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H: We had $m$ (odd number)....
What we will get if we divide $2^{\varphi(m)-1}$ at $m$?
(The answer should be at $m$...)
Thank you!
(Mabye it's something that connect to Euler theorem or Fermat Small Theorem).
AI: You know that
$$2^{\phi(m)} \equiv 1 \pmod{m}$$
Thus
$$2^{\phi(m)-1} \equiv 2^{-1} \pmod{m}$$
Now,
$$2^{-1} \equiv 2^{-1} (m+1) \equiv \frac{m+1}{2} \pmod{n} $$
Where $\frac{m+1}{2} \pmod{m}$ makes sense since $m$ is odd.
Thus, the remainder is $\frac{m+1}{2}$. |
H: Show stretch of numbers are composite, n! + 2, n! + 3 etc
Let $n\in\mathbb{N}$ with $n\geq2$. Consider the numbers $n!+2,n!+3,...,n!+ n$. Show that none of them is prime. Deduce that for each positive integer $N$ there is a stretch of $N$ consecutive composite numbers.
What I don't understand is what happens when n=2 and you work out n! + 3=5 which is prime.
AI: If $n = 2$, then the first number on your list is
$$n! + 2 = 2! + 2 = 4$$
which is not prime. On the other hand, the last number is also
$$n! + n = 2! + 2 = 4$$
which is still not prime. |
H: Grammar derivation
Given these grammar productions:
$$\begin{align*}
&S\to A1B\\
&A\to 0A\mid\lambda\\
&B\to 0B\mid 1B\mid\lambda
\end{align*}$$
And given string $w = 01101$
If I wanted to make a) leftmost derivation and b) rightmost derivation, I thought it would work like this:
a) $S \rightarrow A1B \rightarrow 01B \rightarrow 011B \rightarrow 0110B \rightarrow 01101$
b) $S \rightarrow A1B \rightarrow A11B \rightarrow A110B \rightarrow A1101 \rightarrow 01101$
I'm learning about left/right derivations and not sure if they properly applied here. Maybe someone can tell me where I'm misunderstanding this concept.
EDIT:
Thanks for the great answer, if I'm understanding correctly then I should be doing it like this:
a) $S \rightarrow A1B \to 0A1B \rightarrow 01B \rightarrow 011B \rightarrow 0110B \to 01101B \rightarrow 01101$
b) $S \rightarrow A1B \rightarrow A11B \rightarrow A110B \to A1101B \rightarrow A1101 \to 0A1101 \rightarrow 01101$
AI: In (a) the step $A1B\Rightarrow 01B$ is impossible, because there is no production $A\to 0$. Two steps are required: $A1B\Rightarrow 0A1B\Rightarrow 01B$. The step $0110B\Rightarrow 01101$ is also impossible; again you need two steps, $0110B\Rightarrow 01101B\Rightarrow 01101$.
In (b) you have the same problem with $A110B\Rightarrow A1101$ and $A1101\Rightarrow 01101$: the first should be $A110B\Rightarrow A1101B\Rightarrow A1101$, and the second $A1101\Rightarrow 0A1101\Rightarrow 01101$. |
H: Topological Group $G$ totally disconnected $\Rightarrow$ $G$ hausdorff?
On Wikipedia, I read that a topological group is necessarily Hausdorff if it is totally disconnected. Is that true?
I read it on this page:
http://en.wikipedia.org/wiki/Totally_disconnected_group
If not, does anyone has an example for a totally disconnected group, which is not Hausdorff?
Thanks for help.
AI: It's true. If $G$ is a totally disconnected topological group, then the connected component of the identity $e$ is $\{e\}$ (because components of a totally disconnected space are just points). Since connected components of any topological space are closed, $\{e\}$ is closed. This implies that $G$ is Hausdorff: the the diagonal $\Delta\subseteq G\times G$ is the inverse image of $\{e\}$ under the continuous map $(g,h)\mapsto gh^{-1}$, and is therefore closed. |
H: Is $S^0$ a manifold?
Consider a singleton space $\{x\}$, it is a manifold and it is locally euclidean as there is a homeomorphism to $\mathbb{R}^0$. However, consider $S^0=\{-1,1\}$ with the discrete topology, there does not exist any homeomorphism to any open neighbourhood of $\mathbb{R}^k$ for $k \ge 0$, so it cannot be locally euclidean and cannot be therefore a manifold.
So is $S^0$ the most trivial example of a topological space that is not a manifold?
AI: 'Locally Euclidean' means each point has a neighborhood that is homeomorphic to $\mathbb R^k$. Each point in $S^0$ does have this property for $k = 0$.
More generally, any countable discrete space is a $0$-manifold. |
H: Total k combinations with at least one element from each set.
Given $n$ sets the total amount of ways k of the elements can be combined is given by $$C(|S_1|+|S_2|+...+|S_n|,k)$$
Now suppose we wanted to find the total combinations possible when at least one element from each set must be included. This imposes the following restriction $n \le k$.There are $|S|$ ways of choosing a single element out of $S$. Then through the product rule we know that if we choose only one element from each set all the possible combinations is:
$$\prod_{x=1}^{n}{|S_x|}$$
After selecting one element from each set the total amount of elements that are left to select from is given by: $$\sum_{x=1}^{n}{|S_x|-1}$$
Selecting k elements is thus given by:
$$C\left(\sum_{x=1}^{n}{|S_x|-1},k\right)$$
To ensure that each combination will have at least one element from each set we know each combination out of the remaining elements will only fit into $k - n$. Therefore total combinations of the remaining elements is:
$$C\left(\sum_{x=1}^{n}{|S_x|-1},k - n\right)$$
The next step I'm unsure of I would think that the total combinations possible is equal to:
$$\left(\prod_{x=1}^{n}{|S_x|}\right)C\left(\sum_{x=1}^{n}{|S_x|-1},k - n\right)$$
Because this is the total combination of one element from each set multiplied by the total combinations of the remaining elements, however this seems to be incorrect. If anyone could point me in the right direction that would be great! Thanks.
AI: suppose that the sets are {1,2},{3,4},{6,7,8} and we want to make combinations of 4 elements. Then one way to assign one element to each one is 1,3,6 and then we additionally select 7. Another way to construct the same set is to first take 1,3,7 and then add 6. In other words this construction "constructs the some elements more than once.(what is true is that it constructs all of the combinations).
let $ W_k(f)$ be the number of combinations with at least one element from each set of a family of sets called f. Then we have that $W_k(f)=\frac{x!}{(x-k)!}{k!}- \sum_{i=1}^{2^n-1}W_k(Pi)$ where the union of all $P_i$'s is all of the proper subsets of f. |
H: Runge-Kutta-Fehlberg Method Problem
Suppose all infected individuals remained in the population to spread the disease. A more realistic proposal is to introduce a third variable $z(t)$ to represent the number of individuals who are removed from the affected population at a given time $t$ by isolation, recovery and consequent immunity, or death. This quite naturally complicates the problem, but it can be shown that an approximate solution can be given in the form
$x(t) = x(0)e^{\frac{-k1}{k2} z(t)}$ and $y(t)=m-x(t)-z(t)$,
where $k_1$ is the infective rate, $k_2$ is the removal rate, and $z(t)$ is determined from the differential equation
$z'(t)=k_2(m-z(t)-x(0)e^{\frac{-k1}{k2} z(t)})$.
Find an approximation to $z(30)$,$y(30)$, and $x(30)$, assuming that $m=100,000$,$x(0)=99,000$, $k_1$ = $2\times 10^{-6}$, and $k_2=10^{-4}$.
Now, the question would've been easier for me if there's only one variable, but I don't know how to make it work with three variables on Maple.
So can anyone help me with this?
AI: The differential equation you wrote for $z(t)$ only has one dependent variable $z$. Everything else is a constant. Once you have $z(30)$, you can plug that into the equations for $x(t)$ and $y(t)$. |
H: Discrete math sets help?
How would I do this question?
suppose U = {1,2...,9}, A= all multiples of 2, B = all multiples of 3, and C = {3,4,5,6,7}. Find C-(B - A).
I really don't know how I would approach this question so can anybody help with this?
AI: The multiples of $3$ in $U$ are $3,6$, and $9$, so $B=\{3,6,9\}$. The multiples of $2$ in $U$ are $2,4,6$, and $8$, so $A=\{2,4,6,8\}$. For any sets $X$ and $Y$, the set difference $X\setminus Y$ is the set of objects that are in $X$ but not $Y$. Thus, $B\setminus A$ is the set of digits that are in $B$ but not in $A$. Look at the members of $B$: $3$ and $9$ are in $B$ but not $A$, so they’re in $B\setminus A$. $6$, on the other hand, is in $B$ and $A$, so it’s not in $B\setminus A$. And nothing else is in $B$, never mind whether it’s in $A$ or not, so nothing else is in $B\setminus A$. Thus, $B\setminus A=\{3,9\}$.
Now see if you can apply the same sort of reasoning to find $C\setminus\{3,9\}$. |
H: Show the ring $R = \mathbb{Z}/2\mathbb{Z}[x]/(x^4+x+1)$ is a field
Show that the ring $R = \mathbb{Z}/2\mathbb{Z}[x]/(x^4+x+1)$ is a field. Find the multiplicative inverse of the element [x^2+1] in that field.
From what I know,
$$\mathbb{Z}/2\mathbb{Z}[x]/(x^4+x+1) = \{ a_0 + a_1x+a_2x^2+a_3x^3 + x^4 : a _i = \{0,1\}\}$$
To show that $R$ is a field, I would need to show that it is a commutative division ring.
How can I show that formally? Especially the fact that every non-zero element of $R$ has an inverse with respect to multiplication.
AI: Why don't prove that $x^4+x+1$ is irreducible over $\Bbb Z/2\Bbb Z$?
Denote $a=[x]$. We have $a^4=a+1$, so $a^8=a^2+1$. But $a^{15}=1$ (since $R$ is a field with $2^4=16$ elements and $a\in R^{\times}$), so $a^7$ is the inverse of $a^2+1$. Now, from $a^4=a+1$ we find $a^7=a^3+a+1$. |
H: matrices equation
Let A, B, J 4x4 matrcies, such that:
$\eqalign{
& {A^t}JA = J \cr
& {B^t}JB = J \cr} $
Prove that:
${(AB)^t}J(AB) = J$
Any help will be appreciated.
AI: Hint:
$(AB)^t = B^tA^t$
$(AB)^tJ(AB)=B^t(A^tJA)B$ |
H: Singularities in (Elementary) Real Algebraic Geometry
I've taken an introductory course in algebraic geometry, and am currently studying Sumio Watanabe's book, "Algebraic Geometry and Statistical Learning Theory". In this book Watanabe gives a, possibly antiquated, definition of real algebraic singularities:
A point $P$ on a curve $A$ is said to be nonsingular if there exist
open sets $U,V\in\mathbb{R}^d$ containing $P$ and an analytic isomorphism (bijective
function analytic both ways) $f:U\rightarrow V$ such that $$f(A\cap
U)=\{(x_1,x_2,...,x_r,0,0,...,0);x_i\in\mathbb{R}\}\cap V,$$ where $r$
is a nonnegative integer.
So to gain some intuition for this definition, I took a look at the two curves in $\mathbb{R}^2$:
\begin{align}
&y^3-x^2=0\\
&x^2-(y+1)^3+(y+1)^2=0.
\end{align}
The first of these curves has a singularity at $(0,0)$, while the second one doesn't. The strategy the book uses in its examples is basically to apply the identity function to $x$, and then project $y$ down onto the $x-$axis, in the following way:
\begin{align}
&f(x,y)=0\\
&u=x\\
&v=f(x,y)
\end{align}
Of course this depends on being able to invert the $v=f(x,y)$ function and solve for $y$. And in fact this is what fails for $y^3-x^2=0$, since solving $v=y^3-x^2$ for $y$ gives us $y=\sqrt[3]{u^2+v}$, which is not analytic at $(0,0)$.
So turning to the second equation, which is nonsingular at $(0,0)$, our second analytic function is $v=x^2-(y+1)^3+(y+1)^2$. Solving this equation for $y$ involves finding the roots of a cubic polynomial, which is not feasible for me. Thus if this second equation is to be nonsingular, then should I expect that one of these three roots will be a real-valued analytic function inside some open ball containing the origin, while the other two solutions will be complex-valued always with non-zero imaginary part?
If not, then how would I go about finding this analytic isomorphism?
AI: I think the point is that you don't need to find an explicit formula for the roots. You don't need to worry about the roots that are not near $(0,0)$. A (complex) zero in $y$ of an analytic function $f(y,z)$ is an analytic function
of $z$ in a neighbourhood of any point where the zero has multiplicity $1$. |
H: Why don't I get $e$ when I solve $\lim_{n\to \infty}(1 + \frac{1}{n})^n$?
If I were given $\lim_{n\to \infty}(1 + \frac{1}{n})^n$, and asked to solve, I would do so as follows:
$$\lim_{n\to \infty}(1 + \frac{1}{n})^n$$
$$=(1 + \frac{1}{\infty})^\infty$$
$$=(1 + 0)^\infty$$
$$=1^\infty$$
$$=1$$
I'm aware that this limit is meant to equal to $e$, and so I ask:
why don't I get $e$ when I solve $\lim_{n\to \infty}(1 + \frac{1}{n})^n$?
AI: The main problem here is that you use $\infty $ as if it were a number, and then you just substitute $\infty $ for $n$ and 'compute'. But, since infinity is not a number, you can't just substitute it and the computation you make is meaningless.
I'm assuming from the way you ask the question that you already know how to derive the correct value and that you are just wondering what is wrong with your approach. So, you don't get $e$ when you compute the limit they way you did precisely because your computation is invalid since you treat infinity as a number.
This may be extra confusing since sometimes substituting $\infty $ does lead to the correct answer, but this should be regarded as a fluke. For instance, the limit $\lim _{n\to \infty }\frac {1}{n}$ is $0$, which is what you would get by substituting $\infty $. But this is just coincidence that a completely faulty line of argument using entirely wrong 'computations' leads to the correct answer. Unfortunately, often when teaching calculus such substitutions and manipulations with $\infty $ are glossed over, or worse even encouraged. It is good practice to never ever substitute $\infty $ and compute with it. |
H: N tosses of a coin ,what is the probability that number of heads are even?
We have a coin with probability of head is p and probability for tail is (1-p).
We toss the coin N times, what is the probability that the number of tosses that show head is even?
What I've tried is to sum over all even k's (k= 0,2,4,...) and to sum up the probability that the number of heads is k. Is there a way to elinimate the sum and to give a closed formula?
AI: Let $a_N$ be the probability that $N$ Bernoulli trials result in an even number of successes.
This occurs if an initial failure is followed by an even number of successes, or an initial success
is followed by an odd number of successes. Therefore $a_0=1$ and for $N\geq 1$
$$a_N=q a_{N-1}+p(1-a_{N-1}).$$
Multiplying by $s^N$ and adding over $N$ we see that the generating function
satisfies
$$H(s)=1+qsH(s)+ps(1-s)^{-1}-psH(s)$$
or
$$2H(s)=[1-s]^{-1}+[1-(q-p)s]^{-1}.$$
Expanding the right hand side using geometric series we find that the coefficients
satisfy $$a_N={1\over 2}+{(q-p)^N\over 2}.$$
Reference: Chapter XI of An Introduction to Probability Theory and Its Applications (Volume 1) (3rd edition) by William Feller. |
H: Why discriminant method gives wrong answer when searching parabola that touches x axis?
