In this question ec numbers are introduced, formed by the concatenation of two consecutive Mersenne numbers ($157$ for example is denoted by $ec(4)$).
The ec prime $ec(7)=12763$ divides ec numbers $ec(7717)$, $ec(14259)$, $ec(15906)$,...
Does $ec(7)$ divide an infinite number of ec-numbers?
Is $255127$ the largest ec prime dividing at least one ec number besides itself?
Let $D_n$ be the number of derangements of n objects. Find a combinatorial proof of the following identity:
$n! = D_n + \dbinom{n}{1} \cdot D_{n-1}+ \dbinom{n}{2} \cdot D_{n-2} + \cdots + \dbinom{n}{n-1} \cdot D_1 + \dbinom{n}{n} \cdot D_{0}$
This question has been somewhat asked and answered before here
Finding a combinatorial proof of this identity: $n!=\sum_{i=0}^n \binom{n}{n-i}D_i$
However, I need to come up with a combinatorial proof. I don't think the answer provided in the linked page is a combinatorial proof. When I learned about combinatorial proofs I was told they need to be more 'real life.' For example, in a committee of $n$ people, a subcommittee and a president must be chosen...
How can I write a true combinatorial proof for this identity?
Answer
All permutations of $N$ objects have between $0$ and $N$ objects in their original position. Then count each case, and add them all up:
All of these cases have the same form: "Pick $n$ objects out of $N$" has $_NC_n$ choices. "Scramble $m$ objects" has $D_m$ choices.
If in a normed space an converges to $a$, prove that $||a_n||$ converges to $||a||$
I have seen this proof: for all $\varepsilon>0$ there exists and $N$ s.t. $n>N$ implies
that the $||a_n-a||<\varepsilon$
by the reverse triangle inequality for the same $N \varepsilon > ||an-a|| > |\ ||an||-||a||\ |$
and therefore it converges.
But this is the absolute value of the difference of norms. Don't you need the norm of the difference of norms to be less than $\varepsilon$?
As in
$||\ ||an|| - ||a||\ || < \varepsilon$
not
$|\ ||an|| - ||a||\ | < \varepsilon$
Or am I confused
Thanks!
Is there a way to evaluate this limit:
$$\lim_{x \to 0} \frac{\sin(e^{\tan^2 x} - 1)}{\cos^{\frac35}(x) - \cos(x)}$$
without using de l'Hôpital rule and series expansion?
Thank you,
Answer
You should know the following limits:
$$\begin{split}\lim_{y\to 0} \frac{\sin y}{y} &= 1\\ \lim_{y\to 0} \frac{e^y-1}{y} &= 1\\ \lim_{y\to 0} \frac{\tan y}{ y} &= 1\\ \lim_{y\to 0} \frac{(1+y)^\theta -1}{y} &= \theta \qquad \text{(}\forall \theta \in \mathbb{R}\text{)}\\
\lim_{y\to 0} \frac{1-\cos y}{y^2} &= \frac{1}{2}\end{split}$$
which can be proved using only elementary Calculus tools (i.e. without any Differential Calculus technique).
These five limits are usually written as asymptotic relations in the following manner:
$$\tag{1} \sin y \approx y$$
$$\tag{2} e^y-1 \approx y$$
$$\tag{3} \tan y \approx y$$
$$\tag{4} (1+y)^\theta -1 \approx \theta\ y$$
$$\tag{5} 1-\cos y \approx \frac{1}{2}\ y^2$$
as $y\to 0$. Using asymptotics (1) - (5) you find:
$$\begin{split} \sin(e^{\tan^ 2 x} - 1) &\approx e^{\tan^2 x}-1 &\quad \text{by (1)}\\
&\approx \tan^2 x &\quad \text{by (2)}\\
&\approx x^2 &\quad \text{by (3)}\end{split}$$
$$\begin{split}\cos^{3/5}(x) - \cos(x) &= \Big(\big(1+(\cos x-1)\big)^{3/5} -1\Big) + \Big(1-\cos x\Big)\\ &\approx \frac{3}{5}\ (\cos x-1) + (1-\cos x) &\text{by (4) with } \theta =3/5\\
&= \frac{2}{5}\ (1-\cos x)\\
&\approx \frac{1}{5}\ x^2 &\text{by (5)}\end{split}$$
hence:
$$\lim_{x \to 0} \frac{\sin(e^{\tan ^ 2 {x}} - 1)}{\cos^{\frac{3}{5}}(x) - \cos(x)} = \lim_{x\to 0} \frac{x^2}{\frac{1}{5}\ x^2}=5\; .$$
Can anyone please help me with this random variable question I've stumbled across.
