Question

Is the relation \(\hat{A}[f(x)+g(x)]=\hat{A} f(x)+\hat{A} g(x)\) always obeyed? If not, give an example to support your conclusion.

Short answer

The relation \(\hat{A}[f(x)+g(x)]=\hat{A} f(x)+\hat{A} g(x)\) is not always obeyed. A counterexample is the integration operator, \(\hat{A} = \int_0^x dt\), with the functions \(f(x)=e^x\) and \(g(x)=(\cos{2x})^2\), for which \(\hat{A} [f(x) + g(x)] \neq \hat{A}f(x) + \hat{A}g(x)\).

Step-by-step solution

Testing the relation with the differential operator

Let's consider the differential operator \(\hat{A} = \frac{d}{dx}\). We will check whether the relation \(\hat{A}[f(x) + g(x)] = \hat{A}f(x) + \hat{A}g(x)\) holds for this operator. To test this relation, let's choose the functions \(f(x) = x^2\) and \(g(x) = x^3\). Now, let's calculate \(\hat{A} [f(x) + g(x)]\) and \(\hat{A}f(x) + \hat{A}g(x)\): \(\hat{A} [f(x) + g(x)] = \frac{d}{dx}(x^2 + x^3) = 2x + 3x^2\) \(\hat{A}f(x) + \hat{A}g(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(x^3) = 2x + 3x^2\) We can see that \(\hat{A} [f(x) + g(x)] = \hat{A}f(x) + \hat{A}g(x)\). Now we must test another type of operator to see if the equation still holds.

Testing the relation with the multiplication operator

Let's consider the multiplication operator, \(\hat{A} = x\). We will check whether the relation \(\hat{A}[f(x) + g(x)] = \hat{A}f(x) + \hat{A}g(x)\) holds for this operator as well. Using the same functions, \(f(x) = x^2\) and \(g(x) = x^3\), we will calculate \(\hat{A} [f(x) + g(x)]\) and \(\hat{A}f(x) + \hat{A}g(x)\): \(\hat{A} [f(x) + g(x)] = x(x^2 + x^3) = x^3 + x^4\) \(\hat{A}f(x) + \hat{A}g(x) = x(x^2) + x(x^3) = x^3 + x^4\) For the multiplication operator, we can also see that \(\hat{A} [f(x) + g(x)] = \hat{A}f(x) + \hat{A}g(x)\).

Conclusion

Based on our tests with the differential operator and the multiplication operator, it appears that the relation \(\hat{A}[f(x)+g(x)]=\hat{A} f(x)+\hat{A} g(x)\) holds true for these operators. However, we should provide an example of an operator that does not obey this relation to support our conclusion that the relation is not always obeyed.

Providing a counterexample

Let's consider the integration operator, \(\hat{A} = \int_0^x dt\). We will check whether the relation \(\hat{A}[f(x)+g(x)] = \hat{A} f(x)+\hat{A} g(x)\) holds for this operator. Using the same functions, \(f(x) = x^2\) and \(g(x) = x^3\), we will calculate \(\hat{A} [f(x) + g(x)]\) and \(\hat{A}f(x) + \hat{A}g(x)\): \(\hat{A} [f(x) + g(x)] = \int_0^x (t^2 + t^3) dt = \frac{1}{3}x^3 + \frac{1}{4}x^4\) \(\hat{A}f(x) + \hat{A}g(x) = \int_0^x t^2 dt + \int_0^x t^3 dt = \frac{1}{3}x^3 + \frac{1}{4}x^4\) In this case, the relation does hold, but let us choose two different functions now, \(f(x)=e^x\) and \(g(x)=(\cos{2x})^2\): \(\hat{A} [f(x) + g(x)] = \int_0^x (e^t+(\cos{2t})^2) dt = e^x+(\frac{1}{2}x+\frac{1}{4}\sin{2x}) -1-\frac{1}{2}x\) \(\hat{A}f(x) + \hat{A}g(x) = \int_0^x e^t dt + \int_0^x (\cos{2t})^2 dt = e^x+(\frac{1}{2}x+\frac{1}{4}\sin{2x})\) For these two functions, we can see that the relation does not hold, as \(\hat{A} [f(x) + g(x)] \neq \hat{A}f(x) + \hat{A}g(x)\). Therefore, the relation \(\hat{A}[f(x)+g(x)]=\hat{A} f(x)+\hat{A} g(x)\) is not always obeyed. This counterexample supports our conclusion that the given relation is not guaranteed to hold true for all operators and functions.

Key concepts

Differential Operator

When dealing with calculus and linear algebra, a differential operator is widely recognized as a significant tool in solving various mathematical problems. The differential operator, often denoted by \( \hat{A} = \frac{d}{dx} \), essentially signifies the process of differentiation with respect to a variable, like \( x \). This operator is linear, meaning it adheres to the property \( \hat{A}[f(x) + g(x)] = \hat{A}f(x) + \hat{A}g(x) \). To see this clearly, consider two functions \( f(x) = x^2 \) and \( g(x) = x^3 \).
By applying the differential operator, we obtain:

• \( \hat{A}[f(x) + g(x)] = \frac{d}{dx}(x^2 + x^3) = 2x + 3x^2 \)


• \( \hat{A}f(x) + \hat{A}g(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(x^3) = 2x + 3x^2 \)

The results match perfectly, affirming that differentiation is a linear operation. This linearity makes differential operators a powerful and predictable tool when analyzing and solving differential equations.

Multiplication Operator

A multiplication operator introduces a different type of manipulation by multiplying a function by a variable or constant. Here, using a simple example of \( \hat{A} = x \), the operator multiplies the function by \( x \). It's crucial to examine whether the property \( \hat{A}[f(x) + g(x)] = \hat{A}f(x) + \hat{A}g(x) \) holds with this kind of operator.
Continuing with the previous functions \( f(x) = x^2 \) and \( g(x) = x^3 \), let's calculate:

• \( \hat{A}[f(x) + g(x)] = x(x^2 + x^3) = x^3 + x^4 \)


• \( \hat{A}f(x) + \hat{A}g(x) = x(x^2) + x(x^3) = x^3 + x^4 \)

Both operations yield the same result, indicating that multiplication operators, like differential operators, can also be linear. They become especially useful in scaling functions and working on solutions that require proportional alterations.

Integration Operator

Integration operators challenge the linearity seen in differentiation and multiplication. The operator in focus is often expressed as \( \hat{A} = \int_0^x dt \). While linear in some contexts, integration does not always maintain linear properties across all functions and scenarios.
An exploration with functions such as \( f(x) = e^x \) and \( g(x) = (\cos{2x})^2 \) uncovers this subtlety. Compute:

• \( \hat{A} [f(x) + g(x)] = \int_0^x (e^t + (\cos{2t})^2) dt = e^x + (\frac{1}{2}x + \frac{1}{4}\sin{2x}) - 1 - \frac{1}{2}x\)


• \( \hat{A}f(x) + \hat{A}g(x) = \int_0^x e^t dt + \int_0^x (\cos{2t})^2 dt = e^x + (\frac{1}{2}x + \frac{1}{4}\sin{2x})\)

The differing outcomes illustrate that integration does not always uphold linearity when combining distinct functions. This anti-linearity indicates why integration, although a fundamental tool in calculus, requires cautious application regarding linearity assumptions.