Chapter 13: Problem 284
Determine whether the set \(\left\\{\mathrm{e}^{\mathrm{x}},
\mathrm{e}^{-\mathrm{x}}\right\\}\) is linearly independent for
\(-\infty
Short Answer
Expert verified
The set \(\left\\{\mathrm{e}^{\mathrm{x}}, \mathrm{e}^{-\mathrm{x}}\right\\}\) is linearly independent for \(-\infty<x<\infty\) since the only possible values for the constants \(a\) and \(b\) in the linear combination \(a\mathrm{e}^{\mathrm{x}} + b\mathrm{e}^{-\mathrm{x}} = 0\) are \(a = 0\) and \(b = 0\).
Step by step solution
01
Write down the linear combination of the given functions
We want to test whether the set \(\left\\{\mathrm{e}^{\mathrm{x}}, \mathrm{e}^{-\mathrm{x}}\right\\}\) is linearly independent. First, we consider a linear combination of these functions: $$a\mathrm{e}^{\mathrm{x}} + b\mathrm{e}^{-\mathrm{x}}.$$ Here, \(a\) and \(b\) are constants, and \(x\) is the variable.
02
Set the linear combination equal to zero
Next, we want to determine if there exist any non-zero constants \(a\) and \(b\) that make this linear combination equal to zero: $$a\mathrm{e}^{\mathrm{x}} + b\mathrm{e}^{-\mathrm{x}} = 0$$.
03
Solve for the constants
To find the values for \(a\) and \(b\), we now solve the following equation: $$a\mathrm{e}^{\mathrm{x}} + b\mathrm{e}^{-\mathrm{x}} = 0$$
Divide both sides by \(\mathrm{e}^{\mathrm{x}}\), we get: $$a + b\mathrm{e}^{-2\mathrm{x}} = 0$$
Now, let's differentiate both sides of this equation with respect to \(x\): $$0 - 2b\mathrm{e}^{-2\mathrm{x}} = 0$$.
Now, divide both sides by \(-2b\): $$\mathrm{e}^{-2\mathrm{x}} = 0$$
Since the exponential function can never be equal to zero, this equation has no solution for \(x\), which means that \(b = 0\).
04
Verify that the constants are zero
Now, we plug the value of \(b = 0\) into our original linear combination: $$a\mathrm{e}^{\mathrm{x}} = 0$$. The only possible solution is \(a = 0\).
05
Determine if the set is linearly independent
Since we have shown that the only possible values for constants \(a\) and \(b\) are \(a = 0\) and \(b = 0\), the given set \(\left\\{\mathrm{e}^{\mathrm{x}}, \mathrm{e}^{-\mathrm{x}}\right\\}\) is linearly independent for \(-\infty<x<\infty\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Combination
In mathematics, a linear combination involves combining a set of functions or vectors using constants, usually denoted as coefficients. Specifically, it is expressed as a sum of the functions each multiplied by a constant. For example, the linear combination of the functions \(\mathrm{e}^{\mathrm{x}}\) and \(\mathrm{e}^{-\mathrm{x}}\) is written as:
- \(a\mathrm{e}^{\mathrm{x}} + b\mathrm{e}^{-\mathrm{x}}\)
- \(a\mathrm{e}^{\mathrm{x}} + b\mathrm{e}^{-\mathrm{x}} = 0\)
Exponential Functions
Exponential functions are a crucial part of mathematics and appear in various fields like calculus, engineering, and the natural sciences. An exponential function is of the form \(\mathrm{e}^{\mathrm{x}}\), where \(\mathrm{e}\) is the base of natural logarithms, approximately equal to 2.71828.
- These functions have unique properties, such as their rates of change being proportional to their values.
- The function \(\mathrm{e}^{\mathrm{x}}\) is known for its growth, while \(\mathrm{e}^{-\mathrm{x}}\) showcases decay.
Differentiation
Differentiation is a calculus technique used to find the rate at which a function is changing at any given point—commonly known as the derivative. When applied to exponential functions, it reveals interesting features:
- The derivative of \(\mathrm{e}^{\mathrm{x}}\) is \(\mathrm{e}^{\mathrm{x}}\), retaining its form after differentiation.
- For \(\mathrm{e}^{-\mathrm{x}}\), the derivative is \(-\mathrm{e}^{-\mathrm{x}}\), indicating a flip in the sign.
Solutions to Differential Equations
Differential equations involve unknown functions and their derivatives, and they model how quantitative systems change. Solving them involves finding a function that satisfies the given equation, which could be ordinary or partial in nature.
- Exponential functions are often part of the solution to such equations because they naturally describe growth and decay processes.
- Linear combinations of exponential functions, like \(a\mathrm{e}^{\mathrm{x}} + b\mathrm{e}^{-\mathrm{x}}\), are typical solutions.