Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Discrete Math/Calculus (Given summation formula, find value of x) Textbook Question:
Given $\sum\limits_{i=0}^\mathbb{50}{50\choose i}8^i $ = $x^{100}$ . Find the values of x ?
Textbook Solution:
$\sum\limits_{i=0}^\mathbb{50}{50\choose i}8^i $ = $(1+8)^{50} = 9^{50} = [(+-3)^2)]^{50}$
$\therefore$ x = +-3
Personal Logical Question:
*
*In the given textbook solution, how is $(1+8)^{50}$ obtained?
|
The book is using Binomial Theorem:
$$\sum_{i=0}^{n}{n \choose i}a^ib^{n-i}=(a+b)^{n}$$
Use $a=8$, $b=1$ and $n=50$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2153435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to show that if $9n^2=a^2+b^2$, $a$ and $b$ are multiples of $3$ To be honest, I don't know where to start with this problem:
Let $n\in \mathbb{N}$. Prove that if $9n^2$ is the sum of two perfect squares $(a^2,b^2)$, then $a$ and $b$ are multiples of $3$.
|
If one of them is not, say wlog $ a $, then $ a^2 + b^2 \equiv 0 \pmod{9} $, and $ (b/a)^2 \equiv -1 \pmod{9} $ so that $ b/a $ has order $ 4 $ in the group $ (\mathbf Z/9 \mathbf Z)^{\times} $ of order $ 6 $, contradicting Lagrange's theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2153527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Why can't we define "the" derivative when mapping from $\mathbb{R}^2 \to \mathbb{R}$? When we map from $\mathbb{R} \to \mathbb{R}$, the derivative is given by $\lim_{x \to a} \dfrac {f(x) - f(a)}{x-a}.$ When we map from $\mathbb{C} \to \mathbb{C}$ (basically $\mathbb{R}^2 \to \mathbb{R}^2$) we also have a definition for the derivative, $\lim_{z \to w} \dfrac {f(z) - f(w)}{z-w}.$
However, when we map $\mathbb{R}^2 \to \mathbb{R}$, we have only partial derivatives, $\dfrac {\partial f}{\partial x}$, $\dfrac {\partial f}{\partial y}$, and directional derivatives. Why is it that we can't define the derivative just as we do in the other $2$ cases? Is it because there is no nice way to define division between a member of $\mathbb{R}$ and a member if $\mathbb{R}^2$?
|
One way to look at is that, as you say, there's no way to "divide" vectors in $\mathbf{R}^2$ that really makes sense.
But you can think of it this way. When you differentiate a function $y = f(x)$ at a point $x_0$, what you're doing is saying that that function can be approximated by a linear function for values of $x$ close to $x_0$, namely
$$y \approx f(x_0) + f'(x_0)(x-x_0).$$
When you differentiate a function $z = f(x,y)$ of two variables, you likewise want to write that the function can be approximated by a linear function near $(x_0,y_0)$:
$$z \approx k + a(x-x_0) + b(y-y_0),$$
whose graph is a plane. In this approximation, the constant $k$ is $f(x_0,y_0)$, and the coefficients $a$ and $b$ are $\frac{\partial f}{\partial x} (x_0,y_0)$ and $\frac{\partial f}{\partial y} (x_0,y_0)$, respectively. In any case you need two coefficients.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2153633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
Union of Intersections I am trying to prove that $S\cup(S\cap T)=S$ and the dual statement $S\cap(S\cup T)=S$ for a class, and have gotten stuck with my proof.
$$
S\cup(S\cap T)
$$
$$
=\{x|x\in S\lor x\in (S\cap T)\}
$$
$$
=\{x|x\in S\lor (x\in S\land ]x\in T)\}
$$
$$
=\{x|(x\in S\lor x\in S)\land (x\in S\lor x\in T)\}
$$
Any help on where to go from here would be appreciated.
|
We have
$$S\cup(S\cap T)$$
Let $R = S\cap T$
We know $R\subseteq S$ since the intersection of some set $S$ with any other set is a subset of $S$.
We also know the union of $S$ and any subset of $S$--including $R$--is $S$, i.e. $R\cup S=S$.
Therefore, $S\cup(S\cap T)=S$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2153762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to convert (B ∨ ¬A) → (C ∨ ¬D) to conjunctive normal form? I have tried solving it. But, I am stuck:
(B ∨ ¬A) → (C ∨ ¬D) // removing implication
¬(B ∨ ¬A) V (C ∨ ¬D) // using De Morgan's Laws
(¬B ∧ A) V (C ∨ ¬D) // This is not CNF. What to do next?
|
$(B \lor \lnot A) → (C \lor \lnot D)\label{1}\tag{1}$
$(\lnot B \land A) \lor C \lor \lnot D\label{2}\tag{2}$
The best way to answer these types of questions is by Karnaugh Map. If you are not familiar with them there is a page here that explains them quite well
https://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/Logic/Logic3.html
Take your simplified expression ($\ref{2}$) and draw it out on a Karnaugh Map. Next; select regions of the map in such a way as to OR statements together. Next choose one or more statements to AND your first statement together with to produce the final expression.
I choose to start by writing ($\ref{3}$).
$A\lor C\lor \lnot D\label{3}\tag{3}$
Then I realized that there was one undesirable region on the map that ($\ref{3}$) picked up (namely ($\ref{4}$) ).
$A\land D\land B\land\lnot C\label{4}\tag{4}$
To remove this case, I AND'ed my previous expression ($\ref{3}$) with the negation of ($\ref{4}$). The result was
$(A\lor C\lor \lnot D)\land(\lnot A\lor \lnot B\lor C\lor \lnot D)\label{5}\tag{5}$
Which is equivalent to your original expression ($\ref{1}$) except in CNF (Conjunctive Normal Form)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2153865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Proving solutions of Linear Variational Problem Let $$a(v,w) = \int_0^1 a_o(x)v(x)w(x)dx + \lambda \int_0^1 v(y)w(x)dxdy, \; \; \forall v,w \in L^2(0,1)$$ be a bilinear, continuous, elliptic functional with $a_0(x) \in C^0[0,1], \; \;a_0(x) > 0$ on $[0,1]$ and $\lambda \in \mathbb{R}.$ Suppose that $f \in L^2(0,1)$ and consider the variational problem:
$$u \in L^2(0,1) \; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$
$$ a(u,v) = \int_0^1f(x)v(x)dx, \; \; \forall v \in L^2(0,1) $$
(i) Suppose that $a_0 = 1$. Show that if $\lambda \ne -1$ then this problem has a closed form solution and find it.
(ii) if $\lambda = -1$, the problem has no solutions unless $\int_0^1f(y)dy = 0$ and if this holds, then it actually has infinitely many solutions of the form $u = f + C$ where $C$ is a constant.
What I have so far:
I have manipulated the formulation of the problem to arrive at:
$$ a_0(x)u(x) + \lambda \int_0^1u(y)dy = f(x), x \in (0,1) $$
Substituting the given conditions, this becomes:
$$u(x) + \lambda\int_0^1u(y)dy = f(x), x \in (0,1) $$
I'm kind of stuck here. I was thinking that this could be differentiated somehow to turn it into an ODE that can be shown to be solvable/not solvable based on what lambda is, but I'm not sure how to make that work... or if it's even a valid idea.
Thanks for your input!
|
i) integrate $u(x) + \lambda \int_{0}^{1} u(y) dy = f(x)$ from $0$ to $1$ to get
\begin{equation}
\int_{0}^{1} u(x) dx + \lambda \int_{0}^{1} \int_{0}^{1} u(y) dy dx = \int_{0}^{1} f(x) dx
\end{equation}
now giving $\int_{0}^{1} u(y) dy$ a name, say, $\int_{0}^{1} u(y) dy = q$, we have
\begin{equation}
q + \lambda \int_{0}^{1} q dx = q (1 + \lambda) = \int_{0}^{1} f(x) dx
\end{equation}
so, provided, $\lambda \neq -1$, we have
\begin{equation}
\int_{0}^{1} u(y) dy = q = \frac{1}{(1 + \lambda)}\int_{0}^{1} f(x) dx
\end{equation}
plugging this back into the original equation gives
\begin{equation}
u(x) = f(x) - \lambda q
\end{equation}
or simply
\begin{equation}
u(x) = f(x) - \frac{\lambda}{(1 + \lambda)}\int_{0}^{1} f(x) dx
\end{equation}
ii)
falls out directly from the Fredholm alternative:
https://en.wikipedia.org/wiki/Fredholm_alternative
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2153959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Partial derivatives turning into regular derivative If $z=z(x,y)$, and $x=x(t),y=y(t)$, write down the expression for $\frac{dz}{dx}$.
I am confused by the difference between partial and regular derivatives when they are used together.
|
Observe that
$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$
$$\implies \frac{dz}{dx}=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\cdot \frac{dy}{dx}=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\cdot \frac{dy}{dt}\cdot \frac{dt}{dx}$$
$$\implies \frac{dz}{dx}=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\cdot \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
Tell me if you don't get the first line.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2154104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $ I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in Z $)
Here is what I've done
$$\frac {\sin \theta}{\cos \theta} + \frac {\sin 2\theta} {\cos 2\theta} + \frac{\sin 3\theta}{\cos 3\theta}= \frac {\sin \theta}{\cos \theta} \frac {\sin 2\theta} {\cos 2\theta} \frac{\sin 3\theta}{\cos 3\theta}$$
$$ \frac {\sin \theta \cos 2\theta \cos 3\theta + \cos \theta \sin 2\theta \cos 3\theta + \cos \theta \cos 2\theta \sin 3\theta - \sin \theta \sin 2\theta \sin 3\theta }{\cos\theta \cos 2\theta \cos 3\theta} = 0 $$
$$\cos 2\theta \{\sin\theta \cos 3\theta + cos \theta \sin 3\theta \} + \sin 2\theta \{\cos \theta \cos 3\theta - \sin \theta \sin 3\theta \} = 0 $$
$$\cos 2\theta \sin(3\theta + \theta) +\sin2\theta \cos(3\theta + \theta) = 0 $$
$$ \cos 2\theta \sin 4\theta + sin 2\theta cos 4\theta = 0$$
$$ \sin (2\theta + 4\theta) = 0$$
$$\sin 6\theta = 0 $$
$$ \theta = \frac {n\pi}{6}, n \in Z$$
I understand from this question that whatever mistake I am making is in the third step, where I remove $\cos \theta \cos 2\theta \cos 3\theta $ from the denominator. However, despite reading through the aforementioned post, I couldn't really get the intuition behind why this is wrong.
I'd like :
*
*To understand the intuition behind why removing $\cos \theta \cos 2\theta cos 3\theta $ is a mistake.
*To know how to solve this question correctly
*How do I avoid making these types of mistakes when solving trigonometric equations
|
We have
$$\tan (A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1-(\tan A \tan B + \tan B \tan C + \tan C \tan A)}$$
and the given condition implies that $\tan (6\theta) = 0$. Thus $\theta = \frac{n\pi}{6}$, $n \in \mathbb{Z}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2154221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
How to take the integral? $\int \frac{x^2-3x+2}{x^2+2x+1}dx$ $$\int \frac{x^2-3x+2}{x^2+2x+1}dx$$
So after all I had
$$ \frac{-5x+1}{(x+1)^2} = \frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
and of course $$ \int xdx $$
but it is easy to solve, I do not know how to act with devided things, probably solve the system, or is there easier way to find A and B?
After all steps I finally got:
$$-5x + 1 = Ax + A + B$$
|
\begin{align*}
\frac{x^2 - 3x + 2}{x^2 + 2x + 1} &= 1 + \frac{1-5x}{(x+1)^2} \\
&= 1 + \frac{A}{x+1} + \frac{B}{(x+1)^2}
\end{align*}
Then
\begin{align*}
\frac{1-5x}{(x+1)^2} &= \frac{A}{x+1} + \frac{B}{(x+1)^2} \\
&= \frac{A(x+1)+B}{(x+1)^2} \\
1-5x &= Ax+(A+B)
\end{align*}
So $A = -5$ and $B=6$.
Therefore
\begin{align*}
\frac{x^2 - 3x + 2}{x^2 + 2x + 1} &= 1 - \frac{5}{x+1} + \frac{6}{(x+1)^2} \\
\int\frac{x^2 - 3x + 2}{x^2 + 2x + 1}\,dx &= \int \left( 1 - \frac{5}{x+1} + \frac{6}{(x+1)^2} \right) dx \\
&= x - 5 \ln \lvert x+1 \rvert - \frac{6}{x+1} + C
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2154334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Factorise an ideal into a product of prime ideals So for the ideal $I = (20+\sqrt{-5})_{R}$ when $K=\mathbb{Q}(\sqrt{-5})$, how do I factorise this into a product of prime ideals. Do you start by taking the norm of $I$ and decomposing it into a product of primes? So
$$N(I) = 405 = 3^{4} \times 5.$$
If so, where do I go next?
|
Recall that the primes $ 3, 5 $ split as $ 3R = \mathfrak p \mathfrak p' $ and $ 5R = (\sqrt{-5})^2 $ in $ R = \mathcal O_K = \mathbf Z[\sqrt{-5}] $. The norm suggests that the ideal $ I = (20 + \sqrt{-5}) $ factors as $ \mathfrak p^i \mathfrak p'^j (\sqrt{-5}) $, where $ i + j = 4 $. It follows upon division that
$$ \mathfrak p^i \mathfrak p'^j = (1 - 4\sqrt{-5}) $$
This is not divisible by $ 3 $, therefore either $ i = 0 $ or $ j = 0 $. The primes lying over $ 3 $ are $ (3, 1 \pm \sqrt{-5}) $; by calculation we have
$$ (3, 1 - \sqrt{-5})^2 = (9, 3 - 3 \sqrt{-5}, -4 - 2\sqrt{-5}) = (9, 7 - \sqrt{-5}, -4 - 2\sqrt{-5}) = (9, 7 - \sqrt{-5}) = (2 + \sqrt{-5}) $$
$$ (2 + \sqrt{-5})^2 = (1 - 4 \sqrt{-5}) $$
It follows that
$$ (20 + \sqrt{-5}) = (3, 1 - \sqrt{-5})^4 (\sqrt{-5}) $$
is the desired factorization.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2154454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Difference between two products Let $q$ be a square free natural number. Can the difference
$$
\prod_{p \mid q} (p^2+1) - \prod_{p \mid q} (p-1)^2
$$
be estimated in terms of $q$? What would be the correct order of the difference in terms of $q$? Is the difference $\asymp q^2$? Can something be said about
$$
\prod_{p \mid q} \frac{p^2+1}{p+1} - \prod_{p \mid q} \frac{(p-1)^2}{p+1},
$$
in the same spirit?
|
Such difference equals
$$ q^2\left[\prod_{p\mid q}\left(1+\frac{1}{p^2}\right)-\prod_{p\mid q}\left(1-\frac{1}{p}\right)^2\right]\tag{1} $$
and by Euler's product
$$ \prod_{p\in\mathcal{P}}\left(1+\frac{1}{p^2}\right)=\frac{\zeta(2)}{\zeta(4)}=\frac{15}{\pi^2}\tag{2}$$
while
$$ \prod_{p\leq x}\left(1-\frac{1}{p}\right)^2\approx\frac{C}{\log(x)^2}\tag{3}$$
hence your difference is by $\frac{15}{\pi^2}q^2$, but can be as small as $2q$ if $q$ is a prime.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2154579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find $\frac{\sum_\limits{k=0}^{6}\csc^2\left(x+\frac{k\pi}{7}\right)}{7\csc^2(7x)}$ Find the value of
$\dfrac{\sum_\limits{k=0}^{6}\csc^2\left(x+\dfrac{k\pi}{7}\right)}{7\csc^2(7x)}$
when $x=\dfrac{\pi}{8}$.
The Hint given is: $n\cot nx=\sum_\limits{k=0}^{n-1}\cot\left(x+\dfrac{k\pi}{n}\right)$
I dont know how it comes nor how to use it
|
For future reference with this problem being tagged complex-numbers we
show how to evaluate the sum using residues. Suppose we are interested
in
$$S(n) = \sum_{k=0}^{n-1} \csc^2\left(x+\frac{k\pi}{n}\right)
= \sum_{k=0}^{n-1} \frac{2}{1-\cos\left(2x+\frac{2k\pi}{n}\right)}.$$
where we take $x$ to be a real number.
With $$f(z) = \frac{4}{2-\exp(2ix)z-1/\exp(2ix)/z}
\frac{nz^{n-1}}{z^n-1}$$
or alternatively
$$f(z) = \frac{4}{2-\exp(2ix)z-1/\exp(2ix)/z} \frac{1}{z}
\frac{n}{z^n-1}
\\ = \frac{4}{2z-\exp(2ix)z^2-1/\exp(2ix)}
\frac{n}{z^n-1}
\\ = -\frac{4\exp(-2ix)}{z^2 - 2z\exp(-2ix) + \exp(-4ix)}
\frac{n}{z^n-1}
\\ = -\frac{4\exp(-2ix)}{(z-\exp(-2ix))^2}
\frac{n}{z^n-1}$$
we get for the sum with $\zeta_k = \exp(2\pi i k/n)$
$$\sum_{k=0}^{n-1} \mathrm{Res}_{z=\zeta_k} f(z)$$
which means we can evaluate the sum using the negative of the residues
at $z=\exp(-2ix)$ and at infinity. Note however that with $R$ the
radius of a circle going to infinity we get that $f(z)$ is
$\theta(1/R^{n+2})$ and $2\pi R \times 1/R^{n+2} = 2\pi \times
1/R^{n+1}$ vanishes so the residue at infinity is zero. That leaves
for the other residue
$$\left.
\left(-\frac{4n\exp(-2ix)}{z^n-1}\right)'\right|_{z=\exp(-2ix)}
= \left.\frac{4n\exp(-2ix)}{(z^n-1)^2} \times n z^{n-1}
\right|_{z=\exp(-2ix)}
\\ = \frac{4n^2\exp(-2inx)}{(\exp(-2inx)-1)^2}
= -\frac{(2i)^2 n^2}{(\exp(-inx)-\exp(inx))^2}.$$
We obtain
$$S(n) -\frac{(2i)^2 n^2}{(\exp(inx)-\exp(-inx))^2} = 0$$
or
$$\bbox[5px,border:2px solid #00A000]{
S(n) = n^2 \csc^2(nx).}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2154639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Finding a constant so that this is a martingale
Let $Y_{i}$ be IID with $P(Y_{i}=1)$= $\frac{1}{2}=P(Y_{i}=-1)$.
Define $K_{n}=\sum_{i=1}^{n}{Y_{i}}$.
Find the constant $r$ so that so that $$\Lambda_{n}=exp(K_{n} + rn)$$ is a Martingale.
I'm not sure that I have the right answer on this one.
