Question

Show that the functions \(x^{r} e^{\lambda x}\), where \(r=0,1,2, \ldots, k\), are linearly independent. Hint: Apply appropriate powers of \(\mathbf{D}-\lambda\) to a linear combina tion of \(x^{r} e^{\lambda x}\) for all possible \(r\) 's.

Short answer

The functions in the set \(\{x^{r} e^{\lambda x}\}\) are linearly independent. This is proven by applying the operator \((\mathbf{D}-\lambda)\) multiple times to a linear combination of the functions and showing that all the coefficients must be zero for the combination to yield zero.

Step-by-step solution

Formulate Linear Combination

Let's start by creating a linear combination of the functions \(x^{r} e^{\lambda x}\) for \(r=0,1,2,\ldots,k\). We can do this by adding together the functions, each multiplied by some constant \(c_r\), i.e.: \(S(x)=\sum_{r=0}^{k} c_{r} x^{r} e^{\lambda x}\).

Apply Operator to Linear Combination

Now, we apply the operator \((\mathbf{D}-\lambda)\) to the sum \(S(x)\) repeatedly until we reach only constants (ideally \(0\)). To do this, note that \((\mathbf{D}-\lambda) x^{r} e^{\lambda x} = r x^{r-1} e^{\lambda x}\) for \(r>0\) and \((\mathbf{D}-\lambda) e^{\lambda x} = 0\). So, applying \((\mathbf{D}-\lambda)\) \(r+1\) times, you get: \((\mathbf{D}-\lambda)^{r+1} x^{r} e^{\lambda x} = (r!) e^{\lambda x} \) if \(r\) is nonnegative and \(0\) if \(r<0\). Thus \((\mathbf{D}-\lambda)^{k+1} S(x)= \sum_{r=0}^{k} c_{r} (\mathbf{D}-\lambda)^{k+1} x^{r} e^{\lambda x} = \sum_{r=0}^{k} c_{r} (r!) e^{\lambda x} = 0.\) This last equality must hold for all \(x\), but the exponential function is never zero, so the only way the equality can hold is for \(c_{r} (r!) = 0\) for all \(r\).

Conclude Linear Independence

From the previous step, we've that \(c_{r} (r!) = 0\) for all \(r\). Since \(r! \neq 0\) for all \(r\geq0\), it must be \(c_{r}=0\) for all \(r\). This means no non-zero linear combination of the functions \(x^{r} e^{\lambda x}\) can evaluate to zero, thus proving the functions are linearly independent.

Key concepts

Eigenfunctions

The concept of eigenfunctions bridges the gap between algebra and differential equations. An eigenfunction of a differential operator is a non-zero function that, when the operator is applied, results in a scalar multiple of itself. This scalar is known as the eigenvalue.

For instance, considering the differential operator \(\mathbf{D}\), which represents differentiation with respect to \(x\), when we have a function such that \(\mathbf{D}f = \lambda f\), \(f\) is an eigenfunction of the operator \(\mathbf{D}\) with eigenvalue \(\lambda\). These functions are fundamental solutions to differential equations that arise in physics and engineering, such as the time-independent Schrödinger equation in quantum mechanics.

In the context of the exercise, the functions \(x^{r} e^{\lambda x}\), upon which the operator acts, behave like eigenfunctions since they yield a result directly proportional to the original function, preserving the exponential term \(e^{\lambda x}\) after differentiation.

Exponential Functions

Exponential functions, written as \(e^{\lambda x}\), are vital in the realm of mathematics, especially when solving differential equations. Their most noteworthy property for our purposes is that they never equal zero for any real value of \(x\).

When we differentiate \(e^{\lambda x}\) any number of times, the function remains in the form of \(e^{\lambda x}\), possibly multiplied by constants. In the exercise, the exponential function \(e^{\lambda x}\) is paired with various powers of \(x\) to show linear independence. Since \(e^{\lambda x}\) does not vanish, it's assured that if the linear combination of the functions equals to zero, the coefficients associated with these exponential terms also must be zero, corroborating the linear independence.

Differential Operators

Differential operators are tools we use to signify the action of taking derivatives. The differential operator \(\mathbf{D}\) implies taking the derivative with respect to a variable, often \(x\). More complex operations can be constructed using differential operators, such as \(\mathbf{D}-\lambda\), which instructs us to differentiate and then subtract \(\lambda\) times the original function.

In our exercise, the operator \(\mathbf{D}-\lambda\) is applied repeatedly to a linear combination of functions. By applying it enough times, we eliminate variables and are left with constants which helps us to derive conclusions about the linear independence of the given functions. The operation showcases a key feature of differential operators: the ability to decompose the behavior of complex functions into simpler, understandable parts.

Linear Combination

A linear combination in the context of functions involves adding together multiple functions, each multiplied by a constant coefficient. Mathematically, for a set of functions \(f_i\) and constants \(c_i\), the linear combination is \(\sum_{i} c_{i} f_i\).

In the step-by-step solution provided, a linear combination of the functions \(x^{r} e^{\lambda x}\) for \(r=0,1,2,\ldots,k\) is considered. The claim of linear independence asserts that the only solution for this linear combination to equal zero over all \(x\) is that all coefficients \(c_r\) must be zero. If there can be found non-zero coefficients making the combination vanish, the functions would be linearly dependent; here, that's not the case.

Following the prescribed steps demonstrates each \(c_r\) must indeed be zero due to the never-zero property of the exponential function, thus the functions \(x^{r} e^{\lambda x}\) are proved to be linearly independent through their unique contributions to the linear combination.