Exploring Functions Whose Inverse is 1 Over the Original Function

In mathematics, the concept of a function whose inverse is precisely the reciprocal of the original function can be intriguing and complex. This article delves into this phenomenon, providing a detailed analysis through examples and insights.

Introduction to Inverse Functions

In the realm of functions, the idea of an inverse function is fundamental. A function f(x) and its inverse f^{-1}(x) satisfy the property that f(f^{-1}(f(x))) x. In this piece, we explore a special class of functions where f^{-1}(x) 1/f(x).

Simple Example: f(x) x^n

Let's start with a simple exponential function f(x) x^n. This function has a corresponding inverse given by:

f^{-1}(x) x^{1/n}

Substituting this into the inverse relation, we get:

$$x f^{-1}(f(x)) (f(x))^{1/n} (x^n)^{1/n} x$$

However, what we're looking for is a situation where f^{-1}(x) 1/f(x). For this, we need:

$$x^{1/n} frac{1}{x^n}$$

By comparing exponents, we find that:

$$1/n -n$$

This implies that:

$$n i$$

While n i seems to be a solution, it doesn't straightforwardly provide a well-behaved function in the real number system. Let's explore this further with Euler's formula.

Using Euler's Formula

To find a well-defined function, we use the exponential function:

We start with the function:

$$f(x) e^{i log x}$$

Applying the function again, we get:

$$f^{-1}(x) frac{1}{f(x)} frac{1}{e^{i log x}} e^{-i log x} x^{-i}$$

This confirms that:

$$f^{-1}(x) frac{1}{f(x)}$$

However, the function defined over the complex plane is essentially:

$$f(x) e^{i log x}$$

This is well-defined on mathbb{C} backslash {0}, but not unique, as we can use other roots of unity.

Geometric Interpretation and Mouml;bius Transformation

The geometric interpretation of the reciprocal function can be modeled using Mouml;bius transformations. A Mouml;bius transformation is a linear fractional transformation that can be represented by a 2x2 matrix:

For the function 1/x, the corresponding Mouml;bius transformation is:

$$T begin{pmatrix} 0 1 1 0 end{pmatrix}$$

To find a function that squares to this transformation, we need:

$$M^2 T$$

Through calculations, we find:

$$M frac{1}{2} begin{pmatrix} 1-i i i 1-i end{pmatrix}$$

Applying this transformation, we get:

$$f(x) frac{(1 i)x}{(1-i)x}$$

This function, when applied twice, gives the reciprocal:

$$ff(x) frac{1}{x}$$

This function is defined on the entire Riemann sphere, providing a comprehensive solution to the problem.

Conclusion

The exploration of functions whose inverse is the reciprocal of the original function reveals complex yet fascinating properties. Using Euler's formula and Mouml;bius transformations, we can find such functions that are well-defined on various domains, including the complex plane and the Riemann sphere.