Determine whether the set $H$ of all matrices form a subspace of $M_{2 \times 2}$ Determine if the set $Z$ of all matricies form
$ \left[ \begin{array}{cc}
a & b \\
0 & d
\end{array} \right] $
is a subspace of $M_{2 \times 2}$ (the set of all $2 \times 2$ matrices).

% This is something I came up with. Can someone look at it and let me know any useful corrections/suggestions to the question please.

Answer:

Without specification as to the nature of $a,b$ and $d$, it is assumed that $a,b,d \in \mathbb{R}$

Hence, $H$ is determined to be a subspace of $M_{2 \times 2}$ because it is closed under scalar addition and scalar multiplication and contains the zero vector when $a=b=d=0$.

What you have done looks reasonable except for one tiny point.  When you say "because it is closed under scalar addition and scalar multiplication", you should delete the first (but not the second) scalar, as you are trying to show that adding two elements give a third member of the subspace.

So you should end with

  Hence, $H$ is determined to be a subspace of $M_{2 \times 2}$ because it is closed under  addition and scalar multiplication and contains the zero vector when $a=b=d=0$.

Let $A=
\left[
\begin{array}{cc}
a & b \\
0 & d
\end{array}
\right]$ and
$B=
\left[ \begin{array}{cc}
x & y \\
0 & z
\end{array} \right]$.
Then $(AB)_{2,1}=dz$ and not $0$. It doesn't seem to be closed under multiplication on its own set.


Proving that an equation doesn't have integer solutions I need to prove that there are no integer solutions for a bunch of equations like the following:
$$15x^2 - 7y^2 = 9$$
I was able to solve some simpler ones by picking a dividend and looking into it's remainder table. But it's not working for the others. How should I start thinking this kind of problem? It's from my algebra class and we are looking into divisibility and congruence. Thanks!

Hint: Maybe note first that $3$ must divide $y$. Let $y=3t$. Then we are looking at $15x^2-(7)(9t^2)=9$. So $3$ must divide $x$. Let $x=3s$. We end up with
$$15s^2-7t^2=1.$$
Now work modulo $3$.

Call $x^2=a$ and $y^2=b$ then we have $15a-7b=9$ which is a diaphante equation. It's solution can be found by Euclid's algorithm, but also there is a theorem that no solutions exist iff $9$ isn't a multiple of the $gcd$ of $15$ and $7$.

In this case there are solutions, so you can find them, then show if they can be or can't be 2 squares.


Expectation of norm of a random variable Let $x_k$ be a random vector such that its expectation
$$
E[\Vert x_k \Vert]<a
$$
for some $a>0$. Then can we say that
$$
E[\Vert x_k \Vert^2]<a^2 ?
$$

$\mathbb{E}[\lVert X\rVert^2]$ need not even be defined in general. Consider the random variable on $\mathbb{N}^\ast$ (vector of dimension 1) with probability mass function
$$
\mathbb{P}\{ X = n\} = \frac{1}{\zeta(3)}\cdot\frac{1}{n^3}.
$$
(for which you do have $\mathbb{E}[X] = \sum_{n=1}^\infty \frac{1}{\zeta(3)}\cdot\frac{1}{n^2} = \frac{\pi^2}{6\zeta(3)}$)

No, not in general, but you can say that $\left[\mathbb{E}\left(\biggr|\biggr|\vec{X_k}\biggr|\biggr|\right)\right]^2 <a^2$ .


Reversing the usual inequality involving the determinant of the sum of positive definite matrices Given positive definite matrices $A$ and $B$, of dimension $n$, is it possible to derive an inequality of the form $$\det(A+B)\le f(\det(A),\det(B)),$$ where $f$ is some linear function (perhaps involving n)?.

The Minkowski inequality goes in the other direction, with $f(X,Y)=X+Y$. How about this one, though?

EDIT: I'm also open to allowing $f$ to contain information about the spectral norms of $A$ or $B$, or information of this kind.

This is false even for $2\times2$ diagonal matrices. For these, what
you want reduces to
\begin{eqnarray*}
\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) & = & \det\left(\left(\begin{array}{cc}
a_{1}\\
 & a_{2}
\end{array}\right)+\left(\begin{array}{cc}
b_{1}\\
 & b_{2}
\end{array}\right)\right)\\
 & \overset{!}{\leq} & \alpha\cdot\det\left(\begin{array}{cc}
a_{1}\\
 & a_{2}
\end{array}\right)+\beta\cdot\det\left(\begin{array}{cc}
b_{1}\\
 & b_{2}
\end{array}\right)\\
 & = & \alpha a_{1}a_{2}+\beta b_{1}b_{2}
\end{eqnarray*}
with suitable $\alpha,\beta \in \Bbb{R}$ and all $a_1,a_2,b_1,b_2 >0$.

Now consider $a_{1}=b_{2}=n$ and $a_{2}=b_{1}=\frac{1}{n}$. Then
the desired inequality becomes
$$
n^{2}\leq\left(n+\frac{1}{n}\right)^{2}\leq\alpha+\beta
$$
for all $n\in\mathbb{N}$, which is absurd.

You cannot. Basically, you want some constants $a,b,c$, which are possibly dependent on $n$, such that $$\det(A+B)\leq a\det (A) + b\det (B) + c .$$

However, take $A=\begin{bmatrix}1 & 0 &\dots &0\\
0&0&\dots &0\\
\vdots &\vdots &\ddots &\vdots\\
0&0&\dots&0\end{bmatrix}$
and $B=I_n - A$.

Then, $\det(\alpha (A+B)) = \alpha$, and $\det(A)=\det(B)=0$, which means that for every real value $\alpha$, you have $\alpha \leq c$. Obviously, no such $c$ exists.


A Problem on Time Complexity of Algorithms I want no know if the following problem is solved or not, or how can I solve it?

Problem: For every integer $t$, Is there any problem that can be verified in $O(n^{s})$ but its solution can be found in $T(n)=\omega(n^{st})$, i.e., its solution can not be found in $O(n^{st})$.

By verifying, I mean that given a candidate solution $y$, we can judge whether $y$ is correct or not in time $O(n^s)$.

I think what you want to know is covered by the answers to this question and this question on cstheory stackexchange.

It seems that the answer is likely to be yes, since it follows from $\mathbf{P} \neq \mathbf{NP}$ by a padding argument, but it isn't currently proven.

To elaborate, if I am understanding things correctly what you asking about is this statement:

For all $t$ and all $s$, there is a problem in $\mathbf{NTIME}(n^s)$ that is not in $\mathbf{TIME}(n^{s t})$.

There is a slight subtlety that you may be asking about search problems, but this is about decision problems, but I think that shouldn't matter.

According to the links, we don't even know that $\mathbf{NTIME}(n^s) \neq \mathbf{TIME}(n^s)$ so we couldn't even prove the weaker statement of finding a lower bounded of, say, $n^{s + 1}$, let alone a lower bound of $n ^{s t}$ (for $t>1$).

In other words, if we can't even manage to prove the much weaker statement that $\mathbf{NTIME}(n^s) \neq \mathbf{TIME}(n^s)$, we don't have much hope of proving the much harder statement that $\mathbf{NTIME}(n^s) \not \subseteq \mathbf{TIME}(n^{st})$. You can be fairly sure therefore that it is an open problem.

I also clamied that the statement follows from $\mathbf{P} \neq \mathbf{NP}$. To prove this, we shall assume that the statement fails and use it to prove $\mathbf{P} = \mathbf{NP}$.

If the statement is false, then in particular there is some $t$ and $s$ such that $\mathbf{NTIME}(n^s) \subseteq \mathbf{TIME}(n^{st})$. Let $f$ be a function that has a nonderministic algorithm that runs in polynomial time.

Let $p$ be a padding function, that concatenates the string $11\ldots110$ to its input so that $f(n)$ terminates in less than $|p(n)|$ steps. Then there is a linear time nondeterministic algorithm for $f'$ such that $f'(p(n)) = f(n)$ and $f'(m) = 0$ if there is no $n$ such that $m = p(n)$ (essentially the same algorithm as for $f$ but it is now linear time because the input is longer).

Since $f'(m) \in \mathbf{NTIME}(m)$ we know by assumption that $f'(m) \in \mathbf{TIME}(m^{st})$. So $f'$ has a deterministic algorithm that runs in polynomial time. By composing $f'$ with $p$, we also get a deterministic polynomial time algorithm for $f$, showing that $\mathbf{NP} \subseteq P$.

You have an array of $n^{st}$ integers and want to check if there is a zero. You check it in $O(1)$, if the index is given. You find it just by the brute-force search in $O(n^{st})$.


Normal subgroup of Normal subgroup If $H$ is a normal subgroup of $K$ and $K$ is a normal subgroup of $G$ can we say that $H$ is a normal subgroup of $G$.I could not prove it and cannot find a suitable counter example

Will the results holds for $G$ abelian ?If else what will be counter example ?

It is not true that $H \lhd K \lhd G$ implies $H \lhd G$ (although the stronger condition that $ H \text{ char } K\ \lhd G$ will force $H \lhd G$).

A good tactic would be to choose a group $G$ with an abelian normal subgroup $K$. Any subgroup of $K$ must be abelian, hence normal in $K$. Your job is to find a subgroup of $K$ that's not normal in $G$.

The alternating group on $4$ letters, $A_4$ is a good choice for $G$. There, you can find such a chain $H \lhd K \lhd G$ with $H \not\lhd G$.

If $G$ is abelian then every subgroup its subgroup is normal: $gK = Kg$. Because $H$ also is subgroup of $G$ the answer is yes and proof is straightforward.


Prove ${\forall x \; \forall y \; (x + y = y + x)}$ Question:

  Determine the truth value of the statement if the universe of
  each variable consists of all the integers.

  Give reason to your answer if the statement is true and provide a
  counterexample for the false statement.

  ${\forall x \; \forall y \; (x + y = y + x)}$
  $$\tag*{$(2\;marks)$}$$

Answer:

True.

${Suppose\;x = m,\;y = n,\;m,\;n \in Z}$

${By\;defination\;of\;commutativity,\; m + n = n + m}$

${Then\;x + y = y + x}$

${\;\;\;\;\;\;\;\;\;m + n = n + m}$

Can I prove like this?

If you are given
commutivity,
the result follows
immediately.

If you are given
the Peano axioms
and the definition of addition
in terms of
the successor function,
you should look at
Landau's "Foundations of Analysis"
(do a Google search).

maybe you can use the successor function for the integers

where " s " is the successor function

  x  = s(x-1)    and  y = s(y-1)

  x + y = s(x-1) + y = s(x + y -1)

  y + x = s(y-1) + x = s(y + x -1)

Now it remains to know if "s" is a function that is commutative

I think you can not prove it so because this is a particular case where x = y


How to find the solution of a quadratic equation with complex coefficients? I know how to find the solution for a quadratic equation with real coefficients. But if the coefficient changes to complex numbers then what is the change in the solution? Want an example of such equation with solution.

It's no different.  The quadratic formula works regardless of whether the coefficients are real or complex.

Consider the example
$$(3+i)x^2 + (2-i)x + (5+2i) = 0$$

The quadratic formula gives

$$x = \frac{-(2-i) \pm \sqrt{ (2-i)^2-4(3+i)(5+2i) } }{2(3+i)}$$

Simplifying this is kind of a pain, of course.  Under the radical you have to multiply everything out and combine terms.  Eventually you get the radical into the form $\sqrt{M+Ni}$ where $M$ and $N$ are some constants -- in this example, they will be integers.  Then the question is, how do you simplify the square root of a complex number?  (For that, see How do I get the square root of a complex number?).

Let's see if this works.  Kind of unclear how to use these tools, so I just was going to snapshot a quick writeup.


Are the following graph isomorphic? I have to prove or disprove that the following graph are isomorphic?

If I will give the following isomorphism,

\begin{align*}
  a & \mapsto 1\\
  b & \mapsto 2\\
  e & \mapsto 5\\
  d & \mapsto 8
\end{align*}
then we see that $h$ has only one uncommon vertex to $a$ which is $g$ but the vertex $8$ has two vertices which is not directly connected to the vertex $1$. Hence the two graphs are Not isomorphic.

Am I correct?

You have a few problems. First, you haven't given the full map between the vertex sets.
Second, you've only argued that this map isn't an isomorphism. You haven't clearly ruled out the possibility that some other map is an isomorphism. I think you're trying to argue that the partial map you've shown is perfectly general, but you haven't started anything to that effect.

Typically, showing that two graphs are not isomorphic involves showing some invariant differs between them. Common invariants are degree sequence, diameter, connectivity, chromatic number, and Hamiltonicity.

As a hint, notice that every edge in the left graph is contained in two otherwise disjoint 4-cycles.

The given graphs are Not isomorphic.
Let me call the left graph as $G$ and right graph as $H$.
We can see that graph $G$ has no odd length cycle, so $G$ is bipartite graph (actually $G$ is a hypercube graph $Q_3$).
Graph $H$ has cycles of length $5$ (for example, take $1-2-3-4-5-1$), So, H is Not bipartite graph.
So, $G$ and $H$ are Not isomorphic.


Let $f:[0,2]\to \mathbb{R}$ be a continuous function with no roots. Prove that the function is not surjective and $f(0)\cdot f(2)>0$ Let $f:[0,2]\to \mathbb{R}$ be a continuous function with no roots
i) Prove that the function is not surjective

ii) Show that $f(0)\cdot f(2)>0$
Got no ideas, maybe $y=f(x)$? Doesn't seem helpful though.
Also, is it possible to change the interval from $[0, 2]$ to $[a, b]$? I remember asking about a problem here for a particular 'case' and it ended up being true for every 'case': How to prove the following integral equation? $\int_{0}^{c}x^2f(x)=0$ (hope it makes sense what I'm saying)
Thanks in advance!

