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What is the *exact* definition of a bounded subset in a metric space? I see quite a lot of different definitions of a bounded space. For instance, from nLab:
Let $E$ be a metric space. A subset $B⊆E$ is bounded if there is some real number $r$ such that $d(x,y)<r$ for all $x,y∈B$.
From Wiki:
A subset $S$ of a metric space $(M, d)$ is bounded if it is contained in a ball of finite radius, i.e. if there exists $x$ in $M$ and $r > 0$ such that for all $s$ in $S$, we have $d(x, s) < r$.
If I understand it correctly, the first definition requires that the origin of the "open disc" must be inside the subset, whereas the second definition does not have this restriction. Are these definitions somehow the same, or are they different? If not which one is correct?
The motivation for this question is because I want to understand what the Heine-Borel Theorem means.
|
Assuming the subset is nonempty, the two definitions are equivalent. Each implies the other.
|
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|
How to solve this Diff Eq (with multiple terms) $$\frac{dy}{dx}=7xy$$
I know this turns into
$$\frac{dy}{y}=7xdx$$
.....etc.
But, how do you solve the following:
$$\frac{dy}{dx}=7x+y$$
Not sure how to seperate the parts to respective sides.
|
Here is another approach (guessing, otherwise known as undetermined coefficients).
We call this part the complementary solution, we have
$$y' - y = 0$$
Do you know a function that will provide a zero when differentiated and subtracted from itself?
Choose $y_c(x) = c e^x$.
We now have what is called the particular part of $y' - y = 7x$
Lets guess at $y_p = ax +b$ and substitute back in:
$$y' - y = a - (a x + b) = 7 x \implies a = b = -7$$
The final solution is the linear combination of complementary and particular, hence:
$$y(x) = y_c(x) + y_p(x) = ce^x -7x -7$$
As a check, substitute back into ODE and see you get equality.
We could have used various methods to solve this that include integrating factor, linear substitution, Laplace Transform, exact equation, series methods, Picard Iteration...
|
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|
Finding the Limit in: $\lim\limits_{x\rightarrow1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}$ Need some help finding this limit:
$$\lim_{x\rightarrow1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}$$
Here is what I have so far:
$$\lim_{x\rightarrow1}\dfrac{\dfrac{1-\sqrt{x}}{\sqrt{x}}}{x-1}$$
$$\lim_{x\rightarrow1}\dfrac{1-\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{x-1}$$
$$\lim_{x\rightarrow1}\dfrac{1-\sqrt{x}}{x\sqrt{x}-\sqrt{x}}$$
At this point I keep getting results I don't like, I have tried multiplying by the conjugate but I keep getting denominators of $0$. What am I missing here?
Thanks
|
Another approach is to make the change of variables $y=\sqrt{x}$; because this is a continuous function, we know that the limit as $x\rightarrow 1$ is the same as the limit as $y\rightarrow 1$, so that we have $$\lim_{x\rightarrow 1}{ \frac{1-\sqrt{x}}{x\sqrt{x}-\sqrt{x}}}= \lim_{y\rightarrow 1}{\frac{1-y}{y^3-y}}=\lim_{y\rightarrow 1}{\frac{-(y-1)}{y(y-1)(y+1)}}=\lim_{y\rightarrow 1}{\frac{-1}{y(y+1)}}=-\frac{1}{2}$$
|
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Proving $\big(n!^{\frac1n}\big)_{n\in\mathbb N^*} \to \infty$ By definition, for any $a\in\mathbb R$, there exists $k\in\mathbb N^*$ such that, if $n\in\mathbb N+k$, then $n!^{\large\frac1n}>a$. Therefore, by induction, I must:
*
*find some $k$ that satisfies $k!>a^k$;
*show that, for each $n\ge k$, if $n!>a^n$, then $(n+1)!>a^{n+1}$.
For the 2nd part, it's enough that $k\ge a-1$. I noticed the smallest adequate $k$ is $4$ for $a=2$ and $7$ for $a=3$, and I'm convinced that one can always find a $k$, but I would like a function of $a$ to make an actual argument.
Definition aside, methods other than this one are welcome. I tried picking a sequence dominated by $\big(n!^{\large\frac1n}\big)$ that went to infinity as well, but couldn't find any.
|
Using Stirling's approximation, we have $n!> \sqrt{2\pi n}(n/e)^n$. Then $(n!)^{1/n}> \sqrt[n]{2\pi}n^{1/n} \frac{n}{e}$. Since $\sqrt[n]{2\pi}\rightarrow 1$ and $n^{1/n}\rightarrow 1$ but $\frac{n}{e}\rightarrow \infty$, we see that $(n!)^{1/n}\rightarrow \infty$.
|
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Prove $\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$
Prove the following integral
$$I=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+\sin^2{(\tan{x})}}=\dfrac{\pi}{2\sqrt{2}}\left(\dfrac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\right)$$
This integral result was calculated using Mathematica and
I like this integral. But I can't solve it.
My idea:
Let $$\tan{x}=t\Longrightarrow dx=\dfrac{1}{1+t^2}dt$$
so
$$I=\int\limits_{0}^{\infty}\dfrac{dt}{1+\sin^2{t}}\cdot \dfrac{1}{1+t^2}$$
then I can't proceed. Can you help me? Thank you.
|
Let $f : [-1,1] \to \mathbb{R}$ be any continuous even function on $[-1,1]$.
Consider following integral
$$I_f \stackrel{def}{=}
\int_0^{\pi/2} f(\sin(\tan x)) dx
= \frac12 \int_{-\pi/2}^{\pi/2} f(\sin(\tan x)) dx
= \frac12 \int_{-\infty}^{\infty} f(\sin y)\frac{dy}{1+y^2}
$$
where $y = \tan x$. Since $f(\cdot)$ is even, $f(\sin y)$ is periodic in $y$ with period $\pi$. We can rewrite $I_f$ as
$$
I_f = \frac12 \int_{-\pi/2}^{\pi/2} f(\sin y)g(y) dy
$$
where
$$g(y)
= \sum_{n=-\infty}^\infty \frac{1}{1+(y+n\pi)^2}
= \frac{1}{2i}\sum_{n=-\infty}^\infty \left(\frac{1}{n\pi + y - i} - \frac{1}{n\pi+y + i}\right)$$
Let $z = \tan y$ and $T = -i\tan i = \tanh 1 = \frac{e^2-1}{e^2+1}$. Recall following expansion of $\cot y$,
$$\cot y = \frac{1}{y} + \sum_{n \in \mathbb{Z} \setminus \{0\}}\left( \frac{1}{y+n\pi} - \frac{1}{n\pi}\right)$$
We find
$$g(y) = \frac{1}{2i}(\cot(y-i) - \cot(y+i))
= \frac{1}{2i} \left(\frac{1 + izT}{z - iT} - \frac{1 - izT}{z + iT}\right)
= \frac{(z^2+1)T}{z^2+T^2}
$$
As a result, we can get rid of the explicit trigonometric dependence in $I_f$:
$$
I_f = \frac12 \int_{-\infty}^\infty f\left(\frac{z}{\sqrt{1+z^2}}\right) \frac{(z^2+1)T}{z^2+T^2} \frac{dz}{1+z^2}
= \frac{T}{2}\int_{-\infty}^\infty f\left(\frac{z}{\sqrt{1+z^2}}\right) \frac{dz}{z^2+T^2}
$$
For any $a > 0$, let $b = \sqrt{a^2+1}$ and apply above formula to
$f_{a}(z) \stackrel{def}{=} \frac{1}{a^2+z^2}$, we have
$$\begin{align}
I_{f_a}
&= \frac{T}{2}\int_{-\infty}^\infty \frac{z^2+1}{a^2+b^2z^2}\frac{dz}{z^2+T^2}\\
&= \frac{T}{2(a^2-b^2T^2)}\int_{-\infty}^\infty \left(\frac{1-T^2}{z^2+T^2} - \frac{1}{a^2+b^2z^2}\right) dz\\
&= \frac{T}{2(a^2-b^2T^2)}\left((1-T^2)\frac{\pi}{T} - \frac{\pi}{ab}\right)
= \frac{\pi}{2ab}\frac{b + aT}{a + bT}
\end{align}
$$
When $a = 1$, $b$ becomes $\sqrt{2}$ and $I_{f_1}$ reduces to the integral $I$ we want to compute, i.e.
$$I = I_{f_1} = \frac{\pi}{2\sqrt{2}}\frac{\sqrt{2}+\frac{e^2-1}{e^2+1}}{1+\sqrt{2}\frac{e^2-1}{e^2+1}}
= \frac{\pi}{2\sqrt{2}}\frac{(\sqrt{2}+1)e^2 + (\sqrt{2}-1)}{(\sqrt{2}+1)e^2 - (\sqrt{2}-1)}
= \frac{\pi}{2\sqrt{2}}\frac{e^2 + (\sqrt{2}-1)^2}{e^2 - (\sqrt{2}-1)^2}
$$
|
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Describing asymptotic behaviour of a function
For question B!
x^2+x+1/x^2
= 1+ [x+1/x^2]
shouldnt the answer be
asymptote at x=0 and y=1 ??
i dont understand the textbook solution
|
You are right : there is a vertical asymptote for $x=0$ and an horizontal asymptote at $y=1$.
But the answers given in the textbook correpond to the analysis of the behavior of $y$ when $x$ goes to $0^+$ or $0^-$ as well as when $x$ goes to $+ \infty$ or to $- \infty$. $$y=\frac{x^2+x+1}{x^2}=1+\frac{1}{x}+\frac{1}{x^2}$$ When $x$ goes to infinite values, we can ignore $\frac{1}{x^2}$ which is very small compared to $\frac{1}{x}$; so, for this case, $y \simeq 1+\frac{1}{x}$. If $x$ goes to $+ \infty$, $\frac{1}{x}$ is a small positive number which is added to $1$; then $y$ goes to $1^+$. If $x$ goes to $-
\infty$, $\frac{1}{x}$ is a small negative number which is added to $1$; then $y$ goes to $1^-$
Now, when $x$ goes to zero, the system is dominated by $\frac{1}{x^2}$ which the largest; since it is a square, it is positive. So, if $x$ goes to $0$ by positive or negative values, it does not matter and $y$ goes to $+ \infty$.
Is this making things clearer to you ? If not, just post.
|
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Distribution of different objects into different boxes We want to put n different objects into n different boxes. In how many ways can we do this if we want that exactly two boxes remain empty?
|
*
*Select the $2$ of $n$ boxes you wish to remain empty. $\binom{n}{2}$ arrangements.
*Put the remaining $n$ objects, at least $1$ object into each of the $n-2$ remaining boxes and in either:
*
*1 box will have $3$ objects. Giving $\,^{n-2}C_1 \,^{n}P_{3}$ permutations.
*2 boxes will have 2 objects, giving $\,^{n-2}C_2 \,^{n}P_{2,2}$ permutations
$$\,^{n}C_{2}\cdot \left(\,^{n-2}C_1 \,^{n}P_{3}+\,^{n-2}C_2 \,^{n}P_{2,2}\right) \\ = \frac{n!}{2(n-2)!}\left(\frac{(n-2)!}{1!(n-3)!}\frac{n!}{6}+\frac{(n-2)!}{2!(n-4)!}\frac{n!}{4}\right) \\ = \frac{n!^2(4n-13)}{48(n-4)!}
$$
[edit: distinct objects]
|
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Closest distance between two quadratic curves I'm having trouble with the following problem:
Find the closest distance between $x^2+4y^2=4$ and $xy=4$.
I tried to solve using the properties of ellipse and hyperbola, but the relatively tilted axes makes it hard, i think. I also thought about using a circle that has its center on $xy=4$ and is tangent to $x^2+4y^2=4$, but the method seems to make equations too messy.
Curves generated at https://www.desmos.com/
I'd really appreciate some hints, and I'm also curious if the problem can be generalized to finding the closest distance between two quadratic curves.
Thanks in advance.
|
My try : I selected two arbitrary points $(x_e,y_e)$ and $(x_h,y_h)$ and I computed the distance which I say to be minimum.
$y_e$ can be eliminated (expressed as a function of $x_e$ since the point is along the ellipse).
$y_h$ can be eliminated (expressed as a function of $x_h$ since the point is along the hyperbole).
Now, I want that the derivatives of the distance with respect to both $x_e$ and $x_h$ be equal to zero. This leads to a terrible system of two equations for the two unknowns $x_e$ and $x_h$ but, fortunately, we only need to consider the first quadrant because of the symmetry. Moreover, complex solutions can be discarded.
I cannot get an analytical solution but using numerical methods, the only solution corresponds to $(1.62723,0.581406)$ along the ellipse and $(2.39098,1.67296)$ along the hyperbole.
Using Lagrange multipliers (then $6$ variables) took me to the same result.
The results were later confirmed building a contour plot of the distance as a function of parameters $x_e$ and $x_h$.
Added later to this answer
The problem can reduce to a single variable (say $x_e$) problem, eliminating $x_h$ as a function of $x_e$ from one of the derivative of the distance with respect to the variables. What is then left is to solve the remaining derivative for $x_e$ (the advantage of keeping $x_e$ as the single variable of the problem is that it is bounded between $0$ and $2$).
So, the problem is now reduced to the solution of a single equation for a single unknown. This works and leads to identical results. Newton methods works quite well except if the iterations start at $x_e=0$ or $x_e=2$ which correspond to infinite branches. But, for example, starting iterations at $x_e=1$, the successive iterates are $1.32085$, $1.57018$, $1.62728$, $1.62723$.
This approach could be used for finding the closest distance between two conics.
|
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Find the least value of x which when divided by 3 leaves remainder 1, ... A number when divided by 3 gives a remainder of 1; when divided by 4, gives a remainder of 2; when divided by 5, gives a remainder of 3; and when divided by 6, gives a remainder of 4. Find the smallest such number.
How to solve this question in 1 min?
Any help would be appreciated. :)
|
Hint $\ $ Apply the ubiquitous constant case optimization of CRT
$$\ x\equiv m_i\!-\!2\!\!\!\pmod{m_i}\iff x\!+\!2\equiv 0\!\!\pmod {m_i}\iff m_i\mid x\!+\!2\iff {\rm lcm}\{m_i\}\!\mid x\!+\!2$$
|
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Compute $\int_0^{\pi/2} \frac{1}{(\sinh t)^{2}+(\sin{\theta)^{2}}} d\theta$ by residues We want to compute $\int_0^{\pi/2} \frac{1}{(\sinh t)^{2}+(\sin{\theta)^{2}}} d\theta$ with $t>0$ using residues.
The first thing I want to do is using $z=e^{i\theta}$ to transform the integral to an integral over the unit circle. But I don't know how to do it.
Then I want to use the residue theorem to compute the integral. I have been searching and I found that the solution equals $\pi/\sinh(2t)$.
Please, can anyone help me to solve this integral?
|
If you substitute $z = e^{i\theta}$ directly, you get a polynomial of degree $4$ in the denominator, and finding its roots may be a bit difficult. If you first transform the $(\sin\theta)^2$ using a trigonometric identity,
$$\sin^2\theta = \frac{1}{2}(1 - \cos (2\theta)),$$
the substitution leads to a quadratic polynomial in the denominator, and that is easier to handle. So
$$\begin{align}
\int_0^{\pi/2} \frac{d\theta}{(\sinh t)^2 + (\sin\theta)^2} &= \int_0^{\pi/2} \frac{2\,d\theta}{2(\sinh t)^2 + 1 - \cos (2\theta)}\\
&= \int_0^\pi \frac{d\varphi}{1+2(\sinh t)^2 - \cos \varphi}\\
&= \int_0^\pi \frac{d\varphi}{(\cosh t)^2 + (\sinh t)^2 - \cos\varphi}\\
&= \int_0^\pi \frac{d\varphi}{\cosh (2t) - \cos\varphi}.
\end{align}$$
By parity ($\cos$ is even), we obtain
$$\int_0^{\pi/2} \frac{d\theta}{(\sinh t)^2 + (\sin \theta)^2} = \frac{1}{2}\int_{-\pi}^\pi \frac{d\varphi}{\cosh (2t) - \cos\varphi}.$$
Now we can substitute $z = e^{i\varphi}$ and obtain an integral over the unit circle,
$$\begin{align}
\int_0^{\pi/2} \frac{d\theta}{(\sinh t)^2 + (\sin \theta)^2}
&= \frac{1}{2}\int_{\lvert z\rvert = 1} \frac{1}{\cosh (2t) - \frac{1}{2}(z + z^{-1})}\frac{dz}{iz}\\
&= \frac{1}{i} \int_{\lvert z\rvert = 1} \frac{dz}{2\cosh (2t)z - z^2 - 1}\\
&= i \int_{\lvert z\rvert = 1} \frac{dz}{z^2 - 2\cosh (2t)z + 1}\\
&= i \int_{\lvert z\rvert = 1} \frac{dz}{(z-\cosh (2t))^2 - (\sinh (2t))^2}\\
&= i \int_{\lvert z\rvert = 1} \frac{dz}{(z-\cosh (2t) - \sinh (2t))(z-\cosh (2t) + \sinh (2t))}.
\end{align}$$
The zero $\cosh (2t) + \sinh (2t)$ of the denominator lies outside the unit disk, hence by Cauchy's integral formula/the residue theorem,
$$\begin{align}
\int_0^{\pi/2} \frac{d\theta}{(\sinh t)^2 + (\sin \theta)^2}
&= i\frac{2\pi i}{(\cosh (2t) - \sinh (2t)) - (\cosh (2t) + \sinh (2t))}\\
&= \frac{-2\pi}{-2\sinh (2t)}\\
&= \frac{\pi}{\sinh (2t)}.
\end{align}$$
|
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3 Questions concerning the fundamental group of $S^1\vee S^1\vee S^1$ I have the following questions:
1) What is the fundamental group of $S^1\vee S^1\vee S^1$ and why?
2) Is $S^1\vee S^1\vee S^1$ homeomorphic to the bouquet of 3 circles
3) Are $S^1\vee S^1\vee S^1$ and the bouquet of 3 circles homotopic to one another?
