Show: $\gcd(9k+4,2k+1)=1 ~~ \forall k\in \mathbb Z$
Indirect proof.
If $1\neq d=\gcd(9k+4,2k+1)~\exists k\in \mathbb Z$,
then $d$ has to be of the form $2m+1$ for an integer $m$.
That somehow throws me back at the beginning.
Answer
Please excuse me for this messed up question,
but I've come up with an answer as compensation:
Just apply the Euclidean algorithm, and in 3 lines you get 1 as GCD.
Pratt is given a fair die. He repeatedly
throw the die until he get
at least each number (1 to 6).
Define the random variable $X$ to be the total number of trials that
pratt throws the die. or
How many times he has to throw a die so that each side of the die appears at least once.
Determine the expected value $E(X)$.
If $\theta$ is an angle lying between $90^{\circ}$ and $180^{\circ}$ and $\cos(\theta) = - \frac{4}{5}$, why does this mean that $\sin(\theta) = \frac{3}{5}$?
Can someone explain, in simple terms, why the following limit doesn't exist?
$$\lim \limits_{x\to0}\sin\left(\left|\frac{1}{x}\right|\right)$$
The function is even, so the left hand limit must equal the right hand limit. Why does this limit not exist?
Answer
The function is indeed even, as we will see, this does not prove the existence of the limit.
Assume that the limit exists.
$$ \lim_{x\to 0^+} \sin\left(\frac1x \right) $$
Which is the same as your limit because the left-hand limit and right-hand limit will be equal (if the limit exists). Then since $x > 0$, I removed the absolute value sign.
$$ \lim_{u\to\infty} \sin u $$
Using $u = 1/x$, we see that the limit certainly does not exist.
Suppose that $S^1$ is the unit circle in $\mathbb{C}$ and suppose that $g\colon [0,2\pi]\rightarrow \mathbb{C}$ is continuous such that $g(2\pi) = g(0)$.
I have to show that $h\colon S^1 \rightarrow \mathbb{C}$, $x \mapsto g(t(x))$ is continuous. where $t(x)$ is the number such that $x = e^{it(x)}$.
I guess that I have to show that if $y \in B(x,\epsilon)$, then $y \in B(t(x),\epsilon)$ (if $\epsilon$ is small enough). This directly implies the result. Can anyone help me to show this?
The problem is to show the function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ given by
$$f(x,y)=(\tfrac{1}{2}x^2+y^2+2y,\,x^2-2x+y^3)$$ is injective on the set
$$M=\{(x,y)\in\mathbb{R}^2:|x-1|+|y+1|<\tfrac{1}{6}\}.$$
My idea is to consider the following map (here, $u,v\in\mathbb{R}^2$):
$$\phi_v(u)=u-f(u)+v,\quad v\in M$$
If I manage to show that
$\phi_v:D\rightarrow D$ is well-defined for some closed sets
$\phi_v$ is a contraction (Lipschitz constant $<1$) on $D$
then by the Contraction Mapping Theorem, $\phi_v$ has a unique fixed point. Hence, $v$ has a unique preimage $u$ for each $v$. i.e. $f$ is injective as desired.
But I ran into troubles when I attempted to find a suitable closed set $D$. Obviously it depends on the domain $M$. $M$ given here is really weird so I am not too sure how to proceed.
Answer
I'll present an approach along the lines of my comment. First, I'll normalize the derivatives at $(1,-1)$ by dividing the second component by $3$:
$$\tilde f(x,y) = (x^2/2+y^2+2y, (x^2-2x+y^3)/3)$$
This is done so that the Jacobian matrix of $\tilde f$ at $(1,-1)$ is the identity. Now split $\tilde f=L+g$ where $L(x,y)=(x-3/2,y+1/3)$ is the linear part and $g$ is the rest. If we can show that $g$ is Lipschitz with a constant less than 1, we are done.
The first component of $g$ is $g_1(x,y)=x^2/2+y^2+2y-x+3/2$, with the gradient $\nabla g_1=\langle (x-1),2(y+1)\rangle$. We estimate the gradient by $|\nabla g_1|< \sqrt{5}/6<1/2$.
The second component of $g$ is $g_2(x,y)= (x^2-2x+y^3-3y-1)/3$, with the gradient $\nabla g_2=\langle 2(x-1)/3,y^2-1\rangle$. Since $|y^2-1|\le (|y+1|+2)|y+1|<13/36$, we obtain $|\nabla g_2|\le \sqrt{1/81+(13/36)^2}<1/2$.
Since both components have Lipschitz constant $<1/2$, the map $g$ has Lipschitz constant $<1$. (Of course a more precise bound can be given, but this suffices.)
If I have the integral: $\displaystyle\int_{0}^{\infty}t^{-\frac{1}{2}}e^{-t} dt$
Am I allowed to make the substitution $t=x^2$, because I am then not sure what the limits of integration would be as for $t$ positive $x$ could be negative or positive?
Answer
Hint. You may rather perform the change of variable $x=\sqrt{t}>0$, giving $dx=\frac{1}{2}t^{-\frac{1}{2}}dt$ to get
$$
\int_{0}^{\infty}t^{-\frac{1}{2}}e^{-t} dt=2\int_{0}^{\infty}e^{-x^2} dx
$$ then you may use the standard gaussian result.
