Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a calculation, I've come across a relation along the lines of this: $${a}^{1/2}+{b}^{1/2}$$

My presumption would be that this is somewhat related to the Pythagorean relation: $${a}^{2}+{b}^{2}$$

I can understand the Pythagorean relation, but not the importance of the square root relation. Is there a hidden geometric significance to the first relation, just as there is one for the Pythagorean formula?

Note: the other side of the equation in my calculations could be anything, it's not limited to ${c}^{1/2}$. Actually, in my calculations the full relation is: $${c}={d}^{1/2}\bigg[{a}^{1/2}+{b}^{1/2}\bigg]$$

share|improve this question
    
Without knowing the details, that is probably more geometrically meaningful written as $c=\sqrt{ad}+\sqrt{bd}$. –  Henning Makholm Oct 14 '12 at 14:59
    
where did you come across this relation? –  john mangual Nov 7 '12 at 17:49
    
I was doing original research. But to answer it fully without revealing my work, I was relating two bodies of mass and this relation came about because of that. It's really as simple as that; relating two masses. –  Zchpyvr Nov 7 '12 at 23:29

1 Answer 1

up vote 1 down vote accepted

For positive $a$ and $b$, the generalized mean with exponent 1/2: $M_{1/2}\left(a,b\right)=\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^{2}$ is equal to $a$ if $a=b$, and lies between the geometric mean and the arithmetic mean (the "average"): $\sqrt{ab}\le\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^{2}\le\frac{a+b}{2}$.

If $a$ and $b$ are areas of squares, then $M_{1/2}\left(a,b\right)$ is the area of a square with side length equal to the average of the original side lengths, and your expression is the sum of the side lengths, but without more geometric context I don't know how to get a better geometric picture of these things.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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