Any help will be appreciated.
|
The definition of a Martingale
$$
\mathbb{E}\left[\Lambda_{n+1}|\mathcal{F}_{n}\right] = \Lambda_{n}
$$
we have
$$
\mathbb{E}\left[\mathrm{e}^{K_{n+1} + r(n+1)}|\mathcal{F}_{n}\right] = \mathbb{E}\left[\mathrm{e}^{Y_{n+1} + K_{n} + r(n+1)}|\mathcal{F}_{n}\right]
$$
we can re-write the argument as
$$
K_{n} + rn + Y_{n+1} + r
$$
or
$$
\mathbb{E}\left[\mathrm{e}^{K_{n+1} + r(n+1)}|\mathcal{F}_{n}\right] = \mathbb{E}\left[\Lambda_{n}\mathrm{e}^{Y_{n+1} + r}|\mathcal{F}_{n}\right] = \Lambda_{n}\mathbb{E}\left[\mathrm{e}^{Y_{n+1} + r}|\mathcal{F}_{n}\right]
$$
we can extract the $Y_{n}$ since we know this up to the filtration time $n$. so we need to compute
$$
\mathbb{E}\left[\mathrm{e}^{Y_{n+1} + r}|\mathcal{F}_{n}\right]
$$
For our condition to hold true we need
$$
\mathbb{E}\left[\mathrm{e}^{Y_{n+1} + r}|\mathcal{F}_{n}\right] = 1
$$
can you compute the expectation of the above? You have a simple calculation since we have discrete binary system!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2154787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Proof that $a(x − y) = ax − ay$ for x and y in vector spaces. Hi i'm confused about this homework question:
"Let $V$ be a vector space over the field $F$ and let $a ∈ F$ and $x, y ∈ V$ .
Show that $a(x − y) = ax − ay$ in $V$."
What I did is below but i'm confused because I feel as though it was too simple.
So from the vector space axioms there is a distributivity property that states that $a(\alpha+\beta)=a\alpha+a\beta$ where $a ∈ F$ and $\alpha, \beta$ $ ∈ V$ so I thought if I set $ \alpha=x$ and $\beta=-y$ then the equality above just becomes: $a(x+(-y))=ax+a(-y))$$=ax+(-a)y)$$=ax-ay$.
Is that ok what I've done?
Many thanks
|
(Responding to the suggestions in the comments.)
Recall that anything multiplied by $0$ is $0$.
If unfamiliar, you may observe that $0x + 0x = (0 + 0)x = 0x$, and use a right cancellation law on the equation $0x + 0x = 0x$ to conclude $0x = 0$.
Next, observe that $(-1)y + y = (-1)y + 1y = (-1 + 1)y = 0y = 0$.
In particular, $(-1)y + y = 0$, so adding $-y$ on the right for both expressions, and again using a right cancellation law, we have $(-1)y = -y$ as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2154873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Proving Saalschutz Theorem I saw this in a pdf, and I'm wondering
Questions:
*
*How do you prove Saalschutz Theorem:
$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}\tag{1}$$
I'm somewhat relatively new to Hypergeometrical Series. I understand that the general Hypergeometrical series takes the form$$_pF_q\left[\begin{array}{c,c}\alpha_1,\alpha_2,\ldots,\alpha_p\\\beta_1,\beta_2,\ldots,\beta_q\end{array};x\right]=\sum\limits_{k=0}^{\infty}\dfrac {(\alpha_1)_k(\alpha_2)_k\ldots(\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\ldots(\beta_q)_k}\dfrac {x^k}{k!}\tag{2}$$
So therefore, by $(2)$, we should have$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\sum\limits_{k=0}^{\infty}\dfrac {(-x)_k(-y)_k(-z)_k}{(n+1)_k(-x-y-z-n)_k}\tag{3}$$
However, I'm not sure how to manipulate the RHS of $(3)$ to get the RHS of $(1)$.
EDIT: Since $(a)_k=\Gamma(a+k)/\Gamma(a)$, the RHS of $(3)$ becomes$$\dfrac {(-x)_k(-y)_k(-z)_k}{(n+1)_k(-x-y-z-n)_k}=\dfrac {\Gamma(k-y)\Gamma(n+1)\Gamma(k-x)\Gamma(-x-y-z-n)\Gamma)k-z)}{\Gamma(n+k+1)\Gamma(-x)\Gamma(-y)\Gamma(-z)\Gamma(-x-y-z-n+k)}$$Now, I need to figure out how$$\Gamma(k-y)\Gamma(k-x)\Gamma(k-z)\Gamma(-x-y-z-n)=\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+z+n+1)$$$$\Gamma(n+k+1)\Gamma(-x)\Gamma(-y)\Gamma(-z)\Gamma(-x-y-z-n+k)=\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)$$
Extra: I also believe that using the same general approach, we can prove$$\begin{align*} & _7F_6\left[\begin{array}{c,c}n,\frac 12n+1,-x,-y,-z,-u,x+y+z+u+2n+1\\\frac 12n,x+n+1,y+n+1,z+n+1,u+n+1,-x-y-z-u-n\end{array}\right]\\ & =\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(u+n+1)\Gamma(x+y+z+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+u+n+1)\Gamma(z+u+n+1)}\\ & \times\dfrac {\Gamma(y+z+u+n+1)\Gamma(x+u+z+n+1)\Gamma(x+y+u+n+1)}{\Gamma(x+z+n+1)\Gamma(y+u+n+1)\Gamma(x+y+z+u+n+1)}\end{align*}\tag{4}$$
|
The development of the Saalschütz's identity that I know proceeds along the following path.
Start with the known identity about the sum of the product of three binomials:
$$ \bbox[lightyellow] {
\begin{gathered}
F(m,n,r,s)\quad \left| {\;0 \leqslant \text{integers}\,m,n} \right.\quad = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
m - r + s \\ k \\
\end{gathered} \right)\left( \begin{gathered}
n + r - s \\ n - k \\
\end{gathered} \right)\left( \begin{gathered}
r + k \\ m + n \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\;k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)
\end{subarray}} {\left( \begin{gathered}
m - r + s \\ k \\
\end{gathered} \right)\left( \begin{gathered}
n + r - s \\ n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m + n - j \\
\end{gathered} \right)\left( \begin{gathered}
k \\ j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\;k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)
\end{subarray}} {\left( \begin{gathered}
m - r + s \\ j \\
\end{gathered} \right)\left( \begin{gathered}
m - r + s - j \\ k - j \\
\end{gathered} \right)\left( \begin{gathered}
n + r - s \\ n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - r + s \\ j \\
\end{gathered} \right)\left( \begin{gathered}
n + m - j \\ n - j \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - r + s \\ j \\
\end{gathered} \right)\left( \begin{gathered}
n + m - j \\ m \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - r + s \\ j \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m \\
\end{gathered} \right)\left( \begin{gathered}
r - m \\ n - j \\
\end{gathered} \right)} = \hfill \\
= \left( \begin{gathered}
r \\ m \\
\end{gathered} \right)\left( \begin{gathered}
s \\ n \\
\end{gathered} \right) \hfill \\
\end{gathered} \tag{1}
} $$
where the steps are:
- inverse convolution on last b.;
- trinomial revision on 1st and 4th b.;
- convolution in $k$ on 2nd and 3rd b.;
- symmetry on 2nd b.;
- trinomial revision on 2nd and 3rd b.;
- convolution in $j$ on 1st and 3rd b.
Then consider that the addenda in the sum are
$$
\begin{gathered}
t_{\,k} = \left( \begin{gathered}
m - r + s \\
k \\
\end{gathered} \right)\left( \begin{gathered}
n + r - s \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r + k \\
m + n \\
\end{gathered} \right) = \hfill \\
= \frac{{\left( {m - r + s} \right)^{\,\underline {\,k\,} } }}
{{k!}}\frac{{\left( {n + r - s} \right)^{\,\underline {\,n - k\,} } }}
{{\left( {n - k} \right)!}}\frac{{\left( {r + k} \right)^{\,\underline {\,m + n\,} } }}
{{\left( {m + n} \right)!}} \hfill \\
\end{gathered}
$$
where $x^{\,\underline {\,k\,} }$ indicates the falling factorial.
The value of the initial term is
$$
t_{\,0} = \frac{{\left( {n + r - s} \right)^{\,\underline {\,n\,} } }}
{{n!}}\frac{{r^{\,\underline {\,m + n\,} } }}
{{\left( {m + n} \right)!}} = \left( \begin{gathered}
n + r - s \\
n \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n \\
\end{gathered} \right)
$$
and the ratio of consecutive terms is a rational function in $k$
$$
\begin{gathered}
\frac{{t_{\,k + 1} }}
{{t_{\,k} }} = \frac{{\left( {m - r + s} \right)^{\,\underline {\,k + 1\,} } \left( {n + r - s} \right)^{\,\underline {\,n - k - 1\,} } \left( {r + k + 1} \right)^{\,\underline {\,m + n\,} } }}
{{\left( {m - r + s} \right)^{\,\underline {\,k\,} } \left( {n + r - s} \right)^{\,\underline {\,n - k\,} } \left( {r + k} \right)^{\,\underline {\,m + n\,} } }}\frac{{k!\left( {n - k} \right)!\left( {m + n} \right)!}}
{{\left( {k + 1} \right)!\left( {n - k - 1} \right)!\left( {m + n} \right)!}} = \hfill \\
= \frac{{\left( {m - r + s - k} \right)\left( {r + k + 1} \right)}}
{{\left( {r - s + 1 + k} \right)\left( {r + k - m - n + 1} \right)}}\frac{{\left( {n - k} \right)}}
{{\left( {k + 1} \right)}} = \hfill \\
= \frac{{\left( {r - m - s + k} \right)\left( {r + 1 + k} \right)\left( { - n + k} \right)}}
{{\left( {r - s + 1 + k} \right)\left( {r - m - n + 1 + k} \right)\left( {1 + k} \right)}} \hfill \\
\end{gathered}
$$
so that we can write
$$
\begin{gathered}
F(m,n,r,s) = t_0 \;{}_3F_2 \left[ {\left. \begin{gathered}
r + 1,\;r - m - s,\; - n \hfill \\
r - s + 1,\;r - m - n + 1\; \hfill \\
\end{gathered} \right|\;1} \right] = \hfill \\
= \left( \begin{gathered}
n + r - s \\
n \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n \\
\end{gathered} \right)\;{}_3F_2 \left[ {\left. \begin{gathered}
r + 1,\;r - m - s,\; - n \hfill \\
r - s + 1,\;r - m - n + 1\; \hfill \\
\end{gathered} \right|\;1} \right] \hfill \\
\end{gathered}
$$
that is, changing $n$ to $q$ so not to get confused with your notation:
$$ \bbox[lightyellow] {
\begin{gathered}
{}_3F_2 \left[ {\left. \begin{gathered}
r + 1,\;r - m - s,\; - q \hfill \\
r - s + 1,\;r - m - q + 1\; \hfill \\
\end{gathered} \right|\;1} \right]\quad \left| {\;0 \leqslant \text{integers}\,m,q} \right.\quad = \hfill \\
= \left( \begin{gathered}
s \\
q \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m \\
\end{gathered} \right)\;\mathop /\limits_{} \;\left( {\left( \begin{gathered}
q + r - s \\
q \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + q \\
\end{gathered} \right)} \right) = \hfill \\
= \frac{{\Gamma \left( {s + 1} \right)\Gamma \left( {r + 1} \right)\Gamma \left( {r - s + 1} \right)\Gamma \left( {q + 1} \right)\Gamma \left( {r - m - q + 1} \right)\Gamma \left( {m + q + 1} \right)}}
{{\Gamma \left( {s - q + 1} \right)\Gamma \left( {q + 1} \right)\Gamma \left( {r - m + 1} \right)\Gamma \left( {m + 1} \right)\Gamma \left( {q + r - s + 1} \right)\Gamma \left( {r + 1} \right)}} = \hfill \\
= \frac{{\Gamma \left( {s + 1} \right)\Gamma \left( {r - s + 1} \right)\Gamma \left( {r - m - q + 1} \right)\Gamma \left( {m + q + 1} \right)}}
{{\Gamma \left( {s - q + 1} \right)\Gamma \left( {r - s + q + 1} \right)\Gamma \left( {r - m + 1} \right)\Gamma \left( {m + 1} \right)}} = \hfill \\
= \frac{{s^{\,\underline {\,q\,} } \left( {r - s} \right)^{\,\underline {\, - \,q\,} } }}
{{\left( {r - m} \right)^{\,\underline {\,q\,} } m^{\,\underline {\, - q\,} } }} \hfill \\
\end{gathered} \tag{2}
} $$
which you can compare with your notation ( not forgetting the final $1/k!$ term) to get
$$
\left\{ \begin{gathered}
q = z \hfill \\
m = y + n \hfill \\
r = - x - 1 \hfill \\
s = - x - 1 - n \hfill \\
\end{gathered} \right.
$$
and finally
$$ \bbox[lightyellow] {
\begin{gathered}
{}_3F_2 \left[ {\left. \begin{gathered}
- x,\; - y,\; - z \hfill \\
n + 1,\; - x - y - z - n\; \hfill \\
\end{gathered} \right|\;1} \right]\quad \left| {\;0 \leqslant \text{integers}\,z,\left( {y + n} \right)} \right.\quad = \hfill \\
= \left( \begin{gathered}
- x - 1 - n \\
z \\
\end{gathered} \right)\left( \begin{gathered}
- x - 1 \\
y + n \\
\end{gathered} \right)\;\mathop /\limits_{} \;\left( {\left( \begin{gathered}
z + n \\
z \\
\end{gathered} \right)\left( \begin{gathered}
- x - 1 \\
y + n + z \\
\end{gathered} \right)} \right) = \hfill \\
= \frac{{\Gamma \left( { - x - n} \right)\Gamma \left( {n + 1} \right)\Gamma \left( { - x - y - n - z} \right)\Gamma \left( {y + n + z + 1} \right)}}
{{\Gamma \left( { - x - n - z} \right)\Gamma \left( {n + 1 + z} \right)\Gamma \left( { - x - y - n} \right)\Gamma \left( {y + n + 1} \right)}} = \hfill \\
= \frac{{\left( { - x - 1 - n} \right)^{\,\underline {\,z\,} } \,n^{\,\underline {\, - \,\,z\,} } }}
{{\left( { - x - 1 - n - y} \right)^{\,\underline {\,z\,} } \;\left( {n + y} \right)^{\,\underline {\, - \,z\,} } }} = \hfill \\
= \frac{{\Gamma \left( { - x - n} \right)\Gamma \left( {n + 1} \right)\Gamma \left( { - x - y - n - z} \right)\Gamma \left( {y + n + z + 1} \right)}}
{{\Gamma \left( { - x - y - n} \right)\Gamma \left( {y + n + 1} \right)\Gamma \left( { - x - n - z} \right)\Gamma \left( {n + 1 + z} \right)}} = \hfill \\
= \frac{{\left( { - x - 1 - n} \right)^{\,\underline {\,y\,} } \,n^{\,\underline {\, - \,\,y\,} } }}
{{\left( { - x - 1 - n - z} \right)^{\,\underline {\,y\,} } \;\left( {n + z} \right)^{\,\underline {\, - \,y\,} } }} \hfill \\
\end{gathered} \tag{3}
} $$
---- Addendum -------
Identity (1) is famous because it is one of the very few involving the product of more than two binomials,
and can be found in good collections of binomial identities .
A hint on how it could be related to binomial expansion can be got from considering
how it works for the last step, i.e.
$$
\begin{array}{l}
\left( {1 + x} \right)^{\,r} \left( {1 + y} \right)^{\,s} = \sum\limits_{\left( {0\, \le } \right)\,m} {\sum\limits_{\left( {0\, \le } \right)\,n} {\left( \begin{array}{c}
r \\
m \\
\end{array} \right)\left( \begin{array}{c}
s \\
n \\
\end{array} \right)x^{\,m} y^{\,n} } } = \\
= \sum\limits_{\left( {0\, \le } \right)\,m} {\;\sum\limits_{\left( {0\, \le } \right)\,n} {\sum\limits_{\left( {0\, \le } \right)\;j\,\left( { \le \,n} \right)} {\left( \begin{array}{c}
m - r + s \\
j \\
\end{array} \right)\left( \begin{array}{c}
r - m \\
n - j \\
\end{array} \right)z^{\,j} y^{\,n} \left( \begin{array}{c}
r \\
m \\
\end{array} \right)x^{\,m} } } } = \\
= \sum\limits_{\left( {0\, \le } \right)\,m} {\;\left( {1 + yz} \right)^{\,m - r + s} \left( {1 + y} \right)^{\,r - m} \left( \begin{array}{c}
r \\
m \\
\end{array} \right)x^{\,m} } = \\
= \left( {1 + yz} \right)^s \sum\limits_{\left( {0\, \le } \right)\,m} {\left( {\frac{{1 + y}}{{1 + yz}}} \right)^{\,r - m} \left( \begin{array}{c}
r \\
m \\
\end{array} \right)x^{\,m} } = \\
= \left( {1 + yz} \right)^s \left( {\frac{{1 + y}}{{1 + yz}} + x} \right)^r \\
\end{array}
$$
and then put $z=1$.
Note: the bounds put in brackets are to indicate that, from an algebraic point of view, they are superfluous because inherent to the binomial coeff. (which is null outside them); this is important to carry on in order to correctly apply the convolution.
A combinatorial proof can be found in this interesting paper
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2155025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Riemann-integrability of function with countably many discontinuities Enumerate the rationals in $[0,1]$ by $\{q_k\}_{k\in \mathbb{N}}$. Define for each $k$, $f_k(x) = 1$ if $q_k<x\leq1$ and $0$ if $0\leq x \leq q_k$.
I WTS $F(x) = \sum \frac{f_k(x)}{2^k}$ is Riemann integrable on $[0,1]$. I can't find an easy uniform convergence argument. Also, I'm having trouble calculating what the values of $F(x)$ even are. This is obviously my biggest issue. I'd appreciate some help.
|
Hint:
$$\left| \frac{f_k(x)}{2^k} \right| \leqslant \frac{1}{2^k}, \\ \sum_{k=1}^\infty \frac{1}{2^k} = 1, $$
and $\sum \frac{f_k(x)}{2^k}$ is uniformly convergent by the Weierstrass test.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2155181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Analysis - Prove that there exists a smallest positive number $p$ such that $\cos(p) = 0$
Prove that there exists a smallest positive number $p$ such that $\cos(p) = 0
$.
I think I'm supposed to use either Rolle's theorem or the Mean Value Theorem but I'm not sure how, any help would be appreciated. Thanks!
|
How about you restrict the domain to I=(0,pi) and show that since cos(x) is monotonic on I, it is one to one, and therefore only one value, p=pi/2, would be the solution on I since cos(a)=cos(b)=0 --> a=b, the solution p in I is unique and since any other positive solution "q" would have to be outside the interval I, pi/2=p
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2155270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Product of Finite Abelian Group I need to proof for a finite abelian group $G$ that for all $x\in G$ we have $\prod_{g\in G}xg=\prod_{g\in G}g$.
I figured that using the commutative property $\prod_{g\in G}xg=x^n\prod_{g\in G}g$. Which would leave us to proof $x^n=e$, where $e$ is the identity element.
Hopefully I'm just missing something obvious here, because I feel like smashing my head against the table now.
|
The map $g\mapsto xg$ is a permutation of $G$. hence $\prod xg$ is the same as $\prod g$, just in a different order. As $G$ is abelian, different order doesn't matter.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2155374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Is it always possible to map $n$ distinct vectors to $n$ distinct scalars? Recently while reading a paper I came across the claim that if we have $n$ distinct vectors $\vec{x}_i \in \mathbb{R}^d$ (they didn't clarify what they mean by 'distinct' but I assume it means no two vectors have all the same values) that it is always possible to select another vector $\vec{a} \in \mathbb{R}^d$ such that $i \neq j$ implies $\langle \vec{x}_i, \vec{a}\rangle \neq \langle \vec{x}_j, \vec{a}\rangle$, where $\langle \cdot , \cdot \rangle$ denotes the inner product (i.e., dot product).