Since it has no roots there is no $x \in [0,2]$ such that $f(x) = 0 \in \mathbb R$
(Updated) Assume $f(0)\cdot f(2) \le 0$, but this implies either at least one of $f(0)$ or $f(2)$ is $0$ (contradiction) or they have different signs which means there is a zero between $0$ and $2$ (contradiction). Therefore $f(0)\cdot f(2) > 0$.

i) $f^{-1}(0) = \emptyset$.
ii) Is not true, consider $f \equiv 1$.


Dual of $l^\infty$ is not $l^1$ I know that the dual space of $l^\infty$ is not $l^1$, but I didn't understand the reason.
Could you give me a example of an $x \in l^1$ such that if $y \in l^\infty$, then $ f_x(y) = \sum_{k=1}^{\infty} x_ky_k$ is not a linear bounded functional on $l^\infty$, or maybe an example of a $x \notin l^1$ such that if $y \in l^\infty$, then $ f_x(y) = \sum_{k=1}^{\infty} x_ky_k$ is a linear bounded functional on $l^\infty$?

The point is the following: There are bounded functionals on $\ell^\infty$, which are not of the form $$ f(y) = \sum_k x_k y_k $$
for some $x$. I do not know if such a functional can be given explicitly, but they do exist. Let $f \colon c \to \mathbb R$ (where $c \subseteq \ell^\infty$ denotes the set of convergent sequences) be given by $f(x) = \lim_n x_n$. Then $f$ is bounded, as $|\lim_n x_n| \le \sup_n |x_n| = \|x\|$. Let $g \colon \ell^\infty \to \mathbb R$ be a Hahn-Banach extension. If $g$ where of the above mentioned form, we would have (with $e_n$ the $n$-th unit sequence)
$$ x_n = g(e_n) = f(e_n) = 0 $$
hence $g = 0$. But $g \ne 0$, as for example $g(1,1,\ldots) = 1$.

Counterexample:
Consider the linear functional $\phi$, defined on $l_\infty$ and given by  $\phi(x) = \lim_{N\rightarrow\infty} \frac{1}{N}\sum_{n=1}^N x_n$. Now, $\phi$ (which you might call the average functional) is bounded since $|\phi(x)|\le \lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^N ||x||_\infty = ||x||_\infty$.
Assume $\phi(x) = \sum_{n=1}^\infty x_n y_n$, for some $y\in l_1$.
Then, for $\delta = \{1, 0, 0, ... \}\in l_\infty$, you have $\phi(\delta)=0=y_1$. Similarly, for each $e^n = \{0, ... 0, 1, 0, ...\}$  with a "single $1$" at position $n$ (thus $\delta=e^1$), you have that $\phi(e^n)=0=y_n$. Then $y_n = 0_n$ is the zero sequence. Then $\phi$ is the zero functional. But, since, e.g. for $x_n = (-1)^n$ you have $\phi(x)=0.5\ne 0$, you have a contradiction.


If the roots of the equation $p(q-r)x^2+q(r-p)x+r(p-q)=0$ are equal, show that $\dfrac {1}{p}+\dfrac {1}{r}=\dfrac {2}{q}$. If the roots of the equation $p(q-r)x^2+q(r-p)x+r(p-q)=0$ are equal, show that $$\dfrac {1}{p}+\dfrac {1}{r}=\dfrac {2}{q}$$

My Attempt:
$$p(q-r)x^2+q(r-p)x+r(p-q)=0$$
Comapring with $ax^2+bx+c=0$, we get:
$$a=p(q-r)$$
$$b=q(r-p)$$
$$c=r(p-q)$$.
Since the roots are equal:
$$b^2-4ac=0$$
$$(q(r-p))^2-4p(q-r)r(p-q)=0$$
Multipying and simplfying a bit;
$$q^2r^2-2pq^2r+p^2q^2-4p^2qr+4pq^2r+4p^2r^2-4pr^2q=0$$.

Let A=p(q-r), B=q(r-p) & C=r(p-q). A+B+C=0 => B=-(A+C). Since roots are equal, B²-4AC=0 => (A+C)²-4AC=0 => (A-C)²=0 => A=C
=> p(q-r)=r(p-q)
=> (q-r)/qr=(p-q)/pq [Dividing throughout by pqr]
=> 1/r-1/q=1/p-1/q
=> 2/q=1/p+1/r
Hence QED


Number of possible pairs I have a problem counting all the possible way of "pairing" 2 people in a group of N people (let's assume N is even).
Example:

The professor wants the students to work in pairs (groups of two).
In how many ways the students could pairs ?

I have seen that the answer is $$\frac{N!}{2^{N/2}*\frac{N}{2}!} $$
So, the way i understand it: it is the way of all possible combination (N!)
divided by $$2^{N/2}$$  because ... I don't know (that is the reason of the question) ...
And divided by $$(N/2)!$$ because of the permutation.
So, could you explain me why this division by $$2^{N/2}$$ ?
Also, what are the probability for 2 people (let's say student1 and student 2), to be in the same group ?

Since $N$, the number of students, is even, let's call it $2M$.

The formula you quote can be obtained using the following reasoning. First let us divide the set of people not into pairs, but into pairs with designated leader, who gets to wear a special leader shirt.

We can choose the set of leaders in $\dbinom{2M}{M}$ ways. Now assign a follower to each leader. This can be done in $M!$ ways.

So the number of pairings with designated leader is $\dbinom{2M}{M}M!$, which simplifies to $\dfrac{(2M)!}{M!}$.

But if we want to divide into plain groups of two, we have overcounted by a factor of $2^M$.  For if $D$ is the number of democratic groups of $2$, then the number of groups-with-leader is $2^MD$. This is because for any division into democratic groups,  there are $2^M$ ways to choose the persons in the democratic division who will wear the leader shirts. Thus
$$2^M D=\frac{(2M)!}{M!},$$
and now solving for $D$ we have our count.

Remark: The following is another way to count the number of divisions into groups, this time directly. Line up the students in order of student number, or weight, or beauty.

For concreteness, assume there are $12$ people in the group. The first student in the list can select her partner in $11$ ways.  For each such way, the first still unpartnered person in the lineup can select her partner in $9$ ways. Once this has been done, the first unpartnered person in the list can select her partner in $7$ ways.  And so on. So the number of divisions into groups is
$$(11)(9)(7)(5)(3)(1).$$

To get back to the formula of the OP, multiply top and bottom by the "missing" numbers, so by $(12)(10)(8)(6)(4)(2)$. On top, we get $12!$.

At the bottom, we get $(12)(10)(8)(6)(4)(2)$. Taking out the $2$'s, we get $(2^6)(6!)$.

The same idea works in general. The number of groups of $2$ when there are $2M$ students is $(2M-1)(2M-3)(2M-5)\cdots(5)(3)(1)$.

If we have group from n students, the first has (n-1) possible partner, the second has (n-2) because we already accounted the possibility to be a partner with the first, and so on until the later one don't has any possibility because we already accounted all the possibilities to be  a partner with all the group members, So the result of all possibilities is (n-1)+(n-2)+...+1 or 1+2+3+4+....+(n-1) which is a simple sum of terms of arithmetic serie: the first term + the later term cros the number of terms by two:
((n-1)+1)*(n-1)/2=n(n-1)/2
and the general case of k from n is n!/((n-k)!k!)
note : here AA can't be a pair and AB can not be accounted if BA is accounted


Can I define a function as the set of these points? Can I define a function as the set of these points as $k$ goes to infinity?

$$ \lim_{k\to \infty}\bigg (\frac{k}{k-n},\frac{k-n}{k}\bigg)$$

Where $n$ is the ordered set of natural numbers less than $k$ from least to greatest. $n=1,2,3,...,k-1.$ A point is plotted for each $n.$

For example for $k=10$ it would look like:

$$ \bigg (\frac{10}{10-n},\frac{10-n}{10}\bigg) $$

and there would be $9$ points because $n=1,2,3,4,...,9.$

A function consists of two sets (the domain and the range) and a graph between them.
I think you have a certain relation between the points but the concept of a limit has nothing to do with this.
Examine the definition of a function until it is clear for you.

The function $f(x)=1-\frac{1}{x}$ passes through all the desired points and therefore is the correct one.


Proof of intersection and union of Set A with Empty Set I need to prove the following:

Prove that $A\cup \!\, \varnothing \!\,=A$ and $A\cap \!\, \varnothing \!\,=\varnothing \!\,$

It's my understanding that to prove equality, I must prove that both are subsets of each other.
So to prove $A\cup \!\, \varnothing \!\,=A$, we need to prove that $A\cup \!\, \varnothing \!\,\subseteq \!\,A$ and $A\subseteq \!\,A\cup \!\, \varnothing \!\,$.

However, I found an example proof for $A \cup \!\, A$ in my book and I adapted it and got this:

$A\cup \!\, \varnothing \!\,=$ {$x:x\in \!\, A  \ \text{or} \ x\in \!\, \varnothing \!\,$}
= {$x:x\in \!\, A$} = A

$A\cap \!\,  \varnothing \!\,=$ {$x:x\in \!\, A  \ \text{and} \ x\in \!\, \varnothing \!\,$}
= {$x:x\in \!\, \varnothing \!\,$} = $\varnothing \!\,$

Do my proofs look ok?

This looks fine, but you could point out a few more details. For instance, $x\in \varnothing$ is always false. Therefore
$x \in A \text{ or } x\in \varnothing
$
is logically equivalent to
$
x \in A
$
and therefore the two set descriptions
$$
\{x \mid x \in A \text{ or } x \in \varnothing\},\quad \{x\mid x \in A\}
$$
must describe the same set, since the conditions are true for exactly the same elements $x$.

Similarily, because $x \in \varnothing$ is trivially false, the condition $x \in A \text{ and } x \in \varnothing$ will always be false, so the two set descriptions
$$
\{x \mid x \in A \text{ and } x \in \varnothing\},\quad \{x\mid x \in \varnothing \}
$$
must describe the same set.

Of course, for any set $B$ we have
$$
B = \{x \mid x \in B\}
$$

we need to proof that A U phi=A,
it can be written as,
A U PHI={X:X e A OR X e phi}
to do it in a simpleast way I will use a example,
write in roaster form
A={1,2,3}
PHI={4,2,5}
WHEN YOU WRITE THE UNION IT COMES OUT TO BE {1,2,3,4,5}
THEREFORE AUPHI=A


A Problem on Improper Integrals Let $f(x)$ be continuous except at $x = 0$ and let $a > 0$. Assume that the improper integral $$\int_{0}^{a}f(x)\,dx = \lim_{\epsilon \to 0+}\int_{\epsilon}^{a}f(x)\,dx$$ exists and let $$g(x) = \int_{x}^{a}\frac{f(t)}{t}\,dt$$ Show that $$\int_{0}^{a}g(x)\,dx = \int_{0}^{a}f(x)\,dx$$

I tried integration by parts noting that $g'(x) = -f(x)/x$ and obtained for $0 < \epsilon < a$ the following $$\int_{\epsilon}^{a}g(x)\,dx = [xg(x)]_{x = \epsilon}^{x = a} - \int_{\epsilon}^{a}xg'(x)\,dx$$ or $$\int_{\epsilon}^{a}g(x)\,dx = -\epsilon g(\epsilon) + \int_{\epsilon}^{a}f(x)\,dx$$

The problem is solved if we can somehow show that $\lim_{\epsilon \to 0+}\epsilon g(\epsilon) = 0$. Looking at the definition $g(x)$ we see that we have no information of the behavior of $f(t)$ at $t = 0$ and the $t$ in denominator complicates the analysis of $g(\epsilon)$. Please suggest some hints which can lead to the solution.

Note: This problem is taken from G. H. Hardy's "A Course of Pure Mathematics" 10th ed. Page 397.

Hint: Define
$$h(x)=\int_x^af(t)dt,\ \forall x\in[0,a].$$
Then for every $x\in(0,a]$,
$$g(x)=-\int_x^a\frac{h'(t)}{t}dt=\frac{h(x)}{x}-\int_x^a\frac{h(t)}{t^2}dt=\frac{h(x)}{a}+\int_x^a\frac{h(x)-h(t)}{t^2}dt.$$

Given  $\delta\in(0, a]$, denote
$$M_\delta=\max_{x,y\in[0,\delta]}|h(x)-h(y)|.$$
When $x\in(0,\delta)$,
$$|g(x)-\frac{h(x)}{a}|\le \int_x^\delta\frac{|h(x)-h(t)|}{t^2}dt+\int_\delta^a\frac{|h(x)-h(t)|}{t^2}dt\le \frac{M_\delta}{x}+\frac{M_a}{\delta}.$$

find g'(x)=-f(x)/x
x*g'(x)=f(x)
integrate  both side with limit 0 to a using integration by parts
{x*g(x)|0 to a } - { integration of g(x) limit 0 to a} = -{integration of  f(x) limit 0 to a }
now use given equation and put x =a than u will get g(a)=0
now just solve u will get the desired result .


Sylow Theory: $gHg^{-1} \leq P$ Let $G$ be a finite group and $P$ be a Sylow $p$-subgroup of $G$. If
   $H$ is an arbitrary $p$-subgroup of $G$, then there exists a $g \in G$
  such that $gHg^{-1} \leq P$

Proof:

Using the $1$st Sylow theorem, we have that every $p$-subgroup of $G$ is contained in a Sylow $p$-subgroup.

Also, from the $2$nd Sylow theorem, for every $P'$ Sylow $p$-subgroup there exists a $g \in G$ such that $gP'g^{-1} = P$.

Hence

$$
gHg^{-1} \subset gP'g^{-1}=P
$$

Any suggestions for showing $gHg^{-1}$ is a subgroup of $P$?