Define a so that parabola y = a*x^2 - 5*a*x + 5*a+5 touches x axis.
Touching x axis means that discriminant is 0 (parabola has one double zero on x axis - that is a touch).
Or it can also mean that the vertex is on x axis, which means vertex's coordinates are (x,0).
Now if I calculate a by first clue, discriminant equal to 0, I get:
D=0 a=a b=-5a c=5*a+5
D=b^2 - 4 * a * c
0 = b^2 - 4 * a *c
0=25a^2-20a^2+20a
0=5a^2+20a
-5a^2=20a
a=-4
This gives us parabola y=-4x^2 + 20x -15. This parabola in wolfram alpha for some reason doesn't touch x axis, but passes it by very little.
Now if we calculate a by second clue, that is the vertex, we get:
y coordinate of vertex is calculated with the next formula:
y= -(b^2 - 4*a*c)/4*a
y coordinate is 0 so:
0 = -(5a^2 + 20a)/4a
...
a=4
This gives us parabola y= 4x^2 -20x + 25 which is in wolfram alpha a correct graph that touches axis x.
My question is, why does discriminant method give wrong answer?
Thanks in advance!
AI: The parabola seems to be $\;y=ax^2-5ax+5a+5\;$ , and we must assume $\;a\neq 0\;$ otherwise this is not a quadratic, so
$$\Delta=(5a)^2-4a(5a+5)=5a^2-20a=5a(a-4)=0\iff a=\begin{cases}0\;\;\color{red}X\\4\;\;\color{red}\checkmark\end{cases}$$
(Can you see your mistake in the discriminant?)
The vertex coordinates of a parabola $\;y=ax^2+bx+c\;$ are given by
$$V=\left(-\frac b{2a}\;,\;-\frac\Delta{4a}\right)$$
so in our case we get that the vertex is (remember: $\;a\neq 0\;$ ) :
$$V=\left(\frac{5a}{2a}\;,\;-\frac{5a(a-4)}{4a}\right)=\left(\frac52\;,\;-\frac54(a-4)\right)$$
and the above point is on the $\;x$-axis iff $\;a=4\;$ ...same result! |
H: Why $\mathbb{Z}[\alpha]/23\mathbb{Z}[\alpha] \cong \mathbb{F}_{23}/(x^3-x-1)$ where $\alpha$ satisfies $\alpha^3=\alpha+1$?
This a step in my notes which I can't seem to understand clearly. Why $\mathbb{Z}[\alpha]/23\mathbb{Z}[\alpha] \cong \mathbb{F}_{23}/(x^3-x-1)$ where $\alpha$ satisfies $\alpha^3=\alpha+1$? I see that the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $f(x)=x^3-x-1$ but why does this isomorphism hold?
AI: Because $\mathbb{Z}[\alpha]\simeq \mathbb{Z}[x]/(x^3-x-1)$ for $\alpha$ being a root of $x^3-x-1=0$. |
H: Where does the function $f(z)=z\bar z+z/\bar z$ satisfy the Cauchy-Riemann equations?
Where does the function $f:\mathbb{C}\setminus\{0\}\to\mathbb{C}, f(z)=z\bar z+z/\bar z$ (where $\bar z$ is the complex conjugate of $z$) satisfy the Cauchy-Riemann differential equations?
I tried to write $f(z)$ as: $$f(z)=x^2+y^2+(x^2-y^2)/(x^2+y^2)+(2xyi)/(x^2+y^2)$$ but if I work out the jacobian, it's not of the vorm $u_x=v_y$ and $u_y=-v_x$ (Cauchy Riemann equations). Can somebody help me out?
AI: It looks like you have $f(z)$ right. That is,
$$
f(x+iy) = x^2+y^2+ \frac{x^2-y^2}{x^2+y^2}+i\frac{2xy}{x^2+y^2}\implies\\
u(x,y)= x^2+y^2+ \frac{x^2-y^2}{x^2+y^2}\\
v(x,y) = \frac{2xy}{x^2+y^2}
$$
Now, we find the partial derivatives to be
$$
u_x = 2x + \frac{4xy^3}{(x^2+y^2)^2}\\
u_y = 2y - \frac{4x^3y}{(x^2+y^2)^2}\\
v_x = \frac{2y^3-2x^2y}{(x^2+y^2)^2}\\
v_y = \frac{2x^3-2xy^2}{(x^2+y^2)^2}
$$
Now, in order to find the points where $f$ satisfies the CR equations, find the solution to the simultaneous set of equations
$$
u_x(x,y) = v_y(x,y)\\
u_y(x,y) = -v_x(x,y)
$$
As Daniel's solution shows, the only points for which the above equality holds should be $(1,0)$ and $(-1,0)$. |
H: Find the remainder of $7^{2002}$ divided by 101.
This is what I have so far:
Since 101 is a prime and does not divide 7, we can apply Fermat's Little Theorem to see that $$7^{100} \equiv 1 \ (mod \ 101)$$
We can then reduce $7^{2002}$ to $7^{2} (7^{100})^{20} \equiv 7^{2}(1)^{20} \ (mod \ 101)$ which is where I'm stuck at.
$7^2=49 \equiv 150 \ (mod \ 101)$. How do I reduce $7^2$ in a way that is constructive towards my solution since $(mod \ 101)$ is such a large modulus to operate in?
AI: You had the solution and broke it :)
At
$$7^{2002} \equiv 49\pmod {101} \,,$$
you are done, the remainder must be $49$. Indeed, if you denote the remainder by $r$ then $0 \leq r \leq 100$ and
$$ r \equiv 49 \pmod{101} \,.$$
This means that $101|r-49$, and since $-49 \leq r-49 < 52$ you get $r-49=0$. |
H: normal distribution expected value
In this derivation:
http://www.sonoma.edu/users/w/wilsonst/Papers/Normal/default.html
$$f(x) = \sqrt{\frac{k}{2\pi}}e^{-\frac{k(x-\mu)^2}{2}}$$
they let
$x-\mu = v$
$dx = dv$
and conclude that
$$E(v) = \sqrt{\frac{k}{2\pi}}\int_{-\infty}^\infty ve^{-\frac{kv^2}{2}}dv$$
Why? Shouldn't it be
$$E(v) = \sqrt{\frac{k}{2\pi}}\int_{-\infty}^\infty (v+\mu)e^{-\frac{kv^2}{2}}dv$$
AI: Look further down the web page. You will see
$$ E(x) = E(v) + \mu = \mu .$$
In other words, the problem is with awkward notation rather than any genuine mistake. The web page really shouldn't use the same letter for the random variable and the variable being integrated over. A common notation is to use capital letters for the random variable - then it would have said let $V = X-\mu$ and $v=x-\mu$, then
$$ E(V) = \text{(what the web page wrote on R.H.S.)} = 0$$
and I think it would have been clearer. |
H: When is $\Bbb Q(\sqrt p)\subseteq\Bbb Q(\sqrt[3]q)$, with $p,q$ prime?
Given primes $p, q$, when do we have
$$\Bbb Q(\sqrt p)\subseteq\Bbb Q(\sqrt[3]q)$$
?
At first I tried a linear algebra approach, since in linear algebra, determining when one subspace is inside another is easy, just leading to a system of linear equations. I need to find $x,y,z$ such that $xp^{1/3}+yp^{2/3}+z=\sqrt p$. But to apply linear algebra to this, I'd need to find a basis for the ambient space, say $\Bbb Q(\sqrt p, \sqrt[3]q)$, and trying to do that led me right back to the original question.
Next I tried squaring $xp^{1/3}+yp^{2/3}+z=\sqrt p$, to obtain a linear combination of $(1,p^{1/3},p^{2/3})$ that was equal to $0$. Setting each of the coefficients to $0$ (since $1,p^{1/3},p^{2/3}$ are free vectors) leads to a system of three polynomial equations in $x,y,z,p,q$, but it doesn't look very promising.
How can I tackle this problem? I'm more interested in general methods and ideas on what approach to use. If there's a magic bullet "you just have to see the answer" solution, I'm less interested at the moment (unless there's no other way).
AI: If $\Bbb Q(\sqrt p)\subseteq \Bbb Q(\root 3\of q)$, then since $[\Bbb Q(\root 3\of q)\colon \Bbb Q]=3$, it would follow that $[\Bbb Q(\sqrt p):\Bbb Q]$ divides $3$. But $[\Bbb Q(\sqrt p):\Bbb Q]=2$, therefore this isn't possible. |
H: Question on Rudin's Proof of the Residue Theorem
The Theorem in question is Theorem 10.42.: If $f$ is meromorphic in $U$, $A$ is the set of poles of $f$ and $\Gamma$ is a cycle in $U-A$ so that $Ind _{\Gamma}=0$ in $U^c$ then
\begin{equation}\frac 1{2\pi i}\int_{\Gamma}f=\sum_{a\in A}Ind_{\Gamma}(a)Res_f(a)
\end{equation}
My problem is with the first part of the proof, i.e. proving that the sum, or equivalently the set $B=\left\{a\in A:Ind_{\Gamma}(a)\neq 0\right\}$ is finite. The proof goes like this:
Every component $V$ of $(\Gamma^*)^c$ (that is a component of the interior, or the exterior of $\Gamma$) that is unbounded or intersects $U^c$ is in the exterior of $\Gamma$ ($Ind_{\Gamma}=0$ in $V$). Since $A$ has no limit points in $U$, we conclude that $B$ is a finite set.
How exactly do we conclude that? The arguments made before that are pretty obvious but this one is not. I am thinking that $B$ is bounded and with no limit points, but that does not make it finite right?
AI: Any set $B$ which is bounded and has no limit points must be finite.
Otherwise you can find a sequence $x_n \in B$ with $x_n \neq x_m, \forall m \neq n$.
Since $B$ is bounded, its closure is bounded and closed, thus compact. Hence $x_n$ has a subsequence which is convergent in $\bar{B}$. But then, the limit is a limit point for $B$. |
H: Prove that $\frac{n^2+(-1)^nn+2}{7n^2+3}$ converges to $\frac{1}{7}$
I want to show that $\frac{n^2+(-1)^nn+2}{7n^2+3}$ converges to $\frac{1}{7}$ using the definition of convergence.
Skratch work:
I need $\mid\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}\mid<\epsilon$.
So I take $\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}=\frac{7(-1)^nn+11}{7(7n^2+3)}$.
For $n=1$, $\frac{7(-1)^nn+11}{7(7n^2+3)}=\frac{-4}{70}$
For $n=2$, $\frac{7(-1)^nn+11}{7(7n^2+3)}=\frac{24}{196}$
For $n=4$, $\frac{7(-1)^nn+11}{7(7n^2+3)}=\frac{39}{805}$.
So I think I can safely say that $\frac{7(-1)^nn+11}{7(7n^2+3)}<\frac{1}{n}$.
Now I can start the proof:
Let $\epsilon>0$. And choose $N>\frac{1}{\epsilon}$. Hence, $\frac{1}{N}<\epsilon$. So $n>N \implies \mid\frac{n^2+(-1)^nn+2}{7n^2+3}-\frac{1}{7}\mid < \frac{1}{n}< \frac{1}{N} < \epsilon$.
Does this proof look correct? I think I may have made a mistake in choosing $N$. Please let me know. Thanks
AI: you can choose n in better way
you can use sup of a(n) or a bit more to find n>N ,which satisfy that relation |
H: A dense subset of a Hilbert space
I am curious about the following problem:
Consider the Hilbert space (a weighted $L^2(\mathbb{R})$ space):
$$\mathscr{H}=\bigg\{f: \mathbb{R}\to\mathbb{R}\text{ Lebesgue measurable}\,\bigg|\,\int_\mathbb{R} \big|\,f(x)\,\big|^2\exp\Big(-\frac{x^2}{2}\Big)\,\mathrm{d} x<\infty\bigg\}$$
on which the inner product is defined as: for every $f,g\in\mathscr{H}$,
$$\langle f, g\rangle=\int_\mathbb{R} f(x)g(x)\,\exp\Big(-\frac{x^2}{2}\Big)\,\mathrm{d}x.$$
I want to find a dense subset of $\mathscr{H}$. Obviously, $\mathscr{H}$ contains $L^2(\mathbb{R})$. My question is: is any of the following three a dense subset of $\mathscr{H}$? Can anyone give me some hints about the proof? Or if none of them is, what could be a good candidate? Thanks a lot !!
(1) The set of simple functions in $\mathscr{H}$;
(2) The set of continuous functions in $\mathscr{H}$;
(3) The set of polynomial functions in $\mathscr{H}$.
AI: Since the measure for $\mathscr{H}$ is the Lebesgue measure with a (nice) weight function, you get an isometric isomorphism by multiplication with the square root of the weight:
$$T \colon \mathscr{H} \to L^2(\mathbb{R}); \quad T(f) \colon x \mapsto f(x)\exp \left(-\frac{x^2}{4}\right).$$
That isometry gives you a bijection between the dense subsets of $\mathscr{H}$ and $L^2(\mathbb{R})$. Since the weight is continuous, $T$ and $T^{-1}$ map continuous functions to continuous functions. Since the set of continuous functions is dense in $L^2(\mathbb{R})$, the set of continuous functions in $\mathscr{H}$ is also dense. The space of simple functions is dense in every $L^p(X,\mu)$ for $p < \infty$ (at least for $\sigma$-finite $\mu$), the space of simple functions is dense in $\mathscr{H}$. I think - but I'm not sure about it - that the set of polynomials is also dense in $\mathscr{H}$. |
H: Differential Equation Question $(\frac{2x^3}{y}-4x^2\cdot e^{4y})dy-(2x^2lny-2x\cdot e^{4y}) dx$
I am trying to find the solution for the following equation
$$(\frac{2x^3}{y}-4x^2\cdot e^{4y})dy-(2x^2lny-2x\cdot e^{4y}) dx$$
what I tried to do is, set
$P(x,y) = (2x^2lny-2x\cdot e^{4y})$
$Q(x,y)=(\frac{2x^3}{y}-4x^2\cdot e^{4y})$
$$\frac{dp}{dx}=\frac{2x^2}{y}-8xe^{4y},\frac{dq}{dy}=\frac{6x^2}{y}-8xe^{4y}$$
any suggestions how to continue?
thanks!