While I believe them that it's true, I can't figure out how to prove it for myself. Note that $n$ is finite, and that we know all $n$ vectors before we have to select $\vec{a}$.
|
For any pair of distinct vectors $x_i \not= x_j \in \mathbb{R}^d$, the probability that $\langle x_i, a\rangle = \langle x_j, a\rangle$ for a uniformly random $a \in \mathbb{R}^d$ on the unit sphere is $0$.
You can see that this is true intuitively by recalling the relationship between the dot product and the angle between two vectors. It can be proved formally by integrating over the unit sphere.
Thus, you can choose a uniformly random vector $a$ on the unit sphere and it will work with probability $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2155496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Bound on 1st Order Taylor Expansion Consider a differentiable function $f$ with the property that for any points $x,y$, we know that $\frac{|f'(x)-f'(y)|}{|x-y|} \leq C.$ How do I show that the following inequality holds for this function: $$f(x) - f(y) - f'(y)(x-y) - \frac{C}{2}(x-y)^2 \leq 0$$
Note the similarities to first order Taylor expansion around $y$ which is: $f(y) + f'(y)(x-y)$
|
I'll write this out here for clarity:
From the Lagrange form of Taylor's theorem, we have that
$$f(x) = f(y) + f'(y) (x-y) + \frac{f''(\xi)}{2}(x-y)^2$$
for some $\xi \in (x, y)$. So if we can show that $f''(\xi) \leq C$ for any $\xi \in (x, y)$, we are done.
Using the inequality given in the question, since it holds for any $x$ and $y$, simply take the limit as $y \to x$ and observe that $|f''(x)| \leq C$ for all $x$, so $f''(x) \leq C$, as required.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2155643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What is the sum of the solutions to $6x^3+7x^2-x-2=0$ What is the easy way to solve the problem?
The sum of the solutions to $6x^3+7x^2-x-2=0$ is:
$$A) \ \frac{1}{6}$$ $$B) \ \frac{1}{3}$$ $$C) \ \frac{-7}{6}$$ $$D) -2$$ $$E) \text{ none of above}$$
|
For a cubic equation of the form $ax^3 + bx^2 + cx +d$, the sum of all the roots is given by $\frac{-b}{a}$. Here, $b=7,a=6$ so sum of the roots $= \frac{-7}{6}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2155713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Sample space and total possible events Taken from my book:
How many events are there associated with the roll of one die? Solution: Each event corresponds to a subset of {1,2,3,4,5,6}. there are $2^6$ subsets, so there are $2^6$ possible events.
Now my question is: why $2^6$ events? If you roll only one dice shouldn't the possible events be only 6? {1}, {2}, {3}, {4}, {5} and {6}?
|
You have the terms "event" and "outcome" confused.
The outcomes of the sample space are: $1, 2, 3, 4, 5,$ and $6$. These are the elements of the sample space: $\{1, 2, 3, 4, 5,6\}$ .
An event is a set of outcomes. Such as the event of rolling an even number: $\{2,4,6\}$, the event of rolling a number greater than four: $\{5,6\}$, and such.
There are $2^6$ subsets of the sample space, which are all the plausible events: $\{\}, \{1\}, \{2\}, \{3\}, \{4\}, \{5\}, \{6\}, \{1,2\}, \ldots, \{5,6\}, \{1,2,3\},\ldots,\ldots,\{1,2,3,4,5,6\}$.
The set of these form a sigma-algebra, but that's another topic.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2155961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Prove the equivalence of norms on the Hardy space $H^2(\mathbb{D})$. Let $H^2(\mathbb{D})$ be the space of all functions $f$ holomorphic on the open unit disk $\mathbb{D}$ such that the Hardy norm, given below, is finite:
$$||f||_H^2 = \sup_{0<r<1}\frac{1}{2\pi}\int_0^{2\pi} |f(re^{i\theta})|^2 \ \mathrm{d}\theta.$$
I have already shown that the evaluation $f\mapsto f(z) \ (z\in\mathbb{D})$ is continuous with respect to the norm $||\cdot||_H$.
Now let $||\cdot||$ be any other norm with respect to which $H^2(\mathbb{D})$ is a Banach space, and for which the evaluations $$f\mapsto f(1/(n+1)) \ (n=1,2,...)$$ are continuous. How do I prove that $||\cdot||$ is equivalent to $||\cdot||_H$?
I know that, due to a consequence of the Open Mapping Theorem, one need only show one inequality involving these two norms. Moreover, using the Uniform Boundedness Principle, there is a constant $K$ such that, for all $n\in\mathbb{N}, f\in H^2(\mathbb{D})$, we have $$|f(1/(n+1))|\le K<\infty.$$
|
The key tools here are the closed graph theorem, the open mapping theorem (for Banach spaces), and the identity principle for holomorphic functions.
So suppose $f_m \to f$ in $(H^2,\|\,\|_H)$ and $f_m \to g$ in $(H^2,\|\,\|).$ Because we know point evaluation is continuous on the first space, we have
$$f_m(1/(n+1)) \to f(1/(n+1)), \, n = 1,2,\dots $$
Now we are given the point evaluations at each $1/(n+1)$ are continuous in the second space, so we also have
$$f_m(1/(n+1)) \to g(1/(n+1)), \, n = 1,2,\dots $$
It follows that $f(1/(n+1))= g(1/(n+1)) \, n = 1,2,\dots.$ But $f,g$ are both holomorphic in $\mathbb D$ and $\{1/(n+1): n = 2,3,\dots \}$ is a set with limit point in $\mathbb D.$ By the identity principle, $f=g$ on $\mathbb D.$
By the closed graph theorem, we have shown the identity map
$$I:(H^2,\|\,\|_H)\to(H^2,\|\,\|)$$
is continuous. Now $I$ is obviously a bijection between these two spaces. In particular it is surjective. Thus $I$ is an open map by the open mapping theorem. Thus $I^{-1} = I$ is also continuous. So $I$ is a Banach space isomorphism between $(H^2,\|\,\|_H)$ and $(H^2,\|\,\|).$ This gives the desired equivalence of norms.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2156072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Distribution of the squared sum of a categorical distribution Given $k\in\mathbb{N}^+$ and $P = \{p : p \in [0, 1]^k \wedge \sum_{i=1}^k p_i = 1 \}$, we choose a random element $q$ of $P$. What is the probability distribution of $S = \sum_{i=1}^k q_i^2$?
I have made some plots with synthetic examples and all I have figured out is that the mean seems to be something similar to $1/k$ (although I could be wrong) and the kurtosis appears to decrease with $k$.
|
A common distribution on the simplex is the Dirichlet Distribution. See that your example we set all $\alpha_i=1$ to get $\mathbb{E}[p_i^2]=\frac{1}{k^2}+\frac{k-1}{k^2(k+1)}=\frac{2}{k(k+1)}$, and so $\mathbb{E}[S]=\frac{2}{k+1}$. As far as distributions go the $p_i$ are beta which means their squares don't have an elementary distribution. However if you choose your distribution on the simplex so that $\alpha_i=\frac{1}{k-1}$ then $p_i\sim\text{Beta}(\frac{1}{k-1},1)$ so that $p_i^2\sim\text{Beta}(\frac{1}{2(k-1)},1)$. However sums of betas are not well-understood.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2156228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Minimize $P=5\left(x^2+y^2\right)+2z^2$ For $\left(x+y\right)\left(x+z\right)\left(y+z\right)=144$, minimize $$P=5\left(x^2+y^2\right)+2z^2$$
I have no idea. Can you make a few suggestions?
|
HINT: prove that $$P=5(x^2+y^2)+2z^2\geq 72$$ and the equal sign will atained at $$(x,y,z)=(2,2,4)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2156334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
How to minimize a sum of absolute values using linear programming? I am having trouble understanding the logic behind optimization of cost function of the form $$\min (|x| + |y| + |z|) \,$$ subject to constraints $$Ax \le b \qquad Cx = d $$ such as $$ x + y \le 1 \qquad 2x + z = 3.$$
I have seen methods involving representing absolute values as a new variable and putting constraints on them, $i.e$ $$|x| = a \qquad -a \le x \le a$$ but I don't understand why should we represent an equality as inequality. Aren't we changing the equation itself?
I have tried to think but am unable to grasp it. Please help.
The method is available on Wikipedia as a numerical example:-
https://optimization.mccormick.northwestern.edu/index.php/Optimization_with_absolute_values
EDIT:-
|
On that page, they are solving another problem. They are trying to make the constraint |x| < b, so it's true that they can split that inequation in those two. In your case, you can't, because you need it to be either x, or -x, not any value in between.
What you can do, is using a bivalent variable. Those are also called logical variables.
With them, you can define another variable, like u, and restrain them like this.
Let's say V is a bivalent variable. And M a big number.
x - MV ≤ u ≤ x + MV
-x - M*(1-V) ≤ u ≤ -x + M*(1-V)
This way, when V is 0, the second constraint does nothing. And the first one forces u to be x.
When V is 1, the first constraint does nothing. And the second one forces u to be -x.
Then, when the problem is solved, u can only take one of those values.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2156439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
A Tic-Tac-Toe Variant Let's imagine a game like tic-tac-toe, but you have to have 3 of the same sign in the same line or colon... So the diagonals don't count!
Then I have to prove that the second player to play can always be sure not to lose... But how do I do that?
I am searching a simple argument, not just like constructing a game tree...
|
Suppose X plays first. The only thing O needs to do to avoid losing is to manage to have one mark in each row and one mark in each column. It's easy to find a strategy that can achieve that.
For example, place the first two Os in the same column and the same row as X's first play. Before O's next move, the board looks like this, up to symmetries
X O .
O . .
. . .
plus two Xs that the first player has put down in the mean time. If the bottom right corner is still free, O plays that and has now forced a draw. Otherwise the situation (still up to symmetry) is one of
X O X X O . X O .
O . . O . X O X .
. . X . . X . . X
In each of these cases, O can now play in the last column, and X is then powerless to prevent O from occupying the bottom row the next time, again forcing a draw.
Of course, if you already know that ordinary tic-tac-toe always ends in a draw with perfect play, you don't even need this much analysis: Second player simply pretends he's playing ordinary tic-tac-toe. Since this is enough to prevent the first player from winning, it will also prevent the first player from meeting the strictly harder victory condition of the variant game.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2156604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to find an equation for this sine-ish wave I'm programming a game. Part of the game involves a spinning rectangle, so I'd like to keep track of two points on that rectangle. The center, and the bottom right corner of it.
I want to draw a sine wave of the relative vertical distance between, the center and the corner.
At 0, the distance is -16.
At $\frac{\pi}{2}$, the distance is 32
At $\pi$, the distance is 16
At $\frac{3\pi}{2}$, the distance is -32
At $2\pi$, this distance is -16
This isn't like to periodic functions I studied in school, where the distance between $f(0)$ and $f(\frac{\pi}{2})$ is the same as the distance between $f(\frac{\pi}{2})$ and $f(\pi)$, etc.
I'm having trouble figuring out the equation of this function.
Would appreciate any help.
|
Your corner has the initial coordinates $$\pmatrix{x_0\\y_0} = \pmatrix{32\\-16}.$$ In order to get the new coordinates, you have to apply the rotation, which can be expressed in terms of a matrix-vector multiplication:
$$
\pmatrix{x\\y} = \pmatrix{\cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi} \pmatrix{x_0\\y_0}
$$
So your $y$ can be calculated as
$$
y=32\sin\varphi -16\cos\varphi
$$
If you want to have only one trigonometric function, this can be written as
$$
y=16\sqrt{5}\sin(\varphi-\varphi_0)\;\;\mbox{with}\;\;
\varphi_0=\arctan\left(\frac{1}{2}\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2156718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to find the Laplace Transform of $t^2sin(t)$ Using the rule: $$\mathcal{L}(t^nf(t))=(-1)^n\frac{d^n}{ds^n}F(s)$$
where in this case
$$f(t)=\sin(t),\,\,\,\,\,\,\,\,\,\mathcal{L}(\sin(t))=F(s)=\frac1{s^2+1},\,\,\,\,\,\,\,\,\,\,n=2.$$
Find the 2nd derivative of F(s):
$$\frac{d^2}{ds^2}\Big(\frac1{s^2+1}\Big)=\frac{6s^2-2}{(s^2+1)^3}$$
The transform:
$$\mathcal{L}(t^2sin(t))=(-1)^2\frac{6s^2-2}{(s^2+1)^3}$$
$$= \frac{6s^2-2}{(s^2+1)^3}$$
|
$L\{tf(t)\} = \int_0^{\infty} tf(t) e^{-st} dt$
if we integrate both sides with respect to $s.$
$\int L\{tf(t)\} ds = -\int_0^{\infty} f(t) e^{-st} dt = -L\{f(t)\}$
And then we can differentiate both sides (with respect to $s$).
$L\{tf(t)\} = -\frac {d}{ds} L\{f(t)\}$
$L\{t^2\sin t\}=\frac{d^2}{ds^2} \frac {1}{s^2+1}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2156907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Probability problem with $n$ keys and $k$ locks Let's say we have $n$ keys and $k$ locks on a door, with $n \ge k$. The $n$ keys are different, i.e. only $1$ key goes into a specific lock.
Also every lock has a key that goes to it (but some of the keys may be "duds").
What is the probability of unlockin gthe door on the $M$th attempt?
If we had $n$ keys and $1$ lock then obviously the probability to unlock on $M$-th attempt would be $\frac1n$, but now we have more locks.
So let's say we have $11$ keys and $2$ locks, what the probability that while we unlock we use up $3$ keys that are bad?
Is it $\frac1{11}\cdot\frac1{11}$ because we try the key on both doors before discarding it? If it doesn't unlock any door we don't use it again.
What if we have more locks?
|
Note: I think the question may be defining "attempt" as trying the key on a single lock. In this answer, I took "attempt" to be trying one key (on every lock).
The question is: if there are $n$ keys and $k$ locks, and a random ordering of the $n$ keys, what is the probability that the locks are all unlocked in $m$ steps, where $k \le m \le n$?
Overall, we randomly choose $k$ of the $n$ keys to be the correct ones, and this can be done in $\binom{n}{k}$ ways. The question is then the probability that the last correct key is the $m$th one. The number of ways to do that is $\binom{m-1}{k-1}$, since we have to choose which of the $m-1$ first keys are the $k-1$ remaining correct ones.
Therefore the answer is
$$
\frac{\binom{m-1}{k-1}}{\binom{n}{k}}
= \frac{(m-1)! \cdot (n-k)! \cdot k}{(m-k)! \cdot n!}.
$$
Note the special cases:
*
*When $k = 1$, this gives ${(n-1)!}/{n!} = 1/n$, which is what you computed.
*When $m = k$, this gives $1 / \binom{n}{k}$, because the correct keys must all come at the front.
*When $n = k$, this gives $1$ if $m = n$, $0$ otherwise.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2157045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
A self inverse function $f(x)=\frac{ax+1}{x-b}$
If $$f(x)=\frac{ax+1}{x-b} \forall x \in\mathbb{R}-b,ab\neq1,a\neq1$$ is a self inverse function such that $$\frac{f(4)}{4}=\frac{f(12)}{12}={f\left(\frac{1+b}{1-a}\right)}$$The question is to find out $a$ and $b$
For a self inverse function $f(f(x))=x$.So I tried to put $f(x)$ in place of $x$ and solve the resulting equation but it didnot helped me .Is there a more logical way to solve this problem in limited time?Any ideas?Thanks.
|
Write this as
$$\begin{pmatrix}
a && 1 \\
1 && -b
\end{pmatrix}\begin{pmatrix}x \\ 1\end{pmatrix}$$
Then this is self inverse iff
$$\begin{pmatrix}
a && 1 \\
1 && -b
\end{pmatrix}^2= \lambda I_2$$
i.e.
$$\begin{pmatrix}
a^2+1 && a-b \\
a-b && 1+b^2
\end{pmatrix}= \lambda I_2$$
So $a=b$.
The transformation is projective so all we need now is to check $a,b$ so that your conditions follow.
$$12f(4)=4f(12)\iff 3(4a+1)/(4-a)=(12a+1)/(12-a)$$
$$\iff 3(4a+1)(12-a)=(12a+1)(4-a)$$
$$\iff 143a+36=47a+4$$
$$\iff 96a=-32\iff a=-1/3.$$
Finally
$$\displaystyle f\left({1+b\over1-a}\right)=f(1/2)=1$$
$$\ne -1/52=f(4)/4.$$
So no such transformation exists.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2157238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
The deep reason why $\int \frac{1}{x}\operatorname{d}x$ is a transcendental function ($\log$) In general, the indefinite integral of $x^n$ has power $n+1$. This is the standard power rule. Why does it "break" for $n=-1$? In other words, the derivative rule $$\frac{d}{dx} x^{n} = nx^{n-1}$$ fails to hold for $n=0$.
Is there some deep reason for this discontinuity?
|
The so-called "deep reason" is not deep at all. The term $\displaystyle \log(x)$ is simply the constant term in the expansion of $\displaystyle \frac{x^{n+1}}{n+1}$ around $n=-1$. To see this, we simply write
$$\begin{align}\frac{x^{n+1}}{n+1}&=\frac{e^{(n+1)\log(x)}}{n+1}\\\\&=\frac{1}{n+1}\sum_{k=0}^\infty \frac{(n+1)^k\,\log^k(x)}{k!}\\\\&=\sum_{k=0}^\infty\frac{(n+1)^{k-1}\log^k(x)}{k!}\\\\&=\frac{1}{n+1}+\log(x)+\frac12(n+1)\log^2(x)+O((n+1)^2)\end{align}$$
whence we see the leading terms in the asymptotic ($n\sim -1$) expansion of $\displaystyle \frac{x^{n+1}}{n+1}$. Obviously, we see that $\lim_{n\to -1}\frac{x^{n+1}-1}{n+1}=\log(x)$.
More simply, let $f(x,n)$ be given by the integral
$$\begin{align}
f(x,n)&=\int_1^x t^n\,dt\\\\
&=\frac{x^{n+1}-1}{n+1}\tag 1
\end{align}$$
Note that $f(x,n)$ is continuos on $(0,\infty)\times \mathbb{R}$ (i.e., $n$ need not be restricted to the integers).
Then, note that the limit as $n\to -1$ of $f(x,n)=f(x,-1)$ is
$$\begin{align}
\int_1^x \frac1t\,dt&=f(x,-1)\\\\
&=\lim_{n\to -1}f(x,n)\\\\
&=\lim_{n\to -1} \frac{x^{n+1}-1}{n+1}\\\\
&=\lim_{n\to -1}\frac{e^{(n+1)\log(x)}-1}{n+1}\\\\
&=\log(x)
\end{align}$$
So, we can recover the expected result, $\int_1^x \frac{1}{t}\,dt=\log(x)$, by taking the limit in $(1)$ for $n\ne -1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2157338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "116",
"answer_count": 10,
"answer_id": 1
}
|
Find the limit of the following log terms
I thouht about it alot but could not able to getany idea .
Can anybody provide me a hint
|
Hint:
$$\lim_{x \to 0} \frac{f(3+x) - f(3-x)}{x} = \lim_{x \to 0} \frac{f(3+x) - f(3)}{x} + \lim_{x \to 0} \frac{f(3-x) - f(3)}{-x} = f'(3) + f'(3)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2157481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
If $a^2+b^2+c^2+d^2+e^2=5$ so $\sum\limits_{cyc}\frac{1}{7-2a}\leq1$.