Hit: consider the double coset decomposition of $G$ with respect to $H$ and $P$


Question about the graphical sequences A sequence $( d_1, d_2,...,d_p)$ is said to be graphic if and only if it is the degree sequence of some simple graph with p vertices. Show that the sequences $(7,5,5,5,3,2,1)$ and $(6,6,5,4,2,2,1)$ are not graphic. So for the first one 7 can't work because there are only 6 other vertices to connect to.  But what if you have one degree of multiplicity 2?  Then 7 works and I don't think this is what the question is looking for as an answer anyways.  Using graph theory in Discrete Math, how would you solve this?

I think you're arguement for the first one holds. Usually in these kinds of problems graphs don't have two edges connecting the same vertices.

For the second, two $6$s means that every vertex must has at least two neighbors, since both $6$s connect to every other vertex. But then you can't have a $1$ in the sequence.

It is impossible to draw this graph. A simple graph has no parallel edges nor any loops. There are only 7 vertices in first sequence, so each vertex can only be joined to at most six other vertices, so the maximum degree of any vertex would be 6. Hence, you can’t have a vertex of degree 7


Limit of $x^ny^n$ when $n \to \infty$ Please, someone could help me in solving the following limit.

Let $x \in \mathbb{R}$, $y \in \mathbb{R}$ and $n \in \mathbb{N}$. Also consider that $0<x<1$ and $1<y<\infty$.

$$ \lim_{n \to \infty}x^ny^n = L $$.

What is the value of $L$ ?

$$x^ny^n = (xy)^n$$

If $xy<1$ , Limit = $0$

If $xy=1$, Limit = 1

If $xy>1$, Limit $\to \infty$

$L=\infty$ since the function is in form $\frac{\infty}{0}$ where $n\to \infty$


Distance between ellipse and line What is the distance between the ellipse $$\frac{x^2}{9}+\frac{y^2}{16}=1$$ and the line $y=6-x$.

I think I need to be using Lagrange Multipliers but don't know how.

Just another approach, for the sake of variety.

You can also observe that at the point of closest approach, the ellipse must be parallel to the line $y = 6-x$, which has slope $-1$.  Implicit differentiation of the ellipse gives us

$$
\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx} = 0
$$

$$
16x+9y\frac{dy}{dx} = 0
$$

$$
\frac{dy}{dx} = -\frac{16x}{9y}
$$

which must equal $-1$ at the point of closest approach (with $x, y > 0$—a quick sketch will show why), so

$$
y = \frac{16x}{9}
$$

Plug this back into the equation for the ellipse, solve for the point, and obtain the distance.

Yes, you can use Lagrange multipliers minimize $x^2+ y^2$ (which is equivalent to minimizing $\sqrt{x^2+ y^2}$) subject to the constraints $\frac{x^2}{9}+ \frac{y^2}{16}= 1$ and $x+ y= 6$. $\nabla x^2+ y^2= <2x, 2y>$, $\nabla \frac{x^2}{9}+ \frac{y^2}{16}= <2x/9, y/8>$, and $\nabla x+ y= <1, 1>$.

So you have $<2x, 2y>= \lambda<2x/9, y/8>+ \mu<1, 1>$ together with the constraints to solve for x and y, $\mu$, and $\lambda$.


How many possible combinations in 8 character password? I need to calculate the possible combinations for 8 characters password. The password must contain at least one of the following: (lower case letters, upper case letters, digits, punctuations, special characters).

Assume I have 95 ascii characters (lower case letters, upper case letters, digits, punctuations, special characters).

lower case letters = $26$
upper case letters = $26$
digits = $10$
punctuations & special characters = $33$

The general formula for the possible passwords that I can from from these 95 characters is: $95^8$.

But, accurately, I feel the above formula is incorrect. Please, correct me.
The password policy requires at least one of the listed above ascii characters. Therefore, the password possible combinations = $(26)*(26)*(10)*(33)*(95)*(95)*(95)*(95)$

Which calculation is correct?

EDIT: Please, note that I mean 8 characters password and exactly 8. Also, There is no order specified (i.e. it could start with small letter, symbol, etc.). But it should contain at least one of the specified characters set (upper case, lower case, symbol, no., etc.).

Start with all $8$-character strings: $95^8$

Then remove all passwords with no lowercase ($69^8$), all passwords with no uppercase ($69^8$), all passwords with no digit ($85^8$) and all passwords with no special character ($62^8$).

But then you removed some passwords twice. You must add back all passwords with:

no lowercase AND no uppercase: $43^8$
no lowercase AND no digit: $59^8$
no lowercase AND no special: $36^8$
no uppercase AND no digit: $59^8$
no uppercase AND no special: $36^8$
no digit AND no special: $52^8$

But then you added back a few passwords too many times. For instance, an all-digit password was remove three times in the first step, then put back three times in the second step, so it must be removed again:

only lowercase: $26^8$
only uppercase: $26^8$
only digits: $10^8$
only special: $33^8$

Grand total: $95^8 - 69^8 - 69^8 - 85^8 - 62^8 + 43^8 + 59^8 + 36^8 + 59^8 + 36^8 + 52^8 - 26^8 - 26^8 - 10^8 - 33^8 = 3025989069143040 \approx 3.026\times10^{15}$

There's a simple flaw with the original equation:

It was stated that you have:

26 lowercase letters (a-z)
26 uppercase letters (A-Z)
10 digits (0-9)
33 punctuations and special characters

How many total choices can each character within the password use?

ADD the above numbers to answer that: $26+26+10+33 = 95$
How many characters is the password in question?

I believe we identified 8 in this scenario.
How many combinations for this password are there?

$$\text{(Possible choices)}^\text{(How many characters long)}=\text{(How many combinations)}$$

Or per this example, $96^8=6634204312890625$
Based on 8 characters of anything you can type, the answer is as simple as above.

As stated in a more convoluted, albeit more descriptively accurate, the number changes based on password requirements.
From a hacker / pentester perspective, entropy is stronger than mental complexity.

If people use all lowercase because rules don't force them to use something else, yes, their password is weaker being all lowercase, because I can probe the password based on just lowercase ($26^8$).

Wow, no rule against using humanly recognized words? (That was a common rule early-mid 2000s) The password rules themselves actually make a weaker password than the mathematically possible $96^8$.
Password entropy makes this even more fun as we demonstrate that the entropy of password

  A#1WepOjII95&^2!

is actually weaker than the password

  OMGmathMakesMyHeadWant2EXPLODE

If you're looking at this from a security standpoint, use a long run-on phrase for a more challenging time being cracked. Using rainbow tables, it's now possible to crack a 64-character password within 4 minutes on a single computer.

No, your $120 Atom laptop isn't likely to meet that kind of hacking efficiency. It's simply saying you don't need a cluster of computers anymore.


$\int_0^{\pi}{x \over{a^2\cos^2 x+b^2\sin^2 x}}dx$ How to solve
$$\int_0^{\pi}{x \over{a^2\cos^2 x+b^2\sin^2 x}}dx$$

The answer is ${\pi}^2\over{2ab}$ but I cant prove it.

$$\text{As }\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$

$$I=\int_0^\pi{x \over{a^2\cos^2 x+b^2\sin^2 x}}dx=\int_0^\pi\frac{\pi-x}{a^2\cos^2 x+b^2\sin^2 x}dx$$ as $\displaystyle\sin(\pi-x)=\sin x,\cos(\pi-x)=-\cos x$

$$I+I=\pi \int_0^\pi\frac1{a^2\cos^2 x+b^2\sin^2 x}dx$$

$$=\frac\pi{b^2} \int_0^\pi\frac{\sec^2x}{\left(\frac ab\right)^2+\tan^2 x}dx$$

Put $\tan x=u$

Set $a/b=\tan (\alpha)$ and the intgrand simplifies to the product of $x$ and $\sec (x-\alpha)$ which is readily integrable


The autocovariance function of ARMA(1,1) So I am reading Brockwell and Davis introduction to Time Series analysis on page 89 where he derives the ACVF of an $ARMA(1,1)$ given by:

$X_t - \phi X_{t-1}=Z_t+\theta Z_{t-1}$ with ${Z_t}$ is $WN(0,\sigma^2)$ and $\mid \phi \mid < 1$

What is first told is that by causality assumption, the autocovariance at lag $h$ is:

$\gamma(h)=\sigma^2\sum_{j=0}^\infty\psi_j \psi_{j+\mid h \mid}$

So this at lag $h = 0$ is becomes:

$\gamma (0) = \sigma^2 \sum_{j=0}^\infty \psi_j^2$

How can this then be shown that $\sigma^2 \sum_{j=0}^\infty \psi_j^2 = \sigma^2 \Big[ 1 + \frac{(\theta+\phi)^2}{1-\phi^2} \Big]$ ?

And in the same way for $\gamma(1) = \sigma^2 \Big[ \theta + \phi + \frac{(\theta+\phi)^2\phi}{1-\phi^2} \Big]$?

I know that there is a definition of the function $\psi (z) = \sum_{j=0}^\infty\psi_j z^j = \frac{\theta(z)}{\phi(z)}$, $\mid z \mid\leq 1$. In what can this be applied here?

Eventhough this question is old, here is the answer:

The Autocovariance function for a causal time series is: $\\$

$\gamma(h) = \sigma^2 \sum_{j=0}^{\infty}\psi_j \psi_{j+|h|}$ $\\$

The MA($\infty)$ representation of $X_t$ is:

$X_t = Z_t + \sum_{j=1}^{\infty} (\phi+\theta)\phi^{j-1} Z_{t-j}$ $\\$
,where $\psi_j = (\phi+\theta)\phi^{j-1}$

Note that I will write the white noise $Z_{t-j}$ and not directly $\sigma^2$ to hopefully generate a better understanding of what is going on

$\\$

Case h = 0:

$Cov(X_t,X_t) = E(X_t X_t) = \gamma(0) = \sigma^2 \sum_{j=0}^{\infty}\psi_j^2$ $\\$

$E(X_t X_t)$ $= E(( Z_t + (\phi+\theta)\sum_{j=1}^{\infty} \phi^{j-1} Z_{t-j}) ^2) = E(Z_t^2) + E( ((\phi+\theta)\sum_{j=1}^{\infty} \phi^{j-1} Z_{t-j})^2 ) +$
$2(\phi+\theta) \sum_{j=1}^{\infty} \phi^{j-1}E( Z_{t-j}Z_t)$

since $Z_t$ is a White noise; $E( Z_{t-1}Z_t) = 0$, hence:

$E(X_t X_t)$ $= E(Z_t^2) + (\phi+\theta)^2 \sum_{j=1}^{\infty} \phi^{2j-2} E(Z_{t-j}^2)=
 \sigma^2 + (\phi+\theta)^2 \sum_{j=1}^{\infty} \phi^{2j-2} \sigma^2=
 \sigma^2 ( 1 + (\phi+\theta)^2 \sum_{j=1}^{\infty} \phi^{2j-2})=
 \sigma^2 ( 1 + \frac{(\phi+\theta)^2}{1-\phi^2})$

where $\sum_{j=1}^{\infty} \phi^{2j-2} = \sum_{j=0}^{\infty} \phi^{2j} = $$ \sum_{j=0}^{\infty} (\phi^{j})^2$ is a geometric series converging to $\frac{1}{1-\phi^2}$

$\\$

Case h = 1:

$Cov(X_t,X_{t-1}) = E(X_t X_{t-1}) = \gamma(1) = \sigma^2 \sum_{j=0}^{\infty}\psi_j \psi_{j-1}$ $\\$

$E(X_t X_{t-1})$ $ = E( (Z_t + (\phi+\theta)\sum_{j=1}^{\infty} \phi^{j-1} Z_{t-j}) \cdot ( Z_{t-1} + (\phi+\theta)\sum_{j=2}^{\infty} \phi^{j-2} Z_{t-j}) ) =
E( (Z_t + Z_{t-1}(\phi+\theta) + (\phi+\theta)\sum_{j=2}^{\infty} \phi^{j-1} Z_{t-j}) \cdot ( Z_{t-1} + (\phi+\theta)\sum_{j=2}^{\infty} \phi^{j-2} Z_{t-j}) ) =
E( (Z_t + Z_{t-1}(\phi+\theta) + (\phi+\theta)\phi\sum_{j=2}^{\infty} \phi^{j-2} Z_{t-j}) \cdot ( Z_{t-1} + (\phi+\theta)\sum_{j=2}^{\infty} \phi^{j-2} Z_{t-j}) ) =
(\phi+\theta)E(Z_{t-1}^2) + (\phi+\theta)^2\phi \sum_{j=2}^{\infty} \phi^{2j-4} E(Z_{t-j}^2)=
(\phi+\theta)E(Z_{t-1}^2) + (\phi+\theta)^2\phi \sum_{j=2}^{\infty} \phi^{2j-4} E(Z_{t-j}^2)=
(\phi+\theta)\sigma^2 + (\phi+\theta)^2\phi \sum_{j=2}^{\infty} \phi^{2j-4} \sigma^2)=
\sigma^2(\phi+\theta + \frac{(\phi+\theta)^2\phi}{1-\phi^2} )$

$\\$

Case h = 2:

$Cov(X_t,X_{t-2}) = E(X_t X_{t-2}) = \gamma(2) = \sigma^2 \sum_{j=0}^{\infty}\psi_j \psi_{j-2}$ $\\$

$E(X_t X_{t-2})$ $ = E( (Z_t + (\phi+\theta)\sum_{j=1}^{\infty} \phi^{j-1} Z_{t-j}) \cdot ( Z_{t-2} + (\phi+\theta)\sum_{j=3}^{\infty} \phi^{j-3} Z_{t-j}) ) =
E( (Z_t + Z_{t-1}(\phi+\theta) + Z_{t-2}(\phi+\theta)\phi + (\phi+\theta)\phi^2\sum_{j=3}^{\infty} \phi^{j-3} Z_{t-j}) \cdot ( Z_{t-2} + (\phi+\theta)\sum_{j=3}^{\infty} \phi^{j-3} Z_{t-j}) ) =
(\phi+\theta)\phi E(Z_{t-2}^2) + (\phi+\theta)^2\phi^2\sum_{j=3}^{\infty} \phi^{2j-6} E(Z_{t-j}^2)) =
\sigma^2( (\phi+\theta)\phi  + \frac{(\phi+\theta)^2\phi^2}{1-\phi^2})$

We see the pattern and follow that:

$\gamma(h) = \sigma^2 ( 1 + \frac{(\phi+\theta)^2}{1-\phi^2}) ~$ if h=0

$\gamma(h) = \sigma^2( (\phi+\theta)\phi^{h-1}  + \frac{(\phi+\theta)^2\phi^h}{1-\phi^2}) ~$ if h>0

To be honest, you can get that variance just from the way they defined the model at the start. If you rearrange to make Xt the subject, and then use the fact we have no mean, one can get to the result in a few lines if we use the assumptions that future errors do not depend on past outcomes. I'll walk through it in more detail later as I'm on my phone now.


