When are spectral values a pole of the resolvent? Let $(T_t)_{\geq 0}$ be a $C_0$-semigroup on a Banach space $X$ with the generator $A.$

  What are some conditions sufficient for $\lambda \in \sigma(A)$ to be a pole of resolvent?

I'm looking for conditions on the semigroup or $A.$

Background: I'm trying to prove some stability results for $(T_t)_{t \geq 0}$ and it seems to me that not knowing beforehand that the spectral bound $s(A)$ is a pole of $R(\cdot, A)$ is a roadblock in most cases. Hence I'm looking for conditions that ensure it is a pole, so I can somehow try to prove those conditions.

One sufficient condition for this is that the semigroup is eventually compact. This is proved in Corollary V.3.2 in the book by Engel & Nagel.

Section 5.8 of the book by Taylor is a good source for other conditions.

I'm aware of $2$ sufficient conditions in the case $X$ is a Banach lattice and the resolvent set of $A$ is non-empty:

$1.)\, u \in E_+$ and $D(A^n)\subseteq E_u$ for some $n \in \mathbb{N}.$

$2.)\, u \in E_+$ and $T(t_0)E\subseteq E_u$ for some $t_0>0.$

The proof of these two can be found in Corollary 2.4 and 2.5 of a paper by D. Daners and J. Glück.

A $C_0$ semigroup $S(t)$ on a Banach space is uniformly bounded in operator norm near $0$. That's enough to be able to prove there there is a $\rho > 0$ such that $e^{-\rho t}S(t)$ is a bounded $C_0$ semigroup. A bounded $C_0$ semigroup $S(t)$ has a generator $A$ with resolvent
$$
       -(A-\lambda I)^{-1}= \int_0^{\infty}e^{t(A-\lambda I)}dt=\int_0^{\infty}e^{-\lambda t}S(t)dt.
$$
That gives you a bound on the resolvent in terms of the uniform bound on $e^{-\rho t}S(t)$, and the bound is applicable for $\Re\lambda > \rho$. If there is a pole for the resolvent on the line $\Re\lambda=\rho$, then it is a pole of order $1$ because of the bound that is obtained by the above integral.


determining orientation of a sphere I have a sphere and I am trying to find out its orientation with respect to ground frame.
The sphere is as follows:-

As can be seen from the image, different colors are painted on the sphere with equal thickness of each color along the circumference. I am comparing the sphere with our earth and these colors signify the longitude. The gradually decreasing thickness of each color as one travels towards the poles, is a measure of latitude. Now the angle of the edges (between two colors) with the horizontal of the world frame is what I think to be a measure of tilt somehow (however, the tilt for our earth is fixed at 23.5 degrees with vertical)

My question is that using these images of the sphere captured from a single camera, can I uniquely specify the orientation of the sphere with respect to the ground frame?

I do not know much about complex geometry, so please help.

I'd say that you cannot distinguish between North and South pole. So a machine learning algorithm trained on a dataset of similar pictures can only give you two possible solutions.


Iterative calculation of $\log x$ Suppose one is given an initial approximation of $\log x$, $y_0$, so that:
$$y_0 = \log x + \epsilon \approx \log x$$

Here, all that is known about $x$ is that $x>1$. Is there a general method of improving that estimation using only addition & multiplication, i.e. without exponentiation or logarithms?
$$y_1 = f(y_0, x)=\ ?$$

Instead of solving $y - \ln(x) = 0$ for $y$ you can solve $g(y) = e^y - x = 0$

Given initial approximation $y_0 \approx \ln(x)$ you can try to solve $y$ using Newton's medhod:

$$y_{n + 1} = y_n - \frac{g(y_n)}{g'(y_n)}= y_n - \frac{e^{y_n} - x}{e^{y_n}} = y_n - 1 +\frac{x}{e^{y_n}}$$

When exponentiation isn't allowed you can approximate $e^{y}$ with $\left(1 + \frac{y}{2^m} \right)^{2^m}$ using repeated multiplication: $$\begin{aligned} \left(1 + \frac{y}{2^m} \right)^{2^m} &= \left(1 + \frac{y}{2^m} \right)^{2^{m - 1}} \cdot \left(1 + \frac{y}{2^m} \right)^{2^{m - 1}} \\ \\2^{-m} &= \frac{1}{2} \cdot 2^{-m + 1}\end{aligned}$$

A long time ago, in the last century, I had an early calculator with no log function.  To do a log, I would hit the square root button ten times, subtract $1$, and multiply by $1,024$.  The answer was good to $4$ decimal places, which was better than my two calibrated sticks could deliver.  Square root is a powerful function that can be used to approximate many others.

Do not ridicule calibrated sticks; they did most of the engineering that got us to the moon!  I recently built a slide rule from Starbucks sticks that does multiplication modulo $23$.  I used logs to base $7$.


System of equations $x^3+y=y^2\ \& \ y^3+z=z^2\ \& \ z^3+x=x^2$ Solve over reals:

  $$ \begin{cases} x^3+y=y^2\\ y^3+z=z^2\\ z^3+x=x^2\\ \end{cases} $$

I think the only solution is $x=y=z=0$. If the variables are non-zero multiplying the equations:

$$(x-1)(y-1)(z-1)=x^2y^2z^2 > 0$$

Here, there are two cases: all three variables are greater than $1$ or only one.

If $x,y,z>1$, it's easy to prove there are no solutions because $y^2>x^3$ a.s.o. gives $xyz<1$.

How can I prove there are no solutions when only variable, say $x$, is greater than $1$?

Since only $x>1$
Then $x> y$

Using the first equation,

$x^3=y(y-1)$

For the case $0\leq y <1$, the RHS is negative,

So $y<0$

Taking the third equation,

$z^3=x^2-x$

The RHS is always positive hence $0<z<1$

Taking the second equation,

$y^3=z^2-z$

Differentiate RHS to find stationary point,

$2z-1=0$

$z=1/2$,

$-1/4\leq y^3 <0$

Using the first equation again,

$x^3=y(y-1)$

Differentiating RHS to find min/max for $x$

$y=-1/4$,

$x^3=5/16$, which contradicts that $x>1$


Sum of continuous functions is continuous I am studying for a test and then came this question: $X$ and $Y$ are topological spaces and $f:X \longrightarrow \mathbb{R} , g: Y \longrightarrow\mathbb{R} $ are continuous. Prove that $h: X\times Y \longrightarrow \mathbb{R} $ where $h(x,y)=f(x)+g(y)$ is continuous.

I know that exists a similar question on Math StackExchange but I don't understand that proof.

Hint: $h(x,y) > z$ if and only if there are rational numbers $r, s$ such that $f(x) > r$ and $g(y) > s$ and $r + s > z$.


The convergence of $(n\sin n)$ I  think $\lim_{n\to \infty} n\sin n$ does not exist, for $\sin x$ is a wave function and when $n\to \infty$, then so does factor $n$. But I am confused that how to give a strict proof for the divergence.

I would appreciate it if someone could give some suggestions and comments.

Every convergent sequence is bounded.

Edit: Let $m \in \mathbb{Z}.$ Then $$\sin x\geq \sin\left(\frac12\right)>0\,\, \forall x \in \left[\frac12+2m\pi,\frac32+2m\pi\right].$$ Moreover there exists $n_m \in \left[\frac12+2m\pi,\frac32+2m\pi\right] \cap \mathbb{N}.$ Therefore $$n_m\sin(n_m)\geq \left(\frac12+2m\pi\right)\sin\left(\frac12\right).$$It follows that $(n \sin n)_n$ is unbounded.

We have that $a_n = n \sin{(n)}$. Rewrite it during that $\sin{(x)}\in\left[-1,1\right]$:
$$-n\leq n \sin{(n)} \leq n$$
Than, using the lim:
$$-lim_{n\rightarrow\infty}n \leq lim_{n\rightarrow\infty}n \sin{(n)} \leq lim_{n\rightarrow\infty}n$$
$$-\infty \leq lim_{n\rightarrow\infty}n \sin{(n)} \leq \infty$$
This shows, that $a_n$ is unbounded and thats because limit does not exists:
$$\nexists lim_{n\rightarrow\infty} n \sin{(n)}$$


Tensor invariants under isometries UPD Equivallent formulation --- how do you find all the independent isometric invariants of a tensor?

In what follows $V$ is a real inner product space.

I want to understand how does one find all (scalar) invariants of tensors under isometries. Wikipedia has a relevant article, but it confines itself only with tensors of type $V^* \otimes V$ or equivalent.

What I mean by an invariant of a tensor under the isometry. Invariant of a $V \otimes V^*$ tensor is a function $I : V \otimes V^* \to \mathbb R$ if

$$I\left( Q(A) \right) = I(A)$$

for any isometry $Q$ (transformation) properly lifted to act on tensors.

I've asked a question about the invariants of tensor of type $V$, but couldn't get the intuition to extend the reasoning to higher rank tensors

So far as I understand vectors have invariants of the form

$$f(\langle v, v \rangle)$$

whereas tensors in $V^* \otimes V$ have

$$f(\operatorname{Tr} A, \operatorname{Tr} A^2, \operatorname{Tr} A^3)$$

$f$ is meant to be an arbitrary function to $\mathbb R$ of appropriate type.

I don't even see why there are three independent invariants for rank two tensors.

One might be be puzzled of the importance of the question and why I'm asking it. The above mentioned wikipedia article provides a good example. Potential energy (scalar) in an elastic material is a function of the strain tensor. The coupling between them should be natural in the sense that it should not involve arbitrary tensors or vectors (because there are simply none), one have to construct a scalar just from the strain tensor. Theories in continuum mechanics abounds in dozens of various tensor fields of different ranks and one need to seek the forms of natural couplings between them.

This seems too large for a comment, so I'll post it as an answer.

First of all for matrices (aka tensors of type $V\otimes V^*$). If the tensor/matrix is symmetric, then it is diagonalizable by an isometry (this is spectral theorem), so any isometric invariant is a function of eigenvalues. Moreover as permutations are isometries, it has to be a symmetric function of eigenvalues. If one further restricts to polynomial symmetric functions then all such are well understood - they are all expressible interms of either $a_1+a_2+\ldots a_n$, $a_1^2+a_2^2+\ldots a_n^2$,$\ldots$, $a_1^n+a_2^n+\ldots a_n^n$ - these are the traces in your question - or in terms of coefficients of $(x-a_1)(x-a_2)\cdots (x-a_n)$ - those are the invariants in the wikipedia article (see also http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial). When written not in terms of eigenvalues, but as traces or coefficients of the characteristic polynomial they become invariants of the matrix entries.

For not-necessarily symmetric matrices there could be other invariants - indeed there must be, for the space of all matrices is of dimension $n^2$ and the space of isometries is of dimension $n(n-1)/2$, so the quotient is of dimension $n(n+1)/2>n$. In particular in 3-D one expects 6 invariants). I don't know the answer, but what one is doing is taking the space (variety) of all matrices and quotienting it by the action of the group of isometries, and then asking what are the functions on the resulting space. In the symmetric case, the quotient is actually $\mathbb{R}^n$ coodinatized in this funny way. In the general case I'm not sure, but I suspect people who do invariant theory or representation theory know this.

Note: In the scalar case the quotient $V$ divided by isometries is $\mathbb{R}_+$ coordinatized by the square length.

Note 2: For skew-symmetric matrices one can bring them to canonical block form by isometric transformations. http://en.wikipedia.org/wiki/Skew-symmetric_matrix so the the invariants are given in terms of squares of the eigenvalues, well expressed as invariants of the $A^2$ - so one again gets either traces of $A^2$, $A^4$ and so on or coefficients of the characteristic polynomial of $A^2$. Of course when $n=3$ there is only one invariant - trace $A^2$.

For "higher" tensors things seem less obvious still. You agin take some vector space (of tensors) and quotient it by the action of the group of isometries and ask for the functions on the resulting quotient.

One other comment - if I understand correctly in some theories you get scalars that depend not only on the tensor itself, but also on its derivatives of various orders.  If you allow this generalization the problem becomes  harder.

A second rank tensor in 3 dimensions has only trhee eigenvalues lambda1, lambda2 and lambda3, its characteristic polynomial has 3 different coefficients J1, J2 and J3 and it has only 3 independent traces of its powers Tr(A), Tr(A^2) and Tr(A^3). All this reduces to say that it has only three independent invariants, that can be given in different formats: eigenvalues, principal invariants or traces. The relation between eigenvalues and principal invariants is very direct because the first are the roots of the characteristic polynomial and the second are its coefficients. The third relation is much more difficult to see and its demonstration is called Cayley Hamilton Theorem. In a few words it says that "a second rank tensor satisfies its own characteristic equation". So you have J0*A^3+J1*A^2+J2*A^1+J3*A^0=0 (with J0=1), taking traces you find the last set of independent invariants. Cayley Hamilton theorem is valid in n-dimensions, it is to say J0*A^n+J1*A^(n-1)+...+Jn=0. The demonstration is really amazing and surprising. Note that the independence of the traces seems natural because we have only one equation: the characteristic polynomial. On the other side, Cayley Hamilton theorem is surprising because from this unique equation it gives n^2 equations between the components of the tensor.