AI: Hint: The equation is in the form $P(x,y)dx+Q(x,y)dy=0$ and it has an integrating factor $$\mu(x)=\frac{1}{x^4}.$$ This factor has been found by solving $\frac{P_y-Q_x}{Q}=-\frac{4}{x}=\frac{d\mu}{\mu}$, where $P(x,y) = -2x^2lny+2xe^{4y}$ and $Q(x,y)=\frac{2x^3}{y}-4x^2 e^{4y}$. Here is the calculation: $$P_y=-2x^2y^{-1}+8xe^{4y}$$ and $$Q_x=6x^2y^{-1}-8xe^{4y}$$$$\Rightarrow P_y-Qx= -8x^2y^{-1}+16xe^{4y}=-8(x^2y^{-1}-2xe^{4y})\Rightarrow$$$$\frac{P_y-Q_x}{Q}=\frac{-8(x^2y^{-1}-2xe^{4y})}{2x(x^2y^{-1}-2x e^{4y})}=-\frac{4}{x}.$$ Thus $$\mu(x)=e^{-\int\frac{4}{x}dx}=\frac{1}{x^4}.$$ |
H: Find T10(x): the Taylor polynomial of degree 10
I keep thinking that all you do is plug in 0 into the problem for c0 and then solve. And then you plug in 2 for c2...and so on.
I feel dumb...
Thanks!
AI: Hint: The general form of a Taylor series about $0$ is
$$f(x) = \sum_{n= 0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
The $c_n$ refers to this coefficient; that is,
$$c_n = \frac{f^{(n)}(0)}{n!}$$
So for example, $f^{(0)}(0) = \frac{7x}{x + 14} |_{x = 0} = 0$. Likewise, take the first derivative to find $f^{(1)}(0)$, and so on. |
H: Continuity- Image of a function
I have the following topological space:
$\tau=$ {$U\subseteq R: 1\notin U$} U {$R$}
and the following application:
$f: (R, \tau)\to (R, \tau)$
I have to see that if f(1)=1, then f is continuous.
Is $f^{-1}(R)=R$??
AI: The preimage of the codomain is always the entire domain, so $f^{-1}[\Bbb R]=\Bbb R\in\tau$.
Your function is continuous since the preimage of a closed set, i.e. of a set containing $1$, always contains $1$. Since the image of set containing $1$ also contains $1$, it is also a closed map. |
H: Smallest values of y for values of x
What are the largest and smallest values of $y=x^3 - 12x + 1$ for values of $x$ in the range $-3$ to $+5$?
So I find the question a bit odd as it is stated, so I'm wondering if anyone can clear up what is meant? What I just did is:
$$y' = 3x^2 - 12$$
$$ x^2 = 4, x = \pm 2$$
$$y(2) = -15, y(-2) = 17$$
But I have a feeling that this is a misinterpretation of the question.
AI: The question is asking for the maximum and minimum of $$x^3 - 12x + 1$$ subject to the condition that $-3 \le x \le 5$. What you've done (setting the derivative to $0$) is correct for finding local extrema, but don't forget to check the endpoints - that is, find $y(5)$ and $y(-3)$ and compare. |
H: Proof this function is constant
I have the following topological space:
$\tau= \{U\subseteq R: 1\notin U\} \cup \{R\}$
and the following application:
$f: (R, \tau)\to (R, \tau)$
I have already proved that if $f(1)=1$, then $f$ is continuous.
Now, I have to see that if $f$ is continuous and $f(1)=y$ where $y$ is not $1$ then $f$ is constant.
How can I prove that?
AI: Let $V:=f^{-1}(\{y\})$, this is an 'open' set, as $\{y\}\in\tau$, i.e. $V\in\tau$. By $f(1)=y$, we know that $1\in V$. So, by definition of $\tau$, we have $V=\Bbb R$. |
H: Perfect Fourth Power - Pigeon Hole Principle
Let $a_1, a_2, ..., a_n$ be positive integers all of whose prime divisors are $\le$ 13.
Show that if $n \ge 193$ then there exists four of these integers whose product is a perfect fourth power.
I tried getting many pairs of numbers which multiply to a square but did not get far. For part b, it seems like I want to get two disjoint pairs $a, b$ and $c, d$ such that $\sqrt{ab}\sqrt{cd}$ is a square.
AI: HINT: Consider the exponents of the primes in the factorization of the $a_i$, and think of them $\pmod 4$.
Each of the $a_i$, as well as any product of $a_i$, can be written in the form $$2^{p_2}3^{p_3}5^{p_5}7^{p_7}11^{p_{11}}13^{p_{13}}\cdot n^4$$ where each of the $p_i$ is one of {0, 1, 2, 3}. For example, $$5,244,319,080,000 = 2^23^15^07^311^213^1\cdot(6^4),$$ and $$135 = 2^03^35^17^011^113^0\cdot(1^4).$$
You want to find four of the $a_i$ whose product has each of the $p_i$ equal to zero$\pmod 4$. |
H: Find all the values of x such that the given series would converge.
$\sum_{n=1}^{\infty} (5^n (x-9)^n) / (n+9) $
So this is a homework question and we have unlimited tries to check our answer, however, the answer I got as well as my friend who has been helping me with my Calc hw got is considered wrong.
The result of the ratio test was $5(x-9)$
So I got $-9.2 < x < 9.2$ but this is wrong.
Where am I going wrong or what am I doing wrong?
Thanks.
AI: What about the $\;n$-th root test?
$$\sqrt[n]{\left|\frac{5^n(x-9)^n}{n+9}\right|}=\frac{5|x-9|}{\sqrt[n]{n+9}}\xrightarrow[n\to\infty]{}5|x-9|$$
and now
$$5|x-9|<1\iff -\frac15<x-9<\frac15\iff 9-\frac15<x<9+\frac15\iff$$
$$\frac{44}5=8.8<x<9.2=\frac{46}5$$
Now check end points:
$$x=\frac{44}5\implies \sum_{n=1}^\infty\frac{5^n\left(-\frac15\right)^n}{n+9}=\sum_{n=1}^\infty\frac{(-1)^n}{n+9}$$
and the last one is a Leibniz Series and thus converges.
Check now the other point (and get convinced there is not convergence there), and thus the original series converges for
$$8.8\le x<9.2$$ |
H: Question about one of the first problems in Spivak's Calculus
It's about Chapter I, Problem 21 from Spivak's Calculus:
Prove that if:
$|x - x_0| < \frac{\epsilon}{2}$ and $|y - y_0| < \frac{\epsilon}{2}$
then
$|(x + y) - (x_0 + y_0)| < \epsilon$
$|(x - y) - (x_0 - y_0)| < \epsilon$
I tried somehow making $2\times$ the first expressions equal $1\times$ the second expressions, but ended up in a big hairy mess, not sure what to do. I feel real dumb for not being able to do this, it's just chapter 1. Anyone who's done this have a tip? Maybe try to represent absolute values as $\sqrt{x^2}$ everywhere?
AI: Hint: The triangle inequality states that
$$|a + b| \le |a| + |b|$$
for all choices of $a$ and $b$. This is an extremely important inequality to know. |
H: Expected Value of Y = MIN(X, 100) where X~Geometric
Let X = Geometric$(\theta)$ and Y = Min(X, 100). Compute E(Y)
My thoughts are:
Y = g(X) = Min(X, 100).
E(Y) = E(g(X))
= $\sum_{x=0}^{\infty} g(X)P(X=x)$
$$= \sum_{x=0}^{\infty} Min(X, 100) P(X=x)$$
$$= \sum_{x=0}^{100} x P(X=x) + \sum_{x=101}^{\infty} 100 P(X=x)$$
$$= \theta (\sum_{x=0}^{100} x (1-\theta)^x) + 100 \theta \sum_{x=101}^{\infty} (1- \theta)^x$$
$$= \theta [\frac{100(1-\theta)^{102}) - 101(1-\theta)^{101}+1-\theta}{\theta^2}] - 100 \theta(\frac{(1-\theta)^{101}-1}{(1-\theta)-1}) + 100 \theta \sum_{x=0}^{\infty} (1- \theta)^x$$
$$= \theta [\frac{100(1-\theta)^{102}) - 101(1-\theta)^{101}+1-\theta}{\theta^2}] - 100 [1 - (1-\theta)^{101}] + 100 \theta(\frac{1}\theta{})$$
$$= [\frac{100(1-\theta)^{102}) - 101(1-\theta)^{101}+1-\theta}{\theta}] +100(1-\theta)^{101}$$
Is this right? It looks really ugly.
AI: $$ \sum_{x=0}^n (x - n) r^x = \dfrac{(n+1) r - n - r^{n+1}}{(r-1)^2}$$ |
H: Calculate the confidence interval for a single variance given two variances (using pooled variance)
I am having trouble trying to interpret why I'm given some value for a homework exercise. It goes like this:
Suppose we're studying the time it takes for a certain industrial process to complete. A recent study, which measured the time it took to complete 51 processes, gave a mean time of 2.396 minutes with standard deviation of 1.967 minutes. From past studies it has been observed that the standard deviation of the time it takes for the process to complete is 2.1 minutes. Calculate a confidence interval of 98% for the standard deviation of the time it takes to complete the process.
I don't understand why would they give me the value of the standard deviation of past processes, if I already have a value for $s^2$ (namely $1.967$). By the way, I build the confidence interval by using the fact that $(n-1)S^2/\sigma^2$ follows a Chi squared distribution and then calculating the probability of $\sigma^2$ being between the inferior and superior limits of the $\chi^2$ distribution for which the area under the curve is equal to $(1-0.98)/2$. I do not know why is that past value there if the interval can be calculated using only the first value.
I remember that in class, the professor said that $S_p^2$ is the single best estimator of the variance of a single population in terms of the variance of another population, though I didn't understand that. I suspect I can use the pooled variance in order to calculate a more accurate confidence interval taking in account both values of the standard deviation that they give me?
And in general, what sense does it make to calculate the confidence interval for a variable whose value I already know?
AI: If I'm interpreting the problem correctly, I believe the point of including the true variance is so that you may compare your confidence interval on the standard deviation with the true value. |
H: experimental sequence of number
I'm doing a small numerical experiment. I got, from the first simulations, the following sequence of numbers. I'm trying to imagine a mathematical law behind this sequence. It could be a geometric progression?
1.333333333
1.75
2.491107286
3.835656425
6.219456109
10.44686695
17.99815841
31.6031634
56.31814911
101.5516622
...
AI: Your data looks like a simple exponential growth. To a rough approximation:
y = A*exp(x/t) + y0
y0 0.95321
A 0.26613
t 1.68515
...which makes it a geometric progression with an added constant term. |
H: Mathematical induction... divisible by 4
Hello I need to proof that the expression $(9^{n}+3)$ is divisible by $4$.
It is true if I calculate it for $n=1$
for $n + 1$ I got stuck in here:
$9 \cdot 9^n + 3$
I don't know how to continue. Can anyone help me please?
AI: Suppose $9^n + 3$ is a multiple of $4$. Then
$$ (9^{n+1}+3)-(9^n+3)=9^{n+1}-9^n =9^n(9-1)=8*9^n, $$
which is a multiple of four. The difference of $9^{n+1}+3$ and $9^n+3$ is a multiple of 4, and $9^n + 3$ is a multiple of 4, so $9^{n+1}+3$ is a multiple of 4. |
H: Quotient space of S1 is homeomorphic to S1
$S^1=\{z\in\mathbb{C}\mid |z|=1\}$, let $w\sim z$ iff $w=z\vee w=-z$ (identifying antipodal points). Prove $S^1/\sim$ is homeomorphic to $S^1$. Which function should be used to construct a homeomorphism? I am not good at analysis. Thanks!
AI: $f:S^1\to S^1,\ e^{it}\mapsto e^{2it}$, where $t\in[0,2\pi]$, is the function you should use. Note that this function itself is continuous because it is induced by the map $t\mapsto e^{2it}$ which gives the same value for $t=0$ and $t=2\pi$. Once seen that it is well-defined, you can of course write it as $f(z)=z^2$. |
H: Derivative of $-2e^{-x^2}x$
$$f(x) = -2e^{-x^2}x$$
Find $f'(x)$:
I have making a small mistake some where in my calculation and I cannot find it.
The answer is stated as:
$$e^{-x^2}(4x^2-2)$$
However I got something different from my work:
$$f'(x)=-2[(e^{-x^2})'(x)+(e^{-x^2})(x)']$$
$$=-2[(e^{-x^2})(-2x)+e^{-x^2}]$$
$$=-2[-2xe^{-x^2}+e^{-x^2}]$$
$$=4xe^{-x^2}-2e^{-x^2}$$
$$=e^{-x^2}(4x-2)$$
AI: (1) $f(x)=-2e^{-x^2}x$ (apply product rule)
(2) $-2 (e^{-x^2}(1)+x(e^{-x^2})\frac{d}{dx}(-x^2)$)
(3) $-2(e^{-x^2}-2x^2e^{-x^2})$
(4) $-2e^{-x^2}+4x^2e^{-x^2}$
(5) $e^{-x^2}(4x^2-2)$ |
H: Can we assume independence of random variables?
I have the following problem I want to solve: Let $\delta,\varepsilon>0$ and $X_n$ a sequence of non-negative random variables such that $P(X_n\geq \delta)\geq \varepsilon$, show that with probability one $\sum_{n=1}^{\infty}X_n = \infty$.
I solved this problem very easily by assuming the $X_n$ to be independent. Essentially the argument is to consider $\sum_{n=1}^{\infty} P(X_n\geq \delta) = \infty$ and so by Borel-Cantelli Lemma we have that $P(X_n\geq \delta ~ \text{i.o.})=1$. From where it follows that $\sum_{n=1}^{\infty} X_n = \infty$ since infinitely many of the $X_n\geq \delta$ with probability one.
But in the problem there is no mention of independence. I believe one can do the following, tell me if this is a legitimate step. Construct a probability space $(\Omega,\mathcal{A},\mu)$ with a sequence $Y_n$ of independent random variables on $\Omega$ so that $X_n$ and $Y_n$ induce the same distributions i.e. $P(X_n\in A) = \mu(Y_n\in A)$ for any set $A$ of real numbers. Thus, even though the problem never says to assume the $X_n$ are independent we may assume that anyway.
Thank you.
AI: No, your approach does not work. While you can indeed construct such a sequence of independent random variables, the distribution of their partial sums will depend on their joint distributions, which you would change.
But the result is wrong. Let $X$ be a random variable that is $0$ with probability $1/2$ and $1$ with probability $1/2$. Let $X_n=X$ for all $n$, so that they are perfectly correlated. Let $\delta<1$ and $\epsilon<1/2$. |
H: DFA Transition Function Inductive Proof
Show for any state $q$, string $x$, and input symbol $a$, $\hat\delta(q, ax) = \hat\delta(\delta(q, a), x)$, where $\hat\delta$ is the transitive closure of $\delta$, which is the transition function for some DFA.
I think the best way to proceed is by induction and that the following is the basis step:
Basis: $\hat\delta(q, a) = \hat\delta(\delta(q, a), \epsilon)$
But I am not sure how to proceed to the inductive step as I'm a little unsure how to use induction in these cases: it seems much different from numerical induction. Can someone provide a suggestion?
AI: I think, $\hat\delta$ is not 'transitive closure', at least not as it is meant among relations, rather it is just an extension of $\delta$ to input also strings not just characters.
And it seems that it is just the recursive definition of this, so nothing to prove.
Let $\hat\delta(q,\varepsilon):=q$.