Let $a$, $b$, $c$, $d$ and $d$ be non-negative numbers such that $a^2+b^2+c^2+d^2+e^2=5$. Prove that:
$$\frac{1}{7-2a}+\frac{1}{7-2b}+\frac{1}{7-2c}+\frac{1}{7-2d}+\frac{1}{7-2e}\leq1$$
The equality occurs also for $a=2$ and $b=c=d=e=\frac{1}{2}$.
I tried TL:
$$1-\sum_{cyc}\frac{1}{7-2a}=\sum_{cyc}\left(\frac{1}{5}-\frac{1}{7-2a}\right)=\frac{2}{5}\sum_{cyc}\frac{1-a}{7-2a}=$$
$$=\frac{2}{5}\sum_{cyc}\left(\frac{1-a}{7-2a}+\frac{a^2-1}{10}\right)=\frac{1}{25}\sum_{cyc}\frac{(a-1)^2(3-2a)}{7-2a}$$
and I don't see what is the rest.
|
We must to prove that $f(a_1)+f(a_2)+f(a_3)+f(a_4)+f(a_5) \le 1$ ,
for non-negative $a_1+a_2+a_3+a_4+a_5=5$
$$f(x)=\dfrac{1}{7-2\sqrt{x}}$$
Since $f''(x)=\dfrac{(7-6\sqrt{x})}{2(2x-7\sqrt{x})^3}$ , we only need to consider the inequality in case $0< a_1=a_2=a_3=a_4=t^2 \le 1 \ , \ a_5=5-4t^2$
Clearly
$g(t)=\dfrac{4}{7-2t}+\dfrac{1}{7-2\sqrt{5-4t^2}} $
$ g'(t)=\dfrac{8}{(7-2t)^2}-\dfrac{8t}{\sqrt{5-4t^2}(7-2\sqrt{5-4t^2})^2}\ge 0 \Leftrightarrow $
$\sqrt{5-4t^2}\cdot(69-16t^2) \ge 4t^3-140t^2+49t+140>0 \Leftrightarrow$
$(t-1)(2t-1)(t+1)(104t^3-60t^2+1062t-841) \le 0 \ ,\ t\in [0,1]$
So maximum of $g(t)$ is attained at $t_1=\dfrac{1}{2}$ and $t_2=1$.
$$g(t_1)=g(t_2)=1$$
Equality holdes for : $(a=b=c=d=e=1) $ and $\left( a=b=c=d=\dfrac{1}{2}, e=2\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2157579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Why does $[f'(a)(a+\Delta x-a)+b]-[f'(a)(a-a)+b]=f'(a)\Delta x$ Why does $[f'(a)(a+\Delta x-a)+b]-[f'(a)(a-a)+b]=f'(a)\Delta x$? If I try to calculate this I get $f'(a)(\Delta x-1) \neq f'(a)\Delta x$.
|
Some of the terms cancel. Look:
$$\begin{align}
[f'(a)(\color{red}{a}+\Delta x\color{red}{-a})+b]-[f'(a)\color{blue}{(a-a)}+b]&= [f'(a)(\Delta x)+b]-[f'(a)\cdot \color{blue}{0}+b] \\
&= f'(a)(\Delta x) \color{green}{+b-b} \\
&= f'(a)(\Delta x).\end{align} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2157741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Gradient and transpose I have simple question, when taking the Jacobian of a linear vector function, do you have to transpose the matrix $A$?
$b\in\mathbb{R}^{n},x\in\mathbb{R}^{m}\\J(Ax+b) = A^T, A\in\mathbb{R}^{n\times m}$
|
$J_x(f)$ is the, in case of existence, uniquely determined matrix with the property
$$f(x+h)=f(x)+J_x(f) h+\mathrm o(\Vert h\Vert).$$
When plugging in $f(x)=Ax+b$, you are looking for a $J$ with
$$ A(x+h)+b=Ax + b + J_x(f)h+\mathrm o(\Vert h\Vert).$$
Since the left side is equal to $Ax + Ah + b$ (because of linearity), you see, that the unique solution is $J_x(f)=A$, and you can even drop the $\mathrm o$-term.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2157816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Number of conjugacy classes are there over $\mathbb{F}_p$ I am struggling with the next exercise of my HW:
How many conjugacy classes are in $GL_3(\mathbb{F}_p)$? And how many in $SL_2(\mathbb{F}_p)$?
It's on the topic of Frobenius normal form of finitely generated modules over $\mathbb{F}_p$.
I'd appreciate any idea.
|
For the general linear group I suggest that you count the (irreducible) linear polynomials with non-zero constant term, the irreducible quadratics, and finally the irreducible cubics. You then get a conjugacy class for each cubic, $C(f(X)$; a conjugacy class for each pair (linear, irreducible quadratic), $C(X-\alpha)\oplus C(q(X)$; and then you are left dealing with the elements where the characteristic polynomial is the product of linear factors. These last will give you classes for types $C((X-\alpha)^3)$, $C((X-\alpha)^2)\oplus C(X-\beta)$, $C(X-\alpha)\oplus C(X-\beta) \oplus C(X-\gamma)$. (Note, $\alpha=\beta$ is possible.)
For the special group you now need to identify which of these classes is in the group, and then investigate the relative sizes of the centraliser of an element in the special group and the general group to see whether the $GL$-orbit splits into smaller $SL$-orbits.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2157914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Can a group always be written as a Zappa-Szep product of its Sylow subgroups? In particular, can a simple group (so that we don't have any proper normal subgroups) have an exact factorization, i.e. do there always exist $H, K<G$ with $H\cap K = \{1_G\}$ and $|H|\cdot|K|=|G|$?
|
To answer the question in the title, the group ${\rm PSU}(3,3)$ of order $6048 = 2^5.3^3.7$ cannot be wriiten as a product $P_1P_2P_3$ of $3$ Sylow subgroups in any order. That is proved in a paper I wrote with Petey Rowley a while ago:
D. Holt and P. Rowley, On products of Sylow subgroups in finite groups,
Arch. Math. 60 (1993), 105-107.
The same group is a counterexample to the question in the body because you can check by computing all subgroups of ${\rm PSU}(3,3)$ on a computer that there are do not exist inters $a,b>1$ with $ab=6048$ such that this group has subgroups of orders $a$ and $b$. There may be smaller counterexamples.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
|
Note: $7^2<2\cdot 5^2$ and $5>2^2$
$7^{55}<7\cdot 5^{54}\cdot 2^{27}<5^{55}\cdot 2^{28}<5^{69}<5^{72}$ as required
With an extra jink into factors of $3$, we can show $7^{55}<5^{67}$
Extra notes: $3^3>5^2$ and $5^5>3\cdot2^{10}$
$7^{55}<7\cdot 5^{54}\cdot 2^{27}<5^{54}\cdot 2^{30}<5^{52}\cdot 2^{30}\cdot 3^{3}<5^{67}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
}
|
Studying properties of an integral $f:\mathbb R \to \mathbb R, \: f(x) =\int _0^x\:e^{-t^3}\left(t^3+t^2-t-1\right)dt$
Firstly, I have to find $min(f(x))$. The possible answers are:
A) $f(0)$
B) $f(1)$
C) $f(-1)$
D) $f(3)$
E) $f$ tends to $- \infty$
In order to do that, I have rewritten the function like this
$f\left(x\right)=\int _0^x\:e^{-t^3}\left(t+1\right)^2\left(t-1\right)dt$
and I studied the sign of the function $e^{-t^3}\left(t+1\right)^2\left(t-1\right)$ but there was nothing conclusive. Can someone teach me how to think this kind of problem ?
Secondly, for the same function, I have to mark the correct answer for" $G_f$ has":
A) a horisontal asymptote to $\infty$
B) asymptotes to $\infty$ and $-\infty$
C) a horisontal asymptote and an oblique asymptote
D) two oblique asymptotes
E) no asymptotes
Any hint to this problems would be greatly appreciated.
|
Apply Newton-Leibnitz theorem. Then note that max or min is achieved at $0$ for derivative . Now differentiate the derivative to get second derivative by product rule. See for which of the root of first derivative the value of second derivative is positive(because thats the condition for minima. Hope you know why)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Distribution Function Evaluated With Inequality Let's say you have a distribution function F. According to Intro to Mathematical Statistics (Hogg), the following property holds:
Pr($a < X \leq b$, $c < Y \leq d$) = $F(b,d) - F(b,c) - F(a,d) + F(a,c)$
What is the intuition behind this property, and how can you prove this formula?
Thank you!
|
It is just the Principle of Inclusion and Exclusion.
Consider the quadrants on a Venn Diagram : $~{\bbox[blue]{\Box}~\bbox[white]{\Box}\\\bbox[purple]{\Box}~\bbox[red]{\Box}}$
$$\def\P{\operatorname{\sf P}} \begin{align} \P((X,Y)\in(a;b]{\times}(c;d]) & =\P(\Box) \\[1ex] & = \P\left({\bbox[blue]{\Box}~\bbox[white]{\Box}\\\bbox[purple]{\Box}~\bbox[red]{\Box}}\right)-\P\left({\bbox[blue]{\Box}\\\bbox[purple]{\Box}}\right)-\P({\bbox[purple]{\Box}~\bbox[red]{\Box}})+\P(\bbox[purple]{\Box}) \\[1ex] & = {\P((X,Y)\in(-\infty;b]{\times}(-\infty;d]) \\-\P((X,Y)\in(-\infty;a]{\times}(-\infty;d]) \\ -\P((X,Y)\in(-\infty;b]{\times}(-\infty;c]) \\+\P((X,Y)\in(-\infty;b]{\times}(-\infty;c])} \\[1ex] & = F(b,d)-F(a,d)-F(b,c)+F(a,c)\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Asymptotic estimate of the form $k=C\log\log n$. Suppose that we have
$$
n=k^{\frac{1}{\beta}k^22^{k+12}\log\frac 1{\alpha}}
$$
where $\alpha,\beta\in(0,1)$. It is claimed that
$$
k\sim C\log\log n
$$
for some constant $C=C(\alpha,\beta)$. How do one deduce such an estimate?
I am totally lost here, any help is very much appreciated.
|
I assume logarithms in base $2$. This would not change anything anyway, besides constants.
First, note that
$$
n=k^{\frac{1}{\beta}k^22^{k+12}\log\frac{1}{\alpha}}
= 2^{\frac{1}{\beta}k^22^{k+12}\log\frac{1}{\alpha} \log k}
= 2^{\frac{2^{12}\log\frac{1}{\alpha}}{\beta}k^22 ^{k} \log k}
$$
so
$$
\log n = \frac{2^{12}\log\frac{1}{\alpha}}{\beta}\cdot k^22 ^{k} \log k
= C'\cdot 2 ^{k + 2\log k+\log\log k}
$$
setting $C'\stackrel{\rm def}{=} \frac{2^{12}\log\frac{1}{\alpha}}{\beta}$.
Taking the logarithm again, we get
$$
\log\log n = k + 2\log k+\log\log k + O(1)
$$
and, when $k\to\infty$, the right-hand side satisfies $$k + 2\log k+\log\log k + O(1)\operatorname*{\sim}_{k\to\infty} k$$
which implies
$$
\log\log n\operatorname*{\sim}_{k\to\infty} k.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Reference request regarding Mackey topology I would like to ask for some request on the relationship between Mackey topology and the weak topology. Recently, I have read quite some statements where a property in weak topology (such as compactness) implies that in Mackey topology. The basic connection I have now is that the topology is identitical if we are talking about an equicontinuous subset. However, I have a feeling that there must be something more. The reference I got is schaefer, Topological Vector Space and Bourbaki, Topological Vector Space. However, both of the books are quite brief on this matter.
Thanks in advance.
|
The basic reference is Topological Vector Space, the second is Bourbaki's classic. Other books on Topological Spaces also can be mentioned. However, I want to mention another direction which might interest you.
The Banach Lattice Theory (or more general form), makes it very easy to find Mackey topology. The most important space $L$-space and $M$-space can be identify with $L_1$ and $C(X)$, therefore, it is extremely easy get from weak to Mackey thanks to Banach–Alaoglu theorem
, the basic reference for lattice theory is enter link description here. However, it doesn't reveal the full power. For the full power, see this notorious hard-to-read book, from chatper 1 to chapter 10, as well as the appendix.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What is the number of rooted planar decreasing trees on n vertices? By planar, I mean embedded in the plane (I think sometimes the terms, ordered and plane, are used interchangeably with planar).
Assume the vertices are labeled 1,2,...,n. By decreasing, I mean the labels on the vertices along a path from the root to any other vertex will decrease. Note that this implies the root is labeled with the integer n.
The journal article I am reading states that there are (2n - 3)*(2n - 1)***(3)*(1) such trees. But the article does not give a derivation.
Can someone explain how to count these trees?
|
Let's call these ordered decreasing trees as the term planar is also
used for the cyclic group acting at the root. Now we have the
following recursive combinatorial construction. To assemble one of
these we need a root node, which receives the label $n$ and an ordered
sequence of subtrees, each of some size ranging from one to $n-1$ with
a total of $n-1$ nodes (composition into one to $n-1$ parts). We then
partition the remaining $n-1$ labels into an ordered sequence of sets,
one for each subtree, having the matching number of labels. The key
observation here is that these subtrees say of some size $q$
correspond bijectively to ordered decreasing trees on $q$ nodes where
the elements of the set of labels for these subtrees are placed
according to the ordering induced by the source tree, which is ordered
and decreasing and has labels from $1$ to $q.$ E.g. if we select the
labels $4,7,11$ for one of the subtrees then $4$ will replace $1$, $7$
will replace $2$ and $11$ will replace $3$ in the source tree that is
being attached recursively. At this point we win because this is the
canonical construction that supports all cartesian products of
exponential generating functions.
This yields for $n\ge 2$ (we have $T_1=1$) the recursive relation
$$T_n = \sum_{k=1}^{n-1} \sum_{q_1+q_2+\cdots+q_k = n-1}
{n-1\choose q_1, q_2, \ldots, q_k} \prod_{p=1}^k T_{q_p}.$$
These are standard compositions with no zero elements. We introduce
the EGF as promised and obtain for
$$T(z) = \sum_{q\ge 1} T_q \frac{z^q}{q!}$$
with $n\ge 2$ the relation
$$n! [z^n] T(z) = (n-1)! [z^{n-1}] \frac{T(z)}{1-T(z)}
\quad\text{or}\quad
n [z^n] T(z) = [z^{n-1}] \frac{T(z)}{1-T(z)}$$
which yields
$$[z^{n-1}] T'(z) = [z^{n-1}] \frac{T(z)}{1-T(z)}.$$
Multiply by $z^{n-1}$ and sum over $n\ge 2$ to get
$$\sum_{n\ge 2} z^{n-1} [z^{n-1}] T'(z) =
\sum_{n\ge 2} z^{n-1} [z^{n-1}] \frac{T(z)}{1-T(z)}.$$
Now $T'(z)$ has a constant coefficient which we miss on the left
while $T(z)/(1-T(z))$ does not and we find
$$T'(z) - 1 = \frac{T(z)}{1-T(z)}$$
so that
$$\bbox[5px,border:2px solid #00A000]{ T'(z) = \frac{1}{1-T(z)}.}$$
Solving this by separation of variables we get
$$-\frac{1}{2} (1 - T(z))^2 = z + C_1
\quad\text{or}\quad
T(z) = 1 - \sqrt{C_2-2z}.$$
Since $T(z)$ has no constant coefficient we obtain
$$\bbox[5px,border:2px solid #00A000]{
T(z) = 1 - \sqrt{1-2z}.}$$
Extracting coefficients from this we conclude with (again for $n\ge 2$)
$$n! [z^n] T(z) = - n! {1/2\choose n} (-1)^n 2^n
= - (1/2)^{\underline n} (-1)^n 2^n
\\ = (-1)^{n+1} 2^n \prod_{p=0}^{n-1} (1/2-p)
= (-1)^{n+1} \prod_{p=0}^{n-1} (1-2p)
= (-1)^{n-1} \prod_{p=1}^{n-1} (1-2p)
\\ = \prod_{p=1}^{n-1} (2p-1)$$
which is
$$\bbox[5px,border:2px solid #00A000]{
1\times 3\times\cdots\times (2n-3)}$$
and we have the claim.
Readings. This set of notes by M.
Drmota has the
elementary combinatorial argument for this as well as some additional
generating functions. We also find OEIS
A001147 which offers a considerable number
of references. There is additional material on page 531 of Flajolet /
Sedgewick (page number refers to PDF).
Here is how I approached these concepts.
with(combinat);
T :=
proc(n)
option remember;
local k, comp, res;
if n=1 then return 1 fi;
res := 0;
for k to n-1 do
for comp in composition(n-1, k) do
res := res +
(n-1)!*mul(T(q)/q!, q in comp);
od;
od;
res;
end;
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
How do you know what $\sum_{n=a}^{\infty}p^n$ converges to if you know $\sum_{n=0}^{\infty}p^n = \frac{1}{1-p}$? How do you know what $\sum_{n=a}^{\infty}p^n$ if you know $\sum_{n=0}^{\infty}p^n = \frac{1}{1-p}$ ?
Apparently $\sum_{n=a}^{\infty}p^n$ = $\frac{p^a}{1-p}$, but how can you derive this?
|
Because of the high-school identity:
$$1-p^n=(1-p)(1+p+p^2+\dots+p^{n-1}),$$
rewritten as
$$\frac1{1-p}=1+p+p^2+\dots+p^{n-1}+\frac{p^n}{1-p}, $$
which shows that
$$\Biggl\lvert\frac1{1-p}-(1+p+p^2+\dots+p^{n-1})\Biggr\rvert=\frac{\lvert p^n\rvert}{\rvert 1-p\rvert},$$
and the latter fraction tends to $0$ if $\lvert p\rvert<1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Proving $ 2n^{n-3} = \sum\limits_{i=1}^{n-1}\binom{n-2}{i-1}i^{\ i-2}(n-i)^{\ n-i-2} $ My question is how to show$$ 2n^{n-3} = \sum_{i=1}^{n-1}\binom{n-2}{i-1}i^{\ i-2}(n-i)^{\ n-i-2} $$
I got to this result through a problem of counting labeled trees, but when I try to approach actually computing the sum, I've had no breakthroughs.
|
Start by re-writing as follows:
$$2n^{n-3} = \sum_{k=0}^{n-2}
{n-2\choose k} (k+1)^{k-1} (n-1-k)^{n-3-k}$$
so that we seek to verify that
$$\bbox[5px,border:2px solid #00A000]{
Q_n = \sum_{k=0}^n {n\choose k}
(k+1)^{k-1} (n+1-k)^{n-1-k} = 2(n+2)^{n-1}.}$$
Concerning the exponential generating function for this quantity
$$Q(z) = \sum_{n\ge 0} Q_n \frac{z^n}{n!}$$
we observe that when we multiply two exponential generating functions
of the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the product of the two generating functions is the generating
function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
Therefore what we have here is a convolution of the generating
function
$$A(z) = B(z) = \sum_{n\ge 0} (n+1)^{n-1} \frac{z^n}{n!}$$
with itself.
