8. Two decimal numbers X and Y were approximated with errors \( \mathrm{E}_{1} \) and \( \mathrm{E}_{2} \) respectively. Show that the maximum possible relative error in the approximation \( \mathrm{X}^{2} \mathrm{Y} \) is given by \( 2\left|\frac{E_{1}}{X}\right|+\left|\frac{E_{2}}{Y}\right| \)

To determine the maximum possible relative error in the approximation \( \mathrm{X}^{2} \mathrm{Y} \) when \( X \) and \( Y \) are approximated with relative errors \( \mathrm{E}_{1} \) and \( \mathrm{E}_{2} \) respectively, follow these steps: ### 1. Express the Approximated Values Let's denote the approximated values of \( X \) and \( Y \) as: \[ \tilde{X} = X + e_1 \quad \text{with} \quad |e_1| \leq |E_1 X| \] \[ \tilde{Y} = Y + e_2 \quad \text{with} \quad |e_2| \leq |E_2 Y| \] ### 2. Compute the Approximated Product The approximated product is: \[ \tilde{X}^2 \tilde{Y} = (X + e_1)^2 (Y + e_2) \] Expanding this: \[ \tilde{X}^2 \tilde{Y} = X^2 Y + X^2 e_2 + 2XY e_1 + 2X e_1 e_2 + e_1^2 Y + e_1^2 e_2 \] ### 3. Determine the Absolute Error The absolute error \( \Delta \) in the approximation is: \[ \Delta = |\tilde{X}^2 \tilde{Y} - X^2 Y| = |X^2 e_2 + 2XY e_1 + 2X e_1 e_2 + e_1^2 Y + e_1^2 e_2| \] Using the triangle inequality: \[ \Delta \leq |X^2 e_2| + |2XY e_1| + |2X e_1 e_2| + |e_1^2 Y| + |e_1^2 e_2| \] ### 4. Normalize to Find Relative Error Dividing by \( X^2 Y \) to find the relative error: \[ \frac{\Delta}{X^2 Y} \leq \left|\frac{e_2}{Y}\right| + 2\left|\frac{e_1}{X}\right| + 2\left|\frac{e_1 e_2}{X Y}\right| + \left|\frac{e_1^2}{X^2}\right| + \left|\frac{e_1^2 e_2}{X^2 Y}\right| \] Assuming that the relative errors \( E_1 \) and \( E_2 \) are small, higher-order terms (involving products of errors) can be neglected. This simplifies the expression to: \[ \text{Relative Error} \leq 2\left|\frac{E_1}{X}\right| + \left|\frac{E_2}{Y}\right| \] ### Conclusion The maximum possible relative error in the approximation \( \mathrm{X}^{2} \mathrm{Y} \) is therefore given by: \[ 2\left|\frac{E_1}{X}\right| + \left|\frac{E_2}{Y}\right| \] **Answer:** After expanding, it is equal to twice |E₁⁄X| plus |E₂⁄Y|. Thus, Relative error ≤ 2 |E₁/X| + |E₂/Y|