If it is already defined on word $x$ with all states, then let $\hat\delta(q,ax):=\hat\delta(\delta(q,a),x)$. |
H: Show $\pi\alpha\pi^{-1}$ and $\pi\beta\pi^{-1}$ are disjoint given $\alpha$ and $\beta$ are disjoint for an arbitrary permutation $\pi$
Let $\alpha$ and $\beta$ be disjoint cycles of the same length $s$. Show that for any permutations $\pi$ we have that
$$\pi \alpha \pi^{-1}\quad \text{and }\quad \pi \beta\pi^{-1} $$
are disjoint as well.
$\alpha = (a_1a_2\ldots a_s)$ and $\beta= (b_1b_2\ldots b_s)$. Define the sets of elements $A=(a_1,\ldots ,a_s)$ and $B=(b_1,\ldots ,b_s)$. Now it is possible that there are elements that are unchanged by both permutations. Let these elements be the set $C=(c_1 ,\ldots ,c_r)$. Now for each element $x$, if $\pi^{-1}(x) \in A$ we have that
$$\pi \alpha \pi^{-1}(x) \neq x\quad \text{and }\quad \pi \beta\pi^{-1}(x) = x=e(x) $$
Similarly if $\pi^{-1} \in B$ we find that
$$\pi \alpha \pi^{-1}(x) =x=e(x)\quad \text{and }\quad \pi \beta\pi^{-1}(x)\neq x$$
Lastly, if $\pi^{-1}(x) \in C$ we have that
$$\pi \alpha \pi^{-1}(x) = \pi \beta\pi^{-1}(x) = x=e(a) $$
I feel like this is leading to a proof of the desired fact... I am not sure how to rigorously finish the argument though. Any help would be greatly appreciated. Thanks!
AI: Hint: if $\alpha = (a_1 \; a_2 \; a_3 \;\cdots\; a_s)$, then $\pi \alpha \pi^{-1} = (\pi(a_1) \; \pi(a_2)\; \cdots \; \pi(a_n))$ (do you see why?) |
H: Can even-dimensional complex skew-symmetric matrices be made block diagonal by orthogonal matrices?
Suppose an even dimensional matrix Q is complex and skew-symmetric. Can Q be written as $Q= O^T \Sigma O$, where O is orthogonal (i.e. $O^T O = O O^T=1$) and $\Sigma$ is block diagonal with $2 \times 2$ blocks?
AI: With such a $\Sigma$, the size of every Jordan block of $Q=O^\top \Sigma O=O^{-1}\Sigma O$ must also be at most $2\times2$. However, according to Lemma 5.2.1 on p.35 of Olga Ruff's Master thesis, for any natural number $k$ and any $\lambda\in\mathbb{C}$, the matrix $J_k(\lambda)\oplus J_k(-\lambda)$ is similar to some complex skew-symmetric matrix. Hence there exists a complex skew-symmetric matrix $Q$ whose Jordan form contains a $3\times3$ Jordan block. In other words, the answer to your question is negative when the dimension of $Q$ is at least $6\times6$. |
H: if $\mu(A)>0$, and $f
here is my doubt:
Let $(X,\mathcal{T}, \mu)$ be a measure space, $f,g:X\to \overline{\mathbb{R}}$ measurable functions (indeed with finite integral), and $A \in \mathcal{T}$ such that $\mu(A)>0$ and
$$
f(x)<g(x)\quad \forall x \in A
$$
then $\int_A f\ d\mu<\int_A g\ d\mu$.
Any hint in order to prove (or disprove) it would be wellcome.
Thanks in advance!
AI: Suppose that $\mu(A) > 0$ and $f > 0$ on A. Put
$$E_n = \{\omega\in A| f(\omega) \ge 1/n\}.$$
Then $E_n \uparrow A.$ Since $\mu(A) >0$ there is some $n\in\mathbb{N}$ so that
$\mu(E_n) > 0$. What can you say about
$$\int_{E_n} f\, d\mu?$$ |
H: If $p$ is prime, then $n\mid\varphi(p^n-1)$
How can I prove that the value of $\varphi(p^n-1)$ (where $p$ is prime and $n$ is some positive integer) is some multiple of $n$? The purpose of this is to prove that $n$ divides $\varphi(p^n-1)$.
AI: Consider the group of units $\Bbb Z/D\Bbb Z^\times$ where $D=p^n-1$, $p$ a prime.
What is the order of $p$? |
H: If $r, s, t \in R$, then $r \gcd (s, t)$ is associate to $\gcd(rs, rt)$.
I seem to be stumped on this question. For the setting, let $R$ be an integral domain and let $r, s, t \in R$. The question asks
Show that $r \gcd(s, t)$ is associate to $\gcd (rs, rt)$
To start, let $d$ be some $\gcd$ of $s$ and $t$, and let $\overline{d}$ be some $\gcd$ of $rs$ and $rt$. Since $d \mid s$ and $d \mid t$, we see that $rd \mid rs$ and $rd \mid rt$. By definition, then $rd \mid \overline{d}$. It is here that I am stuck. I would like to show that $\overline{d} \mid rd$ to conclude that $rd$ and $\overline{d}$ are associate, but I cannot see how to get there. I know that I can write $s = di$ and $t = dj$ for some $i, j \in R$. I can also write $rt = \overline{d}n$ and $rs = \overline{d}m$ for some $m, n \in R$. But all this gets me is $rd(i-j) = \overline{d}(m-n)$, which is not quite what I am after. Any helpful hints would be greatly appreciated.
AI: It suffices to show $r\mid\bar{d}$, since then using cancellation ($R$ is an integral domain) we get
$$\bar{d}\mid rs,rt\implies (\bar{d}/r)\mid s,t\implies (\bar{d}/r)\mid d\implies \bar{d}\mid rd.$$
How to show $r\mid\bar{d}$? Easy: $r\mid rs,rt\implies r\mid\bar{d}$. |
H: Quotient ring of a matrix ring
Let $F$ be a field. Consider the set
$$R = \left\{ \begin{bmatrix}
a & b \\
0 & c
\end{bmatrix} : a, b, c \in F \right\}$$
Think of a non-trivial two-sided ideal of $R$ and describe in a concrete way the quotient ring $R/I$. (Your description should be explicit enough to make the fact that $R/I$ is a commutative ring completely transparent.)
I can think of some ideals of $R$, however I don't understand the meaning of a quotient ring when it comes to matrices nor how to describe them.
AI: The meaning of a ring is no different when the elements of the ring are matrices (or vectors, or elephants, or ...).
Given a ring $R$ and an ideal $I \trianglelefteq R$ the quotient $R/I$ is given by
$$R/I = \{ x + I : x \in R \}$$
with addition and multiplication defined by $(x+I)+(y+I) = (x+y)+I$ and $(x+I) \cdot (y+I) = (x \cdot y) + I$.
Here it's exactly the same. If $I \trianglelefteq R$ is an ideal then
$$R/I = \left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} + I : a,b,c \in F \right \}$$
To make it really explicit, notice that
$$\begin{bmatrix} a & b \\ 0 & c \end{bmatrix} + I = \left\{ \begin{bmatrix} a + x & b + y \\ 0 & c + z \end{bmatrix} : \begin{bmatrix} x & y \\ 0 & z \end{bmatrix} \in I \right\}$$ |
H: Is it possible to simplify the following combination?
$$C(n, r-1) = \frac{(n)!}{(n - (r - 1))! (r - 1)!} $$
Could I simplify any further or break up the $(n - (r - 1))!$ part? I'm having a hard time following a problem in the book and if I could understand this it would really help out.
AI: Note that $$\frac{n!}{(n-(r-1))!} = n (n-1) (n-2) \cdots (n-(r-1)+1)$$
and that in particular, the right hand side has $r-1$ terms. Maybe this helps?
For example, $C(10,3) = \frac{10!}{7!3!}=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}$.
This form is a little more intuitive if you "reason" through the counting: to count the number of groups of three you can form out of ten people, there are $10$ choices for the first, $9$ choices for the second, and $8$ choices for the third person, but you need to divide by $3!$ because $10\cdot 9 \cdot 8$ is the number of ordered teams (say, assigning them to president, vice president, and secretary), and we don't care about order. |
H: Prove that $\lim_{n \to \infty} \int_0^2 e^{ x^2 / n}\,{\rm d}x$ exists and evaluate it.
I need to show that this limit exists and then evaluate it. It is from a section on uniform convergence of sequences. I know that if $f_n \rightarrow f$ uniformly and each $f_n$ is integrable, then I can bring the limit inside of the integral. I'm not sure if this will be the right way to approach this, and if it is how to show that it is all of those things that would allow me to bring the limit in. Looking for some advice. Thanks
AI: Hint.
If $0\le x\le 2$ then
$$1\le e^{x^2/n}\le e^{4/n}.$$ |
H: Why is $L^{1}(G)$ unital if and only if $G$ is discrete?
I've seen it stated in several sources and lecture notes for Abstract Harmonic Analysis that for a locally compact group $G$, $L^{1}(G)$ is unital if and only if $G$ is discrete.
What about the locally compact group $\mathbb{T} = \{\lambda\in\mathbb{C}: |\lambda| = 1\}$, which is not discrete because the arclength measure of a point on the unit circle is $0$.
But since it is compact, the constant function $1\in L^{1}(G)$.
AI: The multiplication in the algebra $L^1(G)$ is convolution, not the pointwise product. For groups like $\mathbb{T}$ or $\mathbb{R}^n$, the pointwise product of $L^1$ functions is generally not in $L^1$.
You can extend the convolution to the space of Borel measures of bounded variation, and then you get a unital algebra, where the unit is the Dirac measure (point mass) in the unit of $G$. You can represent a point mass as an $L^1$ function with respect to the Haar measure only if singletons have positive measure. That is the case only if $G$ is discrete. |
H: write formula to predict nth term of sequence $1, 1\cdot3, 1\cdot3\cdot5, 1\cdot3\cdot5\cdot(2n-1)$
How can I write a formula for a sequence with the following behavior:
{$1, 1\cdot3, 1\cdot3\cdot5, 1\cdot3\cdot5\cdot7, 1\cdot3\cdot5\cdot7\cdot9$}
1st term is $1$
2nd term is $1 \cdot 3 = 3$
3rd term is $1 \cdot 3 \cdot 5 = 15$
4th term is $1 \cdot 3 \cdot 5 \cdot 7 = 105$
Please show steps on how to arrive at the answer:
Ted
AI: The classic trick is to multiply and divide by the even terms:
$$
a_n=1\cdot3\cdot\dots\cdot(2n-1)=\frac{1\cdot2\cdot3\cdot \dots\cdot 2n}{2\cdot 4\cdot (2n)}
$$
The numerator is $(2n)!$. As for the denominator, we can factor each of the $n$ terms by $2$, so the denominator is $2^nn!$.
Finally, $a_n=\frac{(2n)!}{2^nn!}$ |
H: If $G=H\rtimes K$ where $H$ is cyclic, and $K$ is abelian, why is $G$ abelian?
This is a curious problem I've been stuck on.
Suppose $G=H\rtimes K$, where $H$ is cyclic of order $n$, $K$ abelian with $\gcd(|K|,\varphi(n))=1$, $\varphi$ being the totient function. Why is $G$ actually abelian?
I let $h$ be a generator of $H$. Computing the product of two elements in two ways
$$
(h^i,k_1)(h^j,k_2)=(h^ik_1h^jk_1^{-1},k_1k_2)
$$
and
$$
(h^j,k_2)(h^i,k_1)=(h^jk_2h^ik_2^{-1},k_2k_1).
$$
Since $K$ is abelian, I just need to show the first entries are equal,
$$
h^ik_1h^jk_1^{-1}=h^jk_2h^ik_2^{-1}.
$$
I don't know where to go after that, and I don't see how to use the fact that $\gcd(|K|,\varphi(n))=1$. Does anyone see what to do? Thanks.
AI: Hint:
If $H\rtimes K$, then we know that it comes from some homomorphism $K\to\text{Aut}(H)$. Now since $H$ is cyclic we know that $|\text{Aut}(H)|=\varphi(|H|)$. So what has to be true about this homomorphism, and what does this tell us about the semidirect product? |
H: Question about Euler's approach to find $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$
For a freshman calculus project, I used Euler's approach to find $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$, and noted from Wikipedia's explanation that the infinite product representation of $\frac{\sin x}x=\prod_{n=1}^\infty(1-\frac{x^2}{n^2\pi^2})$ is unjustified without Weierstrass' factorization theorem. I'm finding it very difficult to follow the article about Weierstrass' theorem.
Can someone explain to me what's unjustified about Euler's infinite product representation? Since $\frac{\sin(x)}x$ has a Taylor polynomial representation, and I think all polynomials have roots (in the set of complex numbers), shouldn't it also have a infinite product of roots representation?
Can someone explain to me what Weierstrass' theorem does to justify Euler's representation, and if it's within the ability of a freshman calculus student, can someone show me a proof that is more accessible that the ones I've found by Googling?
Thanks for your time. This is a very interesting problem and very different from the ones I'm used to doing in my calculus class.
AI: If you believe Euler's method is totally rigorous, you could also argue in the same way that
$$ e^x \frac{\sin x}x = \text{the same product formula you wrote above} $$
because $e^x$ has no complex roots. |
H: Power series representation
I'm trying to find the series representation of $ f(x)=\int_{0}^{x} \frac{e^{t}}{1+t}dt $. I have found it using the Maclaurin series, differentiating multiple times and finding a pattern. But I think there must be an easier way, using the power series of elementary functions. I know that $e^{x}=\sum_{0}^{\infty}\frac{x^{n}}{n!}$ and $\frac{1}{1+x}=\sum_{}^{\infty}(-1)^{n}x^{n}$ but I don't know how to use it here. Thanks
(Don't hesitate to correct my English)
AI: $$e^x=\sum_{n\ge 0}\frac{x^n}{n!}$$
$$\frac{1}{x+1}=\sum_{n\ge 0}(-1)^nx^n$$
$$\frac{e^x}{x+1}=\sum_{b\ge 0}\sum_{a\ge 0} \frac{(-1)^b}{a!}x^{a+b}$$
$$\frac{e^x}{x+1}=\sum_{n\ge 0}\sum_{a+b=n}_{a,b\ge 0} \frac{(-1)^b}{a!}x^{a+b}$$
$$\frac{e^x}{x+1}=\sum_{n=0}^\infty x^n(-1)^n(\sum_{k=0}^n (-1)^{k}\frac{1}{k!})$$
Now the interesting thing about these coefficients above that are given in the form of partial sums is that it turns out that we have:
$$\sum_{k=0}^n\frac{(-1)^k}{k!}=\frac{1}{e}+\frac{1}{n!}(\frac{(-1)^n+1}{2}-\{ \frac{n!}{e}\})$$
Where {.} is the fractional part function.