The combinatorial class of labelled trees has the specification
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\mathcal{T} =
\mathcal{Z} \times \textsc{SET}(\mathcal{T})$$
which gives the functional equation
$$T(z) = z \exp T(z).$$
Extracting coefficients via Lagrange inversion we find
$$n! [z^n] T(z) = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z) \; dz.$$
Put $T(z)=w$ so that $z=w/\exp(w) = w\exp(-w)$ and
$dz = \exp(-w) - w\exp(-w)$
to get
$$\frac{n!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp(w(n+1))}{w^{n+1}}
\times w\times (\exp(-w) - w\exp(-w)) \; dw
\\ = \frac{n!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp(wn)}{w^n} (1 - w) \; dw.$$
But we have
$$n! [w^{n-1}] \exp(w n) =
n! \times \frac{n^{n-1}}{(n-1)!} = n^n$$
and
$$n! [w^{n-2}] \exp(w n) =
n! \times \frac{n^{n-2}}{(n-2)!} = n (n-1) n^{n-2} = (n-1) n^{n-1}$$
which means that $T(z)$ is the exponential generating function
of $$n^n - (n-1) n^{n-1} = n^{n-1}
\quad\text{i.e.}\quad
T(z) = \sum_{n\ge 1} n^{n-1} \frac{z^n}{n!}.$$
Obviously this follows by inspection from Cayley's theorem.
Repeating the same calculation to extract coefficients on $T(z)^2$ we
obtain
$$n! [z^n] T(z)^2 =
n! \times \left(\frac{n^{n-2}}{(n-2)!} - \frac{n^{n-3}}{(n-3)!}\right)
\\= n(n-1) n^{n-2} - (n-1)(n-2) n^{n-2}
= (2n - 2) n^{n-2}$$
so that
$$\sum_{n\ge 1} n^{n-2} \frac{z^n}{n!}
= T(z) - \frac{1}{2} T(z)^2.$$
Hence
$$A(z) = T'(z) - T(z) T'(z).$$
Observe that
$$z T'(z) = z \left(\exp T(z) + z \exp T(z) T'(z) \right)
= T(z) + z T(z) T'(z)$$
which implies that
$$T'(z) = \frac{1}{z} \frac{T(z)}{1-T(z)}.$$
This yields for $A(z)$ that
$$A(z) = \frac{1}{z} \frac{T(z)}{1-T(z)} (1-T(z))
= \frac{1}{z} T(z).$$
On seeing this we realize in retrospect that we could have obtained it
by inspection. Continuing we obtain the coefficient extractor
$$n! [z^n] A(z)^2 =
\frac{n!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp(w(n+3))}{w^{n+3}}
\times w^2\times (\exp(-w) - w\exp(-w)) \; dw
\\ = \frac{n!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp(w(n+2))}{w^{n+1}} (1 - w) \; dw.$$
We find
$$n! \times
\left(\frac{(n+2)^n}{n!} - \frac{(n+2)^{n-1}}{(n-1)!}\right)
= (n+2) (n+2)^{n-1} - n (n+2)^{n-1}$$
which is
$$\bbox[5px,border:2px solid #00A000]{Q_n = 2 (n+2)^{n-1}}$$
and we have the claim.
The labelled tree function recently appeared at this
MSE link.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How smooth is the generalized power mean function? Consider the function of power mean of $1$ and $e$ (so that the computation is simpler). That is, define $\displaystyle f(x)=(\frac{1+e^x}{2})^{1/x}$ when $x\neq 0$ and $f(0)=\sqrt{e}$.
It's not difficult to use L'Hopital rule with std. exponential trick to verify the function is continuous.
The question is, do we know how smooth (in which $C^k$) is this function (and a closed form formula of $f^{(n)}(0)$ if possible)?
The only way I can think of is trying to compute $\displaystyle \frac{f^{(n)}(h)-f^{(n)}(0)}{h}$ to verify if $f^{(n+1)}(0)$ exists, but the formula becomes extremely complicated and it's practically impossible to compute that limit even for $n=1$.
|
I will stick to $x\in \mathbb R.$ We can write
$$\tag 1 f(x) = \exp [(1/x)\ln ((e^x+1)/2)]$$
for $x\ne 0.$ Now $\ln ((e^x+1)/2)$ is real analytic on $\mathbb R.$ Why? Because $(e^x+1)/2$ is real analytic and positive on $\mathbb R,$ $\ln x$ is real analytic on $(0,\infty),$ and compositions of real analytic functions are real analytic. Furthermore, $\ln ((e^x+1)/2)$ equals $0$ when $x=0.$ It follows that $(1/x)\ln ((e^x+1)/2)$ is real analytic on $\mathbb R,$ once we define $f(0)$ to be the value you found, namely $e^{1/2}.$ Again using "compositions of real analytic functions are real analytic", we see from $(1)$ that $f$ is real analytic on $\mathbb R.$ (Since real analytic functions are $C^\infty,$ we have $f\in C^\infty(\mathbb R).$)
Sorry, I don't have any insight towards a nice formula for $f^{(n)}(0).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
A question on roots of a Quadratic. If roots of the equation ${ax}^2+bx+c=0$ are of the form $$\frac{\alpha}{\alpha-1},\frac{\alpha+1}{\alpha}$$
Then the value of $$({a+b+c})^2$$
I have no clue how to approach this one, any help is appreciated!
|
From the form:
$\frac{1}{x} + x = 2$ => $x^2 + x -2 = 0$, and first equation is $kx^2 + kx - 2k = 0$.
So, $(a+b+c)^2=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
The derivative of $(\log_2 n)^5$? The derivative of $ (\log_2 n)^5$? (log base 2)
Hey everyone,
I am not sure how to go about this question because I am not sure what to do with the power $5$ ( or any other power ) in the log? Should I try to convert it to a natural log ? How ? Any help ?
(Note: this is not a HW problem, I just want to learn how to solve such a problem where the log is raised to a power)
Thanks in advance
|
Note that I'm implicitly assuming that $n \in \mathbb R$. First, we observe that
$$\log_2 n=\log(n)/\log(2)$$
where $\log$ is the natural logarithm. Hence,
$$(\log_2n)^5=\left(\frac{\log n}{\log 2}\right)^5=\frac{1}{\log(2)^5} \cdot \log(n)^5.$$
To find the derivative, we use the chain rule: $g(f(x))^\prime=g^\prime(f(x))\cdot f^{\prime}(x)$.
Note that $g=(\log_2n)^5=\frac{1}{\log(2)^5} \cdot x^5$ and $f=(\log(n))$ here.
Can you find these derivatives seperately and use the chain rule to finish?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Distribution of two random variable generating schemes Goal: Generate five numbers from 0 to 1 with sum 1.
Method 1: Generate four numbers in range $(0,1)$ (by uniform distribution) to be the cuts of the interval, i.e. say the four random numbers generated is $a_1<a_2<a_3<a_4$, then the five random numbers are $a_1, a_2-a_1, a_3-a_2, a_4-a_3,1-a_4$
Method 2: Generate five numbers in range $(0, n)$ (by uniform distribution), where $n$ is arbitrary number bigger than zero. And then normalise the sum to be 1, i.e. divide all the numbers by their sum.
My question is, are the distribution of the two methods equivalent? Is any of the two correspondents to some well-known distributions?
|
Let's play this game in a simpler case. Let $X_1$ and $X_2$ be two independent and $U(0,1)$ distributed random variables.
FOR METHOD 1
Generate random variables $A_1$ and $A_2$ such that $A_1+A_2=1$ the following way:
$$A_1=\begin{cases}X_1&\text{ if }& X_1<X_2\\
X_2&\text{ if }& X_2\le X_1\end{cases} \text{ and let }\ A_2=1-A_1$$
Let's see the distribution of $A_1$. That is calculate the following probability
$$P(A_1<x)=P(A_1=X_1\cap X_1<x)+P(A_1=X_2\cap X_2<x)=$$
$$=P(X_1<X_2\cap X_1<x)+P(X_2\le X_1\cap X_2<x).$$
So,
$$P(X_1<X_2\cap X_1<x)=\int_0^xP(u<X_2)\ du=\int_0^x1-u\ du=x-\frac12x^2$$
Because of symmetry reasons
$$P(X_2\ge X_1\cap X_2<x)=x-\frac12x^2.$$
Hence, the cdf and the pdf of $A_1$ are
$$2x-x^2\ \text{ and } 2-2x$$
if $0\le x\le1$.
Which is the pdf of the first element of the order statistic of two independent uniformly distributed random variables.
If we have five such variables then the pdf of $A_1$ can be calculated as well. It will not be of uniform distribution either.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Which of the following $u=\sqrt{c+1} - \sqrt{c}$ and $v=\sqrt{c} - \sqrt{c-1}$ is greater? If
$$u=\sqrt{c+1} - \sqrt{c}$$
and
$$v=\sqrt{c} - \sqrt{c-1}$$
then, which among $u$ and $v$ is greater?
Please help without substituting values.
|
Note that $\sqrt x$ is concave function, so $$2\sqrt x \gt \sqrt {x-1} + \sqrt {x+1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Requesting suggestion on geometry and calculus books i am an engineering student , i had little interest in mathematics but my interest in mathematics inadvertently( exactly because i am loving that) rises to extreme level, although i have little knowledge (subject-syllabus oriented) on mathematics, i am eagerly trying to learn the mathematics as art ( can be apllied anywhere .....). so please suggest books on topics geometry and calculus to like beginner with interests to learn the contents as art(passion).
|
O'Neil's differential geometry is probably a great place to learn both calculus and some elementary differential geometry at the same time! It has been posted to the internet archive. It isn't so easy to find a physical copy of this book, unless you're at a large university. https://archive.org/details/ElementaryDifferentialGeometry
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
How to solve this calculus problem This question was on my quiz for Calc AB, everyone was confused on how to solve it and I did some random substitution but was still getting nowhere. My teacher said it's possible to do it but not possible with the knowledge we have currently and she's going to throw out the problem. The problem was as following:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=y-3x$$
|
Multiply by $e^{-x}$:
$$y'e^{-x}-ye^{-x}=-3xe^{-x}$$
Integrating we obtain: $$ye^{-x}=3(x+1)e^{-x}+C$$
$$y=3x+3+Ce^{-x}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
}
|
Evaluate: $\int_{0}^{\pi}\frac{\cos 2017x}{5-4\cos x}dx$ Evaluate:
$\int\limits_{0}^{\pi}\dfrac{\cos 2017x}{5-4\cos x}~dx$
I thought of using some series but could not get it
|
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\pi}{\cos\pars{2017x} \over 5 - 4\cos\pars{x}}\,\dd x =
\left.\Re\int_{0}^{\pi}{z^{2017} \over 5 - 4\pars{z+1/z}/2}
\,{\dd z \over \ic z}\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left.-\,{1 \over 2}\,\Im\int_{0}^{\pi}{z^{2017} \over z^{2} - \pars{5/2}z + 1}
\,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} =
\left.-\,{1 \over 2}\,\Im\int_{0}^{\pi}{z^{2017} \over
\pars{z - 1/2}\pars{z - 2}}\,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
{1 \over 2}\,\Im\lim_{\epsilon \to 0^{+}}
\int_{\pi}^{0}{\pars{1/2 + \epsilon\expo{\ic\theta}}^{2017} \over
\epsilon\expo{\ic\theta}\pars{1/2 + \epsilon\expo{\ic\theta} - 2}}
\epsilon\expo{\ic\theta}\ic\,\dd\theta =
{1 \over 2}\pars{-\pi}{\pars{1/2}^{2017} \over 1/2 - 2}
\\[5mm] = &\
\bbx{\ds{{2^{-2017} \over 3}\,\pi}} \approx 6.9587 \times 10^{-608}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
}
|
Solve $\lim_{x \rightarrow e} \frac{e-x}{\ln x -1}$ without using L'Hopital's rule I tried:
$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1} = \frac{e-x}{\ln(x)-\ln(e)} = \frac{e-x}{\ln(\frac{x}{e})} = ???$$
What do I do next? I think I could use $\lim \frac{\log_a x}{x} = 0, a>1$, but I'm not sure how.
|
To avoid the derivative (since this is effectively L'Hospital's rule), notice that
$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1} = -e\lim_{x \rightarrow e} \frac{\frac{x}{e} - 1}{\ln \frac{x}{e} } = -e \lim_{y \to 1} \frac{y-1}{\ln y}.$$
Using the well-known inequality $(y-1)/y \leqslant \ln y \leqslant y - 1$ it follows that
$$1 \leqslant \frac{y-1}{\ln y} \leqslant y,$$
and applying the squeeze theorem
$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1} = -e \lim_{y \to 1} \frac{y-1}{\ln y} = -e \cdot 1 = -e.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Find the value of $\arctan(1/3)$ How can I calculate $\arctan\left({1\over 3}\right)$ in terms of $\pi$ ? I know that $\tan^2(\frac{\pi}{6})= {1\over3}$ but don't know if that helps in any way.
|
The numerical computation of $\arctan\frac{1}{3}$ is pretty simple from the Maclaurin series of $\arctan$:
$$\arctan\frac{1}{3}=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)3^{2n+1}} \tag{1}$$
To get $\arctan\frac{1}{3}\approx 0.321751$ we just need to consider the partial sum up to $n=4$.
By multiplying this constant by $\frac{180}{\pi}$ we get
$$ \arctan\frac{1}{3}\approx 18^\circ 26'6'' \tag{2}$$
and the Shafer-Fink inequality provides the algebraic approximation $\arctan\frac{1}{3}\approx \frac{3}{3+2\sqrt{10}}$ with four correct figures. On the other hand it is not difficult to prove that $\arctan\frac{1}{3}\not\in\pi\mathbb{Q}$. Assuming that $\arctan\frac{1}{3}$ is a rational multiple of $\pi$ we have that $\alpha=\frac{3+i}{\sqrt{10}}$ is a root of unity, but its minimal polynomial over $\mathbb{Q}$ is $5x^4-8x^2+5$, which is not a monic polynomial, leading to a contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
}
|
What will happen to the roots of $ax^2 + bx + c = 0$ if the $a \to 0$? Exercise:
What will happen to the roots of the quadratic equation
$$ax^2 + bx + c = 0$$
if the coefficient $a$ approaches zero while the coefficients $b$ and $c$ are constant, and $b \neq 0$?
Attempt:
$\lim\limits_{a \to 0}{(ax^2 + bx + c)} = bx + c = 0 \longrightarrow x = -\frac{c}{b}$
However, I don't think my solution is complete; shouldn't I end up with $2$ roots? (I've only found $1$.)
Request:
Is there indeed another root to find? If so, how to I find it?
|
If $b=0$, then the roots are (if real), $\pm\sqrt{-c/a}$. If $c\ne0$, both roots tend to infinity (positive and negative).
If $b\ne0$, it's not restrictive to assume $b>0$ (otherwise multiply by $-1$).
If $c=0$, the roots are $0$ and $-b/a$, the latter tending to infinity ($\infty$ if $a$ approaches $0$ from the negative side, $-\infty$ if $a$ approaches $0$ from the positive side).
Assume $c\ne0$. The discriminant will be positive when $a$ belongs to a suitable (punctured) neighborhood of $0$, namely for $0<|a|<b^2/|4c|$, so we have two roots.
Now we can rationalize the expression for the roots.
First root:
$$
\frac{-b+\sqrt{b^2-4ac}}{2a}=
\frac{b^2-b^2+4ac}{2a(-b-\sqrt{b^2-4ac})}=
-\frac{2c}{\sqrt{b^2-4ac}+b}
$$
For $a\to0$ this has limit $-c/b$. This is to be expected, because the polynomial will become $bx+c$ for $a=0$.
Second root:
$$
\frac{-b-\sqrt{b^2-4ac}}{2a}=
\frac{b^2-b^2+4ac}{2a(-b+\sqrt{b^2-4ac})}=
\frac{2c}{\sqrt{b^2-4ac}-b}
$$
If $a\to0$, then this has limit $\pm\infty$ (according to the sign of $c$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
}
|
Solving the homogeneous heat equation using the method of fundamental solutions I'm trying to solve the PDE $u_t=Du_{xx}$ using the method of fundamental solutions. I've used the ansatz $u(x,t)=t^{\alpha/2}f(x/\sqrt{t})$ and this has given me the following ODE: $$\frac{\alpha}{2}f(\phi)-\frac{1}{2}\phi f'(\phi)=Df''(\phi)$$ where $\phi=x/\sqrt{t}$.
I'm then told to let $\alpha=-1/2$, leaving me with the ODE $$-\frac{1}{4}f(\phi)-\frac{1}{2}\phi f'(\phi)=Df''(\phi)$$
I'm not sure how to solve this though.
Note: $D \in \mathbb{R}_{>0}$
|
Let's start from the top. Make the ansatz
$$
u(x,t) = t^\alpha f\left(\frac{x}{2\sqrt{Dt}}\right)
$$
Then after some massaging of the differential equation and substituting $\phi = x/(2\sqrt{Dt})$, you end up with
$$
f''(\phi) + 2\phi f'(\phi) - 4\alpha f(\phi) = 0
$$
The form of the differential operator suggests the substitution $f(\phi) = e^{-\phi^2}h(\phi)$, which then gives
$$
h''(\phi) - 2\phi h'(\phi) - 2(1+2\alpha)h(\phi) = 0
$$
This is a well-known differential equation called Hermite's equation. These solutions normally blow up extremely quickly for $\phi \rightarrow\pm \infty$, but for certain special values of $\alpha$ it will have polynomial solutions. Those values are $\alpha = -n-1/2, n\in \mathbb N$, and the polynomials are denoted by $H_n(\phi)$. These will be the only solutions for which $f(\phi)$ remains finite for all $\phi$. So, substituting back into the original expression, we find the fundamental solutions are
$$
u(x,t) = A_n t^{-n-1/2}H_n\left(\frac{x}{2\sqrt{Dt}}\right)\exp\left(\frac{x^2}{4Dt}\right)
$$
for some normalization constant $A_n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
The derivative of $x^0$ For some reason I have not been able to find a straight answer to this.
We know that $\frac{d}{dx}x^n=nx^{n-1}$
And this is true for $n=-1$ and $n=1$ $\implies$ $\frac{d}{dx}x^{-1}=-1x^{-2}$ and $\frac{d}{dx}x^1=1$
We also know that $\frac{d}{dx}C=0$ where $C$ is a constant.
Suppose that $f(x)=x^0$.
Obviously any number to the power of zero is $1$, i.e. $x^0=1$, and $\frac{d}{dx}1=0$, but $x$ is not a constant. So, $$\frac{d}{dx}x^0=x^{-1}$$
Is this true? My thought is possibly. Based on the fact that if $\frac{d}{dx}x^1=1$ and obviously any value to the power of one is equal to that value. I.e. $x^1$ simplifies to be $C$ a constant but $\frac{d}{dx}x^1\not=0$, and we know that $\frac{d}{dx}C=0$. So is it true that $f'(x)=x^{-1}$? Hopefully this is not way more simple than I am making it.
UPDATE:
I obviously made an error by saying that $\frac{d}{dx}x^0=x^{-1}$ It actually would evaluate directly as $0\times x^{-1}$
|
You're incorrectly applying the power rule. We have that $\frac{d}{dx}x^n = nx^{n-1}$. For $x^0$, $n = 0$. So $\frac{d}{dx}x^0 = 0x^{-1} = 0$.
$x^1$ does not simplify to a constant. $x^1$ is just $x$, which is anything but constant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
}
|
Proof by contradiction: There are infinitely many primes I need some help with a proof. I just need to be pointed in the right direction, because I've been looking at this for ages and it's not clicking.
I need to prove that there are infinitely many prime numbers, by contradiction. The original statement is:
For all $n$ in $\mathbb{N}$ where $n > 2$, there exists a $p$ in $\mathbb{P}$[prime] such that $n < p < n!$.
