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In the article Exploring Inverse Functions we made first approach to the topic avoiding formal definition of inverse function. Let's study thoroughly the concept of inverse function.
From set-theoretical point of view a function is a set of ordered pairs. For example, y=2x is {(x, 2*x) | x ϵ R}. Take any one-to-one function f = {(x, y) | x ϵ D, y=f(x)}, where f denotes the function as a set of ordered pairs and as a rule of computing y for given x. The domain of f is D and the range is R. Consider set of ordered pairs g = {(y, x) | x ϵ D, y=f(x)}. So, we switched x and y. Is g a function? Since f is one-to-one, yes, g is a function. For each given y we have exactly one corresponding x. The letters x and y are just symbols used inside curled brackets. If we write g = {(a, b) | x ϵ D, a=f(b)}, the logic does not change. So we can write g = {(x, y) | y ϵ D, x=f(y)} to follow convention that argument is denoted by x and value is denoted by y. Thus we can formulate a definition:
Definition 1.
Let f = {(x, y) | x ϵ D, y=f(x)} be a one-to-one function. Then g = {(x, y) | y ϵ D, x=f(y)}is called inverse function for f.
By convention an inverse function of f is denoted by f -1 . So g = f -1 .
It's possible to give a strict set-theoretical proof that there is exactly one inverse function for given f. Let's accept it as obvious.
Note that if g is the inverse function for f then f is the inverse function for g. Indeed, f = {(x, y) | x ϵ D, y=f(x)} = {(x, y) | y ϵ R, x=g(y)} . Thus the logical roles of f and g are symmetrical. Neither f nor g have any preference over each other. Thus we can say that for any one-to-one function there is exactly one pair of mutually inverse functions. Indeed, switching x and y results in switching between two functions. Let's call it a pair of mutually inverse functions.
Among standard functions we have following pairs of mutually inverse functions: (e^x, ln(x)), (sinh(x), asinh(x)) . We can also construct more trivial pairs of mutually inverse functions: (2x, x/2), (x^3, x^(1/3)) . During such construction we "solve" original function for y. For example, y = 1 + 2x is solved for y : x = (y -1) / 2 . Thus we get a pair of mutually inverse functions (1 + 2x, (x - 1) /2) . We should use such "solution" carefully, since it works only for one-to-one functions (one-to-one functions are called bijections in Set Theory).
We used to think of arcsine as inverse function of sine. Is it not? In sense of Definition 1 the answer is No. Arcsine is a bijection and has an inverse function, that is a restriction of sine to interval [-π/2, π/2] . Thus arcsine and restriction of sine to interval [-π/2, π/2] form a pair of inverse functions. We could call sine and arcsine a conventional pair of inverse functions. Another example of conventional pair of inverse functions is a pair of x squared and square root of x. But we should remember that facts, which are true for pairs of inverse functions, are not necessarily true for conventional pairs of inverse functions.
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