As an alternative proof that the functions er1x,er2x, . .

Chapter 6, Problem 26E

(choose chapter or problem)

As an alternative proof that the functions \(e^{n x}, e^{r 5 x} \ldots e^{t r x}\) are linearly independent on \((-\infty, \infty)\) when \(r_{1}, r_{2}, \ldots . r_{n}\) are distinct, assume
(33) \(C_{1} e^{r_{1} x}+C_{2} e^{r_{2} x}+\cdots+C_{n} e^{r_{0} x}=0\) holds for all  in \((-\infty, \infty)\) and proceed as follows:

(a) Because the \(r_{i}\) 's are distinct we can (if necessary) relabel them so that
\(r_{1}>r_{2}>\cdots>r_{n}\)

Divide equation (33) by \(e^{r_{1} x}\) to obtain
$$C_{1}+C_{2} \frac{e^{r^{x} x}}{e_{1}^{r} x}+\cdots+C_{n} \frac{e^{r n x}}{e^{\eta_{1} x}}=0$$

Now let \(x \rightarrow+\infty\) on the left-hand side to obtain \(C_{1}=0\).

(b) Since \(C_{1}=0\), equation (33) becomes \(C_{2} e^{r_{2}^{2}}+C_{3} e^{t_{3} x}+\cdots+C_{n} e^{r_{k} x}=0\)
for all \(x\) in \((-\infty, \infty)\) . Divide this equation by \(e^{t_{2} x}\)  and let \(x \rightarrow+\infty\) to conclude that \(C_{2}=0\)

(c) Continuing in the manner of (b), argue that all the coefficients, \(C_{1}, C_{2}, \ldots, C_{n}\) are zero and hence \(e^{n x}, e^{r_{2} x} \ldots, e^{r_{\epsilon} x}\) are linearly independent on \((-\infty, \infty)\).

Equation Transcription:

Text Transcription:

e^nx,e^rsx…. e^tr⁡x

(-infinity,infinity)

r_1,r_2,……r_n

C_1 e^r_1 x+C_2 e^r_2 x+⋯+C_n e^r_theta x=0

x

r_i

r_1>r_2>⋯>r_n

e^r_1 x

C_1+C_2 e^r_2 x/e_r^1 x+⋯+C_n e^r_n x/e^eta_1 x=0

C_1=0

C_ 2 e^r^2 2+C_3 e^t_3 x+⋯+C_n e^r_n x=0

e^t_2 x

x right arrow+infinity

C_2=0

C_1,C_2,…,C_n

 e^nx,e^r_2 x…,e^r_n x