We were given the hint that we're supposed to use cases to solve this. Case one is that $n!-1$ is prime, whereby obviously the statement holds.
case two is that $n!-1$ is composite, which is somehow also supposed to prove the statement, and I don't understand how. I know that every natural number $> 1$ has at least one prime factor, but I don't understand how we know that that prime factor is greater than n.
I also don't really understand how to do these cases using contradiction. Maybe I wrote the contradiction wrong, but I thought it came out to :
For all $p$ in $P$, there exists an $n$ in $\mathbb{N}$ where $n > 2 $, such that $n \le p \le n!$.
But maybe I did that wrong?
Can anyone give me a pointer on how to tie this together or where to start? I'd really appreciate it, thank you.
|
Case 1, as you already said, is trivial.
Case 2 is a little more tricky. Suppose that n!-1 is composite. Then it must be divisible by at least two primes, as you have already stated. But since n! is divisible by all numbers less than n, consider- what numbers less than n could go into n!-1?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Calculate using Euler integrals $\int \limits_0^{+\infty} \frac{1}{x^{n+1}} e^{-\frac{\alpha}{2x^2}}dx$ Find existence region and express the function in terms of Euler's integrals
$$\int \limits_0^{+\infty} \frac{1}{x^{n+1}} e^{-\frac{\alpha}{2x^2}}dx, \alpha > 0, n \in N$$
I have tried $t = \frac{1}{x}, dx = - \frac{1}{t^2}dt$
So I got
$$\int \limits_0^{+\infty}t^{n-1}e^{-\frac{\alpha t^2}{2}}dt$$
Is it correct substitution, how to make the next step?
|
Well, we have that:
$$\mathscr{I}\left(\text{a},\text{n},x\right):=\int\frac{\exp\left(-\frac{\text{a}}{2x^2}\right)}{x^{1+\text{n}}}\space\text{d}x\tag1$$
Substitute:
$$\text{u}=\frac{\text{a}^\frac{\text{n}}{2}}{2^\frac{\text{n}}{2}\cdot x^\text{n}}\tag2$$
So, we get that:
$$\mathscr{I}\left(\text{a},\text{n},x\right)=-\frac{2^\frac{\text{n}}{2}}{\text{n}\cdot\text{a}^\frac{\text{n}}{2}}\int\exp\left(-\text{u}^\frac{2}{\text{n}}\right)\space\text{d}\text{u}\tag3$$
And, for the integral we use the incomplete gamma function:
$$\int\exp\left(-\text{u}^\frac{2}{\text{n}}\right)\space\text{d}\text{u}=\text{C}-\frac{\text{n}\Gamma\left(\frac{\text{n}}{2},\text{u}^\frac{2}{\text{n}}\right)}{2}\tag4$$
So, for $(1)$ we get:
$$\mathscr{I}\left(\text{a},\text{n},x\right)=\frac{2^{\frac{\text{n}}{2}-1}}{\text{a}^\frac{\text{n}}{2}}\cdot\Gamma\left(\frac{\text{n}}{2},\frac{\text{a}}{2x^2}\right)+\text{C}\tag5$$
Now, for the boundaries:
*
*When $x=0$:
$$\mathscr{I}\left(\text{a},\text{n},0\right)=\frac{2^{\frac{\text{n}}{2}-1}}{\text{a}^\frac{\text{n}}{2}}\cdot\lim_{x\to0}\Gamma\left(\frac{\text{n}}{2},\frac{\text{a}}{2x^2}\right)\tag6$$
*When $x\to\infty$:
$$\lim_{x\to\infty}\mathscr{I}\left(\text{a},\text{n},x\right)=\frac{2^{\frac{\text{n}}{2}-1}}{\text{a}^\frac{\text{n}}{2}}\cdot\lim_{x\to\infty}\Gamma\left(\frac{\text{n}}{2},\frac{\text{a}}{2x^2}\right)\tag7$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Proof for Property of Complex Numbers Is the inequality $\lvert z_1 + z_2 \rvert \ge \lvert z_1 \rvert - \lvert z_2 \rvert$ incorrect, where $z_1$ and $z_2$ are any two complex numbers? I need an example to prove that it is. And in case it is correct, can you please give the proof? Thanks for any help.
|
This inequality is always true. It follows easily by the triangle inequality.
Start with $$\vert z_1\vert\le\bigl\vert z_1-(-z_2)\bigr\vert+\left\vert-z_2\right\vert\;.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Prove that every translation or rotation of $\Bbb E^2$ is a composite of rotations about $P$ and $Q$. Prove that every translation or rotation of $\Bbb E^2$ is a composite of rotations about $P$ and $Q$.
|
Lemma 1. For $d$ with $0\le d\le 2\operatorname{dist}(P,Q)$, there exists a translation by $d$ in a suitable direction, that can be achieved.
Proof. Rotating around $P$ by $\alpha$ and around $Q$ by $-\alpha$ produces a translation where the distance (and direction) depends continuously on $\alpha$. For $\alpha=0$, the distance is $0$, and for $\alpha=\pi$ it is $2\operatorname{dist}(P,Q)$. The claim follows from the intermediate value theorem. $\square$
Lemma 2. Every translation can be achieved.
Proof. Let $d$ be the distance of the translation $\tau$. For $n$ big enough, $\frac dn$ is between $0$ and $2\operatorname{dist}(P,Q)$. By lemma 1, there exists a translation $\tau'$ by a distance $\frac dn$ that can be achieved. Then so can the $n$-fold repetition of $\tau'$, which is a translation by $d$. If the directions of $\tau'$ and $\tau$ differ by $\alpha$, one obtains $\tau$ by rotating around $P$ by $\alpha$, then applying $\tau'$ repeatedly $n$ times, then rotating around $P$ by $-\alpha$. $\square$
Lemma 3. Every rotation can be achieved.
Proof. A rotation by $\alpha$ around $X$ is obtained by first translating $X$ to $P$ (as in lemma 2), then rotating around $P$ by $\alpha$, then translating $P$ to $X$ again. $\square$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
If the Gateaux derivative is not linear does this mean the Frechet derivative doesn't exist? It seems to be true as if the frechet exists then it is the same as the gateaux which would then be nonlinear (contradiction) but this seems useful yet I can't really find it anywhere and the lecturer hasn't mentioned it. Thanks.
|
I don't when if you are working in Euclidean spaces or normed spaces, but
anyway, Frechet differentiability at a point $x_{0}$ for a function
$f:E\rightarrow Y$, where $E\subseteq X$, means
$$
\lim_{x\rightarrow x_{0}}\frac{\Vert f(x)-f(x_{0})-L(x-x_{0})||_{Y}}{\Vert
x-x_{0}||_{X}}=0
$$
where $L:X\rightarrow Y$ is linear and continuous. Now if $x_{0}$ is an
interior point of $E$, then taking a direction $v\in X$ with norm $\Vert
v||_{X}>0$ and $x=x_{0}+tv$, you get that
\begin{align*}
0 & =\lim_{x\rightarrow x_{0}}\frac{\Vert f(x)-f(x_{0})-L(x-x_{0})||_{Y}%
}{\Vert x-x_{0}||_{X}}=\lim_{t\rightarrow0}\frac{\Vert f(x_{0}+tv)-f(x_{0}%
)-L(x_{0}+tv-x_{0})||_{Y}}{\Vert x_{0}+tv-x_{0}||_{X}}\\
& =\lim_{t\rightarrow0}\frac{\Vert f(x_{0}+tv)-f(x_{0})-tL(v)||_{Y}}{|t|\Vert
v||_{X}}=\frac{1}{\Vert v||_{X}}\lim_{t\rightarrow0}\left\Vert \frac
{f(x_{0}+tv)-f(x_{0})}{t}-L(v)\right\Vert ,
\end{align*}
which implies that there exists the Gateaux derivative in the direction $v$
with$$
\frac{\partial f}{\partial v}(x_{0})=L(v).
$$
Since $L$ is linear, it follows that $v\mapsto\frac{\partial f}{\partial
v}(x_{0})$ has to be linear.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to map a square to a triangle? I need the opposite direction, I was not quite seeing it. I then decided to try and find a transformation from square to triangle in a hope to invert it.
I am trying to find an invertible transformation between the unit square $x,y \in[0,1] \times [0,1]$ and the triangle $T$ bounded by the coordinate axes and the line $y = -\frac{b}{a}x + b$, ($\xi,\eta \in T$) .
My way of thinking:
If we fix an $x$ then run though $y$, we want the $\eta $ ordinate to end at $b$ and the $\xi$ ordinate to end at 0.
$$\xi = ax(1-y)$$
$$\eta = by$$
After a few diagrams I am confident this is correct. I am going to try to use it to simplify an integral so I would like to be sure that this transformation is good before moving on.
EDIT
Putting this in a different way.
I would like an invertible transformation between the unit square defined above and a right-angled triangle bounded by the lines $y = 0$, $x = 0$ and $$y = -\frac{b}{a}x + b$$
This triangle will have vertices $(0,0),(0,b)$ and $(a,0)$.
From the constructive comments, I might have to have the restriction that the area of the triangle and the square must be equal. It is not important that I am mapping from the unit square it just seemed the most simple to begin with.
The transformation that I found is not invertible either. If we consider the determinant of the Jacobian of the transformation we find that
$$|J| = ab(1-y)$$ which is $0$ when $y = 1$. So I have failed in that respect.
|
From this page !!!.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Physical meaning of one dimensional heat equation. Consider heat conduction in a $1$-D material. The temperature distribution $u$ after sufficiently long time can be modeled by $$-(a(x)u_x)_x = f(x), \qquad x \in [0,1]$$ where $a$ is heat conductivity, $f$ denotes internal heat sources.
Now i want to understand how this equation is modeled? Can anyone explain it plz? Also what is the difference between this equation and general heat equation?
|
You can start from the derivation of the general heat equation in one dimension. For example, you can look at the https://en.wikipedia.org/wiki/Heat_equation. You just need to account for the fact that the heat conductivity is position dependent. You should get:
$$\partial_tu-\partial_x(a(x)\partial_xu)=f(x)$$
Now all you need to do is assume that the system reaches an equilibrium when $t\rightarrow\infty$. If the system is in equilibrium, it does not change with time, so $\partial_tu=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2161072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
The natural topology on a $C^k$-manifold Lang- Differential Manifolds p.21
First Question. As it is written above, the author says it is trivial to prove that one can give $X$ a topology in a unique way such that $U_i$ is open, and the $\phi_i$ are topological embeddings, but how?
Second Question. Let $\mathscr{A},\mathscr{B}$ be $C^k$-atlases on $X$ and let $T_1,T_2$ be the natural topologies on $X$ (as in the first question) induced by $\mathscr{A},\mathscr{B}$. If these two atlases are equivalent, then is $T_1=T_2$?
Third Question. Let $\mathscr{A}$ be a $C^k$-atlas on $X$ and let $T$ be the natural topology on $X$ induced by this atlas. Let $\mathscr{T}$ be the topology generated by the domains of charts of all the $C^k$-atlases equivalent to $\mathscr{A}$. Then, is $\mathscr{T}=T$?
Thank you in advance!
|
The topology you want is defined as follows: $V \subset M$ is open iff $\varphi_i(V \cap U_i)$ is open in $\mathbf E_i$ for all $i$.
With this definition, the answers to the next two questions are clearly "yes".
My favourite textbook on this sort of stuff is Berger and Gostiaux - see 2.2.6, 2.2.7 and 2.2.8, which provide proofs of your three claims.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2161186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to find linear programming constraints given a 2 player zero sum matrix Given a problem like the following:
A startup want to build talking washing machines spending the least possible. There are three ways of building them: manually, semi-automatically and automatically. The manual production demands 1 minute of qualified work, 40 minutes of non-qualified work and three minutes of assemblage. The work times are 4, 30 and 2 minutos for the semi-automatic method and 8, 20 and 4 minutos for the fully automatic method. A startup has a pool of 4500 minutes of qualified work, 36000 minutos of non-qualified work and 2700 minutos of assembly. The costs of the production are 70, 80 and 85 euros for the manual, semi-automatic and automatic methods.
There are constraints regarding the number of machines to be produced (999) and the capacity of the factory
I understand how to set up a linear program like this:
minimize $f(x) = 70x_1 + 80x_2 + 85x_3 \ s.t.$
$x_1 + x_2 + x_3 = 999$
$x_1 + 4x_2 + 8x_3 \leq 4500$
$40x_1 + 30x_2 + 20x_3 \leq 36000$
$3x_1 + 2x_2 + 4x_3 \leq 2700$
$x \geq 0$
However, I am confused on how to set up the constraints for a linear program when all you are given is a payout matrix for a 2 player zero sum game. Given the following matrix (row player payouts listed):
\begin{bmatrix}
1 & -1\\ -1 & 1
\end{bmatrix}
I can get to this setup for the row player:
minimize $z \ s.t.$
$x_1 - x_2 = a$
$-x_1 + x_2 = b$
$x_1 + x_2 = 1$
$x_1, x_2 \geq 0$
So $a$ and $b$ are what I'm confused about finding here. I'm wondering how to find these values given this type of input in a general sense, not just for this problem.
|
Your setup is perhaps confusing because it's not clear what $a,b,z$ represent, and there's nothing relating them to each other or constraining them. As it stands, you could pick any feasible solution (e.g., $(x_1,x_2,a,b) = (1,0,1,-1)$), and then an arbitrarily low value for $z$. This shows that the problem is unbounded, and isn't doing what you want it to do.
Instead try:
$$\text{minimize } z \text{ s.t.}$$
$$\begin{array}{c}
x_1 &- &x_2 &\le &z\\
-x_1 &+ &x_2 &\le &z\\
x_1 &+ &x_2 &= &1\\
\end{array}$$
$$x_1, x_2 \geq 0$$
Now take the instances of $z$ to the left-hand side.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2161316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Factors of integers of the form $p-2^\lambda n$. Here $p$ is an odd prime, $n$ is uniform on $[0, 2^\lambda]$, and $\lambda$ is a constant. We define distribution $\mathcal{D}$ by:
$$x \xleftarrow{\$} p-2^\lambda n$$
Assume $p \approx 2^{4\lambda}$, $\lambda \in \{128, 256\}$, and $0 \leq k \leq \log_2 \lambda$. Do a non-negligible fraction of samples from $\mathcal{D}$ contain a (possibly composite) factor within $\left[2^{3\lambda+k}, 2^{3\lambda+k+2}\right]$? A sample contains an appropriate factor if any subset product of the prime factors satisfies the range requirement.
I'd like to obtain some lower bound on the density of choices of $n$ which lead to an appropriate subset of factors. I'm not sure how to attack this problem.
|
So I've been thinking this all day, read an article on wired. https://www.wired.com/2014/12/mathematicians-make-major-discovery-prime-numbers/
If you consider the prime number theorem, $ \pi(n)~ N/logN $ it says that the distribution of primes increases with the $log(N)$.
Because you have a static distribution over your domain that is not exponential, the probability of the range being coprime will have even distribution, would be my hypothesis.
And we have from wikipedia
"the probability of two randomly chosen numbers being relatively prime is $6/π^2$."
So if I were to choose a value for $E>φ(n)$, I would try $N/loglog(N)$ or $N/2loglogN$.
You may be able to look at the probable distribution and infer your conclusions from that. Is it even feasible to factor $ 2^λ$ (very large) integers? Thinking about it, if you go to the trouble of factoring each integer then there is no need for a totient.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2161395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Sequence Limit: $\sin(n^3)$ How can I prove that this sequence does not converge, using the definition?
$$W_n = \sin(n^3)$$
For $n \in \mathbb{N}$. I tried to do a proof by reduction to the absurd but without result.
|
Reference: https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/
As a special case of Corollary 6, we have
Let $P(n)=a_s n^s + \cdots + a_0$ be a polynomial with real coefficients. If $a_s$ is irrational, then $P(n)$ mod $1$ is equidistributed.
With $P(n)=\frac1{2\pi} n^3$, we have $P(n)$ mod $1$ is equidistributed. Thus, $n^3$ mod $2\pi$ is equidistributed. Then it follows that $\sin n^3$ is dense in $[-1,1]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2161491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Solving or approximating infinitely nested integral Let $f$ given by
$$f(x) = g(x) + \int_0^x\left(g(x_1) + \int_0^{x_1} \left( g(x_2) + \int_0^{x_2} \ldots d_{x_n} \ldots \right) d_{x_2} \right) d_{x_1}$$
where $n \rightarrow \infty$ and $g(x)$ is strictly decreasing in $x$.
How can such an integral be solved or approximated for?
|
If $$h_n(x):=g(x) + \int_0^x\left(g(x_1) + \int_0^{x_1} \left( g(x_2) + \int_0^{x_2} \ldots d_{x_n} \ldots \right) d_{x_2} \right) d_{x_1} $$ converges to a limit function $h$, then the self similar nature of this expression implies that
$$g(x)+\int_0^x h(t) \mathrm{d}t=h(x). $$
Differentiating gives
$$g'(x)+h(x)=h'(x), $$
and using the integrating factor $e^{-x}$ gives
$$e^{-x} g'(x)=(e^{-x} h(x))'. $$
Thus
$$e^{-t} h(t) \big|_0^x=\int_0^x e^{-t} g'(t) \mathrm{d}t $$
or
$$h(x)=e^x g(0)+e^x \int_0^xe^{-t} g'(t) \mathrm{d} t .$$
Finally, an integration by parts gives
$$h(x)=g(x)+\int_0^x g(t) e^{x-t} \mathrm{d}t. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2161682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
}
|
Prove that $(\sum_{i=1}^n i)^2$ = $\sum_{i=1}^n i^3$ by induction Prove that: $(\sum_{i=1}^n i)^2$ = $\sum_{i=1}^n i^3$
I can use the fact that $\sum_{i=1}^n i$ = n(n+1)/2 after the inductive hypothesis is invoked.
I'm not sure where to start, I would usually break down one side but there isn't usually two sums, so I'm not sure.
|
If $n = 1$, the hypothesis is true.
Assume that the hypothesis is true for $n =k$ i.e.
$(\sum_{i=1}^{k} i)^2 = \sum_{i=1}^{k} i^3$.
Now try to prove that the hypothesis is true if $n =k+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2161802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
what is the definition of local base Hi guys I am facing problem while understanding the definition of a local base.
Can anybody tell me the difference/relation between the following two definitions?
Def $1$: "Let $p$ be any arbitrary point in a topological space $X$. Then a class $B_p$ of OPEN SUBSETS of $X$ is called a local base at $p$ iff for each open set $U$ containing $p$ there exists $G_p$ in $B_p$ such that $p$ $\in$ $G_p$ $\subseteq$ $U$. The second definition which I found in other books includes NEIGHBOURHOODS OF $P$ instead of open sets.
|
If we have an "open local base", it is of course a "neighbourhood local base" as well.
This is because $N$ is a neighbourhood of $x \in N$ iff there exists an open set $O$ such that $x \in O \subseteq N$. So by definition (take the set itself every time), an open set is a neighbourhood for each of its points (and this is indeed the definition of open, if you define a topology via so-called neighbourhood systems). For given $N$, the subset of its points that $N$ is a neighbourhood of, is called $\operatorname{int}(N)$, the interior of $N$, which is an open set, often also defined as $\operatorname{int}(N) = \cup\{O: O \text{ open, } O \subseteq N\}$, which makes its openness apparent.