Thanks for the help.
|
*
*The fundamental group of $S^1\lor S^1\lor S^1$ is a free group on three generators: each generator is the homotopy class of a loop about one of the three circles. More generally, if $X$ and $Y$ are path-connected spaces with fundamental group $\pi_1(X)$ and $\pi_1(Y)$ then the fundamental group $\pi_1(X\lor Y)$ is isomorphic to the free product of $\pi_1(X)$ and $\pi_1(Y)$. This follows from the Seifert-van Kampen Theorem.
*$S^1\lor S^1\lor S^1$ is homeomorphic to the bouquet of $3$ circles -- in fact, it is usually taken as the definition of that bouquet.
|
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About $ \lim_{x\rightarrow 0}\frac {\sin x}{x} = 1$ I do not understand how $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$ As if $$ x = 0, \frac{\sin (0)}{0} = \frac {0}{0} $$
So if someone could explain this I would appreciate it!
Thanks!
|
Easy way to view: L'Hospital Rule:
$$ \lim_{x\rightarrow 0} \frac {\sin x}{x} = \lim_ {x \rightarrow 0} \frac {\cos x}{1} = \cos 0 = 1 $$
Another way: first take $ 0 <\delta <\frac {\pi}{2} $. Thus, $ \forall x \in (0, \delta) $ we have the following chain of valid inequalities:
$$ \sin x \leq x \leq \tan x $$
As $ 0 <x <\frac {\pi}{2} $, dividing this chain of inequalities for $ \sin x \neq $ 0, we obtain
$$ 1 \leq \frac {x}{\sin x} \leq \frac {1}{\cos x}. $$
Taking the inverse,
$$ \cos x \leq \frac {\sin x}{x} \leq 1 $$
Now, for the "sandwich theorem", making the limit as $ x \rightarrow 0^+ $, we conclude that
$$ \lim_ {x \rightarrow 0^+} \frac {\sin x}{x} = 1. $$
Similarly it is shown that $$ \lim_{x \rightarrow 0^-} \frac{\sin x}{x} = 1. $$
For this, take $ x \in (- \delta, 0) $, in this case $ \sin x <$ 0, etc..
|
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every subset of a measurable set is measurable Is it true that every subset of a measurable set is measurable? for any measure. So if A is a measurable set then, B as a subset of A must be measurable wrt the same measure.
|
The answer to that question depends on a choice of the consistent system of axioms in which we prefer to work.
For example, if we consider the question asking whether every subset of the real axis $R$ is Lebesgue measurable, then we will get the following different pictures:
Theorem 1 (Vitali(1903)). (ZFC) There exists a subset of the real axis $R$ which is not measurable in the Lebesgue sense.
Theorem 2 (Mycielski-Swierczkowski(1964)). (ZF+DC+AD) Every subset of the real axis $R$ is measurable in the Lebesgue sense.
|
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Probability of choosing a subset of elements where each element has a different probability I am trying to write a C++ program to do this but nobody on Stackoverflow can seem to help me so I thought I'd try to do it myself with some help from you guys.
My post on Stackoverflow can be found here:
https://stackoverflow.com/questions/22727196/calculating-a-conditional-probability-that-is-similar-to-a-binomial-sum
My exact question would be, if I had 6 people where the probabilities of each person saying yes is given respectively by $(\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7})$ respectively, what is the closed form summation form of the probability that 2 of them say yes?
I can calculate the probability of this directly by hand but I am having difficulty finding a closed form expression.
|
Let $p_i$ be the probability of person $i$ saying yes, and let $q_k$ be the probability of exactly $k$ out of $n$ persons saying yes. Then,
$$
q_k = \left[\prod_{i=1}^n(1-p_i + p_ix)\right]_k,
$$
where $[f(x)]_k$ is the coefficient of $x^k$ in the series expansion of $f(x)$.
In your question, $p_i = \frac{1}{i}$, and $k=2$. So,
$$
q_2 = \left[\prod_{i=2}^{n+1}\left(1-\frac{1}{i} + \frac{x}{i}\right)\right]_2.
$$
One can find $q_2$ by taking the second derivate of the product with respect to $x$ and setting $x=0$, which gives
\begin{align*}
q_2 &= \sum_j\frac{1}{j}\sum_{i\neq j}\frac{1}{i}\prod_{l\neq i,j}\left(1-\frac{1}{l}\right)
&=\frac{1}{n+1}\sum_{j=2}^{n+1}\sum_{i=2,i\neq j}^{n+1}\frac{1}{(i-1)(j-1)}.
\end{align*}
|
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|
cyclic subgroup elements I'm having hard time finding elements of the cyclic subgroup $\langle a\rangle$ in $S_{10}$, where $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$
This is my attempt:
\begin{align}
a^2 &= (1\ 8\ 5\ 10)(4\ 6\ 9) \\
a^3 &= (1\ 3\ 5\ 10)(4\ 7\ 9\ 6) \\
a^4 &= (1\ 5\ 10)(4\ 9\ 7) \\
a^5 &= (1\ 3\ 8\ 2\ 10)(7\ 6) \\
a^6 &= (1\ 8\ 10)(4\ 6\ 9) \\
a^7 &= (1\ 3\ 10)(4\ 7\ 9\ 6) \\
a^8 &= (1\ 10)(4\ 9\ 7) \\
a^9 &= (1\ 3\ 8\ 2\ 5\ 10)(7\ 6) \\
a^{10} &= (1\ 8\ 5)(4\ 6\ 9) \\
a^{11} &= (1\ 3\ 5\ 10)(4\ 7\ 9) \\
a^{12} &= (1\ 5)(4\ 9\ 7\ 6)
\end{align}
I suspect I already went wrong somewhere. I understand I need to get to $e = (1)$ at some point. Is there a way to check and make sure there are no mistakes when you calculate this?
|
If you have $$a = (1\,3\,8\,2\,5\,10)(4\,7\,6\,9)$$
that means that $a$ is the permutation that takes 1 to 3, 3 to 8, 8 to 2, and so on.
The permutation $a^2$ is obtained by applying $a$ twice. Since $a$ takes 1 to 3, and then 3 to 8, $a^2$ takes 1 to 8. $$\begin{array}{ccc}
a^0 & a^1 & a^2 \\ \hline
1 & 3 & 8 \\
2 & 5 & 10 \\
3 & 8 & 2 \\
4 & 7 & 6 \\
5 & 10 & 1 \\
6 & 9 & 4 \\
7 & 6 & 9 \\
8 & 2 & 5 \\
9 & 4 & 7 \\
10 & 1 & 3
\end{array}$$
Reading off the first and last column of the first row, we have that $a^2$ takes 1 to 8, so it begins $a^2 = (1\,8\ldots)\ldots$. Reading the first and last column of the 8th row, we see that $a^2$ takes 8 to 5, so $a^2 = (1\,8\,5\ldots)\ldots$. Reading off the rest of the rows similarly, we get: $$a^2 = (1\,8\,5)(2\,10\,3)(4\,6)(7\,9).$$
Perhaps you can take it from here.
|
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|
Are integers mod n a unique factorization domain? I am trying to learn abstract algebra from scratch, jolly stuff, but in the process of doing so this puzzles me:
Having a ring of integers mod $n$, where $n=pq$ is composite, as I understand we have that $\mathbb{Z}/n\mathbb{Z}$ is a Principal Ideal Domain (PID) (by this question). Therefore by the pretty chain of inclusions located here, it is also a unique factorization domain.
And this is where I am lost, as I keep thinking of for example $\mathbb{Z}/8\mathbb{Z}$ where I can have $4\equiv 2\cdot2 \equiv 2\cdot2\cdot5 \bmod 8$. Also, $p\cdot q \equiv 0 \bmod n$ which gives two non-zero divisors of zero. In my world, this means that $\mathbb{Z}/8\mathbb{Z}$ is not a UFD and not even integral domain.
I feel like I am missing something very simple yet crucial :-).
|
When $\,n\,$ is composite $\,\Bbb Z/n\,$ is not an integral domain. Factorization theory is much more complicated in non-domains, e.g. $\rm\:x = (3+2x)(2-3x)\in \Bbb Z_6[x].\:$ Basic notions such as associate and irreducible bifurcate into a few inequivalent notions, e.g. see
When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles.
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.
Factorization in Commutative Rings with Zero-divisors.
D.D. Anderson, Silvia Valdes-Leon.
Rocky Mountain J. Math. Volume 28, Number 2 (1996), 439-480
|
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|
Ergodic Rotation of the Torus Consider the measure preserving dynamical system $(\mathbb{R}^2 / \mathbb{Z}^2, \mathcal{B} \otimes \mathcal{B}, \lambda \otimes \lambda, R_{(\alpha, \beta)})$. This is the torus with the borel $\sigma$-algebra, the Lebesgue measure, and the rotation defines as
$$
R_{(\alpha, \beta)}(x,y) = (x + \alpha \ (mod1), y + \beta \ (mod1))
$$
I'm searching for a necessary and suficient condition for this system to be ergodic.
I have come by using Fourier Analysis to a sufficient condition for the system to be ergodic:
$$
\forall (k_1, k_2) \in \mathbb{Z}^2 \backslash \{(0,0)\}, \ k_1 \alpha + k_2 \beta \notin \mathbb{Z}
$$
And by a more geometric analysis to a necessary condition for the system to be ergodic:
$$
\forall \lambda > 0, \ \lambda \cdot (\alpha, \beta) \neq 0
$$
But I haven't been able to unify them into one.
Any help or good references on this topic?
Thanks in advance.
|
Note that the system is ergodic if and only if the orbit of each point is equidistributed in
$\mathbb{R}^2 / \mathbb{Z}^2$.
The first condition $$\forall (k_1, k_2) \in \mathbb{Z}^2 \backslash \{(0,0)\}, \ k_1 \alpha + k_2 \beta \notin \mathbb{Z}$$ is a necessary and sufficient condition. You can use Weyl criterion for equidistribution to show this. The Weyl criterion says that a sequence $x:\mathbb{N}\to(\mathbb{R}/\mathbb{Z})^d$ is equidistributed in $(\mathbb{R}/\mathbb{Z})^d$ if and only if $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}e(k\cdot x(n))=0 \text{ for all } k\in\mathbb{Z}^d\backslash\{0\},$$ where $e(y):=e^{2\pi iy}$ and $(k_1,\ldots,k_d)\cdot(x_1,\ldots,x_d)=k_1x_1+\ldots+k_dx_d$. It is not hard to see that the first condition is equivalent to the above formula.
For a reference, you can find the proof of Weyl criterion in Terence Tao's book "Higher order Fourier analysis".
|
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|
Inequality of Class operators H S and P First few definitions:
$A \in I(K)$ iff $A$ is isomorphic to some member of $K$
$A \in S(K)$ iff $A$ is a subalgebra of some member of $K$
$A \in H(K)$ iff $A$ is a homomorphic image of some member of $K$
$A \in P(K)$ iff $A$ is a direct product of a nonempty family of algebras in $K$
$A \in P_s(K)$ iff $A$ is a subdirect product of a nonempty family of algebras $in$ K.
I have proved that $SH \leq HS$ I am able to do this. But I need to prove that $HS \neq SH$.
|
The only homomorphic images of a field $F$ (when it is considered as a ring) are the field itself and the zero ring $\{0\}$.
If we consider the ring $(\mathbb{Q},+,\cdot,-,0,1)$ we see that $\mathsf{SH}(\mathbb{Q})$ will just contain $\{0\}$ and all the subrings of $\mathbb{Q}$. Note that every subring of $\mathbb{Q}$ will have characteristic 0.
If we look at $\mathsf{HS}(\mathbb{Q})$, we can see that $\mathbb{Z}$ is a subring of $\mathbb{Q}$ and $\mathbb{Z}/p\mathbb{Z}$ is a homomorphic image of $\mathbb{Z}$. But $\mathbb{Z}/p\mathbb{Z}$ has characteristic $p$, so it is not in $\mathsf{SH}(\mathbb{Q})$. Thus $\mathsf{SH}(\mathbb{Q})\neq \mathsf{HS}(\mathbb{Q})$.
|
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|
Generalization of $n$ mod $2 = \dfrac{1-(-1)^n}{2}$ I had this idea that
$n$ mod $2 = \dfrac{1-(-1)^n}{2}$ for $n \in \mathbb{N}$.
Are there any generalizations for this? For example for $n$ mod $3$ etc.?
I would prefer some answers containing basic arithmetic operations, although the use of analytic functions is also accepted.
|
To give an alternate approach to Hurkyl's answer. If you let $\zeta$ be a primitive $n$-th root of unity, the function
$$f_n(x)=\sum_{i=0}^{n-1} i\frac{(\zeta^x-\zeta^0)...(\zeta^x-\zeta^{i-1})(\zeta^x-\zeta^{i+1})...(\zeta^x-\zeta^{n-1})}{(\zeta^i-\zeta^0)...(\zeta^i-\zeta^{i+1})(\zeta^i-\zeta^{i+1})...(\zeta^i-\zeta^{n-1})}.$$
Gives you $f_n(k) = (k\mod n)$, for all integers $k$.
As an example, if we let $n = 2$ we get
$$f_2(x)=0\frac{(-1)^x-(-1)^1}{(-1)^{0}-(-1)^1}+1\frac{(-1)^x-(-1)^0}{(-1)^1-(-1)^0}=0+\frac{1-(-1)^x}{2}$$
which is identical to the formula you have.
Essentially what we did here is found a polynomial $p_n$, using the Lagrange Interpolation Formula, where $p_n(\zeta^k)=k$ for all $0 \leq k < n$. Then we let $f_n(x)=p_n(\zeta^x)$.
|
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|
Dividing 100% by 3 without any left In mathematics, as far as I know, you can't divide 100% by 3 without having 0,1...% left.
Imagine an apple which was cloned two times, so the other 2 are completely equal in 'quality'. The totality of the 3 apples is 100%. Now, you can divide those 3 apples for 3 persons and you will get 100% divided by 3 and none left.
Is this because 1: mathematics is not real 2: there is no 1 or 2, and it's in fact just an invention for measurements? So, in the division of 100% by 3 WITHOUT any left, is NOT accurate?
|
Here, in a practical way – I understand all your concepts, sirs – I was using a simple calculator on my computer, and looks like mine's got a 10 decimal digit precision. Which make this possible:
9 decimal "3s"
33.333333333 * 3
= 99.999999999
10 decimal "3s"
33.3333333333 * 3
= 100
If I divide 100 by 3, it'll give me 10 decimal "3s". And this expression was valid:
100/3 * 3 = 100
|
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|
the continuous functions with norm I'm having trouble trying to understand what does means the first expression in particular
the last term in it
should we add $\|f\|_{\infty} \leq \infty$ or what i can't see what is his role ($\|f\|_{\infty}$) here
\begin{align*}
{C}(\mathbb{R}^{n})&=\{f:\mathbb{R}^{n}\longmapsto \mathbb{R}: \text{continue},\lim_{\|{x}\|\to \infty }f(x)=0,\|{f}\|_{\infty}\}\\
{C}_{0}(\mathbb{R}^{n})&=\{ f \in {C}: \text{of compact support } \mathbb{R}^{n} \} \\
Lip_{0}(\mathbb{R}^{n})&=\{f \in C_{0}: \exists M: |f(x)-f(y)|\leq M|x-y| \} \\
\overline{Lip_{0}}(\mathbb{R}^{n})&={C}(\mathbb{R}^{n}) \\
\mathcal{D}(\Omega)=\mathcal{C}_{c}^{\infty}(\mathbb{R})&=\{f \in \mathcal{C}^{\infty} : \text{ of compact support } \}
\end{align*}
would you please also explain how $\lim_{\|{x}\|\to \infty }f(x)=0,$ implies $\|{f}\|_{\infty} \leq \infty$
|
Well, to answer your question:
let $$f:\Bbb R^n\to \Bbb R, \quad \text{continuous},\quad \lim_{|x|\to\infty}f(x)=0.$$
Fix $\varepsilon >0$; then by definition of the limit there exists $R>0$ such that $|f(x)|<\varepsilon$ whenever $|x|>R$. On the compact set $\{|x|\le R\}$ the function $|f|$ is continuous, hence attains its supremum and infimum (Veierstrass theorem). Therefore, $$\|f\|_\infty\le \max(\varepsilon, \sup_{|x|\le R}|f(x)|)<\infty.$$
As for notations, it's a bit weird to see $C(\Bbb R^n)$ defined as a space of continuous functions vanishing at infinity. In most cases, this notation is reserved for the space of continuous functions. Then you can define $C_b$ as the space of continuous bounded functions, $C_0$ as a space of continuous vanishing functions, $C_c$ as a space of continuous function with compact support, etc.
|
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|
Taylor series of $\sqrt{1+x}$ using sigma notation I want help in writing Taylor series of $\sqrt{1+x}$ using sigma notation
I got till
$1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\ldots$ and so on.
But I don't know what will come in sigma notation.
|
The generalized binomial theorem says that
$$
(1+x)^{1/2}=\sum_{k=0}^\infty\binom{1/2}{k}x^k
$$
where $\binom{1/2}{0}=1$ and for $k\ge1$,
$$
\begin{align}
\binom{1/2}{k}
&=\frac{\frac12(\frac12-1)(\frac12-2)\cdots(\frac12-k+1)}{k!}\\
&=\frac{(-1)^{k-1}}{2^kk!}1\cdot3\cdot5\cdots(2k-3)\\
&=\frac{(-1)^{k-1}}{2^kk!}\frac{(2k-2)!}{2^{k-1}(k-1)!}\\
&=\frac{(-1)^{k-1}}{k2^{2k-1}}\binom{2k-2}{k-1}
\end{align}
$$
Thus,
$$
\begin{align}
(1+x)^{1/2}
&=1-\sum_{k=1}^\infty\frac2k\binom{2k-2}{k-1}\left(-\frac x4\right)^k\\
&=1-\sum_{k=0}^\infty\frac2{k+1}\binom{2k}{k}\left(-\frac x4\right)^{k+1}
\end{align}
$$
|
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|
Show that $\,\,\, \lim_{n\to\infty} \sin\bigl(\pi\sqrt{n^2+1}\bigr)=0 $ Can anyone help me to solve this problem?
