Recent content by Krizalid1

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...

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.

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!

For one second, I almost got you serious, haha.

Are you guys learning any new languages? I did a small course in japanese and now I'm trying to learn a bit by myself. :D

I think you have some typos then, so you're watching $a_n+1$ instead of $a_{n+1}.$ Since $a_n>0$ (this fact is very important in order these things work), we have $\dfrac{{{a}_{n}}}{{{a}_{n}}+1}\le {{a}_{n}},$ so the converse implication is trivial. As for the other, again, since $a_n>0,$ you...