Question

As an alternative proof that the functions $e^{r_{1}x},e^{r_{2}x},e^{r_{3}x},\mathrm{...},e^{r_{n}x}$are linearly independent on (∞,-∞) when $r_1,r_2,...r_n$ are distinct, assume $C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}+C_3{e}^{r_{3}x}+...+C_n{e}^{r_{n}x}$holds for all x in (∞,-∞) and proceed as follows:

(a) Because the ri’s are distinct we can (if necessary)relabel them so that $r_1>r_2>r_3>...>r_n$.Divide equation (33) by to obtain $C_1+\frac{C_2{e}^{r_{2}x}}{e^{r_{2}x}}+\frac{C_3{e}^{r_{3}x}}{e^{r_{3}x}}+...+\frac{C_n{e}^{r_{n}x}}{e^{r_{n}x}}=0$Now let x→∞ on the left-hand side to obtainC1 = 0.(b) Since C1 = 0, equation (33) becomes

$C_2{e}^{r_{2}x}+C_3{e}^{r_{3}x}+...+C_n{e}^{r_{n}x}$= 0for all x in(∞,-∞). Divide this equation by$e^{r_{2}x}$

and let x→∞ to conclude that C2 = 0.

(c) Continuing in the manner of (b), argue that all thecoefficients, C1, C2, . . . ,Cn are zero and hence $e^{r_{1}x},e^{r_{2}x},e^{r_{3}x},...,e^{r_{n}x}$are linearly independent on(∞,-∞).

Short answer

The answer to this problem is:

$C_1=0$, $C_2=0$

Step-by-step solution

Basic differentiation

The Sum rule says the derivative of a sum of functions is the sum of their derivatives. The Difference rule says the derivative of a difference of functions is the difference of their derivatives.

Solving by basic differentiation:

We will do the following question on the basis of basic differentiation ;

$\begin{array}{l}\{e^{r_{1}x},e^{r_{2}x},e^{r_{3}x},\mathrm{...},e^{r_{n}x}\}\\ r_i\ne r_j\forall i,j\notin \{1,2,3,\mathrm{...},n\}\\ \\ C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}+C_3{e}^{r_{3}x}+\mathrm{...}+C_n{e}^{r_{n}x}\end{array}$

(a)

$\begin{array}{l}{C}_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}+C_3{e}^{r_{3}x}+\mathrm{...}+C_n{e}^{r_{n}x}\\ r_1>r_2>r_3>\mathrm{...}>r_n\\ \frac{C_1{e}^{r_{1}x}}{e^{r_{1}x}}+\frac{C_2{e}^{r_{2}x}}{e^{r_{2}x}}+\frac{C_3{e}^{r_{3}x}}{e^{r_{3}x}}+\mathrm{...}+\frac{C_n{e}^{r_{n}x}}{e^{r_{n}x}}=0\\ C_1+C_2\left(e^{(r_2-r_1)x}\right)+C_3\left(e^{(r_3-r_1)x}\right)+\mathrm{...}+C_n\left(e^{(r_n-r_1)x}\right)=0\\ \\ e^{(r_2-r_1)x}=e^{-m_{2}x}\\ e^{(r_3-r_1)x}=e^{-m_{3}x}\\ e^{(r_n-r_1)x}=e^{-m_{n}x}\\ \\ \underset{x\to \infty }{\lim }(C_1+\frac{C_2}{e^{m_{2}x}}+\frac{C_3}{e^{m_{3}x}}+\mathrm{....}+\frac{C_n}{e^{m_{n}x}})=0\\ e^x\to \infty \\ x\to +\infty \\ C_1+C_2\left(0\right)+\mathrm{....}+C_n\left(0\right)=0\\ C_1=0\end{array}$

(b)

$\begin{array}{l}{C}_2{e}^{r_{2}x}+C_3{e}^{r_{3}x}+\mathrm{...}+C_n{e}^{r_{n}x}\\ C_2+C_3\frac{e^{r_{3}x}}{e^{r_{2}x}}+\mathrm{...}+C_n\frac{e^{r_{n}x}}{e^{r_{3}x}}=0\\ C_2+C_3\left(e^{(r_3-r_1)x}\right)+\mathrm{...}+C_n\left(e^{(r_n-r_1)x}\right)=0\\ e^{(r_3-r_1)x}=e^{-s_3}\\ e^{(r_n-r_1)x}=e^{-s_n}\\ \\ \underset{x\to \infty }{\lim }\left(C_2+C_3\left(e^{-s_{3}x}\right)+C_4\left(e^{-s_{4}x}\right)+\mathrm{...}+C_n\left(e^{-s_{n}x}\right)\right)=0\\ e^x\to \infty \\ x\to +\infty \\ C_2+C_3\left(0\right)+\mathrm{....}+C_n\left(0\right)=0\\ C_2=0\\ \end{array}$

(c)

$C_1=C_2=C_3=\mathrm{...}=C_n=0$

Hence, the final answer is:

$C_1=0$, $C_2=0$