So that we get:
$$\frac{e^x}{x+1}=\sum_{n=0}^\infty x^n(-1)^n(\frac{1}{e}+\frac{1}{n!}(\frac{(-1)^n+1}{2}-\{ \frac{n!}{e}\}))$$
$$=\frac{1}{e}\sum_{n=0}^\infty x^n(-1)^n+\sum_{n=0}^\infty
x^n(-1)^n\frac{1}{n!}(\frac{(-1)^n+1}{2}-\{ \frac{n!}{e}\})$$
$$=\frac{1}{e(x+1)}+\frac{1}{2}\sum_{n=0}^\infty\frac{x^n}{n!}(1+(-1)^n)-\sum_{n=0}^\infty\frac{x^n(-1)^n}{n!}\{ \frac{n!}{e}\}$$
$$=\frac{1}{e(x+1)}+\frac{e^x+e^{-x}}{2}-\sum_{n=0}^\infty\frac{x^n(-1)^n}{n!}\{ \frac{n!}{e}\}$$
$$=\frac{1}{e(x+1)}+\frac{e^x+e^{-x}}{2}+\sum_{n=0}^\infty\frac{x^n}{n!}\{ \frac{n!}{e}\}(-1)^{n+1}$$
Now integrating term by term gives,
$$\int_{0}^x \frac{e^t}{1+t} dt=\frac{\ln(x+1)}{e}+\frac{e^x-e^{-x}}{2}+\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}\{ \frac{n!}{e}\}(-1)^{n+1}$$
Thus for all $|x|<1$ the following expansion holds:
$$f(x)=\int_{0}^x \frac{e^t}{1+t} dt=\frac{\ln(x+1)}{e}+\frac{e^x-e^{-x}}{2}+\sum_{n=1}^\infty\frac{(-x)^n}{n!}\{ \frac{(n-1)!}{e}\}$$
Also pretty off topic, but the positive $n^{\text{th}}$ coefficients that took the form of partial sums in our first list of series manipulations above correspond to solutions to the hat-check problem when $n$ hats are being considered. |
H: If $\alpha$ and $\beta$ are algebraic integers then any solution to $x^2+\alpha x + \beta = 0$ is also an algebraic integer.
An algebraic integer is a complex number that is a root of a monic polynomial with coefficients in $\mathbb{Z}$.
Let $\alpha$ and $\beta$ be algebraic integers. Then any solution to $x^2+\alpha x + \beta = 0$ is also an algebraic integer.
What we have so far is that $2x+\alpha$ is an algebraic integer but since the set of algebraic integers is a ring I can't divide by 2 so I can't have an algebraic integer and also the polynomial wouldn't be monic.
I appreciate the help.
AI: If I well remember, there's a characterization of integral elements over a ring R (in our case $R=\mathbb{Z}$), in particular $x$ is integral over $R$ if and only if it is contained in a ring $ C$ such that $R \subset C$ and $C$ is a finite algebra over $R$.
In our case a rooth of $x^2+\alpha x+\beta$ is integral over $\mathbb{Z}[\alpha,\beta]$ and so there exist a ring $C$ with the property above.
Now $\mathbb{Z}[\alpha,\beta]$ is a finite $\mathbb{Z}$-module, because $\alpha$ and $\beta$ are algebraic integers. Then $C$ is a finite $\mathbb{Z}$-module and $x$ is an algebraic integer. |
H: How can I further simplify $(a \le b) \lor (b \le a)$ to prove that it is a tautology?
Over $\mathbb{Z}$, $aRb \iff a \le b \lor a = 3b$. Determine if it is total.
I think it is:
Have arbitrary elements $a,b \in \mathbb{Z}$.
We have to prove that $aRb \lor bRa$, which can be written as:
$$(a \le b \lor a = 3b) \lor (b \le a \lor b = 3a)$$
Further simplified:
$$(a \le b) \lor (b \le a)$$
Clearly, this should be a tautology. But I'm failing to further simplify this. What else can I do with it?
AI: Your simplified $(a \le b) \lor (b \le a)$ is indeed a tautology, because its negation is the following contradiction $(x > y) \land (y > x)$. |
H: Combinations question confusing
There are 8 men and 7 women from which a group of 4 with at least 2 men must be selected. Find how many possible groups there are.
There are 2 ways which both make sense to me but only one method is right.
Correct method:
(No. of groups with 2 men, 2 women) + (No. of groups with 3 men, 1 woman) + (No. of groups with 4 men)
$_8C_2 \times _7C_2 + _8C_3 \times _7C_1 + _8C_4=1050$
Incorrect method:
No. of ways of choosing 2 men from 8 = $_8C_2$
This leaves the remaining 2 spots unrestricted so 2 people can be chosen from the remaining 13 people.
$_8C_2 \times _{13}C_2 = 2184$
Any idea why this second method is incorrect?
AI: The incorrect method counts some possible groups more than once. One possible group (using tradition male/female names to indicate sex) is {Juan, Mary, Patrick, Luke}. The second method will count it three times: once where Juan and Luke are the 2 men chosen first, once with Juan and Patrick, and once with Patrick and Luke. |
H: Probability density function of a summation of continuous random variables
Let $Z_{i} = \tau + X_{i}$, where $X_{i}$ is a exponential random variable ($X_i \sim \varepsilon(\lambda)$), $0<\tau, \lambda < \infty$
Assume $X_i$ are independent random variables.
Suppose $T_{n} = Z_{1} + Z_{2} + \cdots + Z_{n}$, find the PDF of $T_{n}$, $h(t_n)$.
My approach:
$T_{n} = \sum_{i=1}^{n}\left ( \tau + X_{i} \right )$
$T_{n} = n\tau+\sum_{i=1}^{n}\left (X_{i} \right )$
Let $Y$ be $\sum_{i=1}^{n}\left (X_{i} \right )$, I find the PDF of $Y$, $g(y)$
$T_{n} = n\tau+ Y$
$Y = T_{n} - n\tau$
Using the Jacobian of transformation, I find PDF of $T_{n}$: $h(t_n) = \left | \frac{\mathrm{d} }{\mathrm{d} t_{n}}(t_{n} - n\tau) \right |\times g(t_{n} - n\tau)$
I am stuck on step 3, as I have no idea how to find the PDF of $\sum_{i=1}^{n}\left (X_{i} \right )$. Any ideas?
By the way, if there's any wrong with my approach, please say so.
Thanks a lot.
AI: The main thing you need to know here is that the probability distribution of a sum of independent exponentially distributed random variables with the same scale parameter (or, equivalently, with the same rate parameter) has a gamma distribution, whose "shape parameter" is the number of random variables being added. That can be proved by at least (estimating conservatively) two different methods.
(I don't know whether you intended $\lambda$ to be the scale parameter and $1/\lambda$ the rate parameter or vice-versa. It's usually a good idea to specify that when writing about such things, but in this case the paragraph above stands either way.)
PS: Here is a proof. The gamma distribution is the distribution whose density is
$$
\text{constant}\cdot x^{n-1} e^{-x/\lambda} \text{ on the interval }0\le x<\infty
$$
where $\lambda>0$ is the scale parameter and $n>0$ is sometimes called the shape parameter because it determines the shape of the graph of the density function.
Notice that an exponential distribution is a gamma distribution with $n=1$.
Suppose two independent random variables $X$ and $Y$ have gamma distributions with the same scale parameter $\lambda>0$ and shape parameters $m$ and $n$. Then their joint density is
$$
\text{constant}\cdot x^{n-1} e^{-x\lambda} y^{m-1} e^{-y/\lambda}.
$$
in the quadrant $0\le x$, $0\le y$. Then
$$
\Pr(X+Y\le w) = \iint\limits_{\{(x,y)\,\mid\,x+y\,\le\,w\}} \text{constant}\cdot x^{n-1} e^{-x\lambda} y^{m-1} e^{-y/\lambda} \, dx\,dy.\tag1
$$
We will use the substitution
\begin{align}
u & = x+y, \\
v & = y/(x+y), \\[10pt]
\text{hence }
u\,du\,dv & = dx\,dy, \\
x & = u(1-v), \\
y & = uv.
\end{align}
Then $0\le u<\infty$ and $0\le v\le 1$. The integral $(1)$ becomes
$$
\begin{align}
& \phantom{={}}\text{constant}\cdot\int_0^w \int_0^1 u^{n-1}(1-v)^{n-1} u^{m-1} v^{m-1} e^{-u/\lambda} (u\,dv \,du) \\
& = \text{constant}\cdot\int_0^1 (1-v)^{n-1}v^{m-1} \, dv \cdot \int_0^w u^{n+m-1} e^{-u/\lambda} \,du \\
& = \text{constant} \cdot \int_0^w u^{n+m-1} e^{-u/\lambda} \,du.
\end{align}
$$
The integral involving $v$ is a "constant" since it does not depend on $w$. What we get is therefore a gamma distribution with the same scale parameter $\lambda$ and shape paramter $n+m$.
Thus we have: The distribution of the sum of two independent gamma-distributed random variables with the same scale parameter is another gamma-distributed random variable with the same scale paremeter, and the shape parameter is just the sum of the two shape parameters.
Now apply this repeatedly adding one more independent gamma-distributed random variable at each step. |
H: Extended Proof of the Theorem that a bounded analytic function is constant.
I am having trouble feeling convinced by my proof and more importantly - feeling confident in my working out. The question reads
(a) Let $f$ be an entire function such that there exist real constants $M$ and $N$ such that $|f(z)|<M|z|+N$ for all $z$. Prove that for any three pairwise different complex numbers $a,b,c$, $\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}=0$.
(b) Deduce that there are constants $A,B\in\mathbb{C}$ such that $f(z)=Az+B$ for all $z$.
So to begin, I realise that the LHS of the equality that I am required to prove is simply the sum of residues (without $2\pi i$) of the integral $\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz$. That is,
$$\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz=2\pi i\left(\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}\right),$$ for $\Gamma$ being the circle contour of radius $R$. Furthermore, for the equality to hold, singularities at $z=a,b,c$ must be inside $\Gamma$.
So the next chain of thought would be show that $$\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz=0.$$This will obviously yield the result I want. I relate to limits and the M-L Lemma to do this. So,
$$\left|\frac{f(z)}{(z-a)(z-b)(z-c)}dz\right|\leq\frac{MR+N}{(R-|a|)(R-|b|)(R-|c|)},$$ by the equality given in the question, reserve triangle inequality and since complex numbers $a,b,c$ lie inside the contour thus $|a|,|b|,|c|<R$. Thus by the M-L lemma we have that $$lim_{R\to\infty}\left|\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz\right|\leq \lim_{R\to\infty}\frac{(MR+N)2\pi R}{(R-|a|)(R-|b|)(R- |c|)}$$ Clearly the RHS of the inequality converges to $0$. Thus $$lim_{R\to\infty}\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz=0.$$But we know that $$lim_{R\to\infty}\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz=lim_{R\to\infty}2\pi i\left(\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}\right).$$ So $$2\pi i\left(\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}\right)=0$$$$\implies\left(\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}\right)=0.$$
With part (b), I begin by using the result I proved. Simplifying the expression I have that $$f(a)(b-c)+f(b)(c-a)+f(c)(a-b)=0.$$ It's clear that if I let $f(z)=Az+B$ and substitute it into the equation above then I am done. Is this valid though?
Thank you to all for your time in advanced.
AI: It's clear that if I let $f(z)=Az+B$ and substitute it into the equation above then I am done. Is this valid though?
No, that would be a petitio principii. That's the form you want to deduce.
But you have, for any three pairwise different $a,b,c \in \mathbb{C}$, that
$$\frac{f(a)}{(a-b)(a-c)} + \frac{f(b)}{(b-a)(b-c)} + \frac{f(c)}{(c-a)(c-b)} = 0.$$
Now, what if you rename the three, say you use $z = a$, $1 = b$, $0 = c$? |
H: Show that this set of polynomials is ideal in F[x]
In $\mathbb{F}[x]$, where $\mathbb{F}$ is a field, let $J$ be the set of elements of polynomials that have coefficients that add to zero (so $a_0 + a_1 + ... + a_n = 0$). Show that $J$ is an ideal of $\mathbb{F}[x]$. I know that the proof of this statement is meant to be very short, but I don't know how to go about it.
AI: Consider the map $\eta:F[x]\to F$ where $P$ is mapped to the sum of its coefficents, that is $P\mapsto P(1)$. You need to show it is a ring homomorphism, and all will follow. |
H: Is it true that $(4+\sqrt{14})(4-\sqrt{14}) = (4+\sqrt{14})^2$ in $\mathbb{Q}(\sqrt{14})$?
Is it true that $(4+\sqrt{14})(4-\sqrt{14}) = (4+\sqrt{14})^2$ in $\mathbb{Q}(\sqrt{14})$? I am going through the solution of a problem I'm working on and this seems to be what they are saying. If its true, why so? I see that the left hand side is $(2)$ and the right hand side is $(2(15+4\sqrt{14}))$ but then what?
AI: It's true, if you are talking about ideals.
Clearly, $(2(15+4\sqrt{14})) \subset (2)$. For the other inclusion, note that $(15+4\sqrt{14})(15-4\sqrt{14}) = 1$. |
H: How to prove that $P[X_1=X^{(i)}]=\frac{1}{n}$?
Let $X_1$, $X_2$ be two samples draw from a continuous distribution, then I think there is no reason to say that $X_1\leq X_2$ or $X_1\geq X_2$, so we may have
$$P[X_1\leq X_2]=P[X_1>X_2]=\frac{1}{2}$$
more generally, let $X^{(1)},\ldots,X^{(n)}$ denote the $X_1,\ldots,X_n$ ordered, then
$$P[X_1=X^{(i)}]=\frac{1}{n}$$
but I don't know how to prove this rigorously.
AI: For any $k_1$ and $k_2$, by symmetry $\Pr(X_{k_1}=X^{(i)}) = \Pr(X_{k_2}=X^{(i)})$. Also, if $k_1 \ne k_2$, then $\Pr(X_{k_1} = X_{k_2}) = 0$ (since the distribution is continuous). So by the exclusion-inclusion formula
$$ 1 = \sum_{k=1}^n \Pr(X_k = X^{(i)}) = n \Pr(X_1 = X^{(i)}) .$$ |
H: If $f(x) = \cos x$, explain, without taking the derivative, how you would find the $f^{(99)}(x)$?
My theory:
derivative of $\cos x = - \sin x $
derivative of $-\sin x = -\cos x $
derivative of $-\cos x = \sin x.$
cycle occurs three times but then what do you do??
Is there a good way to solve this?
AI: The cycle occurs, not three times but 4 times. Thus every 4 derivatives taken you return to $\cos(x)$. Therefore, since $99=96+3$, you need the $3$rd derivative of $\cos(x)$ which is $\sin(x)$. |
H: Proving that the closed unit square in the plane is compact.
My thoughts on proving this statement is as follows:
Suppose $G_a$ is an open cover of $Q= [0,1] \times [0,1]$. For each $x$ in $[0,1]$, there is some ball around $x$ with radius $r_x$ such that it covers $[x-r_x, x+r_x] \times [0,1]$. Since $[0,1]$ is closed and bounded, it is compact. Since $[0,1]$ is compact, this vertical strip described above is also compact, i.e., there is a finite collection of these balls of $G_a$ that covers this vertical strip.
Next we do the same thing except for horizontal strips.
This is where I am drawing a blank, how am I to show that the union of these balls are finite, and how would I know that it covers all of the unit square?
AI: You're on the right track to prove this. For each $x$, let $G_x$ be a finite subcover of $G_a$ that covers $\{x\}\times[0,1]$.
Claim: There exists a positive number $r_x$ for which $G_x$ covers the entire strip $[x-r_x,x+r_x]\times[0,1]$. You need to prove this. One way uses the fact - which also requires proof - that you can assume $G$ consists of rectangles.