Show that
$$
\lim_{n\to\infty} \sin\bigl(\pi\sqrt{n^2+1}\bigr)=0
$$
|
We have by the Taylor series:
$$\sqrt{n^2+1}=n\sqrt{1+\frac1{n^2}}=n\left(1+O\left(\frac1{n^2}\right)\right)=n+O\left(\frac1{n}\right)$$
hence
$$\sin\left(\pi\sqrt{n^2+1}\right)=\sin\left(n\pi+O\left(\frac1{n}\right)\right)=(-1)^n\sin\left(O\left(\frac1{n}\right)\right)\xrightarrow{n\to\infty}\ 0$$
|
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|
Prove $x^3-3x+4$ is irreducible in $\mathbb{Q}[x]$ I want to prove $x^3-3x+4$ is irreducible in $\mathbb{Q}[x]$. Eisenstein's criterion doesn't apply here, so I think the simplest method would be to use the Rational Roots Test, right?
If I can use the rational roots test here, then since it is monic I simply check factors of the constant term:
\begin{align*}
(1)^3-3(1)+4&=2 \\
(2)^3-3(2)+4&=6 \\
(4)^3-3(4)+4&=56 \\
(-1)^3-3(-1)+4&=6 \\
(-2)^3-3(-2)+4&=2 \\
(-4)^3-3(-4)+4&=-48
\end{align*}
Therefore if $x^3-3x+4$ is reducible, it would have a degree 1 monomial factor $(x-a)$ for one of the $a=1,2,4,-1,-2,-4$ I tested above. However since none of these $a$ are roots, so then it does not have a degree 1 factor and is therefore irreducible.
Unfortunately I can't find any reference to the Rational Roots Theorem in Artin's Algebra text (2nd edition), not even in the index (surprising!). So I'm referring to Wikipedia, and it does not specifically say -- does this test hold in $\mathbb{Q}[x]$? Hopefully it is applicable here and I have used it correctly...
Thanks!
|
Try showing that it has no roots in $\mathbb{Q}$, that is a good criteria for irreducibility.
Another trick, is that if it is irreducible, it is irreducible for all $x\in \mathbb{R}$ so let $x=\frac{1}{y}$ then try eisenstein on the new polynomial.
|
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|
Compute this factor group: $\mathbb Z_4\times\mathbb Z_6/\langle (0,2) \rangle$
So I'm going through example 15.10 in Fraleigh, which is computing $G/H$, where $G = \mathbb Z_4\times\mathbb Z_6$ and $H = \langle (0,2) \rangle$.
We have $H =\{(0,2), (0,4), (0,0)\}$, so the subgroup generated by $H$ has order $3$.
Since $G$ has $24$ elements and each coset has the same number of elements as $H$, there are $24 / 3 = 8$ cosets in $G/H$.
Fraleigh says,
The first factor $\mathbb Z_4$ of $G$ is left alone. The $\mathbb Z_6$ factor is essentially collapsed by a subgroup of order $3$, giving a factor group in the second factor of order $2$ that must be isomorphic to $\mathbb Z_2$. So $G/H$ is isomorphic to $\mathbb Z_4\times\mathbb Z_2$.
The bolded is what confuses me. Here are the elements of our factor group (the cosets of $H$ in $G$):
$(0,0) + H = H$
$(0,1) + H = \{(0,1), (0,3), (0,5)\}$
$(1,0) + H = \{(1,0), (1,2), (1,4)\}$
$(1,1) + H = \{(1,1), (1,3), (1,5)\}$
The process continues and we have $4$ more cosets.
I don't see how Fraleigh computes this factor group so quickly / without writing out the cosets. Even with writing out the cosets, I'm not sure why it's clear that $G / H$ is isomorphic to $\mathbb Z_4\times\mathbb Z_2$.
Any help much appreciated,
Mariogs
|
Note: direct product of quotients is isomorphic to quotient of direct products.
More clearly: $(G_1\times G_2)/(H_1\times H_2) \cong G_1/H_1\times G_2/H_2$ where $H_i\leq G_i$.
From that point; $$\mathbb Z_4\times \mathbb Z_6/\langle(0,2)\rangle\cong \mathbb Z_4/\langle(0)\rangle\times \mathbb Z_6/\langle(2)\rangle\cong \mathbb Z_4 \times \mathbb Z_2.$$
|
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|
Irreducible Variety of Irreducible Polynomial Prove that if $f \in k[x_1,...x_n]$ is an irreducible polynomial, then the variety $V(f) \subseteq A^n_k$ is an irreducible variety.
Basically, I think that I want to prove that the ideal which corresponds to the variety is prime (since there is a bijective correspondence between the two if we are algebraically closed and ideal is radical (which I am assuming, because otherwise, I do not think this will be true)).
However, I do not think that a principal ideal generated by an irreducible polynomial is necessarily prime. Is that true? And I am I taking this in the right direction?
|
Since $k[x_1, \ldots, x_n]$ is a UFD, irreducible elements are prime. Thus any principal ideal generated by an irreducible polynomial is in fact a prime ideal, and so the corresponding variety will be irreducible.
|
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Ratio and root test
Hi! I am working on some ratio and root test online homework problems for my calc2 class and I am not sure how to completely solve this problem. I guessed on the second part that it converges, but Im not sure how to solve of the value that it converges to. If someone could possibly help me with this problem it would be greatly appreciated.
|
You aren't expected to figure out what the value of the series is (although in time you might figure out it has something to do with $e+e^{-1}$). Did you actually compute $\rho$? What is $$\lim_{n\to\infty} \frac{(2n)!}{(2n+2)!}?$$
|
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|
Probability of rolling a 1 before a 6 on a dice what is the probability that I roll a "1" on a dice before rolling a "6". I do not know how to tackle this problem. I was thinking that this is a Geometric random variable but I do not know how to solve it.
Any help is appreciated.
|
By symmetry, the probability is $\frac{1}{2}$.
Remark: We could also do it the long way. We win if we roll a $1$ before we roll a $6$. This can happen in several ways.
*
*We roll a $1$ immediately. The probability that happens is $\frac{1}{6}$.
*We roll something that is neither a $1$ nor a $6$, and then roll a $1$. The probability that happens is $\frac{4}{6}\cdot\frac{1}{6}$.
*We roll something that is neither a $1$ nor a $6$ twice, and then a $1$. The probability that happens is $\left(\frac{4}{6}\right)^2\cdot\frac{1}{6}$.
Continue.
Finally, sum the infinite geometric series.
|
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|
Problem with solving a complicated Integral I need to determine the $ \int \frac{\sin^3(x)}{8-\cos^3(x)} dx$. It's an indefinite integral.
|
Hint: $\sin^3(x) = \sin(x)(\sin^2(x)) = \sin(x)(1-\cos^2(x))$
Take $u = \cos(x) \Rightarrow du = -\sin(x) \ dx$
So we have, $- \int \frac{1-u^2}{8-u^3} du = -\int \frac{1}{8-u^3} du - \int \frac{u^2}{8-u^3} du $
|
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Prove for each $a ∈ G, aHa^{-1}$ is a subgroup of G. Question:
Let $H$ be a subgroup of $G$. For any $a \in G$, let $aHa^{-1} = \{axa^{-1} : x \in H\}$; $aHa^{-1}$ is called a conjugate of $H$. Prove: For each $a \in G$, $aHa^{-1}$ is a subgroup of $G$.
I know in order to prove something is a subgroup it needs to be nonempty, closed under operation, contains it's identity, and inverse.
I do believe to prove H contains it's inverse it goes as follows:
$aea^{-1} = aa^{-1}e = e$ for any $a \in G$. Therefore $e \in H$.
but after that I am at a loss.
|
One slick way to prove a subset $K$ of a group $G$ is a subgroup is to show:
1) The set $K$ is nonempty.
2) For every $x,y \in K, xy^{-1} \in K$.
(See if you can prove this criterion to be a subgroup!)
You've shown 1), since the identity $e \in K = aHa^{-1}$. Now let $x,y \in aHa^{-1}$. Then $x = aha^{-1}$ and $y = ah'a^{-1}$ for some $h,h' \in H$. But $y^{-1} = ah'^{-1}a^{-1}$ (check if you aren't familiar with this formula). Hence $xy^{-1} = aha^{-1}ah'^{-1}a^{-1} = ahh'^{-1}a^{-1}$. Since $H$ is a subgroup of $G$, $hh'^{-1} \in H$, so $xy^{-1} \in aHa^{-1}$, which completes the proof.
|
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|
How to find the order of the elements of $A_4$? Just wondering how to find the order of each element in this group:
$A_4 = \{e,(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}$
I tried writing each elements not in disjoint cycle but it didn't look right to me. I got 3 for all the cycles with 3, and 4 for the last cycles
|
Notice for any 3-cycle $(abc)$, $(abc)^2=(abc)(abc)=(acb)$, and $(abc)^3=(abc)^2(abc)=(acb)(abc)=e$, the identity. Thus the order of any 3-cycle is 3.
Noting that disjoint cycles commute, it is easy to see that $((ab)(cd))^2=(ab)(ab)(cd)(cd)=e*e=e$, so the order of any product of two disjoint transpositions is 2.
|
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|
proof using a recursive definition I am doing a 2-part question. Thus far, I have finished the first part, requiring me to make a recursive definition of a set "S" of all binary strings, starting with a 1. I have:
Base: 1
Recursion: S1 or S0
Restriction: nothing else in the set
Now, the next question is to show, using my recursive definition, that any element in "S" encodes a number > 0.
I know how to write this in words. I would say "Since the base case, 1, is greater than 0, then concatenating 1's and 0's to the end will only make the value larger. Thus, the encoded value will never be less than 1. However, how do I state this using my recursive definition so that's it's valid?
Thanks!
|
Take your recursive strings definition
and append a "value" function for
each part of the definition:
Base: 1 - (val(1) = 1)
Recursion: S1 (val(S1) = 2val(S)+1) or S0 (val(S0) = 2 val(S))
This usually works for any recursive definition
that you want to assign additional properties or interpretions to.
|
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|
I'm looking for the name of a transform that does the following (example images included) I'm in the usual situation that if I would know what the name of the thing was, then I could find the answer. Since I dont know the name, here is what I'm looking for:
Suppose I have the following "snake" of 10 quadrilaterals:
I now want to apply a transformation to each of these quadrilaterals such that the "snake" is straightened. To that end I calculated a perspective transformation according to what I read here. This also seems to work "somehow" as I end up with the following result:
As one can see, the line that was in the center of the "snake" is not in the center of the straightened object anymore.
What I need is a transformation which linearly scales all the content of the "warped" quadrilaterals into straight rectangles. If something was at the center of an edge in the snake, then it should also be at the center of the transformed edge. Apparently a perspective transform does not do this.
My question is: which transformation am I looking for that fulfills this property?
|
From the tag "linear algebra" I guess you are looking for a linear transformation.
Then it is impossible. Because, the inverse, which is again a linear transformation can only map a rectangle to a parallelogram and not to a 'snake'.
|
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|
How to show that this real function is not periodic?
How can one prove that $$\cos\left(\frac{\pi}{2} t \right)+\cos\left(t \right)$$ is not periodic?
This question is motivated by the harmonic spectral representation of time series. Indeed, it is easy to show that a path of a time series given by $$ \cos(\lambda_1 t) + \cos(\lambda_2 t)$$ is periodic if $\frac{\lambda_1}{\lambda_2} \in \mathbb{Q}$. In the above example this is not the case since $\frac{\lambda_1}{\lambda_2} = \pi/2.$
So the case above could serve as an example for the statement: if $\frac{\lambda_1}{\lambda_2} \not\in \mathbb{Q}$ then the path of a harmonic time series is in general not periodic.
|
The function
$$f(t)=\cos\left(\frac{\pi}{2} t \right)+\cos\left(t \right)$$
is even; so $f(t+T)=f(t) \Leftrightarrow f(t-T)=f(t),~~ \forall t$.
Adding the equations
$$f(t+T)=f(t), $$
$$f(t-T)=f(t), $$
we arrive at
$$\cos t \cos T + \cos\left(\frac{\pi}{2} t \right)\cos\left(\frac{\pi}{2} T \right)=
\cos t+ \cos\left(\frac{\pi}{2} t \right),~~ \forall t. $$
Then, for $t=0$ we have
$$ \cos T + \cos\left(\frac{\pi}{2} T \right)= 2 $$
which implies $T=2m\pi$ and $T=4n$, for $n,m\in\mathbb Z$. In summary, there exists no solution to the periodicity problem ($T=0$ is no period in our conventions).
|
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|
Showing $2^{n_2} + 3^{n_3}+\cdots+9^{n_9}$ is dense in $\mathbb{R}^+$ I encountered this problem via a friend. He asked me to prove that
$$ \left\{u: u= \sum_{k=2}^9 k^{n_k} \quad n_k \in \mathbb{Z} \right\}$$
is dense in $\mathbb{R}^+$.
I was able to show that $0$ is approachable(can get arbitrarily close to) by numbers of this type. However to proceed ahead, I need to know what operations preserve the structure of these numbers. For instance, to show $m+n\sqrt{2}$ is dense for all integers $m,n$, I used the fact that product of any two numbers in this form yields a number in the same form. My problem is I don't know what pattern to exploit in the given problem.
I appreciate any hints / patterns you can provide. One idea I have been working on is as follows:
If we could partition $2^{n_2}, 3^{n_3},\ldots, 9^{n_9}$ into two parts, one that would determine the integer part, the other that would approximate the fractional part, we would be done.
|
I don't think this can be done. Think of it like this: Suppose you have $u \in \Bbb{R}_+$ and $u_n \to u$ with
$$ u_n = \sum_{k=2}^9 k^{m_k^n} $$
Now each $(u_n)$ is completely characterized by $(m_2^n,...,m_9^n)$. If $u$ is not zero then not all sequences $(m_k^n)_n$ go to $-\infty$, so some of them are bounded, which means that we will have some values repeated an infinity of times. Then we can extract a diagonal sequence such that every $(m_k^n)_n$ which does not go to $-\infty$ is constant. This means that you approximate $u=2^{a_2}+...+9^{a_9}$ where $a_1,...,a_9$ are integers or $-\infty$ (with convention $a^{-\infty}=0$ for $a>1$.)
Obviously not every $u$ can be written in this form.
In the same way you can prove that the colsure of the set you mention is precisely
$$ \{ u =2^{a_1}+...+9^{a_9} : a_i \in \Bbb{Z} \cup \{-\infty\}\} $$
|
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|
Visually stunning math concepts which are easy to explain Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.
Do you know of any other concepts like these?
|
Ulam Spiral:
Discovered by Stanislaw Ulam, the Ulam Spiral or the Prime Spiral depicts the certain quadratic polynomial's tendency to generate large number of primes.Ulam constructed the spiral by arranging the numbers in a rectangular grid . When he marked the prime numbers along this grid, he observed that the prime numbers thus circled show a tendency to occur along diagonal lines.
A 150x150 Ulam Spiral is shown below where the dots represent the occurance of prime numbers. The high density along the diagonal lines can be seen as represented by the darker shade of blue.
|
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|
Visually stunning math concepts which are easy to explain Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.
Do you know of any other concepts like these?
|
I would like to add some explorations of the concept asked by the OP of my own:
*
*Visualization of the set of real roots of quadratic equations $ax^2+bx+c=0$, for the specific values of the intervals $a \in [-a_i,a_i]$, $b \in [-b_i,b_i]$, $c \in [-c_i,c_i]$, $a,b,c \in \Bbb N$.
By Cartesian coordinates $(x,y)=(x_1,x_2)$. E.g. $a_i,b_i,c_i=75$:
By Polar coordinates $(\theta, r)=(x_1,x_2)$. E.g. $a_i,b_i,c_i=75$:
Due to the symmetries the opposite patterns $(x,y)=(x_2,x_1)$ and $(\theta,r)=(x_2,x_1)$ are similar.
*The Chaos Game on the metric space $S^{1} \times [0,\infty)$ with the metric $d((\theta_1,x_1),(\theta_2,x_2)) = d_{S^1}(\theta_1,\theta_2) + |x_1-x_2|$. The distance in $S^1$ is given by the smallest angle measure between $\theta_1$ and $\theta_2$ (this is actually a scaled Euclidean metric on the unit circle itself).
In this version, the points are $(\theta, r)$, (the angle in radians and the radius). And the three attractor points are $A=(0,0)$,$B=(\frac{5\pi}{4},1)$ and $C=(\frac{7\pi}{4},10^4)$.
This is another example locating the attractor points in the same axis: $A=(0,0)$,$B=(\pi-\frac{\pi}{8},1)$ and $C=(\frac{7\pi}{4},10^4)$.
*Contruction step by step of the Voronoi diagram of the points generated by a classic Chaos Game Sierpinski gasket.
*And my favorite so far, visualization of the $4$-tuples of the extended Euclidean algorithm in a four dimensional tesseract. The projection of the four dimensional points are shown into a $3D$ visualization adding as a reference a tesseract or hypercube:
|
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|
Verify why this is not a metric $d(x,y)=\|x-y\|_p$ ( $\|x\|_p=p^{-h}$ if $x=p^h\dfrac{m}{n}$). $d(x,y)=\|x-y\|_p$
$p$ is prime and $\|x\|_p=p^{-h}$ if $x=p^h\dfrac{m}{n}$, where $m, n$ are coprimes with $p$.
This is not a metric because
if $x=y=p^k\dfrac{m}{n}$, then $x-y=0=p^0\dfrac0n$.
Hence $d(x,y)=p^{-0}=1 \ne0$.