On the other hand, if we have a "neighbourhood local base" $\mathcal{N}$ for $x$, the set $\mathcal{B} = \{\operatorname{int}(N): N \in \mathcal{N}\}$ is an "open local base" for $x$: if we have any open set $U$ containing $x$, we have $x \in N \subseteq O$ for some $N \in \mathcal{N}$, and then also $x \in \operatorname{int}(N) \subseteq N \subseteq U$ as well, as $N$ is a neighbourhood of $x$ by assumption.
So one is a restricted version of the other (demanding the sets to be open is slightly more restrictive than mere neighbourhoods), but the more liberal version can be handy in some more compact formulations, e.g.:
a Hausdorff space $X$ is locally compact iff it has a local base of compact neighbourhoods.
A space is regular iff every point has a local base of closed neighbourhoods.
But as you saw, a "neighbourhood local base" can be made into an open one, taking interiors , so properties like being first countable are not affected by the definition (if you're first countable in one form then also in the other).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2161924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Primality test for numbers of the form $2p+1$ Is this proof correct ?
Claim
Let $p \equiv 5 \pmod 6$ be prime , then $2p+1$ is prime iff $2p+1 \mid 3^p-1$ .
Proof
Suppose $q=2p+1$ is prime. $q \equiv 11 \pmod{12}$ so $3$ is quadratic residue module $q$
and it follows that there is an integer $n$ such that $n^2 \equiv 3 \pmod{q}$ . This shows
$3^p=3^{(q-1)/2} \equiv n^{q-1} \equiv 1 \pmod{q}$ showing $2p+1$ divides $3^p-1$ .
Conversely, let $2p+1$ be factor of $3^p-1$.
Suppose that $2p+1$ is composite and let $q$ be its least prime factor.
Then $3^p \equiv 1 \pmod{q}$ and so we have $p=k \cdot \operatorname{ord_q(3)}$ for some integer $k$ .
Since $p$ is prime there are two possibilities $ \operatorname{ord_q(3)} =1 $ or $ \operatorname{ord_q(3)} =p $ .
The first possibility cannot be true because $q$ is an odd prime number so $ \operatorname{ord_q(3)} =p $ .
On the other hand $\operatorname{ord_q(3)} \mid q-1$ , hence $p$ divides $q-1$ .
This shows $q>p$ and it follows $2p+1>q^2>p^2$
which is contradiction since $p>3$ , hence $2p+1$ is prime .
Q.E.D.
|
Your proof is pretty much correct (the other answers have already mentioned how you could improve the last line), so let me give a somewhat easier proof of the second part. It only uses the Fermat-Euler theorem and some properties of the $\phi$ function.
Let $d$ be the smallest positive integer with:
$$3^d\equiv 1\pmod {2p+1}$$
If $2p+1\mid 3^p-1$, it follows that $d\mid p$ and thus $d=1$ or $d=p$, but $2p+1>2$ and hence $d=p$. This means that $p\mid\phi(2p+1)$ and since $\phi(2p+1),<2p+1$, we have $\phi(2p+1)=2p$ or $\phi(2p+1)=p$. In the second case we have $p\mid 2p+1$, which is impossible. It follows that $\phi(2p+1)=2p$ and therefore $2p+1$ is prime.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2162033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
}
|
What meaning can be given to a permutation to the power of another permutation? Say you have 2 permutations, f and g, how would one calculate $g^f$? Also can you multiply these permutations and how?
|
Usually in group theory $g^f$ is used notation for conjugate of $g$ by $f$, i.e. $g^f := f^{-1}gf$. You can calculate this product in the following manner. If $g = (a_1a_2\cdots a_n)$, then $g^f = f^{-1}gf = (f^{-1}(a_1)f^{-1}(a_2)\cdots f^{-1}(a_n)$).
On the otherside multiplication of permutations is composition of them. To find the actual product, you would need to do term by term caclulations, i.e. $g \cdot f (a) = g(f(a))$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2162181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Show that $2^a|(x-1)$ iff $2^a|(y-1)$. Suppose that $2^n + 1 =xy$, where $x$ and $y$ are integers > $1$ and $n>0$.
Show that $2^a|(x-1)$ iff $2^a|(y-1)$.
Here $a|b$ implies $a$ divides $b$.
|
WLOG let $x-1=p2^a,y-1=q2^b$ where $p,q$ are odd and $a\ge b>0$
$$1+2^n=xy=(p2^a+1)(q2^b+1)=pq2^{a+b}+p2^a+q2^b+1$$
$$q=2^{n-b}-pq2^a-p2^{a-b}$$
As $n-b>0$
$q$ will be even if $a>b$
and $q$ will remain odd if $a=b$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2162298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $Z(S_n) =$ {$e$} I've read some proofs of that statement but I would like to receive a feedback on my own proof:
Let $x \in Z(S_n)$, then it particularly exchanges with a cycle of order n.
Now let $\phi$ be a cycle of order n, then:
$x\phi = \phi x$
Multiplying each side by $x^{-1}$ gives :
$x\phi x^{-1} = \phi$
Then, if $\phi = (a_1,a_2,a_3,....,a_n)$, then $x\phi x^{-1}=(x(a_1),x(a_2),....,x(a_n))$
That gives us that $\forall i \in [1,n]: x(a_i) = a_i$, hence $x = id$.
Is it complete? Am I missing something?
Btw, sorry for my english :)
|
You are missing something. Consider for example $x=\phi$ - you can't conclude that $x=id$.
What you can actually conclude is that $\phi^m$ is a cycle in $x$ for some $m\in\mathbb{Z}$.
You can use this on any two transpositions $(a,b)\ne(b,c)$ to show that $x$ fixes $a,b,c$. Alternatively you can use those transpositions to show that $x$ fixes setwise $\{a,b\}$ and $\{b,c\}$, so fixes $b$. Notice that this assumes $n\ge 3$ - this is necessary as $Z(S_2)=S_2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2162385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Necessary and sufficient condition for the continuity of a function Let $E$ be a topological space and $A\subset E$ how to find a necessary and sufficient condition for the continuity of the function $\chi_A: E\rightarrow\mathbb{R}$ where
$$
\chi_A(x)=
\begin{cases}
1, ~x\in A\\
0,~x\not\in A
\end{cases}
$$
If i suppose that $\chi_A$ is continuous then $$\forall\varepsilon>0, \exists V\in \mathcal{V}_x, \chi_A(V)\subset ]\chi_{A}(x)-\varepsilon, \chi_A(x)+\varepsilon[$$ or
$$\forall\varepsilon>0, \chi_A^{-1}(]\chi_{A}(x)-\varepsilon, \chi_A(x)+\varepsilon[)\in \mathcal{V}_x $$
How to find a condition on $A$?
Thank you
|
Necessary and sufficient condition for $\chi_A$ to be continuous is that $A$ is both open and closed.
Necessity. If $\chi_A$ is continuous, then $\chi_A^{-1}(-\infty,1/2)=X\setminus A$, $\chi_A^{-1}(1/2, \infty)=A$ are open.
Sufficiency. Assume that $A$ and $X\setminus A$ are both open, and $U\subset\mathbb R$ open. Then
$$
\chi_A^{-1}(U)=\left\{
\begin{array}{ccc}
X &\text{if}& 0,1\in U,\\
A&\text{if}& 1\in U \,\&\, 0\not\in U\\
X\setminus A&\text{if}& 0\in U \,\&\, 1\not\in U\\
\varnothing&\text{if}& 0,1\not\in U.
\end{array}
\right.
$$
and hence
$\chi_A$ is continuous.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2162485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Construct the tangent to a point on the parabola Problem:
Given the following parabola, construct a tangent to point $P$. Justify the construction.
Solution:
Draw the line $PA$ which is perpendicular to the axis and intersects that axis at $B$. Mark off the distance $OB$ and use it to find point $C$. The line $PC$ is tangent to the parabola.
Basically my only justification that CP is tangent is that it looks really tangent. I have tried justifying my claim but can't seem to get anywhere. I have noticed that a circle can be constructed through points $PAC$. Let point $D$ be where $\circ PAC$ intersects the axis of the parabola. Then $\triangle CPD$ is an inscribed right triangle. This fact seems like it could be useful, but I'm not sure.
|
EternusVia. Here is another, but NON calculus approach: Consider point of tangency $P(t,at^2)$. Tangent PC passing through P is then of the form $y=mx+at^2-mt $ (verify!).Now intersecting the tangent with the parabola gives us the equation $ax^2=mx+at^2-mt,$ which simplifies to $ax^2-mx-at^2+mt=0.$ This equation is to have only one solution (at $P$) and so the Discriminant $D=b^2-4ac,$ of this equation must be zero. This gives $m^2-4a(-at^2+mt)=0$. Simplify gives $m^2-4amt+4a^2t^2=0$. Surprise...surprise, this equation is factorable as a perfect square: $(m-2at)^2=0$ from which $m=2at$ , etc...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2162577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
}
|
Direct Sum of Projective and Free Modules In Kevin Ventullo's answer on Mathoverflow it is mentioned that if $F$ is a free module on infinitely many generators and $P$ is a projective module, then $P \oplus F$ is again free.
However, I cannot see why this should be true.
Could you please show me why this holds?
Thank you very much for your help!
|
I wasn't aware of this result before, and I believe as you have stated it it's false, but for some specific pairs $P$, $F$ it makes sense using the fact that a projective is a summand of a free. Specifically, let $F$ be free on $\kappa$ generators, where $\kappa$ is infinite, and suppose also that $P\oplus P' = F'$, where $F'$ is free on a set of generators with cardinality $\kappa' \leq \kappa$. Then because $\kappa' \leq \kappa$ and $\kappa$ is infinite, we may write $F$ as an infinite sum of copies of $F'$:
\begin{align*}
F = \bigoplus_iF' = \bigoplus_i(P\oplus P') = (\bigoplus_iP)\oplus(\bigoplus_i P')
\end{align*}
so that
\begin{align*}
P\oplus F = P\oplus\left((\bigoplus_iP)\oplus(\bigoplus_i P')\right) = \left(P\oplus\bigoplus_iP\right)\oplus\bigoplus_i P' = (\bigoplus_iP)\oplus(\bigoplus_i P') = F.
\end{align*}
This trick doesn't work if $\kappa' > \kappa$, and I would guess that the result is actually false in that case.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2162714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How does the binomial expansion helps understanding graph connectedness? I read this post about graph connectedness and adjacency matrix. But I do not understand how the last comment about using the binomial expansion would help. I know this approach is correct, but just do not fully grasp it.
Would you mind to elaborate a little bit more? I did the binomial expansion and got $\binom{n-1}{0}A^{0}+\binom{n-1}{1}A^{1}+.....+\binom{n-1}{n-1}A^{n-1}$ which is a linear combination of all the As. but I still don't know how this helps.
|
As far as I understand $$(I+A)^{n-1} = \sum_{k=0}^{n-1} \binom{n-1}k A^k$$
Only shows how to calculate $(I+A)^{n-1}$ and there is no some extra meaning in such expansion.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2162912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
find key problem Suppose that my keys are in the kitchen with probability $1/3$ and in the living room with probability $2/3$. If I search a room that contains the keys, then I find them with probability $0.8$. Given that I have searched the kitchen once unsuccessfully, what is the probability of the keys being in the kitchen?
|
Hint: this is simply an application of Bayes' Rule .
If $K$ is "the key is in the kitchen" and $S_k$ "a search in the kitchen finds the key", then you know $\def\P{\operatorname{\mathsf P}}~\P (K)=1/3~$ and $~\P(S_k\mid K)=0.8$ (and also $~\P(S\mid K^\complement)=0~$ obviously).
Find: $~\P(K\mid S_k^\complement)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2163035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Need hints for solving 2008 A6: $f\left(x+\frac{1}{y}\right)=f\left(y+\frac{1}{x}\right)$
Let $f(x)$ be function that satisfies
$f\left(x+\frac{1}{y}\right)=f\left(y+\frac{1}{x}\right) \Big| f:\mathbb{R} \to \mathbb{N} $.
Prove that there exists a positive integer that is not in the range of the function.
I'm aware that I can easily access the solution on aops, but I wish to try and solve it on my own and would like some hints in doing so.
Some of the possibly nontrivial stuff I've managed to come up with:
*
*For any open interval of 2, there exists infinitely many $x$ (cardinality continuum) such that $f(x) = n$
For all integers $n$.
*Assuming axiom of choice, if we choose any interval of real numbers there exists an integer $n$ such that there are infinitely (cardinality continuum) many $x$ such that $f(x) = n$.
*It is possible to segregate (aka disjoint sets) real numbers such that there exists no $f(x) = f\left(x+\frac{1}{n}\right) \big| n\in \mathbb{N}$
Would appreciate if noone spoils the solution for me, thanks.
|
Here's a hint: you can show that in fact the function is constant on
$ \mathbb R \backslash \{ 0 \} $!
To show that, substitute $ \frac 1 y $ for $ y $ in the original equation to
get $ f ( x + y ) = f \left( \frac { x + y } { x y } \right) $
(for nonzero $ x $ and $ y $, of course). Now note that
you can choose different $ x $'s and $ y $'s with equal sum but different
product.
Please inform me if you think I should elaborate more.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2163304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Closed form for the sums $S(n)={(2n-1)!\over \sqrt2}\cdot{\left(4/ \pi\right)^{2n}}\cdot\sum\limits_{k=0}^{\infty}(-1)^{k(k+1)/2}(2k+1)^{-2n}$ Consider the sums
$$S(n)={(2n-1)!\over \sqrt2}\cdot{\left(4\over \pi\right)^{2n}}\cdot\sum_{k=0}^{\infty}{(-1)^{k(k+1)\over 2}\over (2k+1)^{2n}}$$
We have $S(1)=1$, $S(2)=11$, $S(3)=361$, $S(4)=24611$.
I cannot spot any pattern within this sequence.
How can we work out a closed form for $S(n)$?
|
We may notice that
$$ T(n)=\sum_{k\geq 0}\frac{(-1)^{k(k+1)/2}}{(2k+1)^{2n}} = \sum_{m\geq 1}\frac{\chi(m)}{m^{2n}} = L(\chi,2n)\tag{1}$$
is a Dirichlet $L$-function associated with the multiplicative function $\chi(m)$, that equals $0$ if $m$ is even, $1$ if $m\equiv \pm 1\pmod{8}$ and $-1$ if $m\equiv \pm 3\pmod{8}$. In particular
$$ T(n) = \prod_{p>2}\left(1-\frac{\left(\frac{2}{p}\right)}{p^{2n}}\right)^{-1}\tag{2} $$
where $\left(\frac{2}{p}\right)$ is Legendre's symbol. From Hazem Orabi's integral representation
$$ S(n) = \frac{1}{\sqrt{2}(2\pi)^{2n}}\int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\left(e^{x/8}-e^{3x/8}-e^{5x/8}+e^{7x/8}\right)\,dx \tag{3} $$
we also have:
$$ S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\int_{1}^{+\infty}\log(x)^{2n-1}\frac{x^2-1}{1+x^4}\,dx \tag{4}$$
or
$$ S(n) = \frac{1}{\sqrt{2}}\left(\frac{4}{\pi}\right)^{2n}\left.\frac{d^{2n-1}}{d\alpha^{2n-1}}\int_{1}^{+\infty}\frac{x^{2+\alpha}-x^{\alpha}}{1+x^4}\,dx\, \right|_{\alpha=0^+}\tag{5}$$
so $S(n)$ depends on $\psi^{(2n-1)}(z)$ (the $(2n-1)$-th derivative of the digamma function, i.e. the $2n$-th derivative of $\log\Gamma$) evaluated at $z=\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}$. By the reflection formulas for the $\psi^{(2n-1)}(z)$ function, everything boils down to the derivatives of the function $\cot(\pi z)$ at $z=\frac{1}{8}$ and $z=\frac{3}{8}$. This proves G.H.Hardy's claim:
$$ S(n) = (2n-1)!\cdot [z^{2n-1}]\frac{\sin(z)}{\cos(2z)}.\tag{6} $$
In particular, $S(n)$ can be expressed in terms of Euler's polynomials $E_n(z)$ evaluated at $z=\frac{1}{4}$ and $z=\frac{3}{4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2163401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
Inequality for positive real numbers less than $1$: $8(abcd+1)>(a+1)(b+1)(c+1)(d+1)$ If $a,b,c,d$ are positive real numbers, each less than 1, prove that the following inequality holds:
$$8(abcd+1)>(a+1)(b+1)(c+1)(d+1).$$
I tried using $\text{AM} > \text{GM}$, but I could not prove it.
|
First, note that $2^0(a+1) \ge a + 1$.
Then we prove by induction that $2^{k-1}(x_1\dots x_k+1) > \prod_{i=1}^k(x_i + 1)$. For the ease of notation, there I will focus on the case $k = 4$.
$$(d+1)(a+1)(b+1)(c+1)\le 4(d+1)(abc+1)< 4(abcd+abc+1+d)< 8(abcd+1)$$
The last inequality is provided by
$$abc+d< abcd+1 \Leftarrow Q+d<Qd+1 \Leftarrow (Q-1)(d-1)>0,$$
where $0 < Q = abc < 1.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2163467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to proceed further.
Please help.
|
By $AM \ge GM$ inequality,$$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge \left(4\sqrt[4]{abc}\right)\left(4\sqrt[4]{\frac{1}{abc}}\right)=16$$and equality holds when $1=a=b=c$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2163597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Find the area of a spherical triangle made by the points $(0, 0, 1)$, $(0, 1, 0)$ and $(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$.
Calculate the area of the spherical triangle defined by the points $(0, 0, 1)$, $(0, 1, 0)$ and $(\dfrac{1}{\sqrt{2}}, 0, \dfrac{1}{\sqrt{2}})$.
I have come up with this:
From the spherical Gauss-Bonnet Formula, where $T$ is a triangle with interior angles $\alpha, \beta, \gamma$. Then the area of the triangle $T$ is $\alpha + \beta + \gamma - \pi$.
How do I work out the interior angles in order to use this formula?
Any help appreciated.
|
Preliminary remark: (see figure below) It is visible that the area of spherical triangle $ABC$ is a half of the area of spherical triangle $ABD$, whose area is the eighth part of the area of the sphere with radius $r=1$ ( triangle $ABD$ is the part of the sphere situated in the positive orthant). Thus, the final result is:
$$\dfrac{1}{16}(4\pi r^2)=\dfrac{\pi}{4}.$$
As the objective is to use Gauss-Bonnet formula, more precisely Girard's theorem (http://www.princeton.edu/~rvdb/WebGL/GirardThmProof.html), instead of using rather opaque spherical trigonometry formulas, it is clearer IMHO to use (double) cross-products. Here is how:
Let $\vec{A}=(0,0,1), \vec{B}=(0,1,0), \vec{C}=(a,0,a)$ with $a=\dfrac{1}{\sqrt{2}}$.
all of them being on the unit sphere.
Cross product $\vec{C_1}=\vec{A} \times \vec{B}=(-1,0,0)$ is a normal vector to plane $OAB$.
In the same way, cross product $\vec{A_1}=\vec{B} \times \vec{C}=(a,0,-a)$ is a normal vector to plane $OBC$.
Thus, cross product $\vec{C_1} \times \vec{A_1}$ is equal to $(0,-a,0)$ with norm $\dfrac{1}{\sqrt{2}}$ is equal to $\|C_1\|\|A_1\|\sin(\pi-\beta)=1 * 1 * \sin(\beta)$.
Thus $\sin(\beta)=\dfrac{1}{\sqrt{2}}$, whence $\beta=\dfrac{\pi}{4}$.
