Recent content by life is maths

Can I use the limit definition? I see no way out... :(

Ah, really? So that was my mistake... I have no idea about contraction mapping theorem. Thanks for enlightening me :biggrin: Is there a more practical way to show a sequence is convergent? :confused:

Homework Statement A sequence {an} defined recursively by a1=1 and an+1=\frac{1}{2+a subn}, n\geq1. Show that the sequence is convergent. Homework Equations If a sequence is bdd below and decreasing or it is bdd above and increasing, then it is convergent. The Attempt at a Solution...

That was a mistake, sorry. I meant IntA\cupIntB.

Thanks, micromass, but I don't understand why I need an extra argument. Is it for the case that A and B are not both open or closed? Or neither open nor closed? Spherics, I had found this myself while thinking before sleep :) cl(0,1)=[0,1] and cl(1,2)= [1,2]. Their intersection is 1, but the...

Hi, my instructor left this as an exercise, but I got confused in the second part. Could you please help me? cl(A\capB)\subseteqcl A \cap cl B But the reverse is not true. Prove this and give a counterexample on the reverse statement. My attempt: If x\in A\capB, then x\in cl(A\capB) x\in A...

Wow, drawing their graphs shows everything clearly. sqrt(2+x) is always smaller than 2+x/4. Hence, this is true for every interval that they are defined (x>=-2). Thanks for your time and effort, Dick and Mark :) You were a great help.

Thanks, Dick. If I take x= -2, then I get \sqrt{-2+2}< 2+ \frac{(-2)}{4}, which is 0 < \frac{3}{2}. It also holds for 0, then \sqrt{2+x} < 2+\frac{x}{4} in the interval [-2,0]. Now how should I proceed? How can I show this holds for every number in this interval? I'm a bit confused :confused:

Thanks a lot, I had thought I should solve for a specific case, but this is also true :)

x2/16+2 must always be positive for all intervals of x. If we put the inequality sign instead of 0, the statement is true. Should I express it like: 'Since x2/16+2>0 is true for all x, then it also holds for x\in[-2,0]' ? Is this a formal statement?

Thanks, Dick, but when I try finding roots of \sqrt{2+x}-(2+\frac{x}{4})= 0, what I get is: 2+x = [\frac{x}{4}]2+4+x x2/16+2=0 but it is not possible. How should I proceed from here? Is there a mistake?

Showing the inequality holds for an interval (?) Homework Statement Hi, my homework question is: Show that the inequality \sqrt{2+x}<2+\frac{x}{4} holds \forallx\in[-2,0] Homework Equations The Attempt at a Solution I tried using IVT or bisection method, but they are just for...

Thanks again :) Wow, I haven't thought of it before... One last question, if you don't mind me :) What is an almost periodic function? It is a term I came across today, and would be grateful if you could explain this, too.

Thanks, Bhaskar. You're right, I should be able to write it formally.

Thanks, LCKurtz. Could you please explain what happens if the periods are not the same? Is it too complicated for a freshman in maths? :)