Show that the powerset partial order is a cartesian closed category. I'm trying to get started learning category theory.  A problem I'm working on is to show that for a set $S$, the partial order $(\mathcal{P}(S),\subseteq)$ viewed as a category is cartesian closed.

So far, I was thinking that $S$ is the terminal object, and that the product of any two subsets $A \times B$ always exists (the greatest lower bound or intersection of $A$ and $B$), but am having trouble completing the proof by showing that the category has exponentiation.

If I understand right, the goal is to show that for any $A$ and $B$, there is a $C$ such that $C \times A \to B$ (i.e., $(C \cap A) \subseteq B$) if and only if there is a $B^A$ such that $C \subseteq B^A$ and $(B^A \cap A) \subseteq B$. But then this doesn't seem quite right.

I'm sorry for such a beginner question. I'd really appreciate any help pointing me in the right direction or maybe clarifying what/how one would go about showing that this category has exponentiation - or what does "$B^A$" mean in terms of these subsets? Does some sequence of operations construct it?

Let's prove something stronger: every infinitary-distributive complete lattice is a complete Heyting algebra (and so cartesian closed as a category).

Recall, a complete lattice is a partial order in which all meets and joins exist. The powerset of a set is certainly a complete lattice, and it moreover satisfies the infinite distributive law:
$$\left( \bigcup_{i \in I} A_i \right) \cap B = \bigcup_{i \in I} (A_i \cap B)$$
Now the claim is that such a complete lattice has a Heyting operation, i.e. a binary operation $\to$ such that $A \subseteq (B \to C)$ if and only if $A \cap B \subseteq C$ for all $A$. How do we do this? Let's apply wishful thinking. Suppose we had such a $(B \to C)$; then $(B \to C) \subseteq (B \to C)$, and in particular, $(B \to C)$ must be the greatest $A$ such that $A \cap B \subseteq C$. On the other hand, we are in a complete lattice, so we can certainly define
$$(B \to C) = \bigcup \{ A : A \cap B \subseteq C \}$$
and this works: if $A \subseteq (B \to C)$, then
$$A \cap B \subseteq (B \to C) \cap B = \left( \bigcup \{ A : A \cap B \subseteq C \} \right) \cap B = \bigcup \{ A \cap B : A \cap B \subseteq C \} \subseteq C$$
by the infinite distributive law; the converse is automatic by construction of $(B \to C)$.

The nice thing about this proof is that it is completely constructive and can be internalised in any topos. But in the classical setting, we can be a little bit more explicit about what $(B \to C)$ is: writing $B^{\mathsf{c}}$ for the complement of $B$, we have $(B \to C) = B^{\mathsf{c}} \cup C$. Clearly, $(B^{\mathsf{c}} \cup C) \cap B = C \cap B \subseteq C$, so $(B^{\mathsf{c}} \cup C) \subseteq (B \to C)$; on the other hand, if $A \cap B \subseteq C$, then $B^{\mathsf{c}} \cup (A \cap B) \subseteq B^{\mathsf{c}} \cup C$, and clearly
$$A = A \cap (B^{\mathsf{c}} \cup B) = (A \cap B^{\mathsf{c}}) \cup (A \cap B) \subseteq B^{\mathsf{c}} \cup (A \cap B)$$
so $A \subseteq B^{\mathsf{c}} \cup C$, and therefore $(B \to C) \subseteq B^{\mathsf{c}} \cup C$, as required. (Exercise: Where have I used the law of excluded middle in this argument?)

I found this question in Pierce's book and have been working on it today. Here's what I've got:

The Problem: The exponent $B^A$ needs to be a member of the powerset. Moreover it has to satisfy the following constraints:

There is a map from $B^A \times A$ to $B$.

in terms of sets, $B^A \times A \subseteq B$

For every other powerset object $C$ such that $g: C \times A \rightarrow B$ exists, we need a unique map from $C$ to $B$

again in terms of sets, we must show that for all $C$ such that $C \times A \subseteq B$, the relation $C \subseteq B$ must hold.

Solution: If $B \subset A$ then $B^A = B$. Otherwise $B^A = S$ (the final object in our powerset category).

Justification: If $B \subset A$ then $B^A$ must be a subset of $B$ for $B^A \times A \subseteq B$ to hold. We deduce that $B^A$ must be $B$ exactly because it is the only choice that guarantees $C \subseteq B^A$ for every $C$ such that $C^A \times A \subseteq B$.

If $A \subseteq B$ then $B^A \times A \subseteq B$ holds for any choice of $B^A$. Likewise, we can use any powerset element $C$ to satisfy $C^A \times A \subseteq B$. Because of this, $B^A$ has to be the entire set $S$. It is the only choice that guarantees $C \subseteq B^A$ for every valid choice of $C$.


Independence of $(X,Y)$ with joint PDF $f_{X,Y}(x,y)=8xy$ on $ 0<y<x<1$ I know that as a general property, if

$f_{X,Y}(x,y)=g(x)h(y)$, then we say X and Y are independent random variables.

However, I am having trouble to accept this statement. Take the case:

$f_{X,Y}(x,y)=8xy, 0<y<x<1$

Then, $f_{X}(x)=4x$ and $f_{Y}(y)=4y$, where $f_{X}(x).f_{Y}(y)\ne f_{X,Y}(x,y)$

Maybe I am getting something wrong over the limits of integration, I often trip on simple things... Or maybe, in this case, since the variables are dependent in the limits of integration, then they cannot be independent. But I havent seen any book point that as a requirement... So I suppose I got something wrong in my calculations.

If someone could please clarify this for me, I would be grateful.

Take the case: $f_{X,Y}(x,y)=8xy, 0<y<x<1$.

Yet again an excellent example of the fact that densities should include restrictions on their domain.

Recall that any density of the couple of random variables $(X,Y)$ should be a function $f:\mathbb R\times\mathbb R\to\mathbb R_+$ such that what-you-know holds. In the present case, $f:\mathbb R\times\mathbb R\to\mathbb R_+$, $(x,y)\mapsto8xy$ is (obviously) not a density. Rather, a density of $(X,Y)$ is  $f:\mathbb R\times\mathbb R\to\mathbb R_+$, $(x,y)\mapsto8xy\mathbf 1_{0<y<x<1}$ (for example $f(1,2)=0$, not $16$).

Now the result you mention is correct:

  The random variables $X$ and $Y$ with density $f$ are independent if and only if there exist $g$ and $h$ such that $f(x,y)=g(x)h(y)$ for (almost) every $(x,y)$ in $\mathbb R\times\mathbb R$.

Which does not hold for the density $f$ in the example.

This remark is also useful when computing marginals. In general,
$$
f_X(x)=\int_\mathbb Rf(x,y)\mathrm dy,
$$
hence in the present case,
$$
f_X(x)=\mathbf 1_{0<x<1}\int_\mathbb R8xy\mathbf 1_{0<y<x}\mathrm dy=x\mathbf 1_{0<x<1}\int_0^x8y\mathrm dy=4x^3\mathbf 1_{0<x<1}.
$$
Note the absence of cases and the automaticity of the computations, thanks to adequate notations.

To sum up:

  When a joint distribution is given by its PDF, a détour by the joint CDF is useless (and frankly often cumbersome) provided one uses the true PDF, which should include indicator functions if need be.

I think you're slightly confused with this characterization of independence. If we assume that $X$ and $Y$ have density functions $f_X$ and $f_Y$ and joint distribution $f_{X,Y}$, then $X$ and $Y$ are independent iff $f_{X,Y} = f_x \cdot f_y$. This is not the same as the property you listed above, the $g(x)$ and $h(y)$ must be the marginal density functions and not just any function of $x$ and $y$.


Collection of independent sets equal to downset? I have been struggling with the following problem.

Every set here is supposed to be finite.

If we have a closure $\lambda$ on $X$, we define the collection of independent sets of $X$ as $$ I_\lambda = \{ A \subseteq X : x \notin\lambda(A\setminus \{ x \})\forall x\in A \} $$

Supose now we have a downset $I \subseteq 2^X$. Is there a closure $\lambda$ on $X$ such that $I_\lambda = I$?

I first tried a pretty straightforward closure, by setting $\lambda(A) = A$ for elements in $I$ and $\lambda(A) = X$ for the rest. This of course does not work but only on a few sets, namely the ones that are not in $I$ but that any $A \setminus \{x\}$ is in $I$ for every $x \in A$, I tried to refine this idea to get rid of those cases but I haven't been able to do it.

This lead to my believe that it may not be possible and probably the trouble sets are those I just mentioned, but on a few small example I have been able to find the closure even when there are sets like those.

Edit: Now I know this is not possible, but I need help finding a counterexample, or even better, trying to understand when it exactly fails, and when it works.

Any help is appreciated

Definitons

A closure $\lambda$ on $X$ is a function $\lambda: 2^X \to 2^X$, such that is increasing (i.e $A \subseteq \lambda(A)$), monotone and idempotent.

An $I$ is said to be a downset if for any $A \in I$ and $B \subseteq A$, then $B \in I$

My first idea was to try
$$
\lambda(A)=\begin{cases}A,&\text{if $A\cup\{x\}\in I$ for some $x\in X\setminus A$}\\X,&\text{otherwise,}\end{cases}
$$
but this does not work (see alejopelaez's comment). Another idea I had was to try something like
$$
\lambda(A)=\{x\in X:\text{$x\in A$ or $A\cup\{x\}\notin I$}\},
$$
but this is not idempotent.

It is possible that an idea from matroids might work. For example, we could take
$$
\lambda(A)=\{x\in X:r(A)=r(A\cup\{x\})\},
$$
where
$$
r(A)=\max\{\lvert B\rvert:\text{$B\subseteq A$ and $B\in I$}\}
$$
is the size of the largest subset of $A$ that is in $I$. The idea is that $r$ is some kind of "rank" function, having the property that $r(A)=\lvert A\rvert$ if and only if $A\in I$.

Try the closure that have the exchange property.


Fiber bundles of $G$-spaces So if $G$ is a topological group and $H,\ J$ are  closed subgroups such that $H\lhd J$, then the principal bundle $G/H\to G/J$ is trivial iff it has a global section.

I have   questions about the general case:

If $H$ and $J$ are closed subgroups of $G$ such that $H<J$ (not necessarily normal), then is the bundle $G/H\to G/J$, $\ \ gH\mapsto gJ$ trivial iff it has a global section?
If $G$ has a closed normal subgroup $N$ that acts transitively on the base $G/J$ and trivially on the fiber $J/H$. Is the bundle $G/H\to G/J$ trivial?

For 1, of course if the bundle is trivial it has a global section.  But the converse need not hold.

Consider the chain of subgroups $SO(2n)\subseteq SO(2n+1)\subseteq SO(2n+2)$.  Then the bundle in 1 is the unit tangent bundle $T^1S^{2n+1}\rightarrow S^{2n+1}$.  A section of this bundle is essentially a non-vanishing vector field on $S^{2n+1}$, so exists for all $n$.  On the other hand, Adams showed this bundle is trivial only for $n=0,1,3$.

For 2, I'm not sure what you're asking.  How does $N$ act on $J/H$?

To answer your first question, the fiber bundle being trivial is actually a more general result about principle fiber bundles.  The group structure doesn't matter since you essentially only need an arbitrary manifold as your base.

  Suppose $(P,\pi, M, G)$ is a principle fiber bundle where $P\times
 G\to P$ is a right action and $\pi:P \to M$ is projection.  Then $P$
  is trivial if and only if there exists a smooth global section.

Proof: The first direction is trivial (if $P$ is trivial then it is a product bundle $P = M\times G$).  Conversely, suppose there is a smooth global section $\sigma:M\to P$.  Then we can define a diffeomorphism $\Phi:P \to M\times G$ as
$$
\Phi(p) \;\; =\;\; \left (\pi(p), \hat{g}\right )
$$
where $\hat{g} \in G$ is the unique group element that has $\sigma(\pi(p)) \cdot \hat{g} = p$.  Such a group element is unique since the action of $G$ is required to be free.  $\Phi$ is clearly smooth and also has a smooth inverse $\Phi^{-1}:M\times G\to P$ given by
$$
\Phi^{-1}(x,g) \;\; =\;\; \sigma(x)\cdot g.
$$
$$\tag*{$\Box$}$$


Showing that the set of all finite subsets of $\mathbb{N}$ is a countable set. I'm working through Understanding Analysis by Abbott and was wondering about the correctness of my solution to a problem in section 1.4. Exercise 1.4.10 asks:

  Show that the set of all finite subsets of $\mathbb{N}$ is a countable set.

So my idea for a solution is to use induction. The base case is simple, with each element of $\mathbb{N}$ mapping to a singleton of itself. Clearly we have a bijection and thus all the subsets of size 1 are countable. From here we assume that the set of all finite subsets of $\mathbb{N}$ of size $k-1$ is a countable set. Let $\{B_1, B_2, B_3, B_4, \dots \}$ be an enumeration of the subsets of size $k-1$ of $\mathbb{N}$. For a given $i \in \mathbb{N}$, the family $\{B_k \cup \{i\} : k \in \mathbb{N}\}$ is similarly countable. The family on subsets of $\mathbb{N}$ of size $k$ can then be written as the union

$$\bigcup_{i=1}^\infty{\left\{B_k \cup \{i\} : k \in \mathbb{N}\right\}}.$$

This is a countable union of countable families of sets and is thus countable. So, the set of all subsets of $\mathbb{N}$ of size $k$ is countable and thus by induction this statement holds true for all $n \in \mathbb{N}$.

