Showing that exponential functions are linearly independent

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SUMMARY

The discussion focuses on proving the linear independence of the functions \(e^{r_1t}\), \(e^{r_2t}\), and \(e^{r_3t}\) for distinct real numbers \(r_1\), \(r_2\), and \(r_3\). The proof employs a contradiction approach, starting with the assumption of linear dependence and leading to the conclusion that \(e^{(r_1 - r_3)t} = C\) cannot hold true for all \(t\) due to the distinctness of \(r_1\) and \(r_3\). The final steps demonstrate that substituting values for \(t\) yields a contradiction, confirming the linear independence of the exponential functions.

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  • Basic knowledge of calculus, particularly differentiation
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Students of mathematics, particularly those studying linear algebra and differential equations, as well as educators seeking to explain the concept of linear independence in the context of exponential functions.

Mr Davis 97
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Homework Statement


If ##r_1, r_2, r_3## are distinct real numbers, show that ##e^{r_1t}, e^{r_2t}, e^{r_3t}## are linearly independent.

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The Attempt at a Solution


By book starts off by assuming that the functions are linearly dependent, towards contradiction. So ##c_1e^{r_1t} + c_2e^{r_2t}+ c_2e^{r_3t} = 0##. After differentiating and doing some manipulations, the book finds that ##e^{(r_1 - r_2)t} = C e^{(r_3 - r_2)t}##, where C is just some constant. It then states that this is a contradiction, so the original statement must be true. I am, however, a little confused as to why this is a contradiction. Is it a contradiction based on some previously shown result that two exponential functions with different powers can never be linearly dependent?
 
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We need a few more steps to get a formal contradiction:

If we multiply both sides by ##e^{(r_2-r_3)t}## the equation becomes ##e^{(r_1-r_3)t}=C##. Substituting successively 0 and 1 for ##t## we get
$$1=e^{(r_1-r_3)\cdot 0}=C=e^{(r_1-r_3)\cdot 1}=e^{r_1-r_3}\neq 1$$
where the last inequality follows from the fact that ##r_1\neq r_3##.
 
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