Are (f(g(x))) and (f(x)g(x)) Always Equal?

The question of whether (f(g(x))) (function composition) is always equal to (f(x)g(x)) (function multiplication) is a topic that often arises in the realm of advanced mathematics. The answer is complex and depends on the specific definitions and domains of the functions (f) and (g).

Context and Definitions

To explore this topic, it is essential to understand the context in which the functions (f) and (g) are defined. In mathematics, functions are typically defined with a domain and a codomain. The domain is the set of input values the function can take, while the codomain is the set of potential output values.

Mathematically, we denote a function (f) as (f: D_f rightarrow C_f), where (D_f) is the domain of (f) and (C_f) is the codomain. For (f(g(x))) to be defined, (x) must be in the domain of (g), and (g(x)) must be in the domain of (f). This is formally written as (g(x) in D_f).

Function Composition vs. Function Multiplication

Function Composition ((f(g(x)))) involves applying function (g) first, followed by function (f). This operation is not commutative, meaning in general, (f(g(x)) eq g(f(x))).

Function Multiplication ((f(x)g(x))) involves the product of the values of (f(x)) and (g(x)). This operation can be defined pointwise, meaning for each (x) in the common domain, the product (f(x)g(x)) is computed.

For these two operations to be equal, (f(g(x)) f(x)g(x)), the functions (f) and (g) must satisfy specific conditions.

Conditions for Equality

Let's consider an example to illustrate when (f(g(x)) f(x)g(x)) might be true:

Example 1: Suppose (f(x) 1) and (g(x) 1). In this case, (f(g(x)) f(1) 1) and (f(x)g(x) 1 cdot 1 1). Clearly, (f(g(x)) f(x)g(x)). Example 2: Suppose (f(x) x) and (g(x) c) where (c) is a constant. Then (f(g(x)) c) and (f(x)g(x) x cdot c). These are equal only if (c 0) or (x 0). Example 3: Consider (f(x) x^2) and (g(x) x). Then (f(g(x)) f(x) x^2), but (f(x)g(x) x^2 cdot x x^3). Here, (f(g(x)) eq f(x)g(x)).

From these examples, it is evident that the equality (f(g(x)) f(x)g(x)) is not a general property of all functions (f) and (g).

Domain and Codomain Considerations

The domains and codomains of (f) and (g) also play a crucial role in determining the equality. If (f) and (g) have different domains, then (f(g(x))) is only defined for values of (x) in both domains. For instance, if (f: mathbb{R} rightarrow [-1, 1]) and (g: mathbb{R} rightarrow [-1, 1]), with (f(x) cos(x)) and (g(x) sin(x)), then:

(fg(pi/4) f(g(pi/4)) f(sin(pi/4)) cos(pi/4) sqrt{2} / 2 otin [-1, 1]) (f(pi/4)g(pi/4) cos(pi/4) cdot sin(pi/4) (sqrt{2} / 2) cdot (sqrt{2} / 2) sqrt{2} / 2 otin [-1, 1])

In such cases, the equality (f(g(x)) f(x)g(x)) does not hold because the resulting values do not lie within the codomains of the respective functions.

Conclusion

In summary, (f(g(x))) is not always equal to (f(x)g(x)). The equality holds in specific cases and depends on the definitions of the functions involved, their domains, and codomains. It is essential to carefully consider these aspects when evaluating the equality of these operations.