Now let $A_i$ be the family of subsets of $\mathbb{N}$ of size $i$. By above each of these families is countable. Thus

$$\bigcup_{i=1}^\infty{A_i}$$

is the set of all finite subsets of $\mathbb{N}$ and since it is a countable union of countable families of sets our result follows.

Is this a valid proof? The only thing I'm a little unsure of is writing the family of subsets of size $k$ as I do in that first union. Any other comments are welcome as well.

With the minor correction made in your comment (the family of $k$-subsets is contained in the union you wrote; it's not equal to that union, because the union contains some $(k-1)$-subsets as well), your proof is correct.  I would probably write it in a way that makes no use of enumerations of the countable sets involved... other answers show that there are very clean ways to prove the countability of $\cal{P}(\mathbb{N})$ specifically, so it's nice for you to have a slightly more abstract approach:

  Thm: If $X$ is a countable set, then so is ${\cal{P}}(X)$.

Proof (OP):

Let $A_k(X)$ be the set of $k$-subsets of $X$.   We will show that $A_k(X)$ is countable for each $k$, by induction.  Clearly $A_0(X)=\{\emptyset\}$ is countable.  For $k\ge 1$, assume $A_{k-1}(X)$ is countable.  We have
$$
A_{k}(X)=\bigcup_{x\in X}\bigcup_{Y\in A_{k-1}(X) : x\notin Y}\{Y\cup\{x\}\}.
$$
This is a countable union (since $X$ is countable) of countable unions (since $A_{k-1}(X)$ is countable by hypothesis) of finite sets; hence it is a countable union of countable sets; hence it is countable.  By induction, then, $A_k(X)$ is countable for all $k$.  We conclude that $${\cal{P}}(X)=\bigcup_{k=0}^{\infty}A_k(X)$$
is also a countable union of countable sets, and hence countable.
$\;\square$

Let $n \in \mathbb{N}$ and $b_m b_{m-1}\ldots b_2 b_1 b_0$ be its expansion in binary. There are technically an infinite number of $0$'s to the left of the expansion, so it makes sense to talk about bits $b_r$, for $r > m$.

The binary expansion of a number provides a natural way to talk about the subsets of $\mathbb{N}$. Each bit is either $1$ or $0$, and each element of $\mathbb{N}$ is either in or not in a given subset.

For $n$ as described above with a binary expansion $b_m b_{m-1}\ldots b_2 b_1 b_0$, construct the subset of $\mathbb{N}$ specified by $\{a \in \mathbb{N} \vert b_{a-1} = 1\}$. This subset is finite since $n$ is finite, and therefore has a finite binary expansion. Moreover, since binary expansions are unique, no two natural numbers can specify the same subset of $\mathbb{N}$. Every subset of $\mathbb{N}$ can be associated with a unique binary expansion (and consequently, a unique natural number), showing that this construction is bijective.

Since $\mathbb{N}$ is countable and the set of finite subsets is bijective to $\mathbb{N}$, this set is also countable.


How to solve IVP using Laplace transform(of matrix)? $$x' = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & -2 \\ 3&2 & 1\end{bmatrix} x, ~~ x{(0)} = \begin{bmatrix} 2 \\ -1 \\ 1\end{bmatrix}$$

I am having very hard time solving this question using Laplace transform. I don't know how to do it for $3\times 3$ matrix. Would anyone be able to show me step by step how to do it for one of the cases? I have no clue how to do the inverse.

We are given:

$$X'(t) = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & -2\\ 3 & 2 & 1\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\\ z(t)\end{bmatrix}, ~~ X(0) = \begin{bmatrix} 2 \\ -1\\ 1\end{bmatrix}$$

We can write this as:

$$\tag 1 \begin{align} x' &= x  \\ y' &= 2x+y-2z \\ z' &= 3x +2y + z \end{align}$$

Taking the Laplace transform of $(1)$ yields:

$$\begin{align} s x(s) - x(0) &= x(s) \\ s y(s) - y(0) &= 2x(s) + y(s) - 2z(s) \\ s z(s) - z(0) &= 3 x(s) + 2y(s) + z(s)\end{align}$$

This reduces to the system:

$$\begin{bmatrix} s-1 & 0 & 0 \\ -2 & s-1 & 2\\ -3 & -2 & s-1\end{bmatrix} \begin{bmatrix} x(s) \\ y(s)\\ z(s)\end{bmatrix} = \begin{bmatrix} 2 \\ -1\\ 1 \end{bmatrix}$$

All that is needed is to solve for $x(s),y(s),z(s)$ and then find the inverse Laplace Transform.

You could have also used many methods to solve this system, including eigenvalues/eigenvectors, matrix exponential, etc.

Update

The inverse of the matrix is:

$$\begin{bmatrix} \frac{1}{s-1} & 0 & 0 \\
 \frac{2 (s-4)}{s^3-3 s^2+7 s-5} & \frac{s-1}{s^2-2 s+5} & -\frac{2}{s^2-2 s+5} \\
 \frac{3 s+1}{s^3-3 s^2+7 s-5} & \frac{2}{s^2-2 s+5} & \frac{s-1}{s^2-2 s+5} \end{bmatrix}$$

The inverse matrix times the column vector yields:

$$\begin{bmatrix} x(s) \\y(s)\\z(s) \end{bmatrix} =
\begin{bmatrix} \frac{2}{s-1} \\ \frac{2 s}{s^2-2s+5}-\frac{3}{s-1} \\
 \frac{s (s+2)+5}{(s-1) (s^2-2s+5)}\end{bmatrix}$$

Next, just find the inverse Laplace Transform of $x(s), y(s), z(s)$ and you are done.

I will assume you can take it from here.

Since now it is not clear why I would post this (partial) answer, I would like to mention that I strongly feel that the accepted answer is not useful because it was quite probably derived using a computer algebra system. This is not a sin, however, most of the students are supposed to solve problems like this with just pencil and paper within quite limited time. In my answer I show how to do it without the extra unnecessary steps presented in the accepted answer.

When someone solves the problem of the form $\dot x=Ax$ by the Laplace transform method, it immediately implies that the application of the Laplace transform yields
$$
s X-x(0)=AX,
$$
where $X=\mathcal L(x)$, or system
$$
(sI-A)X=x(0),
$$
which theoretically is solved by simply
$$
X=(sI-A)^{-1}x(0).
$$
This is a nice formula, but should never be used for actual calculations by hand. Much better to use usual elimination of variables procedure. In the given example, the system actually reads
$$
(s-1)x\quad \quad\quad\quad\qquad\qquad\quad=2\\
-2x+(s-1)y+\qquad 2z=-1\\
-3x\quad\,-2y+(s-1)z=1.
$$
From the first equation we immediately find
$$
x=\frac{2}{s-1}.
$$
Now, what is left is
$$
-\frac{4}{s-1}+(s-1)y+\qquad 2z=-1\\
-\frac{6}{s-1}\quad\,-2y+(s-1)z=1.
$$
Multiply the first equation by 2 and the second by $(s-1)$ and add, you get
$$
(s^2-2s+5)z=\frac{s^2+2s+5}{s-1}.
$$
To find $y$ multiply the first equation in the same system by $(s-1)$ and the second by $-2$, add, and you find
$$
(s^2-2s+5)y=\frac{-s^2+4s-15}{s-1}.
$$
Now the whole system is solved (it took me around 5 minutes on paper).


Is powers of rationals dense in $\mathbb R$ Consider $\mathbb {\tilde {Q}} = \{ x^n : x \in \mathbb Q \} $ $n$ is fixed odd integer.

I have two questions here.

Is this set dense in $\mathbb R$ and
Is there any bijection exists between $\mathbb Q$ and $\mathbb {\tilde Q}$

For the first question, I think the set is dense. Consider $ a,b \in \mathbb R$ WLOG assume $ a,b \ge 0$ and $ b \ge a$. We can find $c \in \mathbb Q$ such that $ a^{1/n} \le c \le b^{1/n}$. And now $ c^n \in \mathbb {\tilde Q}$ and $ a \le c^n \le b$.

So the set is dense.

Is there a way to define bijection between $\mathbb Q$ and $\mathbb {\tilde Q}$

Here is a simpler proof of density. Consider the map $f(x)=x^n$, $n>0$ is odd, $f: {\mathbb R}\to {\mathbb R}$. This map is clearly continuous. The intermediate value theorem implies that this map is surjective. The set of rational numbers is dense in ${\mathbb R}$. Therefore, its image under the continuous map $f$ is also dense in $f({\mathbb R})={\mathbb R}$.

The set of powers of the rationals, to any power, including even, is infinitely dense.  What happens of the even numbers is that the range of the dense region is not less than zero.

One can demonstrate this by noting that any irrational number must be bracketed by two rational numbers, for example, numbers written in base 10.  So we know, for example, that $\sqrt[3]{2}$ lies variously between $1$ and $2$, between $1.2$ and $1.3$, and $1.25$ and $1.26$, and so forth for each decimal of the cube root of $2$.

For this reason, we would then note that there are always rational powers of a given value, between any previous rational value, and a number not of that particular power.  That is, there must exist a cube between $2.197$ and $2$, such as $2.000376$, and even closer ones.

Thus, the powers are infinitely dense on the real number line.

Since every cube must has exactly one real cube root and likewise true for all odd powers, then there is a bijection between the two sets.  For even numbers, one must restrict the tests to ${R}^+$, the positive reals.


If I buy 2 lottery tickets do I double my chance of winning? There's a lottery.  There are 6 balls chosen randomly from 49 and you have to match all the balls to win.

I buy one ticket.   If I buy two tickets with different numbers for the same draw, do I double my chance of winning the jackpot.

What's the correct formula here?

Yes, you do double your chance of winning. There are $\binom{49}{6}$ different possible draws. The probability of winning with one ticket is the chance that the 6 drawn balls are one specific combination, i.e. $\frac{1}{\binom{49}{6}}$. With two tickets, there are two winning combinations, so the probability is $\frac{2}{\binom{49}{6}}$

Basically, the reason it is doubled is because winning one ticket and winning the other ticket are mutually exclusive, i.e. you can't win both tickets (note the problem statement says "different tickets"). If there was a chance of winning both, the chance would be less than doubled.

Understanding lottery.  Let us say there is a lottery where there is 100 tickets but there will be only 1 prize. This is a game of chance, nothing to do with probability. Buy 1 ticket and your chance of winning the prize is 1 in 100. Buy 2 tickers and your chance of winning is 1 in 99 and not 1 in 50 as most people think.

The rational is as follows: of the 2 tickets 1 has to be a looser but you do not know which. Therefore it is unmistakable that you have a looser. It follows then that you may have the winner, so mentally you discard 1 as a loser and maybe the other is the winner reducing your chance to only 1 in 99 and not 1 in 50 as many go on believing.

Think about it - you have to be bright enough to have posed the question.


Parameter of $S^2\times S^2$ I want to compute the connection and curvature of $S^2\times S^2$ with induced metric in $R^n$. But I dont know how to represent it by equation. So, I can't begin. May I have some hint ?

$$
(\sqrt{y_1^2+y_2^2+y^2_3}-3)^2+y_4^2+y_5^2 =1    ~~~~~~~\text{or}   \\
((3+\sin C)\cos B \sin A , (3+\sin C)\cos B\cos A, (3+\sin C)\sin B,\cos C\sin D,\cos C \cos D)
$$

The big radius is 3 ,and little radius is 1. The way is likely to parameter of $T^2$.


Why is the negation of $A \Rightarrow B$ not $A \Rightarrow \lnot B$? The book I am reading says that the negation of "$A$ implies $B$" is "$A$ does not necessarily imply $B$" and not "$A$ implies not $B$". I understand the distinction between the two cases but why is the first one considered true?

Here is a more intuitive explanation. Suppose that $A$ and $B$ are unrelated. For example, $A$ could be "France is a country in Europe" and $B$ could be "I will win the lottery". It is certainly the case that we know $A$ does not imply $B$ for these sentences: knowing that France is in Europe tells me nothing about the future! But we also do not know that $A \Rightarrow \lnot B$, for the same reason.

So in this case, we only know that $A \not  \Rightarrow B$.  This shows there is a difference between $A \not  \Rightarrow B$ (which just says that $A$ does not imply $B$) and $A  \Rightarrow \lnot B$ (which says $A$ does imply the negation of $B$).

This kind of intuition is important when you move from propositional logic to more general mathematics. For example, suppose we are looking at natural numbers. Let $A$ say "$x$ is an even natural number" and let $B$ say "$x$ is a natural number that is a multiple of $6$". Neither $A$ nor $B$ is true or false on its own, because the $x$ has no fixed value. Sometimes $A$ is true and sometimes it is false. But we still have that $A \not \Rightarrow B$ (e.g., $2$ is even but not a multiple of 6) and $B \Rightarrow A$ (because every multiple of 6 is even). We also do not have $A \Rightarrow \lnot B$ (because some even numbers are multiples of 6).

