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.)

  • 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

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, $3(x_1 - x_2)$.


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

Throw in the absolute values and you're done!


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