Kac-Moody algebras/groups for reductive groups The question in brief is: can $\mathfrak{gl}_n$ be constructed as a Kac-Moody algebra, and can $\operatorname{GL}_n$ be constructed as a Kac-Moody group in a compatible way? In general, given any root datum, can we construct the corresponding Lie algebra and reductive group?

Background: In the books I've read on Kac-Moody algebras and groups (Infinite Dimensional Lie Algebras, by Kac, and Kac-Moody Groups, their Flag Varieties and Representation Theory by Kumar), it seems that the option of constructing a reductive (not just semisimple) Lie algebra or group is ruled out, since the only input for the construction is a generalised Cartan matrix rather than a root datum. For example, all of the reductive groups $\operatorname{SL_3}, \operatorname{GL}_3, \operatorname{PSL}_3$ have the Cartan matrix
$$ A = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} $$
however they are all different reductive groups. Can these all be constructed as Kac-Moody groups associated to that Cartan matrix? Looking at the corresponding Kac-Moody algebra, we have $\mathfrak{g}(A) \cong \mathfrak{sl}_3 \cong \mathfrak{psl}_3$, however it seems that $\mathfrak{gl}_3$ cannot be constructed since these books require that the dimension of the Cartan subalgebra is $n + \operatorname{corank}(A) = 2$, while $\mathfrak{gl}_3$ has a three-dimensional subalgebra.

It seems that as input to the Kac-Moody construction, one could instead put a root datum. For example, to construct $\operatorname{GL}_n$ or $\mathfrak{gl}_n$, we would use the root datum
$$\begin{aligned}
\text{coweight lattice} = Y &= \mathbb{Z}\{\varepsilon_1^\vee, \ldots, \varepsilon_n^\vee\}, \\
\text{weight lattice} = X &= \mathbb{Z}\{\varepsilon_1, \ldots, \varepsilon_n\}, \\
\text{perfect pairing}: \quad& \langle \varepsilon_i^\vee, \varepsilon_j \rangle \mapsto \delta_{ij}, \\
\text{coroots} = \Pi^\vee &= \{\alpha_1^\vee = \varepsilon_1^\vee - \varepsilon_2^\vee, \ldots, \alpha_{n-1}^\vee = \varepsilon_{n-1}^\vee - \varepsilon_n^\vee \}, \\
\text{roots} = \Pi &= \{\alpha_1 = \varepsilon_1 - \varepsilon_2, \ldots, \alpha_{n-1} = \varepsilon_{n-1} - \varepsilon_n \}.
\end{aligned}$$
Then the matrix $a_{ij} = \langle \alpha_i^\vee, \alpha_j \rangle$ is the type $A_{n-1}$ Cartan matrix, and we should be able to define the Cartan subalgebra $\mathfrak{h} = Y \otimes_{\mathbb{Z}} \mathbb{C}$, making it $n$ (rather than $n-1$) dimensional, and identifying $\mathfrak{h}^*$ with $X \otimes_{\mathbb{Z}} \mathbb{C}$ by extending the perfect pairing up to $\mathbb{C}$. Would most of the theory of Kac-Moody algebras and groups go through if the construction allowed this? If not, what would fail? The most extreme example of this would be constructing the Lie algebra corresponding to a torus, which would simply be the $n$-dimensional abelian Lie algebra for any $n \geq 0$ fixed in advance.

Update: I should point out some of the reasons for asking this. Firstly, there is a very good off-the-self theory for Kac-Moody algebras, and it would be nice to know if this theory could be applied to any root datum (for example, to $\mathfrak{gl}_n$ and $\hat{\mathfrak{gl}_n}$). Secondly, it seems that some authors believe that Kac-Moody algebras are defined in this generality (or perhaps to them, the term "Kac-Moody algebra" means something more general). For example, in Lusztig's book Introduction to Quantum Groups, he refers to "the Kac-Moody Lie algebra attached to the root datum".

Joppy said in comment:
The answer is yes, you can use pretty much whatever realisation of the root data you want and have the Kac-Moody construction go through with minor changes (dimension of the centre, etc). The most important thing for the theory is that the simple roots and simple coroots each remain linearly independent: without this, significant parts of the theory change.


Show that $\varphi$ is injective. $\varphi : F \rightarrow R$ is a ring homomorphism, $F$ is a field and $R$ is a ring that is not the zero ring. Let $F$ be a field and $R$ a commutative ring where $R$ is not the zero ring. Suppose $\varphi : F \rightarrow R$ is a ring
homomorphism. Show that $\varphi$ is injective.

Any help would be appreciated. I feel pretty lost on this.

$\varphi$ is zero homomorphism, or it is injective:

Assume that $\varphi$ is not injective. Let  $0\neq x\in F$, such that $\varphi (x)=0$. Since $F$ is a field, $x$ has an inverse. So
$$1=\varphi (1)=\varphi (xx^{-1}) =\varphi (x)\varphi (x)^{-1}=0.$$

Hence $\varphi$ is zero homomorphism.

Do you know or can you show that $\phi(0_F) = 0_R$? (EDIT: see the comment below, this isn't actually sufficient. You have to show $\phi(a) = 0_R \implies a = 0_F.$)

Once you have that, say $a,b \in F$. To show that $\phi$ is an injection, go to the definition. You just need to show that $\phi(a) = \phi(b) \implies a=b$. So let's suppose $\phi(a) = \phi(b)$. Now use the definition of a ring homomorphism to show $\phi(a-b) = 0$. Now if you can logically piece this together with the first statement, you've got it.


Prove that $\sum\limits_{n=1}^{\infty}a_n< \infty $ imply $\lim_{x \rightarrow \infty} \frac{1}{x} \sum_{n \leq x} n a_n=0$ Let $ \{a_n\} \ n \in \mathbb{N} $ a sequence of non-negative real numbers.

Prove that
$\displaystyle\sum_{n=1}^{\infty}a_n< \infty $ imply $\displaystyle\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n \leq N} n a_n=0$

Would this be accomplished without the condition $a_n \geq 0$?

You can prove this with Lebesgue's dominated convergence theorem. I'll prove it in a different setting, and I'll let you adapt the proof. Suppose that $g\in L^1_+[0,\infty)$. Then $$\frac 1n \int_0^n xg(x)dx\to 0$$ as $n\to\infty$.

Proof Consider $g_n(x)= \dfrac{xg(x)}n {\bf 1}_{[0,n]}$. Then $|g_n|\leqslant |g|$ and $g_n(x)\to 0$. By DCT, we conclude.

$\dfrac{\displaystyle \sum_{n=1}^N na_n}{N}=\dfrac{a_1+2a_2+3a_3+\cdots +Na_N}{N}=\dfrac{a_1+\cdots+a_N}{N}+\dfrac{N-1}{N}\cdot\dfrac{a_2+\cdots+a_N}{N-1}+\cdots + \dfrac{a_N}{N}\to 0+0+\cdots + 0 = 0$ by the well-known Cesaro's theorem as $a_N \to 0$ when $N \to \infty$ since the first series converges imply the general terms tend to $0$. The result still holds without $a_n \geq 0$.


The range of the function $f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}}$ Find the range of the function $f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}}$.

By simply looking at the problem and simplifying trigonometrically,it looks as if range is zero but not.I think there is some trick to solve this.I am perplexed.I tried to find the domain first but not successful.Please help....

Using $\displaystyle |\sin x|=\left\{\begin{matrix}
\displaystyle \sin x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\
\displaystyle \sin x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\
 \displaystyle -\sin x\;, & \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\
 \displaystyle -\sin x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi
\end{matrix}\right.$ and $\displaystyle |\cos x|=\left\{\begin{matrix}
\displaystyle \cos x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\
\displaystyle -\cos x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\
 \displaystyle -\cos x \;,& \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\
 \displaystyle \cos x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi
\end{matrix}\right.$

So $$\displaystyle f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}} = \sin x\cdot \left|\cos x\right|-\cos x\cdot \left|\sin x\right|$$

Here Function $f(x)$ is Periodic With Time Period $= 2\pi.$

So we will Calculate for Only one Time Period.

In $\bullet \displaystyle \; 0 \leq x\leq \frac{\pi}{2}\;,$ We get $f(x) = \sin  x\cdot \cos x-\cos x\cdot \sin x = 0$

In $\bullet \displaystyle \; \frac{\pi}{2} \leq x\leq \pi\;,$ We get $f(x)=-\sin 2x .$ So $0 \leq f(x)\leq 1$

In $\bullet \displaystyle \; \pi \leq x\leq \frac{3\pi}{2}\;,$ We get $f(x)=0 .$

In $\bullet \displaystyle \; \frac{3\pi}{2} \leq x\leq 2\pi\;,$ We get $f(x)=\sin 2x .$ So $-1 \leq f(x)\leq 0$

So Here We get $\displaystyle -1 \leq f(x)\leq 1$


Ways of getting a number with $n$ dice, each with $k$ sides Assume the dice are numbered from $1$ to $k$. My hunch is that this will form a normal distribution with a median at $n\cdot\frac{k}{2}$. However, I have no idea as to

turn this fact into an answer (I have a minimal knowledge of stats, but I know that I am missing the standard distribution)
and this is probably the wrong approach

How can I approach and solve this problem? (Aside, this is not for a class, stats or other, so any and all approaches welcome).

*Edit: * I want to find the number of ways that the sum of the numbers that are rolled has a particular value, if $n$ dice are rolled, and each has $k$ sides, numbered $1$ to $k$.

The exact probability is a bit complicated: the chance of getting a total of $p$ when you throw $n$ $k$-sided dice is:
$${1\over k^n}\sum_{j=0}^{\lfloor (p-n)/k\rfloor} (-1)^j {n\choose j}{p-kj-1\choose n-1}$$
You can see an explanation and examples here, see equation (10).

You are correct that for moderately large $n$ the distribution is well approximated by a
normal curve with mean $n(k+1)/2$ and standard deviation $\sqrt{n(k^2-1)/12}$.

I am not certain what you mean to ask but based on your comment. The odds of ending up with a particular side out of your k sided die is 1/k, that's assuming your die is unbiased. Now you can throw it over and over the odds of a particular side showing up would be the same 1/k.

your answer can be found at en.wikipedia.org/wiki/Dice#Probability

Cheers


Weak convergence of Hilbert Schmidt operators Let $\{T_i\}$ be a sequence of positive operators in $B_2(H)$ converging to $T$ with respect to the weak topology on $B_2(H)$ (space of Hilbert Schmidt operators on $H$). Does $\{T_ix\}$ converges to $Tx$  in norm (or weakly) for each $x\in H$?

I know that $T$ must be positive since positive operators form a convex set and convex weakly closed sets are norm closed.

Fix $x,y\in H$. Then the functional $T\mapsto \langle Tx,y\rangle$ is continuous.
Then the weak convergence $T_i\rightharpoonup T$ implies
$$
\langle T_i x,y\rangle \to \langle Tx,y\rangle \quad \forall x,y\in H,
$$
which is equivalent to
$$
T_ix \rightharpoonup Tx \quad \forall x\in H.
$$


Uniformly Choosing a number from a range May you please help me how I can choose uniformly a number from a range. I have to use this for trust evaluation in social networks such as the following clause:

  Each user has a quality measurement qi ∈ [0, 1]. For the experiments
  in this paper, the quality of a user is chosen from a normal
  distribution. For any pair of users, i and j, the trust rating from
  node i to node j, denoted as T (i, j), is uniformly chosen from
  [max(qj − δij, 0), min(qj + δij, 1)]. In addition, δij = (1 − qi)/2 is
  a noise parameter that determines how accurate users are at estimating
  the quality of the user that they are trusting

Any of the common programming languages has such a function build in, otherwise you can google it and there are some websites that provide for you.


Determine whether the set is a vector space. So I have a final tomorrow and I have no clue how to determine whether a set is vector space or not. I've looked online on how to do these proofs but I still don't understand how to do them. Can any one help me with a question like this?

Let V be the set of all positive real numbers. Determine whether V is a vector space with the operations below.

$x + y = xy$

$cx=x^c$

If it is, verify each vector space axiom; if not, state all vector space axioms that fail.

Edit: Turns out I'm going to fail the exam based on what you guys are saying.

Choose another notation $x \oplus y := xy$ and $c \otimes x := x^c$. Then the exponential map gives an isomorphism of structures $(\mathbb{R},+,*) \cong (\mathbb{R}^+,\oplus,\otimes)$. Since the first is a vector space, the same is true for the latter. And this way the creator of this "exercise" came up with this artificial vector space (he wanted that you waste your time with computations ...).

What about closure under scalar multiplication? For example, if c = 1/2 and x = -1, we have:  1/2*(-1) = (-1)^(1/2) which is not real. So, it's not closed under scalar multiplication, correct? Therefore, not a vector space.


Integer value for a polynomial for values of $x \in \mathbb{R/Z}$ $P(x)$  be a polynomial with real coefficients.It is given that $\text{deg}P \ge 2$.prove that it is not possible that whenever $P(x)$ is an integer, $x$ is an integer or equivalenty there exist $x_0 \in \mathbb{R/Z}$ such that $P(x_0) \in \mathbb{Z}$.

I could make no considerable progress on this problem

Without loss of generality we can suppose that $\lim_{x \to \infty} P(x)= + \infty$.

This assumptions tells us that $\lim_{x \to \infty} P'(x)= + \infty$ as well: that's because the derivative of $P$ is a polynomial of degree at least $1$.

Pick two consecutive integers $n$  and $n+1$ large enough, say with the property that $P'(x) \ge 3$ in the interval $[n,n+1]$. Then by MVT $$|P(n+1) - P(n)| = |P'(\zeta)| |n+1-n|\ge 3$$ so there is an integer $k$ between $P(n)$ and $P(n+1)$.

By the intermediate value theorem you can find some real number $x_0 \in (n, n+1)$ such that $P(x_0)=k$, and you are done.