Rather than trying to analyze this more complicated kind of implication in terms of truth values, it is helpful to develop the right intuition: $A \Rightarrow B$ says that knowing $A$ alone is enough to know $B$. From that intuition, you can work out the formalities involving truth values, but more importantly you can do mathematics from that intuition.

Try building both $\lnot$(A$\rightarrow$B) and (A$\rightarrow$$\lnot$B) up from their component well-formed formulas.  Or write them in a prefix notation like the following:

$\lnot$($\rightarrow$(A, B)) and

$\rightarrow$(A, $\lnot$(B))

In the former we have "A" and "B" and then a conditional.  Then we negate the conditional.

In the latter we have "A" and "B", then B gets negated, and finally we have a conditional.

This sentence " the negation of "A implies B" is "A does not necessarily imply B"" gets considered true, because "not necessarily" applies to "imply", it does NOT apply to B.


Find $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 ,\ n\ge 1$ has $n$ roots $x_1,x_2,\ldots,x_n \le -1$ and such that $a_0^2+a_1a_n=a_n^2+a_0a_{n-1}.$ Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 ,\  n\ge 1$ have $n$ roots $x_1,x_2,\ldots,x_n \le -1$ and  $a_0^2+a_1a_n=a_n^2+a_0a_{n-1}$.
  Find all such $P(x)$. (Poland 1990).

I used Viete Theorem and get: $$(x_1x_2\cdots x_n)^2 \pm \sum_{j=1}^n\prod_{i \not=j} x_i=1\pm(x_1x_2 \cdots x_n)(x_1+x_2+\cdots+x_n)$$ but not succeeded.

Unless I misunderstand the problem,
$P(x) = x^2+2x+1$ meets the condition: Both roots are $-1$ and
$$a_0^2+a_1a_2 = a_2^2 + a_0a_1=3$$
Was the problem to find all such $P(x)$?

Let $s_i$ denote the elementary symmetric polynomials in $x_1,x_2,...,x_n$.

Then $a_0/a_n=(-1)^ns_n$, $a_1/a_n=(-1)^{n-1}s_{n-1}=(-1)^{n-1}s_n\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)$, and $a_{n-1}/a_n=-s_1$.

We can re-interpret $a_0^2+a_1a_n=a_n^2+a_0a_{n-1}$ as

$$s_n^2+(-1)^{n-1}s_{n-1}=1-s_ns_{n-1}$$

$$s_n(s_n-s_{n-1})+(-1)^{n-1}s_{n-1}=1$$

Assume $n$ is even. Then, $s_n$ is positive, $-s_{n-1}$ is positive, $(-1)^{n-1}s_{n-1}$ is positive. If at least one of the $x_k<-1$ then $(-1)^{n-1}s_{n-1}>1$ and the equality can't happen. Therefore, all $x_k=-1$.

Assume now that $n$ is odd. Similarly as before, $s_n(s_n-s_{n-1})$ is positive, $(-1)^{n-1}s_{n-1}$ is positive again. If at least one of the $x_k<-1$ then the sum on the left is $>1$. Therefore all $x_k=-1$.

Hence $P(x)=a_n(x+1)^{n}$, or forgetting (dividing out) about the $a_n$, $P(x)=(x+1)^{m}$. This would give us $a_0=1$, $a_1=n$, $a_{n-1}=n$. We need $$1^2+n=1^2+n$$ which holds.


Simple exercise on Galois theory Find the splitting field of $x^6-2x^4-8x^2+16$ over $\mathbb {F}_3$ and list the intermediate fields between the base camp and the splitting field.

My factorization was incorrect, I have understood my error. Now to calculate the Galois group of the extension (which is actually a Galois extension, as $\mathbb {F}_9$ is the splitting field of a separable polynomial, having first derivative different from zero, with coefficients in $\mathbb {F}_3$) I just note that the degree of the extension is $[\mathbb {F}_9:\mathbb {F}_3]=2$, so the group has 2 elements and it is isomorphic to $\mathbb {Z}_2$. Is it right?


Prove that $gNg^{-1} \subseteq N$ iff $gNg^{-1} = N$ I'm reading the solution to one of my homework problems and am stuck on something. Here is the problem:

  Let $N \leq G$ be a finite subgroup. Show that $gNg^{-1} \subseteq N$
  if and only if $gNg^{-1} = N$.

The solution is to suppose that $gNg^{-1} \subseteq N$ and consider the map $\varphi: N \to gNg^{-1}$ defined by $\varphi(n) = gng^{-1}$. Observe that such a map is a bijection and thus $|N| = |gNg^{-1}|$. The equality then follows.

I'm just curious as to why we can assume that $\varphi$ is a bijection (i.e. why can't there exist $n_1, \, n_2 \in N$ such that $gn_1g^{-1} = gn_2g^{-1}$ where $n_1 \neq n_2$)?

The fact that $\varphi$ is a bijection follows from the invertibility of group multiplication.

Start with $gn_1g^{-1} = gn_2g^{-1}$. First multiply both sides on the left by $g^{-1}$.  Then multiply on the right by $g$.

One implication : $gNg^{-1} = N $ implies $gNg^{-1} \subset N $ is trivial. Lets try reverse part:
Suppose $gNg^{-1} \subset N $  and we will show that $gNg^{-1} = N $.
On contrary suppose that $gNg^{-1} \neq N $
so there is $ n_1 \in N $ such that
$gng^{-1} \neq n_1$  for all $n \in N$ then on pre and post multiplication by $g^{-1}$ and g we obtain
$ n \neq g^{-1}n_1g $ for all $ n \in N $
which in turn says that $ n \notin  g^{-1}Ng$  for all $ n \in N$ contradicting our hypothesis. This contradiction proves the result.


Initial value $\left ( \frac{dy}{dt} \right )+3y=11$, $y(0)=1$ I have never done an initial value problem, and would like some help on how to start this please.

The first step is to solve the ODE:

$$y'+3y=11\xrightarrow{\text{Using an appropriate integrating factor }~\mu(x)=\exp(3x)}d\left(e^{3x}y\right)=11e^{3x}$$ then follow the way posted in @Shaaeh'spost.

dy/dt=11-3y

=>dy/(11-3y)=dt     (provided y!=11/3)

=>ln(y-11/3)=-3t+C    (integrating both sides where c is constant of integartion)

=>y=(11/3)+exp(-3t)*C1  (where C1=exp(C) which is another constant)

it is given that y(0)=1 i.e., when t=0; y=1;

now we need to find the exact equation and therefore we substitute the value of t=0 and y=1 in the equation.

we get

C1=-8/3;

therefore the exact solution is y=(11/3)+(-8/3)exp(-3*t)


Is "integrability" equivalent to "having antiderivative"? I am wondering if "a function $f(x)$ is integrable on a domain $D$" this proposition is equivalent to "$f(x)$ has antiderivative on domain $D$". If it is not the case, give me a counter example. Thank you.

There is an even simpler example.  Take, e.g.,

$$
f(x) = \begin{cases}
  1 & x \geq 0 \\
 -1 & x < 0
\end{cases}
$$

This function is Riemann-integrable on $[-1, 1]$, but it cannot be the derivative of any function because derivatives cannot have jump discontinuities.  (See, e.g., this question.)

No, they are two different concepts.
If a function y=f(x) has an antiderivative, it simply means that there exists a function, say F(x), whose derivative F'(x) = f(x). We say f is integrable on the interval [a,b] if the limit of its Riemann sums over this closed interval exits and is equal to some finite value.

Let's consider two different conditions:
1) when f is continuous on [a,b]
   when f is continuous, by the Fundamental Theorem of Calculus, it has antiderivative F(x)=∫xaf(t)dt, and F(x) is also continuous on [a,b].  Also, due to the property of a continuous function, f is integrable on [a, b]. In this case, having an antiderivative and integrable can be interpreted as the same meaning.

2) when f is not continuous on [a,b]
   we need to make this clear first. For a continuous and differentiable function, its derivative need not be continuous, that is, a non-continuous function may have an antiderivative; also, for a function to be integrable, it need not be continuous. And we discuss two questions:
   (a) Does the existence of an antiderivative guarantee that a function is integrable?
       No. As we know, a non-continuous function may have an antiderivative. Thus, the function may not be integrable. That is, although there is an explicitly defined antiderivative F(x) (possibly not, but just expressed as an integral using the Fundamental Theorem of Calculus), the limit of the Riemann sums may not exit or may not equal to some finite value. However, we limit our discussion here to any type of continuous functions, not necessarily bounded on the closed interval [a,b]. For bounded functions, we can expect something else.
   (b) Does an integrable function guarantee the existence of an antiderivative?
       No. We can use the same reasoning as the previous argument. That is, a non-continuous function may be integrable. Thus, the function may not have an antiderivative. A famous example would be the Thomae's function. Which has been mentioned above.


Let $a$ and $b$ belong to a group $G.$ Find an $x$ in $G$ such that $xabx^{-1}=ba.$ Let $a$ and $b$ belong to a group $G$. Find an $x$ in $G$ such that $xabx^{-1}= ba$.

This is what I have done so far, but I am stuck and not sure if I am in the right direction:

$xabx^{-1} = ba$

Multiply both sides on the right by $x$.

$xabx^{-1}x = bax$.

Now the $x^{-1}x$ cancel out: $xab = bax$.

This is where I am stuck because I am not sure how to solve for an $x$ in $G$.

Try letting $x = a^{-1}$, so that $x^{-1} = (a^{-1})^{-1} = a$.

Or, if you prefer, let $x = b$, so $x^{-1} = b^{-1}$.

(Since $a, b \in G$, so are $a^{-1}, b^{-1} \in G$, since $G$ is a group and a group is closed under taking inverses.)

If I put $~X = b~$ or $~X = a^{-1} ~$(consider as self inverse)than what we get

I get that the term     $~xabx^{-1} = ba ~$, become   $~ ba = ba~$

Which gives that it is satisfied.

As my opinion


Write a function as $\sum _{n=0} ^{\infty} a_n x^n$ We have $f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4$.

Now I want to write this as $\sum _{n=0} ^{\infty} a_n x^n$.

What I got:

$f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4 = x^4 (1+ x + x^2 + x^3 + x^4 + x^5)^4$

We know: $\frac{1}{1-x} = \sum _{n=0} ^{\infty} x^n$.

Note that $\displaystyle \frac{d^3}{dx^3} (\frac{1}{1-x})= \frac{6}{(1-x)^4}$.

Then $\frac{1}{(1-x)^4} = \frac{1}{6} \sum n (n-1) (n-2) x^{n-3}$, what gives

$x^4 \frac{1}{(1-x)^4} = \frac{1}{6} \sum n (n-1) (n-2) x^{n+1} = \frac{1}{6} \sum  (n-1) (n-2) (n-3) x^{n}$

If I look now on Wolfram Alpha at the expanded form, this expression is only correct for $4 \leq n \leq 9$. For $n \geq 10$ it gives other values of the coefficients. What goes wrong?


Algorithm to maximize shipment value in linear time, confused about wording Question: A shipment company has the option of sending containers of dollar
value $v_1, v_2, … , v_n$. They have a single container ship that can carry only
$N < n$ containers. Present a linear time algorithm to choose the containers
that maximize the shipment value.
 I don't think I am understanding the question right because cant I simply use a linear time sorting algorithm like bucket sort and then take the first $N$ values?

Basically you need to find the $N-th$ greatest elements in an array: so look for them sequentially using a simple selection alghorithm. However it's not linear in $n$, it's $O(N\cdot n)$, so if $N$ is less enough you are done, but if $N \to n$, then it's $O(n^2)$ ... here is the python version:

def get_containers(values, containers):
#side case: allowed 0 at most 0 containers
if containers == 0:
    return []

#side case: allowed only 1 container -> look for maximum
if containers == 1:
    return max(values)

#side case: allowed n containers -> keep them all!
if containers == len(values):
    return values

good_containers = []
values_copy = copy.copy(values)
for container in range(1, containers):
    maximum = max(values_copy)
    good_containers.append(maximum)
    values_copy.remove(maximum)

return good_containers


I need help with an average question My brother and I are going on vacation. We need to rent a car. We will rent the car for $10$ days. However, I will only be on the vacation for $5$ days. My  brother will be on vacation for the full $10$ days. The cost of the car is $£600$. How much should he pay and how much should I pay?

You are paying $60$ for every day you own the car. For the first $5$ days, you pay $300$ and split the bill, meaning you pay $150$, for the second $5$ days, your brother pays the whole $300.$

You will be on vacation for 5 days, and your brother for 10. So the total amount of vacation days is 15. Split the price (600) over the number of days (15) to get 40 pounds per day. Therefore, you pay 200 while your brother pays 400.


What is the $\mathbb Z$-module $\operatorname{Hom}(\prod \mathbb {Z}/2, \mathbb Z/2)$? In this case I consider infinite direct product.

By the universal property of direct product I can see $\operatorname{Hom}(B, \prod A_i)$ for particular modules $B$ and $A_i$. On the other hand, I really struggle to "see" what is the $\operatorname{Hom}( \prod A_i, B)$. I think it's hard for the general case, so I ask for the case $\operatorname{Hom}(\prod \mathbb {Z}/2, \mathbb Z/2)$.

