1. **Problem Q1a:** Calculate absolute error, relative error, and percentage error for each student given true value $25.3267$ and measurements: A = $25.31$, B = $25.33$, C = $25.00$. 2. **Step 1:** Calculate absolute error $= |\text{measured} - \text{true}|$. - Student A: $|25.31 - 25.3267| = 0.0167$ - Student B: $|25.33 - 25.3267| = 0.0033$ - Student C: $|25.00 - 25.3267| = 0.3267$ 3. **Step 2:** Calculate relative error $= \frac{\text{absolute error}}{\text{true value}}$. - A: $\frac{0.0167}{25.3267} \approx 0.000659$ - B: $\frac{0.0033}{25.3267} \approx 0.000130$ - C: $\frac{0.3267}{25.3267} \approx 0.0129$ 4. **Step 3:** Calculate percentage error $= \text{relative error} \times 100$. - A: $0.0659\%$ - B: $0.0130\%$ - C: $1.29\%$ 5. **Problem Q1b:** Identify most accurate and most precise measurement. - Accuracy relates to closeness to true value: Student B is most accurate (smallest absolute error). - Precision relates to consistency; given only one measurement each, precision cannot be fully assessed, but Student A and B are close; Student C is less precise. 6. **Problem Q1c:** Discuss round-off error with least count $0.01$. - Least count $0.01$ means measurements are rounded to nearest $0.01$. - Maximum round-off error is $\pm 0.005$. - Students' measurements are consistent with this precision. 7. **Problem Q1b:** Using Taylor series, approximate $f(x) = e^x$ at $x=0.1$ up to sixth term. 8. **Step 1:** Taylor series for $e^x$ at $x=0$ is: $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ 9. **Step 2:** Approximate up to sixth term (n=0 to 5): $$e^{0.1} \approx \sum_{n=0}^5 \frac{0.1^n}{n!} = 1 + 0.1 + \frac{0.1^2}{2} + \frac{0.1^3}{6} + \frac{0.1^4}{24} + \frac{0.1^5}{120}$$ 10. **Step 3:** Calculate each term: - $1$ - $0.1$ - $0.005$ - $0.0001667$ - $0.000004167$ - $0.0000000833$ 11. **Step 4:** Sum terms: $$1 + 0.1 + 0.005 + 0.0001667 + 0.000004167 + 0.0000000833 \approx 1.10517095$$ 12. **Step 5:** True value $e^{0.1} \approx 1.105170918$ 13. **Step 6:** Truncation error $= |\text{true} - \text{approx}| \approx |1.105170918 - 1.10517095| = 3.2 \times 10^{-8}$ 14. **Problem Q2a:** Use Newton-Raphson method to find root of $f(x) = x^3 - 2x - 5 = 0$ with initial guess $x_0=2$. 15. **Step 1:** Newton-Raphson formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ 16. **Step 2:** Compute derivative: $$f'(x) = 3x^2 - 2$$ 17. **Step 3:** Iteration 1: - $x_0=2$ - $f(2) = 8 - 4 - 5 = -1$ - $f'(2) = 12 - 2 = 10$ - $x_1 = 2 - (-1/10) = 2 + 0.1 = 2.1$ 18. **Step 4:** Iteration 2: - $f(2.1) = (2.1)^3 - 2(2.1) - 5 = 9.261 - 4.2 - 5 = 0.061$ - $f'(2.1) = 3(2.1)^2 - 2 = 3(4.41) - 2 = 13.23 - 2 = 11.23$ - $x_2 = 2.1 - (0.061/11.23) \approx 2.1 - 0.00543 = 2.09457$ 19. **Step 5:** Iteration 3: - $f(2.09457) \approx 9.182 - 4.189 - 5 = -0.007$ - $f'(2.09457) \approx 3(4.387) - 2 = 13.16 - 2 = 11.16$ - $x_3 = 2.09457 - (-0.007/11.16) = 2.09457 + 0.000627 = 2.0952$ 20. **Step 6:** Root correct to three decimals is $2.095$. 21. **Problem Q4a:** Fit line $y = a + bx$ to data $x = [2,4,6,8,10]$, $y = [50,55,65,70,80]$ using least squares and estimate $y$ at $x=7.5$. 22. **Step 1:** Calculate means: - $\bar{x} = \frac{2+4+6+8+10}{5} = 6$ - $\bar{y} = \frac{50+55+65+70+80}{5} = 64$ 23. **Step 2:** Calculate slope $b$: $$b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$$ 24. **Step 3:** Compute numerator: - $(2-6)(50-64) = (-4)(-14) = 56$ - $(4-6)(55-64) = (-2)(-9) = 18$ - $(6-6)(65-64) = 0$ - $(8-6)(70-64) = 2 \times 6 = 12$ - $(10-6)(80-64) = 4 \times 16 = 64$ - Sum = $56 + 18 + 0 + 12 + 64 = 150$ 25. **Step 4:** Compute denominator: - $(2-6)^2 = 16$ - $(4-6)^2 = 4$ - $(6-6)^2 = 0$ - $(8-6)^2 = 4$ - $(10-6)^2 = 16$ - Sum = $16 + 4 + 0 + 4 + 16 = 40$ 26. **Step 5:** Calculate $b = 150/40 = 3.75$ 27. **Step 6:** Calculate intercept $a = \bar{y} - b \bar{x} = 64 - 3.75 \times 6 = 64 - 22.5 = 41.5$ 28. **Step 7:** Equation of line: $$y = 41.5 + 3.75x$$ 29. **Step 8:** Estimate $y$ at $x=7.5$: $$y = 41.5 + 3.75 \times 7.5 = 41.5 + 28.125 = 69.625$$ **Final answers:** - Q1a: Absolute, relative, percentage errors for each student as above. - Q1b: Most accurate is Student B. - Q1c: Round-off error up to $\pm 0.005$ due to least count. - Q1b (Taylor): Approximate $e^{0.1} \approx 1.10517095$ with truncation error $3.2 \times 10^{-8}$. - Q2a: Root of $x^3 - 2x - 5=0$ is approximately $2.095$. - Q4a: Least squares line $y=41.5 + 3.75x$, estimate at $x=7.5$ is $69.625$.