1
$\begingroup$

Could someone explain to me this transformation? It is used frequently in my uni course, and I do not understand what's happening:

$$\left|\sqrt{2+3x_1} - \sqrt{2+3x_2}\right| = \left| \frac{3(x_1-x_2)}{\sqrt{2+3x_1} + \sqrt{2+3x_2}} \right|\qquad (x_1,x_2 \in [0,3))$$

Is there some obvious rule I don't see? Thanks in advance. (The transformation is used to prove Lipschitz continuity.)

$\endgroup$
  • 2
    $\begingroup$ If $A_i = \sqrt{2+3x_i}$, then what is $(A_1-A_2)(A_1+A_2)$? $\endgroup$ – Don Thousand Jun 3 '19 at 2:58
0
$\begingroup$

Let $A_i = \sqrt{2+3x_i}$ for $i=1,2$ as suggested by Don Thousand in the comments. Then your original expression is $|A_1 - A_2|$.

Notice, then:

$$A_1 - A_2 = (A_1-A_2) \cdot \underbrace{\frac{A_1 + A_2}{A_1 + A_2}}_{=1} = \frac{A_1^2 - A_2^2}{A_1+A_2}$$

$A_i^2 = 2+3x_i$ for both $i$ indices and thus the difference in the numerator is $2+3x_1 - 2 - 3x_2$, or, more simply, $3(x_1 - x_2)$.

Then

$$A_1 - A_2 = \cdots = \frac{A_1^2 - A_2^2}{A_1+A_2} = \frac{3(x_1 - x_2)}{\sqrt{2+3x_1} + \sqrt{2+3x_2}}$$

Throw in the absolute values and you're done!

$\endgroup$

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.