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Let \(f\) and \(g\) be two differentiable functions that are linearly dependent. Show that their Wronskian vanishes.

Short Answer

Expert verified
If \(f\) and \(g\) are linearly dependent, then the Wronskian \(W(f,g) = f(x)g'(x) - f'(x)g(x)\) equals zero.

Step by step solution

01

Define the linear dependence

If \(f\) and \(g\) are linearly dependent functions, there exist constants \(a\) and \(b\), not both equal to zero, such that \(af(x) + bg(x) = 0\).
02

Define the Wronskian

The Wronskian \(W\) of two differentiable functions \(f\) and \(g\) is defined as: \[W(f,g) = \begin{vmatrix} f & g \ f' & g' \end{vmatrix} = f(x)g'(x) - f'(x)g(x)\]
03

Substitute dependence into Wronskian and simplify

Substitute \(f(x) = -b/a * g(x)\) in the expression for the Wronskian: \[W(f,g) = -b/a * g(x) * g'(x) - (-b/a * g'(x)) * g(x) = 0\] Hence we see that the Wronskian vanishes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linearly Dependent Functions
In mathematics, particularly in linear algebra, linearly dependent functions are functions that are related to each other by a linear equation. This means that one function can be expressed as a multiple of another function or as a sum involving multiples of other functions. For example, if we consider two functions, let's call them f and g, being linearly dependent means that there exist constants a and b, not both zero, such that

\(af(x) + bg(x) = 0\)

This equation tells us that no matter what value x takes, the relationship defined by this equation will always hold true. The importance of this concept is especially palpable in systems of differential equations and when dealing with the space of solutions to linear systems. To put it simply, the concept of linear dependence is akin to having a redundancy or repetition in information—a scenario where one piece of data can predict or determine another.
Wronskian of Functions
Let's delve deeper into the concept of the Wronskian, which acts as a litmus test for the linearity and independence of functions. The Wronskian is named after the Polish mathematician Józef Hoene-Wroński and is a determinant used in the analysis of systems of linear differential equations. For two differentiable functions f and g, the Wronskian is defined by the determinant:

\[W(f,g) = \begin{vmatrix} f & g \ f' & g' \end{vmatrix} = f(x)g'(x) - f'(x)g(x)\]

The Wronskian combines the functions and their first derivatives in a manner that reveals their linear independence (or dependence). If the Wronskian is non-zero at some point, the functions are linearly independent. Conversely, if the Wronskian vanishes for all x, it implies that the functions are linearly dependent. In the context of solving differential equations, if one is tasked with ensuring a set of solutions is comprehensive (a fundamental set), the Wronskian can serve as a crucial tool for verification.
Mathematical Physics
The field of mathematical physics often requires solving complex problems that involve functions and their behaviors. Understanding the linear independence or dependence of solutions, such as waves or quantum states, is crucial in formulating physical theories. The Wronskian is a powerful tool in this aspect, assisting physicists in analyzing whether a set of solutions to differential equations indeed forms a valid basis for describing a physical system. If, for instance, the Wronskian of a group of functions vanishes, a physicist could infer that a redundancy exists among the solutions, indicating that the solutions are not all necessary to describe the system comprehensively.

In practical terms, suppose you have two solutions to a wave equation. Checking the Wronskian could tell you if the waves are truly distinctive or essentially the same wave in disguise. This ties back to the concept of linear dependence—if the solutions are dependent, it means you could predict one using the other, an insight that can be both simplifying and powerful when unifying physical laws or reducing the complexity of a model.

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