Are the following two conclusions followed by MVT?

  1. $ |\tan(x)-\tan(y)| \le |x-y|$, $0\le x,y \le \pi/2 $
  2. $ |2^x-2^y| \ge |x-y| $

For (1), by MVT, $$ \frac{\tan(x)-\tan(y)}{x-y}=\sec^2(c) \text{ for some } c \in (0, \pi/2) \tag 3 $$
since $\sec^2(c)\ge1$, $c \in (0, \pi/2) $ and it's no harm to put an absolute value on both side of (3), so (1) makes sense.

I think I should get this one correct, though not very confident.

However, for (2), I'm not quite sure:

By MVT, $$ \frac{2^x-2^y}{x-y}=2^c\ln2,$$ $c$ is between $x$ and $y$
if $$c\ge0$$ then $$2^c\ge1$$ therefore $$ 2^x-2^y \ge (\ln2) (x-y)$$ if c<0, then $$0<2^c<1$$ therefore $$ 2^x-2^y < (\ln2) (x-y) <x-y$$
then I don't know what's next.

And I also have a feeling that my argument for (2) is not complete, when I have the range for $2^c$, I didn't know how to assign $x$ and $y$. for example, if $c<0$, and $c$ is the number between $x$ and $y$, and $x$ can be both negative or one positive and one negative, then I got lost....

Can anyone help me? Thanks a lot.


1 Answer 1


The second equation is not true in general. Take $y=0$. If $x$ is small, then $2^x - 2^y = e^{x \ln 2} - 1\approx (1 + x \ln 2) - 1 = x \ln 2 < x = x - y$

But the mean value theorem can be used to show when $|2^x - 2^y| \geq |x-y|$ is true. $\frac{2^x - 2^y}{x-y} = (\ln2) 2^c \geq 1$ when $2^c \geq 1/\ln(2)$ or $c \geq \log_2(1/\ln2) \approx 0.53$. So $|2^x - 2^y| \geq |x-y|$ holds when $x, y \geq \log_2(1/\ln2)$

  • $\begingroup$ agree~~ I think the second one is false, but I don't know if my argument make sense. $\endgroup$
    – Lily
    Oct 6, 2013 at 1:34
  • $\begingroup$ Your argument is good, just modify it to compute the values of $x$ and $y$ for which the inequality holds. I've updated my answer with the details. $\endgroup$ Oct 6, 2013 at 7:03
  • $\begingroup$ i see. thank you so much!! :) $\endgroup$
    – Lily
    Oct 6, 2013 at 15:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .