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My attempt at a solution:

Suppose that there exists a positive $\delta$ such that for each $x$ and each $y$ in $\mathbb R$, $|x-y|<\delta \implies |\sin e^x - \sin e^y|<1$. (Is $1$ a correct choice for $\epsilon$ here?)

We then try to derive a contradiction to show that $\sin e^x$ is not uniformly continuous on $\mathbb R$. Let $x_n=\ln (2n+\frac12)\pi$ and $y_n=\ln (2n+\frac32)\pi \space \space \space \space\space (n \in \mathbb N) $.

Is my attempt headed in the right direction? If so, I would appreciate any help sketching out the rest of the proof. If not, please correct my solution (for example, my choice of $x_n$ and $y_n$ seem wrong since they do not converge to any point in the domain) and perhaps provide some guidance.

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$1$ is a good choice, but $2$ or anything smaller would work. –  Jonas Meyer Jan 30 '13 at 20:10
    
@JonasMeyer Thanks for the tip. –  Ryan Jan 31 '13 at 3:05

2 Answers 2

up vote 4 down vote accepted

Choosing $\epsilon=1$ is a bold move but perfect in this case. Furthermore you have chosen $x_n$ and $y_n$ such that on the one hand $$\bigl|f(y_n)-f(x_n)\bigr|=2>\epsilon\qquad\forall n\geq 1 $$ is guaranteed, and that on the other hand the distances $|y_n-x_n|$ get arbitrarily small. This would imply that there is no $\delta>0$ such that $|f(y)-f(x)|<1$ whenever $x$ and $y$ are real numbers with $|y-x|<\delta$.

But in your post you have not proven that in fact $\lim_{n\to\infty}(y_n-x_n)=0$. The proof is easy: One has $$y_n-x_n=\log{2n+{3\over2}\over 2n+{1\over2}}=\log\left(1+{1\over 2n+{1\over2}}\right)\to0\qquad(n\to\infty)\ .$$

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Thank you! The crucial (and now obvious) insight I had missed was that $x-y$ needs to converge to $0$. I had been stuck trying to figure out how to show that they converge to the same arbitrary limit... –  Ryan Jan 30 '13 at 21:24
    
Btw there's a small error in your working: it should instead be $\ln (1+\frac 1{2n+\frac 12})$. Also, if desired to rigorously prove from first principles instead of using limits, we can write: "For each positive $\delta$, when $n>\frac 1{2(e^\delta-1)}-\frac 14$, $|x_n - y_n|=\ln (1+\frac 1{2n+\frac 12})<\delta$ but blah blah." –  Ryan Jan 31 '13 at 5:01

Your choice of $\epsilon, x_n$ and $y_n$ are fine. To conclude, note that $|\sin e^{x_n}-\sin e^{y_n}|=2$ for all $n$ and $$|x_n-y_n|=\ln((2n+3/2)\pi)-\ln(2n+1/2)\pi)=\int_{(2n+1/2)\pi}^{(2n+3/2)\pi}\frac{1}{n}< \frac{\pi}{(2n+1/2)\pi}$$ which goes to $0$ as $n\to \infty$, so in particular for any $\delta>0$ we have $n$ such that $|x_n-y_n|<\delta$.

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@JonasMeyer Whoops, yes. I was looking at the 1 in the OP, but as you point out that's smaller than necessary. –  Alex Becker Jan 30 '13 at 20:16
1  
@Ryan You have to be a little more specific. To begin, you understand why $|\sin e^{x_n}-\sin e^{y_n}|=2$, right? (I assume that's why you chose $x_n$ and $y_n$ as such) –  Alex Becker Jan 30 '13 at 20:35
    
Sorry Alex, I should've considered the problem and your answer more carefully. I was thrown off by the integral sign (and still am). Thanks. –  Ryan Jan 30 '13 at 21:25
    
@Ryan The integral is just a way of working with logarithms. $\ln a$ is defined as the integral of $1/x$ from $1$ to $a$, so it is a good way to bound logarithms. –  Alex Becker Jan 30 '13 at 22:05
    
oic! Thank you! –  Ryan Jan 31 '13 at 5:12

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