|u|=x+C$$

$$\implies \ln(-u)=x+C$$

$$\implies -u=Ae^x$$

$$\implies u=-Ae^x$$

for $A=e^C > 0$

Altogether we can merge the three cases to have $u(x)=Ae^x, x \in \mathbb R, A \in \mathbb R$ where $sgn(A)=sgn(u(x))$


How to find all the positive integral solutions of $5x+7y=100$? How to find the number of all the positive integral solutions and the solutions itselves of $$5x+7y=100?$$

Please help me!!

Assume we have $5x+7y=100$. Let us solve for the variable with the smallest coefficient $x=20-\frac{7y}{5}$. This means that $\frac{7y}{5}$ is integer. Therefore $y=5y_2$ for some integer $y_2$.

Plugging this into the original equation we get $x=20-7y_2$. You see now, that for every integer you put in $y_2$ you get an integer value for $x$, and an integer value for $y=5y_2$ that satisfy the equation $5x+7y=100$.

Since we want positive solutions you can put $y_2>0$ such that $20-7y_2>0$, i.e. $20/7>y_2$. So, $0<y_2\leq 2$.

We have then two possibilities: $y_2=1$ or $y_2=2$, for which we get $y=5$, or $y=10$ respectively. The corresponding values for $x$ are $x=13$, or $6$.

Thank you for pointing out my mistake, postmortem.
Through diophantine equation, I get

$$x = 300 -7r,$$
for all $r-$ integers,
$$y = -200 +5r $$

To solve the inequalities above. So, to get the positive solution, for the above inequalities,

we need to set certain boundaries such that,
$r\ge40$ , and $r<42$,

sorry for my mistake, due to I am having fever. Thank you so much for pointing out my mistake.

Reference：
How to find solutions of linear Diophantine ax + by = c?


$3\times 3$ matrix always has determinant $0$. Must $7$ of the elements be $0$? Let $M$ be a $3\times 3$ real matrix with at least $3$ distinct elements and the property that any permutation of it's elements gives a matrix with determinant $0$.
Must $M$ contain exactly seven $0$s?

This question is a special case of my previous question: Exactly $n-1$ nonzero elements if $\det(A)=0$ for every arrangement
Thanks to user Holonomia, we know a counterexample to the analogous question for $2\times 2$ matrices: Any $2\times 2$ matrix with two $1$s and two $-1$s has determinant $0$.
However, I have not been able to make much progress on the $3\times 3$ case. Theoretically the property gives us a system of $9!$ equations (with some symmetries and repeats), but I haven't found a good way to compute with this idea.

Let $a, b, c$ be three distinct elements in the entries of the matrix. We will be using this lemma a lot:

Woolfitt's lemma: Let the matrix be given by
$$(*)\ \ \ \ \begin{bmatrix} a & b& c \\ \cdot & \cdot & \cdot \\ \cdot& \cdot& \cdot \end{bmatrix},$$

then $A=B=C$, where $A, B, C$ are the minors with respect to $a, b, c$ respectively.

(See the comment for the proof)

Now we split into two cases.

First case: there is yet another entry $d$ not equals to $a, b, c$. Then consider the matrix

\begin{bmatrix} d & b& c \\ a & \cdot & \cdot \\ \cdot & \cdot & \cdot \end{bmatrix}

and use the lemma. Then one also get $D' = B' =C'$ (Three new minors). But actually $A = D'$. Thus they are all the same. Then there will be two zeroes in the matrix (the $(3, 2)$ and $(3, 3)$ entries). By my comment in the question, there has to be seven zeroes and we are done.

Second case: All the other entries are $a, b, c$: By symmetry, assume that we have at least three $a$'s. By another comment below the question, we can assume $a\neq 0$ (or we are done).

Claim: There are three $b$'s (or $c$): If not, there has to be five $a$'s, but if we expand the first row of

$$\begin{bmatrix} a & b& c \\ a & a& e \\ a& a& f \end{bmatrix},$$

then by the lemma,
$$0 = C = A = \det \begin{bmatrix} a & e \\ a& f\end{bmatrix}\Rightarrow e = f$$

One can check that this $e$ cannot be $a$ (or there will be seven $a$'s). Then the claim is shown.

Now we are almost done: the matrix can be written as

$$\begin{bmatrix} a & b& c \\ a & b& e \\ a& b& f \end{bmatrix},$$

Again using the lemma, we have $e = f$. By the claim, $e=f = c$. (It can't be $a$, $b$). Thus the matrix is

\begin{bmatrix} a & b& c \\ a & b& c \\ a& b& c \end{bmatrix}

But then

$$\det \begin{bmatrix} a & a& a \\ b& b& c \\ c & b& c\end{bmatrix} = -a(bc-c^2 ) + a(b^2 - bc)= a(c-b)^2\neq 0$$

This last contradiction concludes the proof.

For $3 \times 3$ we can use three $+1$, three $e^{2\pi \mathbf{i}/3}$ and three $e^{4\pi\mathbf{i}/3}$ and the determinant is $0$.

$$
A = \det
\left(
\begin{array}{ccc}
1 & e^{2\pi \mathbf{i}/3} & e^{4\pi \mathbf{i}/3}\\
e^{2\pi \mathbf{i}/3} & e^{4\pi \mathbf{i}/3} & 1\\
e^{4\pi \mathbf{i}/3} & 1 & e^{2\pi \mathbf{i}/3}
\end{array}
\right) = 0
$$

So we don't need zeros..


Can a finite ring have a characteristic $0$? I am working on studying Abstract Algebra and doing one of the review problems for the exam.  I do not think a finite ring can have characteristic zero because when I think of, for example, Integers mod $4$, this ring has a characteristic greater than $0$. I also remember learning about infinite rings and I almost naturally associate an infinite ring with having a characteristic of $0$.  Could anyone help me prove the above statement? I am thinking of starting off with a contradiction: let R be a finite ring with characteristic $0$...

I'm not sure whether your ring has unity, so here is a general argument.

If a ring $R$ has characteristic zero, then, for each $n>0$, there is some $a\in R$ with additive order $>n$.  This means the set $\{0,a,2a,\ldots ,na\}$ consists of distinct elements (this requires the existence of additive inverses), so that $|R|>n$.

Since $n$ was arbitrary, $R$ is infinite.

This argument is a little simpler if we can assume that $R$ has unity.  In that case, $R$ must actually contain $\mathbb{Z}$ as an additive subgroup.

No it can't be zero because the characteristic of a finite ring, say (R,+,.), divides the order of (R,+). It's a trivial proof.


Understanding a particular transformation of an integral given in a proof Using the theorem of mean values find the sign of the integral...
  $$\int_{0}^{2 \pi}{\sin x \over x}dx= \int_{0}^{\pi}{\sin x \over x}dx+\int_{\pi}^{2 \pi}{\sin x \over x}dx$$
  Then: $[x-\pi=t ; dx=dt]$
  $$\int_{0}^{\pi}{\sin x \over x}dx+\int_{0}^{\pi}{\sin(t+ \pi) \over t+ \pi}dt \, \,\color{red}{\fbox{=}} \int_{0}^{\pi}{\sin x \over x}dx-\int_{0}^{\pi}{\sin(x)\over x+ \pi }dx$$

How does one see the (boxed, red) equality?

Hint:

if you have $f(x)=F'(x)$, you can change the ''symbol'' for the variable, i.e write $t=x$, and you have: $f(t)=F'(t)$.

Than:

$$
\int_a^b f(x)=F(x)|_a^b=F(b)-F(a)
$$

$$
\int_a^b f(t)=F(t)|_a^b=F(b)-F(a)
$$

In your case you have:
$$
\int_0^\pi\dfrac{\sin (t+\pi)}{t+\pi}dt=-\int_0^\pi\dfrac{\sin t}{t+\pi}dt=-\int_0^\pi\dfrac{\sin x}{x+\pi}dt
$$


Total number of outcomes for flipping 2 coins 4 times? I just wanted to make sure I understand this correctly.

Total number of outcomes for $2$ coins on $1$ flip is $2^2 = 4$.
Therefore, would flipping the $2$ coins $4$ times have $4^4 = 256$ possible outcomes?

The equation for the exponent. The base represents the total outcomes and the exponent is the total number of flips or number of coins?

let's mark $ L_i - $ the number of possibility's for one turn (flipping)
we see that for any $i \neq j$ the event $L_i \ and \ L_j$ are independence and there for the total outcome for n turn is $\Pi_{i=1}^{n} L_i $ and because $L_i = 4 $ for every i:

(h - h , h - t , t -h , t - t)
then the total outcomes number is $4^n$


Chance of a double in three dice What is the chance of rolling doubles in three six sided dice?

The answer I have is:
$$
\frac{1}{6}•\frac{1}{6}•\frac{3}{2} = \frac{1}{24}
$$

The number of elements in the sample space = 216 or (6*6*6)

Of the three numbers two of them to be same is (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) = 6 doublets.
 Every one of this can happen in 3 different ways, that is (1,1,x) (1,x,1) (x,1,1) = 3 ways.
 Now x can be any of the remaining 5 numbers.
 So total number of ways two of them can be same = 6 x 3 x 5 = 90

So probability of the event = 90/216 = 5/12

There are 36 possible permutations when rolling the first and second die.  6 of these produce doubles.  Therefore the probability of rolling doubles with the first and second die is 6/36, or 1/6.  There is also a 1/6 chance of rolling doubles with the first and third die.  And a 1/6 chance of rolling doubles with the second and third die.  Therefore the probability of rolling doubles with all three dice is 1/6 + 1/6 + 1/6 = 3/6 or 1/2 (including triples).  The probability of rolling triples is 1/36 so the probability of rolling doubles with no triples is 18/36 - 1/36 = 17/36.


Does there exist a measurable $A \subseteq [0,1]$ with $\lambda(A \cap [0,a]) = a/2$ $\forall a \in [0,1]$? does there exist a Borel set $A \in [0,1]$ such that, for any $a \in [0,1]$, the Lebesgue measure of the set $A \cap [0,a]$ is $a/2$?