Coherent configurations (example and explanation) A coherent configuration is a
pair $(X, S)$ consisting of a finite set $X $ of size $v$  and a set $S$ of binary relations on $X$ such that

•$ S$ is a partition of $X × X$;

• the diagonal relation $ ∆X$  is the union of some relations in $S$;

• for each $R  \in  S$ it holds that $R^{T}  \in S$;

• there exist integers $p^{R}_{ ST}$ such that
|$\{z ∈ X|(x, z)  \in  S$  and  $(z, y) \in  T \}| = p^{R}_{ ST} \}$
whenever $(x, y) \in  R$, for each $R, S, T ∈ S$.

I am new to this subject, so it is hard for me to understand. I request to explain above four axioms using an example.

I have tried some notes, most of them have not included any example.

The canonical examples arise from a group of permutations acting on $X$. Define two ordered pairs $(a,x)$ and $(b,y)$ to be related if there is an element of $G$ that maps $(a,x)$ to $(b,y)$, i.e., some element $g$ in $G$ maps $a$ to $b$ and $x$ to $y$. The set of ordered pairs related to a given pair is an orbit of $G$ acting on $X\times X$, and these orbits form a partition of $X\times X$. Thus the first axiom holds.

For the second axiom, note that all pairs in the orbit of a pair $(a,a)$ have the form $(b,b)$, and so the diagonal of $X\times X$ has a partition as required.

The orbit of $(a,b)$ is the ``transpose'' of the orbit of $(b,a)$.

Finally, if $x$ is $S$-related to $a$ and $T$-related to $b$ and $g\in G$, then $x^g$ is also $S$-related $a^g$ and $T$-related to $b^g$. So, in what I hope is an obvious notation,
\[ p_{S,T}(a,b) = p_{S,T}(a^g,b^g)\]
for all $g$ in $G$. From this it follows that the intersection numbers are well-defined.

We also know that if each relation R is represented by a matrix A whose rows and columns are indexed by the elements of X with (x,y) entry 1 if (x,y) belongs to R , otherwise 0.
Then the three conditions become:
1. The sum of the matrices is the all-1 matrix.
2.There is a subset of the matrices which sums to the identity matrix.
3.The set of matrices is closed under transposition.

But what will the fourth condition become ? Please explain in detail.


Intuition behind the derivative of dirac delta function Let me first begin what I mean by saying the intuition behind the " $\delta'(x)$ ". For example the smooth approximations of the delta function looks like the following:

(Left:the smooth approximation of $\delta(x)$
Right:the smooth approximation of $\delta'(x)$)

And by using my intuition I can understand why
$$
\int_{-\infty}^{\infty}f( \bar{x} )\delta(x-\bar{x}) \mathrm{d}\bar{x}=f(x)
$$
because I can say that the delta function fires whenever $x=\bar{x}$ and picks up the value of $f(x)$ at that point and when I integrate over all values of x, I get my function f(x) back. In other words it is like building the function $f(x)$ from thin sticks, which has the same hight as the value of the function. (Although I know that this explanation is nowhere near mathematical, it helps me and others to understand -whatever that means- the concept easier.) When I learned about the derivative of the delta function and its following property I was utterly shocked:
$$
\int_{-\infty}^{\infty}f(\bar{x})\delta'(x-\bar{x}) \mathrm{d}\bar{x}=f'(x)
$$

Because no matter how long I think about the subject I was unable to build a correct intuition about this distribution. My question is this: Can you explain me intuitively why the derivative of the delta function gives arise to a derivative?

PS: I know why this is true mathematically (integrating by parts and so on).

Suppose the spikes in the smooth approximation to $\delta'(x)$ are located at $x=-h$ and $x=h$.

When $\bar{x} \approx x+h$, the smooth approximation to $\delta'(x-\bar{x})$ will be large and positive, so the integral will roughly pick up "something large" times $f(x+h)$. Similary, for $\bar{x} \approx x-h$, the integral will pick up the same large factor times $f(x-h)$, but with the opposite sign. So if that large factor turns out to be of the magnitude $\frac{1}{2h}$, the integral will be roughly
$$
\frac{f(x+h)-f(x-h)}{2h} = \frac{\bigl(f(x) + h \, f'(x) + O(h^2)\bigr) - \bigl(f(x) - h \, f'(x) + O(h^2)\bigr)}{2h}
,
$$
which tends to $f'(x)$ as $h \to 0$.

So you might have heard about the discrete difference. It's basically the discrete counterpart for derivatives. Its definition is DF[x] = F[x]-F[x-1]. Now imagine a continuous function f. If you look closely, what the integral of f with delta function's derivative is doing is taking something very similar to a discrete difference for f that is discretized at a really really small step size.


Counterexample for Dirichlet product of two completely multiplicative functions. The below text is the proof of why Dirichlet product of two multiplicative functions are multiplicative:

It's obvious how the assumption of $(m,n)=1$ allowed the proof to be proceeded. However I am trying to find a counterexample to show that the Dirichlet product of two completely multiplicative functions is not be always completely multiplicative, but I couldn't succeed. Considering f and g to be power functions, $n^a$ and $n^b$ respectively, don't help since I can't find a way to decompose the double sum but it doesn't mean that $h(mn) \ne h(n)h(m)$. Is there any elementary counterexample?

If $p$ is a prime and $h = f * g$, $h(p) = f(1) g(p) + f(p) g(1) = g(p) + f(p)$ while $h(p^2) = f(1) g(p^2) + f(p) g(p) + f(p^2) g(1) = g(p)^2 + f(p) g(p) + f(p)^2$, so for $f * g$  to be completely multiplicative we'd need $f(p) g(p) = 2 f(p) g(p)$, and thus $f(p) = 0$ or $g(p) = 0$.

A counterexample is $I: \mathbb N\mapsto \mathbb N$ you have :
$$\forall n\in \mathbb N \quad \quad (I*I)(n)=nd(n) $$
and we know that $n\mapsto nd(n)$ is not completely multiplicative because $d$ the divisor counting function is not completely multiplicative.


How could we show that the set is equal to the empty set? I want to show that the intersection of any inductive set is empty since every inductive set contains the empty set.

I thought that we could do it like that:

We know that $B$ is an inductive set. So:

$$\varnothing \in B \wedge \forall x(x \in B \rightarrow x' \in B)$$

$$y \in \bigcap B \leftrightarrow \forall b \in B: y \in b$$

Since $\varnothing \in B$ we get that $y \in \bigcap B \leftrightarrow y \in \varnothing$.

But since there is no $y$ such that $y \in \varnothing$ we conclude that we cannot find a $y$ such that $y \in \bigcap B$.

But it isn't right, since we cannot just take one set to get the equivalence, right?

How else could we do this?

This looks right. So let me reorder your argument:

Now your argument: we know $\emptyset \in B$. So for an arbitrary $y$ we have two directions to prove:

if $y\in \bigcap B$ it follows $y \in \emptyset$ (By your property for $\bigcap$).
if we have $y\in \emptyset$, then we have a contradictory assumption (by the definition of the emptyset), and so we have $y \in \bigcap B$ in particular.

So in total: $\forall y (y \in \bigcap B \leftrightarrow y \in \emptyset)$. And by the axiom of extensionality, we have $\bigcap B = \emptyset$.

Alternatively you can prove it as follows: We know that the emptyset $\emptyset$ is the unique set with the property $\forall y(y\not\in \emptyset)$. Assume there is an element $y \in \bigcap B$, so we have $y \in \emptyset \in B$ by the property of $\bigcap B$, a contradiction. So $\forall y(y\not\in \bigcap B)$, and $\bigcap B = \emptyset$ by uniqueness.

EDIT: And to answer the title question "How could we show that the set is equal to the empty set?": You show that a set $x$ is equal to the emptyset by proving that $\forall y. (y\not\in x)$.

Counterexample: Consider $N = 29$ and $M = 20$ with $N \cap M = 20$


differentiation of a matrix X with respect to vec(X) Assume $X$ is a $n\times n$ matrix, I am looking for the solution of
$\frac{\partial X}{\partial vec(X)}$
Anyone that can shed light on this?

$\def\bb{\mathbb}\def\v{{\rm vec}}\def\M{{\rm Mat}}\def\d{{\rm diag}}\def\D{{\rm Diag}}\def\e{\varepsilon}\def\o{{\tt1}}\def\vcal#1{\vec{\cal #1}}\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}\def\p#1#2{\frac{\partial #1}{\partial #2}}$The
transformation between a matrix and its vectorized form can be indicated by functions
$$\eqalign{
x &= \v(X) \quad&\iff\quad &X=\M(x) \\
}$$
or by dot products with special third-order tensors
$$\eqalign{
x &= \vec\nu:X \quad&\iff\quad X&=\vec\mu\cdot x \\
x &= X:\vec\mu \quad&\iff\quad X&=x\cdot\vec\nu \\
}$$
The gradient of a matrix with respect to itself yields the fourth-order identity tensor
$$\vcal E = \frac{\partial X}{\partial X}
\quad\implies\quad \vcal E:dX\;=\;dX\;=\;dX:\vcal E$$
just as the gradient of a vector wrt itself yields the second-order identity matrix
$$I = \frac{\partial x}{\partial x}
\quad\implies\quad I\cdot dx\;=\;dx\;=\;dx\cdot I\quad$$
Combining the above ideas yields
$$\eqalign{
\frac{\partial X}{\partial x}
  &= \left(\frac{\partial X}{\partial X}\right):\vec\mu
  &= (\vcal E):\vec\mu  = \vec\mu \\
\frac{\partial x}{\partial X}
  &= \vec\nu:\left(\frac{\partial X}{\partial X}\right)
  &= \vec\nu:(\vcal E)  = \vec\nu \\
}$$
This can also be derived using the identity matrix
$$\eqalign{
\frac{\partial X}{\partial x}
  &= \vec\mu\cdot\left(\frac{\partial x}{\partial x}\right)
  &= \vec\mu\cdot(I) = \vec\mu \\
\frac{\partial x}{\partial X}
  &= \left(\frac{\partial x}{\partial x}\right)\cdot\vec\nu
  &= (I)\cdot\vec\nu  = \vec\nu \\
}$$
For $(m\times n)$ matrices the components of these zero-one tensors are
$$\eqalign{
{\vec\mu}_{jk\ell} &= {\vec\nu}_{\ell jk} \\
{\vec\nu}_{\ell jk} &= \begin{cases}
1\quad{\rm if}\;\;\ell=j+mk-m \\
0\quad{\rm otherwise} \\
\end{cases}
\\
{\vcal E}_{jkpq} &= \begin{cases}
1\quad{\rm if}\;\;j=p\;\;\&\;\;k=q \\
0\quad{\rm otherwise} \\
\end{cases}
\\
I_{\ell r} &= \begin{cases}
1\quad{\rm if}\;\;\ell=r \\
0\quad{\rm otherwise} \\
\end{cases}
\\
}$$
and the index ranges are
$$\eqalign{
1&\le\; j,p  \;&\le m
 &\qquad\big({\rm row\,index}\big) \\
1&\le\; k,q  \;&\le n
 &\qquad\big({\rm column\,index}\big) \\
1&\le\; \ell,r \;&\le mn
 &\qquad\big({\rm vectorized\,index}\big) \\
}$$
In some sense the third-order tensors are the more fundamental quantities,
since given
$\big(\vec\mu,\vec\nu\big)$
the identity tensors can be calculated as
$$\eqalign{
\vcal E &= \vec\mu\cdot\vec\nu
  \qquad&\big({\rm contract\,over\,vectorized\,index}\big) \\
I &= \vec\nu:\vec\mu
  \qquad&\big({\rm contract\,over\,row/col\,indexes}\big) \\
}$$


Convergence in $L^1$ problem. Problem: Let $f \in L^1(\mathbb{R},~\mu)$, where $\mu$ is the Lebesgue measure. For any $h \in \mathbb{R}$, define $f_h : \mathbb{R} \rightarrow \mathbb{R}$ by $f_h(x) = f(x - h)$. Prove that:
$$\lim_{h \rightarrow 0} \|f - f_h\|_{L^1} = 0.$$

My attempt: So, I know that given $\epsilon > 0$, we can find a continuous function $g : \mathbb{R} \rightarrow \mathbb{R}$ with compact support such that
$$\int_{\mathbb{R}} |f - g|d\mu < \epsilon.$$
We can then use the inequality $|f - f_h| \leq |f - g| + |g - g_h| + |g_h - f_h|$ to reduce the problem to the continuous case, so to speak, since the integral of the first and last terms will be $< \epsilon$. But now I'm stuck trying to show that
$$\lim_{h \rightarrow 0} \|g - g_h\|_{L^1} = 0.$$
I tried taking a sequence $(h_n)_{n \in \mathbb{N}}$ converging to $0$ and considering $g_n := g_{h_n}$, but I don't have monotonicity and the convergence doesn't seem to be dominated either, so I don't know what to do.