Recall from calculus that a function $h$ is called non-decreasing if $x \le y$ implies $h(x) \le h(y)$, for every $x, y \in \mathop{\mathrm{dom}} h$.
Q1a) Let $X$ be a continuous random variable with probability density function $f$. Prove
that the probability distribution function of $X$ is non-decreasing.
I'm assuming this means show $F(x) = \int_{-\infty}^x f(y)\,dy$, is a non-decreasing function of $x$ in $\mathbb R$.
Q1b) Show that $\lim_{x\to-\infty} F(x) = 0$ and $\lim_{x\to \infty} F(x) = 1$, and explain the probabilistic meaning of these facts.
Sorry about the layout i'm not used to using this site, hope it makes sense!
Answer
One way to do 1(b):
$F(x) = \int_{-\infty}^x f(t)\ dt$ is an improper integral, which by definition of improper integral means $\lim_{a \to -\infty} \int_a^x f(t)\ dt$. Now $\int_a^x f(t)\ dt = F(x) - F(a)$, so
$$ F(x) = \lim_{a \to -\infty} (F(x) - F(a)) = F(x) - \lim_{a \to -\infty} F(a)$$
and you can solve for $\lim_{a \to -\infty} F(a)$.
As for $\lim_{x \to \infty} F(x) = \lim_{x \to \infty} \int_{-\infty}^x f(t)\ dt$, that is $\int_{-\infty}^\infty f(t)\ dt$, which according to the definition of a probability density function must be $1$.
While solving $$\int \frac{x^2+1}{x^4+1}\,dx,$$ I tried to use partial fractions in the denominator by writing $x^4+1=(x^2+i)(x^2-i)$ And then I got $\Re\left[\sqrt2 \tan^{-1}{x\over \sqrt i}\right]$. In my book they used another method without complex numbers and they got $\frac{1}{\sqrt2} \tan^{-1} \left(\frac{x^2-1}{x\sqrt2}\right)$. How do I prove my answer is equal to theirs? I tried the realtions between log and atan but I couldnt get rid of the $i$. Edit: Please note that this question is not about solving the integral (which I already solved to get $\Re\left[\sqrt2 \tan^{-1}{x\over \sqrt i}\right]$.) but about the simplification of the answer I got using complex partial fractions method to reduce it to the real part only.
Answer
$$\Re\left[\sqrt2 \tan^{-1}{x\over \sqrt i}\right]=\Re\left[\sqrt2 \tan^{-1}{x\cdot e^{-i\pi/4}}\right]=\frac{(\sqrt2 \tan^{-1}{x\cdot e^{-i\pi/4}})+(\sqrt2 \tan^{-1}{x\cdot e^{i\pi/4}})}{2}$$
$$=\frac{1}{\sqrt{2}}\left( \tan^{-1}{x\cdot e^{i\pi/4}}+\tan^{-1}{x\cdot e^{-i\pi/4}} \right)=\frac{1}{\sqrt{2}}\tan^{-1}\frac{x\cdot e^{i\pi/4}+x\cdot e^{-i\pi/4}}{1-x\cdot e^{i\pi/4}\cdot x\cdot e^{-i\pi/4}}$$
$$=\frac{1}{\sqrt{2}}\tan^{-1}\frac{2x\cos \frac{\pi}{4}}{1-x^2}=\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{2}x}{1-x^2}$$
where line $2\to3$ is achieved by using that $\tan(x+y)=\frac{\tan x + \tan y}{1- \tan x \tan y}$.
The differs from the expression you give by a constant, after further noting that $\tan^{-1}x+\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}$ for $x>0$.
I'm stuck in a very simple problem . But, I could not find an answer in the books. That's why I want to ask MSE.
Why is not this correct (according to Wolfram Alpha) ?
$\sqrt[6]{-1}=\sqrt{\sqrt[3]{-1}}=\sqrt {-1}=i$
$\sqrt[3]{-1}=-1$, Because, $(-1)^3=-1.$
Where is the mistake?