Is this correct?
|
You didn't specify what $m,n$ are in the definition of $\|.\|_p$. Supposedly, they integers that are not divisible by $p$. However, $0$ is divisible by $p$, hence you cannot compute $\|0\|_p$ this way. Indeed, one needs to define explicitly that $\|0\|_p=0$; with this addon, $d$ is a metric.
|
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|
Limit of a functional I'd like to find:
$$
\lim_{\varepsilon\rightarrow 0}\frac{\varepsilon}{\varepsilon^2+x^2}\qquad \mbox{ in }\mathcal D'(\mathbb{R})
$$
And I started with the definition:
$$
\left\langle \frac{\varepsilon}{\varepsilon^2+x^2},\varphi\right\rangle
$$
After doing the integration by parts I had some problems with the support of $\varphi$. I think the result could be $0$, but I don't know how to prove it in a proper way. (during the proof I used the substitution $x=\varepsilon y$ or $y=\varepsilon x$).
|
Define $$T_\varepsilon(\varphi):=\int_{-\infty}^{+\infty}\frac{\varepsilon}{\varepsilon^2+x^2}\varphi(x)\mathrm dx.$$
Then, after the substitution $x=\varepsilon t$, we obtain
$$T_\varepsilon(\varphi)=\int_{-\infty}^{+\infty}\frac{1}{1+t^2}\varphi(t\varepsilon)\mathrm dt.$$
Using dominated convergence, we obtain that for each $\varphi\in\mathcal D(\mathbf R)$,
$$\lim_{\varepsilon\to 0}T_\varepsilon(\varphi)=\varphi(0)\cdot\int_{-\infty}^{+\infty}\frac{1}{1+t^2}\mathrm dt =\pi\varphi(0).$$
|
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How find this function $f(x)$ such $f(a+f(b))=f(a+b)+f(b)$ let function $f:R_{+}\to R_{+}$,and such
$$f(a+f(b))=f(a+b)+f(b),\forall a,b\in R_{+}$$
Find $f(x)$.
my try: let $a=b=1$,then
$$f(1+f(1))=f(2)+f(1)$$
$a=1,b=2$,then
$$f(1+f(2))=f(3)+f(2)$$
then I can't find have any regular,so I can't.Thank you
Edit: This is IMO2007 SL, Problem A4.
|
the hint is :
$f(f(2a))=f(a+f(a+f^{-1}(a)))=f(2a+f^{-1}(a))+a$
$f(2f(a))=f(f(a)+a)+f(a)=f(2a)+2a$
$\Longrightarrow$$f(2a)+a=f(2a+f^{-1}(a))$
$\Longrightarrow$$f(x)=2x$
|
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|
Continuity of a partial derivative I have the function
$$f(x,y)=\begin{cases}
x^2ysin(\frac1x) & \text{if $x$ is not 0} \\
0 & \text{if $x=0$}\end{cases}$$
And I need to find the derivative and the partial derivatives, and see if they are continuos.
I´ve already proved that the function is continuos for all (x,y). I´ve also found the derivatives:
$$\frac {\partial f}{\partial x}= \begin{cases}
2xysin(\frac1x)-ycos(\frac1x) & \text{if $x$ is not 0} \\
0 & \text{if $x=0$}\end{cases}$$
$$\frac {\partial f}{\partial y}= \begin{cases}
x^2sin(\frac1x) & \text{if $x$ is not 0} \\
0 & \text{if $x=0$}\end{cases}$$
And I've also proved the continuity of $\frac {\partial f}{\partial y}$,
but I have yet to prove that $\frac {\partial f}{\partial x}$ is continuos (or not), and find the actual derivate of the function.
If someone could give me a hand, it would be great.
Thanks!
|
Notice that $\sin$ and $\cos$ remain bounded, while the factors they have in front tend to zero as $(x,y)$ tends to $(0,0)$.
But now consider the limit to points $(0,y)$ with $y\neq0$.
|
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|
Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b,c > 0 $? Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b> 0$ and $c\geq 0$ ?
Is there a general proof for that relation ?
|
Good observation, these inequalities are quite useful. But you need a little bit more:
*
*if $a\ge b$, then $\dfrac ab\ge\dfrac{a+c}{b+c}$
*if $a\le b$, then $\dfrac ab\le\dfrac{a+c}{b+c}$
You can prove it by multiplying by the common denominator:
*
*$a(b+c)\ge b(a+c)\Longleftrightarrow ac\ge bc$
*$a(b+c)\le b(a+c)\Longleftrightarrow ac\le bc$
|
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|
Summation involving Fibonacci numbers Find:
$$
\sum_{n=0}^\infty \sum_{k=0}^n \frac{F_{2k}F_{n-k}}{10^n}
$$
where $F_n$ is $n$-th Fibonacci number.
|
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{%
\sum_{n = 0}^{\infty}\,\sum_{k = 0}^{n}{F_{2k}\, F_{n - k} \over 10^{n}}} & =
\sum_{k = 0}^{\infty}F_{2k}\, \sum_{n = k}^{\infty}{F_{n - k} \over 10^{n}} =
\sum_{k = 0}^{\infty}{F_{2k} \over 10^{k}}
\sum_{n = 0}^{\infty}{F_{n} \over 10^{n}}
\\[5mm] & =
\bracks{\sum_{n = 0}^{\infty}F_{n}\pars{1 \over 10}^{n}}
\bracks{\half\sum_{k = 0}^{\infty}{F_{k}\pars{1 \over\root{10}}^{k}} +
\half\sum_{k = 0}^{\infty}{F_{k} \pars{-\,{1 \over \root{10}}}^{k}}}
\end{align}
With the Fibonacci generating function
$\ds{\,\mc{F}\pars{z} = \sum_{n = 0}^{\infty}F_{n}\,z^{n} =
{z \over 1 - z - z^{2}}}$:
\begin{align}
\color{#f00}{%
\sum_{n = 0}^{\infty}\,\sum_{k = 0}^{n}{F_{2k}\, F_{n - k} \over 10^{n}}} & =
\half\,\mc{F}\pars{1 \over 10}\bracks{\mc{F}\pars{1 \over \root{10}} +
\mc{F}\pars{-\,{1 \over \root{10}}}} = \color{#f00}{100 \over 6319}
\approx 0.0158
\end{align}
Note that $\ds{\quad\mc{F}\pars{1 \over 10} = {10 \over 89}\quad
\mbox{and}\quad
\,\mc{F}\pars{\pm\,{1 \over \root{10}}} = {10 \pm 9\root{10} \over 71}}$.
|
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|
Induction question help. Let $x$ and $y$ belong to a commutative ring $R$ with prime characteristic $p$.
Show that, for all positive integers $n$
$$ (( x + y )^p)^n = (x^p)^n + (y^p)^n $$
I hope you can can understand notation.
We have to use induction on $n$.
For $n=1$ $ (x + y)^p $ = $ x^p $ + $ y^p $
Assume for $n=k$
I have almost done it. I am having trouble with $n=k+1$.
|
Note that by the binomial theorem
$$
(x+y)^{pn} = \sum_{k=0}^{pn}\binom{pn}{k}x^ky^{pn-k} = \sum_{k=0}^{pn}\frac{(pn)!}{k!(pn-k)!}x^ky^{pn-k} = \sum_{k=0}^{pn}\frac{pn\cdot\ldots\cdot(pn-k+1)}{k!}x^ky^{pn-k} = \sum_{k=0}^{pn}p\frac{n\cdot\ldots\cdot(pn-k+1)}{k!}x^ky^{pn-k} = x^{pn}+y^{pn},
$$
since only when $k=0$ or $pn$ the binomial coefficients dissapear.
|
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|
Show $f$ is a complex polynomial of degree at most 2 Suppose $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire function and
$$\displaystyle\min\{\left|f'(z)\right|,\left|f(z)\right|\}\leq \left|z\right|+2$$ for all $z\in\mathbb{C}$.
How to see that $f$ is a polynomial in $z$ of degree at most 2?
I can only see it when $\left|f(z)\right|\leq \left|z\right|+2$ by using Cauchy estimate. How to handle $f'(z)$ part together?
|
By adding a constant, you can assume that $f(0) = 0$ . For an arbitrary point $z \in \mathbb{C}$ consider the minimal part of the line segment $L_z$ from $z$ to $0$ which connects $z$ to a point $z_0$ with $|f(z_0)| \le |z_0| + 2$. (If this is already satisfied for $z_0 = z$, the line segment will just be the point $z = z_0$.) For any $w \in L_z$ by assumption $|f'(w)| \le |w| + 2$. Integrating $f'$ along $L_z$ and using the fundamental theorem for line integrals, you get by standard estimates that $|f(z)|$ grows at most quadratically in $|z|$. Then proceed using Cauchy's estimates.
|
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|
Are contractive completely positive maps trace decreasing? Are contractive completely positive maps trace decreasing? More precisely, suppose that $f\colon M\to N$ is a normal cpc map between von Neumann algebras with normalised normal traces. (That is $\tau_M(1_M)=\tau_N(1_N)=1$). Do we have that $\tau_N(f(a))\leqslant \tau_M(a)$ for all $a\in M_+$?
|
For von Neumann algebras in general it is false, there is two much leeway in the choice of the traces. For instance let $M=N=\mathbb C\oplus\mathbb C$, with $f$ the identity map. Let the traces be
$$
\tau_M(x,y)=\frac x4+\frac{3y}4, \ \ \tau_N=\frac{3x}4+\frac y4.
$$
Then, for $a=(1,0)$,
$$
\tau_M(a)=\tau_M(1,0)=\frac14<\frac34=\tau_N(1,0)=\tau_N(f(a))
$$
And with almost the same idea, we can make it fail between factors. Let $M=N=M_2(\mathbb C)$. Put
$$
f\left(\begin{bmatrix}x&y\\ z&w\end{bmatrix}\right)=\frac x4+\frac{3w}4.
$$
This is a state, so we can see it as a ucp map $M\to N$. And, for $a=\begin{bmatrix}0&0\\0&1\end{bmatrix}$,
$$
\tau_N(f(a))=\frac34>\frac12\,\tau_N(a)
$$
|
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|
Multiplicative version of the principle of Archimedes
Any clear proof of the above theorem is greatly appreciated.
|
Suppose first that $y\ge 1$. We prove that the set $S=\{x^n\}$, $n=0,1,2,\dots$ is unbounded. Suppose to the contrary that $S$ bounded. Let $b$ be the supremum. Then for some $n$, we have $x^n \gt \frac{b}{(x+1)/2}$. Then $x^{n+1}\gt b\frac{x}{(x+1)/2}\gt b$. This contradicts the fact that $b$ is an upper bound of $S$.
Thus the set of positive integers $k$ such that $x^k\gt y$ is non-empty. Let $n$ be the smallest element of this set. Then $x^{n-1}\le y$, and we are finished.
For $0\lt y\lt 1$, the easiest way to proceed is to show that there is a non-negative integer $m$ such that $x^m \lt \frac{1}{y}\le x^{m+1}$. The argument for this is almost the same as the previous one.
|
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Building a partial injective relation Question : A Partial Injective Relation from $A \rightarrow B$ is maximal if its graph of an injection function from $A$ to $B$ or the graph of an injection function from $B$ to $A$.
Example: $A =[a,b,c]$ and $B=[1,2,3,4]$.
Build a partial injection relation. Are there 3-4 elements or a lot of elements?
Attempt:
Definition 5.4.5 states that a function $f: X \rightarrow Y$ with the property $ \forall x_1, x_2 \in X)[ x_1 \neq x_2] \rightarrow f(x_1) \neq f(x_2)$ is an injection of $X$ into $Y$
We're using the contrapositive of the injection definition which is $ \forall x_1, x_2 \in X)[ x_1 = x_2] \rightarrow f(x_1) = f(x_2)$
Suppose $ f: b \rightarrow a$ is a partial injection.
From the example, we have $A =[a,b,c]$ and $B = [1,2,3,4]$. This is a partial injective relation if we have $S=[(a,1),(b,2),(c,3)]$. There are only three elements because from $B$, the number $4$ isn't matching anything on $A$.
So if I construct a bijection map $f: B \rightarrow A$, I would have $f(1) = a$, $f(2) = b$, and $f(3) =c$. I won't have anything equal to 4 because set $A$ only has three elements and set $B$ has four elements. Therefore, if I take the bijection, I would have three elements all together.
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There are only injections from the smaller set to the larger, namely
$$
f: A \to B
$$
There are $4$ choices for the value $f(a)$, $3$ choices for $f(b)$, and $2$ choices for $f(c)$ for a total of $4 \cdot 3 \cdot 2 = 24$ possible injective functions. Each of these gives a distinct maximal injective relation, consisting of the pairs
$$
\big\{ (a, f(a)), (b, f(b)), (c, f(c)) \big\}.
$$
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Uniform Distribution and Distribution function technique Let $X_1$ and $X_2$ be independent random variables having the uniform density with $\alpha = 0$ and $\beta = 1$. Find expressions for the function $Y =X_1 + X_2$.
(a)$y \le 0$
(b)$0<y<1$
(c)$1<y<2$
(d)$y\ge2$
I'm thinking $f(x_1)=f(x_2) = 1$ for $0 \le x_1\le 1$ and $0 \le x_2\le 1$. I suppose this implies that $f(x_1,x_2) = 1$ so $F(x_1,x_2) = \int_0^y\int_0^{y-x_2} f(x_1,x_2) dx_1dx_2 $ Then I would need $\frac{d}{dy} F(x_1,x_2)$ to find an appropriate $f(y)$
I already have the answers to these in my book but I don't know how to come upon them as they don't correspond at all to what I have here. Where is my fault?
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You have $P(X_1 \leq u) = P(X_2 \leq u) = u$ for $0 \leq u \leq 1$ and $X_1,X_2$ independent. Thus $$
P(X_1 + X_2 \leq u) = \int_{(x_1,x_2) \in A_u} \,d(P_{X_1} \times P_{X_2}) \text{ where } A_u = \{(x_1,x_2) \in [0,1]^2 \mid x_1 + x_2 \leq u\} \text{.}
$$
which yields the CDF (note that the density of $X_1$ and $X_2$ on $[0,1]$ is $1$) for $u \in [0,2]$ $$ \begin{eqnarray}
F_{X_1+X_2}(u) &=& \int_0^u \int_0^{u-x_1} dP_{X_2}(x_2)\,dP_{X_1}(x_1)
= \int_0^{\min\{u,1\}} \int_0^{\min\{u-x_1,1\}} \,d x_1 \,d x_2 \\
&=& \int_0^{\min\{u,1\}} \min\{u-x_1,1\} \,dx_1 \\
&=& \begin{cases}
\int_0^{u-1} 1 \,dx_1 + \int_{u-1}^1 u - x_1 \,dx_1 & \text{$u > 1$} \\
\int_0^u u - x_1 \,dx_1 &\text{$u \leq 1$}
\end{cases} \\
&=& \begin{cases}
(u-1) + (1 - (u-1))u - \frac{1}{2}(1^2 - (u-1)^2) & \text{$u > 1$} \\
u^2 - \frac{1}{2}(u^2 - 0^2) &\text{$u \leq 1$}
\end{cases} \\
&=& \begin{cases}
1 &\text{$u > 2$} \\
-\frac{1}{2}u^2 + 2u - 1 & \text{$u > 1$} \\
\frac{1}{2}u^2 &\text{$u \leq 1$} \\
0 &\text{$u < 0$.}
\end{cases}
\end{eqnarray}$$
Note that the CDF is continuous as expected, so for any particular $u$, $P(X_1 + X_2 = u) = 0$. Using the CDF and this fact, you get
*
*$P(Y \in (0,1)) = P(Y \leq 1) - P(Y \leq 0) = \frac{1}{2} - 0$
*$P(Y \in (1,2)) = P(Y \leq 2) - P(Y \leq 1) = 1 - \frac{1}{2} = \frac{1}{2}$
For the other two it's immediately obvious even without the CDF that
*
*$P(Y \leq 0) = 0$
*$P(Y \geq 2) = 0$
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Recurrence problem for $a_5$ Assume that the sequence $\{a_0,a_1,a_2,\ldots\}$ satisfies the recurrence $a_{n+1} = a_n + 2a_{n−1}$. We know that $a_0 = 4$ and $a_2 = 13$. What is $a_5$?
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Rearranging the recurrence relation gives $a_n = a_{n+1} - 2a_{n-1}$, so that $a_1 = 13 - 2 \cdot 4 = 5$. Now that we know $a_0,a_1$ and $a_2$ you are good to go by simply applying the recursion repeatedly until you get to the fifth term.
|
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Possible difference between $\mathbb{Z}$-modules and vector spaces Suppose $G$ is a free abelian groups, i.e. a free $\mathbb{Z}$-module; we have a set $S \subset G $ such that $S$ spans $G$.
Can we conclude that the rank of $G$ as a $\mathbb{Z}$-module is $ \leq |S| $ as in the vector-space case ? Why ?
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There are also many differences between vector spaces and $\mathbb{Z}$-modules. Every vector space has a basis, but not every $\mathbb{Z}$-module. For example, any finite abelian group is not a free $\mathbb{Z}$-module, and the $\mathbb{Z}$-module $\mathbb{Q}$ is not free.
Furthermore a free $\mathbb{Z}$-module may have a linear independent set which cannot be extended to a basis. It also may have a subset $S$ which spans it, but does not contain a basis.
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I need someone to explain this proof from James Munkres' Topology.
The author writes $(X-C)\cap Y = Y-A,$ and, also, $A=Y \cap (X-U)$. I was wondering how is that something anyone writing an original proof of the theorem saw and if there is an analytic proof that the equalities holds true. Thanks for your help.
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He defined $A = C \cap Y$; To see $(X-C) \cap Y = Y - A$, we can compute (as Frank did) or reason: a point is in the left hand side if it is not in $C$ but it is in $Y$. This means not in $A$ (to be in $A$, one has to be both in $C$ and $Y$), so it is in $Y - A$. On the other hand, if a point is in $Y - A$, it is in $Y$ but not in $A$, and the only way that could happen is if it is not in $C$ (otherwise it would be in $A$ by definition). So it is not in $C$ (so in $X-C$) and in $Y$, so in $Y \cap (X-C)$.
Or one could draw a schematic picture, and see it that way (it's quite obvious then).
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|
Soft Question: Algorithms: Will We (One Day) No Longer Need to Study Algorithms? I'm just now getting into the study of algorithms and it seems like as computers get faster and faster the need to study algorithms may begin to diminish.
How likely is it that in 50 years there won't be much of a demand for the analysis of algorithms (other than for historical purposes, for example).
Edited to include:
The reason that I ask is that I'm interested in studying algorithms so much that I could see myself moving into a graduate level of research. However, if the field isn't even going to be around in 50 years, I'm going to study the basics and move onto something else...
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Euclidean geometry has existed for a long time and we certainly don't only study it for historical purposes.