Let's assume first that all coefficients are integers.

Let $n$ be the degree of $P$, $p > n$ a prime that does not divide the leading coefficient of $P$ and let $\overline{P}$ be the reduction of $P$ mod $p$. Now show two things:

For every integer $a$ with $0 \leq a \leq p-1$, there exists an integer $b$ with $a \equiv b \mod p$ such that there exists an $x_b \in \mathbb{R}$ with $P(x_b) = b$ (hint: make $|b|$ big enough to force roots of a certain polynomial).
If all these $x_b$ are integers, then $\overline{P}$ is the zero polynomial and hence by choice of $p$ also $P$ itself is.

Once you have done that, the general case with real coefficients is not that much harder (as long as you already know algebraic number theory): Let $\alpha_1,\alpha_2,\ldots, \alpha_m$ be the coefficients of $P$. Replace the ring of integers $\mathbb{Z}$ with the integral closure in $\mathbb{Q}(\alpha_1,\alpha_2,\ldots, \alpha_m)$ and repeat the same argument. :)


Find $f^{(302)}(0)$ where $f(z)=\sin(z^{100}+z^{101})$ and $z\in \mathbb{C}$ Find $$f^{(302)}(0)$$ where $$f(z)=\sin(z^{100}+z^{101})~~~and~~~z\in \mathbb{C}$$

Can anyone  help me with this problem I really don't where to start.

please at least any idea to where to start


Can $\Bbb Z_n$l ideals be generated by one element? Let's define the ring, $\Bbb Z_n$ (It is surely a well-known ring.)

set :  $Z_n$ ={$[0]_{n}$, $[1]_{n}$, ... $[n-1]_{n} \}$
operation : addition or multiplication for the $mod n$

Say the $m \in \Bbb Z_n$, It is obvious that $\langle m \rangle$ is an ideal of $\Bbb Z_n$.

Then, $\forall$ ideal J of the $Z_n$,
 can the ideal J be expressed as the $\langle m \rangle$ ?

consider $I\subset \mathbb{Z}_n$ be ideal.Let $\eta:\mathbb{Z}\to \mathbb{Z}_n$ be natural projection then observe  $\eta^{-1}(I)$ is ideal of $\mathbb{Z}$. So it is generated by a single element (say a) as $\mathbb{Z}$ being PID. Now it is easy to see $I$ is generated by $\eta(a)$.

Since $\mathbb{Z}_n\cong\frac{\mathbb{Z}}{\langle n\rangle}$ and $\mathbb{Z}$ is a PID, we have $\mathbb{Z}_n$ is also a PID.


Name for n-nomial distribution (binomial extended to 3+ possible random values)? I'm trying to work out the required number of samples to have an accurate average for a random series which can take a finite (6) number of values with known probability for each value.

I'm guessing the best mathematical tools would come from a generalisation of binomial distribution studies to a greater number of possible values than 2.

However looking on the web and wikipedia I cannot find the relevant ressources and I assume that I'm just not typing the right keywords to find my answers.

Some my questions:

if it exists how is a n-nomial distribution called?
if known, what is the formula to achieve a given accuracy in relation to number of samples (similar what we have for AWGN): $$meanerror \approx \frac{\sigma}{\sqrt{N}}$$

From $n$ trials and if you have $q(=6)$ variables you can calculate your expected sum by a simple sum of the probability of each outcome multiplied by its value.  For each outcome, define the probability of success $P(O_1)$ and the value of that outcome $O_1$.

Now your expected sum over $n$ trials is given by $$n\sum_{q=1}^6 O_q\times P(O_q)$$

And your expected mean is $$\sum_q O_q\times P(O_q)$$

As for the variance a good estimate is the sum of the individual variances, which are given by $O\times np(1-p)$ so your sum over all $O$ is:

$$\sum_qO_q\times P(O_q)(1-P(O_q))$$

Over sufficiently large number of trials this will approach a normal distribution so you may as well look up on Normal distribution tables - but it will depend on the individual values of $P(O_q)$ and $O_q$ so there's no general solution without plugging those values in and getting the particular mean and variance.


Solving two greatest integer function equations If $$x\lfloor x\rfloor =39 \quad \text{and}\quad  y\lfloor y \rfloor=68.$$

What is the value of:

$$\lfloor x\rfloor+\lfloor y \rfloor $$

I don't know how to solve such problems.

I would appreciate an insight regarding the general approach to such problems.

Notice that $\lfloor x\rfloor$ is not very different from $x$, so $x\lfloor x\rfloor $ is not much different from $ x^2$. If you want $x\lfloor x\rfloor  = 39$, you need $x^2$ to be about $ 39$ also, which means $x$ is going to be around 6 or so, and $\lfloor x\rfloor$ will be exactly 6.  Then $x=6\frac12$ does the trick.

Do the same for $y$, and then add the results.

$$ x |x| = 39 $$

Means $x$ is at the very least positive, since otherwise 39 could not be positive:

$$ -x |-x| = -xx = -39 $$

So this is pretty much equivalent, but not equal to:

$$x^2 = 39$$

Likewise for the other expression. So the result should be something like:

$$\sqrt{39} + \sqrt{68}$$


If a smooth map between manifolds is injective, is the induced map on the tangent spaces injective too? If $\phi:M\longrightarrow N$ is an injective smooth map between two manifolds, then is $d\phi_m:M_m\longrightarrow N_{\phi(m)}$, the induced map between the tangent spaces injective too?

I tried the following : If $v\in M_m$ is such that $d\phi_m(v)=0$, then for all $g$, $C^{\infty}$ function in a neighbourhood of $\phi(m)$, $d\phi_m(v)(g)=0$, that is $v(g\circ\phi)=0$ for all such $g$. From this can we conclude that $v$ is the $0$ tangent vector.

I got this doubt when I was trying to understand the definition of an immersion. I was wondering if $\phi$ being injective will automatically make it an immersion.

The answer is no. Let $\phi:\mathbb R\rightarrow\mathbb R^2$ be $\phi(t)=(t^3,t^9)$. Then the derivative at $t=0$ is not injective.

If the map $\phi:M\longrightarrow N$ is an diffeomorphism then $d\phi_m:M_m\longrightarrow N_{\phi(m)}$ is an isomorphism.


A Class of Diophantine Equations in Three Variables This question is inspired by the problem https://projecteuler.net/problem=748
Consider the Diophantine equation
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{k}{z^2}$$
$k$ is a squarefree number. $A_k(n)$ is the number of solutions of the equation such that $1 \leq x+y+z \leq n$, $x \leq y$ and $\gcd(x,y,z)=1$. This equation has infinite solutions for $k=1$ and $k>1$ that can be expressed as sum of two perfect squares.
Let $$A_k=\lim_{n \to \infty}\frac{A_k(n)}{\sqrt{n}}$$
\begin{array}{|c|c|c|c|c|}
\hline
k& A_k\left(10^{12}\right)& A_k\left(10^{14}\right)& A_k\left(10^{16}\right)& A_k\left(10^{18}\right)& A_k \\ \hline
1& 127803& 1277995& 12779996& 127799963& 0.12779996...\\ \hline
2& 103698& 1037011& 10369954& 103699534& 0.1036995...\\ \hline
5& 129104& 1291096& 12911049& 129110713& 0.129110...\\ \hline
10& 90010& 900113& 9000661& 90006202& 0.0900062...\\ \hline
13& 103886& 1038829& 10388560& 103885465& 0.103885...\\ \hline
17& 86751& 867550& 8675250& 86752373& 0.086752...\\ \hline
\end{array}
From these data, it seems that these limits converge. I wonder if it is possible to write them in terms of known constants or some sort of infinite series. For Pythagorean triples, there are about $\frac{n}{2\pi}$ triples with hypotenuse $\leq n$.
The Python code to calculate $A_1(n)$:
def gcd(a, b):
   if b == 0:
       return a
   return gcd(b, a % b)

N = 10 ** 14

cnt = 0
for a in range(1, 22000):
   a2 = a * a
   for b in range(1, a):
       if (a + b) % 2 == 1 and gcd(a, b) == 1:
           b2 = b * b
           x = 2 * a * b * (a2 + b2)
           y = a2 * a2 - b2 * b2
           z = 2 * a * b * (a2 - b2)
           if x + y + z > N:
               continue
           cnt += 1
print(cnt)

The Python code to calculate $A_2(n)$:
def gcd(a, b):
   if b == 0:
       return a
   return gcd(b, a % b)

N = 10 ** 14

cnt = 1     # (1, 1, 1)
for a in range(1, 22000):
   a2 = a * a
   a4 = a2 * a2
   for b in range(1, a):
       if gcd(a, b) == 1:
           b2 = b * b
           b4 = b2 * b2
           x = 2 * a * b * (a2 + b2) - (a4 - b4)
           y = 2 * a * b * (a2 + b2) + (a4 - b4)
           z = 6 * a2 * b2 - (a4 + b4)
           if x > 0 and y > 0 and z > 0 and x + y + z <= N and gcd(x, gcd(y, z)) == 1:
               cnt += 1
           x = (a4 - b4) - 2 * a * b * (a2 + b2)
           y = 2 * a * b * (a2 + b2) + (a4 - b4)
           z = (a4 + b4) - 6 * a2 * b2
           if x > 0 and y > 0 and z > 0 and x + y + z <= N and gcd(x, gcd(y, z)) == 1:
               cnt += 1
print(cnt)

This is a partial answer. Consider the case $k=1$. The parametric solutions are given by $(x,y,z)=\left(2ab(a^2+b^2),a^4-b^4,2ab(a^2-b^2)\right)$, where $\gcd(a,b)=1$, $a>b>0$ and $a+b$ is odd. In the triples $(x,y,z)$ generated by this parametrization, it is not necessarily always $x\leq y$. For example, $(20,15,12)$ is a triple generated by this parametrization, and $(15,20,12)$ will not be generated. This parametrization generates only the unique solutions. So we can have two possible cases, either $0<a^2-b^2\leq 2ab$ or $0<2ab\leq a^2-b^2$.
The sum of the triples = $x+y+z=4a^3b+a^4-b^4 \leq n$.
Let us consider the first case. Denote the number of solutions by $A_1^1(n)$ and $A_1^1=\lim_{n \to \infty}\frac{A_1^1(n)}{\sqrt{n}}$.
$$0<a^2-b^2\leq 2ab \implies \sqrt{2}-1\leq\frac{b}{a}=x<1$$
Number of pairs $(a,b)$ satisfying above constraints and also $4a^3b+a^4-b^4 \leq n$ can be approximated by
\begin{align}
\begin{split}
p(n)&\approx\int_{\frac{b}{a}=\sqrt{2}-1}^{1}\int_{4a^3b+a^4-b^4 \leq n}a \ da\ db\\
&=\int_{x=\sqrt{2}-1}^{1}\int_{0<a^4\leq \frac{n}{1+4x-x^4}}a \ da\ dx \\
&=\frac{\sqrt{n}}{2}\int_{x=\sqrt{2}-1}^{1}\frac{1}{\sqrt{1+4x-x^4}}\ dx
\end{split}
\end{align}
The natural density of coprime numbers is $\frac{6}{\pi^2}$. The condition $gcd(a,b)=1$ will contribute a multiplication factor of $\frac{6}{\pi^2}$. Among all the coprime pairs $(a,b)$ such that $a>b>0$, fraction of which have odd value of $a+b$ is $\frac{2}{3}$. We have to multiply this factor.
\begin{align}
\begin{split}
A_1^1 &=\lim_{n \to \infty}\frac{A_1^1(n)}{\sqrt{n}}\\
&= \lim_{n \to \infty} \frac{2}{3} \cdot \frac{6}{\pi^2} \cdot\frac{p(n)}{\sqrt{n}}\\
&= \frac{2}{\pi^2} \int_{x=\sqrt{2}-1}^{1}\frac{dx}{\sqrt{1+4x-x^4}} \\
\end{split}
\end{align}
Similarly, it can be shown that for the other case exactly the same limit exists. So, $$A_1=2A_1^1=\frac{4}{\pi^2} \int_{x=\sqrt{2}-1}^{1}\frac{dx}{\sqrt{1+4x-x^4}} \approx 0.1277999513464289...$$
Using a similar approach it can be shown that
\begin{align}
\begin{split}
A_2&=\frac{2}{\pi^2} \left[ \int_{x=0}^{\sqrt{2}-1}\frac{dx}{\sqrt{3-6x^2-x^4}} +\int_{x=\sqrt{2}-1}^{1}\frac{dx}{\sqrt{4x+6x^2+4x^3-1-x^4}} \right]\\
& \approx 0.1036994744684913...
\end{split}
\end{align}
\begin{align}
\begin{split}
A_{13}&=\frac{52}{7\pi^2} \int_{x=(\sqrt{13}-3)/2}^{(\sqrt{26}-1)/5}\frac{dx}{\sqrt{20x+36x^2-5-7x^4}}\approx 0.1038855856479065...
\end{split}
\end{align}

To solve the Diophantine equation.
$$ \frac{ 1 }{ x^2 } +\frac{ 1 }{ y^2 } = \frac{ q }{ z^2 } $$
It is necessary to use the solution of the following equation.
$$a^2+b^2=qc^2$$
There are solutions when the coefficient can be represented as the sum of squares. $q=t^2+k^2$
$$a=-tp^2+2kps+ts^2$$
$$b=kp^2+2tps-ks^2$$
$$c=p^2+s^2$$
And the solution itself can be presented in this form.
$x=ac$
$y=bc$
$z=ab$


Why does a function and its inverse always intersect on the line y=x I've been working through a textbook and noticed that if a function intersects with its inverse function it's always on the line y = x. Why is this?

