0
$\begingroup$

If we have to compute this without using calculator, is there a quick way to find the answer approximately of the following problem:

which one is smaller ? $$ A = \frac{\sqrt{2007}-\sqrt{2006}}{\sqrt{2008}-\sqrt{2007}} $$ or $$ B = \frac{\sqrt{2010}-\sqrt{2009}}{\sqrt{2011}-\sqrt{2010}} $$

My thinking is to multiply A with $\displaystyle \frac{\sqrt{2008}+\sqrt{2007}}{\sqrt{2008}+\sqrt{2007}}$ and B with $\displaystyle \frac{\sqrt{2011}+\sqrt{2010}}{\sqrt{2011}+\sqrt{2010}}$, and simplify from fraction into multiplication and subtraction only to become:$A' = (\sqrt{2007}-\sqrt{2006})(\sqrt{2008}+\sqrt{2007})$, and $B' = (\sqrt{2010}-\sqrt{2009})({\sqrt{2011}+\sqrt{2010}})$.

This form is still not easy to calculate for me.

$\endgroup$
  • 1
    $\begingroup$ you can use \sqrt{...} to get the root over the entire ... expression $\endgroup$ – Guest 86 Jan 27 '13 at 13:00
  • 2
    $\begingroup$ So the idea in your last paragraph, how well does it work? Do you run into trouble? $\endgroup$ – Henning Makholm Jan 27 '13 at 13:07
  • $\begingroup$ Sorry, now i completed my last paragraph. $\endgroup$ – kuskus Jan 27 '13 at 13:10
  • 1
    $\begingroup$ @kusg1: But what happens when you do those multiplications? Does it solve the problem? Do you get stuck? If so, where? $\endgroup$ – Henning Makholm Jan 27 '13 at 13:22
3
$\begingroup$

This looks like a job for calculus. Basically the question is whether the function $$ x \mapsto \frac{\sqrt{x}-\sqrt{x-1}}{\sqrt{x+1}-{\sqrt{x}}} $$ is increasing or decreasing around $x=2010$. Since the increases of $1$ are small compared to 2010 (which in this case means that none of the relevant derivatives show much relative change when $x$ varies by $1$) we can probably get away with setting $g(x)=\sqrt x$ and approximating $$ \frac{g(x)-g(x-1)}{g(x+1)-g(x)} \approx \frac{g'(x-1)}{g'(x)} \approx \frac{g'(x)-g''(x)}{g'(x)} = 1 - \frac{g''(x)}{g'(x)} = 1 + \frac{1}{2x}$$

$\endgroup$
  • $\begingroup$ This looks what I look for. Thanks $\endgroup$ – kuskus Jan 27 '13 at 13:57
  • $\begingroup$ From calculator, A will give 1.00024915 and B will give 1.00024871, this agrees with 1/(2*2007) > 1/(2*2010) $\endgroup$ – kuskus Jan 27 '13 at 14:04
0
$\begingroup$

Hint : the square root function is concave.

$\endgroup$
  • $\begingroup$ Could you give further pointer with concave definition in this context ? thanks $\endgroup$ – kuskus Jan 27 '13 at 13:11
  • 1
    $\begingroup$ So ? $x \mapsto 1-e^{-x}$ is also concave, but if you go and replace the square roots with it, you get $A=B=e$ $\endgroup$ – mercio Jan 27 '13 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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