If $y=\left(1+\alpha+\alpha^2+\ldots\right) \mathrm{e}^{\mathrm{nx}}$, where $\alpha$ and n are constants, then the relative error in y is

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Aicharya · Accepted Answer

Given, $y=\left(1+\alpha+\alpha^2+\ldots.ight) e^{n x}$ $\begin{aligned} & \because \frac{d y}{d x}=n\left(1+\alpha+\alpha^2+\ldots . .ight) e^{n x} \ & \Rightarrow \Delta y=n\left(1+\alpha+\alpha^2+\ldots \ldotsight) e^{n x} \Delta x \ & \Rightarrow \Delta y=n y \Delta x \Rightarrow \frac{\Delta y}{y}=n \Delta x \end{aligned}$ So, relative error in $y=n$. (error in $x)$

If $y=\left(1+\alpha+\alpha^2+\ldots\right) \mathrm{e}^{\mathrm{nx}}$, where $\alpha$ and n are constants,…

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