I don't know how many copies of $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ you have, but I assume that the direct product is over an infinite set $I$.  Then, $\prod_{i\in I}\,\mathbb{F}_2=\mathbb{F}_2^I$ is an $\mathbb{F}_2$-vector space of dimension $2^{|I|}$ according to this link.  Thus, $\prod_{i\in I}\,\mathbb{F}_2\cong \mathbb{F}_2^{\oplus 2^{|I|}}$ as $\mathbb{F}_2$-vector spaces.  This means
$$\text{Hom}_\mathbb{Z}\left(\prod_{i\in I}\,\mathbb{F}_2,\mathbb{F}_2\right)\cong\text{Hom}_{\mathbb{F}_2}\left(\mathbb{F}_2^{\oplus 2^{|I|}},\mathbb{F}_2\right)\cong\Biggl(\text{Hom}_{\mathbb{F}_2}\left(\mathbb{F}_2,\mathbb{F}_2\right)\Biggr)^{2^{|I|}}\cong \mathbb{F}_2^{2^{|I|}}\,.$$  Thus, the $\mathbb{Z}$-module $\text{Hom}_\mathbb{Z}\left(\prod_{i\in I}\,\mathbb{F}_2,\mathbb{F}_2\right)$ is isomorphic to the $\mathbb{Z}$-module $\mathbb{F}_2^{2^{|I|}}$, which is a $2^{2^{|I|}}$-dimensional vector space over $\mathbb{F}_2$, again according to the same link.

You write "$\prod$" yet you use the "infinite direct product" name. Usually by "$\prod$" we mean an "infinite product". For infinite direct product we use "$\bigoplus$" symbol.

Anyway since "$\bigoplus$" is a coproduct in the category of abelian groups (or $\mathbb{Z}$-modules) then

$$\hom\big(\bigoplus A_i, B\big)\simeq \prod\hom(A_i, B)$$

In particular

$$\hom\big(\bigoplus\mathbb{Z}_2, \mathbb{Z}_2\big)\simeq\prod\hom\big(\mathbb{Z}_2, \mathbb{Z}_2\big)\simeq\prod\mathbb{Z}_2$$

This doesn't hold for infinite product "$\prod$" on the left. I do not know the solution in this case (probably there is no known general one).


Proving that the point spectrum of the right shift operator on $\mathscr{l}^2(\mathbb{Z})$ is empty. How can I prove that no series $x\in\mathscr{l}^2(\mathbb{Z})$ of the form $$\forall i\in\mathbb{Z}: x_{k-1}=\lambda x_k$$
exists other than the zero sequence?

In particular I want to prove that the point spectrum of the right shift operator $$R:\mathscr{l}^2(\mathbb{Z})\rightarrow\mathscr{l}^2(\mathbb{Z})$$ $$\forall i\in\mathbb{Z}:x_i\mapsto x_{i-1}$$
is empty. I did this for $\mathscr{l}^2(\mathbb{N})$ already but can't prove the first statement, which I think i need for the $\mathscr{l}^2(\mathbb{Z})$ case.

Suppose there exists some $\lambda \in \mathbb{C}$ such that $\lambda \in \sigma_{pt} (R)$. Then, for some nonzero sequence $x \in l^2 (\mathbb{Z})$, $$Rx = \lambda x \leftrightarrow x_{n-1} = \lambda x_n \forall n \in \mathbb{Z} .$$

Let's then divide this into cases. First, suppose that $| \lambda | \geq 1$. In this case, the above equation can be rewritten as $$\frac{x_{n-1}}{x_n} = \lambda \rightarrow \left\vert \frac{x_{n-1}}{x_n} \right\vert = |\lambda| \geq 1.$$ However, this is a contradiction, since the terms of a sequence in $l^2(\mathbb{Z})$ must vanish as $n \rightarrow \pm \infty$.

The second case follows a similar argument. Suppose $|\lambda| < 1$. Then, $$\left\vert \frac{x_{n-1}}{x_n} \right\vert = |\lambda| \leftrightarrow \left\vert \frac{x_{n}}{x_{n-1}} \right\vert = \frac{1}{|\lambda|} > 1.$$ The contradiction follows as before.

Thus, there is no nonzero sequence $x \in l^2 (\mathbb{Z})$ such that $R x = \lambda x$ for some $\lambda \in \mathbb{C}$.

A representation of this operator is multiplication by $e^{ix}$ on $L^{2}[-\pi,\pi]$ with inner product
$$
           (f,g) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt.
$$
This is because $\{ e^{inx} \}_{n=-\infty}^{\infty}$ is an orthonormal basis of $L^{2}[-\pi,\pi]$, and
$$
            e^{ix}\sum_{n=-\infty}^{\infty}f_n e^{inx}=\sum_{n=-\infty}^{\infty}f_{n-1}e^{inx}.
$$
A multiplication operator $M_{a}f=a(x)f(x)$ on $L^{2}$ cannot have an eigenvalue unless what you're multiplying by ('a' in this case) is equal to a constant on a set of non-zero measure because
$$
                 0 = \|(M_{a}-\lambda I)f\|^{2}=\int_{-\pi}^{\pi}|a-\lambda|^{2}|f|^{2}dx \\
             \implies |a-\lambda||f| = 0\;\; a.e.
$$


Why is $-a \times -b = ab$? This might be a bit of a meta-mathematical question (if there is such a thing as meta-mathematics) but it confuses me slightly.

I think of multiplication (whether rightly or wrongly) as recursive adding. For example, $2 \times 4$ is just adding 2 four times i.e $+2 +2 +2 +2 = 8$.

I think of multiplying negative numerals e.g $2 \times -4$ as subtracting 2 four times i.e $-2 -2 -2 -2 = -8$. Unfortunately this analogy seems to break down when multiplying to negative numbers e.g $-2 \times -4$. I don't understand how $-2 \times -4$ can result in $+8$.

What is the reason behind this, if any? Is my analogy incorrect? Is there a flaw in my fundamental understanding of multiplication?

Receiving 4 gifts of 5 dollars gives you a result of $4 \times 5 = +20$ dollars. This means you benefit by 20 dollars.

"Receiving" $-4$ gifts of 5 dollars means giving four gifts of 5 dollars. This results in a loss to you of 20 dollars, so $(-4) \times 5 = -20$.

Receiving 4 "gifts" of $-5$ dollars means someone is making you pay them 5 dollars, and doing it four times. That gives you a result of $4 \times (-5) = -20$ dollars.

"Receiving" $-4$ "gifts" of $-5$ dollars again means giving someone four of these (undesirable) gifts. Thus you are charging someone 5 dollars, and doing it four times. The result is of course good for you (if you don't care about karma), since you benefit by 20 dollars. So $(-4) \times (-5) = 20$.

Here is a second approach.

$4 \times (-5)$ is obtained by adding four copies of $-5$:
$$4 \times (-5) = (-5) + (-5) + (-5) + (-5) = -20.$$

So it is natural that $4 \times (-5)$ should be -20, by the "repeated addition" understanding of multiplication that you mentioned. Of course, we can't add "-4 copies" of -5 together, but pay attention to the pattern in the following list.

$4 \times (-5) = -20$

$3 \times (-5) = -15$

$2 \times (-5) = -10$

$1 \times (-5) = -5$

$0 \times (-5) = 0$

You will surely agree that if we are to multiply negative numbers at all, then the pattern should continue this way:

$(-1) \times (-5) = 5$

$(-2) \times (-5) = 10$

$(-3) \times (-5) = 15$

etc.

If you find natural that $2 \times (-4) = (-4) \times 2 = (-4) + (-4)$ you would agree that $a \times (-b) = - (a\times b)$. This can be made true even if $a<0$ and/or $b<0$ resulting in $(-a)\times(-b) = - ((-a)\times b) = -(-(a \times b)) = a \times b$.


Probability of die rolled 60 times A die is rolled $60$ times.

Find the normal approximation to the chance that the face
with six spots appears between 9, 10, or 11 times.
The exact chance that the face with six spots appears 9, 10, or 11 times?

my solution; in one roll of the die $P(6 \text{ spots}) = \frac{1}{6}$ In $60$ rolls of the die the expected number of times the face with $6$ spots appears is given by $E(X)=\text{Mean}=\frac{60}{6} = 10$ times.

$SE(6 spots)=SD=SQR(\frac{60}{6}\cdot \frac{5}{6})=2.8868$

Applying the normal approximation to the chance that the face with six spots appears 10 times, I obtain the following: mean = 10

$\text{Standard deviation} = 2.8868$

Using the continuity correction of 0.5 the following z-scores are found:
$z_1=(9.5−10)/2.8868=−0.1732$
and
$z_2=(10.5−10)/2.8868=0.1732$
Subtracting the cumulative probabilities for the two z-scores gives $0.5688 - 0.4312 = 0.1376$ which is the approximate chance that the face with six spots appears 10 times.   find the six spots appears between 9,10, or 11?

The exact chance that the face with six spots appears 10 times is found from

$60C10\left(\frac{1}{6}\right)^{10×}\left(\frac{5}{6}\right)^{50}=0.137$ find six spots appears 9, 10 or 11 times?

Your calculations (as far as I'm able to follow them in your somewhat idiosyncratic exposition) all appear correct. However, they calculate (approximately and exactly, respectively) the probability that a $6$ is rolled $10$ times. You can apply the same approach to the probabilities that a $6$ is rolled $9$ times or $11$ times, and then add the three results. (In the case of the Gaussian approximation, you can also calculate the sum in one go by subtracting the values of the cumulative distribution function at $11.5$ and $8.5$.)

For the "Find the normal approximation to the chance that the face with six spots appears between 9, 10, or 11 times". and Find the exact chance that the face with six spots appears 9, 10, or 11 times. can you please check that the answers 0.3967 and .39589 are correct.


Proof of $ \text{Poisson}(\lambda p) $-arrivals. We have a queue where people pass out of it with $ \text{Poisson}(\lambda) $ and they come in with probability $ p $.

I understand that the arrivals follow $ \text{Poisson}(\lambda p) $, but how can I prove it?

Let X be the number of arrivals.
\begin{eqnarray*}
&&P(X=r)=\sum\limits_{k=r}^{\infty}\dfrac{\lambda ^k}{k!}\exp{(-\lambda)}C_k^rp^r(1-p)^{k-r}\\
&&=\sum\limits_{k=r}^{\infty}\exp{(-\lambda)}\dfrac{\lambda ^k k!}{r!(k-r)!k!}p^r(1-p)^{k-r}\\
&&=\dfrac{\lambda^r p^r}{r!}\sum\limits_{k=r}^{\infty}\exp{(-\lambda)}\dfrac{\lambda ^{k-r} }{(k-r)!}(1-p)^{k-r}\\
&&=\dfrac{\lambda^r p^r}{r!}\exp{(-\lambda)}\exp{(\lambda(1-p))}=\dfrac{(\lambda p)^r}{r!}e^{-\lambda p}
\end{eqnarray*}
So $X\sim P(\lambda p)$

Here is a cool proof I thought about.

I am going to split the original process into two parts, one that can arise with probability p and the other with probability 1-p. Here in your example, the original process is passing out, with proba p going in, proba 1-p not going in.

Now what we want to prove is how the splitted process is still a poisson process, and this will give us the proof that your process is indeed poisson with parameter $\lambda p$.

Also, since it is the case that interested me, I prove it for poisson proceses, not poisson random variables (replace $\lambda$ by $\lambda t $.

  We have the poisson process

  $$X(t) = X_1(t) + X_2(t) $$

  First we calculate the joint probability

  $$P[X_1(t) = k, X_2(t) = m] = \sum_{n=0}^{\infty} P[X_1(t) = k, X_2(t)
> = m \mid X(t) = n]P[X(t) = n]$$

  Note that

  $$P[X_1(t) = k, X_2(t) = m \mid X(t) = n] = 0 \:\:\: \text{when}\:\: n
> \neq k+m$$

  Now

  $$P[X_1(t) = k, X_2(t) = m] = P[X_1(t) = k, X_2(t) = m \mid X(t) =
> k+m]P[X(t) = k+m]$$

  $$= P[X_1(t) = k, X_2(t) = m \mid X(t) = k+m]e^{-\lambda
> t}\frac{(\lambda t)^{k+m}}{(k+m)!}$$

  Now, given that $k+m$ events occurred, since each event has
  probability $p$ of being a type $1$ event and probability $1-p$ of
  being a type $2$ event, it follows that

  $$P[X_1(t) = k, X_2(t) = m \mid X(t) = k+m] = \binom{k+m}{k}
> p^k(1-p)^m$$

  Thus,

  $$P[X_1(t) = k, X_2(t) = m] = \binom{k+m}{k} p^k(1-p)^m e^{-\lambda
> t}\frac{(\lambda t)^{k+m}}{(k+m)!}$$

  $$= \frac{(k+m)!}{k!m!} p^k(1-p)^m e^{-\lambda t}\frac{(\lambda
> t)^{k+m}}{(k+m)!}$$

  $$= e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t}
> \frac{[\lambda (1 - p)t]^m}{m!} \:\:\:\:\:\:\:\:\:(1)$$

  Then

  $$P(X_1 = k) = \sum_{m=1}^{\infty} P[X_1(t) = k, X_2(t) = m]$$

  $$=e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t}
> \sum_{m=1}^{\infty} \frac{[\lambda (1 - p)t]^m}{m!}$$

  $$= e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t}
> e^{\lambda(1-p) t}$$

  $$=e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} \:\:\:\:\:\: (2)$$

  which indicates that $X_1(t)$ is a poisson process with rate $\lambda
> p$.