Well, the statement is not true. Take any bijective function for which $f(1) = 2$ and $f(2)=1$ - and just make the graph not be symmetric in the diagonal ($y=x$). Then the graph of $f$ and that of $f^{-1}$ may intersect on the diagonal, but they will certainly also meet at $(1,2)$ and at $(2, 1)$. (Sorry for the poor drawing...) By the way, the condition "not symmetric in the diagonal" is not really necessary; at the extreme, the function $f(x)=-x$ is its own inverse, so it intersects itself at all the points on it. To avoid such brain twisters, just assume the graphs only intersect in a few points.

If $f(5) = 22$ then $f^{-1}(22) = 5$. $\qquad\longleftarrow$This is the main idea, without which this will be incomprehensible.

Similarly:

If $f(9) = 9$ then $f^{-1}(9) = 9$. And the point $(9,9)$ is on the line  $y=x$, since $9=9$.


How to find area of a polygon built on the roots of a given polynomial? How to find the area of a (maximum area convex) polygon, built on the roots of a given polynomial in the complex plane?
For example, consider the equation:
$$2x^5+3x^3-x+1=0$$
It has one real and four complex roots and makes a nice convex pentagon in the complex plane (thanks, Wolfram Alpha):

Using the formula for the area of a convex polygon:
$$A=\frac{1}{2} \left( \begin{array}| x_1 & x_2 \\ y_1 & y_2 \end{array} + \begin{array}| x_2 & x_3 \\ y_2 & y_3 \end{array} + \dots +  \begin{array}| x_n & x_1 \\ y_n & y_1 \end{array} \right)$$
I obtained for this case (using numerical values of the roots):
$$A=1.460144\dots$$

Another simple case - roots of unity. They just make regular polygons and the general formula for the area is well known.

However, I would like to know if it's possible to find out this area without computing the roots, using only the coefficients of the polynomial? (The coefficients are meant to be rational).

I know that polynomials with only real roots will all have $A=0$, and for the polynomials with several real roots some of them will be inside our maximum area polygon.
There is a useful theorem (see Rouche's theorem ), according to which:

For a monic polynomial $$z^n+a_{n-1} z^{n-1}+\dots+a_1 z+a_0$$
All its roots will be located inside the circle $|z|=1+\max |a_k|$.

But this theorem gives relatively large area, and can't be used to approximate the area of the polygon.

The polygon with vertices of roots of a polynomial must be  symmetric about real axis because of the conjugate roots.  Area of the polygon equals to 2 times the region in upper half plane.The formula of area of polygon must be abandoned if polygon area is hoped to be computed with coefficients. This is because that formula requires the vertices of polygon which is impossible to be computed by the coefficients of the polynomial if the degree of the polynomial is equals or higher than 5 due to Abel-Ruffini theorem.It is possible for us to find out area generated by polynomial of degree 3 or 4(1 and 2 do not give a region) with the coefficients.

Here are some hints.


Probability of winning a prize in a raffle My work is having it's annual Christmas raffle today. 1600 tickets have been sold, and there are 40 prizes to win. I have bought ten tickets. What are the odds I will win a prize?

While an initial estimate of 1/160 is probably within a close enough range to suggest I have little chance of winning, I am curious as to what the precise odds would be. Tickets are not put back in once they have been drawn.

You'll be surprised. The correct probability of winning at least one ticket is around $0.2242$.

Assuming exactly one prize is given, your answer of $\frac{1}{160}$ is the probability of winning is correct. That is, you go home empty-handed with probability $\frac{159}{160}$. However, $40$ tickets are chosen for prizes, not just one. So even if you miss out on a prize the first time, you could still end up with the second winning ticket; or the third; or the $40^{th}$. What we need to calculate is the chance of winning at least one of those tickets.

For the moment, assume that the prizes are drawn with replacement. Then in order for you to not get a prize, you need to miss the first time, and the second time, and the third time, and so on, until the $40^{th}$ time. Under our assumption that these are drawn with replacement, all these $40$ events are independent. Therefore, the probability that you miss out on a prize is simply the probability that miss out in any given trial, raised to the power of $40$; i.e.,
$$
\left(\frac{159}{160} \right)^{40} \approx 0.7782.
$$
Hence, the chance that you win a prize is $1 - 0.7782 \approx 0.2218$.

When the prizes are drawn without replacement. Now we are going to compute the exact answer without any assumptions. There are $1600$ tickets, out of which you bought the first ten (say). The judges pick $40$ winners out of the $1600$ tickets; this can be done in $\binom{1600}{40}$ ways. [See binomial coefficients in Wikipedia.] Of these, you will not win a prize if those $40$ tickets are drawn from the $1590$ tickets that you did not buy. That is, there are $\binom{1590}{40}$ possible outcomes in which you will go home empty-handed. That is, you go home empty-handed with probability
$$
\frac{\binom{1590}{40}}{\binom{1600}{40}}.
$$
Therefore, you will win a prize with the complementary probability
$$
1 - \frac{\binom{1590}{40}}{\binom{1600}{40}} = \frac{1420730930795547} {6335978517846620} \approx 0.2242.
$$
As you can see, that the approximate answer is quite close to the exact one.

[I did these calculations in Wolfram Alpha.]

I was just in a company Christmas raffle and was wondering my odds with the single ticket.

this allows you to change the number of tickets you have, # of prizes and # of remaining tickets after each draw. What this does not cover is the "you must be present to win clause" This assumes all drawn tickets are winners.

static void Main(string[] args)
        {
            int myTickets = 0;
            int prizes = 0;
            int ticketsRemaining = 1;

        int.TryParse(args[0], out myTickets);
        int.TryParse(args[1], out prizes);
        int.TryParse(args[2], out ticketsRemaining);

        double numerator = ticketsRemaining - myTickets;

        double odds = 0.000000000000000000000000000000;
        bool first = true;
         for(int i = prizes; i> 0; i--)
         {
             if (first)
             {
                 odds = numerator/ticketsRemaining;
                 first = false;
             }
             else
             {
                 odds = odds*(numerator/ticketsRemaining);
             }
             numerator--;
             ticketsRemaining--;
         }
        odds = 1 - odds;
        Console.WriteLine(odds.ToString(".0000000000000000000000000"));

        Console.Read();
    }


Inverse of Torus parametrization any tips on finding the inverse of the following map:

$$(\theta,\phi)\mapsto ((2+\cos\theta)\cos(\phi),(2+\cos\theta)\sin(\phi),\sin\theta)$$

From doing it with cartesian :

One of the maps is:
I get $\phi^{\pm}(x_1,x_2,x_3)=(x_1,x_2)$ and

$$(\phi^{\pm})^{-1}(x_1,x_2) = \left(x_1,x_2,\pm\sqrt{1-\left(\sqrt{x_1^2+x_2^2}-2\right)^2}\right)$$

I will add more as I go.

thanks

This might help

ParametricTorus[u_, v_, a_,
   b_] := {(a Cos[u] + b) Cos[v], (a Cos[u] + b) Sin[v], a Sin[u]};

InverseParametricTorus[x_, y_, z_, a_, b_] :=
  With[{v = Pi/2 - ArcTan[y, x], v2 = ArcTan[y, x] + 2 Pi},
   With[{u = Pi/2 - ArcTan[z, (x/Cos[v] - b)]},
    {UnitStep[u]u + (1 - UnitStep[u])(u + 2 Pi),
     UnitStep[v]v + (1 - UnitStep[v])(v + 2 Pi)}]];


explaining $|a+b|≤|a|+|b|$ in simple terms I'm struggling to get to grips with the Triangle Inequalities. The problem is I don't really understand what it means. This is what my lecturer has written in the notes:
$$
|a+b|≤|a|+|b|.
$$
First of all, I don't understand why it's less than or equal to.

If I plug in numbers, for example $a=2$ and $b=5$, I struggle to see how $|2+5|≤|2|+|5|$.

If someone could explain this to me I would be very grateful.

OK. I will try to explain why
$$
|a+b|\leq |a| + |b|
$$
is true in a somewhat intuitive/pictorial way.

First off, do you know what $|x|$ means for some number $x$? You probably know that $|x|$ means the absolute value of $x$, but what does that even mean? The absolute value of a number $x$ may be thought of as its "absolute" distance from zero on the real number line. For example, how far away is $3$ from $0$? This is clearly $3$ units away. What about $-4$? Well, this is just $4$ units away from $0$ but in a different direction (remember, the only thing that matters is absolute distance and distances are always positive). Thus, we get the following definition for the absolute value of a number $x$:
$$
|x|=
\begin{cases}
x &\text{if} & x\geq 0,\\
-x &\text{if} & x<0.
\end{cases}
$$
This guarantees us that the distance from $x$ to $0$ will always be positive, regardless of the value of $x$.

Phew. Okay. Now consider what $|a+b|\leq|a|+|b|$ means. What if $a$ is positive and $b$ is negative? What number will be larger: $|a+b|$ or $|a|+|b|$? Can you see how this might confirm the truth of the triangle inequality?

If you need more explanation or want a general proof (although I think a general proof may confuse you more than help you at the moment), then let me know and I'll add it.

$|x|$ means the absolute value of $x$. This means that if the value of $x<0$, then $|x|=-x$ otherwise $|x|=x$. In simple words, $|x|$ is formed by removing the minus sign from $x$, if any. So $|4|=4$ and $|-3|=3$

The equation states an obvious fact:
$$|a+b|\leq|a|+|b|$$

If $a$ and $b$ are both positive, we have $|a+b|=|a|+|b|$, else we have $|a+b|<|a|+|b|$

Try putting these values, and you should be able to figuring it out: $(2,5)$, $(2,-5)$, $(-2,-5)$

What this has to do with a triangle is unclear to me, though.


What is $\nabla X$ in Riemannian geometry? What exactly is $\nabla X$, for a vector field $X$? Can we write it in coordinates?

I'm familiar with $\nabla f$, and also $\nabla_Y X$, where both $X$ and $Y$ are vector fields. However, what would $\nabla X$ be? Should we think of it as a map $TM\to TM$?

I imagine this is in the context of differential geometry (rather than vector calculus, where I would not know what it stands for). Then it is very simply related to $\nabla_YX $ which you say you are familiar with:
\begin{equation}
\nabla X (Y) = \nabla _Y X.
\end{equation}
While the covariant derivative preserves a tensor field rank (that is, the covariant derivative of a vector is a vector, of a 1-form is a 1-form, and so on), the action of $\nabla$ itself increases the covariant (differential form) degree by one. Hence we can characterise it by saying how it acts on vector fields as above.

"nabla", ∇, can be interpreted as the "vector operator", $\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}$.

There are three different kinds of vector multiplication- scalar product (between a scalar and a vector), dot product, and cross product (between two vectors)- so there are three different ways to use nabla:

"gradient": One takes the "grad" of a scalar valued function of x, y, and z and the result is a vector valued function.

$\nabla f= \left(\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}\right) f(x, y, z)= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$.

That is the vector pointing in the direction of fastest increase of f.  It's length is the rate of increase in that direction.

"divergence"  One takes the "dot product" with a vector valued function and the result is a scalar.

$\nabla\cdot f\vec{i}+ g\vec{j}+ h\vec{k}= $$\left(\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}\right)\cdot \left(f\vec{i}+ g\vec{j}+ h\vec{k}\right)=$$ \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial g}{\partial y}\vec{j}+ \frac{\partial h}{\partial z}\vec{k}$.

If we think of the vector function as the velocity vector of a fluid at each point, that measures, roughly, how fast the fluid "spreads out".

"curl":  One takes the "cross product" with a vector valued function and the result is another vector value function.

$\nabla \times \left(f\vec{i}+ g\vec{j}+ h\vec{k}\right)= \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f & g & h \end{array}\right|= \left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}- \left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$

If we think of the vector function as the velocity vector of a fluid at each point, that measures, roughly, the rotation of the fluid.

Also of great importance in physics is the "Laplacian",  the divergence of the gradient of a scalar valued function:

$\nabla^2 f= \nabla\cdot\left(\nabla f\right)= \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2}+ \frac{\partial^2 f}{\partial x^2}$

For some reason physicists tend to write that as "$\triangle f$"


How do you change the order of integration in this double integral? I need to reverse the order of integration for the following problem:

$\int_0^{324}{\int_{\sqrt{y}/2}^9}\sqrt{x^3+5}dxdy$

I plotted the function to find new limits of integration:
$0\leq y\leq 2x^2, 0\leq x \leq 9$

\begin{align}
\int_0^{9}{\int_0^{2x^2}\sqrt{x^3+5}dydx} & = \int_0^{9}{\sqrt{x^3+5}y\Big|_0^{2x^2}dx}\\
& = \int_0^{9}{\sqrt{x^3+5}\cdot2x^2dx}\\
& = \frac{4}{9}(x^3+5)^{2/3}\Big|_0^{9}\\
& = \frac{2936\cdot\sqrt{734}-20\sqrt{5}}{9}
\end{align}

However, this is not the correct answer. I think my limits of integration are wrong.

Edit: There is a mistake in the initial problem. The limits are supposed to be $\sqrt{y}/2 \leq x \leq 9$.

In your first integral, your region is bound by

$x= \sqrt {\frac  y2}\\
x = 9\\
y = 0\\
y = 324$

Graphing of these curves

We get two regions.

If this is indeed what you are trying to integrate.

flipping the order of integration

$\int_0^9 \int_0^{2x^2} f(x,y) \ dy \ dx + \int_9^{9\sqrt{2}}\int_{2x^2}^{324} f(x,y) \ dy \ dx$

Or, as Daniel Fisher points out, your curve is supposed to be $x= \frac {\sqrt y}{2}.$

$\int_0^9 \int_0^{4x^2} f(x,y) \ dy \ dx $


When is co-adjoint orbit a Kähler manifold? Is there a simple condition when the co-adjoint orbit of a semi-simple Lie group $\mathcal{G}$ is a Kähler manifold?

I am particularly interested in the symplectic group, so I do not want to require compactness!