Any help appreciated. Thanks.

By your construction, $g$ is continuous and compactly supported.
Let $K$ be the support of $g$, and let $K_h=K\cup (h+K)$.
Then we have
$$
\int|g(x)-g(x-h)|\,\mathrm{d}x\leq |K_h|\|g-g_h\|_{L^\infty(K_h)}.
$$
For all $h>0$ sufficiently small we have $K_h\subset K_1$, and you can invoke uniform continuity of $g$.

I think you can use the fact that $f_h$ is in $L^1$ by translation invariance and the fact that $|f-f_h|\leq|f|+|f_h|$.  So now you have a function $g_n=|f-f_h|$ which converges to 0 and is bounded by an integrable function.


How can it be proved that a continuous function is bounded? Say $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$, what's the most concise way you know of to show that it's bounded?

I was thinking let $A=\{u : f(x) \text{ is bounded on }x<u\}$
Is there a way to show that $\sup(A)=b$?

Here is a "from scratch" attempt. Suppose $f$ were to be unbounded on $[a,b]$ yet continuous there.  Then $f$ is unbounded on one of $[a,(a + b)/2]$ or $[(a + b)/2, b]$.  Denote this interval by $I_1$.  Keep subdividing in this fashion to obtain a sequence of intervals $I_n$ so that $I_{n+1}\subseteq I_n$ for all $n$ and so that the length of $I_n$ is $$(b-a)/2^n.$$

Now write $I_n = [a_n, b_n]$ for each $n$.  The sequence $a_n$ is increasing and bounded by $b_1$ so it converges to a limit $l$.  Since the lengths of the $I_n$ converge to zero, we have $b_n\rightarrow l$.  By continuity, $f(l) = \lim f(a_n) = \lim f(b_n).$  By continuity, we can choose $\delta > 0$ so that $f(x) < f(l) + 1$ for $l - \delta < x < l + \delta$.

Pick $n$ so that $I_n \subseteq (l - \delta, l + \delta)$.  The function $f$ must be bounded on $I_n$, a contradiction of our construction.

The arabesque executed here has a feel very similar to that of the proof of Heine-Borel theorem.

The distance (function) of points from the set $\{0\}$ is uniformly continuous, so it is bounded on any compact set, so the compact set can be enclosed in a ball with a radius of that distance. Consider the compact set $f([0,1])$.


Can a regular grammar be ambiguous? An ambiguous grammar is a context-free grammar for which there exists a string that has more than one leftmost derivation, while an unambiguous grammar is a context-free grammar for which every valid string has a unique leftmost derivation.

A regular grammar is a mathematical object, $G$, with four components, $G = (N, \Sigma, P, S)$, where

$N$ is a nonempty, finite set of nonterminal symbols,
$\Sigma$ is a finite set of terminal symbols, or alphabet, symbols,
$P$ is a set of grammar rules, each of one having one of the forms:

$A \rightarrow aB$
$A \rightarrow a$
$A \rightarrow \varepsilon$
for $A, B \in N$, $a \in Σ$, and $\varepsilon$ the empty string, and

$S ∈ N$ is the start symbol.

Now the question is: Can a regular grammar also be ambiguous?

There do indeed exist ambiguous regular grammars. Take for example

$S\rightarrow A~|~B$

$A\rightarrow a$

$B\rightarrow a$

We know from Chomsky's Heirarchy of Languages that every regular language is also a context free language.

We also know every regular language is also a regular grammar.

Therefore, every regular grammar is also a context free grammar. Since CFGs can be abmbiguous, therefore by logic, some regular grammars can be ambiguous. (not ALL but there exists, in this case).


Why is identity element required for groups? I would like to know the necessity for having an identity element for every group.
I know the meaning of an identity element.

Most definitions of general objects in mathematics are a consequence of abstracting the properties of certain key examples that the object was originally developed to study so it could be extended to use in other fields. For example, the theory of measure and integration developed from precursor theories that were generalizations of ideas that ultimately had their roots in area and volume formulas that arose in construction problems in Ancient Greece.

Groups originally arose in the study of the symmetry properties of geometric objects under the classical isometry and similarity transformations of Euclidean geometry. It was Felix Klein who recognized that the set of isometries in Euclidean space-rotation,reflection and identity-formed a group under composition of functions since the operation is associative,yields a unique identity function and an inverse for each mapping which yields the identity map when composed with the mapping. For example, a reflection of an object through a mirror plane is it's own inverse since applying the map twice successively leaves the object unchanged and is therefore the identity mapping.

The generalization of the group of transformations of necessity gave an identity element in the definition of group when Leopold Kronecker proposed the abstract definition at the end of the 19th century.

By requiring having a unit $e\in G$ you achieve to things:

Avoiding dealing with the concept of empty sets as the base of the Group.
Avoiding dealing with empty operations (as an operation is a subset of $(G\times G)\times G)$.

Since the empty set and empty operation have all the other axioms of groups there could be an empty Group. The empty group will be contradicting almost every theorem.

Also, requiring a unit element gives a meaning to having an inverse element and to solving equations.

From the comments below I have missed one main thing. A Group is a structure that exists in nature a lot.


Why is a linear transformation of a cauchy sequence in a normed space also cauchy? Suppose we have a cauchy sequence $\{a_n\}$ in a normed vector space $V$. Given a linear transformation $T:V \rightarrow V$, is the sequence $\{T(a_n)\}$ also cauchy? Or is it true only for finite dimensional normed spaces? I'd be much obliged if someone could give a proof for this, preferably an elementary one.
Thanks!

Given $n,m\in \mathbb N; \exists p\in \mathbb N $  such that $||a_n-a_m||<\epsilon \forall m,n\geq p$

Now $||T(a_n)-T(a_m)||=||T(a_n-a_m)||<\epsilon$


How adding a joker makes events dependent? I have a question about probability. Imagine we have a deck of 40 cards with 4 different suits of ten cards. Define two events:

$A \equiv $ We take a card and it's an ace $\Rightarrow P(A)=\frac{1}{10}$

$B \equiv $ We take a card and it's a spade $\Rightarrow P(B)=\frac{1}{4}$

$P(A\cap B) = \frac{1}{40}$

We can see that, since we return the card each time, these events are independent:

$P(A\cap B) = P(A) \cdot P(B)$

If we now add a joker to the deck, which can take any card value, then:

$P(A) = \frac{5}{41}$

$P(B) = \frac{11}{41}$

$P(A\cap B) = \frac{2}{41}$

And we can see that now the events aren't independent!

$P(A\cap B) \ne P(A)\cdot P(B)$

My question is, why does adding the joker make the events independent? How is it different from just having one more ace of spades? I mean, before adding the joker we already had a card that belonged to both events (the ace of spades)...

Thanks!

Note:  the rules aren't entirely clear.  Is the Joker always interpreted as $A\spadesuit$?  If not, what rules determine how it is interpreted?

To be clear:   If you added a second $A\spadesuit$ then the events aren't independent either.  You'd get $$P(A)=\frac 5{41}\quad P(B)=\frac {11}{41}\quad P(A\cap B)=\frac 2{41}$$ just as before.

Indeed, having doubled the $A\spadesuit$, either directly or by the Joker, you make it so that drawing an Ace is evidence that the card is a spade, and drawing a spade is evidence that it is an Ace.

Specifically,  before you draw anything, the probability that a random draw will be a spade is $\frac {11}{41}=0.268$.  If I tell you that you have drawn an ace, however, the probability that it is a spade is now $\frac 25=.4$  so drawing an ace is strong evidence that you have drawn a spade.

We take a card and its an ace

This is not a description of an event to which you can assign probability.
What is the probability I will draw a card today and it will be an ace?
Maybe I will not draw any cards today. How do you compute the probability of that?

Now introduce a second description, "We take a card and its a spades,"
and the situation is even more murky.
Did we put the first card back and then draw another? Or are the two sentences talking about things that would occur simultaneously, that is we draw just one card from the deck one time, and look at that single card, which might be an ace and might be a spade.

A better way to describe probabilities is to set up exactly the experiment/trial (or experiments/trials) that we will perform, and then specify the results as events.
So either there is a single trial ("shuffle these $40$ cards, draw one, and look at it") or two trials ("shuffle these $40$ cards, draw one, look at it, then shuffle the cards again, draw a card, and look at it").

For a double trial, with one joker in the deck, there are not just $41$ possible equally-likely, distinguishable outcomes; there are $41\times41$ of those outcomes.
For each card we could draw the first time there are $41$ possible cards we could draw the second time. The number of ways to draw an ace the first time and a spade the second time is $5\times 11$ ways. So if we actually make two draws then the events are independent.


Area of intersection of a circle with a rectangle I want to find the area of a given circle that comes under the region of a given rectangle. I searched many posts on stackoverflow but they are not satisfying. I followed this post
http://www.eex-dev.net/index.php?id=100
But it doesn't seem accurate. I'd appreciate if someone provides me a good accurate solution atleast upto 10^-6 precision. Thanks.

Decompose your intersection area into polygons which are completely inside the circle, and circular segments formed by a chord and a part of the arc. Use e.g. the shoelace formula to compute the area of the polygoms, and the segment area formula for the segments. Compute these to whatever accuracy you want.

The decomposition will need to make many case distinctions. But choosing integration bounds and integration formulas isn't really any easier, in my opinion. There just is no simple way around all those case distinctions.

You put the circle at the origin. The idea is to divide the rectangle
into four rectangles, and then replace all rectangles
that do not fall on the first quadrant by an equivalent rectangle, but located on the first quadrant. Then you deal with each
case according to the number of vertex inside the circle. At the end all areas are given in terms of a single function F(U,V)
that arises from the evaluation of the intersection area in the second case,
which is the case when only on vertex is inside the ellipse. Then you sum the four areas. The matlab code is:

A=2;        % Circle of radius 2 (or an ellipse of semi axes A,B)
B=2;
L_x=0.5;    % The  width
L_y=2.75;   % The height

x_1=0;     %bottom left corner (x_1,y_1)
y_1=-1;

%This function makes the calculation

int_area(x_1,y_1,L_x,L_y,A,B)

%The function definition
function [suma]=int_area(x_1,y_1,L_x,L_y,A,B)

x(1)=x_1;
y(1)=y_1;
% each one of the remaining vertex

x(2)=x(1);
y(2)=y(1)+L_y;
x(3)=x(1)+L_x;
y(3)=y(1)+L_y;
x(4)=x(1)+L_x;
y(4)=y(1);

% the center of the rectangle

x_m=x(1)+L_x/2;
y_m=y(1)+L_y/2;
suma=0;

% The original rectangle was divided in four rectangles
% with the new vertex coordinate of the closest vertex to the origin given by (a,b)
% according to the article http://www.dtic.mil/dtic/tr/fulltext/u2/410103.pdf
% a (x- coordinate ) b( y coordinate)
% c new width
% d new height

for i=1:4

a(i)=max([ 0 (-1)^( 1/2*(i^2-i) )  x_m - L_x/2 ]);
b(i)=max([ 0 (-1)^( 1/2(i^2+i-2) )y_m - L_y/2 ]);
c(i)=max([ 0 (-1)^( 1/2(i^2-i) )  x_m + L_x/2 - a(i) ]);
d(i)=max([ 0 (-1)^( 1/2(i^2+i-2) )*y_m + L_y/2 - b(i) ]);

%in case the width and the height are different from 0 otherwise it contributes area zero
if (c(i) !=0 && d(i) !=0 )

%this is the ellipse equation evaluated on each vertex
eq_1=(a(i)/A)^2 + (b(i)/B)^2;
eq_2=(a(i)/A)^2 + ((b(i)+d(i))/B)^2;
eq_3=((a(i)+c(i))/A)^2 + ((b(i)+d(i))/B)^2;
eq_4=((a(i)+c(i))/A)^2 + (b(i)/B)^2;

% Area of intersection for each case according to the number of vertex inside the circle (or ellipse)
% S intersection area

if( eq_1 >=1  && eq_2 >=1 && eq_3 >=1 && eq_4 >=1)
S=0;        %case_I:  All vertex outside
end

if( eq_1 <1  && eq_2 >=1 && eq_3 >=1 && eq_4 >=1)
S=AB/2 F(a(i)/A,b(i)/B);   %case_II: vertex I inside
end

if( eq_1 <1  && eq_2 >=1 && eq_3 >=1 && eq_4 <1)
S=AB/2 (F(a(i)/A,b(i)/B)-F((a(i)+c(i))/A,b(i)/B));   %case_III: vertex 1 and 4 inside
end

if( eq_1 <1  && eq_2 <1 && eq_3 >=1 && eq_4 >=1)
S=AB/2 (F(a(i)/A,b(i)/B)-F(a(i)/A,(b(i)+d(i))/B));   %case_IV: vertex 1 and 2 inside
end

if( eq_1 <1  && eq_2 <1 && eq_3 >=1 && eq_4 <1)
S=AB/2 (F(a(i)/A,b(i)/B)-F((a(i)+c(i))/A,b(i)/B) - F(a(i)/A,(b(i)+d(i))/B));   %case_V: vertex 3 outside
end

if( eq_1 <1  && eq_2 <1 && eq_3 <1 && eq_4 <1)
S=c(i)*d(i);     %case_VI: all inside
end

else
S=0;    %in case the width or the height of the new rectangle is zero
end

suma=suma+S;   %the total area is suma
end

end

function [res] =  F (U,V)
  res=asin( sqrt(1-U^2)sqrt(1-V^2) -UV ) -Usqrt(1-U^2)-Vsqrt(1-V^2)+2*U*V;
end

Note: for some reason some '*' (multiplication symbol) doesn't appear on the screen, but matlab uses it. All after the sign '%' is a comment

Best Regards

Ed.


