Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

prove that $$f(x)= \begin{cases} {1-cosx\over x^2}, & \text{if $x\neq 0$} \\ 1/2, & \text{if x= 0} \\ \end{cases}$$ is uniformly continuous in $\mathbb R$ I can´t derivate in this problem I have already asked this question but I can´t do this and here are my attempts:

I need to prove that $\forall \epsilon \gt 0$ $\exists \delta \gt 0$ so that $\forall x,y\in \mathbb R$, $|x-y|\lt \delta\ \Rightarrow |{1-cosx\over x^2}-{1-cosy\over y^2}|\lt\epsilon$ in my first attempt I went backwards in other words i wanted to have an expression that contains$|x-y|$: $|{1-cosx\over x^2}-{1-cosy\over y^2}|=|{{1\over x^2}-{cosx\over x^2}-{1\over y^2}+{cosy\over y^2}}|$ $\le {|{1\over x^2}|+|{-cosx\over x^2}|+|{-1\over y^2}|+|{cosy\over y^2}|}$ $={|{1\over x^2}|+|{cosx\over x^2}|+|{1\over y^2}|+|{cosy\over y^2}|}$ and $|cosx|\le1$ then ${|cosx|\over |x^2|}\le{1\over |x^2|}$ so ${|{1\over x^2}|+|{cosx\over x^2}|+|{1\over y^2}|+|{cosy\over y^2}|}\le {|{1\over x^2}|+{1\over |x^2|}+|{1\over y^2}|+{1\over |y^2|}}$ $={2\over x^2}+{2\over y^2}\le\epsilon$ and after this i dont know how to get the expression that contains $|x-y|$

in my second attempt I use the fact that if a function $f$ is continuous in $[0,\infty)$ and is uniformly continuous in $[k,\infty]$ for some $k\gt 0$ then $f$ is uniformly continuous in $[0,\infty)$, I plot the function in geogebra and in $[{3\pi\over 2}, \infty)$the function ${1-cosx\over x^2}\le{1\over x^2}$ but i dont know that in the definition of uniformly continuous function i can do: $|{1-cosx\over x^2}-{1-cosy\over y^2}|\le|{1\over x^2}-{1\over y^2}|\lt\epsilon$ so I am really desperate i need to this for tommorow and I don´t know how to do this a really need some help

share|improve this question
1  
    
as I said I have already asked the question in I answered that i would tried to do the problem but I can´t do it... –  David Hernandez Nov 30 '13 at 5:13
2  
Why not use commas and periods to separate sentences and paragraghs? –  awllower Nov 30 '13 at 5:15
    
I am sorry :) I am not very good at writing in mathjax –  David Hernandez Nov 30 '13 at 5:24

1 Answer 1

Helpful identity: note that $$ 1 - \cos(x) = 2\frac{1 - \cos(2\frac{x}{2})}{2} = 2 \sin^2\left(\frac{x}{2}\right) $$ so that our function is simply $$ f(x)= \begin{cases} 2 {\sin^2(\frac x2)\over x^2}, & \text{if $x\neq 0$} \\ 1/2, & \text{if x= 0} \\ \end{cases} $$ With that in mind, we can find (assuming $x,y \neq 0$) $$ \begin{align} \left|f(x)-f(y)\right| &= 2\left| \left(\frac{\sin(\frac y2)}{y}\right)^2 - \left(\frac{\sin(\frac x2)}{x}\right)^2 \right|\\ &=2\left| \left(\frac{\sin(\frac y2)}{y}\right) + \left(\frac{\sin(\frac x2)}{x}\right) \right| \left| \left(\frac{\sin(\frac y2)}{y}\right) - \left(\frac{\sin(\frac x2)}{x}\right) \right| \end{align} $$ Now, we just have to work with $\left| \left(\frac{\sin(\frac y2)}{y}\right) - \left(\frac{\sin(\frac x2)}{x}\right) \right|$ to get out a $|x-y|$.

share|improve this answer
    
thank you!! and we also have that ${|xsin(y)-ysin(x)|\over |xy|}\le{|x-y|\over |xy|}$ is this correct? –  David Hernandez Nov 30 '13 at 5:50
    
I think there is a typo. $\frac{2\sin^2(x)}{x^2}$ should be $\frac{2\sin^2(x/2)}{x^2}$ and similarly for the rest –  E.O. Nov 30 '13 at 6:00
    
@E.O. thank you, duly noted. David: I'm not sure you can make that simplification directly; doing so assumes $\sin(y/2)$ and $\sin(x/2)$ have the same sign. Maybe there's some condition on $\delta$ that allows you to do that though –  Omnomnomnom Nov 30 '13 at 6:10
    
Maybe you can say something like "as shown in the text/class, $\sin(x/2)/x = \frac 12 \frac{\sin(x/2)}{x/2}$ is uniformly continuous" –  Omnomnomnom Nov 30 '13 at 6:13
    
thanks a lot!!! I didnt think about simplifying my expression first with a trigonometric identity :D –  David Hernandez Nov 30 '13 at 6:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.