Thanks

If so this would force $\lambda(A \cap [a,b]) = \dfrac{b-a}2$ for every interval $[a,b] \subset [0,1]$.

Whenever $\{I_k\}$ is a cover of $A$ by closed subintervals of $[0,1]$ this would imply $$\lambda (A) = \lambda \left( A \cap \cup I_k \right) \le \sum_k \lambda (A \cap I_k) \le \frac 12 \sum_k \ell(I_k)$$
and by taking the infimum of all such coverings, $\lambda(A) \le \dfrac 12 \lambda(A)$ which forces $\lambda(A) = 0$.

tl;dr NO

Let $f(a) = \int_0^a 1_A$, then the Lebesgue differentiation theorem gives
$f'(a) = 1_A(a)$ for ae. $a$. In particular, we can't have $f'(a) = {1 \over 2}$.


How to integrate $\int \frac{e^x dx}{1\,+\,e^{2x}}$ Ok, I give up, I have tried with $u$-substitution and integration by parts but I can't solve it. The integral is:

$$\int{\frac{e^x dx}{1+e^{2x}}}$$

I have tried $u=e^x$, $u=e^{2x}$ and also integration by parts but I can't solve it. The result should be:

$$\arctan(e^x)$$

Use $u = e^x, du = e^x dx.$

Then you have:

$$\int \frac{du}{1 + u^2} \text{because} \space (e^x)^2 = e^{2x} $$

$$\arctan (u) + C$$

$$\arctan(e^x) + C$$

$$\newcommand{\arctg}{\operatorname{arctg}} \int{\frac{e^x dx}{1+e^{2x}}} = \int{\frac{du}{1+u^{2}}} = \arctg(u)+c = \arctg(e^x)+c$$


Finding 3 variables a,b,c. 31a+30b+28c=365 well, I have this question:
Find positive integers $a,b,c$ that solve
$$31a+30b+28c=365.$$

Well, it was easy to find them, since i noticed that it actually asks for the months in a year. As, how many months have 31 day each? Which is $a$, and how many have 30? Which is $b$, and $c$ is the ones with 28.
It turns to be that $a=7, b=4$, and $c=1$.

My question is that if there's any mathematical way to do it, rather than counting months.

Also, for ANY $a,b,c$ (positive integer) that the statement ($31a+30b+28c=365$) is correct, prove that
$$a+b+c=12$$ must be correct.

How can i do it? (Not just adding $7+4+1=12$; it must be generalized)

Thank you!

To show that in any solution the sum $a+b+c$ must equal $12$:

Write $31a+30b+28c=28(a+b+c)+3a+2b$.

If $a+b+c>12$, then $28(a+b+c)+3a+2b\ge28\cdot 13+3a+2b=364+3a+2b$.  So with positive integer values of $a,b,c$ the sum will exceed $365$ if $a+b+c$ exceeds $12$.

If $a+b+c<12$, then $31a+30b+28c \le 31\cdot 11=341$.  Too small.

Since $31a+30b+28c=365$ and $a,b,c$ are positive integers then:

Roughly:

$a\leq \dfrac{365}{31}$ i.e $0 \leq a \leq 11$

$b\leq\dfrac{365}{30}$ i.e $0 \leq b \leq 12$

A program can easily find  the solutions:

$(5,7,0);(9,1,2);(7,4,1)$

Notice that all solutions are such that: $a+b+c=12$

NB:
Since $31a=365-2(15b+14c)$ and $31$ is odd then $a$ is odd.

PS:
A complete solution without computer:

since $0\leq a\leq 11$ is odd, $a$ belongs to $\{1,3,5,7,9,11\}$

The solutions of the equation $15x+14y=t$ ,with $t>0,x,y$ integers, are $(t+14k,-t-15k)$,$k$ integer.

$t+14k$ and $-t-15k$ are positive integers if only if $k \in \Big[-\dfrac{t}{14};-\dfrac{t}{15}\Big]$

if $a=1$ then $(b,c)$ are solutions of $167=15x+14y$ but there is no integer in
$\Big[-\dfrac{167}{14};-\dfrac{167}{15}\Big]$

if $a=3$ then $(b,c)$ are solutions of $136=15x+14y$ but there is no integer in
$\Big[-\dfrac{136}{14};-\dfrac{136}{15}\Big]$

if $a=5$ then $(b,c)$ are solutions of $105=15x+14y$ but there is only one integer in $\Big[-\dfrac{105}{14};-\dfrac{105}{15}\Big]$, $-7$.

Therefore $(a,b,c)=(5,7,0)$

if $a=7$ then $(b,c)$ are solutions of $74=15x+14y$ but there is only one integer in $\Big[-\dfrac{74}{14};-\dfrac{74}{15}\Big]$, $-5$.

Therefore $(a,b,c)=(7,4,1)$

if $a=9$ then $(b,c)$ are solutions of $43=15x+14y$ but there is only one integer in $\Big[-\dfrac{43}{14};-\dfrac{43}{15}\Big]$, $-3$.

Therefore $(a,b,c)=(9,1,2)$

if $a=11$ then $(b,c)$ are solutions of $12=15x+14y$ but there is no integer in $\Big[-\dfrac{12}{14};-\dfrac{12}{15}\Big]$


What are the units of $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$ I'm struggling with this a lot.
I think it all boils down to my basic understanding of what the ring   $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$ is. As much as I know is that it is a ring of 10 elements, but I have no idea how I would go about finding the units in this.

The way I view it is:

$\mathbb Z/2\mathbb Z  = \{0,1\},\qquad
  \mathbb Z/5\mathbb Z= \{0,1,2,3,4\}$.

So would   $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z= \{0\times0, 0\times1, 0\times2, 0\times3, 0\times4, 1\times0, 1\times1, 1\times2, 1\times3, 1\times4\} = \{0,0,0,0,0,0,1,2,3,4\}$?

This seems totally incorrect to me, so it would be great if someone could clear this up for me.

I understand a unit is an element $u$ such that $uv=vu$, but in this I don't really even know what the elements are...
Would the unit be $0$ and $1$?

Any help would be great.
Thanks

I'll clarify again the product of rings:
For $R, S$ rings, the ring $R \times S$ is defined to be
$$R \times S := \{(r,s) \mid r \in R, s \in S\}$$
with addition and multiplication defined pointwise, additive identity $(0_R, 0_S)$ and multiplicative identity $(1_R, 1_S)$ and additive inverses also defined pointwise.

So, the product $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$ looks like
$$\{(0,0),(0,1),(0,2),(0,3),(0,4),(1,0),(1,1),(1,2),(1,3),(1,4)\}$$
with $10$ elements as you expected.

Now, for $u=(u_1,u_2)$ to be a unit, we need to find a $v = (v_1, v_2)$ such that $uv = 1$. Looking at the definition of the product ring, this means $u_1 v_1 = 1$ and $u_2 v_2 = 1$, i.e. $u_1$ and $u_2$ are units in their respective rings. In $\mathbb{Z}/2\mathbb{Z}$, the only unit is $1$ and in $\mathbb{Z}/5\mathbb{Z}$, the non-zero elements $1,2,3,4$ are all units. So, $u = (1,1),(1,2),(1,3),(1,4)$ are the units of $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$.

Just make the multiplication table.

The ring $\mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$ is made up of all ordered pairs $(p,q)$, where $p \in \mathbb Z/2\mathbb Z$ and $q \in \mathbb Z/5\mathbb Z$.

Multiplication is defined in the natural way: $(p,q) \times (p',q') = (p\times p',q\times q')$.

The units are all those elements with multiplicative inverses.

You want pairs, for which there is another pair with $(p,q) \times (p',q') = (1,1)$, remembering to reduce the first component modulo 2, and the second component modulo 5, e.g.

$$(1,4) \times (1,3) = (1\times 1, 4 \times 3) = (1,12) = (1,2)$$


Is there a 0-1 law for the theory of groups? For each first order sentence $\phi$ in the language of groups, define :

$$p_N(\phi)=\frac{\text{number of nonisomorphic groups $G$ of order} \le N\text{ such that } \phi \text{ is valid in } G}{\text{number of nonisomorphic groups of order} \le N}$$

Thus, $p_N(\phi)$ can be regarded as the probability that $\phi$ is valid in a randomly chosen group of order $\le N$.

Now define $$p(\phi)=\lim_{N \to \infty}p_N(\phi)$$
if this limit exists.

We say that the theory of groups fulfills a first order zero-one law if for every sentence $\phi$, $p(\phi)$ exists and equals either $0$ or $1$. I'm asking myself whether this 0-1 law holds indeed in group theory.

Since it is conjectured that "almost every group is a 2-group", statements like $\exists x: x\ne 1 \wedge x^2=1 \wedge \forall y:xy=yx$ (meaning $2|Z(G)$) or $\forall x: x^3=1 \to x=1$ (no element has order 3) should have probability $1$ and I don't see any possibility to construct any sentence with $p\not \in \{0,1\}$. Am I missing an obvious counterexample, or can you show (under the condition that almost every group is indeed a 2-group) that the theory of finite groups fulfills this 0-1 law?

Sharing the question within MO has brought some interesting comments - I send you there for reference - but in the last 2 years there has been no new contribution.
So I'm wrapping it up for the less technical scope of SE:

As per now, a FO 0-1 law for group theory as defined by Dominik doesn't exist and may never exist.

The consensus seems to be that an accepted and complete view of what group theoretic concepts correspond to in a classical logic framework is needed to address the problem but - as per now - it is not available.

The actual methods available to cross-translate group theoretic concepts into logic ones seem to suggest that such a law cannot exist for the whole group theory or even for finite group theory.
Limiting laws have infact been identified such that group theoretical entities can be partitioned into different classes exhibiting different logical behaviours (i.e. quantification over functions which is ubiquitous in group theory; but also abelianity appears to be relevant).
While yet to be proven, some of these classes can be expected to obey a 0-1 law for FOL while some others can be expected to not obey it.