Usually what happens with a field of mathematics is that it gets incorporated into something bigger and more complex when we attain total mastery of it. For example, from Euclidean geometry we've gotten a lot of tools useful in number theory, topology, and other areas.
With algorithm analysis there's quite a similar situation. In fact, euclid himself came up with an algorithm for calculating GCD's. In some sense, then, algorithms have been studied as long as geometry. The field's not going anywhere.
The last thing I'd like to say is that you should treat algorithms as a tool separate from, but useful in computer science. They've been around since before we had computers and they're still useful outside of computer science.
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Nice book on geometry to gift an undergraduate in mathematics I would like some suggestions on a nice book on geometry to gift an undergraduate. I'm not searching for something that is common: I need something new and exciting. Suggestions?
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I love this one. It is one of the most stimulating books on geometry that I have ever read: it is strikingly innovative and really enjoyable (also, it does not require a particularly advanced background).
From the book description:
New Horizons in Geometry represents the fruits of 15 years of work in
geometry by a remarkable team of prize-winning authors—Tom Apostol and
Mamikon Mnatsakanian. It serves as a capstone to an amazing
collaboration. Apostol and Mamikon provide fresh and powerful insights
into geometry that requires only a modest background in mathematics.
Using new and intuitively rich methods, they give beautifully
illustrated proofs of results, the majority of which are new, and
frequently develop extensions of familiar theorems that are often
surprising and sometimes astounding. It is mathematical exposition of
the highest order.
The hundreds of full color illustrations by Mamikon are visually
enticing and provide great motivation to read further and savor the
wonderful results. Lengths, areas, and volumes of curves, surfaces,
and solids are explored from a visually captivating perspective. It is
an understatement to say that Apostol and Mamikon have breathed new
life into geometry.
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Is it possible to formalize areas such as image processing and computer vision? Is it possible to formalize areas such as image processing?
By formalize I mean setup axioms, then derive theorems, and
reason about image processing concepts and methods formally.
I would say now image processing is pretty informal and ill-defined
without strong foundations.
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I would beg to disagree. Image processing is quite formal. Images are treated as discrete functions in 2D, and are manipulated using very formal methods from calculus, linear algebra, and statistics.
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Question regarding notation involving vector spaces. Let V be the set of all ordered pairs of real numbers, and consider the following addition and scalar multiplication operations on $u+v=(u_1+v_1+1,u_2+v_2+1),$ $ku=(ku_1,ku_2)$.
Show that $(0,0)\neq0$.
I'm confused by their notation. Can anyone explain?
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By $\mathbf{0}$ they mean the zero vector. The zero vector has to be the additive identity in order for $V$ to be a vector space, in other words, $\mathbf{0}$ has to be a vector such that $\mathbf{v}+\mathbf{0}=\mathbf{v}$.
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Which is the topology generated by the neighborhood system $V(x)=\{\{x\}\}$ My question is the following:
Which is the topology generated by the neighborhood system $V(x)=\{\{x\}\}$ ?
I say that is the coarse topology but I don't know how is the mechanism to generate a topology from a basis
sorry if this is so trivial but I always have this doubt
thank you
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By definition a set $U$ is open if and only if for every $x\in U$, there is a neighborhood $B_x\in V(x)$ such that $B_x\subset U$. Particularly, $V(x)\ni\{x\}\subseteq\{x\}$.. You should recognize that this is a standard topology on a set after some thought.
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How prove $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n}}<2$ Let $$A=\{a_{1},a_{2},\ldots,a_{n}\}\subset N$$
Suppose that for any two distinct subsets $B, C\subseteq A$, we have
$$\sum_{x\in B}x\neq \sum_{x\in C}x$$
Then show that
$$\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\dfrac{1}{a_{3}}+\cdots+\dfrac{1}{a_{n}}<2$$
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This is an old conjecture of Erdos, which was subsequently proved (Ryavek from my notes) - though I cannot find a handy online reference just now. The proof goes along the following lines, IIRC:
With $0 < x< 1$, we have by the distinct sum condition:
$$ \prod_{k=1}^n (1+x^{a_k}) < \sum_{k=0}^{\infty} x^k = \frac1{1-x} $$
$$\implies \sum_{k=1}^n \log(1+x^{a_k}) < - \log (1-x)$$
As both sides are positive, we can divide by $x$ and integrate to get
$$\implies \sum_{k=1}^n \int_0^1 \frac{ \log(1+x^{a_k})}x dx < - \int_0^1 \frac{\log (1-x)}x dx $$
$$\implies \sum_{k=1}^n \frac1{a_k} \cdot \int_0^1 \frac{ \log(1+t)}t dt < \frac{\pi^2}6 $$
$$\implies \sum_{k=1}^n \frac1{a_k} \cdot \frac{\pi^2}{12} < \frac{\pi^2}6 \implies \sum_{k=1}^n \frac1{a_k} < 2 $$
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Sum of probabilities or mean of probability My question is about being confused about two way of approaching a problem, which in this case lead me to the same solution. One method is very verbose, the other one is fast and clean.
Let's consider this problem where there are three only components of a computer with probability of being defective 0.1, 0.05, 0.02. Now, I want the expected value E(X), where X is number of defective components.
Method 1
Sum of probabilities, a computer has a piece of each then:
$$E(X) = 0.1+0.05+0.02=0.17 $$
Method 2
Mean of probabilities, I calculate $ \Sigma\ x_i p_i$. Therefore:
$$p_1 = 0.1\times0.95\times0.98+0.9\times0.05\times0.98+ ... = 0.1543$$
$$p_2 = 0.9\times0.05\times0.02+0.1\times0.95\times0.02+... = 0.0077$$
$$p_3 = 0.1\times0.05\times0.02 = 0.0001$$
To then do $E(X) = 1\times0.1543+2\times0.0077+3\times0.0001 = 0.17$
In summary, both the methods give the same answer. The first one seems logical to me, but I don't have in my hands any proper formula, like I have it for the second method. Is there any flow that I am not seeing?
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Both methods are correct!
To see that Method 1 is correct, define the independent Bernoulli random variables
$$X_i=\begin{cases} 1, &\text{component $i$ is defective} \\ 0, &\text{component $i$ is not defective} \end{cases}$$ for $i=1,2,3$. Then $$X_i \sim \mathrm{Bern}(p_i)$$ where the $p_i$'s are the given probabilities. Now, obviously $$X=\sum_{i=1}^{3}X_i$$ so that due to linearity of expectation $$E[X]=E\left[\sum_{i=1}^{3}X_i\right]=\sum_{i=1}^{3}E[X_i]$$ But, since $E[X_i]=p_i$ we have that $$E[X]=\sum_{i=1}^{3}p_i$$ which yields the result that you found with Method 1, i.e. $E[X]=0.17$.
Just to verify that Method 2 yields also the correct result:
Since you have defined your random variable $X$ then - in order to calculate it's expected value - you should proceed as follows
*
*Firstly, determine the sample space of $X$. There can be from $0$ up to $3$ defective components, so that $$X \in \{0,1,2,3\}$$
*Secondly, determine the probability mass function of $X$, i.e. the probability of $X$ taking each of these values. You have that $$P(X=0)=(1-0.1)\cdot(1-0.05)\cdot(1-0.02)$$
and $$\begin{align*}P(X=1)=&0.1(1-0.05)(1-0.02)+(1-0.1)0.05(1-0.02)+(1-0.1)(1-0.05)0.02\end{align*}$$
and
$$\begin{align}P(X=2)=&(0.1)(0.05)(1-0.02)+(0.1)(1-0.05)(0.02)+(1-0.1)(0.05)(0.02)\end{align}$$
and
$$P(X=3)=0.1\cdot0.05\cdot0.02$$
This gives the following probabilities $$P(X=x)=\begin{cases}0.8379, & x=0 \\ 0.1543, & x=1 \\ 0.0077, & x=2 \\ 0.0001, & x=3 \\ \end{cases}$$ (note that these probabilities add up to $1$). Now the expected value of $X$ is given by $$E[X]=\sum_{x=0}^{3}xP(X=x)=0P(X=0)+P(X=1)+2P(X=2)+3P(X=3)=0.17$$
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Solving a non linear ODE problem Please I will like to solve this non linear ODE problem
$$
y'(x) = e^{-b(x)} \times \left( p + \left( \frac{q}{n} y(x) \right)\right)\left( n - y(x) \right).
$$
Can anyone help? Thank you
I made some correction to the equation
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Using the same approach as izoec, we can arrive to$$\frac{n (n (p+q) \log (n p+q y)-q y)}{q^2}= \int e^{-b(x)} \, \mathrm{d} x = \xi (x)$$ and the solution of $y$ is $$y=- \frac {n} {q} \left(p+(p+q) W\left(-\frac{\left(e^{\frac{\xi(x) q^2}{n^2
p}-1}\right)^{\frac{p}{p+q}}}{n (p+q)}\right)\right)$$
I hope and wish that I did not make any mistake.
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Suppose $Y\subset X$ and $X,Y$ are connected and $A,B$ form separation for $X-Y$ then, prove that $Y\cup A$ and $Y\cup B$ are connected Question is :
Suppose $Y\subset X$ and $X,Y$ are connected and $A,B$ form separation for $X-Y$ then, Prove that $Y\cup A$ and $Y\cup B$ are connected.
What i have tried is :
Suppose $Y\cup A$ has separation say $C\cup D$ then all i could see is that
Either $Y\subset C$ or $Y\subset D$ as $Y$ is connected and $
C\cup D$ is separation for a subset that contains $Y$.
Without loss of generality we could assume $Y\subset C$
As $Y\cup A=C\cup D$ and $Y\subset C$ i can say $D\subset A$ (I do not know how does this help)
I have $X-Y=A\cup B$ with $A\cap B=\emptyset$
I have not used connectedness of $X$ till now, So i thought of using that and end up with following :
$X-Y=A\cup B\Rightarrow X=Y\cup A\cup B=(Y\cup A)\cup (Y\cup B)$
I do not know what to conclude from this...
I would be thankful if some one can help me by giving some "hints"
Thank you
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Continuing with your argument, We can prove that $B \cup C$ and $D$ form the partition of $X$.(Assuming your conclusion that $D \subset A$).
We have $X = (B \cup C) \cup D$.
$C$ and $D$ form separation of $Y \cup A$. So, using theorem (23.1), no limit point of $C$ is in $D$. Similarly, no limit point of $B$ is in $D$ since $D \subset A$ and $A$ and $B$ form separation of $X-Y$. From this, we can easily say that no limit point of $B \cup C$ is in $D$ i.e. Every limit point of $B \cup C$ wrt $X$ is in $B \cup C$. So, $B \cup C$ is closed in $X$ and $D$ is open in $X$.
Now, use a similar argument that, no limit point of $D$ is in $B \cup C$ implying that $D$ is closed in $X$ and $B\cup C$ is open in X. Hence, $B \cup C$ and $D$ are $\textit{clopen}$ in $X$ and form partition of $X$.
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Computing $\lim\limits_{n\to\infty}(1+1/n^2)^n$
Why is $\lim\limits_{n\to\infty}(1+\frac{1}{n^2})^n = 1$?
Could someone elaborate on this? I know that $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n = e$.
|
There are some good answers here, but let me share a method that is more computational (in that it doesn't require noticing certain inequalities and using the squeeze theorem), and may help you with other similar limits.
First, one convenient way of taking exponents out of a limit expression is to take a logarithm:
$$\lim_{n \rightarrow \infty} f(n)^{g(n)} = \exp\left[\lim_{n \rightarrow \infty} g(n)\ln(f(n))\right]$$
Provided that $f(n) > 0$ (and either limit exists). In your case, this becomes:
$$\lim_{n \rightarrow \infty} (1 + \frac{1}{n^2})^{n} = \exp\left[\lim_{n \rightarrow \infty} n\ln(1 + \frac{1}{n^2})\right]$$
To evaluate the limit inside the brackets on the right, you can use l'Hospital's rule. You need to change the indeterminate form to $0/0$ and replace $n$ with a 'continuous variable' $x$:
$$\lim_{n \rightarrow \infty} n \ln(1 + \frac{1}{n^2}) = \lim_{x \rightarrow \infty} \frac{\ln(1 + \frac{1}{x^2})}{\frac{1}{x}} = \lim_{x \rightarrow \infty} \frac{(-\frac{2}{x^3})/(1 + \frac{1}{x^2})}{-\frac{1}{x^2}} = \lim_{x \rightarrow \infty} \frac{-\frac{2}{x^3}}{-\frac{1}{x^2}(1 + \frac{1}{x^2})}$$
To simplify this last fraction, multiply the numerator and denominator by $-x^4$:
$$= \lim_{x \rightarrow \infty} \frac{2x}{x^2 + 1} = 0$$
Finally, your answer is:
$$\exp\left[\lim_{n \rightarrow \infty} n\ln(1 + \frac{1}{n^2})\right] = \exp[0] = 1$$
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Proving the Division Algorithm using induction Let $n \in \mathbb{N}$. For every $m \in \mathbb{Z}$, there exist unique $q, r \in \mathbb{Z}$ such that $ m = qn+r$ and $0 \le r \le n-1$. We call $q$ the quotient and $r$ the remainder when dividing into $m$.
I'm having trouble proving this with induction. I believe the idea is to first prove for all $m \in \mathbb{Z} _{\ge 0}$ by induction, then prove for all $m \in \mathbb{Z}$ but $m \notin \mathbb{Z} _{\ge 0}$ using the induction proof on $-m$, since $-m \in \mathbb{N}$.
I'd appreciate any help, thank you.
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Uniqueness doesn't need induction. Suppose $m=qn+r=q'n+r'$, where $0\le r\le n-1$. It's not restrictive to assume $r\le r'$, so we have $0\le r'-r\le n-1$; but $r'-r=n(q-q')$, so
$$
0\le n(q-q')<n
$$
As $n>0$, this implies
$$
0\le q-q'<1
$$
so $q=q'$ and therefore $r'=r$.
The proof of existence can be conveniently split into the cases $m\ge0$ and $m<0$.
The first case is done by induction. The case $m=0$ is obvious: take $q=0$ and $r=0$. Assume you know $m=qn+r$, with $0\le r<n$; then
$$
m+1=qn+r+1
$$
If $r+1=n$, then $m+1=q(n+1)+0$, otherwise $r+1<n$ (using the hypothesis that $r\le n-1$, so $r+1\le n$) and the assert is true.
Now let's prove the case $m<0$. From the first case we get
$$
-m=qn+r
$$
with $0\le r<n$. If $r=0$, then $m=(-q)n+0$ and we're done. Otherwise $0<r<n$ and
$$
m=(-q)n-r=(-q)n-n+n-r=(-q-1)n+(n-r)
$$
where $0<n-r<n$ and we're done.
|
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Lagrange Polynomial Interpolation - Equation Help I understand the concept of Lagrange Interpolation but am having issues understanding how to interpret the following general equation (which I will be provided) for n points. For example, how would you get the equation for n = 4 points from the general equation below?
Thanks!
|
Let $x_1,x_2,x_3,x_4$ be mutually distinct numbers and we need to fit the polynomial of degree 3 passing through $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$. Then
$$p(x)=L_1(x)y_1+L_2(x)y_2+L_3(x)y_3+L_4(x)y_4,$$
where
$$L_1(x)=\frac{(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)},$$
$$L_2(x)=\frac{(x-x_1)(x-x_3)(x-x_4)}{(x_2-x_1)(x_2-x_3)(x_2-x_4)},$$
$$L_3(x)=\frac{(x-x_1)(x-x_2)(x-x_4)}{(x_3-x_1)(x_3-x_2)(x_3-x_4)},$$
$$L_4(x)=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_4-x_1)(x_4-x_2)(x_4-x_3)}.$$
|
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For $(x_n)$ increasing, $\sum_{n=1}^{\infty}\left(1-\frac{x_n}{x_{n+1}}\right)$ if $(x_n)$ is bounded and diverges if it is unbounded Let $\{x_n\}$ be monotone increasing sequence of positive real numbers. Show that if $\{x_n\}$ is bounded, then $\sum_{n=1}^{\infty}\left(1-\frac{x_n}{x_{n+1}}\right)$ converges. On the other hand, if the sequence is unbounded, the series is divergent.
The following is the proof given by my lecturer, but I do not understand the proof:
Let $u_m = x_{m+1} -x_m$ and $d_m = \sum_{k=1}^m u_k = x_{m+1} - x_1$ and hence $x_{m+1} = d_m + x_1$
$$\sum_{n=1}^{\infty}\left(1-\frac{x_n}{x_{n+1}}\right) = \sum_{n=1}^{\infty}\frac{u_m}{d_m +x_1} $$
Then he says that if $d_m$ diverges, then the sum diverges. If $d_m$ converges, then then sum converges.
He says that it has something to do with Gauss Test.
Can someone explain the proof to me. Or does anyone has a better proof.
|
I try to rewrite the proof cleaning some details.
Call $r_n=\frac{x_n}{x_{n+1}}\in(0,1]$. If there's an infinite number of terms such that $1-r_n=1-\frac{x_n}{x_{n+1}}\geq \frac12$ then the series clearly diverges, while by rewriting the condition as $x_{n+1}\geq 2 x_n$ also the sequence diverges.
If instead eventually $\frac12\leq r_n\leq 1$, then we can use that
$$\log r\leq r-1\leq C\log r\qquad\text{for}\quad \frac12 \leq r\leq 1$$
for some positive $C$ to obtain
$$\sum_{n=1}^N\left(\frac{x_n}{x_{n+1}}-1\right)\leq C \sum_{n=1}^N \log \frac{x_n}{x_{n+1}}=C\log\frac{x_1}{x_N}$$
and
$$\sum_{n=1}^N\left(\frac{x_n}{x_{n+1}}-1\right)\geq \sum_{n=1}^N \log \frac{x_n}{x_{n+1}}=\log\frac{x_1}{x_N}.$$
Therefore the series has the same behaviour of $\lim_{N\to\infty}\log\frac{x_1}{x_{N}}$, which diverges iff $x_N\to\infty$.
|
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|
Ordinary Generating functions for $b_n$ Problem
Let $f(x)$ be a ordinary generating function for the sequence $ \{\ a_0, a_1, a_2... \}\ $ Find the ordinary generating function for $b_0 = b_1 = 0, b_2 = 1$ $b_n = a_n$ for $n \geq 3$.