Let us recall that a dual vector $\beta\in\mathfrak{g}$ of dual Lie algebra generates a coadjoint orbit as $O=\{\mathrm{Ad}_g^* \beta|g\in\mathcal
G\}\subset\mathfrak{g}^*$. The Lie bracket naturally equips $O$ with a non-degenerate, closed 2-form, this is to say it $O$ must be even dimensional and is a symplectic manifold.

However, I also studied several examples where $O$ is naturally a Kähler manifold with compatible inner product induced by the Killing form. More precisely, if we define
\begin{align}
\mathfrak{s}=\{A\in\mathfrak{g}|\mathrm{ad}_A^*(\beta)=0\}\,,
\end{align}
we can use the non-degenerate Killing form to define the orthogonal complement $\mathfrak{s}_\perp$ and its dual $\mathfrak{s}_\perp^*$ (using again the Killing form to provide an isomorphism between $\mathfrak{g}$ and $\mathfrak{g}^*$).

In the examples, I studied, it turned out that minus the inverse Killing form restricted to $\mathfrak{s}_\perp^*$ was positive definite and compatible with the symlectic form on $\mathfrak{s}_\perp^*$. Using the group action, we can move this everywhere in $O$, which proves that it is a Kähler manifold. Note that I used here that $\mathfrak{s}^*_\perp\subset\mathfrak{g}^*$ can be naturally understood as the tangent space of $O$ at the point $\beta\in\mathfrak{g}^*$.

I would like to know how general this is and if there are simple conditions onto $\beta$ to ensure that the resulting metric (constructed from the Killing form) gives rise to a Kähler manifold?

I invite you to read:
[A] Della Vedova A.. Special homogeneous almost complex structures on symplectic manifolds. Journal of Symplectic Geometry, 17(5), 1251-1295, 2019
[B] Della Vedova A., Gatti A., Almost Kaehler geometry of adjoint orbits of semisimple Lie groups, arXiv:1811.06958, 2018
[C] Gatti A., Special almost-Kähler geometry of some homogeneous manifolds, PhD of Universi-tà degli Studi di Pavia, supervised by Dr. Alberto Della Vedova, December 2019


Project allocation optimisation Code I've been formulating an integer optimisation model for allocating students to projects where students give their preferences and rank them 1,2,or 3 with one being their best project preference.

Variables

$X_{i,j}$ = 1, if student i is allocated to project j and 0, otherwise

$C_{i,j}$ can take any value in set C where,C ∈ [1,2,3]

Constraints

$\sum_{j=1}^{Tp} X_{i,j} = 1$

Each student is allocated to one project where Tp = total projects

$\sum_{i=1}^{Ts} X_{i,j} = 1$

Each project is allocated to one student where Ts = total students

$\sum_{j=1}^{Tp} C_{i,j} \times X_{i,j} \geq 1$

Each student is allocated a project that is one of their preferences

Objective Function

Minimize $\sum_{i=1}^{Ts}\sum_{j=1}^{Tp} C_{i,j} \times X_{i,j}$

To minimize the sum of the rankings.

How can I code this integer optimisation problem in MATLAB or another way?

Thank you!!

First, a comment: I don't see the purpose of the third constraint ($\sum_{j=1}^{T_p} C_{i,j} X_{i,j} \ge 1$). Since $C_{i,j}\ge 1$, this constraint is implied by the first one (assign every student to a project).

Second,regarding software, you could solve this in Excel (assuming a "reasonable" number of students and projects). Excel has a built-in linear programming solver, with limited capacity (but enough to do this for one class). There is also an open-source plug-in (OpenSolver) that works quite well and handles larger problem instances. If you are comfortable with spreadsheets but not an Excel user, OpenSolver also has a version for Google Sheets. To do it in MATLAB, you might start here: https://www.mathworks.com/help/optim/ug/first-choose-problem-based-or-solver-based-approach.html.

I agree with @prubin in both statements.
There is no need for the third constraint and Excel + Solver may solve your problem.

In case you are looking for something a bit more robust, I strongly recommend Google Colab (enviroment for Python notebooks): https://colab.research.google.com/

Combine it with OR tools (https://developers.google.com/optimization/lp) and you will have a good tool to solve and teach linear programming problems right from the web browser.

It can even have some interactive tools (forms, interface with spreadsheets, etc.).

An short example with your code:
https://colab.research.google.com/drive/1ukeA5Isx9COqHjAuDbUCecAlHhdUid8M

An alternative could be Gusek (http://gusek.sourceforge.net/gusek.html).


Related limits of a function. Let $f:[0,\infty]\rightarrow \mathbb{R}$ a bounded function in each bounded interval. If $\lim_{x\rightarrow \infty} [f(x+1)-f(x)]=L$ then prove that $\lim_{x\rightarrow \infty}\displaystyle\frac{f(x)}{x}=L$

I appreciate any hint to solve this problem.

Thanks a lot!

For any $a > 0$:

  Let $y$ be such that $f(x+1) - f(x) \in L + [-a,a]$ for any $x \ge y$

  Then $f(y+t+k) - f(y+t) \in k ( L + [-a,a] )$ for any $k \in \mathbb{N}$ and $t \in [0,1]$

  Also $f(y+t)$ is bounded over all $t \in [0,1]$

  Thus ${\large \frac{f(y+t+k)}{y+t+k} } \in {\large \frac{f(y+t)+k(L+[-a,a])}{y+t+k} } \to L+[-a,a]$ uniformly for all $t \in [0,1]$ as $k \to \infty$

  Thus ${ \large \frac{f(x)}{x} } \in L + [-2a,2a]$ as $x \to \infty$

Therefore ${ \large \frac{f(x)}{x} } \to L$ as $x \to \infty$

To explain what happens in the last two lines inside "For any $a>0$", we basically have $\large\frac{p(t)+k(L+q(t))}{y+t+k}$ for some $p$ that is bounded on $[0,1]$ and $q$ that is in $[-a,a]$. As $k$ increases, $p(t)$ becomes insignificant because it is bounded, and also $\frac{k}{y+t+k}$ approaches 1 uniformly over all $t \in [0,1]$, hence the expression approaches something in $L+[-a,a]$ uniformly over all $t \in [0,1]$. "Uniformly" means that for any desired margin of error there is a common cutoff point for $k$ beyond which the expression is within the error margin for all $t$. The next line follows from this because eventually the expression will be within $a$ of $L+[-a,a]$.

This is another proof:

Given $\epsilon>0$ there exist $y>0$ such that if $x>y$ then:

$$L-\epsilon<f(x+1)-f(x)<L+\epsilon$$

we can conclude the following inequalytis valid for every $n$

$$L-\epsilon<f(x+2)-f(x+1)<L+\epsilon$$

$$L-\epsilon<f(x+2)-f(x+1)<L+\epsilon$$

$$\vdots$$

$$L-\epsilon<f(x+n)-f(x+n-1)<L+\epsilon$$

adding each of this inequalitys

$$nL-n\epsilon<f(x+n)-f(x)<nL+n\epsilon $$

taken $n=[x]$ and $h(x)=x-[x]$ then

$$[x]L-[x]\epsilon<f(x)-f(h(x))<[x]L+[x]\epsilon $$

since $0\leq h(x) <1$ and $[0,1)$ is bounded, then there exist $M>0$ such that $|f(h(x))|<M$. This implies that $\lim_{x\rightarrow\infty}\displaystyle\frac{f(h(x))}{x}=0$. Dividing by $x$:

$$\displaystyle\frac{[x]}{x}L-\displaystyle\frac{[x]}{x}\epsilon<\displaystyle\frac{f(x)-f(h(x))}{x}<\displaystyle\frac{[x]}{x}L+\displaystyle\frac{[x]}{x}\epsilon $$

and recalling that $\lim_{x\rightarrow\infty}\displaystyle\frac{[x]}{x}=1$. We take  $x\rightarrow\infty$ then

$$L-\epsilon\leq\lim_{x\rightarrow\infty}\displaystyle\frac{f(x)}{x}\leq L+\epsilon$$

since $\epsilon>0$ is aritrary we get the result.


Minimum Number of Nodes for Full Binary Tree with Level $\lambda$ If the level ($\lambda$) of a full binary tree at zero is just a root node, than I know that I can get the maximum possible number of nodes (N) for a full binary tree using the following:

N = $2^{\lambda+1}$- 1

Is the minimum possible number of nodes the following?

N = 2*$\lambda$ + 1

For a full binary tree $T$ of height $\lambda$, I believe that the maximum number of nodes is $N = 2^{\lambda + 1} - 1$ (not $+1$.)

It seems likely that you can prove the minimum number of nodes for a full binary tree of height $\lambda$ inductively. (We can readily verify that the minimum number of nodes for $\lambda = 1$ is $2\times 1 + 1 = 3$, showing the base case to be true.) Assuming (inductively) that for $\lambda = k$ we have a minimum of $N = 2k+1$ nodes, if we add a node, it must branch from one of the leaves. But in order to maintain a full binary tree, we must add an additional node; that is, adding an additional levels requires at minimum two more nodes. So, we will have $N+2$ nodes. Then, by our induction hypothesis $N+2 = (2k+1) + 2 = 2(k+1) + 1$, which is what we wanted.

Not exactly formal, but does that make sense?

Sorry to say but what all of you are discussing. AFAIK for full binary tree nodes = [(2^(h+1)) - 1] (fixed).

For strict binary tree, max node = [2^h + 1] and min node = [2h + 1].


Solve complex equation $z^3=\left(\sqrt3+i\right)^9$ Solve complex equation $z^3=\left(\sqrt3+i\right)^9$

$|z|=\sqrt{3+1}=2\\\sin \phi=\frac{1}{2}\\\cos\phi=\frac{\sqrt3}{2},\phi=\frac{\pi}{6}\\z_0=\sqrt[3]{2}\left(\cos\frac{\frac{\pi}{6}}{3}+i\sin\frac{\frac{\pi}{6}}{3}\right)=\sqrt[3]2\left(\cos\frac{\pi}{18}+i\sin\frac{\pi}{18}\right)$
Wolfram say $z_0=8i$ what I do wrong?

I think you are taking RHS as $(\sqrt 3+i)^{3}$ instead of $(\sqrt 3+i)^{9}$

Also the given cubic has three solutions. $8i$ is just one of them.


Probability that a stick randomly broken in five places can form a tetrahedron Edit (June. 2015) This question has been moved to MathOverflow, where a recent write-up finds a similar approximation as leonbloy's post below; see here.

Randomly break a stick in five places.

Question: What is the probability that the resulting six pieces can form a tetrahedron?

Clearly satisfying the triangle inequality on each face is a necessary but not sufficient condition; an example is provided below.

Furthermore, another commenter kindly points to a reference that may be of help in resolving this problem. In particular, it relates the question of when six numbers can be edges of a tetrahedron to a certain $5 \times 5$ determinant.

Finally, a third commenter points out that since one such construction is possible, there is an admissible neighborhood around this arrangement, so that the probability is in fact positive.

In any event, this problem is far harder than the classic $2D$ "form a triangle" one.

Several numerical attacks can be found below; I will be grateful if anyone can provide an exact solution.

Not an answer, but it might help to progress further. [The derivation that follows provides a strict -but practically useless- bound. The second part of the answer has some results that might be of interest, but they are merely empirical]

Let's consider the (much more restricted) event that the six lengths form a tetrahedron in whatever order. In the linked pdf this set of lengths is called "completely tetrahedral", and a necessary-sufficient condition is given (Theorem 4.2) which is equivalent to the following: $ u \le \sqrt{2}v $, where $u,v$ are the maximum and minimum lengths. This, informally, would correspond to "almost regular" tetrahedra.
Let's then compute the probability that the lengths are completely tetrahedral. Because the points are chosen at random, uniformly, the lengths are probabilistically equivalent to a set of iid exponential variables with arbitrary parameter, conditioned to a constant sum. Because we are only interested in ratios, we can even ignore this conditioning.
Now, the joint probability of the maximum and minimum of a set of $n$ iid variables is given by

$$ f_{u,v}= n(n-1) f(u) f(v) [F(u) -F(v)]^{n-2}, \hskip{1cm} u\ge v$$

In our case: $n=6$, $f(u)=e^{-u}$, and the probability that $u<a \, v$ is a straightforward integral, which gives:

$$P(u<a \, v)= 5 (a -1)\left( \frac{1}{1+5\,a}-\frac{4}{2+4\,a}+\frac{6}{3+3\,a}-\frac{4}{4+2\,a}+\frac{1}{5+1\,a} \right)$$

And $P(u<\sqrt{2} v) \approx 7.46 \times 10^{-5}$
This should be a strict bound on the desired probability, but, surely, far from tight.

[Update: indeed, the bound is indeed practically useless, it corresponds to an irrelevant tail. The probability, as per my simulations, is around $p=0.06528$ ($N=10^9$ tries, $3 \, \sigma \approx 2.3 \times 10^{-5}$), which agrees with other results.]

The only empirical result that might be of interest: It's easy to see that, from the $6!$ possible permutations of the lenghts, we can restrict ourselves to $30$, from symmetry considerations; now, from my simulations, I've found that it's sufficient to consider 7 permutations,  the first two being already enough for more than $90\%$ of the successes;
and the (need to consider the) seventh one is extremely small. These permutations are:

$$p_1  =  [0 ,  1 ,  4 ,  5 ,  3 ,  2]  \hskip{1 cm}  (0.75837)\\
p_2  =  [0 ,  1 ,  4 ,  3 ,  5 ,  2]  \hskip{1 cm}  (0.15231)\\
p_3  =  [0 ,  2 ,  4 ,  1 ,  5 ,  3]  \hskip{1 cm}  (0.08165)\\
p_4  =  [0 ,  1 ,  4 ,  5 ,  2 ,  3]  \hskip{1 cm}  (0.00404)\\
p_5  =  [0 ,  1 ,  4 ,  2 ,  5 ,  3]  \hskip{1 cm}  (0.00245)\\
p_6  =  [0 ,  1 ,  3 ,  5 ,  4 ,  2]  \hskip{1 cm}  (0.00116)\\
p_7  =  [0 ,  1 ,  3 ,  4 ,  5 ,  2]  \hskip{1 cm}  (0.00002)\\
$$

The length indexes correspond to a sorted array (say, ascending), and following the convention of the linked paper: the first three sides have a common vertex, the following three are the corresponding opposite sides (so, for example, in the first permutation, and by far the most favorable one, the longest and shortest sides are opposite). The numbers on the right are the probability that this permutation (when one tries in the above order) is the successful one (given that they form a tetrahedron). I cannot be totally sure if there is some rare case that requires other permutation (very improbable, I'd say), but I'm quite sure (unless I've made some mistake) that the set cannot be further reduced.