  Similarly, we can obtain

  $$P(X_2(t) = m) = \sum_{k=1}^{\infty} P[X_1(t) = k, X_2(t) = m]$$

  $$= e^{-\lambda(1-p) t} \frac{[\lambda (1 - p)t]^m}{m!} \:\:\:\:\:\:
> (3)$$

  and so $X_2(t)$ is a poisson process with rate $\lambda (1-p)$.

  Finally, from equations $(1)$, $(2)$ and $(3)$

  $$P[X_1(t) = k, X_2(t) = m] = P[X_1(t) = k]P[X_2(t) = m]$$

  Hence, $X_1(t)$ and $X_2(t)$ are independent.

Then you have the answer to your problem ! I am not sure that there is a simplier way to prove it, other than intuition.

The original part of the proof comes from here : Splitting Poisson process formal proof and the reward of the work should be attributed to JKnecht. I wanted to use his proof for another interesting similar case, which is yours, because in the end, it is the same problem.


How many ways can you collect six dollars from $8$ people if $6$ people give $0$ or $1$ dollars and $2$ people each give $0$, $1$, or $5$ dollars? I must use a generating function to solve this question:

In how many ways can you collect six dollars from eight people if six people give either $0$ or $1$ dollars and the other two people each give $0$, $1$, or $5$ dollars?

Here is what I have so far.

The generating function is $(1+x)^6(1+x+x^5)^2$. Then let $f(x)= (1+x)^6$, and $g(x) = (1+x+x^5)^2$. We want the coefficient of $x^6$. Then, we note f(x) has the expansion:

$(1+x)^6$ = $1$ + $6 \choose 1$$x$ + $6 \choose 2$$x^2$ + ... + $6 \choose 6$$x^6$

But here is where I get stuck. We can let $h(x) = f(x)g(x)$ where $a_0b_0 + (a_1b_0+a_0b_1)x + (a_2b_0+ a_1b_1 + a_0b_2)x^2 + ... + (a_rb_0+ a_{r-1}b_1 + a_{r-2}b_2 + ... + a_0b_r)x^r$,

But I can't figure out the expansion of $g(x) = $$(1+x+x^5)^2$.

Can someone help me find the expansion of $g(x)$, and help me solve the problem?

Hint its multinomial theorem where $n=x_1+x_2+...x_n$ and coefficient is given by ${n\choose x_1,x_2,x_3..x_n}$


Probability of getting the same 10 cards after a 104 card shuffle Consider this situation:
You have two standard decks merged together (104 cards). You give 10 cards to 4 players. What is the probability that the next time you deal (after the deck is randomly shuffled), everyone gets the same hand they got last turn? Just one person gets the same hand they got last turn?
My approach was something along 104 choose 40 (for everyone getting the same card) or 104 choose 10 (for one person getting the same card), but in my approach I'm not considering the duplicate cards in my deck. I don't think it's 52 choose 40 or 52 choose 10 because then I wouldn't be considering that there is more than one card.
I am very curious about this question and I would love to get an answer. I know its a combinatorics question, but I am unsure of the details.
P.S. If the question is confusing or non-descriptive, please ask for clarification instead of down voting.

This is not a complete answer, but provides some suggestions on how to proceed.

Merge two $n$ card decks, and deal one hand of $k$ cards.  The merged deck has $2n$ cards, but only $n$ kinds of card.  A hand can contain the same kind of card twice.  When a card appears in a hand along with the other card of the same kind, we call it a double.  Otherwise, we call it a single.

There are $\binom{2n}{k}$ ways of dealing a $k$ card hand.  There are $\binom{n}{d}\binom{n-d}{k-2d}$ kinds of hand with $d$ doubles and $k-2d$ singles.  There are $2^{k-2d}$ different ways of dealing a given $k$ card hand with exactly $d$ doubles, giving a probability of $2^{k-2d}\big/\binom{2n}{k}$ of getting such a hand.  There are $2^{k-2d}\binom{n}{d}\binom{n-d}{k-2d}$ ways of dealing some $k$ card hand with exactly $d$ doubles.  If we sum over all possible values of $d,$ we should get $\binom{2n}{k}.$  It can be confirmed that, indeed,
$$
\sum_{d=0}^{\lfloor k/2\rfloor}2^{k-2d}\binom{n}{d}\binom{n-d}{k-2d}=\binom{2n}{k}
$$
by evaluating the $x^k$ term in $(1+x)^{2n}=(1+2x+x^2)^n$ in two different ways.

Define $g(n,k,d)$ to be the probability that a $k$ card hand contains exactly $d$ doubles:
$$
g(n,k,d):=\frac{2^{k-2d}\binom{n}{d}\binom{n-d}{k-2d}}{\binom{2n}{k}}.
$$
Define $h(n,k,d)$ to be the probability of getting a particular $k$ card hand with exactly $d$ doubles:
$$
h(n,k,d):=\frac{2^{k-2d}}{\binom{2n}{k}}.
$$

Then
$$
\begin{aligned}
&\Pr[\text{I get the hand I was dealt last time}]\\
&\quad=\sum_{d=0}^{\lfloor k/2\rfloor}\Pr[\text{hand last time had exactly $d$ doubles}]\\
&\quad\quad\cdot\Pr[\text{I get a particular hand with exactly $d$ doubles}]\\
&\quad=\sum_{d=0}^{\lfloor k/2\rfloor}g(n,k,d)\,h(n,k,d)\\
&\quad=\sum_{d=0}^{\lfloor k/2\rfloor}\frac{2^{k-2d}\binom{n}{d}\binom{n-d}{k-2d}}{\binom{2n}{k}}\frac{2^{k-2d}}{\binom{2n}{k}}.
\end{aligned}
$$
By means of further manipulations, this expression can be written in terms of Gauss's ${}_2F_1$ hypergeometric function:
$$
\begin{aligned}
&\Pr[\text{I get the hand I was dealt last time}]\\
&\ =\frac{1}{\left[\binom{2n}{k}\right]^2}\sum_{d=0}^{\lfloor k/2\rfloor}\frac{4^{k-2d}n!}{d!(k-2d)!(n-k+d)!}\\
&\ =\frac{\binom{n}{k}}{\left[\binom{2n}{k}\right]^2}\sum_{d=0}^{\lfloor k/2\rfloor}\frac{4^{k-2d}k!(n-k)!}{d!(k-2d)!(n-k+d)!}\\
&\ =\frac{4^k\binom{n}{k}}{\left[\binom{2n}{k}\right]^2}\sum_{d=0}^\infty\frac{4^{-2d}k(k-1)(k-2)\ldots(k-2d+1)}{d!(n-k+1)(n-k+2)\ldots(n-k+d)}\\
&\ =\frac{4^k\binom{n}{k}}{\left[\binom{2n}{k}\right]^2}\\
&\ \ \cdot\sum_{d=0}^\infty\frac{4^{-d}\left(-\frac{k}{2}\right)\left(-\frac{k}{2}+1\right)\ldots\left(-\frac{k}{2}+d-1\right)\cdot\left(-\frac{k}{2}+\frac{1}{2}\right)\left(-\frac{k}{2}+\frac{3}{2}\right)\ldots\left(-\frac{k}{2}+d-\frac{1}{2}\right)}{d!(n-k+1)(n-k+2)\ldots(n-k+d)}\\
&\ =\frac{4^k\binom{n}{k}}{\left[\binom{2n}{k}\right]^2}{}_2F_1\left(-\frac{k}{2},-\frac{k}{2}+\frac{1}{2};n-k+1;\frac{1}{4}\right).
\end{aligned}
$$

Now let's suppose that $k$ card hands are dealt to two players.  What is the probability that if this is done twice, both players get the same hands on both deals?  We need to generalize the functions defined in the previous analysis.  Let $G(D,S,k,d,e,i)$ denote the probability that a $k$ card hand dealt from a deck with $D$ doubles and $S$ singles contains $d$ doubles, $e$ singles with no matching card left in the deck, and $k-2d-e$ singles with a matching card left in the deck.  Then
$$
G(D,S,k,d,e)=\frac{2^{k-2d-e}\binom{D}{d}\binom{S}{e}\binom{D-d}{k-2d-e}}{\binom{2D+S}{k}}.
$$
Let $H(D,S,k,d,e)$ be the probability of getting a particular $k$ card hand with exactly $d$ doubles and $e$ singles with no matching card left in the deck.  Then
$$
H(D,S,k,d,e)=\frac{2^{k-2d-e}}{\binom{2D+S}{k}}.
$$

Then the probability that both players get the same hand on the second deal as on the first is
$$
\begin{aligned}
&\sum_{d_1=0}^{\lfloor k/2\rfloor}\sum_{d_2=0}^{\lfloor k/2\rfloor}\sum_{e_2=0}^{k-2d_2}G(n,0,k,d_1,0)G(n-k+d_1,k-2d_1,d_2,e_2)\\
&\qquad\qquad\qquad\times H(n,0,k,d_1,0)H(n-k+d_1,k-2d_1,d_2,e_2).
\end{aligned}
$$
This doesn't appear to simplify.  The upper limits on the sums can be replaced by $\infty$ since the summand will be zero beyond those limits.

To find the probability that exactly one of the two players gets the same hand on the second deal as on the first, subtract the probability that both players do from the probability one player does.  Then double the result.  To find the probability that at least one of the two players gets the same hand, add the probability that both do to the probability that exacty one does.

This doesn't answer your questions, but the method can be extended to handle the four-player situation.

1st Question: Of the $52$ distinct cards in a deck, choose $40$. Given $2$ decks, this is the only way we can duplicate the hands of all players in the second deal. In the second deal, we need the same $40$ cards chosen in the first deal. However, each card can come from either deck, giving us $2^{40}$ ways to choose them. Finally, in dealing groups of $10$ out to our players, we choose $10$ from $40, 30, 20$ in succession, giving the remainder to the final player. In doing so, there is only one way to designate the correct groups of $10$ to each player according to our previous deal. So the probability that all get their original hands would be:$$\frac{\binom{52}{40}}{\binom{104}{40}}\cdot\frac{2^{40}}{\binom{104}{40}}\cdot\frac{1}{\binom{40}{10}\binom{30}{10}\binom{20}{10}}$$


Find the edge chromatic number of $K_n$ when $n$ is a positive integer. Find the edge chromatic number of $K_n$ when $n$ is a positive integer. Note: $K_n$ is
the complete graph on $n$ vertices。

I draw for $n = 1,2,3,4,5,6$ cases but have no general solution or proof about this question.

Let $n>1$. At any vertex $n-1$ edges meet and therefore the chromatic number is at least $n-1$.  If such a colouring exists then consider one particular colour, red say. Each red edge   pairs up vertices and every vertex would be in such a pair. Such a colouring is therefore impossible if $n$ is odd.

Conversely, we can  number the vertices and the colours $0,1,...,n-1$. We  can then colour the edge from $i$ to $j$ with colour $i+j \mod n$ and we thus obtain an edge-colouring with $n$ colours.

We now know that the chromatic index is precisely $n$ when $n$ is odd.

Now consider $n=2k+2$ and delete one vertex, $v$ say. The graph which remains, on $2k+1$ vertices, has a colouring with $2k+1$ colours obtained as in the second paragraph. Then vertex $i$ is incident with edges of all colours apart from colour $2i$ (modulo $2k+1$).  Add vertex $v$ back in and join it to vertex $i$ with colour $2i$ (modulo $2k+1$). These colours are all different since Now $2i\equiv 2j \mod 2k+1$ if and only if  $i\equiv j \mod 2k+1$.

So the chromatic index is precisely $n-1$ when $n$ is even.


Why is the Penrose triangle "impossible"? I remember seeing this shape as a kid in school and at that time it was pretty obvious to me that it was "impossible". Now I looked at it again and I can't see why it is impossible anymore.. Why can't an object like the one represented in the following picture be a subset of $\mathbb{R}^3$?

I can't resist posting an answer based on the Mathematics Stack Exchange logo.

Let's add some more cubes to the logo to make it clear that it's a subset of the Penrose triangle (or would be, if it was a real 3D object)

Now note that the cubes are overlapping, so some must be in front of others. But in fact, each cube is partially obscured by at least one other cube, in such a way that it appears to be some distance behind it. You can go around the hexagon in the original logo, in clockwise order, and see that each cube appears to be located further from the 'camera' than the next one in the cycle - which means that each cube is in front of itself. There's no consistent "z ordering" that you can give to the different parts of the figure, and that's one way to see that it's impossible.

In reply to some of the comments, just to be explicit, the point here isn't just that the cubes all overlap each other. If that was the case it would be incorrect, since it's possible to have mutually overlapping arragements of cubes, as in this image provided by Misha Lavrov.