Also find the generating function for $b_n = 0$ for even $n$, $b_n = a_n$ for odd $n$.
My attempts
For the first problem, $A(x) = 0+0+1x^2+a_3x^3+a_4x^4...$
So $$g(x) = x^2 + \sum_{n=3}^\infty a_n x^n$$
$$\Rightarrow x^2 + \sum_{n=0}^\infty a_n x^{n+3}$$
$$ \Rightarrow x^2 + \sum_{n=0}^\infty a_n x^n \cdot x^{-3}$$
$$\Rightarrow x^2 + f(x) \cdot x^{-3}$$
For the next problem I've got the sequence to be: $$ \{\ 0,a_1 , 0 , a_2 , 0, a_3\ldots \}\ $$
So $$A(x) = \{\ a_1 x + a_3 x^3 + a_5 x^5 \ldots \}\ $$
Which implies $$g(x) = \sum_{n=0}^\infty a_n x^n - \sum_{n=0}^\infty a_{2n} x^n$$
$$g(x) = f(x) - \sum_{n=0}^\infty a_{2n} x^n$$
But I can't get further. I hope my progress so far is correct in these.
Thank you.
|
So for the first one there are a couple of issues, when you go from summing from $n = 3$ to $n=0$ you add 3 on to the power, but not the index of the coefficient.
Also when you factor out the $x^3$ it turns into $x^{-3}$. Fix these things and see how far you can get.
For the second I'd recommend looking at $f(-x)$ and comparing that to $f(x)$.
|
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|
Solving the recurrence relation $T(n) = T(n-\sqrt n) + 1$ I have an algorithm that at each step can discard $\lceil\sqrt(n)\rceil$ possibilities at a cost $1$. The solution to the recurrence relation below is related to the question of complexity of such algorithm:
$$T(n) = T\left(n-\lceil\sqrt n\rceil\right) + 1$$
I know that: $T(0)=0$, $T(1) = 1$.
I tried substituting $n=2^{2^k}$ but it did not get me far.
|
Here is a geometric intuition.
We started with a width of $\sqrt{n}$. At each step, the width reduced by (1/2). Hence within roughly $2\sqrt{n}$ steps we reach the base case.
|
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|
If two cycles are disjoints, then they commute I'm trying to prove that two $\tau$-cycles commute provided that these cycles are disjoints. Hungerford in his book says the following remark about this fact:
Intuitively clearly this is true, but how can we prove this formally?
Thanks in advance
|
You can probably do this more formally by considering the permutations as elements which act on a set, namely the set of letters $\{1, \ldots, n\}$. In such a case, a cycle cyclically permutes a sub-collection of letters $\{\ell_1, \ldots, \ell_r\}$, and the fact that two cycles commute tell us that these sets which are permuted are disjoint. It should be then clear that the order in which we perform this permutation is irrelevant.
I should note that it is important that the representation of the group elements as permutations of letters is a faithful representation, otherwise this doesn't work. But this is certainly the case here.
|
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Linear functional in Banach space Let $X$ be a Banach space, $(f_n)\in X^{*}$ a sequence with $f_n\neq 0 $ $ \forall n\in \Bbb N$. Show that there is a $x\in X$ such that $f_n(x)\neq 0 $ $\forall n$.
Need some help. Thank you!
|
It's something with the kernel of the functionals,right?
Right. What you want is an $x$ with $f_n(x) \neq 0$ for all $n$, so
$$x \in \bigcap_{n\in\mathbb{N}} \left(X\setminus \ker f_n\right) = X \setminus \bigcup_{n\in\mathbb{N}} \ker f_n.$$
The $f_n$ are continuous, so $\ker f_n$ is closed. $f_n \neq 0$, hence $\ker f_n \neq X$, and therefore $(\ker f_n)^{\Large\circ} = \varnothing$. $X$ is a Banach space, hence a complete metric space, hence Baire's theorem ...
|
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Does there exist an $x$ such that $3^x = x^2$? I tried solving for $x$ by using $x \log(3) = \log(x^2)
$$\log(3) = \frac{\log(x^2)}{x}$$
I'm stuck on this part. how do I isolate $x$ by itself?
Any help would be appreciated.
|
The unique real solution is $$-2 \dfrac{W(\ln(3)/2)}{\ln(3)}$$
where $W$ is the Lambert W function. There are also complex solutions, corresponding to the different branches of $-2 \dfrac{W(\pm\ln(3)/2)}{\ln(3)}$
|
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How prove this inequality $\frac{1}{(a+1)^2+\sqrt{2(b^4+1)}}+\frac{1}{(b+1)^2+\sqrt{2(c^4+1)}}+\frac{1}{(c+1)^2+\sqrt{2(a^4+1)}}\le\frac{1}{2}$ let $a,b,c>0$,and such $abc=1$, show that
$$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}+\dfrac{1}{(b+1)^2+\sqrt{2(c^4+1)}}+\dfrac{1}{(c+1)^2+\sqrt{2(a^4+1)}}\le\dfrac{1}{2}$$
My idea : Use Cauchy-Schwarz inequality,we have
$$2(b^4+1)=(1+1)(b^4+1)\ge (b^2+1)^2$$
so
$$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}\le\dfrac{1}{(a+1)^2+(b^2+1)}\le\dfrac{1}{(a+1)^2+\dfrac{(b+1)^2}{2}}$$
then we only prove
$$\Longleftrightarrow\sum_{cyc}\dfrac{2}{2(a+1)^2+(b+1)^2}\le\dfrac{1}{2}$$
$$\Longleftrightarrow\sum_{cyc}\dfrac{1}{2(a+1)^2+(b+1)^2}\le\dfrac{1}{4}$$
Then I can't.Thank you
|
since Use Cauchy-Schwarz inequality and AM-GM inequality,we have
$$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}\le\dfrac{1}{(a+1)^2+b^2+1}\le\dfrac{1}{2ab+2a+2}=\dfrac{1}{2}\cdot\dfrac{1}{ab+a+1}$$
Use follow well know reslut,if $abc=1$,then we have
$$\sum_{cyc}\dfrac{1}{ab+a+1}=1$$
poof: since
\begin{align*}\dfrac{1}{ab+a+1}+\dfrac{1}{bc+b+1}+\dfrac{1}{ac+c+1}&=\dfrac{1}{ab+a+1}+\dfrac{a}{abc+ab+a}+\dfrac{ab}{abc\cdot a+abc+1}\\
&=\dfrac{ab+a+1}{ab+a+1}\\
&=1
\end{align*}
so the inequality we have By Done
|
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Fibonacci trick and proving it. I am trying to learn Fibonacci tricks and I have one that I can not prove. I know it works because Ive tried it multiple times but I have not a clue how to prove. Here it is:
f(0)^2 + f(1)^2 + f(2)^2 + f(3)^2 = f(3)f(3+1)
0 + 1 + 1 + 4 = 2 * 3
= 6 =6
Is there a way to prove this?
|
You've already proven that it works for one example $k=3$. Now write up the general equation
$$
\sum_{k=0}^n F_k^2 = F_n\cdot F_{n+1}
$$
and add $F_{n+1}^2$ on both sides. You get
$$
F_{n+1}^2+ \sum_{k=0}^n F_k^2 =\sum_{k=0}^{n+1} F_k^2 = F_{n+1}^2+ F_n\cdot F_{n+1}=\left(F_{n+1}+ F_n\right)\cdot F_{n+1}
$$
and use $F_{n+2}=F_{n+1}+ F_n$, the definition of Fibonacci numbers.
|
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Proving equality of sigma-algebras Let $C_1$ and $C_2$ are two collections of subsets of the set $\Omega$. We want to show that if $C_2$ $\subset$ $\sigma$[$C_1$] and $C_1$ $\subset$ $\sigma$[$C_2$], then $\sigma$[$C_1$]=$\sigma$[$C_2$] where $\sigma$[$C$] is defined to be the sigma-algebra generated by a collection $C$ of subsets of $\Omega$ (or the smallest sigma-algebra).
I'm having trouble characterizing the smallest sigma-algebra of sets in $C_1$ and $C_2$. Would the smallest sigma-algebra of $C_1$ be {$\emptyset$, $\Omega$, $C_1$, $\Omega$ $\not$ $C_1$} or would we have to characterize them in terms of the sets in $C_1$ which is the countable subcollection of sets in $\Omega$?
|
Use twice the fact that, if $A$ and $B$ are two subsets of $2^\Omega$ such that $A\subseteq\sigma(B)$, then $\sigma(A)\subseteq\sigma(B)$, once for $(A,B)=(C_1,C_2)$, and once for $(B,A)=(C_1,C_2)$.
The fact itself is direct using the identity
$$\sigma(B)=\bigcap_{F\in\mathfrak F} F,
$$
where $\mathfrak F$ is the collection of sigma-algebras $F$ such that $B\subseteq F$.
|
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History of Morse theory. How can I get good references which give many information about history of Morse theory? Now I am interested in how and who found that Hessian have a lot of data. Thank you for your helping!!
|
You may want to try these references:
*
*R. Bott, "Marston Morse and his mathematical works", Bull. Amer. Math. Soc. 3 (1980) 907–950.
*Dieudonné's A history of algebraic and differential topology.
*Morse theory in Tu's The life and works of Raoul Bott [pdf]. Also in Notice of AMS.
|
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if $a_n=\frac{a_{[\frac{n}{2}]}}{2}+\frac{a_{[\frac{n}{3}]}}{3}+\ldots+\frac{a_{[\frac{n}{n}]}}{n}$,then $a_{2n}<2a_{n}$ Question:
Consider the following sequence : $$a_1=1 ; a_n=\frac{a_{[\frac{n}{2}]}}{2}+\frac{a_{[\frac{n}{3}]}}{3}+\ldots+\frac{a_{[\frac{n}{n}]}}{n}$$. Prove that: $$a_{2n}< 2a_{n } (\forall n\in\mathbb{N})$$
where $[x]$ is the largest integer not greater than x
This problem is Kazakhstan NMO 2013 problem:see:link
first I want use Mathematical induction solve this problem.
since
$n=1$,then $a_{2}=\dfrac{1}{2}$.so
$$a_{2}=\dfrac{1}{2}<1=2a_{1}$$
Assume that $n$ have
$$a_{2n}<2a_{n}$$
then
$$a_{2n+2}=\dfrac{a_{n+1}}{2}+\dfrac{a_{[(n+1)/3]}}{3}+\cdots+\dfrac{a_{1}}{n+1}$$
then I can't.Thank you
|
First let's prove that $(a_n)$ is increasing for $n \geq 2$. We have $a_1=1,a_2=0.5, a_3=1/2+1/3$ so $a_3\geq a_2$. We have
$$ a_{n+1}-a_n = \sum_{k=2}^n \frac{a_{[(n+1)/k]}-a_{[n/k]}}{k}+1/(n+1).$$
Suppose that for every $2\leq i,j \leq n$ with $i<j$ we have $a_i\leq a_j$. Therefore in the above sum the only way that the term $a_{[(n+1)/k]}-a_{[n/k]}$ could be negative is to have $[(n+1)/k]=2$ and $[n/k]=1$ for some $k$. This happens precisely if $n < 2k\leq n+1$. Therefore, if $n+1$ is odd the sum is positive and $a_{n+1}>a_n$. If $n+1=2k$ is even we have
$$ a_{n+1}-a_n \geq \frac{1/2-1}{k}+\frac{1}{n+1} =-\frac{1}{2k}+\frac{1}{2k}=0 .$$
By induction $(a_n)$ is increasing for $n \geq 2$.
We have
$$ a_{2n} = \frac{a_n}{2}+\sum_{k=2}^n \frac{a_{[2n/(2k)]}}{2k}+\sum_{k=1}^{n-1}\frac{a_{[2n/(2k+1)]}}{2k+1}=$$
$$a_n+\sum_{k=1}^{n-1}\frac{a_{[2n/(2k+1)]}}{2k+1}\ \ \ (*)$$
When we look at $\sum_{k=1}^{n-1}\frac{a_{[2n/(2k+1)]}}{2k+1}$ we want to use the fact that $[2n/(2k+1)] \leq [2n/(2k)]$ and then use the monotonicity inequality proved above. We need to be careful in the case where $[2n/(2k+1)]=1$ and $[2n/(2k)]=2$, since then the monotonicity does not hold. If the last two equalities hold then
$$ 4k \leq 2n < 4k+2 $$
so we may have problems only when $n=2k$ is even. If $n$ is odd then
$$\sum_{k=1}^{n-1}\frac{a_{[2n/(2k+1)]}}{2k+1} \leq \sum_{k=1}^{n-1} \frac{a_{[n/k]}}{2k}=\frac{a_n}{2}+\sum_{k=2}^{n-1}\frac{a_{[n/k]}}{2k}=a_n -\frac{1}{2n}<a_n.$$
If $n=2p$ is even then we may apply the monotonicity inequality only for $k \neq p$ so
$$\sum_{k=1}^{n-1}\frac{a_{[2n/(2k+1)]}}{2k+1} \leq \sum_{k=1, k \neq p}^{n-1} \frac{a_{[n/k]}}{2k}+\frac{1}{2p+1}=$$ $$=\frac{a_n}{2}+\sum_{k=2}^{n-1} \frac{a_{[n/k]}}{2k}+\frac{1}{n+1}-\frac{0.5}{2p}=$$
$$=\frac{a_n}{2}+\frac{a_n}{2}-\frac{1}{2n}+\frac{1}{n+1}-\frac{1}{2n}=a_n+\frac{1}{n+1}-\frac{1}{n}<a_n$$
Combining this with $(*)$ we get that $a_{2n}<2a_n$.
|
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Proving whether a squared function has double pole at $z_0$ Let D be a domain in $\Bbb C$ and let $z_0 \in$ D. If a function $f:$ D \ {$z_0$} $\rightarrow \Bbb C$ has a simple pole at $z_0$, is it true that $g$ has a double pole at $z_0$, where $g(z) = [f(z)]^2 $ $\forall z \in$ D\ {$z_0$}
Definition of a pole: If there exists $m\in \Bbb Z^+$ such that $b_m \neq 0$ for all $n>m$ (i.e. the principal part has finitely many non-zero terms) so that
$$f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n + \frac{b_1}{z-z_0} + \frac{b_2}{(z-z_0)^2} +...+ \frac{b_m}{(z-z_0)^m}$$
Please help me check my ans:
$z_0$ is simple pole of $f(z) \Rightarrow$ there exist a function h which is analytic (i.e. holomorphic) on some domain D - {$z_0$} s.t. $f(z) = \displaystyle\frac{h(z)}{z-z_0}$, $\forall z \in$ D and $h(z_0) \neq 0$
$g(z) = [f(z)]^2 = \displaystyle\frac{[ h(z)]^2}{(z-z_0)^2}$, $z \in$ D - {$z_0$}
Since the only singular point in $g(z)$ is $z_0$, $g(z)$ is analytic everywhere except at $z_0$ $\Rightarrow$ $g(z)$ is also analytic in D - {$z_0$} $\Rightarrow [h(z)]^2$ is also analytic in D - {$z_0$}, $(h(z_0))^2 \neq 0$
Therefore, $g$ has a double pole at $z_0$
|
$\;z_0\;$ is a simple pole of $\;f(z)\;$ iff there's a holomorphic $\;h\;$ in some domain $\;D_0-\{z_0\}\;$ s.t.
$$f(z)=\frac{h(z)}{(z-z_0)}\;\;,\;\;\forall\,z\in D_0\;\;\;\text{and}\;\;\;h(z_0)\neq 0$$
From here, we have that
$$\forall\,z\in D_0-\{z_0\}\;\;,\;\;\;g(z):=f(z)^2=\frac{h^2(z)}{(z-z_0)^2}$$
Prove now $\;h(z)^2\;$ is holomorphic in $\;D_0-\{z_0\}\;$ and $\;h(z_0)^2\neq 0\;$ and you're done...
|
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|
Demonstrate that $\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} \simeq 1.7 \sqrt{n}$ As in the title, I know that
$\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} = \frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1} \simeq 1.7 \sqrt{n}$
Could you give some hint to prove it?
(should I look the series expansion of $\sqrt{n}$?)
Thank you anyway!
|
Hint Take the square root of the Wallis product...
|
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if $gcd(a,b)=gcd(a,b,c)$ then I need to prove that $ax+by=c$ has solution in $\mathbb Z$ if $gcd(a,b)=gcd(a,b,c)$ then I need to prove that $ax+by=c$ has solution in $\mathbb Z$ that is: $gcd(a,b)|c$
but how can I prove it with the given hypothesis?
|
First divide $a,b,c$ by $\gcd(a,b)$ so that we can assume that $a,b$ (and therefore $a,b,c$) are relatively prime.
Apply now Euclid's algorithm to $a,b$. This will give you equations:
$a=bq_1+r_1$, $b=q_2r_1+r_2$, ..., $r_n=q_{n+1}r_{n-1}+r_{n+1}$, where $r_{n+1}=1$ is the $\gcd(a,b)$.
Substituting backwards these equations we get $ax+by=1$. Multiply the whole equation by $c$ to get $a(cx)+b(cy)=c$.
|
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Counterexample for the normalizer being a normal subgroup Let $G$ be a group, and $H$ a subgroup.
$H$'s normalizer is defined: $N(H):=\{g\in G| gHg^{-1}=H \}$.
Prove $N(H)$ is a normal subgroup of G, or give counterexample.
Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.
Thans in advance for any assistance!
|
Let $G=\langle f,r : f^2 = r^8 = 1, rf =fr^7 \rangle$ be the dihedral group of order 16, and let $H=\langle f \rangle$. Then $N_G(H) = \langle f,r^4 \rangle$ and $N_G( N_G(H) ) = \langle f, r^2 \rangle$ and $N_G( N_G( N_G(H) ) ) = \langle f,r \rangle =G$.
In other words, $H$ is not normal, neither is its normalizer, but the normalizer of the normalizer is normal.
