[narkee]

How is it that inverse seems to be used as "multiplicative inverse" in this context? It seems like a really ambiguous term, because it could also be interpreted as either:

inverse of the square root (which is just the squaring operation), or

the inverse of some other binary operator, like addition or anything else...

[StefanKarpinski]

I think you’ve hit the nail on the head:

> it could also be interpreted as ... [the] inverse of the square root (which is just the squaring operation)

Since the other obvious interpretation is not very useful and has a clearer name—i.e. “the square”—the term “inverse square root” has only one useful meaning, which is therefore how it’s interpreted. (I don’t follow the second option about binary operators.) Mathematical terminology and notation in general are full of ambiguities which are resolved by extensive contextual knowledge. As noted by a sibling comment, calling it the reciprocal square root would be clearer.

[n4r9]

A better phrase would be "reciprocal square root".

[dustmop]

Yes, "inverse" could perhaps be more clearly stated as "inverted". I've heard the term used this way before, but it's not common.

[meta_AU]

It is the inverse of the square root. If you want to normalise a vector, you divide the components by the length. The length is the sqrt of the sum squares (Pythagoras). Divide is more expensive than multiply. So get the inverse sqrt of the sum of the squared components, then multiply the components by the inverse sqrt.

[majewsky]

The point is that "inverse" usually refers to the function that reverses the effect of the original function, i.e.

  f_inv(f(x)) = x for all x in Domain(f)

g(x) = 1 / sqrt(x) is not the inverse of f(x) = sqrt(x) in this sense.