With $r_x$ defined in this way for each $x\in[0,1]$, consider the collection of intervals $[x-r_x,x+r_x]$. These cover $[0,1]$, which is compact, so there's a finite set $X$ for which $[0,1]\subseteq \bigcup_{x\in X}[x-r_x,x+r_x]$. Then $\bigcup_{x\in X}G_x$ (which is a finite subcover of $G_a$) covers $Q$. |
H: Is this a valid way to show a sequence does not split?
Suppose you have
$$
0\to\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to 0
$$
viewed as additive groups. Let $i$ be the inclusion of $\mathbb{Z}$ into $\mathbb{Q}$. I want to say this is not a split sequence, since there is no retract $f:\mathbb{Q}\to\mathbb{Z}$ such that $f\circ i=id_\mathbb{Z}$. If there were, then $f(n)=n$ for all integers $n$. Then
$$
nf(1/n)=f(1/n)\cdots+f(1/n)=f(1/n+\cdots+1/n)=f(1)=1
$$
so $f(1/n)=1/n$, which doesn't make any sense if $n>1$, since $f(1/n)\in\mathbb{Z}$. So no such $f$ exists. Is that all there is to do?
AI: Yes! That is good. You're basically observing that $\text{Hom}(\mathbb{Q},\mathbb{Z})=\{0\}$ and so you clearly can't have a retraction.
You could also note that if it split then $\mathbb{Q}\cong \mathbb{Z}\oplus(\mathbb{Q}/\mathbb{Z})$, but $\mathbb{Q}$ has no torsion :) |
H: Checkers on a Chessboard
Given 2k pieces on a k by k chessboard, prove that there is always a sequence of pieces $K_1, K_2 \ldots K_{2n}$ such that $K_1$ and $K_2$ are in the same row, $K_2$ and $K_3$ are in the same column, $K_3$ and $K_4$ are in the same row ... $K_{2n-1}$ and $K_{2n}$ are in the same row, and $K_{2n}$ and $K_1$ are in the same column. This sequence does not necessarily contain all the pieces on the board.
AI: Another way to do this is motivated by graph theory.
Let $R=\{r_1,r_2,\dots, r_k\}$ and $C=\{c_1,c_2,\dots,c_k\}$, and consider $R\cup C$ as a set of vertices. For each position $(i,j)$ on the chessboard with a checker, draw an edge between the vertices $r_i$ and $c_j$. This will create a graph with $2k$ vertices and $2k$ edges. A graph with at least as many edges as vertices must contain a cycle. Follow the cycle (which must alternate between vertices in $R$ and vertices in $C$ by construction, so must have even length) to get your sequence. |
H: Orthonormal Matrices-Intuition
Why is it, geometrically, that the row space is ALSO orthonormal? What exactly does the transpose LOOK like? Normally the row and column space are two separate things, but in the case of an orthonormal matrix, you have that the row space is actually the inverse of the column space, so also, what does an inverse space look like? I feel there should be a very good geometric interpretation, but I don't see it.
AI: I think your question is non-trivial. One way to see this is to consider the same problem in infinite dimensions. Let $H$ be an infinite dimensional Hilbert space with o.n. basis $e_1,e_2,\dots$. Let $T$ be the operator
$$ T e_i = e_{i+1} .$$
Then in some sense you can say that the columns of the matrix representing $T$ are orthonormal. In other words, $T$ is an isometry.
But now consider the adjoint operator $T^*$. This corresponds to the transpose of $T$. But $T^*$ has a one-dimensional kernel, and hence it is not an isometry, and hence its columns are not orthonormal.
This means that any answer to your question must take into account the fact that the matrix is finite dimensional. |
H: Equivalence Relations Proof dealing 3 dividing x + y
Consider the relation $S$ on the Natural Numbers defined by $\quad x\,S\,y\quad$ if $3$ divides $\quad x + y.\quad$ Prove $S$ is not an equivalence relation.
I know an equivalence relation is one that is reflexive, symmetric, and transitive. I believe that S does not satisfy the reflexive property. For example, the element (2,2) would not be an element of S.
Is this correct? Also I am unsure how to state that formerly.
Thanks for the Help.
AI: I don't have enough points to comment, or I would make this into a comment:
You need to show:
i) Every element in the set is related to itself; in this case, $x\sim x$, meaning
$x+x$ is divisible by $3$.
ii) If $x\sim y$, then $y\sim x$, i.e., if $x+y$ is divisible by $3$, then so is $y+x$.
iii) If $x\sim y$ and $y\sim z$, then $x\sim z$.
Can you take it from here? |
H: Checking if proof is correct
I would like to check if my proof of this proposition has been correctly done. I would also like help on proving part (c). Thanks in advance.
Proposition: Let $H$ be a subgroup of a group $G$, and let $N$ be the normalizer of $H$. Prove that:
(a) $H$ is a normal subgroup of $N$
(b) $H$ is a normal subgroup of $G$ if and only if $N$ = $G$
(c) $|H|$ divides $|N|$ and $|N|$ divides $|G|$.
Proof:
Part (a):
Suppose $H$ is a subgroup of $G$. Suppose $N$ is a normalizer of $H$, meaning $N(H)$ $=$ $\{g \in G: gHg^{-1}=H\}$. We want to show that $H$ is a normal subgroup of $N$. Note that by definition: $g$ $\in$ $N$ $\leftrightarrow gHg^{-1}=H$. Thus $gHg^{-1}=H$ for every $g$ $\in$ $N$. Hence, $H$ is normal in $N$.
Part (b): Suppose $H$ is a normal subgroup of $G$. Thus $gHg^{-1}=H$ for every $g$ $\in$ $G$. Thus for every $g$ in $G$, we have $gHg^{-1} = H$ $\in$ $N$. So $G$ is a subset of $N$. Since $N$ is a subset of $G$, we must have $N=G$.
Now other direction: Suppose $N=G$. Then, for every $g \in G$ and $g \in N$, we have: $gHg^{-1} = H$. Thus, by definition of normal, $H$ is normal in $G$.
How do I proceed with part c?
Thanks.
AI: The proofs for parts ($a$) and ($b$) look great!
Hint for ($c$): Lagrange's Theorem; you already know $H\lhd N$, so can you show $N\leq G$? |
H: Find a basis for $p \in P_2(\mathbb R) $ with $p(7) = 0$?
Find a basis for the subset: $$ S = \{\;p \in P_2(\mathbb R)\;\; |\;\; p(7) = 0\; \} $$
I'm not sure how to approach this question.
$$
p(7) = a_0 + 7a_1 + 49a_2 = 0
$$
$$
a_0 = -7a_1 - 49a_2
$$
$$
a_1(-7 + x) + a_2(-49 + x^2) = 0
$$
Is this even in the right direction? I don't know what to do from here, nor how to proceed by manipulating the standard basis $\{1,x,x^2\}$
AI: Hint: A polynomial $p$ has $p(7) = 0$ if and only if $(x - 7) | p(x)$. That is, if we can write
$$p(x) = (x - 7)(ax + b)$$
since $p$ has degree $2$. Now $a$ and $b$ give two free choices. |
H: What does it mean for a function $u : D \to \mathbb{C}$ to be harmonic, $D \subset \mathbb{R^2}$?
On page 167 of David Ullrich's "Complex Made Simple", he defines $u : D \to \mathbb{C}$ to be harmonic, $D \subset \mathbb{R^2}$, to be harmonic in $D$ if it is twice continuously real differentiable and $u_{xx} = u_{yy}$.
Since $\mathbb{R^2} = \mathbb{C}$, does this mean that $u$ is a function from $D$ to $\mathbb{R^2}$, and so $u(x,y) = (f(x,y),g(x,y))$? Then, what does $u_{xx}$ mean for a vector valued function?
AI: If $u=(f,g)$, then $u_{xx} = (f_{xx},g_{xx})$. That is, do it component-wise. |
H: Compute Left Eigenvectors
How does one compute the left eigenvectors of a matrix? I cannot seem to quite get the answer.. I don't care what the matrix is. Let's just say I have matrix $A$ and have found the 'right' eigenvectors $e$ and I want to compute the left eigenvectors. Do we $A^Te = c$ Then the left eigenvector is $c^T$? I did this for a 2x2 and it seemed to work, but for a 3x3 it did not (according to the answers in the book).
Thanks in advance!
AI: In general, the left (resp. right) eigenvectors of a matrix $A$ are the right (resp. left) eigenvectors of the matrix $A^T$. It follows quite straightforwardly that the eigenvalues of $A$ and those of $A^T$ coincide (including multiplicities) but the same is not true of the left and right eigenvectors. Generally, one simply needs to compute the, say, right eigenvectors even if one already has the left eigenvectors. Of course, for special matrices (like symmetric ones), left and right eigenvectors do coincide. |
H: Find expected value of this discrete distribution
My knowledge on probability topics is a bit rusty, so I was hoping you guys could help me.
Let X be the amount of products a person buys.
The probability that the person buys 1 to 12 is 60%.
From 13 to 20, 35%.
From 21 to 100, 5%.
I need to find the expected number of products he/she buys.
I did (12-1)*0.6 + (20-13)*0.35 + (100-21)*0.05 = 13.
Is that correct?
AI: it should be (1+2+...+12)(0.6/12) + (13+...+20)(0.35/8)+(21+...+100)(0.05/80) |
H: Is the "first nonzero digit" function surjective?
For sets $A= \{x \in \mathbb{R}: 0< x< 1 \}$ and $B=\mathbb{Z_+}$ let $f$ be a function $\space f:A \rightarrow B$ such that $f(x)$ is the position of the first nonzero digit of $x$, ex.g. $\space f(0.2)=1$, $\space f(0.02)=2$. Determine whether $f$ is injective and/or surjective.
Attempt: It is easy to see that $f$ is not injective since it produces same values for distinct x's.
I am having a difficulty with determining whether $f$ is surjective or not. To be surjective the cardinality of $B$ should be less or equal to the cardinality of $A$, $\left\vert{B}\right\vert \le \left\vert{A}\right\vert$. However, both sets have infinite number of elements. So, it might appear they are onto, but on second thought it looks like one set might be bigger than other, namely $\left\vert{B}\right\vert > \left\vert{A}\right\vert$ . I am a little confused here. Clarification would be very helpful. Thank you.
AI: You need to demonstrate with an example that there are distinct $x,y\in(0,1)$ such that $f(x)=f(y)$. Yes, it’s trivial to do, but the unsupported assertion that such values exist isn’t enough.
$B=\Bbb Z_+$ is countable, and $A=(0,1)$ is uncountable, so in fact $|A|>|B|$. However, you don’t need to worry about cardinality at all. Let $k$ be a positive integer; can you write down a specific number $x\in(0,1)$ such that $f(x)=k$? (HINT: You should even be able to specify one that has a terminating decimal expansion.) |
H: Differentiate $f(x)=\frac{2-x^2}{3x+x^2}$
I am double checking a question I was given on a test that was marked wrong and I am not sure how I got it wrong.
Find $f'(x)$:$$\begin{align*} f(x) &=\dfrac{2-x^2}{3x+x^2}\\
f'(x) &=\dfrac{(2x)(3x+x^2)-(2-x^2)(3+2x)}{(3x+x^2)^2}\\
f'(x) &=\dfrac{3x^2+4x-6}{(3x+x^2)^2} \end{align*}$$
However on the test he marked a negative sign out on the second step on the $2x$. Would anyone mind explaining where I went wrong?
AI: By the quotient rule, the derivative is
\begin{align*}
f'(x) &= \frac{(3x + x^2) \frac{d}{dx}(2 - x^2) - (2 - x^2)\frac{d}{dx}(3x + x^2)'}{(3x + x^2)^2} \\
\end{align*}
The derivative of $2 - x^2$ is $-2x$. Everything else seems correct. |
H: Prove that a sequence diverges if and only if its subsequence diverges
Prove that $(x_n)_n$ diverges if and only if for every $a\in\mathbb{R}$, there exists an $\epsilon > 0$ and a subsequence $(x_{n_k})_k$ for $(x_n)_n$ such that for all $k\in \mathbb{N}$, $|{x_{n_k} - a}| \ge \epsilon$.
Thank you!
AI: If a sequence diverges, then since it is a subsequence of itself, it has a subsequence which diverges.
If a sequence converges, then every subsequence converges to the same limit. Hence if you have a divergent subsequence, the sequence cannot converge. |
H: Projected area of a paralelogram over a plane
Let $u=\hat{i} +\hat{j} +\hat{k}$ and $v=\hat { i } - \hat { j } -\hat { k }$ two vectors that are the two coterminal sides of a paralelogram. Compute the projected area by this paralelogram over the plane whose unit normal vector is $n=\hat {i} + 2\dfrac{\hat {j}}{3} +2\dfrac{\hat {k}}{3}$
AI: The projection of $u$ onto the plane is $u - \hat{n}\hat{n}^Tu$ and similarly for $v$. Once you have the two projected sides, take the magnitude of their cross product to get the parallelogram area.
Where are you getting stuck? |
H: Braid group B3 pure group
I will denote b1 as the twist of first two strands and b2 as the twist in the last two strands so that I have my two generators. When finding the kernelof B3 to S3 why is it wrong to list b1inverseb1 for the kernel?is it because by definition it is identity of braid group? What is an example of element in kernel then?
AI: I would imagine that you're not trying to list elements in the kernel, since $b_1^{-1}b_1$ will definitely be in the kernel, since it's already trivial. However, if I were trying to provide generators for the kernel, I would not want to list things that are already trivial. For instance, $b_1^2$ is not trivial in $B_3$, but it becomes trivial in $S_3$. This is possibly something you would want to see in your description of the kernel.
EDIT: In your title, you mentioned the pure braid group, which is by definition the kernel of the homomorphism you describe in the post. So yes, if you wanted to describe the pure braid group, you would need to give a list of things in the braid group that generate the entire kernel. Since something like $b_1^{-1}b_1$ is trivial, it doesn't make any sense to include in the list for the things that generate the kernel, since the trivial element is understood to be included in any list of generators for the kernel. |
H: Prove regular language closed under min and max
Given some regular language $L$, show that $L$ is closed under the following operations:
$$\begin{align*}
\min(L) &= \{w\mid w\in L,\text{ but no prefix of }w\text{ is in }L \}\\
\max(L) &= \{w\mid w\in L,\text{ but for no }x\text{ other than }\epsilon\text{ is }wx\in L \}
\end{align*}$$
It seems that $\min(L)=\max(L)$, is this true? If not, what is the simplest way to show that both $\min(L)$ and $\max(L)$ are regular? I have seen solutions involving homomorphisms, but am hoping that there is a simpler way to approach these problems.
AI: Let $L=\{a^n:n\ge 2\}$; then $\min(L)=\{aa\}$, and $\max(L)=\varnothing$, so $\min(L)\ne\max(L)$.
One fairly easy way to prove that $\min(L)$ or $\max(L)$ is regular is to start with a DFA that recognizes $L$ and modify it to accept $\min(L)$ or $\max(L)$. Let $M=\langle Q,\Sigma,\delta,q_0,A\rangle$ be a DFA that accepts $L$; $Q$ is the set of states, $\Sigma$ is the input alphabet, $\delta$ is the transition function, $q_0$ is the initial state, and $A$ is the set of acceptor states. (You may know them as final states.) I’ll sketch the modifications and leave the detailed verification to you.