Using dihedral groups of order $2^n$ you can create arbitrarily long chains of non-normal normalizers that eventually end up being normal.
|
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Closed and Connected Subset of a Metric Space My English may not be perfect since I'm not a native speaker, so please do point out the grammar mistakes if there are any.
I've been reading Conway's "Functions of One Complex Variable", and encountered following exercise in the book (p. 17). $X$ here is a metric space and $d$ is the metric of it.
Show that if $F\subset X$ is closed and connected then for every pair of points $a,b$ in $F$ and each $\epsilon>0$ there are points $z_{0},z_{1},\dots,z_{n}$ in $F$ with $z_{0}=a, z_{n}=b$ and $d(z_{k-1},z_{k})<\epsilon$ for $1\leq k\leq n$. Is the hypothesis that $F$ be closed needed? If $F$ is a set which satisfies this property then $F$ is not necessarily connected, even if $F$ is closed. Give an example to illustrate this.
Now, I guess the right way to solve this problem is to first fix a point $a$ and consider the subset $S$ of $F$ definded by $$S=\{b\in X;\ for\ all\ \epsilon>0,\ there\ are\ points \ z_{0},z_{1},\dots,z_{n}\ in \ F\ with\ z_{0}=a, z_{n}=b\ and \ d(z_{k-1},z_{k})<\epsilon\ for \ 1\leq k\leq n\}$$and show that this set is nonempty, open and closed, but I'm currently stuck. I'm not sure where I should use the condition that $F$ is closed.
Moreover, the rest of the problem is also challenging, and I can't think of any possible way to get started.
Am I going in the right direction? Please give me a clear motivation on this problem and possibly the solution!! :D
|
Your approach seems correct. Your set contains the point $a$, so it's nonempty. For the openness pick any point and show that an open ball of radius $\varepsilon$ is contained in the set too. For the closedness pick a point in the closure and show there's a point from the set that is $\varepsilon$ close to it.
So your set is a nonempty clopen subset of $F$ and must be the whole $F$ because of connectedness. You didn't need that $F$ is closed in $X$ anywhere.
As for an example of a closed set which satisfies the property, but is not connected consider the graph of $f(x)=\dfrac1{x^2}$.
|
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Are the entries in matrix/vector product independent This question has been confusing me. Choose a random square matrix $M$ with $M_{i,j} \in \{1,-1\}$ so that $M_{i,j} = 1$ with probability $1/2$ and $M_{i,j} = -1$ with probability $1/2$. Say all the $M_{i,j}$ are i.i.d. Now select a random vector $v$ so that $v_i = 1$ with probability $1/2$ and $v_i = -1$ with probability $1/2$ and all the $v_i$ are i.i.d. We also have that $v$ and $M$ are independent. Let $y = Mv$.
Are the entries $y_i$ independent of each other?
|
Since the matrix rows of $M$ are i.i.d. and the column vector $v$ is i.i.d and $y_i$ is the $i$th row of $M$ times $v$, you won't get any dependence among $y_i$, since there is no dependence among the rows of $M$.
So the answer is yes!
|
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Manipulating identities I'm having some trouble deriving certain identities. If
$$S(z) = \prod_{i=1}^n (z-z_i)$$
then how can I write
$$\frac{1}{S(z)}\frac{d^2S}{dz^2} = \sum_{i=1}^n\frac{1}{z-z_i}\sum_{j\neq i}^n\frac{2}{z_i-z_j}$$
and
$$ \frac{1}{S(z)}\frac{dS}{dz}= \sum_{i=1}^n\frac{1}{z-z_i} $$
Truth be told, I'm having some trouble writing the derivative of $S(z)$ in a neat, compact form.
|
Hint for the second:
$${\frac {{
\frac {d}{dz}}S \left( z \right) }{S \left( z \right) }}={\frac {d}{dz}}\ln \left( S \left( z \right) \right) $$
Hint A for the first:
$${\frac {{\frac {d^{2}}{d{z}^{2}}}S \left( z \right) }{S \left( z
\right) }}={\frac {d^{2}}{d{z}^{2}}}\ln \left( S \left( z \right)
\right) + \left( {\frac {d}{dz}}\ln \left( S \left( z \right)
\right) \right) ^{2}$$
Hint B for the first:
$${\frac {1}{ \left( z-z_{{i}} \right) \left( z-z_{{j}} \right) }}={
\frac {1}{ \left( z_{{i}}-z_{{j}} \right) \left( z-z_{{i}} \right) }
}+{\frac {1}{ \left( z_{{j}}-z_{{i}} \right) \left( z-z_{{j}}
\right) }}$$
Where the identities hold for any differentiable function (proof: chain rule).
|
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Probability exercise Bernoulli. Probability random signals. Im late I have no idea to start and this is for tomorrow. I was on training and have no break to do this work. I do this.You are an Internet savvy and enjoy watching video clips of your favorite artists. You normally download video clips from the Web site http://www.coolvideos.com. The probability that you can connect to this site in any one attempt is $p$. Define $$X as the number of successes and $Y$ as the number of failures in $n$ attempts.
(a) Find the probability mass function (PMF) of $Z=X−Y$.
(b) Find $E[Z]$ (in terms of $n$ and $p$).
(c) Find Var[Z] (in terms of n and p).
A) This is what I attempted
Pr [z=k]= Pr [2x-n=k]= Pr [x=((n+k)÷2)]=?
|
You have started essentially correctly. We have $X-Y=X-(n-X)=2X-n$. So
$$\Pr(Z=k)=\Pr(2X-n=k)=\Pr\left(X=\frac{n+k}{2}\right)=\binom{n}{(n+k)/2}p^{(n+k)/2}(1-p)^{(n-k)/2}.$$
This makes sense only when $-n\le k\le n$ and $n$ and $k$ are of the same parity (both even or both odd).
The rest of the problems are more mechanical.
$$E(Z)=E(2X-n)=2E(X)-n=2np-n.$$
$$\text{Var}(X)=2^2\text{Var}(X)=4np(1-p).$$
|
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What type of triangle satisfies the equation $\cos(A)-\cos(B)+\sin(C)=0$? A triangle with angle $A,B,C$ satisfies the equation $\cos(A)-\cos(B)+\sin(C)=0$.
What type of triangle is this? Regular, acute, right, obtuse etc.
I tried using sine and cosine rule, but no result.
|
Using Prosthaphaeresis Formulas,
$$\sin C=\cos B-\cos A=2\sin\dfrac{A+B}2\sin\dfrac{A-B}2$$
Now, $\displaystyle \sin\dfrac{A+B}2=\sin\dfrac{\pi-C}2=\cos\dfrac C2$
Using double angle formula the given relation becomes, $$2\sin\dfrac C2\cos\dfrac C2=2\cos\dfrac C2\sin\dfrac{A-B}2$$
which implies
$(1)$ either $\displaystyle\cos\dfrac C2=0\iff\frac C2=(2n+1)\frac\pi2\iff C=(2n+1)\pi$ where $n$ is any integer
But it is impossible as $0<C<\pi$
$(2)$ or $\displaystyle\sin\dfrac C2 =\sin\dfrac{A-B}2$
$\displaystyle\implies \dfrac C2=m\pi+(-1)^m\dfrac{A-B}2\iff C=2m\pi+(-1)^m(A-B)$ where $m$ is any integer
If $m$ is even $=2r$(say), $\displaystyle C=4r\pi+(A-B)\iff B+C=4r\pi+A$
$\displaystyle\implies \pi-A=4r\pi+A\iff A=\dfrac{(1-4r)\pi}2$
As $\displaystyle0<A<\pi, 0<\dfrac{(1-4r)\pi}2<\pi\iff 0<1-4r<2\iff0>4r-1>-2$
$\displaystyle\implies r=0\implies A=\dfrac{(1-4\cdot0)\pi}2$
Similarly check for odd $m=2r+1$(say)
|
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Can you take off a sweater while wearing headphones? This seems like a graph theory problem, but I'm not sure how to approach it.
To clarify potential ambiguities, let's set up the situation.
You are wearing a sweater (with one arm through each sleeve). You are also wearing a pair of earphones, which are connected to your stationary electronic device. Each ear is connected by one wire to the electronic device.
Can you take off your jacket without taking off your earphones?
|
Theoretically, yes, if your electronic device is in your pocket. While it seemingly forms a closed loop, it doesn't actually.
Thus, if you can pull your sweater from over your head and follow the wire into your pocket and fit the entire sweater into your picket to leave the electronic device stationary and exit the 'loop', then you're done.
Alternatively, taking the electronic device out of your pocket and letting it hang makes doing it 'reasonable'.
|
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Ideal of a polynomial ring, and an isomorphism between $R[x]/I$ and $R$ Let $R$ be a ring. $I\subset R[x]$ is the ideal of all elements with a zero constant term. Show that $I$ is an ideal, and show that $R[x]/I\cong R$.
Attempt: $R[x]=\{a_0+a_1x+...+a_nx^n:a_i\in R\}$. Then $I=\{a_1x+...+a_nx^n : a_i \in R\}$. Consider $f\in R[x]$ and $p\in I$. We want $fp\in I$. So, $fp=a_0a_1x+a_0a_2x^2+...+a_n^2x^{2n}\in I$ so $I$ is an ideal.
Now, for the second part, which isn't clear to me fully:
$R[x]/I$ will have elements of the form $a_i$ which are elements of the ring, since everything but the coefficients will cancel in the quotient. Is there a rigorous way to show this by factoring, or some other way? Also, I think I need to find an isomorphism $\phi:R[x]/I\rightarrow R$. For context, I am roughly 1 week into studying rings.
|
For the first part you also need to show that $I$ is closed under addition.
For the second part, the intuition is that taking the quotient by $I$ removes all $x$ terms from $R[x]$ and results in $R$. To rigorously show that $R[x]/I\cong R$ I recommend constructing a surjective map $R[x]\to R$ whose kernel is $I$, which will do the trick by the isomorphism theorems (let me know if you haven't learned about that yet.) The map you come up with should be very simple.
|
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Continuity and lower bound I was initially presented with the following problem:
Show that $x < \tan{x}$ for all $x \in (0, \pi/2)$.
My solution involved setting $f(x) = \tan{x}-x$ and then showing that $f(x)$ is strictly increasing on $(0, \pi/2)$. However, I need the following lemma proved to complete (I know that there is an alternative solution with the Mean Value Theorem, but I'm interested in whether or not my proof for the lemma is correct):
Lemma: If $f$ is continuous on $[a, b)$ and increasing on $(a,b)$, $f(a) \le f(x)$ for all $x \in (a, b)$.
My proof:
EDIT: I realize this proof is incorrect. Is the lemma correct? What other proof can I use?
Assume that there exists a $x_0 \in (a, b)$ such that $f(a) > f(x_0)$.
Let $\delta = x_0 - a$. Note that $\delta > 0$ since $x_0 > a$.
We may construct a sequence $(x_n)$ such that $x_n = a + \frac{\delta}{n}$.
Since $x_n < x_0$ for all $n \in \mathbb{N}$ and $f$ is increasing on $(a, b)$, $f(x_n) < f(x_0) < f(a)$ for all $n \in \mathbb{N}$.
Notice that
$\lim{x_n} = 0$
However, since $f$ is continous on $[a, b)$,
$\lim f(x_n) = f(a)$
which is a contradiction since $f(x_n) < f(a)$ for all $n \in \mathbb{N}$.
Is that a rigorous proof of the lemma presented?
Thanks
|
First let me give an insight to the proof. I am trying to prove that if there is an $x_0$ such that $f(a)>f(x_0)$ then under those circumstances $f$ cannot be continuous at $a$.
As $f$ is increasing on $\left(a,b\right)$, hence note that for every $x_1,x_0\in (a,b)$ and $x_1 < x_0 \implies f\left(x_1\right)\le f\left(x_0\right)$.
Now if there is an $x_0$ such that $f(x_0)<f(a)$ using the above idea for every $x<a\implies f(x)\le f(x_0)<f(a)$$\implies 0<f(a)-f(x_0)\le f(a) - f(x) \quad\left(\color{blue}{\star}\right)$
As $f$ is continuous at $a$ for $\large \varepsilon =\frac{f(a)-f(x_0)}{2}$ there exists a $\delta$ such that $\large x<a+\delta\implies f(a)-f(x)<\frac{f(a)-f(x_0)}{2}$ which is a contradiction to $\left(\color{blue}{\star}\right)$.
|
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Use convergence theorem to prove the limit Suppose $f:\mathbb{R}\to\mathbb{R}$ is $L^1$ and it is continuous at the origin. Let $g_n(x)=\frac{1}{1+x^2}f(\frac{x}{\sqrt{n}})$. The problem is to prove $$\lim_n\int_0^\infty g_ndx=\frac{\pi}{2}f(0)$$ The method I tried so far is using the dominated convergence theorem but failed finding a uniform upper bound for $g_n$. Any suggestions?
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EDIT: the fact that the sequence is unbounded for $y = 0$ is not a problem since we don't care about sets of measure $0$.
Clearly $$\lim_{n \to \infty} \frac{1}{1 + x^2}f\Big(\frac{x}{\sqrt{n}}\Big) = \frac{1}{1 + x^2}f(0).$$
Moreover we can find $N$ such that if $n \ge N$ then $\Big|f\Big(\frac{x}{\sqrt{n}}\Big) - f(0)\Big| \le 1$, then in particular $$\Big|f\Big(\frac{x}{\sqrt{n}}\Big)\Big| \le 1 + |f(0)|.$$
Then (forgetting about the first $N - 1$ terms of the sequence) we have $$\Big|\frac{1}{1 + x^2}f\Big(\frac{x}{\sqrt{n}}\Big) - \frac{1}{1 + x^2}f(0)\Big| \le \frac{1}{1 + x^2}(1 + |f(0)| + |f(0)|) = \frac{1 + 2|f(0)|}{1 + x^2} \in L^{1}.$$
Finally pass to the limit under the integral sign to get the result ;)
Original post (not of any use anymore!):
(Hint: $\frac{x}{\sqrt{n}} = y$, then $dy = \frac{dx}{\sqrt{n}}$. Substituting we get: $$\int_0^{\infty}\frac{1}{1 + x^2}f\Big(\frac{x}{\sqrt{n}}\Big)\ dx = \int_0^{\infty}\frac{\sqrt{n}}{1 + ny^2}f(y)\ dy.$$
The sequence in the second integral looks like something that can be bounded by an $L^1$ function. I hope it helps a little bit!)
|
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Polar coordinates for $xz$-plane: $z = r\sin\theta$ ? [Stewart P1091 16.7.25]
$1.$ The unit disk is projected onto the xz-plane, so shouldn’t $x = 1\cos \theta$ and $\color{red}{z = 1 \sin \theta} $?
User Semsem below kindly identified the problem: The normal to the disk is on the direction $-j$ so we have to reverse the orientation as follows.
$2.$ Would someone please explain why the orientation must be reversed? By "reverse", does Semsem mean the following, that thee $xz$-plane should be viewed in the direction of the green arrow (instead in that of the red arrow, which was my problem)?
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I will show you how the double integral works out no matter what you select for $z$.
Consider $$\begin{align}I &= \int_0^{2\pi}\!\int_0^1 \!\! r^3 + 2r^3f^2(\theta)\, \mathrm{dr} \, \mathrm{d}\theta \\ &= \int_0^{2\pi}\!\dfrac{1}{4} + \dfrac{1}{2}f^2(\theta)\;\mathrm{d}\theta\end{align}$$
$f(\theta)$ is either $\cos(\theta)$ or $\sin(\theta)$ so $f^2(\theta) = \dfrac{1\pm\cos(2\theta)}{2}$.
$$\begin{align} I &= \int_0^{2\pi}\!\dfrac{1}{4} + \dfrac{1}{2}f^2(\theta)\;\mathrm{d}\theta \\&= \dfrac{\pi}{2} + \dfrac{1}{2}\int_0^{2\pi}\!\dfrac{1\pm\cos(2\theta)}{2}\mathrm{d}\theta \\ &= \dfrac{\pi}{2} + \dfrac{1}{4}\int_0^{2\pi}\!1\pm\cos(2\theta)\;\mathrm{d}\theta \\&= \dfrac{\pi}{2} + \dfrac{1}{4}\int_0^{2\pi}\!1\;\mathrm{d}\theta\pm\int_0^{2\pi}\!\cos(2\theta)\;\mathrm{d}\theta \\ &=\pi\pm \int_0^{2\pi}\!\cos(2\theta) \; \mathrm{d}\theta \\&= \pi\end{align}$$
So you can see that it does work out whether or not you set $z = \sin(\theta)$ or $z = \cos(\theta)$.
|
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A combinatorics question on a $n \times n$ grid In the book 'Foundations of Data Science' by Hopcroft and Kannan, they have the following exercise (Ex. 5.46):
Let G be a $n \times n$ lattice and let $S$ be a subset of $G$ with cardinality at most $\frac{n^2}2$. Define $$N = \{(i,j) \in S \,\, | \text{all elements in row $i$ and all elements in column $j$ belong to $S$}\}$$Show that
$$\displaystyle |N| \leq \dfrac{|S|}2$$
My failed attempt:
I tried to fix $k$ elements in $N$ and calculate the minimum number of elements that should be added to $S$ so that the membership criteria of $N$ is satisfied. I added an element, one at a time to $N$, and kept track of how $|S|$ increases.
The first element in $N$, immediately adds $2n-1$ elements to $S$. The second element adds at least $n-1$ elements and the third element adds at least $n-2$ elements. Sadly, I ran into a problem. As an example consider, adding row 1 and column 1 to $S$, that is, adding $(1,1)$ to $N$. Now subsequently add $(1,2)$ and $(2,1)$ to $N$ and therefore add $n-1 + n-2$ more elements in $S$. But observe how $(2,2)$ can now be freely added to $N$ without adding any elements in $S$. This shows that picking elements for $S$ affects the choice of elements for $N$ and it is hard to keep track. :(
Is there a simple solution to this problem? Or is there a work around for my failed attempt?
Thanks
|
Here's an entirely different approach (although somewhat related to Alex' solution) that can give a much stronger bound
$$
\sqrt{|N|}+\sqrt{n^2-|S|}\le n
$$
from which the result follows.