Sentence proposals:

The cardinality of $G$ is even.

I strongly suspect the limit diverges by having a zero $\lim\inf$ and a unit $\lim \sup$.  Via supermultiplicity, the number of groups of order $2^k n$ is bounded below by the number of order $2^k$ times the number of order $n$, so is infrequently "small" ($n$ prime) and frequently large.  Additionally, the number of groups of a given order is upper bounded (see http://www.jstor.org/stable/2946623 ) by something not too stupendously rapidly growing, so no $N$ is going to miraculously overwhelm the running average.

Let $P(n)$ be the cardinality of the set of isomorphism classes of groups of order $n$.  For all positive integers $n$, $P(n)>0$ since there is at least a cyclic group of order $n$.  Let $C(n) = \sum_{i=1}^n P(n)$, which is clearly positive and monotonically increasing on the positive integers.  Set $a_0=1$ and, for $j>0$, set $a_j = \min_i\{i \mid C(i)-a_{j-1} > 2^{2j}\}$, such an $i$ exists because $C$ is monotonically increasing.  Let $T = \bigcup_{j=0}^\infty [a_{2j}, a_{2j+1})$ and let $\phi$ be the predicate $|G| \in T$.  By construction, $p_N(\phi)$ swings by a factor of two through each interval $[a_j, a_{j+1})$ (from something < $2^{-j}$ to something $> 1- 2^{-j}$ and vice versa).  Therefore, the limit as $N$ goes to infinity does not exist.

It's not clear to me how this sort of see-saw argument fails in any set that has something like a positive semidefinite inverse norm with infinite support.  ($P(n)$ is that inverse norm here, taking cardinalities (a usual norm) to the number of isomorphism classes.)


Another way of show number of exclusive elements in a set Sorry for any confusing. This is the first time I asked on this site, if there is any term that's wrongly used, please let me know

So given a diagram of A and B, to show exclusively A and exlusively B, we have

A or B - A and B

alternatively, we have

~ (A and B)

Then given A, B, C. To show exlusively A,exlusively B and exlusively C, we can do

A or B or C - A and B and C - A and B and not C - A and C and not B - B and C and not A

or

~ (A and B and C) - A and B and ~C - A and C and ~B - B and C and ~A

Then given A, B, C and D, we have

A or B or C or D - A and B and C and ~ D - A and B and D ~ C - A and D and C and ~ B - B and C and D and ~ A - A and B and ~ (C and D) - A and C and ~ (B and D) - A and D and ~ (B and C) - B and C and ~ (C and A) - C and D ~ (A and B)

I came up with above expression by drawing the van diagram. Is the above expression correct? Is there a better way to express this?

There is the operator xor which has the following truth table

Its symbol is $\oplus$, so in general if you have $A_1, A_2, A_3,....,A_n$ , a good expression is :

$$A_1 \oplus A_2 \oplus . . . \oplus A_n $$

:)


Defining Linear Systems from Ratio (word question from MIT multivarible calculus pset2) source:https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/problem-set-2/MIT18_02SC_pset2.pdf

I am working through problem set 2 of 18.02sc on MIT OCW, Multivariable calculus. I don't understand the answer given for problem 2 part A, where you have to solve the linear equation. I understand the explanation but don't get the jump from their to the answer given. Many thanks for any help. Pics are below, of q and a.

The part I don't understand is part A) getting from the word question to the linear system (and is A°p =m), clearly A is in fact the ratios and P and m are just p1, p2, p3 and m1 etc but Im not clear on how they got A setup like they did with ratios as columns ie. Col 1 is the ratios of m1,m2,m3 for p1, col2 is for p2 etc, I would've thought it was A°m=P with the ratios as rows for each P it row 1 is p1?

A manufacturing process mixes three raw materials M1, M2, and M3 to produce three
products P1, P2, and P3. The ratios of the amounts of the raw materials (in the order
M1, M2, M3) which are used to make up each of the three products are as follows:
For P1 the ratio is 1: 2: 3; for P2 the ratio is 1: 3: 5; and for P3 the ratio is 3: 5: 8.
In a certain production run, 137 units of M1, 279 units of M2, and 448 units of M3
were used. The problem is to determine how many units of each of the products P1,
P2, and P3 were produced in that run.

a) Set this problem up in matrix form. Use the letter A for the matrix, and write
down the (one-line) formula for the solution in matrix form.

ANSWER

First you have to notice that the restrictions are raw material restrictions.

If you produce the amount $P_1$ you need $\frac16 P_1$ of raw material $M_1$.

If you produce the amount $P_2$ you need $\frac19 P_2$ of raw material $M_1$.

If you produce the amount $P_3$ you need $\frac3{16} P_3$ of raw material $M_1$.

Is this comprehensible ?

Thus the restriction for raw material 1 is

$\frac16 P_1+\frac19 P_2+\frac3{16} P_3=137$

Similar for the restrictions for $M_2$ and $M_3$

$\frac26 P_1+\frac39 P_2+\frac5{16} P_3=279$

$\frac36 P_1+\frac59 P_2+\frac8{16} P_3=448$

The Matrix equation  is $A\cdot \mathbf P=\mathbf m$

In your case

$\begin{pmatrix} \frac16& \frac19 &\frac3{16} \\ \frac26 &\frac39 &\frac5{16}  \\ \frac36 &\frac59 &\frac8{16}\end{pmatrix}\cdot \begin{pmatrix} P_1 \\ P_2 \\ P_3\end{pmatrix}=\begin{pmatrix} 137 \\ 279 \\ 448\end{pmatrix}$

If you calculate the LHS you will see that you will get the three equations above. The kind of the equation comes from the rules of matrix calculation.

You  would get the equations as well if you would transpose both sides:

$(A\cdot \mathbf P)^T=\mathbf m^T$

$\mathbf P^T \cdot A^T=m^T$

$\begin{pmatrix} P_1 & P_2 & P_3\end{pmatrix}\cdot A^T=\begin{pmatrix} 137 & 279 & 448\end{pmatrix}$

Check both equations.


The convergence of $\sum ( \sqrt{n+1}-\sqrt{n})^ a$. How do I keep it from here? Check when the series $\sum ( \sqrt{n+1}-\sqrt{n})^ a$, $a\in \Bbb{R}$ converges.

$Attempt$: Let us focus on $a_n= ( \sqrt{n+1}-\sqrt{n})^ a$. $a_n=( \sqrt{n+1}-\sqrt{n})^ a=( (\sqrt{n+1}-\sqrt{n})\cdot{\sqrt{n+1}+\sqrt{n}\over \sqrt{n+1}+\sqrt{n}})^ a=({1\over \sqrt{n+1}+\sqrt{n}})^a$.

I know I sort of have to find an $a$ for which $({1\over \sqrt{n+1}+\sqrt{n}})^a\le {1\over n^b}$ where $b>1$. Any suggestion for a way to $resume$ my attempt? Thank you.

Okay it remained unsolved but I solved it so if it means anything:
$({1\over \sqrt{n+1}+\sqrt{n}})^a\le ({1\over 2\sqrt{n}})^a$.

If $\sum ({1\over 2\sqrt{n}})^a$ converges so does the original one.

Therefore, $\sum ({1\over 2\sqrt{n}})^a=({1\over 2})^a\sum ({1\over\sqrt{n}})^a$ converges $\iff$ $\sum ({1\over\sqrt{n}})^a$ converges.

But, $\sum ({1\over\sqrt{n}})^a=\sum ({1\over\ n^{1\over 2}})^a=\sum {1\over\ n^{a\over 2}}$
By $p-series$ test, $\sum {1\over\ n^{a\over 2}}$ converges $\iff$ $a>2.$


Is there any difference between $P$ and $\Pr$ to represent probabilities? I have come across both $P(\dots)$ and $\Pr(\dots)$ being used to represent probabilities. Is there any difference in the meaning of these notations, or are they just different shorthands?

I seem to come by $\Pr(\dots)$ more often in Bayesian probability contexts, though I wouldn't say that's a rule.

They are just different conventions.   They don't signify any different meaning.

I personally find the $\Pr$ notation most useful when the discussion involves combinatorics.   It distinguishes probability somewhat from permutation.  (Unless you use ${^n{\rm P}_r}$  ...)

It also has that convenient LaTeX command \Pr which renders it in times roman font, and with some space padding, which helps it stand out in a line of multiplied probabilities using just a few keystrokes.

In classical approach to probability defined by Laplace, probability is the ratio of number of favourable cases to the total number of equally likely cases. Here probability of occurrence of A is denoted by P(A). Symbolically, if an event A can happen in 'a' ways out to total 'n' ways then the probability of occurrence of the event is denoted by Pr(A)=a/n


Interpolating polynomial with integer coefficients Let $n_1 < \dotsb < n_k$ and $m_1, \dotsc, m_k$ be integers. Can we find conditions for the existence polynomial $p$ with integer coefficients such that $p(n_i) = m_i$ for every $i$?

This isn't always possible: for example, since $p(i) \equiv p(0) \mod{i}$, we cannot have $p(0) = 0$ and $p(2) = 1$.

By this answer, if such a polynomial exists then the Lagrange interpolating polynomial has integer coefficients, so the question may be restated as when does the Lagrange interpolating polynomial have integer coefficients.

The coefficients of the Lagrange polynomial that goes through $(n_1,m_1),\ldots,(n_k,m_k)$, can be found by solving $$\begin{pmatrix}1&n_1&\cdots&n_1^{k-1}\\\vdots\\1&n_k&\cdots&n_k^{k-1}\end{pmatrix}\begin{pmatrix}a_0\\\vdots\\a_{n-1}\end{pmatrix}=\begin{pmatrix}m_1\\\vdots\\m_k\end{pmatrix}$$

Hence the coefficients $a_i$ are integers precisely when the determinant of the Vandermonde matrix above divides its adjugate matrix applied to $(m_1,\ldots,m_k)$.