To get a DFA that accepts $\min(L)$, add a new trap state $q^*$ and modify $\delta$ so that any input to an acceptor state sends $M$ to $q^*$.
To get a DFA that accepts $\max(L)$, you need only change some acceptor states to non-acceptor states. Specifically, say that a state $q\in Q$ is non-terminal if there is some non-empty word $w$ that sends $M$ from $q$ to an acceptor state, and let $N$ be the set of non-terminal states of $M$. If $A'=A\setminus N$, the automaton $\langle Q,\Sigma,\delta,q_0,A'\rangle$ recognizes $\max(L)$. |
H: Prove that $\Bbb R^2 - \{0\}$ is homeomorphic to $S^1 \times \Bbb R$.
No idea where to even begin. There is a hint: this requires construction of an explicit function.
AI: Hint: The plane minus the origin can be written as
$$\Bbb{R}^2 \setminus \{0\} = \{(r \cos{t}, r \sin{t}) : 0 < r < \infty, 0 \le t < 2\pi\}$$
Do you see how this looks like a circle cross a line? |
H: If two matrices both multiplied by the same vector are equal are the matrices equal?
Assume A and B are n x n matrices.
If Av$_k$ = Bv$_k$ then is A = B where v$_k$ is a vector in R$^n$?
AI: No. Counterexample:
$A=\left[ \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]$, $B=\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0\end{array}\right]$, $v_k=(1,0)$.
Notice that for $w=(0,1)$ we have $Aw\neq Bw$. This relates to what was mentioned in the comments: that equality must hold over every element of a basis to conclude that $A=B$. |
H: Ring homomorphism is injective but not surjective
Let $F$ be a field, let $R_1=F[x]$ be the ring of polynomials with coefficients in $F$, and let $R_2$ be the ring of all functions from $F$ to itself, with addition and multiplication defined as the usual operations on functions with values in a ring. The function
$$\phi: R_1 \rightarrow R_2$$
which send the polynomial $f\in F[x]$ to the $F$-valued function $a \rightarrow f(a)$ on $F$ which induces, is a homomorphism.
When $F=\mathbb Q$ or $\mathbb R$, show that the homomorphism $\phi$ is injective but not surjective.
I would start like so
if $\phi(a) = b$ and $\phi(a') = b$, then $a=a'$. Since $a,a' \in F[x]$...
AI: Every function in the image of your map is (or extends to) a continuous function $\mathbb R\to\mathbb R$. Yet there are functions $F\to F$ which are not continuous (or, when $F=\mathbb Q$, do not extend to continuous functions).
This means that the map is not surjective.
Alternatively, every function in the image of your maps is identically zero if it has infinitely many zeroes, yet there exist non-zero functions $F\to F$ which have infinitely many zeroes. This again implies non-surjectivity, and if you look at it correctly, also injectivity. |
H: Partitions on C = {i,-1,-i,1}
Let $ C = \{i, -1, -i, 1\}$ , where $ i^2 = -1 $. The relation $R$ on $C$ given by $xRy$ iff $xy = \pm 1$ is an equivalence relation on $C$. Give the partition of $C$ associated with $R$
I would really appreciate any help. I am not really sure where to start with this. Is it as simple as the partition just containing 1 and -1.
Thank you very much.
AI: The equivalence classes are the classes that fulfill that relationship, then your classes must partition C. Note that $C_i = \{i,-i\}$ and $C_1=\{1,-1\}$ do this so these are your classes. If you took any element in $C_i$ and one in $C_1$ and multiplied them you would not get $\pm 1$ so they are not in the same equivalence class. |
H: Connected $T_3$ space
I wanted to prove that a connected $T_3-$space containing atleast two points must be uncountable.
My attempt is as follows:
Let $X$ be a connected $T_3-$ space and let $x,y\in X$. Then there exist disjoint open sets $G,H$ containing $x$ and $y$ respectively. Since $X$ is connected, therefore, $X\neq G\cup H$. Thus $\exists z\in X$ such that $z\notin G\cup H$.
I couldn't proceed further. Please help.
AI: The only proofs that I can see at the moment are indirect: suppose that $X$ is countable and $T_3$, and show that $X$ cannot be connected. A countable space is clearly Lindelöf, and it’s a standard basic result that a Lindelöf $T_3$-space is $T_4$, so $X$ is $T_4$ and therefore Tikhonov. Now let $x$ and $y$ be distinct points of $X$; since $X$ is Tikhonov, there is a continuous function $f:X\to[0,1]$ such that $f(x)=0$ and $f(1)=1$. Now use the fact that $f[X]$ is countable to find disjoint non-empty open sets $U$ and $V$ in $X$ such that $U\cup V=X$. |
H: Partitioning a set with a relation on it
Let R be a relation on a set A that is reflexive and transitive but not symmetric.
Let R(x) = {y: xRy}. Does the set a = {R(x): x ∈ A} always form a partition of A?
I really don't know where to start with this one. I know that R(x) is the same as x/R except R is not an equivalence relation.
AI: HINT: What happens if $R$ is the relation $\le$ on $\Bbb Z$, say? Start by finding $R(0)=\{n\in\Bbb Z:0\le n\}$ and $R(1)$. |
H: If $|f|$ is Hölder continuous, what about $f$?
Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that $|f|$ is Hölder continuous with exponent $0<\alpha\leq1$. Does it follow that $f$ is also Hölder continuous with the same exponent?
I thought of this statement a few days ago and had trouble proving it or generating a counterexample. If we don't stipulate that $f$ is itself continuous, then the result is false (take $f(0)=1$ and $f(x)=-1$, otherwise). Additionally, the converse follows from the reverse of the triangle inequality, for if $f$ is Hölder continuous, then
$$
\big||f(x)|-|f(y)|\big| \leq |f(x)-f(y)| \leq C|x-y|^\alpha.
$$
AI: Yes, this is corrrect. Let $x,y\in\mathbb R$. If $f(x)$ and $f(y)$ have the same sign, then $$|f(x)-f(y)|=\big||f(x)|-|f(y)|\big|\leq C|x-y|^{\alpha}.$$ Now, suppose that $f(x)$ and $f(y)$ have different signs. Then, from continuity of $f$, there exists $z\in(x,y)$ such that $f(z)=0$. Then, $$|f(x)-f(y)|\leq|f(x)|+|f(y)|=\big||f(x)|-|f(z)|\big|+\big||f(z)|-|f(y)|\big|\leq$$$$C|x-z|^{\alpha}+C|y-z|^{\alpha}\leq C|x-y|^{\alpha}+C|x-y|^{\alpha}=2C|x-y|^{\alpha}.$$ So, if $x,y\in\mathbb R$, you have that $|f(x)-f(y)|\leq 2C|x-y|^{\alpha}$, so $f$ is Hölder continuous with exponent $\alpha$. |
H: Produce two distinct topologies of R such that the first is strictly finer than the second but the two are homeomorphic to each other.
Need to prove that they are homeomorphic to each other.
So if T_1 and T_2 are topologies on set X with T_1 contained in T_2, every element of T_1 will be in T_2 and T_2 is the "finer" (stronger and larger) topology. So to them be homeomorphic is T_1 going to have to be a hole in T_2?
AI: Generate $T_1$ by the intervals $(-2n,2n)$, $n\in\mathbb{N}$. (The topology is generated by taking all unions and finite intersections of these sets - which really doesn't add anything except the whole of $\mathbb{R}$ through an infinite union. And this also includes $(0,0)$ which means $\{\emptyset\}$.)
Generate $T_2$ by the intervals $(-n,n)$, $n\in\mathbb{N}$.
I believe these topologies meets your needs. The homeomorphism is just $x\mapsto2x$. |
H: Proving Convergent Series made by continuous function $f$, $a_n=f(1/n)$
Let $f$ is continuous on an interval around 0, and let $a_n=f(1/n)$
Prove that if $f''(0)$ exists and $f(0)=f'(0)=0$, then $\displaystyle{\sum_{n=1}^{\infty}{a_n}}$ converges.
This is problem from Spivak calculus 4th, chapter 23 Exercise 6.
If "$f''(0)$ exists" were missing, $a_n=\frac{1}{n\log{n}}$ is a counterexample.
Its reverse also false. $a_n=\frac{1}{n^{3/2}}$ is a counterexample.
AI: The existence of the second derivative at $0$ implies that there exists $k,\delta>0$ such that
$$
|f'(x)|\leq kx,\ \ \ x\in[0,\delta)
$$
(just take $\varepsilon=1$, $k=f''(0)+1$, and use that $\lim_{x\to0}f'(x)/x=f''(0)$).
By the Mean Value Theorem,
$$
f(x)=f'(\xi)\,x,
$$
for some $\xi\in(0,x)$. So, for $x\in[0,\delta)$,
$$
|f(x)|\leq kx^2.
$$
Now, if $N>1/\delta$,$$
\sum_{n>N}|f\left(\frac1n\right)|\leq\sum_{n>N}\frac k{n^2}<\infty.
$$ |
H: finding all the roots (including complex)of the equation
Find all the roots of $z^4=16(z+2i)^4$.
Can someone help me teach/ guide to solve this equation?
AI: Just take the square root of both sides. Then affix with +/- to get 2 cases for that equation. Then for each case, get the square root again. Then affix the resulting equation with +/- to get 4 cases. Finally, solve for z for each of the 4 equations. |
H: How does $1^\infty=\infty$?
I remember hearing in school long ago that $1^\infty=\infty$.
I was just wondering if anyone could explain this in laymen's terms?
AI: $1^\infty$ is indeterminate, as this Wikipedia entry explains. Or perhaps you remember something like $\lim_{n\to\infty}\sqrt[n]n=1$ ? |
H: Evaluate $\int_0^e{W(x)}\,\mathrm{d}x$
The function $W(x)$ satisfies $W(x)e^{W(x)}=x$ for all $x$. Evaluate $$\int_0^e{W(x)}\,\mathrm{d}x$$
I tried integrating $xe^{-W(x)}$ but can't see how to do it.
AI: Integrate by parts with $u=W(x)$ and $\mathrm{d}v=\mathrm{d}x$
$$\int{W(x)}\,\mathrm{d}x=xW(x) - \int xW'(x)\,\mathrm{d}x=xW(x)-\int W(x)W'(x)e^{W(x)}\,\mathrm{d}x$$
Integrate by parts again with $u = W(x)$ and $\mathrm{d}v=W'(x)e^{W(x)}\,\mathrm{d}x$
$$\int W(x)W'(x)e^{W(x)}\,\mathrm{d}x=W(x)e^{W(x)} - \int W'(x)e^{W(x)}\,\mathrm{d}x=W(x)e^{W(x)}-e^{W(x)}$$
So $$\int{W(x)}\,\mathrm{d}x=xW(x) -W(x)e^{W(x)}+e^{W(x)}=x\left(W(x)-1+\frac{1}{W(x)}\right)$$
Note that $W(e) = 1$ and $W(0)=0$ and $\displaystyle{\lim_{x\to 0}\frac{x}{W(x)}=1}$
$$\left|x\left(W(x)-1+\frac{1}{W(x)}\right)\right|_0^e=e\left(W(e)-1+\frac{1}{W(e)}\right) - \left(\lim_{x\to 0}\frac{x}{W(x)}\right) = e-1$$ |
H: The number of monomials of a given degree
I'm trying to understand why the number of the monomials of degree $d$ in $n+1$ variables is $C_{n+d,n}$. If someone could help me to remember how to solve this, I would be glad.
Thanks.
AI: The stars and bars argument in combinatorics shows that the number of ways to place $d$ indistinguishable objects (stars) in $n + 1$ distinguishable boxes is to lay the objects in a line and place barriers (bars) to establish boundaries between the boxes.
Here's an example of the bijection with $d = 3$ and $n = 2$.
$$
\begin{align}
x_1^3 & \longleftrightarrow \star \star \star \;| \;| \\
x_1^2 x_2 & \longleftrightarrow \star \star \,| \star | \\
x_1^2 x_3 & \longleftrightarrow \star \star \,| \;| \star \\
x_1 x_2^2 & \longleftrightarrow \star \;| \star \star \;| \\
x_1 x_2 x_3 & \longleftrightarrow \star \;| \star \,| \star \\
x_1 x_3^2 & \longleftrightarrow \star \;| \;| \star \star \\
x_2^3 & \longleftrightarrow \,| \star \star \star | \\
x_2^2 x_3 & \longleftrightarrow \,| \star \star \,| \star \\
x_2 x_3^2 & \longleftrightarrow \,| \star \,| \star \star \\
x_3^3 & \longleftrightarrow \,| \;| \star \star \star
\end{align}
$$
You can see the $\binom{2 + 3}{2} = 10$ monomials of degree $3$ in $2 + 1 = 3$ variables. |
H: Uniformly continuous bijection from $X$ to the Cantor set
Let $X$ be a metric space, and $C$ be the Cantor set (equipped with the standard topology).
Let $f: X\to C$ be a uniformly continuous function. Assume that $f$ is
a bijection. Does it follow that $f$ is a homeomorphism?
I know that if $X$ is compact, then the answer is yes. This follows from the following theorem:
A continuous bijection from a compact set to a Hausdorff space is
homeomorphism.
I am hoping that the condition "uniform continuity" is strong enough to cover the case even when $K$ is not compact.
I appreciate any help :)
AI: Let $\rho$ be the discrete metric on the Cantor set $C$, and let $d$ be the usual metric; then the identity map from $\langle C,\rho\rangle$ to $\langle C,d\rangle$ is a uniformly continuous bijection but not a homeomorphism. |
H: Limit of $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$
What is the limit of this sequence $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$?
Where $a$ is a constant and $n \to \infty$.
If answered with proofs, it will be best.
AI: With $S_n = \frac{1}{a} + \frac{2}{a^2} + \frac{3}{a^3} + \cdots \frac{n}{a^n}$ and using the closed form of geometric sums,
$$
\begin{align}
\lim_{n \to \infty} S_n &= \frac{1}{a} + \frac{2}{a^2} + \frac{3}{a^3} + \cdots \\
&= \frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a}} + \frac{1}{a^2} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a} + \frac{1}{a^2}} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3}} \ddots \\
&= \frac{1}{a - 1} + \frac{1}{a} \cdot \frac{1}{a - 1} + \frac{1}{a^2} \cdot \frac{1}{a - 1} + \cdots \\
&= \frac{1}{a - 1} \left( 1 + \frac{1}{a} + \frac{1}{a^2} + \cdots \right) \\
&= \frac{1}{a - 1} \cdot \frac{a}{a - 1} \\
&= \frac{a}{(a - 1)^2}
\end{align}
$$ |
H: probability problem related to card shuffling
You have five cards numbered 1,...,5. You shuffle them, so that at the end every permutation of them is equally likely to show up. What is the probability that no card 'i' will end up in the i-th card after shuffling?