Let the stick be of unit length. We generate five random numbers uniformly distributed between 0 and 1, and after we sort them in ascending order we have some r1, r2, r3, r4, r5. Let a=r1, b=r2-r1, c=r3-r2, d=r4-r3, e=r5-r4, f=1-r5 be the edges of a possible tetrahedron. The edges should satisfy the triangle inequalities for the four faces of tetrahedron. A program based on this reasoning for 1000000 trials yields an approximate probability p = 0.0185 .

Best regards

Emmanuel.


Dependencies in 'epsilon of room'-style proofs I want to try and prove that

  If $\{a_n\}_{n=1}^\infty$ is a positive sequence converging to zero,
  then there exists $N \in \mathbb{N}$ such that $a_{n+N} \leq a_n$ for
  every $n \in \mathbb{N}$.

Proof. Given $\varepsilon > 0$, it was easy to prove that there exists $N$ that depends on $\varepsilon$, such that $a_{n+N} \leq a_n + \varepsilon$. For as $\{\sup_{n \geq k}a_n\}_{k=1}^\infty$ is decreasing, I pick $N$ such that $\sup_{n \geq k} a_n < \varepsilon$ whenever $k \geq N$. This means that for every $n \in \mathbb{N}$, $a_{n+N} < \varepsilon \leq \varepsilon + a_n$.

Question: I have heard of 'Epsilon of room'-proofs, but I cannot find out whether I am allowed to let $N$ depend on $\varepsilon$ in such proofs, or not. Could you provide and answer to whether such a dependecy is allowed, and also explain what kind of dependencies are allowed and forbidden in such proofs.

What you want to prove is not true. For example, define a sequence $(a_n)_{n \geq 1}$ by
$$\forall n \geq 1 \text{ such that there exists }k \text{ with } 2^k < n < 2^{k+1}, \quad a_n = \frac{1}{k+1}$$
and
$$a_{2^k} = \frac{2}{k+1}$$

For such a sequence, there is no such $N$.


Diagonal entries of an Hermitian matrix are real A matrix $A$ is Hermitian if $A=A^*$ where $A^* = (\bar A)^T$. I read in a book that the diagonal entries of every hermitian matrix is real. I really can not understand why it is true.

Thanks

Take a simple example of a 1-dimensional complex matrix: $[a+ib]$.
To be hermitian, you need to show that $[a+ib]=\overline{[a+ib]}=[a-ib]$, hence $b=0$.

When you take the transpose, the diagonal entries will not change the position. Then after taking the conjugate, the diagonal entries should be the same. That means they must be real.


Solving $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = m \in \mathbb{Z}$, $\frac{a}{c} + \frac{c}{b} + \frac{b}{a }= n \in \mathbb{Z}$ Whether non-zero integers $a, b, c$ with the property that $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = m \in \mathbb{Z}$$ and $$\frac{a}{c} + \frac{c}{b} + \frac{b}{a }= n \in \mathbb{Z}$$ Calculate all possible values for $m + n$.

We can assume that the three numbers do not a share a prime in their factorization. Rearrange the equations a bit:

$$a^2c + b^2a+c^2b = mabc \\
a^2b + c^2a + b^2c = nabc$$

Pick your favorite prime $p$. Suppose $N$ is largest so that $p^N\mid gcd(a,b)$.

If $p^{N+1}\nmid a$, then examine the second equation to see that $p^{2N}\mid a^2b,b^2c, nabc$ and so $p^{2N}\mid c^2a$, and so $p \mid c$.

If $p^{N+1}\nmid b$, then examine the first equation to see that $p^{2N}\mid a^2c,b^2a,mabc$ and so $p^{2N}\mid c^2b$, and so $p \mid c$.

This implies that a prime shared by 2 of the numbers is shared by the third, contradicting our initial assumption.

This is a hint to get you started. I'll add the rest of the solution later.

let integer $a,b,c$ such $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$ and $\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}$ is also integer, then we have
$$|a|=|b|=|c|$$

proof: WLOG:we may assume $gcd(a,b,c)=1$,Indeed,we can consider$\dfrac{a}{d},\dfrac{b}{d},\dfrac{c}{d}$
where $\gcd{(a,b,c)}=d$

a contradiction,then $a,b,c$ least one $\neq\pm 1$,WLOG $a\neq\pm 1$,and let prime number $p|a$,then
$$p|abc\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)=a^2c+b^2a+c^2b$$
so
$p|c^2b$
so $p|b$ or $p|c$
WLOG: $p|b,p\nmid c$,and let $v_{p}(a)=r,v_{p}(b)=s,r\le s$,then we have
$$p^{r+s}|abc\left(\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}\right)=a^2b+c^2a+b^2c$$
But $$p^{s+r}\nmid c^2a,p^{r+s}\nmid b^2c,p^{r+s}|a^2b$$
a contradiction.

so
$$|a|=|b|=|c|$$
then easy to find $m,n$


How to prove $\prod\limits_{i=1}^{n}(x-4i+2)(x-4i+1)>\prod\limits_{i=1}^{n}(x-4i+3)(x-4i)$ for all $x\in\mathbb{R}$? I would like to prove that for $n\in\mathbb{N}$ we have $f_n(x):=\prod\limits_{r=1}^{n}(x-4r+2)(x-4r+1)>\prod\limits_{r=1}^{n}(x-4r+3)(x-4r)=:g_n(x)$ for all $x\in\mathbb{R}$ (actually it would suffice for $n$ even). My attempt was to form pairs on each side in such a way that we always obtain the same function plus a constant, i.e. for $n=2k$ even:
$$
f_n(x)=(x-2)(x-8k+1)\cdot(x-3)(x-4n+2)\cdot...\cdot(x-4k-2)(x-4k+1)=\\((x-4k-0.5)^2-(4k-1.5)^2)\cdot((x-4k-0.5)^2-(4k-2.5)^2)\cdot...\cdot((x-4k-0.5)^2-(1.5)^2)
$$
And similarly for $g_n(x)$. If we then substitute $x=0.5y+4k+0.5$ we see that the inequality is equivalent to:
$$
(y^2-3^2)(y^2-5^2)\cdot...\cdot(y^2-(2k-5)^2)(y^2-(2k-3)^2)>(y^2-1^2)(y^2-7^2)\cdot...\cdot(y^2-(2k-7)^2)(y^2-(2k-1)^2)
$$
for all $y\in\mathbb{R}$. I think now it would be enough to expand both sides and probably we would end up with a polynomial in $y^2$ where all coefficients are positive. However, computing all the coefficients will be quite tedious. How to prove it more elegantly?

Edit:

As several comments and answers use this same idea: simply reducing it to $(x-4r+2)(x-4r+1)=(x-4r)^2+3(x-4r)+2>(x-4r)^2+3(x-4r)=(x-4r+3)(x-4r)$ doesn't work as $a>c,b>d$ doesn't imply $ab>cd$ in general. But it was my first reflex too :)

Let $a_r(x)=(x-(4r-2))(x-(4r-3)), b_r(x)=(x-(4r-1))(x-4r),
c_r(x)=\frac{a_r(x)}{b_r(x)}$,
and $A_n(x)=\prod_{r=1}^{n}a_r(x),B_n(x)=\prod_{r=1}^{n}b_r(x)$.

We wish to show that $A_n(x)>B_n(x)$ for any $n$ and $x$.

Note that when $n=1$, the result follows from $A_1-B_1=2$. Suppose now that
$n\geq 2$.  For $k\in[|1,n|]$, let $I_k=[4k-2,4k-1], J_k=[4k-3,4k]$.

If $x$ is outside all the $J_k(1\leq k \leq n)$, then all the $a_r(x),b_r(x),c_r(x)$ are positive. From the $n=1$ case, we have $a_r>b_r$, whence $c_r>1$
and $\frac{A_n}{B_n}=\prod_{r=1}^n c_r >1$, so we are done.

If $x$ is in some $J_k\setminus I_k(1\leq k \leq n)$, then $A_n\geq 0 \geq B_n$
and $A_n$ and $B_n$ are not both zero, so we are done.

The only case left is therefore when $x$ is in some $I_k(1\leq k \leq n)$. And on that interval, $A_n$ and $B_n$ are both negative ; our job is then to show that
$\frac{A_n}{B_n}=\prod_{r=1}^n c_r$ is  $\leq 1$ on $I_k$.

It is easy to see that the derivative of $c_r$ vanishes at $4r-\frac{3}{2}$ and
nowhere else. It follows that $c_r$ attains its maximum value on $I_k$ at a point
$m_k$ defined by

$$
m_k=\left\lbrace\begin{array}{lcl}
4k-2 & \text{when} & r \lt k, \\
4k-\frac{3}{2} & \text{when} & r=k,\\
4k-1 & \text{when} & r \gt k. \\
\end{array}\right.
$$

So

$$
\begin{array}{lcl}
\prod_{r=1}^n c_r
& = & \bigg(\prod_{r=1}^{k-1} c_r \bigg) c_k \bigg( \prod_{r=k+1}^{n} c_r \bigg) \\
&\leq & \bigg(\prod_{r=1}^{k-1} c_r(4k-2) \bigg) c_k(4k-\frac{3}{2}) \bigg(  \prod_{r=k+1}^{n} c_r(4k-1) \bigg) \\
&\leq & \bigg(\prod_{t=1}^{k-1} c_{k-t}(4k-2) \bigg) c_k(4k-\frac{3}{2}) \bigg( \prod_{t=1}^{n} c_{k+t}(4k-1) \bigg) \\
& = & \bigg(\prod_{t=1}^{k-1}  \frac{8t^2-2t}{8t^2-2t-1}\bigg) \frac{1}{9} \bigg( \prod_{t=1}^{n} \frac{8t^2-2t}{8t^2-2t-1} \bigg) \\
& \leq  &  \frac{1}{9} \bigg( \prod_{t=1}^{\infty} \frac{8t^2-2t}{8t^2-2t-1} \bigg)^2 \
( \ \text{because} \ \frac{8t^2-2t}{8t^2-2t-1} >1 ) \\
& \leq  &  \frac{1}{9} \bigg( \prod_{t=1}^{\infty} \frac{t^2+2t+1}{t^2+2t} \bigg)^2 \
( \ \text{because} \ 8t^2-2t-1\geq t^2+2t) \\
& =  &  \frac{1}{9} \bigg( \prod_{t=1}^{\infty} \frac{t+1}{t}\frac{t+1}{t+2} \bigg)^2 \\
& \leq  &  \frac{4}{9} \ ( \ \text{telescoping product} \ )
 \\
 \\
\end{array}
$$

This concludes the proof.

$\begin{array}\\
(x-4i+2)(x-4i+1)-(x-4i+3)(x-4i)
&=((x-4i)^2+3(x-4i)+2)-((x-4i)^2+3(x-4i))\\
&=2\\
\end{array}
$

so each individual term is greater,
so their product is greater.


Does $f(x)\in\mathbb P$ for polynomial $f$ and some $x>0$ imply infinitely many prime $f(x)$? Let $$f(x):=\sum_{n=0}^{k} a_n x^n$$ be some non-constant polynomial where all coefficients $a_n$ are nonnegative integers except for the constant term $a_0$, which must be positive.

Assuming the Bunyakovsky conjecture, if $f(c)$ is prime for some integer $c>0$, would it follow that $f(x)$ yields infinitely many primes?

Yes.

$f(x)$ must be strictly increasing given the constraints. If $f(c)$ is prime and $c>0$, then since $f(0)<f(c)$, we know $f(c) \nmid f(0)$ and $f(0)$ must be relatively prime to $f(c)$, thus fulfilling the requirements of the Bunyakovsky conjecture.


Integral depending on a path? I need to check whether differential form $\omega$ has, in the domain $G$, such property that it's integral doesn't depend on path. In my exercise: $\omega = \frac{ydx -xdy}{x^2+xy+y^2}$ and $G= R^2 \setminus {(0,0)}$.
So I found the antiderivative, it's equal to $\frac {2}{\sqrt{3}}\arctan\left(\frac{2x+y}{\sqrt{3}y}\right)$. And my problem is that it isn't defined for $y=0$. So what should I do now to answer my question? Define it for $y=0$? Or maybe it will turn out that this integral depends on a path?

It is true that your differential
$$\omega=Pdx+Qdy:={y\over x^2+xy+y^2}\ dx-{x\over x^2+xy+y^2}\ dy$$
satisfies the integrability condition $Q_x-P_y\equiv0$. Therefore in simply connected subsets $\Omega\subset G$ there exist potential functions $f:\ \Omega\to{\mathbb R}$ with $df =\omega$, or $\nabla f=(P,Q)$.