For example, for $n_1<n_2<n_3$ and $m_1,m_2,m_3$, the condition becomes $$(n_1-n_2)(n_1-n_3)(n_2-n_3)\quad | \quad \begin{pmatrix}n_2(n_2 - n_3) n_3 m_1 + n_1 n_3 (n_3-n_1) m_2 + n_1 (n_1 - n_2) n_2 m_3\\
n_3^2 (m_1 - m_2) + n_1^2 (m_2 - m_3) + n_2^2 (m_3-m_1)\\
n_3 (m_2-m_1) + n_2 (m_1 - m_3) + n_1 (m_3-m_2)\end{pmatrix}$$ (by which is meant that the left side divides each of the terms on the right).

I don’t have enough reputation to comment, this is not a complete answer.

I have a necessary and sufficient condition for consecutive ni, which is that the terms of the nth level difference are divisible by n! (We define the kth level difference to be the differences between consecutive terms of the (k-1)th level difference, with the 0th level difference being the sequence itself

Ex:

2nd level differences:  2 2

1st level differences: 2 4 6

The sequence itself: 0 2 6 12

This is proven by observing that any integer polynomial can be written as the sum of integer combinations of the polynomials x(x-1)(x-2)...(x-i)


Sum involving binomial coefficients Exist a closed form for $$\left(-1\right)^{N}\underset{i=1}{\overset{N}{\sum}}\left(-1\right)^{i}\dbinom{N}{i}\dbinom{N+i}{i-1}\,\frac{1}{2i+1}?$$ I think I've to use in some way the formula of the shifted legendre polynomials $$P_{N}\left(x\right)=\left(-1\right)^{N}\underset{i=0}{\overset{N}{\sum}}\dbinom{N}{i}\dbinom{N+i}{i}\left(-x\right)^{i}$$ but I'm not sure about it. I'm wrong? Thank you

For $N\ge 1$ we get the fractions 1/3, 2/15, 1/7, 4/45, 1/11, 6/91, 1/15, 8/153 and so on. So the numerators are 1,2,1,4,1,6,1,8, one's interleaved with the even numbers, and for the denominators 3,15,7,45,11,91,15,153, which are the sequence 3,7,11,15, of numbers == 3 (mod 4) interlaced with the values of 15,45,91,153,.. reproduced in https://oeis.org/A014634. So for odd $N$ the values are $1/(2N+1)$, and for even $N$ they are $N/((N+1)(2N+1))$


What's the midpoint of $[a,b)$? While the midpoint of either $[a,b]$ or $(a,b)$ is $\dfrac{a+b}{2}$, what's the midpoint of $[a,b)$ if it has been defined at all?

It depends on how you want to define the so called "mid point" of an interval. One can simply define the mid point of an interval to be $\frac{a+b}2$ for any interval of the form $(a,b), (a,b], [a,b)$ or $[a,b]$. This definition would serve most general purposes quite well since a single point has measure zero.

However, suppose you want the define the mid point to have the property such that

  for an interval $I$, $m$ is called the midpoint of $I$ if for any $x\in I$ there exist $x'\in I$ such that $x-m=m-x'$

then it's provable that $[a,b)$ has no mid point. (Hint: if such $m$ exists then $I$ must be of the form $(m-d,m+d)$ or $[m-d,m+d]$)

The standard definition of the midpoint is, "The point on a line segment dividing it into two segments of equal length" or the equivalent. As a result, the midpoint is found by the formula $\frac{a + b}{2}$ for all types of intervals.

https://en.wikipedia.org/wiki/Midpoint
http://mathworld.wolfram.com/Midpoint.html


External unit normal field of cylinder Hello, I consider
    $$
A:=\left\{(x,y,z)\in\mathbb{R}^3:x^2+y^2\leq 1,0\leq z\leq 1\right\}
$$
    and
    $$
v(x,y,z):=(x^3,x^2y,zx^2).
$$
    and the task is to calculate the surface of $A$ by using the integral theorem of Gauß.

I guess it is meant to use
$$
\int\limits_{\partial A}\langle v,w\rangle\, dS=\int\limits_A\mbox{div } v\, d^3x,
$$
where $w$ is the external unit normal field of $\partial A$.

Am I right that the external normal field of $\partial A$ is given by $w(x)=e_3$, when $x$ is in the "cover" of the cylinder $A$, $w(x)=-e_3$, when $x$ is in the "ground" of the cylinder "A" and $w(x)=(x,y,0)$ is $x$ is the rest of the cylinder $A$?

$\displaystyle{\vec{r}_{\perp} \equiv x\,\hat{x}\ +\ y\,\hat{y}}$.

\begin{eqnarray*}
\int\left\vert{\rm d}\vec{S}\right\vert
& = &
\int\hat{r}_{\perp}\cdot{\rm d}\vec{S}
=
\int\nabla\cdot\left(\vec{r}_{\perp} \over r_{\perp}\right)\,{\rm d}V
=
\int\left\lbrack
\vec{r}_{\perp}\cdot\nabla\left(1 \over r_{\perp}\right)
+
{1 \over r_{\perp}}\,\nabla\cdot\vec{r}_{\perp}
\right\rbrack\,{\rm d}V
\\
& = &
\int\left\lbrack
\vec{r}_{\perp}\cdot\left({-1 \over r_{\perp}^{2}}\,{\vec{r}_{\perp} \over r_{\perp}}\right)
+
{1 \over r_{\perp}}\,2
\right\rbrack\,{\rm d}V
=
\int{1 \over r_{\perp}}\,{\rm d}V
=
\int_{0}^{r}{\rm d}r_{\perp}\int_{0}^{2\pi}\,{\rm d}\theta\int_{0}^{h}{\rm d}z
\\
& = &
2\pi rh
\end{eqnarray*}


Prove that: $\aleph_0 \cdot \frak{c} = \frak{c} \cdot \frak{c}$ I've been fiddling with this enough.
Found an answer here but didn't quite understand it.

How do I prove that:
$$\aleph_0 \cdot \frak{c} \leq \frak{c} \cdot \frak{c}$$
$$\frak{c} \cdot \frak{c} \leq \aleph_0 \cdot \frak{c}$$

The first inequality I've proven already, the second one I just cannot get it(obvious is it may be).

The only thing I'm allowed to use is the definition of cardinality, the cantor-bernstein-schroder theorem and the fact that:
$$ k \leq p \wedge m \leq n \to k\cdot m \leq p \cdot n $$
(which is helpful through the definition of cardinality)

Any insights are welcomed!

$\mathfrak{c}\times\mathfrak{c} = 2^{\aleph_0}\times 2^{\aleph_0} = 2^{\aleph_0+\aleph_0} = 2^{\aleph_0} = 1\times 2^{\aleph_0} \leq \aleph_0\times 2^{\aleph_0}$.

For the equality $2^{\aleph_0}\times 2^{\aleph_0} = 2^{\aleph_0+\aleph_0}$, note that a pair of functions from $\aleph_0$ to $\{0,1\}$ is equivalent to a function from the disjoint union of two copies of $\aleph_0$ to $\{0,1\}$.

Since $\aleph_0+\aleph_0$ is the cardinality of the disjoint union of two copies of $\aleph_0=\omega$, and $2^{\kappa}$ is the cardinality of the set of all functions from $\kappa$ to $\{0,1\}$, the above argument seems to meet your requirements.

$\mathfrak{c}\times\mathfrak{c} =\text {max}\{\mathfrak{c},\mathfrak{c}\}
=\mathfrak{c}=\text{max}\{\mathfrak{c},\aleph_0\}=\mathfrak{c}\times\aleph_0  $


Assumption of a Random error term in a regression In one of my recent statistics courses, our teacher introduced the linear regression model. The typical $y=\alpha + \beta X + \epsilon$, where $\epsilon$ is a "random" error term. The teacher then proceeded to explain that this error term is normally distributed and has a mean zero.

The error term is what is confusing me. What exactly does random mean? My back ground in statistics is very low level, but I understand that a random variable is defined as a mapping from a sample space to the real numbers. This definition makes sense, but the assumption of a zero mean is what I get tripped up on. How can we assume this fact?

I've been trying to think about it intuitively, and can only think that in regards to the real numbers, zero is in a sense "the middle ground" and splits up the reals into 2 "equal length" parts. However, I know that the reals are uncountable so this may be a case where my intuition is incorrect.

I apologize in advance if my question is confusing.

Here's the general idea - someone who has a better background than I do in statistics could probably give a better explanation. So you have this linear regression model:
$$Y = \alpha + \beta X + \epsilon $$
where $\epsilon$ follows a normal distribution with mean $0$.

  What exactly does random mean?  My back ground in statistics is very low level, but I understand that a random variable is defined as a mapping from a sample space to the real numbers. This definition makes sense, but the assumption of a zero mean is what I get tripped up on. How can we assume this fact?

Personally, I've always taken the idea that $\epsilon$ follows a normal distribution with mean $0$ as an axiom of sorts for the linear regression model. My understanding is that it's just something nice we would like the linear regression model to have and lends itself well to certain properties. Remember:

  Essentially, all models are wrong, but some are useful.

which is attributed to George E.P. Box.

Why would we want such an axiom? Well... on average, it would be nice to have zero error.

In my honest opinion (this is based off the little measure-theoretic probability I have studied), it would be best to approach this idea of "randomness" intuitively, as you would in an undergraduate probability course.

The idea about anything that is random is that you will never know the value of it. So, in an undergraduate probability class, what you do is you assign probabilities to the values your quality of interest can take by creating a probabilistic model. Your model, 99% of the time, won't be perfect, but that doesn't stop anyone from not trying.

The normal distribution with mean 0 is just an example of a probabilistic model that statisticians feel is a suitable model for the error term. It isn't perfect, but it's suitable for most purposes. I worked with a professor whose focus is on assuming a skew-normal error term, which complicates things, but is usually more realistic, since, in reality, not everything looks like a bell curve.

