Question

Show that \(f * g=g * f\). That is, show that \(\int_{0}^{t} f(t-\lambda) g(\lambda) d \lambda=\int_{0}^{t} g(t-\sigma) f(\sigma) d \sigma\).

Short answer

Question: Prove that the convolution of two functions, \(f\) and \(g\), is commutative. Short Answer: To prove that the convolution of two functions \(f\) and \(g\) is commutative, we need to demonstrate that \(f * g = g * f\). By introducing a change of variable, such as \(\sigma = t - \lambda\), we can manipulate the right-hand side integral and make it identical to the left-hand side integral. This shows that the convolution is commutative, as both integrals share the same structure: \(\int_{0}^{t} f(t-\lambda)g(\lambda)d\lambda = \int_{0}^{t} g(\lambda)f(t-\lambda)d\lambda\).

Step-by-step solution

Convolution definitions

First, we must understand the given expressions. The left-hand side (LHS) \(\int_{0}^{t} f(t-\lambda)g(\lambda)d\lambda\) represents the convolution of \(f\) and \(g\), denoted as \(f * g\). Similarly, the right-hand side (RHS) \(\int_{0}^{t} g(t-\sigma)f(\sigma)d\sigma\) represents the convolution of \(g\) and \(f\), denoted as \(g * f\). The goal is to prove that LHS = RHS, which means proving \(f * g = g * f\).

Introduce a change of variable

To show that LHS = RHS, let's introduce a change of variable. We can rewrite \(\sigma\) in terms of \(\lambda\), specifically, we can let \(\sigma = t - \lambda\). Then, we have \(d\sigma = -d\lambda\).

Substitute the new variable

Now, substitute the new variable \(\sigma\) in the RHS integral: $$ \int_{0}^{t} g(t-\sigma)f(\sigma)d\sigma $$ Upon substitution, the integral becomes: $$ -\int_{t}^{0} g(\lambda)f(t-\lambda)(-d\lambda) $$ Notice that the limits of integration are flipped due to the negative sign on \(d\sigma = -d\lambda\).

Reverse the limits of integration

Next, we can reverse the limits of integration, which changes the sign of the integral: $$ -\int_{t}^{0} g(\lambda)f(t-\lambda)(-d\lambda) = \int_{0}^{t} g(\lambda)f(t-\lambda)d\lambda $$ At this point, we can see that both integrals have the same structure: $$ \int_{0}^{t} f(t-\lambda)g(\lambda)d\lambda = \int_{0}^{t} g(\lambda)f(t-\lambda)d\lambda $$ This proves that \(f * g = g * f\) or the convolution is commutative. The integral equation is satisfied, and we have shown that the convolution of two functions is commutative.

Key concepts

Commutative Property

The commutative property is a fundamental principle in mathematics that states that the order in which two elements are combined does not affect the outcome of an operation. This property is quite familiar in basic arithmetic, such as addition, where the equation \(a + b = b + a\) always holds true. Similarly, in multiplication, \(a \times b = b \times a\) is always valid.

When we talk about the convolution of two functions, the commutative property holds as well, which means \(f * g = g * f\). This implies that convolving function \(f\) with function \(g\) yields the same result as convolving \(g\) with \(f\).

This property is particularly useful when dealing with complex problems in signal processing and other fields of applied mathematics, as it adds the flexibility to switch the order of functions without changing the result. Understanding the commutative property in convolution helps simplify calculations and proofs. In the provided exercise, showing the commutative nature of convolution was key to proving that the two integral expressions are indeed equal.

Variable Substitution

Variable substitution is a powerful technique in calculus used to simplify integrals by altering the variable of integration. This method often involves substituting a new variable in place of the original variable, with the primary aim of transforming an integral into a form that is easier to evaluate.

In the context of the exercise, we used substitution to prove that the convolution of two functions is commutative. We substituted \(\sigma = t - \lambda\) in the integral, rewriting \(d\sigma = -d\lambda\). This transformation allowed us to simplify and match the expressions on both sides of the convolution equation.

Through careful substitution and attention to the changes in the limits of integration, we effectively flipped the order, showing that the integrals were equivalent. This process highlights the critical role of variable substitution in integral calculus, allowing complex expressions to be broken down into manageable and solvable forms.

Integral Calculus

Integral calculus deals with the processes of integration and finding the areas under curves. It's a core component of calculus, used for a wide range of applications including calculating volumes, solving differential equations, and understanding continuous distributions.

In this specific exercise, we employed integral calculus to evaluate convolution integrals, which are essentially a way to combine two functions into a new function through integration. The integration limits from \(0\) to \(t\) represent a bounded domain over which the convolution operates.

What's fascinating here is how integral calculus allows us to explore relationships between functions through operations like convolution. By integrating the product of one function with a shifted version of another, it's possible to capture dynamics that are otherwise not obvious—a technique frequently used in engineering and physics.

Overall, integral calculus not only helps in direct computation but also in developing deeper insights into function transformations like convolution.

Differential Equations

Differential equations involve equations that relate a function with its derivatives. These equations are central to describing various phenomena in physics, engineering, biology, and many other fields, where they can model dynamics like motion, growth, decay, or even complex systems.

While the exercise itself focused on proving a property of convolution, understanding its applications in solving differential equations is highly beneficial. Convolution frequently appears in solutions to linear differential equations, particularly when utilizing transform methods such as the Laplace or Fourier transform.

For example, in systems governed by linear differential equations, convolution allows for the superposition of inputs, making the analysis of system responses much more manageable. By leveraging the commutative property of convolution, one can simplify these analyses significantly.

In essence, even though not explicitly used in our exercise, differential equations provide context and show the relevance of convolution in practical scenarios, bridging abstract theory with real-world applications.