So I'm figuring out this problem. I got stuck because the probability of relocating card A changes the probability of moving other cards. For example, if card 1 is moved to location 2 after shuffling, moving card 2 to another location is 4/5, while if card 1 is moved to location 3, moving card 2 to another location will have a probability of 3/5. I'm confused here. How do I solve this problem? Thank you.
AI: Look up "problem of derangements". |
H: Cauchy's Problem Question $x(x^2+1)dy-(3x^2+1)ydx= x(x^2+1)^2dx,\ \ y(1)=2$
I want to solve the following equation
$$x(x^2+1)dy-(3x^2+1)ydx= x(x^2+1)^2dx,\ \ y(1)=2$$
what I chose to do is to order the equation to $()dy+()dx=0$ then to find integrating factor. so what I did :
$$(x^2+x)dy=(x(x^2+1)^2+(3x^2+1)y)dx$$
$$(x^2+x)dy=(x(x^4+2x^2+1)+(3x^2y+y))dx$$
$$(x^2+x)dy=(x^5+2x^3+x+3x^2y+y)dx$$
$$(x^2+x)dy+(-x^5-2x^3-x-3x^2y-y)dx$$
set $Q(x,y)=x^2+x,\ \ P(x,y)=-x^5-2x^3-x-3x^2y-y$
$\frac{dQ}{dx}\neq \frac{dP}{dy}$ so I need to find integration factor.
$$\frac{\frac{dP}{dy}-\frac{dQ}{dx}}{Q}=\frac{-3x^2-2x-2}{x^2+x}$$
this is the right way? I need some advice how to continue, thanks.
AI: Use the quotient rule; first, divide by $x^2 (x^2+1)^2 dx$:
$$\frac{x (x^2+1) y'-(3 x^2+1) y}{x^2 (x^2+1)^2} = \frac{1}{x}$$
The LHS may be rewritten:
$$\frac{d}{dx} \frac{y}{x (x^2+1)} = \frac{1}{x}$$
Integrate both sides with respect to $x$:
$$\frac{y}{x (x^2+1)} = \log{x} + C$$
where $C$ is an integration constant. Using $y(1)=2$ implies that $C=1$. Therefore
$$y(x) = x (x^2+1) (1+\log{x})$$ |
H: How prove this inequality $f(a)\le f(b)$
Suppose $f(x)$ is continous on $[a,b]$,and for any $x_{0}\in [a,b]$.
the limit
$$\varliminf_{x\to x^{-}_{0}}\dfrac{f(x)-f(x_{0})}{x-x_{0}}\ge 0$$
show that
$$f(a)\le f(b)$$
My try: I found this problem is same as
How prove this $f(a)\le f(b)$
But for my problem,this is inflimit,and the limit form is not same,Thank you
AI: Suppose $f(b) < f(a)$. Let $m = \frac{f(b)-f(a)}{b-a} <0$.
Let $\phi(x) = f(x)+(x-a) \frac{f(a)-f(b)}{b-a}$. Note that $\phi(a)=\phi(b) = 0$. Since $\phi$ is continuous, $\phi$ has a maximum and minimum on $[a,b]$. Let $\overline{\phi} = \max \phi([a,b])$ and $\underline{\phi} = \min \phi([a,b])$.
If $\overline{\phi} = \underline{\phi}$, then $\phi(x) = 0$ and $\liminf_{x \uparrow b} \frac{f(x)-f(b)}{x-b} = \frac{f(b)-f(a)}{b-a} = m < 0$.
If $\underline{\phi} < 0$, then let $x_0 \in (a,b)$ be a minimizer. Then $\phi(x) \ge \phi(x_0) $ for all $ x \in [a,b]$. In particular, for $x < x_0$, we have $\phi(x) - \phi(x_0) \ge 0$, and so $\liminf_{x \uparrow x_0} \frac{\phi(x)-\phi(x_0)}{x-x_0} \le 0$. Since
$f(x) -f(x_0)= \phi(x)-\phi(x_0)+(x-x_0) \frac{f(b)-f(a)}{b-a}$, we have
$\liminf_{x \uparrow x_0} \frac{f(x)-f(x_0)}{x-x_0} \le \frac{f(b)-f(a)}{b-a} =m < 0$.
If $\overline{\phi} >0$, let $x_0 \in (a,b)$ be a maximizer. Then $\phi(x) \le \phi(x_0) $ for all $ x \in [a,b]$. In particular, $\phi(x) \le \phi(x_0)$, and so
$f(x) -f(x_0) \le (x-x_0) \frac{f(b)-f(a)}{b-a}$, or in other words,
$\frac{f(x)-f(x_0)}{x-x_0} \le \frac{f(b)-f(a)}{b-a} =m$ for all $x \in (x_0,b]$. Now repeat the above process on the interval $[x_0,b]$.
One of two things happens above. Either the process terminates, and we have found a point $x_0$ such that
$\liminf_{x \uparrow x_0} \frac{f(x)-f(x_0)}{x-x_0} \le \frac{f(b)-f(a)}{b-a} =m< 0$, or the process does not terminate, and we find a sequence of points $b >x_n > x_{n-1}$ such that
$\frac{f(x)-f(x_n)}{x-x_n} \le \frac{f(b)-f(a)}{b-a} =m$ for all $x \in (x_n,b]$.
Since $x_n$ is increasing, we have $x_n \uparrow x^* \le b$. Then the above shows that $\frac{f(x^*)-f(x_n)}{x^*-x_n} \le \frac{f(b)-f(a)}{b-a} =m$. In otherwords,
$\liminf_{x \uparrow x^*} \frac{f(x)-f(x^*)}{x-x^*} \le \frac{f(a)-f(b)}{b-a} =m< 0$.
Consequently, if $f(b)<f(a)$, then there exists some $x_0 \in [a,b]$ such that
$\liminf_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} < 0$.
Hence it follows that $f(a) \le f(b)$. |
H: An inequality in $L^p$-spaces
Let $\{f_k\}_{k=1}^{\infty}$ be a sequence in $L^p(\Omega,\Sigma,\mu)$ for $1\leq p<\infty$. Suppose $0<c=\inf_k \lVert f_k\rVert_p\leq \sup_k \lVert f_k\rVert_p=C<\infty$ and $f_if_j=0$ for $i\neq j$. Let $(a_k)_{k=1}^{\infty}$ be a sequence of real numbers such that $\sum_k \lvert a_k\rvert^p<\infty$. Then by the completeness of $L^p$, $f=\sum_{k=1}^\infty a_k f_k$ is a function in $L^p(\Omega,\Sigma,\mu)$. Moreover,
$$ \lVert f\rVert _p\leq C(\sum_{k=1}^\infty \lvert a_k\rvert ^p)^{1/p}. $$
And when $p=2$, I can obtain
$$ \lVert f\rVert_2\geq c(\sum_{k=1}^\infty \lvert a_k\rvert^2)^{1/2}. $$
In general, how to prove
$$ \lVert f\rVert _p\geq c(\sum_{k=1}^\infty \lvert a_k\rvert^p)^{1/p} $$
???
I have no way to prove it.
AI: Let $S_n:=\{x,f_n(x)\neq 0\}$. Then $\mu(S_i\cap S_j)=0$ if $i\neq j$ (this is the key point). We thus have
$$\lVert f\rVert^p_p=\sum_{i\geqslant 1}\int_{A_i}|f(x)|^p\mathrm d\mu=\sum_{i\geqslant 1}\int_{A_i}|a_i|^p|f_i(x)|^p\mathrm d\mu=\sum_{i\geqslant 1}|a_i|^p\int_{A_i}|f_i(x)|^p\mathrm d\mu\\=\sum_{i\geqslant 1}|a_i|^p\int_{\Omega}|f_i(x)|^p\mathrm d\mu$$
and the result follows. |
H: Let P1 and P2 be path connected. Prove that P1 x P2 is path connected.
So we know that separately they are both path connected.
That in the topological space X $\forall x,y \in X$ $\exists$ $f: I \rightarrow X$ such that $f(0)=x$ and $f(1)=y$.
Given P1 and P2 are path connected.... ok not sure what now....
AI: Let $\left(x_{1},x_{2}\right),\left(y_{1},y_{2}\right)\in P_{1}\times P_{2}$.
If there are continous functions $f_{i}:\mathbb{I}\rightarrow P_{i}$
with $f_{i}\left(0\right)=x_{i}$ and $f_{i}\left(1\right)=y_{i}$
then $f:\mathbb{I}\rightarrow P_{1}\times P_{2}$ defined by $t\mapsto\left(f_{1}\left(t\right),f_{2}\left(t\right)\right)$
is continous and is a path from $\left(x_{1},x_{2}\right)$ to $\left(y_{1},y_{2}\right)$. |
H: Proof via equivalence laws; $(a \lor b) \equiv (b \lor a)$?
Is this a correct progression to prove that $p \rightarrow (q \rightarrow r) \equiv q \rightarrow (p \rightarrow r)$?
$$\begin{align} p \rightarrow (q \rightarrow r) & \equiv p \rightarrow (q \rightarrow r) \\
& \equiv \neg p \lor (q \rightarrow r) \text{ implication law}\\
& \equiv \neg p \lor (\neg q \lor r) \text{ implication law}\\
& \equiv \neg q \lor (\neg p \lor r) \text { associative law}\\
& \equiv \neg q \lor (p \rightarrow r) \text{ implication law}\\
p \rightarrow (q \rightarrow r)& \equiv q \rightarrow (p \rightarrow r) \text{ implication law}
\end{align}$$
For step 4, I've interpreted Associative Laws assuming $(q \lor r) \equiv (r \lor q)$. Is this an assumption I can make?
$$\neg p \lor (\neg q \lor r) \equiv (\neg p \lor \neg q) \lor r \equiv \neg q \lor (\neg p \lor r)?$$
Can I interpret Associative Laws such that $a \lor (b \lor c) \equiv \text{either } b \lor (a \lor c) \text{ or } c \lor (a \lor b)?$
AI: This one doesn't rely on the associativity of disjunction:
If you replace Qs with Ps everywhere you'll have the other direction.
Explanation. To show that $p \rightarrow (q \rightarrow r) \equiv q \rightarrow (p \rightarrow r)$, it will suffice to do two things:
assume $p \rightarrow (q \rightarrow r)$ and derive $q \rightarrow (p \rightarrow r)$
assume $q \rightarrow (p \rightarrow r)$ and derive $p \rightarrow (q \rightarrow r)$
The image shows how (1) is done. We assume $p \rightarrow (q \rightarrow r)$. Now we want to derive something (viz. $q \rightarrow (p \rightarrow r)$) that is a conditional, so we assume its antecedent (viz. $q$) with the hope of deriving its consequent (viz. $(p \rightarrow r)$). But that consequent is also a conditional, so to derive that we assume its antecedent, namely: $p$. Now $p$ with the initial assumption (Premise 1 in the image), by modus ponens gives us $(q \rightarrow r)$ (this is Premise 4 in the image). Next we use the $q$ we assumed (in Premise 2) with that $(q \rightarrow r)$, by modus ponens again, to obtain $r$. Now, to wrap things up. Since we assumed $p$ and obtained $r$, we can conclude that $p \rightarrow r$ (Premise 6). Lastly, since we assumed $q$ and obtained $p \rightarrow r$, we can conclude that $q \rightarrow (p \rightarrow r)$.
The same thing is done for step (2). |
H: is this conjecture true or false?
I want to know if this conjecture istrue or false
$$\Large e^{\frac{ \ln x}{x}} \notin \mathbb{Z} $$
for every $x \in \mathbb{R} \setminus \{1,-1,0\} $
AI: You should ask this only for $x>0$, as the expression is not well defined otherwise. You can rule out the case $x\in(0,1)$ easily, since it implies $\ln(x)/x<0$. Now find the maximum of $\ln(x)/x$ on $(1,\infty)$, and conclude. |
H: How can i get the all digit of a number from 1st to last?
For example:
Given number=1234;
Failing,how can i separate all the digits from 1st to last?like 1,2,3,4
AI: Hint
$\lfloor \log_{10} x \rfloor$ is the number of digits (starting with $0$) and
$\left\lfloor \frac{x}{10^i} \right\rfloor {\rm mod}\ 10$ is the $i$-th digit (the digit with magnitude $10^i$).
Digits of naturals are counted from $0$ to $n-1$ where $n$ is the length of the string representation. |
H: Does $g(f(x))$ imply $g(x)$?
If $g$ is a function of $f(x)$ does this imply that $g$ is a function of $x$?
If yes, am I allowed to write the chain rule as:
$$\frac{{d[g(f[x])]}}{{dx}} = \frac{{d[g(x)]}}{{d[f(x)]}} \cdot \frac{{d[f(x)]}}{{dx}}
% MathType!MTEF!2!1!+-
% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
% hiov2DGi1BTfMBaerbdXwyZnxAJfgBGCuyYfwBV5gzaerbtrvsybst
% HrhAaeXatLxBI9gBaerbnfgDOPuy49vtHrhAaerbd9wDYLwzYbItLD
% harqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea
% 0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0Firpepe
% KkFr0xfr-xfr-xb9adbaGadiWadaqaaqaafeGaayaabaWbaaGcbaWa
% aSaaaeaacaWGKbGaai4waiaadEgacaGGOaGaamOzaiaacUfacaWG4b
% GaaiyxaiaacMcacaGGDbaabaGaamizaiaadIhaaaGaeyypa0ZaaSaa
% aeaacaWGKbGaai4waiaadEgacaGGOaGaamiEaiaacMcacaGGDbaaba
% GaamizaiaacUfacaWGMbGaaiikaiaadIhacaGGPaGaaiyxaaaacqGH
% flY1daWcaaqaaiaadsgacaGGBbGaamOzaiaacIcacaWG4bGaaiykai
% aac2faaeaacaWGKbGaamiEaaaaaaa!664A!
$$
Thanks in advance
AI: It's perfectly fine to write the chain rule as
$$
\frac{\mathrm d[g(f(x))]}{\mathrm dx}=\frac{\mathrm d[g(f(x))]}{\mathrm d[f(x)]}\cdot\frac{\mathrm d[f(x)]}{\mathrm dx}.
$$
The first term just means "differentiate $g(f(x))$ with respect to $f(x)$", i.e. pretend $f(x)$ is the independent variable, then fix your error by multiplying with how $f(x)$ changes with $x$, i.e. $\mathrm d[f(x)]/\mathrm dx$. That's exactly how the chain rule works.
I think you are confusing $g$ with $g\circ f$. Those are two different maps. This confusion may arise from sloppy notation, like the notation often found in physics, where $g$ and $g(y)$ are used interchangably for a quantity $g$ that depends on another quantity $y$. A good notation get's rid of this confusion by focusing on the maps and their respective domains and ranges instead of quantities and what they depend on.
In this case the maps are $f:X\to Y$, $g:Y\to Z$ and $g\circ f:X\to Z$, where $X$, $Y$ and $Z$ are sets. So $f$ takes an element of the set $X$ to the set $Y$ which might then be mapped to an element in $Z$ by applying $g$. Using $f'$ and $g'$ for the derivatives with respect to the only independent variable each of these functions have, the chain rule is written as
$$ (g\circ f)'(x) = g'(f(x))\cdot f'(x) = (g'\circ f)(x)\cdot f'(x).$$ |
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