My two cents. Hopefully I've helped somewhat.

I think the other answers have done a pretty good job of explaining what the assumption means, but I just want to add why this is such an important assumption for any future readers. This implies that outside of the covariates we include in our model, the rest of the variance is completely independent and normally distributed across the observations. However there are actually a lot of times in the real world that this isn't true. Let's say you want to find the relationship between a person's income and their BMI from a dataset of thousands of people from all 50 states in US. Intuitively, we may suspect that people from the same area of the country are probably more similar to each other in both weight and income than people from random parts of the country. Therefore, we can't assume the error is a zero-mean normally independently distributed term.

However, we also probably don't want to add 50 state coefficients or hundreds of county coefficients into our model. Instead we can use random effects hierarchical model where we assume that the states or counties also have their own error distribution and that component of the error term is separated from the individual error term. By doing this we reduce the risk of identifying a spurious relationship between BMI and income which really is due to regional differences in income and BMI. Similar approaches are done when modeling spatial phenomena (i.e. we expect neighboring ZIP codes or counties to be more similar to each other than farther apart ones) and longitudinal analysis (observations within one patient are going to be related to each other).


Greater/lesser search with one false answer allowed It is well known that you can determine the values of $n\geq 2$ bits using $k$ yes/no questions about the bits (for example, "is $x_1 \oplus x_3 = 1$?), even if one (but not more) of the answers obtained may be a lie, if and only if
$$
2^n k \leq 2^k
$$
This is just the observation that you can always construct a Hamming code using $k-n$ extra bits, which corrects single-bit errors, if $k$ is of that size.

(This does not work if $n=1$ since you require $3$ bits rather than $2$ to achieve single error correction.)

For example, you can determine a 4-bit number using seven questions, even allowing for one possible lie, by asking for the values of each bit, and then asking for the values of $x_0 \oplus x_1 \oplus x_2$, then $x_0 \oplus x_1 \oplus x_3$, then $x_0 \oplus x_2 \oplus x_3$.  Note that these questions can be fixed beforehand; later questions need not depend on answers to the earlier ones.

But what if the nature of the yes/no questions allowed is restricted?

In particular, can you determine (with certainty) a specific number $x \in [0,15]$ using no more that seven questions of the form "Is $x > q$", coping with up to one lie?  It is allowable to tailor later questions based on the answers to earlier questions.

As a generalization, what are the conditions on $N$ and  $k$ that suffice to ensure that you can determine a specific $x \in [0,N)$ using up to $k$ questions of the form "Is $x > q$", coping with up to one lie? My real question to be posed is:

What is the largest value of $N$ such that a specific number $x \in [0,N)$ can with certainty be determined in $7$ "Is $x > q$" questions, coping with at most one lie?

Stanislaus Ulam did some work on this in the 1940's and that led to the work that led to Hamming codes, but I don't remember whether he actually presented a condition for general $N$.

For some examples,
$$k(N=2) = 3 \\
k(N=3) = 4 \\
k(N=4) = 5 \\
k(N=5) = 6 \\
k(N=6) = 6 \\
k(N=7) = 7 \\
k(N=8) = 7 \\
$$

As for the case of $N = 15$, I strongly suspect that the "information theory value" of seven questions does not suffice when the ordering restriction is imposed.  But I'd be interested to see a proof.

I have proven an optimal strategy and written a short program to compute for small $N$; see my answer to my generalized question.

This completely answers your question rigorously. Specifically, $7$ queries is enough to distinguish up to at most $12$ possibilities, while $8$ queries is enough to distinguish up to at most $22$ possibilities.

There has been some discussion of this in a newer question, where you will find a detailed exposition and a certain amount of disagreement; but I will summarise the information-theoretic aspects here.

The key is to note the correct information content of the problem. Let us suppose that $Q$ questions are to be asked. I will cause irritation and confusion by supposing that the values to be guessed are in the half-open interval $[0,N)$ rather than the closed one you specified. This means that there are exactly $N$ possibilities to be guessed, rather than $N+1$, and it makes the algebra easier to read. The other change in notation is that I will refer to the questions as being of the form "$x<k$?".

The answerer has to have two pieces of information to do his job. One is the number $n$ to be guessed: $N$ possibilities. The other is the number $q$ of the question which he is to answer with a lie. $q$ does lie in the closed interval $[0,Q]$, because we can take $q=Q$ to mean "the answerer never lies". Thus the total number of possibilities between which the answerer has to choose is $(Q+1)N$.
The questioner can't reach his conclusion without deducing all the information that the answerer needed to produce his answers: thus, $(Q+1)N$ possibilities. Another way of looking at it is that when the questioner has guessed the number $n$, he also knows the number $q$.

Since each question can distinguish between two possibilities only, $Q$ questions are needed to distinguish between $2^Q$ possibilities. Thus, comparing the possibilities distinguishable by $Q$ questions with the possibilities that need to be distinguished, we have $$2^Q\ge(Q+1)N$$

I shall present an algorithm which, I claim, is within one or two units of achieving a successful guess with the smallest integer $Q$ that satisfies the above equation.

The strategy

The questioner's strategy is to play $Q+1$ games simultaneously: one in which the answerer lies at question $0$, one in which he lies at question $1$, and so on up to question $Q-1$; plus one in which the answerer doesn't lie at all (which it is not illogical to describe at "lies at question $Q$", since question $Q$ is never asked).

At each stage, the questioner asks one and only one question, "is $x<k$?", and applies the answer to all $Q+1$ simultaneous games. Thus the answerer only hears $Q$ questions, as required.

At the start, each game has all $N$ numbers as possibilities. Each answer to a question reduces that number. It should be noted that there is no question of lying in these sub-games: each one receives, at each stage, a true answer. All the questioner has to remember is that the answer to question $q$ should be inverted ("Yes" becomes "No" and "No" becomes "Yes") for game $q$, to allow for the fact that game $q$ is the one in which answer $q$ is a lie.

As time goes on, the possible numbers resulting from each of the $Q+1$ games become fewer and fewer. For some games they all disappear, and eventually they disappear for all games except one, and in that game only one number is left. At that point the number is known, and the identity of the game which still contains that number gives the number of the question that was answered falsely (with, as before, game number $Q$ denoting "there were no false answers").

The strategy at step $q$ is therefore:

Define $N_q$ as the number of possible numbers, totalled across all games (thus, the number of possible $\langle n,q\rangle$ pairs). Clearly $N_0=(Q+1)N$.
Ask a question "$x<k$?" such that a Yes answer eliminates precisely half of the possible numbers, totalled across all games. Then a No answer will also eliminate half the possible numbers. We can say that $N_<=N_\ge=\frac12N_q$.
Given the answer, eliminate all numbers in game $q$ which do correspond to the answer, because in this round the answer to game $q$ is a lie so that "$x<k$" means that $x$ is greater than or equal to $k$. In all other games, eliminate all numbers which do not correspond to the answer, because in this round the answer to all games other than $q$ is true.
You now have either $N_{q+1}=N_<$ or $N_{q+1}=N_\ge$: that is to say, $N_{q+1}=\frac12 N_q<$ if $k$ was chosen perfectly.

The possibility of choosing $k$

At any stage, games can be of three kinds:

Games in which the lie has already been told.
Games in which the lie is being told in the current round.
Games in which the lie will be told in the future (or will never be told).

The ranges of numbers covered by class (1) are all disjoint (since each game comes from a different sequence of Yes/No after allowing for lies). Classes (2) and (3) all contain identical ranges to each other, but that range is disjoint from any of the class (1) ones.

The effect of this on the value of $N_<$ for various potential values of $k$ is as follows:

Wherever $k$ is in class (1), $dN_</dk=+1$. That is, each increase of 1 in "$x<k?$" yields an increase of 1 in the number of numbers that remain possible if the answer is Yes.
Wherever $k$ is in class (2) or (3), $dN_</dk=Q-q-2$. That is, each increase of 1 in "$x<k?$" yields an increase of $Q-q-2$ in the number of numbers that remain possible if the answer is Yes. This is because there are $Q-q-1$ games in class (3) and they all share the same range of possibilities and are all affected by answers in the same way, and there is $1$ game in class (2) which shares the same range, but for which the number of possible numbers decreases by 1 unit for each 1-unit increase in $k$. (The reverse direction is because a "Yes" to "$x<k?$" means, for class (2), "No").
If $k$ is not in any of these classes, $dN_</dk=0$.

Leaving aside occasional flat spots, the graph of $N_<$ versus $k$ is therefore monotone. Where the slope is 1, it is easy to find a $k$ such that $N_<=\frac12N_q$. This would yield an explicit algorithm which halves the number of possibilities at each stage and therefore reaches a conclusion in the minimum theoretical time.

The only time where this breaks down is when the slope is not 1 at the point where the horizontal line $N_<=\frac12 N_q$ intersects the graph. In that region $N_<$ jumps by $Q-q-2$ units for each one-unit step in $k$. This means that the attainable value of $N_<$ closest to $\frac12 N_q$ could be $\frac12(Q-q-2)$ units away from it. This, in turn, means that rather than having the ideal $$N_{q+1}=\frac12 N_q$$ we may have $$N_{q+1}=\frac12 N_q+\frac12(Q-q-2)$$

This slightly slows the progress of the game, and it makes explicit analysis difficult; but since the largest value of $(Q-q-2)$ will be of the order of $\log_2{N}$, it will be seen that it does not have an overwhelming effect on progress.

The algorithmic complexity of finding the right value of $k$ is essentially logarithmic. It is a question of sorting the available ranges into order; a binary search to find the range which brackets the desired value of $N_<$; and simple arithmetic to find the desired value in the range.


Infinite product of probability measures is a premeasure This is an exercise from Real Analysis by Stein and Shakarchi (Chapter 6, Exercise 15).