But there is no global $f:\ G\to{\mathbb R}$ of this sort. For a proof consider the elliptic curve
$$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr):=\left(\cos t-{1\over\sqrt{3}}\sin t,\ {2\over\sqrt{3}}\sin t\right)\qquad(0\leq t\leq2\pi)\ .$$
One computes
$$x^2(t)+x(t)y(t)+y^2(t)\equiv1,\quad \dot x(t)=-\sin t-{1\over\sqrt{3}}\cos t,\quad \dot y(t)={2\over\sqrt{3}}\cos t\ .$$
It follows that
$$\int_\gamma \omega=\int_0^{2\pi}\bigl(y(t)\dot x(t)-x(t)\dot y(t)\bigr)\ dt=-{4\pi\over\sqrt{3}}\ne0\ .$$

You can write your $\omega$
$$\omega=\frac{y}{x^2+xy+y^2}dx-\frac{x}{x^2+xy+y^2}dy$$
For integral it has the form
$$\int\bigg(Pdx+Qdy\bigg)$$
which you can test for beeing conservative such as
$$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$
where
$$\frac{\partial P}{\partial y}=\frac{(x-y)(x+y)}{\big(x^2+xy+y^2\big)^2}$$
$$\frac{\partial Q}{\partial x}=\frac{(x-y)(x+y)}{\big(x^2+xy+y^2\big)^2}$$
Which means that the field is conservative and the line integral is path-independent.

EDIT

I must have said in the beginning "Assuming that the path is closed..."

and at the end "This is a necessary requirement but not sufficient"


How prove $g(x)$ is odd function :$g(x)=-g(-x)$ QUestion:

  let $f(x),g(x)$ is continuous on $R$,and such
  $$f(x-y)=f(x)f(y)-g(x)g(y)$$
  and $f(0)=1$

  show that: for any $x\in R$, have $g(x)=-g(-x)$

my try: let $x=y=0$,then
$$f(0)-[f(0)]^2=g(0)g(0)\Longrightarrow g(0)=0$$

and let $x=0$, note $f(0)=1,g(0)=0$,so
$$f(-y)=f(y)$$
since
$$g(x)g(y)=f(x)f(y)-f(x-y)$$
so
$$g(-x)g(y)=f(-x)f(y)-f(-x-y)=f(x)f(y)-f(x+y)$$
since
$$f(x+y)=f(x)f(-y)-g(x)g(-y)\Longrightarrow g(x)g(-y)=f(x)f(y)-f(x+y)$$
so
$$g(-x)g(y)=g(x)g(-y)$$

But I can't have $g(x)=-g(-x)$

Since the proof comes directly from Kelenner's comment, I post this answer in community wiki, so as to not earn undeserved points.

We assume $f(0)=1$ and

$$\forall x,y \in \Bbb R, f(x-y)=f(x)f(y)-g(x)g(y)$$

Notice that we use nowhere the assumption that $f$ or $g$ is continuous.

Then, for all $x\in\Bbb R$,

$$1=f(0)=f(x-x)=f(x)^2-g(x)^2$$

And

$$f(2x)=f(x)f(-x)-g(x)g(-x)$$

Also, $\forall x,y\in \Bbb R$,

$$f(y-x)=f(x)f(y)-g(x)g(y)=f(x-y)$$

Hence $\forall x\in\Bbb R$, $f(x)=f(-x)$, that is, $f$ is even.

Therefore, $f(-x-y)=f(x+y)$, or, $\forall x,y\in \Bbb R$,

$$f(-x)f(y)-g(-x)g(y)=f(x)f(-y)-g(x)g(-y)$$

Hence, $\forall x,y\in \Bbb R$, $g(-x)g(y)=g(x)g(-y)$.

Now, let's suppose $g(y_0)\neq0$ for some $y_0 \in \Bbb R$.

We have then, $\forall x\in\Bbb R$, such that $g(x)\neq0$,
$$\frac{g(-x)}{g(x)}=\frac{g(-y_0)}{g(y_0)}=a$$

Where $a$ is a constant real number.

And if $g(x)=0$, then from $g(-x)g(y_0)=g(x)g(-y_0)=0$ and $g(y_0)\neq0$, we have that $g(-x)=0$ too.

That is, $\forall x\in\Bbb R$, $g(-x)=ag(x)$.

But then $\forall x\in\Bbb R$, $g(x)=ag(-x)=a^2g(x)$, and since $g(y_0)\neq0$, we can simplify and $a^2=1$, or $a=\pm1$. Thus the function $g$ is either even or odd.

If $g$ is even, then $\forall x\in\Bbb R$,

$$f(2x)=f(x)^2-g(x)^2=1$$

So $f$ is constant ($=1$).

But then $\forall x\in\Bbb R$, $g(x)^2=f(x)^2-f(2x)=0$.

Hence, if $g$ is even, it's the null function.

But that means $g$ is always odd, since either it is odd, either it is null (thus also odd).

First note that by using $f(x)=f(-x)$
$$
f(0) = f(x-x) = f(x)^2- g(x)^2  \\
f(0) = f(-x-(-x))=f(x)^2-g(-x)^2 \\
g(x)^2 = g(-x)^2
$$
Thus let one define into two sets (note an element may be in both)
$$
\Gamma^+=\{x \in \mathbb{R} : g(x) = g(-x) \} \\
\Gamma^-=\{x \in \mathbb{R} : g(x) = -g(-x) \} \\
$$
Clearly $\mathbb{R}=\Gamma^+ \cup \Gamma^-$, such that $g$ is even for $x\in\Gamma^+$, and odd for $x\in\Gamma^-$.

Consider $x\in\Gamma^+$, one has
$$
f(x-y)=f(x)f(-y)-g(x)g(-y) \\
\therefore f(x-y)=f(x+y)\Rightarrow f(x-x)=f(x+x) \Rightarrow f(x)=1 \:\forall x\textrm{ as }f(0)=1 \\
\Rightarrow g(x)=0 \:\forall x \in\Gamma^+
$$
Thus $x\in\Gamma^+\Rightarrow g(x)=0 \Rightarrow g(x)=-g(-x)\Rightarrow x\in\Gamma^-$. Thus $\mathbb{R}=\Gamma^-$.


In statistics, why do you reject the null hypothesis when the p-value is less than the alpha value (the level of significance) This is a question that I've always wondered in statistics, but never had the guts to ask the professor. The professor would say that if the p-value is less than or equal to the level of significance (denoted by alpha) we reject the null hypothesis because the test statistic falls in the rejection region. When I first learned this, I did not understand why were comparing the p values to the alpha values. After all, the alpha values were brought in arbitrarily. What is the reason for comparing them to the alpha values and where do the alpha values of 0.05 and 0.10 come from? Why does the statement $ p_\text{value} \leq \alpha$ allow you to reject $H_0$?

Here's the idea: you have a hypothesis you want to test about a given population. How do you test it? You take data from a random sample, and then you determine how likely (this is the confidence level) it is that a population with that assumed hypothesis and an assumed distribution would produce such data. You decide: if this data has a probability less than, say $95$% of coming from this population, then you reject at this confidence level--so $95$% is your confidence level. How do you decide how likely it is for the data to come from a given population? You use a certain assumed distribution of the data, together with any parameters of the population that you may know.

A concrete example: You want to test the claim that the average adult male weight is $170 lbs$ . You know that adult weight is normally-distributed, with standard deviation, say, 10 pounds. You say: I will accept this hypothesis, if the sample data I get comes from this population with probability at least $95$% . How do you decide how likely the sample data is? You use the fact that the data is normally-distributed, with (population) standard deviation=$10$, and you assume the mean is $170$ . How do you determine how likely it is for the sample data to come from this population: the $z-$ value you get ( since this is a normally-distributed variable , and a table allows you to determine the probability.

So, say the average of the random sample of adult male weights is $188lbs$. Do you accept the claim that the population mean is $170$? . Well, the decision comes down to : how likely (how probable) is it that a normally-distributed variable with mean $170$ and standard deviation $10$ would produce a sample value of $188lb$? . Since you have the necessary values for the distribution, you can test how likely this value of $188$ is, in a population $N(170,10)$ by finding its $z-$ value. If this $z-$ -value is less than the critical value, then the value you obtain is less likely than your willing to accept. Otherwise, you accept.

Let’s consider two significant levels ($\alpha$=0.05 and $\alpha$=0.01) for p=0.03.

p=0.03: there is a 3% chance of observing an effect, solely due to the sampling error, when there is actually none.

$\alpha$=0.05: we are up to 5% confident that any observed effect is due to random sampling (or 95% confident that it is not!).

Since we only have 3% chance of observing the effect due to sampling error, we cannot be up to 5% confident that the observed effect is due to random sampling and support $H_0$ at this level. Therefore, we reject $H_0$ at 5% level.

Following the same logic, we are $\alpha$=0.01 confident that the observed effect is random and $H_0$ is true. Therefore, we do not reject $H_0$ at 1% level.

I hope the reader a better understanding of why $H_0$
 is reject when $ p \leq \alpha$

I had the same question and this post too motivated me to write a short piece about it. The example above is taken from there


Do complex numbers really exist? Complex numbers involve the square root of negative one, and most non-mathematicians find it hard to accept that such a number is meaningful. In contrast, they feel that real numbers have an obvious and intuitive meaning. What's the best way to explain to a non-mathematician that complex numbers are necessary and meaningful, in the same way that real numbers are?

This is not a Platonic question about the reality of mathematics, or whether abstractions are as real as physical entities, but an attempt to bridge a comprehension gap that many people experience when encountering complex numbers for the first time. The wording, although provocative, is deliberately designed to match the way that many people actually ask this question.

There are a few good answers to this question, depending on the audience. I've used all of these on occasion.

A way to solve polynomials

We came up with equations like $x - 5 = 0$, what is $x$?, and the naturals solved them (easily). Then we asked, "wait, what about $x + 5 = 0$?" So we invented negative numbers. Then we asked "wait, what about $2x = 1$?" So we invented rational numbers. Then we asked "wait, what about $x^2 = 2$?" so we invented irrational numbers.

Finally, we asked, "wait, what about $x^2 = -1$?" This is the only question that was left, so we decided to invent the "imaginary" numbers to solve it. All the other numbers, at some point, didn't exist and didn't seem "real", but now they're fine. Now that we have imaginary numbers, we can solve every polynomial, so it makes sense that that's the last place to stop.

Pairs of numbers

This explanation goes the route of redefinition. Tell the listener to forget everything he or she knows about imaginary numbers. You're defining a new number system, only now there are always pairs of numbers. Why? For fun. Then go through explaining how addition/multiplication work. Try and find a good "realistic" use of pairs of numbers (many exist).

Then, show that in this system,  $(0,1) * (0,1) = (-1,0)$, in other words, we've defined a new system, under which it makes sense to say that $\sqrt{-1} = i$, when $i=(0,1)$. And that's really all there is to imaginary numbers: a definition of a new number system, which makes sense to use in most places. And under that system, there is an answer to $\sqrt{-1}$.

The historical explanation

Explain the history of the imaginary numbers. Showing that mathematicians also fought against them for a long time helps people understand the mathematical process, i.e., that it's all definitions in the end.

I'm a little rusty, but I think there were certain equations that kept having parts of them which used $\sqrt{-1}$, and the mathematicians kept  throwing out the equations since there is no such thing.

Then, one mathematician decided to just "roll with it", and kept working, and found out that all those square roots cancelled each other out.

Amazingly, the answer that was left was the correct answer (he was working on finding roots of polynomials, I think). Which lead him to think that there was a valid reason to use $\sqrt{-1}$, even if it took a long time to understand it.

In the en they exists as a consistent definition, you cannot be agnostic about it.


Exit Time of an Interval Brownian Motion - Distribution Let $W_t$ be a Brownian motion, fix $a<0<b$ and let $\tau_x=\mathrm{inf}(t\ge0:W_t=x)$.

Show there is an $\alpha<1$: $P(\tau_a \wedge \tau_b>n )\le \alpha^n$ for all $n \in \mathbb{N}$.

Proof-Idea: Use the distribution of the min and max of the brownian motion and their independence, pray and find an estimate:

$$
\begin{eqnarray}
P(\tau_a \wedge \tau_b>n ) &=& (1-P(\tau_a\le n ))(1-P(\tau_b\le n))\\
&=&(1-\Phi(\frac{-a}{\sqrt{n}})(1-\Phi(\frac{b}{\sqrt{n}})\\
\end{eqnarray}
$$

But i am not able to find an estimation such that this expression is dominated be $\alpha^n$.

A number of ways of doing this:

Let $\tau = \tau_a \wedge \tau_b$. Then
\begin{align}
P[\tau > n]
&= \int^b_aP[\tau > n | \tau > n-1, W_{n-1} = x] P[\tau > n-1, W_{n-1} \in dx] \\
&= \int^b_aP[\tau > n | W_{n-1} = x] P[\tau > n-1, W_{n-1} \in dx] \quad(\text{Markov property}) \\
&= \int^b_aP[\tau > 1 | W_0 = x] P[\tau > n-1, W_{n-1} \in dx] \quad(\text{Stationarity of BM segments}) \\
&\leq \max_{x}P[\tau > 1 | W_0 = x] \int^b_a P[\tau > n-1, W_{n-1} \in dx] \\
&= \max_x P[\tau > 1 | W_0 = x] P[\tau > n-1].
\end{align}
Writing $\alpha :=  \max_x P[\tau > 1 | W_0 = x]$, we have
$
P[\tau > n] \leq \alpha^n.
$

Now to show that $\alpha < 1$: Exact calculation of the distribution of $\tau$ requires some heavy machinery (optional stopping & Laplace transforms), but here's an easy but crude upper bound:

Let $x^* := \text{argmax}_x P[\tau > 1 | W_0 = x]$. We know that $x^*$ cannot be $a$ or $b$, and in fact $a < x^* < b$ (strict ineqs.) so that $b - x^* > 0$.
\begin{align}
\alpha &= P[\tau > 1 | W_0 = x^*] \\
&\leq P[\tau_b > 1 | W_0 = x^*] \quad (\tau_b \text{ never comes sooner than }\tau) \\
&= 1 - 2P[W_1 > b | W_0 = x^*] \\
&= 1 - 2 \bar{\Phi}(b - x^*) < 1 \quad \text{(since $b - x^* > 0$ so that $\bar{\Phi}(b - x^*) < 1/2$}).
\end{align}

Let $B$ be a Borel subset of the interval $[a,b]$. Then $\Theta(t,B):= P[\tau > t, W_t \in B]$ is a (family of) probability measure(s) (indexed by $t$) on $[a, b]$ with Lebesgue density $\theta(t,x)dx:= P[\tau > t, W_t \in dx]$. If you know your PDE theory, $\theta(t,x)$ solves the heat equation in the strip $(t,x) \in R^+ \times [a,b]$ with initial heat atom at $(0,0)$ and boundary condition $\theta(t,a) = \theta(t,b) \equiv 0.$ The probability you are interested in is simply $P[\tau > n] = \int^b_a \theta(n, x)dx$. You can then invoke results from PDE theory about the decay rate of temperatures (parabolic functions) in such problems (Widder etc).
If you know some optional stopping techniques, you can calculate the Laplace transform of the distribution of $\tau$, and use a Tauberian theorem to find the exact decay rate of $P[\tau > n]$ in terms of the radius of convergence of the Laplace transform, like here:Burq & Jones, Exact Simulation of Brownian Motion at First Passage Times.