Let $H$ be a subgroup of $G$. Define $N(H) = \{ a \in G | aHa^{-1}=H \}$. Show $H \subset N(H)$. Let $H$ be a subgroup of $G$. Define $N(H) = \{ a \in G | aHa^{-1}=H \}$. Show $H \subset N(H)$.

I was able to show $N(H) $ is a subgroup of $G$ but now I am unsure about showing the relation $H \subset N(H)$.

My attempt:

Let $x \in H$. Let $h_1 \in H$ be some element in $H$.

Then $x h_1 x^{-1} = (x h_1) x^{-1} = h_2 x^{-1}$ where $x h_1 = h_2 \in H$. Now $h_2 x^{-1}= h_3 \in H$.

So we have shown for any $h_1 \in H$, $\exists h_3 \in H$ s.t. $x h_1 x^{-1} = h_3$. Hence, $x H x^{-1} = H$ and thus $x \in N(H) \implies H \subset N(H)$.

Did I make any mistake or leave out any necessary detail?

Edit: as per the comment below, I have incorrectly claimed that $xHx^{-1} = H$ whereas I had only shown $xHx^{-1} \subset H$. So for the other part,

Let $h \in H$. Consider $x^{-1}hx = (x^{-1}h)x = h_1 x $ where $h_1 = x^{-1}h \in H$. Similarly $h_1x = h_2 \in H$.

So $x^{-1}hx = h_2 \implies h = x h_2 x^{-1} \in xHx^{-1} \implies H \subset xHx^{-1}$.

Therefore, $H = xHx^{-1}$ and the rest of the proof follows from here.

see that $a\in H\implies aH=Ha=H\implies aHa^{-1}=H$ so,   $a\in H\implies a\in N(H)$ hence $H\subset N(H)$.


How prove this $ \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}\geq 1.$ Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$, Prove that

  $$ \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}\geq 1.$$

This problem is from  http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=555716

@Calvin Lin  Thank you

Let $\displaystyle A = \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}$ and $\displaystyle B = \sum_{cyc}a^2(a+3b+5bc)$.

Then by Hölder's inequality we have $A^2B \ge (a+b+c)^3 = 27$.

So it is sufficient to prove that $B \le 27$

$$B = \sum_{cyc}a^3 + 3 \sum_{cyc}a^2b+5\sum_{cyc}a^2bc$$

As $\displaystyle \sum_{cyc}ab^2 \ge 3abc$ by AM-GM, we have

$$B \le \left(\sum_{cyc}a^3 + 3 \sum_{cyc}a^2b + 3 \sum_{cyc}ab^2 + 6abc\right) - 15abc + 5\sum_{cyc}a^2bc \\
 = (a+b+c)^3 - 5abc (3-\sum_{cyc}a) = 27$$

We know the one result if sum of two numbers $a$ and $b$ is constant, the max. of $ab$ occurs when both $a$ and $b$ equal. Eg. if $a+b=6$, then $\max\{ab\}=9$. Use this result, we get proof.


Proving that $8^x+4^x\geq 5^x+6^x$ for $x\geq 0$. I want to prove that $$8^x+4^x\geq 6^x+5^x$$ for all $x\geq 0$. How can I do this?

My attempt:

I try by AM-GM: $$8^x+4^x\geq 2\sqrt{8^x4^x}=2(\sqrt{32})^x.$$

However, $\sqrt{32}\approx 5.5$ so I am not sure if $$2(\sqrt{32})^x\geq 5^x+6^x$$ is true.

Also, I try to compute derivatives but this doesn't simplify the problem. What can I do?

Hint. Let $f(t)=t^x$ then by the Mean Value Theorem there is $t_1\in (6,8)$ such that
$$f(8)-f(6)=f'(t_1)(8-6)\Leftrightarrow 8^x-6^x=2xt_1^{x-1}.$$
Similarly there is $t_2\in (4,5)$ such that
$$f(5)-f(4)=f'(t_2)(5-4)\Leftrightarrow 5^x-4^x=xt_2^{x-1}.$$
It remains to show that for $x\geq 0$
$$2xt_1^{x-1}\geq xt_2^{x-1}.$$

The best and the easiest way to prove it is using Induction .

 Base Case :

For $x=1$ , $ 8 + 4 \gt 6+5$

Induction Step :

Let us assume that $8^k +4^k \ge5^k + 6^k$ . This implies that $$8^k \ge (5^k + 6^k) - 4^k$$

Multiplying both the sides by $8$ , we get

$$\color{#f14}{8^{k+1} \ge 8\cdot5^k +8\cdot6^k -8\cdot4^k} \quad \quad \text{(1.)}$$

Now all that is left is to prove that the R.H.S is bigger than $5^{k+1}+6^{k+1}-4^{k+1}$.We prove it by assuming it is true . Then ,

$$\begin{align}8\cdot5^k +8\cdot6^k -8\cdot4^k & \ge 5^{k+1}+6^{k+1}-4^{k+1} \\
5^k(8-5) + 6^k(8-6) +4^k(4-8) & \ge 0  \\
3\cdot 5^k +2\cdot6^k - 4\cdot4^k & \ge 0 \\
\color{#2c0}{\left(\frac 34\right)\cdot \left(\frac 54\right)^k + \left(\frac 24\right)\cdot \left(\frac 64\right)^k }& \color{#2c0}{\ge  0 }\quad \quad \text{(2.)}\end{align}$$

which is obviously true for $k \ge 0$ . Hence our initial assumption was true.

Now combining $(1.)$ and $(2.)$ , we get ,

$$\color{#f14}{8^{k+1}} \ge \color{navy}{8\cdot5^k +8\cdot6^k -8\cdot4^k} \ge \color{#2c0}{5^{k+1}+6^{k+1}-4^{k+1}} $$

or $$ 8^{k+1}  \ge 5^{k+1} + 6^{k+1} -4^{k+1} \implies \boxed { 8^{k+1} + 4^{k+1} \ge 5^{k+1} + 6^{k+1}}$$

Which completes our induction.


cubic polynomial equation $f(x)=0$ has a uniqe solution I would appreciate if somebody could help me with the following problem

Q: Let $f(x)=ax^3+bx^2+cx+d(a,b,c,d\in\mathbb{R},a\neq 0)$

prove that if $f(x)=f'(x)q(x)+r(r\in\mathbb{R})$,$q(x)$ a polynomial then equation $f(x)=0$ has a uniqe solution in $\mathbb{R}$

Letting $q(x)=ex+f$ where $e,f\in\mathbb R$, we have
$$ax^3+bx^2+cx+d=(3ax^2+2bx+c)(ex+f)+r.$$
Then, we get the followings :
$$a=3ae$$$$b=3af+2be$$$$c=2bf+ce$$$$d=cf+r$$
Hence, we get
$$e=\frac 13, f=\frac{4}{9a}, c=\frac{b^2}{3a}, d=\frac{b^3}{27a^2}+r.$$
Hence, since
$$f(x)=ax^3+bx^2+\frac{b^2}{3a}x+d,$$
we have
$$f^\prime(x)=3ax^2+2bx+\frac{b^2}{3a}.$$

Since the discriminant of $f^\prime(x)=0$ is $0$, we now know that we reach what we want.

Otherwise $f$ must has only two distinct roots in $R$ of multiplicity 1 and 2 respectively. Let $x_0$ be the former and $x_1$ the latter. Then $f'(x_1)=0$, and from the equation$$f(x)=f'(x)q(x)+r$$ we find that$$r=0$$ Thus $$f(x)=f'(x)q(x)$$ Since $f'$ has order 2, then there must be another real root that is different from $x_1$of $f'$, let's denote it by $x_2$. Then $x_2$ is also a root of $f$ having multiplicity 2, a contradiction.


How to find the limit $\lim_{x \rightarrow 1^+}\left (1 - \frac{1}{x}\right)^x \left( \log\left(1 - \frac{1}{x}\right) + \frac{1}{x - 1}\right)$ How can I find the limit $\lim_{x \rightarrow 1^+} \left (1 - \frac{1}{x}\right)^x \left( \log\left(1 - \frac{1}{x}\right) + \frac{1}{x - 1}\right)$?

I tried turning into a fraction so that I could apply L'Hopital's rule:
$\left(1 - \frac{1}{x}\right)^x \left(\frac{\log\left(1 - \frac{1}{x}\right)(x-1) + 1}{x - 1}\right)$

But that didn't seem to get me anywhere. Thanks.

Change variables to $y=\frac{x}{x-1}$, then $x=\frac{y}{y-1}$. Then your limit becomes:
$$
\lim_{y\to+\infty}\exp\left[\log (y-1-\log y)-\frac{y}{y-1}\log y\right]
$$
Now subtract $\log y$ from each of the terms to obtain:

$$
\lim_{y\to+\infty}\exp\left[\log \left(1-\frac{1+\log y}{y}\right)-\frac{\log y}{y-1}\right]=\lim_{y\to\infty}\exp(\log (1-0)-0)=\boxed{1}.
$$


entries of incidence matrix of undireccted graph What is the sum of the entries in a row (respectively column) of the incidence matrix for an undirected graph?

I didn't fully understand the question, does it require a number as an answer or just an explanation? I tried to solve it, but all I got is that the rows are the vertices and the columns are the edges and I couldn't proceed

The sum on each column of the matrix is 2 because each edge is connecting two vertices.


What is the value of $(-1)^\frac{4}{3}$? I was trying to plot the graph of $y=x^\frac{4}{3}$. However, I tried two online plotters, both gave me curves only on the right side of the y-axis. There is nothing on the left side of the y-axis. Shouldn't it be a curve symmetric to the y-axis? Similarily for $y=x^\frac{5}{3}$ which I thought is a function symmetric to origin but only has value for nonnegative x.

I tried to use google to compute $(-1)^\frac{4}{3}$ and it automatically gives me $-0.5 - 0.866025404 i$ instead of $1$. And $(-1)^\frac{5}{3}$ got a answer of $0.5 - 0.866025404 i$ instead of $-1$. Why does the result include an imaginary part?

Fractional powers of negative numbers aren't uniquely defined.

There are three cube roots of $-1$:  $-1$, $\frac12+\frac{\sqrt{3}}2i$, and $\frac12-\frac{\sqrt3}2i$.

The answer given by Google for $(-1)^{4/3}$ was the fourth power of the middle one of those three.

The function is defined for positive and negative values of $x$.  The graph shows the real and imaginary parts:

$${\rm Re}[(-1)^{4/3}] = -\frac{1}{2}$$

$${\rm Im}[(-1)^{4/3}] = - \frac{\sqrt{3}}{2}$$

Given that there is no unique way to compute a partial root of a negative number, Mathematica seems to assume the most general complex form.

There is no reason that the function should be (or is) symmetric with respect to the interchange $x \leftrightarrow -x$.


Bounds of $\frac{\ln(x+1)}{x}\ \forall x>0$ $f:(0,\infty)$,
$f(x)=\frac{\ln(x+1)}{x}$
Prove that for $\forall\ x>0$ that $f(x) \in(0,1)$. I calculated the derivative of $f(x)$: $f'(x)=\frac{\frac{x}{x+1}-\ln(1+x)}{x^2}$ which I think simplifies to $\frac{x^3}{x+1}-x^2(\ln(1+x))$. I have no idea what to do next, I can't find the roots of this equation and I don't see any connection as of why it should be bounded by 0 and 1.

I hope I formatted this well, I don't usually post here but I am really curious how could I solve this kind of exercise.

For $x \in (0,\infty)$, you have
$$\ln (x+1) = \int_0^x \frac{dt}{1+t}$$

Hence

$$0 \le f(x) = \frac{1}{x}\int_0^x \frac{dt}{1+t} < \frac{1}{x}\int_0^x \ dt =1$$

As all considered maps are continuous.

It is much simpler to show that the derivative of $\ln(x+1)$ is always in $(0,1)$ (and it is continuous).
Then $\ln(x+1)$ is strictly between $0$ and $x$, think about it.


Find all triples $(x,y,z)\in \Bbb{R}$ that satisfy the following conditions: The question is to find all real numbers solutions to the system of equations:

$y=\Large\frac{4x^2}{4x^2+1}$,
$z=\Large\frac{4y^2}{4y^2+1}$,
$x=\Large\frac{4z^2}{4z^2+1}$,

This seems simple enough, so I tried substituting the values of x, y and z into the different equations but I only ended up with a huge degree 8 equation which clearly doesn't seem like the right approach. I really have no idea on how to go about solving this if substitution is not the answer.
Any help would be greatly appreciated :)

We can note that $4a^2+1 \ge 4a \ \ \forall a \in R$. Also, since $\frac{4a^2}{4a^2+1} \ge 0$, then we know that $x,y,z \ge 0$

Therefore $y=\frac{4x^2}{4x^2+1} \le \frac{4x^2}{4x}=x$ for nonezero values of $x$. Similarly $z \le y$ and $x \le y$. Therefore $x \le y \le z \le x \implies x=y=z$.

Now solving equation $a=\frac{4a^2}{4a^2+1} \implies 4a^2+1=4a \implies a = \frac{1}{2} \implies x=y=z=\frac{1}{2}$.

Finally, we assumed that numbers are non-zero, so we should include solution $(0,0,0)$

$\bullet\; $ Clearly $x=y=z=0$ are the solution of system of eqn

$\bullet\; $ If $x,y,z\neq 0\; $ Then

$\displaystyle y=\frac{4x^2}{4x^2+1}\Longrightarrow \frac{1}{y}=1+\frac{1}{4x^2}$,
$\displaystyle z=\frac{4y^2}{4y^2+1}\Longrightarrow \frac{1}{z}=1+\frac{1}{4y^2},$
$\displaystyle x=\frac{4z^2}{4z^2+1}\Longrightarrow \frac{1}{x}=1+\frac{1}{4z^2}$,

Adding all three

$\displaystyle \bigg(1-\frac{1}{2x}\bigg)^2+\bigg(1-\frac{1}{2y}\bigg)^2+\bigg(1-\frac{1}{2z}\bigg)^2=0$

which is possible when

$\displaystyle 1-\frac{1}{2x}=0,1-\frac{1}{2y}=0,1-\frac{1}{2z}=0$

System of equation have $\displaystyle (x,y,z)=\bigg(\frac{1}{2},\frac{1}{2},\frac{1}{2}\bigg)$ as solution also


Calculating probability of game ending after $n$ flips Two players A and B flip a coin sequentially. The game finishes when the sequence TTH is formed and player A wins or the sequence HTT is formed and player B wins. What is the probability that the game will finish at the $n$-th flip?