However, if we're assuming that the Stack Exchange logo is a subset of the Penrose triangle then we know the cubes aren't arranged like that. Instead, each cube is positioned so that some of its sides are coplanar with those of the next cube, and each cube is separated from the next by some distance in the z direction, where z is perpendicular to the plane of the image. Therefore the cubes' centres of mass can't be given consistent z coordinates.
As an extra bonus point, even if we don't assume that, and instead assume that each cube is as close to the next as it can be (in the z direction) without the surfaces intersecting, the Math.SE logo still can't be made into a consistent 3D shape, as the following animation shows. Note that it doesn't quite form the Math.SE logo, since one cube ends up in front of all the rest. Of the six neighbouring pairs of cubes, three of them can have equal z coordinates, but for the remaining three pairs, one cube unavoidably has to have a greater z coordinate than the next.

As another additional bonus point, although it's not possible to embed the Penrose triangle into normal, flat, Euclidean 3D space, it is possible to embed it into curved three dimensional space. The video below, by @ZenoRogue on Twitter, shows Penrose triangles embedded into something called "nil geometry". I don't pretend to understand the details, but it's a kind of curved space such that Penrose triangles really are possible.
video link: https://www.youtube.com/watch?v=YmFDd49WsrY
screenshot:

Imagine keeping the corners in the same place, but reducing the width of the square cross-section of each side down to zero, until each side is a one-dimensional line segment. You would end up with a triangle with three $90^{\circ}$ angles, which is impossible in Euclidean space $\mathbb{R}^n$.


Macaulay2: How to compute the remainder when dividing a polynomial by a set of polynomials (in some order)? I'm writing Buchberger's Criterion in a program in Macaulay2 to check whether or not the set of polynomials I have form a Grobner basis for the ideal it generates. However, I have not been able to find a method that gives me the remainder when a polynomial, say $f$, is divided by a set of polynomials $G=\{g_1, g_2, ..., g_t\}$ in some order. Would anyone know if such a method exists and, if yes, what is its name?

Although that's not all I have, here is the part of the program where I try to implement the Buchberger's Criterion:

n=0;
for i to #polynomials-2 do
    (
        for j from i+1 to #polynomials-1 do
    (
Spoly := lcm(leadTerm(polynomials#i),leadTerm(polynomials#j))/leadTerm(polynomials#i)*polynomials#i-lcm(leadTerm(polynomials#i),leadTerm(polynomials#j))/leadTerm(polynomials#j)*polynomials#j;
remainder := Spoly%polynomials;
if remainder == 0 then n=n+1;
  );
    );
    if n == binomial(#polynomials, 2) then print "The polynomials form a Grobner basis for the ideal it generates." else print "The polynomials don't form a Grobner basis for the ideal it generates."

On the codes above, I use % as the method I'm looking for - but it doesn't work for a polynomial being divided by a list of polynomials.

It seems this is not built in (at least, not exposed to users), but it is easy enough to achieve:

Given a monomial ordering, dividing a polynomial $f$ by an ordered list of polynomials $f_1,\ldots,f_s$ means to express $$f = q_1f_1 + \ldots + q_sf_s + r,$$ where either $r=0$ or none of the monomials in $r$ are divisible by any of $LT(f_1), \ldots, LT(f_s)$.

Algorithm. To begin, set $q_1:= 0, \ldots, q_s :=0$ and $r := 0$, and introduce $p := f$.

While $p \neq 0$, we remove $LT(p)$ (and possibly more) from $p$ as follows:

Try to find the smallest index $i$ such that $LT(f_i)$ divides $LT(p)$.

If $i$ is found (division step): set $q_i := q_i + LT(p)/LT(f_i)$ and $p := p - (LT(p)/LT(f_i))f_i$.
If no such $i$ exists (remainder step), then set $r := r + LT(p)$ and $p := p - LT(p)$.

Finally, return $(q_1,\ldots,q_s)$ and $r$.

See Ideals, Varieties, and Algorithms 4th ed. by Cox, Little and O'Shea, Chapter 2, $\S$3, Theorem 3, pp. 64 – 66. They give more imperative pseudo-code which should be easy to translate into Macaulay2. It is a nice programming exercise.

If you are given f and g two polynomials the remainder is given by f % g.
suppose now you are given with G ideal and G=(g1,g2,...,gn). then the remainder when f is divided by G should be f % G.
Hope this helps.


Determine position of ellipse that contacts two fixed ellipses I'm working in a vector program with three identical ellipses, all of which additionally have the same angle of rotation. The first two are at the same Y coordinate and are tangent. I would like to position the third ellipse such that it is tangent with the first two. How can I calculate the third ellipse's position?

One solution good enough for my end goal has been to start with two circles, rotate a copy of one 90° along the intersection point, and then decrease the eccentricity of all three to the desired form. This gives me the result I'm looking for, but I'd rather know a more elegant and less procedural solution.

If there is no other constraint on the position of the third ellipse other than bitangency, there are infinitely many solutions.

If you have enough with one solution, stretching three tangent circles is certainly the most elegant.

Ellipses are drawn with major axis parallel to x-axis.

The placement of ellipses results in anti-symmetry about lines drawn parallel to x-axis as shown making contact points at $y= \pm b/2, b$ is minor axis as shown around points P and Q.

A vertical positioning of ellipse would be attempted next and a general tilted position later.

When contacting ellipse has major axis vertical

By a property of ellipse connecting slope $\phi$ and angle at elliptic arc $\psi$ eccentricity $e<1$

$$ \frac{\cos \psi_1}{\cos \phi_1}= e\,;  \frac{\cos \psi_2}{\cos \phi_2}= e\tag1 $$

$$ \phi_1+\phi_2=\pi/2,\,\sqrt{\cos \psi_1^2+\cos \psi_1^2} =e \tag2 $$

$$ \frac{\cos \psi_1}{\cos \psi_2}= \tan\phi_2,\, \quad    \frac{\cos \psi_1}{\sin \phi_2}=e\,  \tag3 $$

The detail at contact area is zoomed at right. $F_1F_2$ is of arbitrary length upto a scale factor of construction. The construction has been made to numerical values

$$ \psi_1=1.2,\,\psi_2=0.47, \phi_1=1.1, \,e= 0.798855 $$

in radians for the three ellipses. If the first of 2) is chosen different from $\pi/2$ then the resulting intruding contact ellipse can be computed and drawn to any inclination.


What algebraic structure do date, temperature, and similar quantities belongs to? I find that some quantities share serveral characteristics.

For date:

"1st July" + "1 day" = "2nd July"
"2nd July" - "1st July" = "1 day"
But "20th August" + "29th August" is nonsense.

For temperature:

An object can be heated up from 25 degree Celsius by 5 degree Celsius to 30 degree Celsius.
The temperature of boiling water is 100 degree Celsius. The temperature of the surrounding is 25 degree Celsius. The difference is 75 degree Celsius.
But we cannot add the temperature of a cup of tea to the temperature of a cup coffee.

Also, we cannot do multiplication on date or temperature. Depending on context, their relative values are useful, but their absolute value are not relavent. E.g., we don't count date from the Big Bang; weather forecast does not involve absolute temperature.

How does abstract algebra describe these quantities? Is there an algebraic structure captures their characteristics?

What you are describing is similar to the concept of an affine space.

An affine space over a vector space $V$ is a space filled with points. You can add a vector to a point to get another point, and you can subtract one point from another to get a vector. But you cannot add two points together.

Calendar dates and temperatures in °C are sets where group operations are defined on. You can add and subtract days as differences in days follow the model of integer numbers ($\mathbb{Z}$). The elements of the group of day differences then operate on the "amorphic" set of calendar dates which has no own algebraic structure (calendar dates are not a vector space or a group; you cannot add 2nd July to 1st July as you correctly noted).

You can add and subtract temperature differences in K. Kelvin, because by convention temperature differences are never measured in °C but in K. Here we see clearly that we don't have an algebraic structure on the temperatures themselves (14°C - 25°C is not a temperature in °C but a temperature difference in Kelvin). The group of temperature differences thus operates on the set of temperatures.

Operations like these are quite important in analysing (multidimensional) data: You are not interested in the numbers themselves but on the information they convey. So if this information is not changed by some operation on the data, you have to use an analysis tool that is invariant to this class of operations. There are also examples where a more specific structure than a group operates on the data.


What is the dimension of the eigenspace corresponsing to eigenvalue lambda = 9? Another homework problem is asking me to find the dimension of the eigenspace corresponding to lambda = 9. From what I understand, I need to subtract original matrix A from 9 times a 4x4 identity matrix, but I'm seeking to understand what to do after that because my matrix has the number 9 in every column:
$$
    \begin{pmatrix}
    3-\lambda & 0 & 0 & 0 \\
    3 & 9-\lambda & 0 & 0 \\
    9 & 0 & 9-\lambda & 0 \\
    3 & 3 & 9 & 3-\lambda \\
    \end{pmatrix}
$$
What should I do now? Am I even on the right track?

Well...you're sort of on the right track. Plug in $9$ for $\lambda$ in your matrix and you get
$$
A - 9I =
    \begin{bmatrix}
    -6 & 0 & 0 & 0 \\
    3 & 0 & 0 & 0 \\
    9 & 0 & 0 & 0 \\
    3 & 3 & 9 & -6 \\
    \end{bmatrix}
$$
And now you have to compute the dimension of the nullspace of that matrix (or you could compute the rank of that matrix and use the rank-nullity theorem).

$$(A-\lambda I)v=0$$
$$\begin{pmatrix}
    -6& 0 & 0 & 0 \\
    3 & 0 & 0 & 0 \\
    9 & 0 & 0 & 0 \\
    3 & 3 & 9 & -6 \\
    \end{pmatrix}\pmatrix {v_1 \\v_2 \\v_3 \\v_4}=\pmatrix {0\\0\\0\\0}$$
You deduce that:
$$
\begin{cases}
v_1=0 \\
v_2+3v_3-2v_4=0
\end{cases}
$$
So that we have :
$$v=\left(0,v_2,v_3,\dfrac 12 v_2+\dfrac 32 v_3\right)^t$$
$$v=c_1\left(0,2,0,1\right)^t+c_2\left(0,0,2,3\right)^t$$
You can deduce the dimenson of the eigenspace now.


What is a root vector? Are "root vectors" and generilised eigenvectors the same thing? Im reading a book written by Nikolski named "Operators, Functions, and Systems - An Easy Reading: Hardy, Hankel, and Toeplitz" and he uses the term every now and then.

A generilised eigenvector is a vector $v$ such that

$(A-\lambda)^{n}v=0$ for some $n$ and $(A-\lambda)^{k}\ne 0$ for $k<n$.

Looking for references.

No. Given a Lie algebra $L$ and a Cartan subalgebra $H$, a root is a non-zero element $\alpha\in H^*$ such that the associated root space $L_{\alpha}=\left\{x\in L\mid [h,x]=\alpha(h)x, \forall h\in H\right\}$ is non-zero. A non-zero $x\in L_{\alpha}$ is called a root vector.

You could see a root $\alpha$ as generalisation of an eigenvalue in the sense that it satisfies the equation $[h,x]=\alpha(h)x$. Here $x$ is an eigenvector of $[h,-]$ with eigenvalue $\alpha(h)$, but unlike standard linear algebra, this holds for all $h\in H$, hence $x$ is simultaneously an eigenvector for all $[h,-]$.

The importance of roots is that given a semisimple complex Lie algebra $L$, you can decompose $L$ as $$L=L_0\oplus \bigoplus_{\alpha\in \Phi}L_{\alpha},$$
where $\Phi$ is the set of roots. It turns out that $L_0=H$ is a Cartan subalgebra of $L$. The set $\Phi$ is called the root system of $L$. One can associate a Dynkin diagram to the root system, and these diagrams are classified. It follows that the root systems can be classified which in turn can be used to classify the semisimple Lie algebras. This is perhaps one of greatest achievements in mathematics and is not very difficult to understand. (In fact, you only need basic linear algebra to understand this).

Generalized eigenvectors on the other hand are very different from elements in root spaces. Given a linear transformation $T:V\rightarrow V$ on a finite-dimensional vector space (over an algebraically closed field such as $\mathbb{C})$, one can choose a basis of $V$ consisting of eigenvectors and generalized eigenvectors. Considering the matrix of $T$ w.r.t. this basis yields he Jordan canonical form of $T$. When $T$ is diagonalizable, there are only eigenvectors in this basis and no (strict) generalized eigenvectors.

A root vector in this context is a generlized eigenvector.


Show that extension is simple Let $E/k$ be a finite extension and for any sub field $F_1$ and $F_2$ of $E$ containing $k$ either $F_1 \subset F_2$ or $F_2 \subset F_1$ then $E=k(a)$ for some $a$

Solution: Assume, $E=k(x_i, 1 \leq i \leq n)$ and $F_i=k(x_i)$. By given condition of comparability we can compare $F_i$ and therefore there will be a “largest” $F_i$ which will be equal to $E$

Is this solution correct? Any other solution?

Let $E= k(x_1,\ldots,x_n)$ be a finite extension of $k$. Since the simple extensions $F_i=k(x_i)$ are pairwise comparable, there is largest element. W.l.o.g.\ let $F_n=k(x_n)$ be the largest one. Then $k(x_i)\subseteq k(x_n)$ for all $1\leq i\leq n$ and so $x_1,\ldots,x_{n-1},x_n\in F_n$. Hence, $E=F_n$ since both are the smallest extensions of $k$ containing $x_1,\ldots,x_n$.