I will use three relations here, which are fairly easy to prove (or to find via googling):

$\mathbb{E}[\tau_a \wedge \tau_b] = -ab$ (related question on MathExchange),
$\sum_n x^n = \frac{1}{1-x}$ for $x < 1$,
$\mathbb{E}[X] = \sum_{n=0}^{\infty} \mathbb{P}(X > n)$ for discrete random variable $X$, taking positive integer values.

Let $\alpha = \frac{ab}{ab-1}$, and suppose that $\mathbb{P}(\tau_a \wedge \tau_b > n) > \alpha^n$, then
$$
\begin{aligned}
-ab &= \mathbb{E}[\tau_a \wedge \tau_b] \geq \mathbb{E}[\lfloor \tau_a \wedge \tau_b\rfloor] =  \sum_{n \in \mathbb{N_0}} \mathbb{P}(\lfloor \tau_a \wedge \tau_b\rfloor > n) \\ &= \sum_{n \in \mathbb{N}} \mathbb{P}( \tau_a \wedge \tau_b > n) > \sum_{n \in \mathbb{N}} \alpha^n = \frac{\alpha}{1-\alpha} = -ab,
\end{aligned}
$$
which is a contradiction.


Prove that $B$ is basis of Topology on $X\times Y$ Please check my proof; this proof is my first proof about topology.

I was asked to prove the following:

Let $B_{1}$ be basis of topology $T_{1}$, $B_{2}$ be basis of topology $T_{2}.$ The Set $X\times Y$ consist of all order pair $(x,y)$ with $x\in X$ and $y\in Y$. Let $B$ be the collection of all subsets of $X\times Y$ consisting of $B_{1}\times B_{2}$ where $B_{1} \in B_{1}$ and $B_{2}\in B_{2}$. Prove that $B$ is basis of a topology on $X\times Y$

1
Consider it satisfy first condition
since $(x,y)\in B$
since all $x\in B_{1}$ and all $y\in B_{2}$ then $B$ is all collection of union of order pair that have component from $B_{1}$ and $B_{2}$

therefore it's satisfy first condition

2.condition 2

Suppose $C_{1}\subseteq B$ $C_{2}\subseteq B$  and let $(x_{1},y_{1})\in C_{1}$ and $(x_{1},y_{1})\in C_{2}$

there exist some $C_{3}$ that contain $(x_{1},y_{1})$

therefore it's satisfy second condition

You check that $\mathcal{B}_1 \times \mathcal{B}_2$ is a base for some topology on $X \times Y$, but you have to check it is a base for the product topology, so for every $O$ open in the product topology and every $(x,y) \in O$ you have to find a set of the form $B_1 \times B_2, B_1 \in \mathcal{B}_1, B_2 \in \mathcal{B}_2$ such that $$(x,y) \in B_1 \times B_2 \subseteq O$$

This you can do because you first note that there must be an "open box" $O_1 \times O_2$ in $X \times Y$ with $O_1$ open in $X$ and $O_2$ open in $Y$, such that

$$(x,y) \in O_1 \times O_2 \subseteq O$$

by the definition of the product topology. Now we apply the fact that $\mathcal{B}_1$ is a base for $X$ to find $B_1 \in \mathcal{B}_1$ with $x \in B_1 \subseteq O_1$. Also, as $\mathcal{B}_2$ is a base for $Y$ we have $B_2 \in \mathcal{B}_2$ such that $y \in B_2 \subseteq O_2$.

Now $B_1 \times B_2$ is as required.

Choose $(x,y)\in X×Y$ then there exists $U\in \mathscr B_1$ and there exists $V\in \mathscr B_2$ such that $x\in U\ and\ y\in V$. Then $(x,y)\in U×V \in \mathscr B$.

Choose two elements $U_1×V_1, U_2×V_2 \in \mathscr B$ both containing the element $(x_1,y_1)\in X×Y$ then $x_1\in U_1\cap U_2$ and $ y_1\in V_1\cap V_2$ , therefore there exists an element of $\mathscr B_1$ say, $ W_1(\subseteq U_1\cap U_2)$ containing $x_1$ and there exists an element of $\mathscr B_2$ say, $W_2(\subseteq V_1\cap V_2)$ containing $y_1$. Therefore $(x_1,y_1)\in W_1×W_2 \subseteq (U_1\cap U_2)×(V_1\cap V_2)$. Also we have $ W_1×W_2\in \mathscr B$.Note that $(U_1\cap U_2)×(V_1\cap V_2)=(U_1×V_1)\cap (U_2×V_2)$.


The limit of truncated sums of harmonic series, $\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}$ What is the sum of the 'second half' of the harmonic series?

$$\lim_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?$$

More precisely, what is the limit of the above sequence of partial sums?

Rewriting the sum as
$$
\sum_{n=k+1}^{2k}\frac1n=\sum_{n=k+1}^{2k}\frac1k\cdot\frac1{n/k}
$$
allows us to identify this as a Riemann sum related to the definite integral
$$\int_1^2\frac1x\,dx=\ln 2.$$
To see that, divide the interval $[1,2]$ to $k$ equal length subintervals, and evaluate the function $f(x)=1/x$ at the right end of each subinterval. When $k\to\infty$, the Riemann sums will then tend to the value of this definite integral.

Here is still another method to calculate the limit

$$s=\lim_{k\to \infty } \, \left(\sum _{n=k+1}^{2 k} \frac{1}{n}\right)=\log (2)$$

First of all, from the definitions, we identify the partial sum as a difference of harmonic numbers

$$s_k=\sum _{n=k+1}^{2 k} \frac{1}{n}= H_{2k}-H_k$$

Now we form the "generating" sum of the partial sums

$$g(x) = \sum_{k=1}^\infty s_k x^k$$

Using the formula for the generating function of the harmonic number

$$h_1(x) = \sum_{k=1}^\infty H_k x^k = -\frac{\log(1-x)}{1-x}$$

and

$$h_2(x) = \sum_{k=1}^\infty H_{2k} x^k=\frac{1}{2} \left(\frac{\log \left(1-\sqrt{x}\right)}{\sqrt{x}-1}-\frac{\log \left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)$$

we find

$$g(x) = h_1+h_2=\frac{1}{2} \left(\frac{\log \left(1-\sqrt{x}\right)}{\sqrt{x}-1}-\frac{\log \left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)+\frac{\log (1-x)}{1-x}$$

Assuming that the limit $s = c$ exists we would have

$$g(x) = \sum_{k=1}^\infty c x^k = \frac{c}{1-x}$$

Hence our limit is

$$s = \lim_{x \to 1 }g(x)(1-x) \, $$

which is easily calculated to be $\log(2)$.


If $5P+7Q= 109$ for primes $P$ and $Q$, find $P$. Suppose $P$ and $Q$ are both prime numbers
  $$5P+7Q= 109$$
  Find the value of $P$.

I thought about examining all terms for parity, and I approached it by this:

Suppose $5P$ was an even number. However, the only even prime number is $2$, so that would make $5P$ = $10$ and $7Q =99$. However, since $7$ doesn't divide fully into $99$, this means that $7Q$ must be even. But how do I approach it from there?

P.S. I'm only an Year 7 and this is Year 9 work. Please explain clearly how you found the solution.

You have already deduced that one of $(P,Q)$ must be $2$.

Also, you’ve tried putting $P=2$ and correctly found that did not give a solution.

The only other option is $Q=2$, so $7Q=14$ and the equation becomes,
$$5P+14=109$$
$$5P=95$$
$$P=19$$

$$5(-2)+7(17)=109$$

Since (-2) is not a prime number, the problem should be:

Suppose $P$ and $Q$ are both prime numbers

$$5P= 109-7Q$$

Find the value of $P$

The answer: $P=2$


How to Solve the following Sum? $\sum_{i=1}^{80} (3^i+3\times 2^i)$

I tried plugging in $i = 1$, and got $9$ as the output, following the same procedure, plugging in $i=2$ gave me $21$. I took note of $21/9$, but noticed the 3rd term divided by the second term $(51/21)$ did not give the same ratio, so somehow this is not a arithmetic or geometric series? Any help would be much appreciated.

$$\sum_{i=1}^{80}\left( 3^i + 3\times2^i\right)=\sum_{i=1}^{80}3^i + 3\sum_{i=1}^{80}2^i$$

So the series has been split into the sum of two geometric series which you can tackle with the standard tactic and combine to get your answer.

Hint:  Write out several terms of a sum on x^i.  Now multiply that by 1-x and see what happens when you cancel terms.  (1 + x + x^2 + x^3 +...+x^N)(1 - x).  Lots of terms will cancel, but there will be terms a 1 and a x^(N+1) that don't cancel.


Trigonometric expression manipulation I'm working on manipulating trig expressions using Khan Academy.

According to the provided answer this:

$$\cos(\theta)\sin(\theta)$$

is supposed to equal:

$$\frac1 2\sin(2\theta)$$

I do not understand how, so could someone please show me.

Hint: $\sin (x + y) =\sin (x) \cdot \cos (y) +\cos (x) \cdot \sin (y) $

What happens when $x=y$ ?


How does one prove that $1$ is a limit point using the definition for $p \in (0,1)$ and some $r>0$, $B_r(x)= \{q \in X: d(x,y) <r\}$? Let's suppose that we're given an open set $(0,1)$. How does one prove that $1$ is a limit point using the definition, for $p \in (0,1)$ and some $r>0$, $B_r(x)= \{q \in X: d(p,q) <r\}$?

Let $B_r(1)$ be a neighborhood of $1$. Assume without loss of generality that $r<1$ (if $r\geqslant 1$ then just take the point, say $\frac12$). Then $|1 - (1-r/2)| = r/2<r$, so $1-r/2\in B_r(1)$. Moreover, $1-r/2>0$, so $1-r/2\in (0,1)$. Therefore $B_r(1)$ contains a point of $(0,1)$ other than $1$, so that $1$ is a limit point of $(0,1)$.

Depends on your definition of a limit point, so I am assuming that you are using the following definition: "A limit point $x$ of a set $S$ is a point such that for every $r>0$, $B_r(x)\bigcap S\neq\emptyset$ and $B_r(x)\bigcap S\neq \{x\}$ ".

Edit: I forgot to say $B_r(x)\bigcap S\neq \{x\}$, but given you are working with real numbers it is impossible anyways.

Now consider $B_r(1) = (1-r,1+r)$. Is it possible that $B_r(1) \bigcap (0,1) = \emptyset$?


Proof of Expected Value for Independent random variables with two possibilities Let ${B_i}$ ($i=$natural number) be a random variable which takes $1$ with probability $p$ and takes $-1$ with probability $1-p$ where $0<p<1$. Assume that ${B_i}$ and ${B_j}$ are independent if $i \neq j$. Let  ${S_N}=\sum_{i=1}^{N} {B_i}$ ($N=$Natural Number) and $E[.]$ denote expectation.

Show that
$E\left[{B_i}^m{B_j}^n\right]=E\left[{B_i}^m\right]E\left[{B_j}^n\right]$ for any natural numbers $m$ and $n$ if $i\neq j$.

I am really new to this subject, so I hope someone experienced can help show me how I should prove this.

If two random variables $X$ and $Y$ are independent then $E(XY)=E(X)E(Y)$. This result does not depend on the type of random variables.

This is because $$E(XY)=\sum_{x,y}xyP(X=x,Y=y)=\sum_{x,y}xyP(X=x)P(Y=y)\\=\Big(\sum_xxP(X=x)\Big)\Big(\sum_yyP(Y=y)\Big),$$
which uses independence to go between the second and third expressions.

For sums, much more is true: $E(X+Y)=E(X)+E(Y)$ even if the variables are not independent.

The $n$ and $m$ in your question are a distraction. $X_i=\pm1$ so if $m$ is odd then $X_i^m\equiv X_i$, and if $m$ is even then $X^m\equiv E(X^m)=1$.