What I did:

A wins iff the sequence is $n-1$ T's followed by a single H : $\frac{1}{2^n}$
B wins iff the sequence ends with HTT and we have no two consecutive T's in the first $n-3$ flips: this happens (I think) with probability $\frac{F_{n-2}}{2^n}$ where $F_{n}$ is the $n$-th Fibonacci number. I proved this by induction.
Thus, the total probability is $\frac{F_{n-2}+1}{2^n}$

Can someone verify that this is correct and/or share how you would solve this problem?

If the answer is correct, then by summing over all $n$ we can obtain an interesting identity involving the Fibonacci numbers!

In the first three tosses, one of the players will win is $\frac{2}{8}$ i.e HTT and THH are the winning flips out of $2^3$.  Thus the proability the game will end in 3 tosses is $\frac{1}{4}$.  In an addtional toss, one of the players will win is $\frac{4}{16}$, i.e THTT, HHTT, TTHH, HTHH are the winning flips out of $2^4$. Thus the probability that the game will end in 4 tosses is equal to game not ending in the first three tosses and ends in the fourth toss and thus  equal to $\frac{3}{4}.\frac{4}{16}$ $=(\frac{3}{4})^{4-3}.\frac{1}{4}$  Extending the logic, you will have the probability that the game will end ( in other words one of the players will win) in n flips is $=(\frac{3}{4})^{n-3}.\frac{1}{4}$. The book is correct in its solution.


Incorrect proof of the infinities between 0 and 1 and 0 and 2 In reading another question (Explaining Infinite Sets and The Fault in Our Stars) it got me thinking about the way that you can prove that the number of numbers between 0 and 1 and between 0 and 2 are the same. (apologies if my terminology is a bit woolly and imprecise, hopefully you catch my drift though).

The way it is proved is that you can show that there is a projection of all the numbers on [0,1] to [0,2] and vice versa. I'm good with this.

However I then got to thinking that you can also create a projection that takes all the numbers from [0,1] and maps them to two numbers from [0,2] by saying for a number x it can go to x or x+1. This is reversible to so you can say that you can find a pair of numbers in [0,2] such that they differ by one and the lowest is a member of [0,1].

Why is it that this doesn't prove that there are twice as many numbers in [0,2] than in [0,1]. It seems to me that this is the crux of why it runs counter to intuition but I can't work out the flaw.

Or is it just in the nature of infinity that infinity*2 is still the same infinity and thus its just that infinite is "weird"?

The crux is the fact that you don't specify how you measure "size" of an infinite set.

In the case of the real numbers, and even more so when we consider intervals, we can measure their length. In which case $[0,2]$ is twice as long as $[0,1]$ and therefore twice as large.

If you want a "raw" measurement of how large a set is, then you reduce to the notion of cardinality, in which case we only care about bijections and therefore $[0,1]$ and $[0,2]$ and in fact $\Bbb R$ itself all have the same size.

There is still a problem with your argument. The fact that you can map each number to two different numbers (or rather, map exactly two numbers to the same number) is not a good argument for "there are twice as many elements" (which implies a strict inequality, to my ears anyway). For example, consider $\Bbb N$ and map every even element $2k$ to $k$, and every odd element, $2k+1$ to $k$ as well, and of course $\Bbb N$ does not have strictly more elements than $\Bbb N$.

You also have that each natural number has exactly two numbers which map to it, but it still doesn't mean that there are twice as many natural numbers as there are natural numbers. That's just not good mathematics.

For finite sets the simplest way is to count the elements to have a proper concept of size.

For infinite sets it gets more tricky. In naive set theory one compares sets by trying to establish one to one mappings, like you wrote, in which case the sets are considered to be of the same size.

Looking at $d = b - a$ to compare intervals $[a,b]$ or $[0,1] \subset [0,2]$ does not help in the context of comparing the number of their elements relative to each other.
They all end up as large as $[0,1]$ (for $a\ne b$).

Regarding the last line of the question:

While $2 \cdot \infty$ might yield just $\infty$ in your case and is counter-intuitive, or $\mbox{card}(\mathbb{N}^n) = \mbox{card}(\mathbb{N})$ which I found remarkable (link), you will find funny results for $\infty - \infty$ (see certain quantum field theoretic calculations) and enlarge already infinite sets $A$ with the power set construction $2^A$, getting into different orders of infinity (which remarkably is the reason why there are uncomputable programs).

Why is it "weird" or counter-intuitive? Personally, I tend to the biological explanation. It is us not the subject. Our brain is working fine for our environment and us living in it, which is

finite,
mostly flat,
rather slow (compared to the speed of light),
has not that much gravity (compared to the conditions on the surface of a neutron star),
is not too small and not too large

So we seem to have more difficulties grasping everything which is not like that, like infinities, theory of relativity and quantumn mechanics.

If we had to grapple the last billion years with infinite objects in our physical world, I believe we wouldn't be that surprised often.


If in a semigroup $S$, $\forall x \exists ! y:xyx=x$, then $S$ is a group If for all $x$ in a semigroup $S$, there exists a unique $y$ such that $x y x=x$, then $S$ is a group. (Not to be confused with inverse semigroup, where only $y$ satisfying both $xyx=x$ and $yxy=y$ is unique)

After tring with no result, I used Prover9 to find a proof. I did get one but it was very hard to understand (possible to go through once but really hard to remember what the point is).

Is there any somewhat comprehensible or conceptual proof to this? Is there a theory underlying this?

Hint: You just need to find unique inverses for every element, and the identity element will arise by making $xx^{-1}$ with any element.

If for all $x$, there is a unique $y$ such that $xyx=x$, then for all $x$ there is a unique $y$ such that $yx$ does nothing to the element... Can you find the identity element and the ivnerses from here?


Probability problem. There are 16 disks in a box. Question: There are 16 disks in a box. Five of them are painted red, five of them are painted blue, and six are painted red on one side, and blue on the other side. We are given a disk at random, and see that one of its sides is red. Is the other side of this disk more likely to be red or blue?

There are $5 \times 2 + 6 \times 1 = 16$ red sides and of the $16$ red sides, $10$ have red on the other side, so the probability the other side is red is $\frac{10}{16}=\frac{5}{8} \gt \frac{1}{2}$, and so the other side of this disk more likely to be red.

Alternatively, if you are given a disk at random, then it is more likely to be same colour on both sides (and red and blue have equal probabilities) , so if you see a red side then the other side is more likely than not red.

there are 5 red discs with red on the other side ... and 6 red discs with blue on the other side ...
makes sense that you are more likely to find blue on the other side ... unless you lucky at blackjack


How to prove that $x \rightarrow e^{1/x}$ is not a restriction of any real distribution to $ \mathbb {R}_+$? This is an excercise 2.2 from Hormander, vol. I:

Does there exist a distribution $u$ on $\mathbb{R}$ with the restriction
$x \rightarrow e^{1/x}$ to $\mathbb{R}_+$?

The answer, provided in the book, is "No". I am trying to "cook up" appropriate test function(s) such that $ \int \phi(x)e^{1/x} \leq C\sum_{\alpha \leq k} \sup\left|\partial^{\alpha}\phi\right|$ for no $k$, and I'm not sure at all what function(s) to take. What is the appropriate function? Is there a general method to come up with just right test functions?

Following Fedja's hint, let $\phi$ be a nonnegative test function supported in $(1,2)$ such that $\phi=1$ on $(5/4,7/4)$. For $b>0$, let $\phi_{b}(x):=\phi(bx)$. Observe that

$$\int_{\mathbb{R}}\phi_{b}(x)e^{1/x}dx=b^{-1}\int_{\mathbb{R}}\phi(x)e^{b/x}dx,\qquad\forall b>0$$

Suppose that there is a distribution $u\in\mathcal{D}'(\mathbb{R})$ of order $k$ whose restriction to $\mathbb{R}^{+}$ is $e^{1/x}$:

$$\left|\langle{u,\psi}\rangle\right|\leq C\sum_{\alpha\leq k}\left\|\partial^{\alpha}\psi\right\|_{\infty},\qquad\forall\psi\in C_{c}^{\infty}(\mathbb{R})$$

and in particular,
$$\left|\langle{u,\phi_{b}}\rangle\right|\leq C\sum_{\alpha\leq k}\left\|\partial^{\alpha}\phi_{b}\right\|_{\infty}\leq C\sum_{\alpha\leq k}b^{\alpha}\left\|\partial^{\alpha}\phi\right\|_{\infty}\leq Cb^{k}\sum_{\alpha\leq k}\left\|\partial^{\alpha}\phi\right\|_{\infty},\qquad\forall b\geq 1$$
You can check that there is a constant $C'>0$ such that

$$\dfrac{e^{b/x}}{b^{k+1}}\geq C'\dfrac{b}{x^{k+2}},\qquad\forall x>0$$

Whence,

$$b^{-k}\left|\langle{u,\phi_{b}}\rangle\right|=b^{-k-1}\int_{\mathbb{R}}\phi(x)e^{b/x}dx\geq C' b\int_{1}^{2}\phi(x)x^{-k-2}dx$$

Since the RHS tends to $\infty$ as $b\rightarrow\infty$, we obtain a contradiction.

I think the argument is simpler than you expect, unless I make a terrible mistake...

Just pick any test function with the property that $\phi(0) >1$ ($\phi(0)\neq 0$ is actually enough).

Since $\phi(0)>1$ there exists some $a$ so that $\phi >1$ on $[0,1]$.

Then

$$\int \phi(x) e^{\frac{1}{x}}> \int_0^a e^{\frac{1}{x}} dx = \int_\frac{1}{a}^\infty \frac{e^u}{u^2}du = \infty $$


What's the difference between an initial value problem and a boundary value problem? I don't really see the difference, because in both case we need to determine y and the values of the constants. The only difference is that we give the value of y and y' in the former and the value of either 2 y or 2 y' in the latter.

I solve both problems the same way. I don't really understand the theory, I guess.

For a simple example (second order ODE), an initial value problem would say $y(a)=p$, $y'(a)=q$.

A boundary value problem would specify $y(a)=p$, $y(b)=q$.

In initial value problem we always want to determine the value of f(x)and f'(x) at initial point it may be 0 or something else but initial like f(1)=3 and f'(1)=2 then we can determine the constant.But in boundry value problem the condition will in form of a interval i.e.f(0)=3,f(2)=5


How to prove a function from $\mathbb N\times \mathbb N$ to $\mathbb N$ is bijective. I am having trouble with this problem:
$f\colon \mathbb N\times \mathbb N \rightarrow \mathbb N$ is defined by

$f(i,j)=\dfrac{(i+j-1)(i+j-2)}{2}+i$.

How do you prove that $f$ is a bijection from $\mathbb N\times \mathbb N$ to $\mathbb N$?

Work: I tried to set $f(i,j)=f(a,b)$. Assuming that $f(i,j)\neq f(a,b)$, then I assume $f(i,j)< f(a,b)$and that $i+j=m$ and $a+b=m+r$ for some remainder $r$. I replaced these into my equation and try to obtain a contradiction thus proving that $i=a, j=b$. However, I get stuck after this part and I do not know of a better way of doing this.

Please help.

Instead, let $a=i+m$ and $b=j+n$. Then $(i+j-1)(i+j-2)+2i=(i+j+m+n-1)(i+j+m+n-2)+2i+2m$. Try multiply this out and collecting terms. You want that $m,n=0$.

You might ask why is this a more natural substitution. When you make a substitution, you usually don't want to substitute for only some instances of a variable. Notice that in your set up you still have an $a$ term on the end, despite substituting for some of the $a$s, and likewise for $i$

$$T(n)=\frac{n(n+1)}{2}$$
$$f(i,j)-T(i+j-2)=i > 0$$
$$T(i+j-1)-f(i,j)=j-1≥0$$
$$T(i+j-2)<f(i,j)≤T(i+j-1)$$
Suppose $f(i,j)=f(a,b)$. Now, $T(i+j-2)<f(i,j)≤T(i+j-1)$ and $T(a+b-2)<f(i,j)≤T(a+b-1)$. Again, $T(n)$ is a strictly increasing sequence, hence $a+b=i+j$. But then the relation $f(i,j)=f(a,b)$ gives $i=a$ and $j=b$, hence f is one to one. I leave the work of proving that $f$ is onto upto you.


If $\gcd(x,y)=1379,\operatorname{lcm}(x,y)=n!$ then find $n$ If $\gcd(x,y)=1379,\operatorname{lcm}(x,y)=n! $ then find $n$  ($x,y,n$ are positive integers).
I tried using the relation: $\gcd(x,y)\cdot\operatorname{lcm}(x,y)=xy$ and also wrote $1379=7\cdot197$ but nothing more.

Since $\gcd(x,y)=7\cdot 197$, we can assume $x=7\cdot 197a$, and $y=7\cdot 197b$, where $\gcd(a,b)=1$. Then $\text{lcm}(x,y)=7\cdot 197\cdot ab=n!$. Thus, $n\geqslant 197$. Actually, we can assign different prime factors of $\frac{n!}{7\cdot 197}$ to $a$ and $b$ respectively to make sure they are coprime.

$n$ can be any integer which is no smaller than $197$.