Why do we require a topological space to be closed under finite intersection? In the definition of topological space, we require the intersection of a finite number of open sets to be open while we require the arbitrary union of open sets to be open. why is this?

I'm assuming this has something to do with the following observation: $\cap_{n=1}^{\infty} (-\frac{1}{n},\frac{1}{n}) = \{0\}$ and there is some reason we don't want singletons to be considered open, I am wondering what this reason is. Am I thinking in the right direction here?

Thanks :)

You need to think about what the intuition behind open sets are. One way to think about it is through neighborhoods: an open set is a set which is a neighborhood of each of its points. What is a neighborhood of a point? A neighborhood of a point $x$ is a set that contains of all points that are "sufficiently close" to $x$ (what does "sufficiently close" mean? It depends on the situation; you think of different neighborhoods perhaps specifying different degrees of closeness). In particular, any set that contains a neighborhood of $x$ is itself a neighborhood of $x$. And specifying two degrees of closeness specifies another degree of closeness that makes sense (the smaller of the two at any given place, say).

So: if you think about open sets as sets that are neighborhoods of all of the points they contain, then it is natural that the union of any family of pen sets will be open: each point in the union is one of the open sets, and that open set is a neighborhood, and the union contains that neighborhood and so is itself a neighborhood. So the arbitrary union of open sets should still be open.

What about intersection? Well, if you take two open sets $O_1$ and $O_2$, and you consider a point $x$ in $O_1\cap O_2$, then $O_1$ contains all points that are $1$-sufficiently close to $x$, and $O_2$ contains all points that are $2$-sufficiently close to $x$ (with "$1$-sufficiently" and "$2$-sufficiently" describing the two degrees of closeness required), so $O_1\cap O_2$ will contains all points that are both $1$-sufficiently and $2$-sufficiently close to $x$, and so it contains all points that are "sufficiently close" to $x$ for some meaning of "sufficiently close", so it is also an open set. This gives you, inductively, any finite intersection.

But what about arbitrary intersections? Then you run into trouble, because specfying two degrees of "closeness" gives you a degree of closeness (the smaller one), but an infinite number of degrees of closeness may end up excluding everything! (Just as in your example, taking the intersection of all $(-\frac{1}{n},\frac{1}{n})$, which specify all points that are $\frac{1}{n}$-close to $0$, but the intersection excludes everything). So we don't want to require that an arbitrary intersection of neighborhoods be a neighborhood, and so we don't want to require that an arbitrary intersection of open sets be an open set.

I think that finite intersection is necessary for a topological space because when you are dealing with infinite intersection then you are leaving almost everything. Lots of open sets in topology will disappear and that's alarming.


How many 11-digit numbers are there which contain all of the digits 0-9 at least once? How many $11$-digit numbers are there which contain all of the digits $0$-$9$ at least once?
Note that in this question a number cannot begin with a digit of $0$.

My instructor explained this as a case problem, but rushed through it without pausing for questions. The answer ended up being $99 \times (10!/2!)$, but I have no idea how to get there!

My thinking:
When you have 11 items in which 9 are unique and 2 are the same numeral repeated, then that can be arranged in $\frac{11!}{2!}$ ways.  Then, there are 10 different choices for which numeral is repeated.  So we have $\frac{10\times11!}{2!}$.  Finally, one tenth of these arrangements should have 0 as its first digit and those don't count.  So we need to multiply by $\frac{9}{10}$

$$\frac{9}{10}\times\frac{10\times11!}{2!}=99\times\frac{10!}{2!}$$

10!×10×11-9!

10! is permutaion of 0,.......9

10 is the number additionally added

11 is the number of place available to add the number

9! is to eliminate numbers begin with 0


Prove that $\frac{x}{e^x}$ tends to zero as $x \to \infty $ As the title states, I want to prove
$$\lim_{x \to \infty} \frac{x}{e^x} =0$$

Clearly, L'Hopital's rule easily solves this. However, I'm curious to see if there's another way to prove it, without involving some differential or integral calculus (that is, by algebraic means).
What I'm really interested about, is to prove that
$$\lim_{x \to \infty} \frac{x}{e^{x^2}}=0 $$
I assume that proving the first limit will provide a way to prove the second one, using the squeeze method. If you know a direct way to prove the second limit, it will be more than perfect.

Thanks in advance!

First show $$\frac{x}{a^x}\to 0$$ for any $a>1$. This means
$$\frac{x}{e^{2x}}\to 0$$ and write this as
$$\frac{x^2}{e^{2x^2}}\to 0$$ now take the square root to get
$$\frac{x}{e^{x^2}}\to 0$$


What is the name for an equation that has both an exponential variable and a variable in the base? I would like to know if there is a name given to equations like the following:

$$-31=\frac{-39.2}{x}(1-e^{\frac{x}{4}})$$

or more generally:

$$a=\frac{b}{x}c^x$$

The formatting doesn't really matter, I'm just talking about an equation where there are variables both in an exponent and in another term in a base. Another example would be this:

$$20x^2 = e^x$$

I've searched for things like "composite exponential functions" but to no avail.

Also, I guess I could solve my example by rearranging so that there is an expression containing x on the left and an exponential one on the right. Then I could graph each side and see where they equal. However, is there a way to do it analytically?

Perhaps I'm not explaining this as clearly as I could be, but I would appreciate any advice you could offer.

Thanks!

Welcome to the world of Lambert function !
If you consider equation $$a=\frac{b \, e^{c x}}{x^d}$$ the solution is
$$x=-\frac{d}{c} W\left(-\frac{c }{d}\left(\frac{b}{a}\right)^{1/d}\right)$$ where $W(z)$, the Lambert function, is such that $z=W(z)\,e^{W(z)}$.
The Wikipedia page will provide you many examples of the series of manipulations to be done.
In the real domain, $W(z)$ exists if $z \geq -\frac 1 e$ and, for $z<0$, it shows two branches.
Considering the example you give $a=20$, $b=1$, $c=1$, $d=2$, the result will be
$$x=-2 W\left(-\frac{1}{4 \sqrt{5}}\right)$$
The Wikipedia page will also provide you series expansions for computing the value of $W(z)$. Using for example
$$W(z)=z-z^2+\frac{3 z^3}{2}-\frac{8 z^4}{3}+\frac{125 z^5}{24}+O\left(z^6\right)$$ amking for your example
$$x=\frac{7936+31321 \sqrt{5}}{307200}\approx 0.253815$$ while the exact value should be $\approx 0.253871$.
In fact, sooner or later, you will learn that any equation which can write $A+Bx+C\log(D+Ex)=0$ has solution(s) in terms of Lmabert function.

Thanks for the detailed answer!

I was a bit confused at first because the Wikipedia page mentioned that it is defined on the complex plane. However, I reread your comment and saw what you said about it being defined for real numbers with domain restrictions, and things are making more sense now.

Hopefully I will get to study Lambert functions in more depth at some point in the future, but, for now, I'm glad that I understand the basic concept.

I think I would still be searching around for an answer had I just continued trying to Google things, so much appreciation for helping out!

-I


Problem From Hungerford (Abstract Algebra) (Ch 1 Section 1) The Question:

  Let $a,b,c,$ and $q$ be as in Exercise 5. Suppose that when $q$ is divided by $c$, the quotient is $k$. Prove that when $a$ is divided by $bc$, the quotient is also $k$.

Exercise 5 Question:

  Let $a$ be any integer and let $b$ and $c$ be positive integers. Suppose that when $a$ is divided by $b$, the quotient is $q$ and the remainder is $r$, such that
  $$a = bq + r \text{ and } 0 \leq r < b.$$
  If $ac$ is divided by $bc$, the quotient is $q$ and the remainder is $rc$.

What I have done so far:
Let $a=bq+r, 0 \leq r < b$, and $q=ck+z, 0 \leq z < c$, where $a,b,c,q,k,z$ are integers. Then it follows that:
$$a = bq + r \implies a = b(ck+z)+r \implies a = bck + bz + r, 0 \leq bz+r < bc.$$
What I am stuck on is how to show the last inequality is true. I have tried manipulating the inequalities given:
$$ 0 \leq r < b, 0 \leq z < c \implies 0 \leq rc+bz < bc \implies 0 \leq bz + r < bz + rc < 2bc $$
$$ 0 \leq bz + r < 2bc $$
I think this is to no avail. Can anyone give me tips on how to proceed?

I know I'm super late I'm sorry. but the problem only want you to show k. so set r = to zero and you are done. you don't need z. when you substitute you will get 2r, then set are to zero.


We have, $$\ln x+\ln x(x+1)=\ln 2$$ $$\implies \ln(x(x+1))=\ln 2$$
$$\implies \frac{\ln(x(x+1))}{\ln 2}=1$$ $$\implies \log_2(x(x+1))=1$$ $$\implies x(x+1)=2$$ $$\implies x^2+x-2=0$$ now, solving above quadratic equation for $x$ as follows $$\implies x=\frac{-1\pm\sqrt{(1)^2-4(1)(-2)}}{2(1)}$$   $$\implies x=\frac{-1\pm\sqrt{9}}{2}$$   $$\implies x=\frac{-1\pm 3}{2}$$ $$\implies x=\frac{-1+3}{2}=\color{}{1}$$ &$$\implies x=\frac{-1-3}{2}=\color{}{-2}$$
Edit: Since log is defined for positive number i.e. $x>0$ hence we have $x=\color{blue}{1}$


quadratic equation form maximum solutions My Pearson intermediate algebra book has a "concept check" question in its section on solving equations by using quadratic methods. These questions are supposed to highlight fundamental concepts that indicate full or poor understanding of the subject. The question asks:

a. True or false? The maximum number of solutions that a quadratic equation can have is 2.
b. True or false? The maximum number of solutions that an equation in quadratic form can have is 2.

The answers are listed as a. true and b. false.

I'm having difficulty searching for information on this point because search results yield explanations of how to determine the number of solutions based on the discriminant, but don't seem to get into why an equation in quadratic form is not necessarily a quadratic equation, or why it wouldn't have the same of maximum number of solutions as a quadratic equation. I'm also not finding an explanation anywhere in the text, which is mostly examples and the phrase "equation in quadratic form" is nowhere to be found.

I googled "quadratic in form" and found the following explanation from Paul's online notes:http://tutorial.math.lamar.edu/Classes/Alg/ReducibleToQuadratic.aspx

For example, an equation like $x^4+12x^2-74=0$ is an equation in quadratic form.

Now, you are asked the following questions:

1) How many solutions can a quadratic equation have? The answer is not more than two. To see this, write down the formula for roots of the quadratic equation $ax^2+bx+c=0$, which is:$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

Therefore there are at most two roots, given by $\dfrac{-b + \sqrt{b^2-4ac}}{2a}$ and $\dfrac{-b - \sqrt{b^2-4ac}}{2a}$. Of course, if the discriminant $b^2-4ac$ is zero, there is only one solution.

1) How many solutions can an equation in quadratic form have? The answer is can be more than two. To see this, look at the equation $x^4-3x^2+2=0$. This is an equation in quadratic form, which we write as $X^2-3X+2=0$, where $X=x^2$,and get $X = 1,2$ and the solutions are given by $1,-1,\sqrt{2},-\sqrt{2}$, which is more than two. I think this answers your question.

The discriminant determines the number of solutions a Quadratic equation can have. Let, D = b^2-4ac ( i.e. the discriminant). If D=0 then one real solution exists. If D>0 then two real solutions exist. If D<0 then two imaginary solutions exist.
Note.
The solution x = 0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation.
So to answer your question the maximum number of solutions is two.


What is the square root of $3 + 2\sqrt{10}i$? I need to compute the square root of $3 + 2\sqrt{10}i$.

I know how to solve it, but for some reason I'm not getting the correct answer. I attempted to solve it like this:

$$
  \sqrt{3 + 2\sqrt{10}i} = x + iy
    \quad \longrightarrow \quad
  3 + 2\sqrt{10}i = x^2 - y^2 +2xyi
$$
and so forth, but my answer isn't correct.

As I explained here. there is  a very simple formula for denesting such radicals, namely

Simple Denesting Rule $\rm\ \ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $

$\ 3+2\sqrt{-10}\ $ has norm $= 49.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 7\,\ $ yields $\ {-}4+2\sqrt{-10}\:$

with $\, {\rm\ \sqrt{trace}}\, =\, 2\sqrt{-2}.\ \ \ \rm \color{brown}{Dividing\ this\ out}\ $ of the above we obtain $\,\ \sqrt{-2} + \sqrt 5$

Checking it we find  $\,\ (\sqrt{-2} + \sqrt 5)^2 =\, -2+5 + 2\sqrt{-2}\sqrt 5\, =\, 3+ 2\sqrt{-10}$

Remark $\ $ Many more worked examples are in prior posts on this denesting rule.

You want $z=a+bi$ where $a,b\in\mathbb{R}$ such that

$$z^2=a^2-b^2+2abi=3+2\sqrt{10}i.$$

Comparing coefficients, you need $a^2-b^2=3$ and $ab=\sqrt{10}$. So $a=\sqrt{5}$ and $b=\sqrt{2}$, or $a=-\sqrt{5}$ and $b=-\sqrt{2}$. Thus the square root of $3+2\sqrt{10}$:

$$\pm(\sqrt{5}+\sqrt{2}i).$$