In probability theory, two random variables are said to independent when joint distribution$ F_{A,B}(a, b) = F_A(a)F_B(b)$. When densities exist then $f_{A,B}(a, b) = f_A(a)f_B(b)$

In your case $E[B^m_i, B^n_j] = \sum_{b_i,b_j \in \{\pm1\}^2} b^m_ib^n_j f_{B_i, B_j}(b_i,b_j) = \sum_{b_i, b_j \in \{\pm1\}^2} b^m_ib^n_j f_{B_i}(b_i)f_{B_j}(b_j)$

$=\sum_{b_i \in\{\pm1\}} b^m_i f_{B_i}(b_i)\sum_{b_j \in \{\pm1\}}b^n_j f_{B_j}(b_j) = E(B^m_i)E(B^n_j)$


Find a matrix $P$ such that $[\overrightarrow{\rm x}]_C$ = $P[\overrightarrow{\rm x}]_B$ for all x in $\mathbb{R}^2$ Let $B$ be the basis of $\mathbb{R}^2$ consisting of the vectors

${\begin{bmatrix}
5\\
-2
\end{bmatrix}
,
\begin{bmatrix}
2\\
3
\end{bmatrix}}$

and let $C$ be the basis consisting of

${\begin{bmatrix}
-2\\
3
\end{bmatrix} ,
\begin{bmatrix}
1\\
-2
\end{bmatrix}}$

I tried doing $ P =C*B^{-1}$ in which I got

$\begin{bmatrix}
\frac{-4}{19} & \frac{9}{19}\\
\frac{5}{19} & \frac{-16}{19}
\end{bmatrix}$

but when I went to check it, it was wrong.

Note that for any $x \in \mathbb R^2$, we can write $x =  x_1(5,-2) + x_2(2,3) = (5x_1 + 2x_2, 3x_2 - 2x_1)$, and it can also be written as $x = x_3(-2,3) + x_4(1,-2) =(x_4-2x_3,3x_3-2x_4)$.

Equating these, we get:$5x_1+2x_2 = x_4-2x_3, 3x_2-2x_1 = 3x_3-2x_4$. Now, solve for $x_3,x_4$, we get :
$$
x_3 = -8x_1-7x_2 , x_4 = -11x_1-12x_2
$$
and therefore,
$$
\begin{pmatrix} x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} -8 \quad -7 \\ -11 \quad -12\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}
$$
where $x_3,x_4$ is the representation of $x$ in basis $C$, and $x_1,x_2$ is the representation of $x$ in basis $B$, both as column vectors.

Since you are given $xC = PxB$

**The Answer is: $P=C^{-1} xB$ **


Is$ \prod_{n=1}^{\infty} (x-n\pi)$ a factor of $\sin x$? As it holds for a polynomial it should be true for a power series too. Why not?

The product is not convergent for $x$ except for positive integer multiples of $\pi$, where it is $0$. So given that it isn't even well defined for most of $\mathbb{R}$, we shouldn't expect it to be a factor of $\sin x$.

A power series is not necessarily a polynomial, nor is an infinite product of factors. And just because something holds in the finite case does not mean it holds in the infinite case. For example, all polynomials have derivatives that are also polynomials, so should we expect this to be true in the limit with power series of $e^x$ or $\sin x$? No, so we shouldn't expect an infinite product of $(x-r_i)$ for roots $r_i$ of a function to equal that function. In fact, why should we expect your product to equal $\sin x$ and not any of the other functions with the same roots like a triangle wave or a square wave?

Euler managed to make the product work by manipulating it into a convergent product but there is more going on behind the scenes than it seems. There is more detail on that product in (Question 674769).

No, because this product does not converge. e.g. for $x = 0$ you get $\pi \cdot 2\pi \cdot 3\pi \rightarrow \infty$.  You also forgot about the negative roots. But you can use this general idea of factoring $\sin(x)$ into linear factors corresponding to the roots.  If you write the linear factors as $1 - x/(\pi n)$, with $x=\pi n$ as the corresponding roots, and group the roots in pairs $\pm \pi n$ to get $$(1 - \frac{x}{\pi n})(1 + \frac{x}{\pi n}) = 1 - \frac{x^2}{\pi^2n^2},$$ and divide away the factor $x$ corresponding to the root $x = 0$, you might guess that $$\frac{\sin(x)}{x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{\pi^2n^2}\right)$$ and this turns out to be true! That is, the right-hand product converges to $\sin(x)/x$ for all $x$.  This was used in Euler's famous first solution to the Basel problem, expressing $\sin (x) /x$ as a Taylor series and extracting the $x^2$ term on both sides to get $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$ Not simple to prove this factorization works, though, without more analysis.  In particular this is justified by the Weierstrass factorization theorem, proved some hundred years later.  But how did Euler know that this would work?  Well, he was Euler.  A genius, and also a bit fast and loose with his logic.


Why is it ok to factor an equation with no limit so it has a limit? I'm just starting out in calculus, so please bear with me if this is not a sensible question.

In the book I'm reading, the author gives the example of the problem of finding the limit of $\lim\limits_{x\to 5}(\frac{x^2 - 25}{x-5})$, because if you substitute in $x=5$, you get a denominator of $0$, so the output of the function at $x=5$ is undefined.

He then goes on to demonstrate how by factoring this equation to $\lim\limits_{x\to 5}(x+5)$, you can now plug in $x=5$ to get $\lim\limits_{x\to 5}(x+5) = 10$.

But is not the fact that 'at $x=5$ the function is undefined' an integral part of the original function? By factoring it, have you not added to the domain of the original function and therefore created a different function? So now you have the limit of a different function?

Consider $\displaystyle f(x)=\frac{x^2-25}{x-5}$ and $g(x)=x+5$.

If $x\neq 5$ then $f(x)=g(x)$, so for all $x$, except for $x=5$, the graphs of $f(x)$ and $g(x)$ are identical.

At $x=5$, $f(x)$ is undefined but $g(5)=10$.

Therefore the graph of $f(x)$ is the same as $g(x)$ except that it has a hole at $x=5$.

If you draw this picture it is clear why the limits are the same.

+1 for asking the question --- it is one that students often don't ask about unless the instructor actually shows that functions that are equal at all points - except perhaps at a single point - have equal limits.

First, I will show you the solution to your problem:

$$\lim_{x\to5}\frac{x^2-25}{x-5}=$$
$$\lim_{x\to5}\frac{(x-5)(x+5)}{x-5}=$$
$$\lim_{x\to5}x+5=5+5=\boxed{10}.$$

To answer you question in short, you are simplifying the continuous function. So essentially, $\frac{x^2-25}{x-5}$ is equivalent to $x+5$ by means of factoring to make the function unchanged. Just simplified.

We can also prove this limit bye performing a delta-epsilon proof:

Proof. Let $\epsilon>0$. Choose $\delta=\epsilon$. If $0<|x-5|<\delta$, then
$$\left|\frac{x^2-25}{x-5}-10\right|=\left|\frac{(x+5)(x-5)}{(x-5)}-10\right|=|x+5-10|=|x-5|<\delta=\epsilon.$$

Thus, we have $\left|\frac{x^2-25}{x-5}-10\right|<\epsilon. \ \ \ \ \ \ \ \blacksquare$


Russell's paradox and the barber example In his reply, Peter Smith provided a clear explanation of the barber's paradox. However, I am having some hard time to understand the link between the established theorem and Russell's paradox. In other words, I do not know how to make the sets U and X fit into the established theorem:$$\neg\exists x\forall y(Rxy \leftrightarrow \neg Ryy)$$

I will be happy to have some help on that.

interpret $Rxy$ as '$x$ contains $y$', I,e '$y$ is an element of $x$'

So where does the $U$ come in?

Well, if there was a universal set of all sets, then we could define a set $X$ as follows:

$$X = \{ Y \in U | Y \not \in Y \}$$

Think of $X$ as 'the set of all normal sets' where a 'normal' set is one that doesn' t contain itself as one of its elements.

But of course since $X$ is a set, and since $U$ is supposed to be the set of all sets, we have $X \in U$, and thus it would be true that:

$$\exists X \in U \ \forall Y \in U (Y \in X \leftrightarrow Y \not \in Y)$$

But since that leads to a contradiction, there cannot be a universal set.

Russell's Paradox itself had nothing to do with the so-called universal set $U.$ It arose from the supposed existence of a set $X$ such that $\forall y: [y\in X \iff y\notin y]$ as required by an early attempt to axiomatize set theory by G. Frege. There is no need to refer to $U$. Specifying $y=X$, as we are permitted by the rules of logic, we obtain the contradiction $X\in X \iff X\notin X.$ It was a genuine paradox since it resulted from faulty axioms. Frege's axioms said that such an $X$ should exist, but its existence results in a logical contradiction, therefore it cannot exist. This is not the case with the Barber "Paradox." It is just a theorem of the rules of logic: $\neg
\exists b:\forall y: [bSy \iff \neg ySy].$ It has nothing to do with any sets.


Cancellation problem: $R\not\cong S$ but $R[t]\cong S[t]$ (Danielewski surfaces) I would like to understand why the two rings
$$
R={\mathbb{C}[x,y,z]}/{(xy - (1 - z^2))} \\ S=\mathbb{C}[x,y,z]/{(x^2y - (1 - z^2))}
$$
are not isomorphic, but $R[t]\cong S[t]$. This example is discussed in this document by Hochster. He references a paper by Danielewski, but it is a preprint from 1989, so I could not locate it anywhere. In fact, this example was also given by Timothy Wagner in this MSE thread, but no proof was supplied.

Finally, this MSE thread seems to refer to this problem (though possibly with different formulation), but unfortunately it hasn't even received a single comment in the last 3.5 years.

Can someone either find a precise reference (preferably online, or at least published in some book/journal) or supply a proof for this claim?

First of all, a good friend of mine wrote a nice exposition of this topic! Also, there is a very nice overview by Kraft: "Challenging problems on affine $n$-space".

R $\not\cong$ S

There are two methods I found of proving this:

Topological method (Danielewski, Fieseler, tom Dieck). Let X, Y be the surfaces in $\mathbf{C}^3$ defined by $xy-(1-z^2)$ and $x^2y-(1-z^2)$, respectively. Then, their fundamental groups at infinity are not isomorphic, in particular
$$\pi_1^\infty(X) \cong \mathbf{Z}/2\mathbf{Z}, \qquad \pi_1^\infty(Y) \cong \mathbf{Z}/4\mathbf{Z}.$$
Danielewsi and Fieseler showed similar statements for the first homology groups at infinity; the former showed $H_1^\infty(X) \cong \mathbf{Z}/2\mathbf{Z}$ and that 4 divides the order of $H_1^\infty(Y)$, while Fieseler was able to compute them. Tom Dieck computed the $\pi_1^\infty$.

Algebraic method (Makar-Limanov). For $A$ a $\mathbf{C}$-algebra, recall that a $\mathbf{C}$-endomorphism $\delta\colon A \to A$ is a derivation if it satisfies the Leibniz rule. Denote by $A^\delta$ the kernel of a derivation $\delta$. We call a derivation $\delta$ locally nilpotent if for every $a \in A$ there is an $n$ such that $\delta^n(a) = 0$, and denote the set of these $\delta$ as $\operatorname{LND}(A)$. Finally, the AK-invariant denoted $\operatorname{AK}(A)$ of $A$ is the intersection of the kernels of all locally nilpotent derivations, i.e.,
$$\operatorname{AK}(A) := \bigcap_{\delta \in \operatorname{LND}(A)} A^\delta.$$
Then, Makar-Limanov showed that $\operatorname{AK}(R) = \mathbf{C}$, while $\operatorname{AK}(S) = \mathbf{C}[x]$. The former is simple to show, since
$$x \mapsto 0, \quad y \mapsto -2z, \quad z \mapsto x$$
is a locally nilpotent derivation with kernel $\mathbf{C}[x]$, and since we can replace $x \leftrightarrow y$ to get a locally nilpotent derivation with kernel $\mathbf{C}[y]$. The other computation is a bit more difficult; see Makar-Limanov's "On the group of automorphisms of a surface $x^ny = P(z)$". There is a nice textbook discussion of this in Rowen's Graduate algebra: commutative view, starting at p. 201.

R[t] $\cong$ S[t]

The idea is as follows, following Rem. 1.5 in Fieseler. Both $X$ and $Y$ can be realized as $\mathbf{G}_a$-principal bundles over the line $V$ with two origins; see one of Prof. Speyer's answers on MathOverflow (which is for a slightly different defining equation, but a similar glueing argument should work). Then, in the cartesian square
$$\require{AMScd}
\begin{CD}
X \times_V Y @>>> Y\\
@VVV @VVV\\
X @>>> V
\end{CD}
$$
all maps are projections of $\mathbf{G}_a$-principal bundles. Since $X$ and $Y$ are both affine, $H^1(X,\mathbf{G}_a) = H^1(Y,\mathbf{G}_a) = 0$, and so by Hilbert's theorem 90 (see Serre's "Espaces fibrés algébriques," §2.3), $X \times \mathbf{C} \cong X \times_V Y \cong Y \times \mathbf{C}$ is a trivial $\mathbf{G}_a$-principal bundle over both $X$ and $Y$.

I would love to know if the last isomorphism can be written out explicitly!

M. Hochster, Nonuniqueness of coefficient rings in a polynomial ring, Proc. Amer. Math. Soc. 34 (1972), 81-82, online


Arrangement of the word 'Success' Number of ways the word 'Success' can be arranged, such that no two S's and C's are together.

These problems quickly
get out of hand if the words are long and there are lots of multiple letters.
Here is a sophisticated solution that uses ideas from algebraic combinatorics.
I learned it from Jair Taylor's wonderful answer here.
See this question also.

Define polynomials for $k\geq 1$ by  $q_k(x) =
\sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$. Here are the first few
polynomials:
$$q_1(x)=x,\quad q_2(x)=x^2/2-x,\quad q_3(x)=x^3/6-x^2+x.$$

The number of permutations with no equal  neighbors, using
an alphabet with frequencies $k_1,k_2,\dots$ is:

  $$\int_0^\infty \prod_j q_{k_j}(x)\,  e^{-x}\,dx.$$

For the "success" problem, the product of the $q$ functions is
$$ q_3(x)\, q_2(x)\, q_1(x)^2=(x^3/6-x^2+x)(x^2/2-x)x^2 =
x^7/12-2x^6/3+3x^5/2-x^4,$$

and performing the integral gives the answer 96.

In my opinion the answer would be as follows
$$
\frac{7!}{3!\cdot 2!} = 420
$$ ways we can arrange word success

by: Javed Masood - FUUAST