Given infinitely many measure spaces $(X_i, \mathcal M_i, m_i)$, each of which has measure 1, one can define an algebra on the product space consisting of all finite unions of the “cylinders”, by which we mean rectangles of the form $E_1 \times E_2 \times \cdots$, where $E_i$ belong to $\mathcal M_i$ and all but finitely many of $E_i$ are equal to $X_i$. Then define $m(E_1 \times E_2 \times \cdots) = m_1(E_1)\,m_2(E_2)\cdots$. How does one prove that $m$ is a premeasure on the algebra defined above?

One only needs to check the equality in the definition of premeasure, but it seems a subtle problem of the exchange of summation and limit progress is involved, which can be easily ignored without carefulness.

I would like some hints or any reference book about it.

This is proven as Theorem 3.5.1 in Measure Theory, Volume 1 by Vladimir Bogachev (WorldCat link if you want to find a library copy).

An overarching theme of the proof is that each cylinder set is essentially an element of finite product $\sigma$-algebras; the countably infinitely many factors $X_i$ tacked on at the end act essentially as filler.

Hint.

First prove the following lemma: Let $(X,\mathcal M, \mu)$ and $(Y,\mathcal N, \nu)$ be measure spaces. Suppose $A \in \mathcal M \otimes \mathcal N$ satisfies $(\mu \times \nu)(A) > t > 0$. For $x \in X$, we have the sections/slices $A^x := \{y \in Y : (x,y) \in A\}$. Let $B := \{x \in X : \nu(A^x) > t/2\}$. Then $B \in \mathcal M$ and $\mu(B) > t/2$.

To establish the main result, it suffices to show that if $(A_k)_{k=1}^\infty$ is a sequence of cylinder sets for which there exists a constant $\epsilon > 0$ such that $m(A_k) > \epsilon$ for all $k$, then $\bigcap_{k=1}^\infty A_k \subset \prod_{j=1}^\infty X_j$ has at least one element. The lemma is helpful here.

There's another proof outlined in Sam Drury's class notes for his Math 355 course at McGill under Section 4.6, "Infinite products of probability spaces". You can find my commentary on the proof in an old revision of this answer.

The trick is that we know the union must have only finitely many non-trivial co-ordinates and that the unions are increasing.

To prove $m$ is a pre-measure, you only need to prove the summation condition for disjoint events $\{A_i\}_{i \ge 1}$ where $\cup_i A_i$ lies in the algebra. Only way $\cup_i A_i$ lies in the algebra is if $$\cup_i A_i = E_{i_1} \times E_{i_2 } \times \ldots \times E_{i_k}$$

Let $E_{i_j}^{(N)}$ be the $i_j$th coordinate for $\cup_{i=1}^N A_i$.

which means for all $\epsilon >0$, for all large enough $N$ , $$m (E_{i_j} \setminus E_{i_j}^{(N)}) < \epsilon$$
for all $j=1,\ldots, k$.Hence
$$ m(\cup_iA_i) -  m(\cup_{i=1}^N A_i) = m (\cup_iA_i \setminus \cup_{i=1}^N A_i)  < \prod_{j=1}^k m(E_{i_j} \setminus E_{i_j}^{(N)}) < \epsilon^k$$

since in $\cup_iA_i \setminus \cup_{i=1}^N A_i$, there are finitely many non-trivial co-ordinates and we can replace the measures of all co-ordinates except $i_1,\ldots, i_k$ by $1$ for the upperbound.


Proof using Pigeon hole principle correct? Here is problem 33 from "Putnam and Beyond" :

Given 50 distinct positive integers strictly less than 100, prove that some two of
them sum to 99.

Now I answered the problem differently than the official solution in the book and I was hoping you could tell me if my answer is still correct, or if it is not, explain to me where I went wrong. So here is my answer:

As the book suggests I try and use the pigeon hole principle. In my answer the "pigeons" are all two element subsets of the given set and the "holes" are the possible values the sum of the two elements of any two element subset can take. There are 1225 two-element subsets of any such given set and the sum of the two members of these subsets can take any value from 1+2=3 to  99+98=197. Hence there are 1225 "pigeons" and only 195 "boxes". There must be more than two pigeons in every box and hence at least two subsets with sum of 99.

Thats my answer. Thanks for any help!

The $49$ pairs $(1/98) , (2/97) , (3/96) ,\cdots (49,50)$ sum up to $99$. If you choose $50$ distinct positive integers, at least one pair must be hit twice.

The simple answer to your question is that your choice of "pigeons" and "holes" didn't support a proof of the claim. That's not to say it didn't do something. As the commenters correctly pointed out, you showed that you can't have a set of $50$ numbers less than $99$ such that any two numbers in the $50$ sum to a unique number, relative to any other pair. If this is what the question had been asking, then your logic is absolutely sound.

But the quesiton wasn't asking that. It was asking for the existence of $99$ as one of the pairwise sums. And, in this case, the choice of pigeons and holes given in the book works, and yours doesn't.

The thing about the pigeon hole principle, or any other proof technique, is that it's a tool. And like any tool, it has to be applied the right way to solve the problem at hand. This is a great lesson for you as a mathematician- to see that there are many ways to apply a single tool to a single problem. It's not enough to say "The pigeon hole principle didn't work". You must say "The pigeon hole principle, with this choice of pigeons and holes, gives me this result, which doesn't seem to help." All the same, you mentally note it in case you need it later, and then you either change your choice of tool or try to apply the tool in a different way. Problem solving is an art, as much as a science.


determination of probability Recently I have faced a problem related to probability.I have tried it by bayseian theorem but failed.Here is the problem:

A doctor knows that pneumonia causes a fever 95% of the time. She knows that if a person is selected
randomly from the pneumonia. 1 in 100 people suffer from fever. You can go to the doctor complaining
about the symptom of having a fever (evidence).  What is the probability that pneumonia is the cause of
this symptom?

Can anyone give some hints or how can i solve it?

There are many ambiguities in the question itself and the question can be interpreted in two ways.

At first it says that "pneumonia causes a fever 95% of the time" and later it is being contradictory/(false in terms of stats) by saying " if a person is selected randomly from the pneumonia. 1 in 100 people suffer from fever." So, how come only 1 in 100 person are suffering from fever when pneumonia itself causes fever 95% of the time. If we go by this, then it should have been "...95 in 100 people suffer from fever".

Another way to interpret the question is,

Since there a 'fullstop' between "if a person..." and "1 in 100...". Therefore it in this way, the line "She knows that if a person is selected randomly from the pneumonia" will make no sense at all.

Putting everything in one and somehow trying to get the answer, may return multiple different answers because different people will interpret this question as differently.

But the answer would be:

Consider that there are only 100 people in that town. Out of which it is given that 1 person suffers from "fever". We are also provided that pneumonia causes fever 95% of the times. Therefore, for that "1" person, who suffers from fever, there would be $ 95/100 * 100 = 95\% $ person chances that it is caused by pneumonia.


Integral, Definite Integral $ \int_{-\infty}^\infty \exp{\big(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon}\big)dx, \ \alpha <0. $ Calculate  the integral

$$
I=\int_{-\infty}^\infty \exp{\big(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon}\big)dx, \ \alpha <0.
$$
The answer can be expressed analytically in terms of a series which will have a $\Gamma$ function in it.  I am not sure how to approach it since I do not know how to work with the argument of the exponential.  The other constants we can assume to be real.  The answer can be expressed as
$$
I=e^{\epsilon} \sum_{n,m,p=0} \frac{\beta^{4n}}{(4n)!}\frac{\gamma^{2m}}{(2m)!}\frac{\delta^{4p}}{(4p)!}\frac{\Gamma(3n+m+p+\frac{1}{4})}{\alpha^{3n+m+p+\frac{1}{4}}}
$$

OP mentions in a comment that the integral

$$ \tag{1} I~:=~\int_{-\infty}^{\infty} \! dx~\exp \left(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon \right), \qquad \alpha ~<~0,$$

appears on the Superstring theory Wikipedia page.
The $\epsilon$ dependence is trivial, so let's put $\epsilon=0$ in what follows.

Define
$$ a~:=~\sqrt[4]{-\alpha}>0, \qquad b~:=~\frac{\beta}{a^3}, \qquad c~:=~\frac{\gamma}{a^2}, \qquad d~:=~\frac{\gamma}{a}, $$
$$ \tag{2} y~:=~ax, \qquad t~:=~y^4. $$

Then

$$\begin{align} I &= \frac{1}{a}\int_{-\infty}^{\infty} \! dy~\exp \left(- y^4+by^3+cy^2 +dy\right)\\
&=\frac{1}{a}\sum_{n,m,p\in \mathbb{N}_0}  \frac{b^n}{n!}\frac{c^m}{m!}\frac{d^p}{p!}\int_{-\infty}^{\infty} \! dy~y^{3n+2m+p}e^{- y^4}\\
&\stackrel{(4)}{=} \frac{1}{a}\sum_{k,n,m,p\in \mathbb{N}_0} \frac{b^n}{n!}\frac{c^m}{m!}\frac{d^p}{p!} ~\delta_{2k,3n+2m+p} ~\frac{1}{2} \Gamma\left(\frac{2k+1}{4}\right)\\
&\tag{3}~=~ \frac{1}{2}  \sum^{n\equiv p \pmod 2}_{n,m,p\in \mathbb{N}_0}  \frac{\beta^n}{n!}\frac{\gamma^m}{m!}\frac{\delta^p}{p!} \frac{\Gamma\left(\frac{3n+2m+p+1}{4}\right)}{a^{3n+2m+p+1}}\\
\end{align}$$

where we used that

$$\tag{4}  2\int_{0}^\infty \! dy~y^{2k} e^{- y^4}
~=~\frac{1}{2}\int_{0}^{\infty}  \! dt~t^{\frac{2k-3}{4}}e^{-t}~=~\frac{1}{2}\Gamma\left(\frac{2k+1}{4}\right).$$