Is every proper subset of the real numbers bounded by the set of real numbers? Trying to see if I understand the definition of bounded. Are these true:

$(-1, 1)$ is bdd above by $\mathbb{R}$ (and below)
$(-1, 1)$ is bdd above by $(-1, 1]\, $,  and $1$ is the supremum of $(-1,1)$
$[-1, 1]$ is bdd above and below by itself
$(-1, 1)$ is not bdd above by $[-1,1)$
$\mathbb{R}^+$ is not bdd above by $\mathbb{R}$ but it is bounded below, with an infimum of $0$.

Any mistakes or other elementary cases I should consider?

I have up-voted the answer by Xander Henderson. But since the subject line of the question says "Is every proper subset of the real numbers bounded by the set of real numbers?" it seems like a good idea to add that the set $\mathbb Z$ of all integers is a proper subset of $\mathbb R$ that is not bounded either above or below within $\mathbb R.$ (And I wrote "within $\mathbb R,$" not "by $\mathbb R.$")

That is why the real number system is 'Complete'.
In case of open interval, say (a,b) we can take the deleted neighbourhood of a and b. Each member of the neighbourhood system lies in the set R. But, we cannot 'fix' a upper bound or lower bound.
Sorry for giving a rather informal answer, but I hope it'll be of use.


Is it possible to eliminate a contradiction without recourse to the principle of explosion? I'd like to derive the following inference rule:

$$
\frac{p\lor(q\land\neg q)}{p}\quad\text{[ContradictionElimination]}
$$

I assumed that I could do this minimally somehow, however it turns out I need an alternative form of the principle of explosion. My derivation is:

Rule (ContradictionElimination)
  Premise
    P∨(Q∧⌐Q)
  Conclusion
    P
  Proof
    Suppose
      P
    Hence
      P
    P=>P

    Suppose
      Q∧⌐Q
    Then
      Q
      ⌐Q
    Hence
      P by PrincipleOfExplosionAlternativeForm
    Q∧⌐Q=>P

    P by DisjunctionElimination

My alternative form of the principle of explosion is, by the way:

$$
\frac{p\quad\neg p}{q}\quad\text{[PrincipleOfExplosionAlternativeForm]}
$$

This is easy enough to derive from the standard principle of explosion and modus ponens.

Without a way to eliminate contradictions minimally, so to speak, all my minimal proofs of De Morgan's laws become intuitionsitic. This seems wrong to me.

I'm going to try to answer this question although I'm not entirely sure the answer is right. Maybe it will encourage debate, though. I think that the answer is no, you cannot derive this rule without some form of the principle of explosion. My reasoning goes as follows...

The premise is a disjunction $p\lor(q\land\neg q)$. I have chosen to arrive at the conclusion by proof by cases, disjunction elimination in other words, which means I have to deal with $q\land\neg q$ somehow. The only other way that I can see to make progress from the premise is to use the rule that distributes disjunction over conjunction. This gives $(p\lor q)\land(p\lor\neg q)$. Now I have two disjunctions to deal with. Taking the rightmost $p\lor q$, I can only proceed by proof by cases again. Assuming $p$, I'm done. Assuming $q$ on the other hand, I must make use of the other side of the conjunction, namely $p\lor\neg q$ and I must then proceed to $q\land(p\lor\neg q)$. Using the rule that conjunction distributes over disjunction I get, unfortunately, $(p\land q)\lor(q\land\neg q)$ and I'm no better of than where I started.

My line of reasoning is based on there only really being one way that the derivation can proceed. However it seems exhaustive, so I think there's something in it. It also seems, from an intuitive standpoint, that you can't eliminate $q\land\neg q$ without something like the principle of explosion, although admittedly this intuition is tainted by my findings above. Perhaps some other interpretation would shed more light on this.

In Polish notation you might use the rule of inference

CN$\alpha$K$\beta$N$\beta$ $\vdash$ $\alpha$

which I'll call No.

I'll also use

Kol: K$\alpha$$\beta$ $\vdash$ $\alpha$

Kor: K$\alpha$$\beta$ $\vdash$ $\beta$

Ki: $\alpha$, $\beta$ $\vdash$ K$\alpha$$\beta$

and

Ao: A$\alpha$$\beta$, C$\alpha$$\gamma$, C$\beta$$\gamma$ $\vdash$ $\gamma$

The proof then can go:

premise   1 ApKqNq
suppose   2 | p
Ci 2-2    3 Cpp
suppose   4 | KqNq
suppose   5 || Np
Kol 4     6 || q
Kor 4     7 || Nq
Ki 6 7    8 || KqNq
Ci 5-8    9 | CNpKqNq
No 9      10 | p
Ci 4-10   11 CKqNqp
Ao 1 3 11 12 p


Metric over a Lie algebra $\mathfrak{u}(n)$ Let $\mathfrak{u}(n)$ be the Lie algebra of the Lie group $U(n)$. I can define a positive-definite inner product over $\mathfrak{u}(n)$ in this way: if $A,B \in \mathfrak{u}(n)$ I define $\langle A,B \rangle := \Re(\operatorname{Tr}(AB^*))$, where by $\Re$ I denote the real part and by $B^*$ the conjugate transpose of $B$. Why does this inner product over $\mathfrak{u}(n)$ define the unique left-invariant metric on the Lie group $U(n)$?

Let $A$ and $B$ be complex $n \times n$ matrices with respective $(j, k)$ entries $A_{jk}$ and $B_{jk}$, and note that $B^*$ (the conjugate transpose) has $(j, k)$ entry $\bar{B}_{kj}$. By definition,
$$
\langle A, B\rangle
  = \Re\bigl(\text{Tr}(AB^*)\bigr)
  = \Re \sum_{j,k=1}^n A_{jk} \bar{B}_{jk},$$
which is precisely the Euclidean inner product of $A$ and $B$ if these matrices are identified with complex vectors in $\mathbf{C}^{n^2}$. The resulting pairing on $\mathfrak{u}(n)$ is the restriction of this inner product.

Generally, if $G$ is a Lie group and $g \in G$, then the left multiplication map $\ell_g:G \to G$ is a diffeomorphism sending $e$ to $g$, so the push-forward $(\ell_g)_*:\mathfrak{g} \to T_gG$ is an isomorphism of vector spaces. An inner product on $\mathfrak{g}$ thereby determines an inner product on each tangent space $T_gG$, and since multiplication is smooth (as a function of $g$) these inner products constitute a Riemannian metric on $G$.

(In case it matters, this left-invariant metric is only "unique" in the sense that it is completely determined by the choice of inner product on $\mathfrak{g}$.)

Because Tr(AB*) is invariant under transformations of the orthogonal base of the algebra u(n). That's why define invariant metric on group U(n) and is left because B are tensors and must respect the sequence of < A,B >.


Complex Number solutions for a Circle I have a circle of radius 5 with its center at the origin represented as $X^2+Y^2=25$. I get that it has a solution for all values ranging from $-5$ to $+5$.

My question is what does it mean when the equation returns a complex number? Example for $x=6$, I get $y = \pm i\sqrt{11}$. In doing this for all real number greater than $+5$ and less than $-5$, what is being returned/plotted and what plane is this plot on? Is this another circle on the imaginary plane, all values on the plane beyond a circle?

When you graph the solutions in "the plane", you are restricting yourself to look at solutions to the equation where both $x$ and $y$ are real.  You could, for example, restrict further to only allow $x$ and $y$ to be rational numbers, and think about how those points fit in with all the real solutions.

To think geometrically about non-real complex solutions, you will need more (real) dimensions!  You could restrict yourself, as it sounds like you are doing in the question, to just solutions where $x$ is real and $y$ is allowed to be complex.  Then you will need another dimension/direction for the imaginary part of $y$.  You could graph this in a "$z$" direction, so that solutions where $x>5$ not lie in the plane but above/below it.  You will find that for $|x|>5$ the solutions will be points where $x^2 - z^2 = 25$ so that if you just look at the $(x,z)$ plane the solution set will look like a hyperbola.

Probably the most interesting thing to look at is when you allow both $x$ and $y$ to be complex... but graphing this would require more dimensions.

All the points that lie inside or on the boundary of the circle
$$ \left\{ \ (x, y) \in \mathbb{R}^2 \ \colon \ x^2+ y^2 = 5^2 \ \right\} $$
lie to the right of the line
$$ \left\{ \ (x, y) \in \mathbb{R}^2 \ \colon \  x = -5  \ \right\} $$
and to the left of the line
$$ \left\{ \ (x, y) \in \mathbb{R}^2 \ \colon \  x = 5  \ \right\}. $$
Moreover, the circle intersects the former line at the point $(-5, 0)$ and the latter one at the point $(5, 0)$.


How many integer numbers between 0 and 9999 are there that have exactly one digit 1 and exactly one digit 3? What do I need to think about this problem?

  How many integer numbers between 0 and 9999 are there that have exactly one digit 1 and exactly one digit 3?

The only thing I know that the total configurations is $10^4$ so if I want to count the numbers which at least have 1 three firstly I get the numbers which not have three's $9^4$ and then subtract and the same for four but how can I count with more restrictions?

The position of the digit $1$ can be chosen in $4$ ways, the position of the digit $3$ can be chosen in $3$ ways. The remaining two digits should belong to the set $\{0,2,4,5,6,7,8,9\}$ which has $8$ elements.

Hence the number of integers between 0 and 9999 that have exactly one digit 1 and exactly one digit 3 is
$$4\cdot 3\cdot 8\cdot 8=768.$$

I get a different answer...while I understand your logic here.
Here is mine:
Suppose $A = \{1,\} B =\{ 3\}, C = \{0,2,4,5,6,7,8,9\}$ and $D = \{0,2,4,5,6,7,8,9\}$
In a 4-digit number, we can place: $A-B-C-D$ in $4!$ ways.
Now, we know that there is only 1 way to choose $A$, 1 way to choose $B$, but 8 ways to choose $C$ and 8 ways to choose $D$.
So, by product rule, we have:
$$4! * 1 * 1 * 8 * 8 = 1536$$.
I get $1536 = 2*768$, but I guess it is because of the way I think of it, the order of $C$ and $D$ matters $(=2!)$.
As I understand this problem to be referring to k-permutations (the order of each digit matters), shouldn't the answer be $1536$ and not $768$?


The Bent Washer Problem -- divide a shape into 2 pieces of the same volume. Fans of the Ham sandwich theorem know that any set of points can be divided by a plane into two equal halves.

Consider instead a 3-D shape that must be divided into 2 equal pieces by a single cut.  A sufficiently bent spring washer or keyring cannot be divided into 2 pieces by a plane.  But it's possible to make a simpler cut that works -- a partial plane cut.

Is that the simplest shape that cannot be split into 2 pieces by a plane?

Is there a simple shape that cannot be split into 2 equal pieces by a simple cut?

The Ham Sandwich Theorem says that given three measurable subsets of $\mathbb{R}^3$ can be cut into two equal (with respect to measure) pieces by a single plane. In particular, we can choose two of our sets to be empty. So any measurable subset of $\mathbb{R}^3$ can be cut in half by a plane.

EDIT: I originally misunderstood. You're interested in when the cut pieces are connected.


Find the asymtotes to the function: $f(x)={1+ln|x|\over x(1-ln|x|)}$ Find the asymtotes to the function: $f(x)={1+ln|x|\over x(1-ln|x|)}$. I have difficulties with this one when i try to find the limits, i get indefinites that look like (infinity over infinity)over infinity.. Can anyone help me with this?

The important moments in your function's life include $x = 0$ and $\ln |x| = 1$ (so $x = \pm e$), and $x \rightarrow \pm \infty$.

As we go to $\pm \infty$ we get $1 \pm \ln |x| \approx \ln |x|$ so we reduce the expression to approximately $\ln |x| / (-x \ln |x|)$ which divides to simply $-1/x$, so these asymptotes are the horizontal asymptote $y = 0$.

At $x = 0$ the logarithms again overpower the nearby constants, creating something which goes like $-1/x$. But at $0$ this function has a vertical asymptote.

At $x = \pm e$ you'll have vertical asymptotes as the numerator does not go to zero while the denominator does.

It may be helpful to look at a graph of $x f(x)$ as that has the essential "weird" behavior spelled out for you: WolframAlpha graph. This function has a vertical asymptote at $y = -1$ for $x \rightarrow \pm \infty$, and also has a value of $y=-1$ at $0$.


What does George E. Martin mean by "The belief that geometries can be classified by their symmetry groups is no longer tenable"? In the preface to George E. Martin's Transformation Geometry: An Introduction to Symmetry, he writes (emphasis mine)

Transformation geometry is a relatively recent expression of the successful venture of bringing together geometry and algebra. The name describes an approach as much as the content. Our subject is Euclidean geometry. Essential to the study of the plane or any mathematical system is an understanding of the transformations on that system that preserve designated features of the system.

...

The belief that geometries can be classified by their symmetry groups is no longer tenable. However, the correspondence for the classical geometries and their groups remains valid. Undergraduates should not be expected to grasp the idea of Klein's Erlanger program before encountering at least the projective and hyperbolic geometries. Therefore, although the basic spirit of the text is to begin to carry out Klein's program, little mention of the program is made within the text.

What does he mean in the bolded statement? From the Erlangen program Wikipedia page, what I understand is that there are geometric objects that have "the same symmetries" (I don't know how else to put it), but are nevertheless distinct.
Comments would be appreciated!

I suppose that he means that we now know that the symmetry group is not sufficient to capture the space. The idea of the Erlanger Program is that the group of symmetries determines what are the geometric figures and geometric properties within that space. If you have two subsets of the space that can be transformed into each other with a symmetry, then they are instances of the same type of geometric figure, and if a property is invariant under symmetry transformations, then that is a meaningful property for that given geometry.
The Lens example given by Jason De Vito in the comments illustrates how the symmetry group doesn't characterize the space