Problem 19 - Advanced Engineering Mathematics

Question

Determine whether the given set of functions is linearly dependent or linearly independent on the interval \((-\infty, \infty)\). $$ f_{1}(x)=x, \quad f_{2}(x)=x-1, \quad f_{3}(x)=x+3 $$

Step-by-Step Solution

Answer

The functions \(f_1(x) = x\), \(f_2(x) = x - 1\), and \(f_3(x) = x + 3\) are linearly dependent.

Step 1: Set Up the Equation

To determine if the functions \(f_1(x) = x\), \(f_2(x) = x - 1\), and \(f_3(x) = x + 3\) are linearly dependent, we start by assuming a linear combination of these functions equals zero. This is represented as: \(c_1 \cdot f_1(x) + c_2 \cdot f_2(x) + c_3 \cdot f_3(x) = 0\). Substitute the functions, we have \(c_1 \cdot x + c_2 \cdot (x - 1) + c_3 \cdot (x + 3) = 0\).

Step 2: Simplify the Equation

Expand and combine the terms of the linear combination: \(c_1 \cdot x + c_2 \cdot x - c_2 + c_3 \cdot x + 3c_3 = 0\). Combine like terms: \((c_1 + c_2 + c_3)x + (-c_2 + 3c_3) = 0\).

Step 3: Solve the System of Equations

The above expression must hold for all \(x\), leading to the system of equations: \(c_1 + c_2 + c_3 = 0\) for the coefficient of \(x\), and \(-c_2 + 3c_3 = 0\) for the constant term. Solve these equations to determine \(c_1\), \(c_2\), and \(c_3\).

Step 4: Analyze the Trivial Solution

From \(-c_2 + 3c_3 = 0\), we solve \(c_2 = 3c_3\). Substituting in \(c_1 + c_2 + c_3 = 0\), we get \(c_1 + 3c_3 + c_3 = 0\) or \(c_1 + 4c_3 = 0\). Solve \(c_1 = -4c_3\). Thus, \(c_1 = -4k\), \(c_2 = 3k\), \(c_3 = k\), where \(k\) is any constant. The only way for these coefficients to all be zero is if \(k = 0\), meaning there's only a trivial solution.

Key Concepts

Linear Dependence

Linear dependence is a fundamental concept in linear algebra. It involves determining whether a set of functions or vectors can be expressed as a linear combination of one another. In simpler terms, if you can write one function in the set as a combination of the others, they are linearly dependent. This situation generally implies redundancy since one of the functions can be removed without losing any essential information.
For the given functions: \(f_1(x) = x\), \(f_2(x) = x - 1\), and \(f_3(x) = x + 3\), we need to see if we can express any of them using the others. When this occurs, there's a certain combination of coefficients (other than zero), which, when applied to these functions, the sum equals zero. This is precisely what having a linearly dependent set means. If such coefficients exist, the set is dependent. If not, it's independent.

System of Equations

A system of equations is essentially a set of equations with multiple variables. In our context, it involves determining how the coefficients of the linear combination relate to one another. When checking for linear dependence, we assumed a linear combination equal to zero: \(c_1 \cdot x + c_2 \cdot (x - 1) + c_3 \cdot (x + 3) = 0\).
This expands into a system of equations:

• \(c_1 + c_2 + c_3 = 0\) for the terms involving \(x\)
• \(-c_2 + 3c_3 = 0\) for the constant terms

Solving this system helps us find the coefficients that satisfy this equation. It's important since these equations test if there's a specific linear combination (other than using zeros) that equals the zero function on the entire interval.

Linear Combination

A linear combination refers to constructing a new expression by multiplying each function by a constant and adding the results. For example, given functions \(f_1(x) = x\), \(f_2(x) = x - 1\), and \(f_3(x) = x + 3\), a linear combination looks like this:

• \(c_1 \cdot f_1(x) + c_2 \cdot f_2(x) + c_3 \cdot f_3(x)\)

Here, \(c_1, c_2,\) and \(c_3\) are constants (called coefficients).
In the context of linear dependence, we investigate if there's a non-zero combination of \(c_1, c_2,\) and \(c_3\) that makes the entire expression equal to zero for all \(x\). Finding such a set of coefficients means that one or more of the functions can be constructed through others, confirming their linear dependence.

Trivial Solution

A trivial solution in a linear system refers to a solution where all the coefficients are zero. It's a basic, but essential concept in linear algebra. In the exercise context, we assumed a linear combination \(c_1 \cdot f_1(x) + c_2 \cdot f_2(x) + c_3 \cdot f_3(x) = 0\), and solved, deriving the equations:

• \(-c_2 + 3c_3 = 0\)
• \(c_1 + 4c_3 = 0\)

Solving these, we get \(c_1 = -4c_3\), \(c_2 = 3c_3\), and \(c_3 = k\), where \(k\) can be any number.
This leads us to the trivial solution when \(k = 0\), meaning all constants are zero, which confirms that the set of functions has no other dependencies than the obvious one. This reassures that the set is linearly independent, maintaining their individuality in any linear combination.