The formula (3) is not the formula from the Superstring theory Wikipedia page. Without having checked the physics behind the integral $I$, it seems that there might be an implicit procedure assumed, where one e.g. should somehow replace $a\to \omega a$, where $\omega =\sqrt[4]{1}\in\{1,i,-1,-i\}$ is a fourth root of unity, and then somehow average over these four roots. Or something along these lines. We do not plan to pursuit or check such loose speculations here.

why not use the definition of $\Gamma$ function ?

you can use the  $\Gamma$ function at first :

$
I=e^{\epsilon} \sum_{n,m,p=0} \frac{\beta^{4n}}{(4n)!}\frac{\gamma^{2m}}{(2m)!}\frac{\delta^{4p}}{(4p)!}\frac{\Gamma(3n+m+p+\frac{1}{4})}{\alpha^{3n+m+p+\frac{1}{4}}}
$$=e^{\epsilon} \sum_{n,m,p=0} \frac{\beta^{4n}}{(4n)!}\frac{\gamma^{2m}}{(2m)!}\frac{\delta^{4p}}{(4p)!}\frac{\int{e^{-t}t^{{3n+m+p+\frac{1}{4}}-1}dt}}{\alpha^{3n+m+p+\frac{1}{4}}}
$

then, you can expand the follwoing two series :

$ \exp{\big(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon}\big), \ \alpha <0$
${e^{-t}t^{{3n+m+p+\frac{1}{4}}-1}}$

and compare their value nextly, a point you should be careful is that the integral doman of the two formulas above are different !

the hint is to assume :

$y=\alpha^{1/4}x$

then, $ \exp{\big(\alpha x^4+\beta x^{3}+\gamma x^2 +\delta x}\big)$$ =\exp{\big(y^4+\frac{\beta y^3}{\alpha^{3/4}}+\frac{\gamma y^2}{\alpha^{1/2}} +\frac{\delta y}{\alpha^{1/4}}}\big)$

therefore, we can let $n=1/4$$,$$m=1/2$$,$$p=1/4$

which lead that :

$ =\exp{\big(y^4+\frac{\beta y^3}{\alpha^{3/4}}+\frac{\gamma y^2}{\alpha^{1/2}} +\frac{\delta y}{\alpha^{1/4}}}\big)=$

$ =\exp{\big(y^4+\frac{(\beta^{1/4}y^{3/4})^{4}}{\alpha^{3/4}}+\frac{({\gamma^{1/2} y})^2}{\alpha^{1/2}} +\frac{（\delta^{1/4} y^{1/4})^{4}}{\alpha^{1/4}}}\big)$

consequently, we can put the following terms out :

$\sum_{n,m,p=0} \frac{\beta^{4n}}{(4n)!}\frac{\gamma^{2m}}{(2m)!}\frac{\delta^{4p}}{(4p)!}\frac{1}{\alpha^{3n+m+p+\frac{1}{4}}}$

then we can make our assumption :

$z=(ay)^{4}$

so the only thing you should compare is the relation below :

${e^{-t}t^{{3n+m+p+\frac{1}{4}}-1}}$$=e^{-t}{t^{3/4}}=e^{zy+z^{3/4}+z^{1/2}+z^{1/4}}$$=exp$$(z^{3/4}(z^{1/4}+1+z^{-1/4}+z^{-1/2}))$$\Longrightarrow$$(z^{3/4}(z^{1/4}+1++z^{-1/4}+z^{-1/2}))$$=e^{-t}{t^{3/4}}-1$
$\Longrightarrow$$\infty=\infty$

the condition is :

$t,z\longrightarrow\epsilon$$\Longrightarrow$$t^{3/4}=z^{3/4}$$\Longrightarrow$$t^{-3/4}=z^{-1/4-1/2}$

then your solution is right , thank you !


Model the constraints of an integer linear program I have to model the costraints of an integer linear program. The objective function is not important here.
Suppose we have $C$ configurations (indexed by $i = 1, \dots, C$) and $T$ time intervals (indexed by $j = 1, \dots, T$). I wish that:
a) only one configuration is active at each time interval;
b) when a configuration has been chosen I must keep it for at least $\bar{t}$ time intervals (with $\bar{t} \leq T$).
So I define the decision variables:
$y(i,j) :=
\begin{cases}
1 & \text{if configuration $i$ is active at time $j$}; \\[2ex]
0 & \text{otherwise}.
\end{cases}$
For request a) I write the constraints:
\begin{equation}
\sum_{i=1}^C y(i,j) = 1 \ \ \ \forall \ j = 1, \dots, T.
\end{equation}
Now, I don't know how to proceed with the request b). Any help or suggestions are appreciated. Thanks in advance.

For part b), you can introduce another binary variable $x_{i,j}$ to indicate the start, together with constraints:
\begin{align}
x_{i,j} &\le y_{i,k} &&\text{for $k\in\{j,\dots,j+\overline{t}-1\}$} \tag1\\
x_{i,j} &\le 1 - y_{i,j-1} \tag2\\
(1 - y_{i,j-1}) + y_{i,j} - 1 &\le x_{i,j} \tag3
\end{align}
Constraint $(1)$ enforces $x_{i,j} = 1 \implies \land_{k=j}^{j+\overline{t}-1} (y_{i,k} = 1)$.
Constraint $(2)$ enforces $x_{i,j} = 1 \implies y_{i,j-1} = 0$.
Constraint $(3)$ enforces $(y_{i,j-1} = 0 \land y_{i,j} = 1) \implies x_{i,j} = 1$.

Alternatively, you can omit $x$ and just impose
$$(1 - y_{i,j-1}) + y_{i,j} - 1 \le y_{i,k} \quad\text{for $k\in\{j+1,\dots,j+\overline{t}-1\}$},$$
equivalently,
$$y_{i,j} - y_{i,j-1} \le y_{i,k} \quad\text{for $k\in\{j+1,\dots,j+\overline{t}-1\}$}, \tag4$$
(where $y_{i,j}$ is treated as $0$ if $j<0$ or $j>T$),
which enforces
$$(y_{i,j-1} = 0 \land y_{i,j} = 1) \implies \bigwedge_{k=j+1}^{j+\overline{t}-1} (y_{i,k} = 1).$$

Yet a third approach is to enforce
$$\left(y_{i,j-1} \land \lnot \bigwedge_{k=1}^{\overline{t}} y_{i,j-k}\right) \implies y_{i,j},$$
equivalently,
$$\left(\lnot y_{i,j-1} \lor \bigwedge_{k=1}^{\overline{t}} y_{i,j-k}\right) \lor y_{i,j},$$
which has conjunctive normal form
$$\bigwedge_{k=1}^{\overline{t}} \left(\lnot y_{i,j-1} \lor y_{i,j-k} \lor y_{i,j}\right),$$
yielding linear constraints
$$1 - y_{i,j-1} + y_{i,j-k} + y_{i,j} \ge 1 \quad \text{for $k \in \{1,\dots,\overline{t}\}$}.$$
Rearranging this, we obtain
$$y_{i,j} \ge y_{i,j-1} - y_{i,j-k} \quad \text{for $k \in \{1,\dots,\overline{t}\}$}. \tag5$$
Now we can recover @toronto hrb's formulation by aggregation:
$$y_{i,j} \ge y_{i,j-1} - \frac{1}{\overline{t}}\sum_{k=1}^{\overline{t}} y_{i,j-k} \tag6$$
So $(6)$ is correct but weaker than $(5)$.

Writing variables as $y(i,j)$ is not conventional. Following what you've written, define
$y_{ij}=
\begin{cases}
1 & \text{if configuration $i$ is active at time $j$}; \\[2ex]
0 & \text{otherwise}.
\end{cases}$
You also need to define binary variables to indicate if a configuration is chosen or not as
$z_i =
\begin{cases}
1 & \text{if configuration $i$ is chosen}; \\[2ex]
0 & \text{otherwise}.
\end{cases}$
Now the constraint of part b) can be written as
$$\sum_{j=1}^{T}y_{ij}\ge \bar{t} \cdot z_i, i=1,\cdots, C$$
If configuration $i$ is chosen, $z_i=1$, at least $\bar t$ time intervals are forced; otherwise, $z_i=0$, which places no constraints on time invervals.
Edit for consecutive time intervals. No need to use $z_i$ any more.
Let's use Configuration 1 with $\bar{t}=3$ as an example. The constraints should be like
$$y_{12}\ge y_{11}-\frac{1}{3}y_{11}$$
$$y_{13}\ge y_{12}-\frac{1}{3}(y_{11}+y_{12})$$
$$y_{14}\ge y_{13}-\frac{1}{3}(y_{11}+y_{12}+y_{13})$$
$$y_{15}\ge y_{14}-\frac{1}{3}(y_{12}+y_{13}+y_{14})$$
$$\cdots$$
$$y_{1,T}\ge y_{1,T-1}-\frac{1}{3}(y_{1,T-3}+y_{1,T-2}+y_{1,T-1})$$
Does this work? If $y_{11}=1$, then the first three constraints force $y_{12}=y_{13}=1$, but $y_{14}$ is flexible.
Putting it in a compact form, it becomes
$$y_{ij}\ge y_{i,j-1}-\frac{1}{\bar{t}}\sum_{k=1}^{\bar{t}}y_{i,j-k}, j=2, \cdots, T; i=1,\cdots, C.$$
$$y_{i,0}=y_{i,-1}=0, i=1, \cdots, C.$$


Why $ \|x^* x \| = \|x\|\|x^*\|$ is equivalent to $\|xx^*\| = \|x\|^2$ in the definition of $C^*$ algebra? I read the definition of  $C^*$ algebra in Wikipedia where it says $\|x^* x \| = \|x\|\|x^*\|$ is equivalent to $\|xx^*\| = \|x\|^2$ but I do not know why. Can you show me how to derive $\|xx^*\| = \|x\|^2$ from $\|x^* x \| = \|x\|\|x^*\|$?

