Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

As an alternative proof that the functions $e^{r_{1} x}, e^{r_{2} x}, e^{r_{3} x}, ..., e^{r_{n} x}$ 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 obtain C1 = 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 (∞,-∞).

The answer to this problem is:

$C_{1} = 0$ , $C_{2} = 0$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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.

Step 2: Solving by basic differentiation:

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

$\begin{array}{l} {e^{r_{1} x}, e^{r_{2} x}, e^{r_{3} x}, ..., e^{r_{n} x}} \\ r_{i} \neq r_{j} \forall i, j \notin {1, 2, 3, ..., n} \\ C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x} + C_{3} e^{r_{3} x} + ... + C_{n} e^{r_{n} x} \end{array}$

(a)

$\begin{array}{l} C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x} + C_{3} e^{r_{3} x} + ... + C_{n} e^{r_{n} x} \\ r_{1} > r_{2} > r_{3} > ... > r_{n} \\ \frac{C_{1} e^{r_{1} x}}{e^{r_{1} x}} + \frac{C_{2} e^{r_{2} x}}{e^{r_{2} x}} + \frac{C_{3} e^{r_{3} x}}{e^{r_{3} x}} + ... + \frac{C_{n} e^{r_{n} x}}{e^{r_{n} x}} = 0 \\ C_{1} + C_{2} (e^{(r_{2} - r_{1}) x}) + C_{3} (e^{(r_{3} - r_{1}) x}) + ... + C_{n} (e^{(r_{n} - r_{1}) x}) = 0 \\ e^{(r_{2} - r_{1}) x} = e^{- m_{2} x} \\ e^{(r_{3} - r_{1}) x} = e^{- m_{3} x} \\ e^{(r_{n} - r_{1}) x} = e^{- m_{n} x} \\ \lim_{x \to \infty} (C_{1} + \frac{C_{2}}{e^{m_{2} x}} + \frac{C_{3}}{e^{m_{3} x}} + .... + \frac{C_{n}}{e^{m_{n} x}}) = 0 \\ e^{x} \to \infty \\ x \to + \infty \\ C_{1} + C_{2} (0) + .... + C_{n} (0) = 0 \\ C_{1} = 0 \end{array}$

(b)

$\begin{array}{l} C_{2} e^{r_{2} x} + C_{3} e^{r_{3} x} + ... + C_{n} e^{r_{n} x} \\ C_{2} + C_{3} \frac{e^{r_{3} x}}{e^{r_{2} x}} + ... + C_{n} \frac{e^{r_{n} x}}{e^{r_{3} x}} = 0 \\ C_{2} + C_{3} (e^{(r_{3} - r_{1}) x}) + ... + C_{n} (e^{(r_{n} - r_{1}) x}) = 0 \\ e^{(r_{3} - r_{1}) x} = e^{- s_{3}} \\ e^{(r_{n} - r_{1}) x} = e^{- s_{n}} \\ \lim_{x \to \infty} (C_{2} + C_{3} (e^{- s_{3} x}) + C_{4} (e^{- s_{4} x}) + ... + C_{n} (e^{- s_{n} x})) = 0 \\ e^{x} \to \infty \\ x \to + \infty \\ C_{2} + C_{3} (0) + .... + C_{n} (0) = 0 \\ C_{2} = 0 \end{array}$

(c)

$C_{1} = C_{2} = C_{3} = ... = C_{n} = 0$

Hence, the final answer is:

$C_{1} = 0$ , $C_{2} = 0$

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