Question

If $$ y = (1 + a + a^2 + \dots)e^{nx} $$ then the relative error in \( y \) is: 

• A. Error in \( x \)
• B. Percentage error in \( x \)
• C. \( n \times \) (error in \( x \))
• D. \( n \times \) (relative error in \( x \))

Hint:

Use the formula for the sum of an infinite geometric series and take logarithms to simplify the expression for \( y \). Then differentiate to find the relative error.

Answer 1

The Correct Option is C

Step 1: Simplify the expression for \( y \).
The expression \( 1 + a + a^2 + \dots \) is an infinite geometric series with first term 1 and common ratio \( a \).
Since the series is infinite, we assume \( |a|<1 \) for the series to converge.
The sum of the infinite geometric series is \( \frac{1}{1-a} \).
Therefore, \( y = \frac{e^{nx}}{1-a} \). Step 2: Find the relative error in \( y \).
The relative error in \( y \) is given by \( \frac{\Delta y}{y} \), where \( \Delta y \) is the error in \( y \).
Taking logarithms on both sides of \( y = \frac{e^{nx}}{1-a} \), we get: \( \ln y = \ln \left(\frac{e^{nx}}{1-a}\right) = \ln e^{nx} - \ln (1-a) = nx - \ln (1-a) \).
Differentiating both sides, we get: \( \frac{dy}{y} = n dx \).
Therefore, the relative error in \( y \) is \( n \) times the error in \( x \).