Remark: in fact, $\beta=3\dfrac{\pi}{4}$ is forbidden because points $A,B,C$ are situated in the positive orthant. Had the elimination of ambiguity not been possible between $\dfrac{\pi}{4}$ and $3\dfrac{\pi}{4}$, we would have used the value of $\cos(\beta)$ obtained using the dot product $\vec{C_1}.\vec{A_1}=\|C_1\|\|A_1\|\cos(\beta)$.
The same process can be used to find angles $\alpha=\dfrac{\pi}{2}$ and $\gamma=\dfrac{\pi}{2}$ (see figure), and then use Girard's formula.
(this figure has been built with a Matlab program).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2163683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
equation $x^4+ax^3-6x^2+ax+1 = 0$ has two distinct positive roots Finding parameter of $a$ for which equation $x^4+ax^3-6x^2+ax+1 = 0$
has two distinct positive roots
Attempt: writing equation $\displaystyle \bigg(x^2+\frac{1}{x^2}\bigg)+a\bigg(x+\frac{1}{x}\bigg)-6=0\;,$ where $x\neq 0$
So $\displaystyle \bigg(x+\frac{1}{x}\bigg)^2+a\bigg(x+\frac{1}{x}\bigg)-8=0$
put $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t\;,$ where $|t|\geq 2$
so $t^2+at-8=0.$ So for real roots $D\geq 0.$ So $a^2+32\geq 0$
So $\displaystyle t = \frac{-a\pm \sqrt{a^2+32}}{2}.$
could some help me how to solve it, thanks
|
With $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t$ the roots occur when $t^2+at-8=0$.
$f(x)=x^4+ax^3-6x^2+ax+1 = 0$ shows $f(0)=1>0$ and $\displaystyle \lim_{x\to+\infty} f=+\infty$, which show $f$ has two distinct positive roots if there is $x_0>0$ for which $f(x_0)<0$. Let $\displaystyle \bigg(x_0+\frac{1}{x_0}\bigg)=t_0\geq2$ and $f(x_0)=t_0^2+at_0-8<0$ or
$$4+2a-8\leq t_0^2+at_0-8<0$$
Shows $a<2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2163770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Is there an equivalent to the Intermediate Value Theorem for $\mathbb{R}^2$ scalar fields? I am to prove some sort of mean value theorem for double integrals. That is,
if $f: R \subset \mathbb{R}^2 \rightarrow \mathbb{R}$ is continuous on some rectangle $R$, then there exists $c \in R$ such that $\iint_R f\, dA = f(c) \mu(R)$, where $\mu(R)$ is the area of the rectangle $R$.
My only idea so far is to prove this theorem the same way I proved it for real valued functions of real variables: using the Intermediate Value Theorem thanks to the hypothesis that $f$ is continuous. However I am unsure if such theorem exists for several variables. Is there an equivalent to it?
|
Yes, here it is! Mean value theorem for several variables and also for vector-valued functions.
https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorem_in_several_variables
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2163922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Number of subsets size N. So I have a problem where I have a set of ordered objects size N. For example,
A = {1, 2, 3, 4, 5}
How do I find how many ordered subsets there are of A that are size N? For instance, for N = 2, there are four total subsets of size N.
{1, 1}, {1, 2}, {2, 1}, {2, 2}
But when you make the subsets ordered, there are now only three subsets.
{1, 1}, {1, 2}, {2, 2}
For N = 3, there are 27 possible subsets, but 10 ordered subsets. For N = 4, there are 256 possible subsets, but only 34 unique ones.
We can see the for a set size N, there are N^N subsets of the set size N. What is the pattern for the number of ordered subsets there are of size N, where the original set is size N?
|
Viewing this as a problem of stars and bars there are $N$ distinct boxes and $N$ identical balls.
$\binom{2+2-1}{2}=\binom{3}{2}=3$
$\binom{3+3-1}{3}=\binom{5}{3}=10$
$\binom{4+4-1}{4}=\binom{7}{4}=\color{red}{35}$ (you must have missed one when you were counting by hand, remember counting by hand is usually a terrible idea)
In general there are $\binom{2N-1}{N}$ objects that you describe.
I say "objects" because what you describe are not technically sets, subsets, or ordered sets. A more correct phrase would be ordered multisets, but I would prefer to just call them "monotonic increasing sequences of length $N$ (with each entry taken from $\{1,2,\dots,N\}$)"
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2164211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Eigenvalues and corresponding eigenspace I know that $λ$ is an eigenvalue of a square matrix $A$
$\iff \exists X \ne0: AX=λX$
$\iff λ$ is a root of the characteristic polynomial of $A$.
Given a matrix $A\in \Bbb F^{n\times n}$, we find its characteristic polynomial and so its eigenvalues.
So by continuing to find corresponding eigenvectors of an eigenvalue $λ$, we discover that it corresponds only to the zero vector.
So now we should stop calling it an eigenvalue?
|
As I explained before, if it is a root of the characteristic polynomial, then there exists an eigenvector; it does not matter if you can find it or not. However, if you can prove that there is no eigenvector, then it cannot be a root of characteristic polynomial.
To elevate the confusion, let me hint at a proof.
Suppose $\lambda$ is a root of $\det(A- x I)$. This means that the matrix $(A - \lambda I)$ is not invertible (because a matrix is invertible iff its determinant is non-zero).
Now is a matrix $B$ is non-invertible, there exists a non-zero vector $X$ such that $BX = 0$. Can you prove this?
Now, since $(A - \lambda I)$ is not invertible, there exists a non-zero $X$ such that $(A-\lambda I)X = 0 \implies AX = \lambda X$ whence $X$ is an eigenvector for $\lambda$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2164342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
nCr and nPr: Order of Operations Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
Where would a counting concept like nCr and nPr fall into this mix?
|
There is no strong convention for these notations.
In most cases the $C$ and $P$ notations are self-delimiting, because the arguments are typeset as subscripts or superscripts. They can be written as ${}_nC_r$ or ${}^nC_r$ or $C^n_r$, where in each case it is unambiguous that the argument expression is whatever is written in smaller type -- or as $C(n,r)$ where the parentheses and comma make the structure explicit.
There are a few cases where ambiguity does arise, such that if you write the functions as ${}_nC_r$ and want to multiply two of them. In that case it is strongly advisable to use parentheses to disambiguate rather than try to rely on a convention: $({}_aC_b)({}_dC_e)$.
In any case, in more complex situations it is often preferred to use binomial-coefficient notation $\binom nr$ instead of ${}_nC_r$, and that is completely self-delimiting. The counts that are written ${}_nP_r$ in some elementary combinatorics texts appear to be rare enough in advanced material that it is generally feasible to write $\frac{n!}{(n-r)!}$ for them.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2164465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Implicit differentiation : two or four tangents?
My Question is: The equation of a curve is $$ x ^ 2 - 2 x y + 2 y ^ 2 = 4 $$ Find the coordinates of each point on the curve at which the tangent is parallel to the $ x $-axis.
So I established that $$ \frac { \mathop { \sf d } y } { \mathop { \sf d } x } = 0 \implies y = x $$ Therefore $ x = \pm 2 $ by subbing $ x = y $ in to the original equation. The problem I'm having is that if I sub $ x = \pm 2 $ in to the original equation I get $ 4 $ coordinates. But when I sub it in to $ \frac { \mathop { \sf d } y } { \mathop { \sf d } x } $ I get $ 2 $ coordinates?
The correct answer is $ ( 2 , 2 ) $ and $ ( 2 , - 2 ) $ which is from subbing $ x $ in to $ \frac { \mathop { \sf d } y } { \mathop { \sf d } x } = 0 $.
Why isn't subbing $ x $ in to the original equation correct?
|
Subbing $x$ into the original equation isn't wrong, it's just not enough. This is because $x^2 - 2xy + 2y^2 = 4$ does not express $y$ as a function of $x$.
With implicit functions, it's not enough to just plug the $x$-coordinates back into the original function, because that doesn't tell you which $x$- and $y$-coordinates correspond to points with the desired slope. This is why you also need to evaluate $dy/dx$ at the points you find. This is not an issue with explicit functions because each $x$-coordinate corresponds to at most one $y$-coordinate.
Another way of looking at it: For this problem, you've found that $dy/dx = 0$ implies $x = \pm 2$. This is correct. But this does not mean that $x= \pm 2$ implies $dy/dx = 0$. And again, this is not an issue with explicit functions because for explicit functions it isn't possible for one $x$-coordinate to correspond to more than one point on the graph (and therefore potentially more than one derivative or slope).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2164599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Finding The Derivative Using $\frac{d}{dx}x^n=nx^{n-1}$ So I am learning how to differentiate now,
and I came across this problem
$$f(x)=\frac{1-x}{2+x}$$
We are wanted to find $f'(x)$.
When I use $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
I find that $f'(x)=\frac{-3}{(2+x)^2}$ but, when I try to find $f'(x)$ the easy way.
i.e $\frac{d}{dx}x^=nx^{n-1}$. I cannot do it for some reason. Is it not possible to use that derivatives property when dealing with quotients? I know we can use it polynomial addition, subtraction and multiplication but I am struggling with quotients. Can someone please explain what it is I am not seeing?
My Attempt:
$$\frac{d}{dx}\frac{1-x}{2+x}=\frac{d}{dx}(1-x)(2+x)^{-1}=(1)(-1)(2+x)^{-2}$$
Which is obviously wrong so can someone please break this down for me.:)
|
You can try this
$\frac{d}{dx}f(x) = \frac{d}{dx}(-1+\frac{3}{2+x}) = 3\frac{d}{dx}(2+x)^{-1} = \frac{-3}{(x+2)^2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2164704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
Diffeq question: method of undetermined coefficients I have a question here from Differential Equations. We are learning how to solve second order nonhomogenous equations using the method of undetermined coefficients. The equation at hand is $$y''+4y=3\sin(2t)$$
I understand that the solution to the corresponding homogenous equation is $$y_c(t)=c_1\cos(2t)+c_2\sin(2t)$$
And I understand that to solve the equation using this method, I have to find an equation with "undetermined coefficients" that I then add to $y_c(t)$. But what form does this part of the equation take?
p.s. The answer in the back of the book is $y=2\cos2t-(1/8)\sin2t-(3/4)t\cos2t$. The initial conditions were $y(0)=2, y'(0)=-1$, but I don't need help with initial conditions.
|
Your problem is $(D^2+4)[y] = 3 \sin (2t)$. Operate by $D^2+4$ once again to find $(D^2+4)^2[y] = (D^2+4)(3\sin(2t)) =0$. Then,
$$ y = c_1 \cos( 2t)+c_2\sin( 2t) + c_3t\cos (2t) + c_4t \sin (2t) $$
is the general solution. Identify the first two terms as the homogeneous solution to the given differential equation. Hence form:
$$ y_p = t(A \cos (2t)+ \sin (2t)) $$
via the $c_3$ and $c_4$ terms. Next,
$$ (D^2+4)[y_p] = 3 \sin (2t)$$
will yield equations to fix $A$, $B$. Finally, apply the initial conditions to fix $c_1,c_2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2164800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Solving $\int_0^1 \frac{e^{-t} }{1+t} dt\,$ using Exponential Integral I'm having a really hard time figuring out how to solve the following integrals:
$$
\int_0^1 \frac{e^{-t} }{1+t} dt\, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\int_0^1 \frac{e^{-t} }{(1+t)^2} dt\,
$$
According to my professor, the m-th Exponential Integral is suppose to be used. I'm not that familiar with the Exponential Integral and am really unsure how to get the above integrals in the right format, particularly for the bounds (Does a transformation need to be done?), or solve it. The equation we were given for the Exponential Integral is $E_m(x)=\int_1^\infty \frac{e^{-xt} }{t^m} dt\,$.
Any insight would be incredibly helpful (especially steps!!). Thank you!
|
How about using the geometric series? That is
$$\frac{1}{1+t}=\sum_{n=0}^\infty (-t)^n$$
Then we have
$$
\int^1_0\frac{e^{-t}}{1+t}\,dt = \sum_{n=0}^\infty \int^1_0 e^{-t}(-t)^n\,dt.
$$
This isn't quite the exponential integral, since the power of $t$ is in the numerator instead of the denominator, and the bounds are $0\leq t\leq1$ instead of $1\leq t\leq\infty$. So let's invert $t$ to transform that. Let $u=\frac{1}{t}$ and we have:
$$
\int^1_0e^{-t}t^n\,dt = -\int^1_\infty \frac{e^{-1/u}}{u^n}\frac{1}{u^2}\,du = \int^\infty_1 \frac{e^{-1/u}}{u^{n+2}}\,du
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2164882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Calculate the following limit I need to evaluate the following limit, however, in doing so, I let $\sqrt{1+c}=1$ which I came to undertand that it's not valid. My procedure was the following:
$$\lim_{c\to0}\left(-\ln(c)\sqrt{1+c}-\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right)\right)$$ $$=\lim_{c\to0}\ln\left(\frac{\left(1-\sqrt{1+c}\right)\left(1+\sqrt2\right)}{c\left(1+\sqrt{1+c}\right)\left(1-\sqrt2\right)}\right)=\ln\left(\lim_{c\to0}\frac{\left(1-\sqrt{1+c}\right)\left(1+\sqrt2\right)}{c\left(1+\sqrt{1+c}\right)\left(1-\sqrt2\right)}\right)$$
$$$$We can use L'Hôspital's rule on the fraction to obtain the simplified version:
$$\ln\left(\lim_{c\to0}\left(\frac{3+2\sqrt2}{2+3c+2\sqrt{1+c}}\right)\right)=\boxed{\ln\left(\frac{3+2\sqrt2}4\right)}$$
The answer that I obtained matches with the answer of Wolfram Alpha even though there is a mistake. If I don't let $\sqrt{1+c}=1$ the limit becomes really difficult and I don't know how to solve it. Does someone have any ideas on how to solve it? And why is my answer correct even though the procedure is not?
|
You can write this as
$$-\ln(c)\sqrt{1+c}-\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right) \\
= -\ln(c)\sqrt{1+c} + \ln c- \ln c -\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right) \\ = \ln c ( 1 - \sqrt{1 + c}) + A(c) $$
where
$$A(c) = \ln\left(\frac{\left(1-\sqrt{1+c}\right)\left(1+\sqrt2\right)}{c\left(1+\sqrt{1+c}\right)\left(1-\sqrt2\right)}\right).$$
As you have shown
$$\lim_{c \to 0} A(c) = \ln\left(\frac{3+2\sqrt2}4\right)$$
You will get the correct limit for the original expression now because $\lim_{c \to 0} \ln c ( 1 - \sqrt{1 + c}) = 0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2164990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Is any multiplicative linear functional $ \Phi $ on $ C([a,b]) $ bounded? Denote by $ X $ the function space
$$
C([a,b])
= \{
u: [a,b] \to \mathbb{C} \mid u ~ \text{is continuous at each} ~ t \in [a,b]
\}
$$
over the scalar field $ \mathbb{C} $ of complex numbers, as usual, in which the norm of $ u \in X $ is given by
$$
\| u \| = \max_{t \in [a,b]} |u(t)|.
$$
Suppose that $ \Phi: X \to \mathbb{C} $ is a linear functional satisfying
$$
\forall x,y \in X: \qquad
\Phi(x y) = \Phi(x) \Phi(y).
$$
Can we conclude that $ \Phi $ is bounded?
At this stage, I guess that there exists $ t_{0} \in [a,b] $ such that $ \Phi(x) = x(t_{0}) $ for all $ x \in X $.
|
Here is a more direct proof of your original question without proving your guess, or using measure theory.
*
*$\Phi(1)=1$, where $1$ denotes the constant function with value $1$.
*For any function $x\in C[a,b]$ without zeroes, it is $\Phi(x)\neq 0$, because $\Phi(x)\Phi(x^{-1})=\Phi(xx^{-1})=\Phi(1)=1$, where $x^{-1}$ is the pointwise inverse of $x$.
*For any $x\in C[a,b]$, the function $(x-\Phi(x))$ has a zero because $\Phi(x-\Phi(x))=0$. This implies $|\Phi(x)|\le ||x||$, i.e. $||\Phi||\le 1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2165065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Prove a relation between functions from an inequality relating limits Given two functions $f$ and $g$ satisfying the inequality $$\lim_{h \to 0} \frac{f(h)}{h} > \lim_{h \to 0} \frac{g(h)}{h} \tag{1},$$
prove that there exists $h>0$ such that $f(h) > g(h)$. I was thinking that intuitively, if $f(0) = g(0)$ and $f'(0) > g'(0)$, $f(h) > g(h)$ for some $h>0$.
I have tried the following using the epsilon-delta definition of limits.
Define the following quanities
\begin{align}
A = \lim_{h \to 0} \frac{f(h)}{h} \tag{2}\\
B = \lim_{h \to 0} \frac{g(h)}{h} \tag{3}
\end{align}
where $A>B$. For any $\epsilon > 0$, there exists $\delta_1$ and $\delta_2$ such that
\begin{align}
|h| < \delta_1 \implies \left| \frac{f(h)}{h} - A\right| < \epsilon \tag{4}\\
|h| < \delta_2 \implies \left| \frac{g(h)}{h} - B\right| < \epsilon \tag{5}
\end{align}
From equation $(4)$ and $(5)$, I derived the following results for $h>0$:
\begin{align}
A &< \frac{f(h)}{h} + \epsilon \tag{6} \\
B &> \frac{g(h)}{h} - \epsilon \tag{7} \\
A > B \implies f(h) &> g(h) - 2 \epsilon h \tag{8}
\end{align}
However, I am unsure how to proceed from equation $(8)$ to get $f(h) > g(h)$.
|
Intuitively, you have two intervals,
$$\left(A - \varepsilon, A + \varepsilon\right) \text{ and } \left(B - \varepsilon,B + \varepsilon\right),$$
and if you take $|h|$ sufficiently small (in your case, if you take $|h|<\delta_1$ and $|h|<\delta_2$),
$$\frac{f(h)}{h} \in \left(A - \varepsilon, A + \varepsilon\right) \text{ and } \frac{g(h)}{h} \in\left(B - \varepsilon,B + \varepsilon\right).$$
We are done if the two intervals were disjoint; if so,
$$\frac{f(h)}{h} > A-\varepsilon \geq B+\varepsilon > \frac{g(h)}{h}.$$
But we are free to choose $\varepsilon$. In this case
$$A-\varepsilon \geq B + \varepsilon \iff \varepsilon \leq \frac{A-B}{2}$$
so just take such one such $\varepsilon>0$ (then we have the necessary $\delta_1$, $\delta_2>0$, so take any $|h| < \min\{\delta_1, \delta_2\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2165175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Series rapresentation $f(t)=\frac{1}{(2t+1)^{2}}$ How to find the series rapresentation in $t=0$ of the following expression
$$f(t)=\frac{1}{(2t+1)^{2}}$$
Can someone show me also the steps.
Thank you so much!
|
$\displaystyle \frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n\enspace$ for $\enspace-1<x<1$ , proof by multiplication with $1-x$ .
One derivation for $x$ gives $\enspace\displaystyle \frac{1}{(1-x)^2}=\sum\limits_{n=1}^\infty nx^{n-1}$ .
With $x:=-2t$ and therefore $\enspace -\frac{1}{2}< t< \frac{1}{2}$ follows:
$$\displaystyle \frac{1}{(1+2t)^2}=\sum\limits_{n=1}^\infty n(-2t)^{n-1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2165286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.