You can take a look at this book by Doran and Belfi. The proof you are looking for takes most of chapter 3, some 15+ pages. Unfortunately, not all of it is shown in Google Books.

On a brief historical note, the question you are asking was asked by Gelfand and Naimark in 1943. Glimm and Kadison proved it for unital algebras in 1960, using deep results. The non-unital case is from the 1970s.

For one direction, if we start by assuming $\|xx^*\| = \|x\|^2$ then as $C^*$-algebras are Banach algebras we get $ \|x\|^2 = \|xx^*\| \le \|x\|\|x^*\|$ so that $\|x\| \le \|x^*\|$. Then $\|x\| \le \|x^*\| \le \| (x^*)^*\| = \|x\|$ so that $\|x\| = \|x^*\|$. It then follows that $\|xx^*\| = \|x\|^2 = \|x\|\|x^*\|$.

The other direction is a fair bit tricker and I'm trying to remember how it goes.


Taylor expansion in elementary symmetric polynomials Assume that I have an analytic, symmetric function $F(x,y,z)$. Such a function should have an expansion of the form
$$  F(x,y,z)  = \sum_{k,m,n\geq 0} c_{k,m,n} e_1^ne_2^m e_3^k$$
where $e_1,e_2,e_3$ are the elementray symmetric polynomials. How would I go about calculating these $c_{m,n,k}$ in an efficient way (by efficient I mean something better than rewriting the polynomials $$ P_{n_1,n_2,n_3}(x_1,x_2,x_3) = \sum_{\sigma \in S_3} x^{n_{\sigma(1)}}y^{n_{\sigma(2)}}z^{n_{\sigma(3)}}$$
 in terms of elementary polynomials by hand...).

More concretely I would be interested in knowing whether the function
$$ G(x_1,x_2,x_3)  =  \sum_{\sigma \in S_3} \operatorname{exp}\Big[ Ax_{\sigma(1)}+Bx_{\sigma(2)}+Cx_{\sigma(3)}\Big] $$
can be rewritten as a function in elementary symmetric polynomials in a nice form.

I hope you are deligent enough to do heavy duty algebraic computation. I use the Cauchy itegration form of remainder calculation,
$\mathop {\lim }\limits_{x\; \to \;0} \left\{ {f\;\bmod g} \right\} = \frac{1}{{2\pi \sqrt { - 1} }}\oint {\frac{{{x^{\deg f - 1}}}}{{f\left( x \right)}}g\left( x \right)dx} $
The contour includes all the roots of $f$.
Then answer to your question is obtainable in two steps
$\Phi \left( z \right) = \mathop {\lim }\limits_{x\; \to \;0} \left\{ {F\left( {x,{e_1} - z - x,z} \right)\;{{\bmod }_x}\left( {{x^2} - \left( {{e_1} - z} \right)x + {e_2} - z\left( {{e_1} - z} \right)} \right)} \right\}$
then
$F\left( {{e_1},{e_2},{e_2}} \right) = \mathop {\lim }\limits_{z\; \to \;0} \left\{ {\Phi \left( z \right){{\bmod }_z}\left( {{z^3} - {e_1}{z^2} + {e_2}z - {e_3}} \right)} \right\}$
I let Mathematica do the messy computation, for
$F\left( {x,y,z} \right) = \sum\limits_{{m_1} + 2{m_2} + 3{m_3} = {n_1} + {n_2} + {n_2}} {{C_{{n_1},{n_2},{n_3}}}\left({{m_1},{m_2},{m_3}} \right)\left\{{\frac{1}{{{n_1}!{n_2}!{n_3}!}}\frac{{{\partial^{{n_1}+{n_2}+{n_2}}}\;F}}{{\partial{x^{{n_1}}}\partial{y^{{n_2}}}\partial{z^{{n_2}}}}}\left({0,0,0}\right)} \right\}{e_1^{{m_1}}}{e_2^{{m_2}}}{e_3^{{m_3}}}}$
I got
${C_{{n_1},{n_2},{n_3}}}\left( {{m_1},{m_2},{m_3}} \right) = \sum\limits_{{k_1} + 2{k_2} = {n_1} + {n_2}} {\left( {\begin{array}{*{20}{c}}{{n_2} - {m_2}}\\
{{m_1}}\end{array}} \right)\frac{{{{\left( { - 1} \right)}^{{n_1} + {n_3} + {m_1} + {m_3} + {k_2}}}}}{{{m_3}!}}\sum\limits_{r = 0}^{{m_1} \wedge {k_2}} {\left( {\begin{array}{*{20}{c}}{{k_2}}\\r\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{m_1} + {m_2} + {m_3} - {k_1} - {k_2}}\\{{m_1} - r}\end{array}} \right)\frac{{\left( {r + {m_2} + {m_3} - {k_2}} \right)!}}{{\left( {r + {m_2} - {k_2}} \right)!}}} } $
I hope my coding is correct.


Open cover, finite subcover Find an open cover of $(-\infty,0]$ that does not have a finite subcover.

Can anyone help me with this? Please explain the intuition of finding it too.

Thank you so much!

A very easy solution is just $$\bigcup_{n=1}^\infty(-n,0].$$ Clearly it can't be finite, and it's open.

If the space isn't to be taken as $(-\infty,0]$, then take the union to be $$\bigcup_{n=1}^\infty(-n,1).$$

Maybe you want $C = \{(-n-2, -n+2), n \in \mathbb{N}\}$

Let $V_n = (-n-2, -n+2)$

Then $[0, \infty) \subseteq \bigcup\limits_{n = 1}^\infty V_n $

Try to find an finite subcover $U$ such that $[0, \infty)  \subseteq U$, you can't!


Fibers of a morphism Let $f:X\to Y$ be a birational morphism of projective varieties, with $Y$ non-singular.

Consider a fiber $X_y=f^{-1}(y)$ for a closed point $y\in Y$. Is $X_y$ also a variety, or at least a finite union of projective varieties?

If $X$ is projective, it is obvious that the fiber $X_y$ is a closed subset of $X$, therefore it is a projective variety, obviously not necessarily irreducible.
I do not know if under your hypotheses ($f$ birational, $Y$ projective and smooth) it is true that the fibers are irreducible.

The fibre needs not be irreducible: consider a singular curve $C$ and its normalization $C^\prime\rightarrow C$ within the function field $K(C)$. Then the fibre over a double or ordinary multiple point of $C$ consists of finitely many closed points.


Cardioid: converting parametric form into polar coordinates I am interested in converting parametric equations:

$$\varphi=\left(\varphi_{1},\varphi_{2}\right)=\left(2\cos{t}-\cos{2t},2\sin{t}-\sin{2t}\right)$$

which describe a cardioid, into polar coordinates. What would be a good beginning for this type of an operation? I would be thankful for some hints.

The parametric equations describe $(x,y)(t) = (2 \cos(t) - \cos(2t), 2 \sin(t) - \sin(2t))$:

In order to convert this into polar coordinates, express the radius, and the angle in terms of $x$ and $y$ first:
$$
    r(t)^2 = x(t)^2 + y(t)^2
$$
this would be a simple expression in terms of $\cos(t)$. You could them express the polar angle using atan2 as follows $\theta(t) = \arctan(x(t), y(t))$. The radial representation would be obtained if it were possible to solve for $t = t(\theta)$, and substituted into $r(t)$ to obtain $r(\theta)$. But the equation is not linear and does not admit a simple closed-form.

However, seeing that $r(t)$ is much simpler, we can invert it to find $t=t(r)$ and then $\theta(r) = \theta(t(r))$. Doing so will only allow us to parametrize half of the cardoid, due explicit symmetry $r(t) = r(2\pi - t)$.

If you permit the use of software, here is a way to get the radial representation:

In GeoGebra you could try something like this:

https://i.imgur.com/6GQr31V.png

There is still a minor issue with that attempt (near the points where the graph crosses the y-axis), but with a little tweaking it should work out ok.

To elaborate a little: The rectangular form of the curve is given in terms of horizontal and vertical displacement in a parameter. If you use the Pythagorean theorem, you can obtain the displacement along a ray starting at the origin. Then you use the arctan of the ratio of vertical displacement over horizontal displacement to transform the parameter in a suitable fashion. This can sometimes result in a graph that is a mirror image of the curve that you're after, hence there is an if command that applies a transformation when the parameter is in a certain range and the chosen range of the parameter can also make a difference. By tweaking those aspects, you can obtain the desired parametric curve in polar form. With Wolfram Alpha you can find the root of 2 cos(x)-cos(2x) (the point where the graph intersects the x-axis) and this can be used to set the appropriate range.

https://i.imgur.com/j5onZrr.png

https://i.imgur.com/SAxavZl.png


Why do we use the word "scalar" and not "number" in Linear Algebra? During a year and half of studying Linear Algebra in academy, I have never questioned why we use the word "scalar" and not "number". When I started the course our professor said we would use "scalar" but he never said why.

So, why do we use the word "scalar" and not "number" in Linear Algebra?

So first of all, "integer" would not be adequate; vector spaces have fields of scalars and the integers are not a field. "Number" would be adequate in the common cases (where the field is $\mathbb{R}$ or $\mathbb{C}$ or some other subfield of $\mathbb{C}$), but even in those cases, "scalar" is better for the following reason. We can identify $c$ in the base field with the function $*_c : V \to V,*_c(v)=cv$. Especially when the field is $\mathbb{R}$, you can see that geometrically, this function acts on the space by "scaling" a vector (stretching or contracting it and possibly reflecting it). Thus the role of the scalars is to scale the vectors, and the word "scalar" hints us toward this way of thinking about it.

Using the term "scalar" in the context of Linear Algebra we emphasize that it is an atomic element rather than a collection of elements like "vector" or "matrix". This the main point. In this context we do not care about a nature of such element. Particularly in this context we do not care whether it is a number or not. In this context we do not care if the set of all the scalars is a field.