If $S$ contains $p$ full rows and $q$ full columns, then $|N=pq|$. However, with $p$ rows and $q$ columns full, if $k$ of the remaining $(n-p)(n-q)$ grid points (from the rows and columns other than these $p$ and $q$ full ones) are in $S$, that makes
$$
|S|=n^2-(n-p)(n-q)+k\ge np+nq-pq.
$$
Now, we may use $\frac{x+y}{2}\ge\sqrt{xy}$ for any $x,y\ge0$ to get
$$
n^2-|S|\le(n-p)(n-q)\le\left(n-\frac{p+q}{2}\right)^2.
$$
On the other hand,
$$
|N|=pq\le\left(\frac{p+q}{2}\right)^2.
$$
Taking square roots and adding these two inequalities, we get
$$
\sqrt{|N|}+\sqrt{n^2-|S|}\le n,
$$
which, entering $s=|S|/n^2$, makes
$$
\frac{|N|}{n^2}\le \left(1-\sqrt{1-s}\right)^2
=\left(\frac{s}{1+\sqrt{1-s}}\right)^2
$$
which is $\le s/2$ whenever $s\le8/9$. If $s\le 1/2$, then
$|N|/|S|\le3-2\sqrt{2}=0.17157\ldots$
follows.
|
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cartesian product $A^2 = A$, possible? Do there exist non-empty sets $A$ such that $A\times A = A$?
$A\times A = A$ looks a little strange to me, since $A\times A$ seems somehow more complicated than $A$, hence it is unlike that they are equal, but then, I cannot think of any reason why there should not be such a (non-empty) set.
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Under the usual axioms of set theory, the answer is negative. The reason is that one of these axioms, the axiom of foundation, allows us to assign to each set a rank $\alpha$ (an ordinal number) that indicates at what stage of the transfinite construction of the universe the set appeared. Ordinals have the property that any non-empty collection of them has a first element, and if a set $x$ belongs to a set $y$, then the rank of $x$ is strictly smaller than the rank of $y$ (intuitively, a set cannot be constructed until all its elements are present).
Now, if $A$ is non-empty, let $a\in A$ be an element of smallest rank. Then we cannot have $a=(b,c)$ for some $b,c\in A$, since $(b,c)=\{\{b\},\{b,c\}\}$, so both $b$ and $c$ would have to have rank smaller than the rank of $a$, contradicting its minimality. This shows that $A$ cannot be a subset of $A\times A$.
The argument works for just about any reasonable definition of ordered pair, not just the usual one.
Note, however, that the other containment $A\times A\subset A$ is possible, and can be easily arranged by a recursive construction: Start with any set $A_0$. Let $A_1=A_0\cup(A_0\times A_0)$, $A_2=A_1\cup(A_1\times A_1)$, etc. Then $A_\omega=\bigcup_n A_n$ satisfies that $A_\omega\times A_\omega\subset A_\omega$.
(In fact, in the transfinite construction of the universe of sets, for any limit stage $\alpha$ we have that $V_\alpha\times V_\alpha \subset V_\alpha$.)
On the other hand, it is consistent with the negation of the axiom of foundation that there are sets $\Omega$ such that $\Omega=\{\Omega\}$, and this implies that $\Omega=\Omega\times\Omega$.
|
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Demonstrate the following .... The diagonals of a trapezoid (ABCD where AD parallel to BC ) divide it into four triangles all having one vertice in O. Knowing that the area of triangle $\triangle AOD$ is $A_1$ and of $\triangle BOC$ is $A_2$, demonstrate that the area of the trapezoid is
$$(\sqrt{A_1} + \sqrt{A_2})^2$$
Thank you very much if you are now looking at my problem!
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Note that $\triangle ABD$ and $\triangle ADC$ have the same area (same base, equal heights). They have $\triangle AOD$ in common, so the areas of the two parts where they differ are equal, That is, $\triangle AOB$ has the same area as $\triangle DOC$.
Call the area of each by the name $x$.
Look at $\triangle ABC$. It is divided into two parts, with bases $AO$ and $OC$, and the same heights. Thus
$$\frac{x}{A_2}=\frac{OA}{OC}.\tag{1}$$
A similar calculation using $\triangle ACD$ shows that
$$\frac{x}{A_1}=\frac{OC}{OA}.\tag{2}$$
Multiply the left sides of (1) and (2), and the right sides. We get
$$\frac{x^2}{A_1A_2}=1,$$
which tells us that $x=\sqrt{A_1A_2}$ and we are finished.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determining if a number is an nth root I am working on a proof that depends on if an adversary can determine if a number is an $nth$ power for some large prime $p$. My intuition tells me that for a sufficiently large value of $n$ this is impossible. However I am unaware of any theorems that state this. Could you point me in the right direction?
Thanks
|
Here is a simple way to test whether $r$ is an $n$th power modulo $p$.
Given $n,p,r$, compute the least common multiple $s=nt$ of $n$ and $p-1$, then compute $r^t$. $r$ is an $n$th power modulo $p$ if and only if $r^t\equiv1\pmod p$.
Proof. Let $g$ be a primitive root mod $p$. Then $r=g^u$ for some $u$, and $r^t=g^{tu}\equiv1\pmod p$ if and only if $p-1$ divides $tu$, which makes $u$ a multiple of $n$.
|
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|
Calculating time to arbitrary points of distance with initial velocity and non-uniform acceleration I've been trying to tackle this question for a while now, but I'm afraid it's not going anywhere without some outside help.
So let's say we have some body falling towards a planet without any atmosphere, and let's assume that the initial velocity is not equal to zero. Given these conditions I'm trying to derive a formula which would allow me to determine the exact time of impact or, in fact, time to any given point along the trajectory of the falling body .
Let's define $r_0$ as initial separation; $r$ as a destination point on the trajectory of the falling body(surface, for example); $v_0$ as initial velocity; $v$ as instantaneous velocity at point $r$; $t_0$ as initial time; and $T$ as time at point $r$.
Now acceleration as a function of distance is given by:
$$\frac{d^2r}{dt^2} = \frac{GM}{r^2}$$
Using the chain rule we arrive at:
$$\ a(r) = \frac{dv}{dt}= \frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr}$$
Therefore
$$a(r)dr= vdv$$
Equivalently:
$$\frac{GM}{r^2}dr=vdv$$
Now in order to get $v$:
$$v = \frac{dr}{dt} = \sqrt{\frac{2G(m_1 + m_2)}{r} - \frac{2G(m_1 + m_2)}{r_0} + v_0^2}$$
$$v = \frac{dr}{dt} = \sqrt{ \frac{2 G r_0 (m_1 + m_2) - 2 G r (m_1 + m_2)+ v_0^2r_0 r} {r\ r_0}}$$
$dt$ is calculated as:
$$dt=\sqrt\frac{{r_0 \ r} \ dr} {{ 2 G r_0 (m_1 + m_2) - 2 G r (m_1 + m_2)+(v_0^2rr_0)}}$$
Now I have absolutely no idea how to calculate $dr$, but I hear it has something to do with gravitational potential energy. Some guidance on this matter would be much appreciated. I also have some doubts regarding the next step of my strategy. Let's say I manage to find $dr$, and what then? Intuitively, I feel this is a right path to take but in fact I have no idea how to convert $dr$ and $dt$ into $T$.
|
Let me try to help you, but I cannot assure you that my approach is correct. Let $m$ be the mass of falling body and $M$ be the mass of planet. First, we start from the relation of the Newton's second law and Newton's law of universal gravitation.
$$
\begin{align}
ma&=-\frac{GMm}{r^2}\\
\frac{dv}{dt}&=-\frac{GM}{r^2}\\
\frac{dv}{dr}\cdot\frac{dr}{dt}&=-\frac{GM}{r^2}\\
v\ dv&=-\frac{GM}{r^2}\ dr\\
\int_{v_o}^vv\ dv&=-\int_{r_o}^r\frac{GM}{r^2}\ dr\\
\frac{v^2-v_o^2}{2}&=\frac{GM}{r}-\frac{GM}{r_o}\\
v^2&=\frac{2GM}{r}-\frac{2GM}{r_o}+v_o^2\\
v&=\sqrt{\frac{2GM}{r}+v_o^2-\frac{2GM}{r_o}}\\
\frac{dr}{dt}&=\sqrt{\frac{2GM}{r}+v_o^2-\frac{2GM}{r_o}}\\
dt&=\frac{dr}{\sqrt{\frac{2GM}{r}+v_o^2-\frac{2GM}{r_o}}}.
\end{align}
$$
Now, let $a=v_o^2-\dfrac{2GM}{r_o}$ and $b=2GM$, then
$$
\begin{align}
dt&=\frac{dr}{\sqrt{\frac{b}{r}+a}}\\
dt&=\sqrt{\frac{r}{ar+b}}\ dr\\
\int_0^T\ dt&=\int_{r_o}^r\sqrt{\frac{r}{ar+b}}\ dr\\
T&=\int_{r_o}^r\sqrt{\frac{r}{ar+b}}\ dr\\
\end{align}
$$
Let
$$
x^2=\frac{r}{ar+b}\;\quad\Rightarrow\;\quad r= \frac{bx^2}{1-ax^2}\;\quad\Rightarrow\;\quad dr=\frac{2bx}{(1-ax^2)^2}\ dx,
$$
then
$$
T=\int_{r=r_o}^r\frac{2bx^2}{(1-ax^2)^2}\ dx.
$$
The last form integral can be evaluated either using partial fraction expansion, trigonometry substitution, or simply looking the following list of integrals.
$$\\$$
$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
|
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|
Doubly infinite sequence limit I have a 2-indexed sequence $F^n_m$ where $n,m$ are natural numbers and I am concerned about the behavior as $n\to\infty$ and/or $m\to\infty$. The sequence is expressed as
$$F^n_m=\prod_{k=1}^na_{mk}$$
For each finite $n$ and $k$, as $m \to \infty, a_{mk}$ converges to a finite nonzero limit, say $a_{\infty k}$ and therefore
$$F^n_\infty=\lim_{m \to \infty}F_m^n=\prod_{k=1}^na_{\infty k}$$
exists and is finite as a finite product of converging sequences. Now if we let $n \to \infty$ I have shown that
$$F=\lim_{n \to \infty} F_{\infty}^{n}$$
exists and is expressible as the corresponding infinite product (which is by the way the product representation of Riemann's zeta function). My question is what happens when we approach the limit by letting $n \to \infty$ first and then $m \to \infty$. In fact I haven't found a closed form for
$$F_m^\infty=\lim_{n \to \infty}F_m^n=\prod_{k=1}^{\infty}a_{mk}$$
and I don't know if it even converges. Are there any conditions under which we can do that, then take the limit as $m \to \infty$ and come up with the same answer?
And what about if we approach $\infty$ "diagonally"? By that I mean given a strictly increasing sequence $p(m)$ of integers and consider
$$\lim_{m \to \infty} F_m^{p(m)}$$
Are we going to get a value that depends on $p$ or the same value for all such sequences? And under what conditions? There is a theorem that states that the infinite product
$$\prod_{k=1}^\infty(1+b_k)$$
converges or diverges according to the sum
$$\sum_{k=1}^\infty b_k$$
but it didn't quite do it...
I am looking for rigorous as well as not-so-rigorous approach, anything really!
|
Looks like an iterated limit, right?
There has not been given an accepted (i.e. satisfactory) answer to the following question:
Commutativity of iterated limits
Please read the question, the answer and the comments in there and make up your mind.
Anyway, it is not a standard result in common mathematics - don't ask me why.
In your case the Commutativity of iterated limits Theorem (see above reference) says:
$$
\lim_{n \to \infty} \left[ \lim_{m \to \infty} F^n_m=\prod_{k=1}^na_{mk} \right]
= \lim_{m \to \infty} \left[ \lim_{n \to \infty} F^n_m=\prod_{k=1}^na_{mk} \right]
$$
|
{
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|
Evaluating: $ \lim\limits_{x\to 1^{-}}\left[\ln\left(1-x \over 1 + x\right) - \ln\left(1 - x^{2}\right)\right] $ I would like to evaluate the following limit:
$$
\lim_{x\to 1^{-}}\left[%
\ln\left(1-x \over 1 + x\right) - \ln\left(1 - x^{2}\right)\right]
$$
I put this into a limit calculator and the calculator gave as answer $-\log\left(4\right)$. Why ?.
Should the answer not be?:
$$
\lim_{x \to 1^{-}}\ln\left(1 - x \over 1+x\right)
-\lim_{x \to 1^{-}}\ln\left(1 - x^{2}\right)
=-\infty - \left(-\infty\right) = -\infty + \infty$$
My second, and to me more important question, what is the value of $$-\infty + \infty$$
Thanks for your help.
|
Your expression $-\infty + \infty$ is an indeterminate form, much like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. So, on its own, it has no meaning.
In your specific case, using the property of logs that $\ln(AB) = \ln(A) + \ln(B)$ will likely be helpful here.
|
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|
Functions of the form $\int_a^x f(t) dt$ that are commonly used. I am a graduate student and teaching assistant, and I am teaching Calc 1 for the first time. In a few weeks I will be covering the Fundamental Theorem of Calculus. I'm using James Stewart's Calculus textbook, and I was hoping to give students several "real world" examples of functions of the form $g(x)=\int_a^x f(t) dt$ to make the first part of the FTC more accessible. Stewart gives one example, the Fresnel function
$$S(x)=\int_0^x \sin (\pi t^2/2)$$
which is (apparently) used in the theory of the diffraction of light waves. But I was hoping for more examples from physics, chemistry, etc. Any thoughts or ideas?
|
In cartography in the year 1599 the following problem arose. Mercator wanted a map of the world on which compass bearings on the earth correspond to those on the map. E.g. $13^\circ$ east of north on the surface of the earth corresponds to $13^\circ$ counterclockwise from straight up on the map, at every point. Going through a bit of geometry, it is found that at latitude $\theta$, the ratio of distance on the map to distance on the earth has to be proportional to $\sec\theta$. That implies that the location of points on the map at latitude $\theta$ should be at the following distance from the equator:
$$
\int_0^\theta \sec\alpha\,d\alpha.
$$
Evaluating this integral remained a famous unsolved problem for (I think?) more than half a century.
|
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|
Connections between prime numbers and geometry This might be a little open-ended, but I was wondering: are there any natural connections between geometry and the prime numbers? Put differently, are there any specific topics in either field which might entertain relatively close connections?
PS: feel free to interpret the term natural in a broad sense; I only included it to avoid answers along the lines of "take [fact about the primes] $\to$ [string of connections between various areas of mathematics] $\to$ [geometry!]"
|
A simple example: number p>2 is prime iff any equiangular p-gon with rational side lengths is regular, see, e.g., http://www.cut-the-knot.org/Outline/Geometry/EquiangularP-gon.shtml
|
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|
Prove that if $f(x) = Ax^2 + Bx + C$ is an integer whenever $x$ is an integer, then $2A$, $A+B$ and $C$ are also integers.
Prove that if $f(x) = Ax^2 + Bx + C$ is an integer whenever $x$ is an integer, then $2A$, $A+B$ and $C$ are also integers.
I've tried a lot to do it, but can't get it exactly right.
|
We have
$f(x) = Ax^2 + Bx + C; \tag{1}$
set $x = 0$, an integer. Then we have
$f(0) = C, \tag{2}$
so $C$ is an integer. Set $x = 1$; then
$f(1) - C = A + B, \tag{3}$
showing $A + B$ is an integer. Set $x = -1$; then
$f(-1) - C = A - B \tag{4}$
is also an integer. Since
$2A = (A + B) + (A - B), \tag{4}$
we see that $2A$ is an integer as well. Noting
$2B = (A + B) - (A - B) \tag{5}$
we also see $2B$ is an integer, "for free". It appears to me that lab bhattacharjee was onto the right idea, though I did work this out for myself, concurrently as it were. Cute, I must say . . .
Hope this helps! Cheerio,
and as always,
Fiat Lux!!!
|
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|
a spider has 1 sock and 1 shoe for each leg. then find out the the total possibilities. a spider has one sock and one shoe for each of its 8 legs.in how many different orders can the spider put on its shocks and shoes; assuming that on each leg ;the shock must be put on before the shoe? i have tried the problem in following processes-
*
*we consider in k th step it has completed covering exactly its k legs with socks.now we find the total possibilities in (k+1) th step.
*in (k+1) th step either one of the remaining (8-k) legs will be covered by shocks or one of the (k-r) legs will be covered by shoe. where r denote the number of legs among the k legs which were already covered by shoes. so 0$\leq$r$\leq$k and 1$\leq$k$\leq$8
*so in the 1st case there are (8-k) possibilities and in the 2nd case ..........if 1 leg is covered with shoe then (k-1) possibilities; if 2 legs are covered with shoes then (k-2) possibilities; if no leg then k possibilities. so total possibilities is k.(k-1).(k-2)(k-3).........1=k!
*in (k+1)th step the total possibilities is (8-k).k!.[in each step it puts on either a shock or a shoe]
*so the solution is $\Sigma$(8-k).k! ;k=1 to 16
*if i am wrong in any step then show me please. if i am not then give an alternate solution of this.
|
Let the act of putting on a sock be $1,2,3,4,5,6,7,8$, each corresponding to a leg. Let the act of putting on a shoe be $A,B,C,D,E,F,G,H$, each corresponding to a leg as well.
This problem is now bijective to counting the number of strings such that $A,B,C,D,E,F,G,H$ comes after $1,2,3,4,5,6,7, 8$ respectively.
We start with $1$ leg. Then there is only $1$ way: $1A$. Now we want to add the next leg. There are $3 \times 4$ ways to add $2$ and $B$ into the string, without any order. Why? Because there are $3$ ways to insert $2$ into $1A$, and $4$ ways to insert $B$ into each of the the resultant string.
But there are as many ways to insert $2$ and $B$ such that $2$ comes after $B$ as there are ways of insertion such that $B$ comes after $2$. So there are $\frac{1}{2} \times 3 \times 4$ ways such that $B$ comes after $2$, i.e. second shoe comes after second sock.
Generalize this to get a recursion: Let $a_n$ be the number of ways for $n$ legs. Then,
$$a_{n + 1} = \frac{1}{2}\times a_n \times (2n + 1)\times(2n + 2)$$
|
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