Recent content by Krizalid1

I'm pretty rusted with this stuff. How about a hand? Let (U,V) be a random vector such that f_V(u)=\dfrac{3}{v^3}I_{[3,\infty)}(v) and U/V=v has uniform distribution over the interval ]0,3v[. Find: a) A density for the random vector (U,V). b) A density for the random variable U. c) A...

Following Opalg's hint, you have [0,1] is a compact set and f is clearly continuous on [0,1]. Does that ring any bell? Besides, proving uniform continuity on [1,\infty) follows easily by definition.

Why doing the second step? Since \dfrac{d}{dx}{{\sin }^{2}}x=2\cos (x)\sin(x)

I'll be posting Analysis questions. :D If you submit a solution, please use Spoiler. If you have an alternate one, use the Spoiler anyway. If this goes right, I'll be making a big PDF containing all solutions so that it may be useful for students that are taking this course. 1. Let $(X,d)$ be...

Okay so I need some opinions. Which is the best multivariable calculus book you may guys recommend?

I know, it's just that I know the solution but I want other peoply to try it. :D

Let $A_{n\times n}$ be a matrix with real and distincts eigenvalues. Let $u(t,x_0)$ be a solution for the initial value problem $\overset{\cdot }{\mathop{x}}\,=Ax$ with $x(0)=x_0,$ then show that for each fixed $t\in\mathbb R,$ we have $$\lim_{y_0\to x_0}u(t,y_0)=u(t,x_0).$$

Hint:

The misuse of the $\implies$ sign is common and makes no sense when writing things like $1+1\implies2.$ It's used when you already stated an equality and want to "imply" some consequence.

Let $a_n$ and $b_n$ be sequences in $\mathbb R.$ Show that if $\displaystyle\sum b_n$ converges and $\displaystyle\sum|a_n-a_{n+1}|<\infty,$ then $\displaystyle\sum a_nb_n$ converges.

Give it a try! It's a nice problem!

Classic computation of the following derivative and integral as follows: $f(x)=\sin(2x)\implies f'(x)=\cos(2x),$ since most of students forget to apply the chain rule. $\displaystyle\int{\frac{dx}{1+2{{x}^{2}}}}=\tan^{-1}(2x)+k,$ again the same error since they don't consider the fact the...

Now, more generally: Let $f:\mathbb R\to\mathbb R$ be a uniform continuous function, then exist $a,b>0$ such that $|f(x)|\le a|x|+b.$ Let $\delta > 0$ be such that $|x-y|\le \delta \implies |f(x)-f(y)| < 1$. Claim: $|f(x)| \le \frac{2}{\delta}|x| +|f(0)|$ for all $x$. Proof: WLOG, $x>0,$...

Actually what follows, it's the converse: Let $f:A\subset\mathbb R\to\mathbb R$ be with bounded derivative then $f$ is Lipschitz on $A,$ hence, uniformly continuous.

Thank you! You're learning a beautiful language!