Let's break down the solution step by step as requested in the assignment: --- ## **Given** \[ I_n = \frac{1}{e} \int_{0}^{1} x^n e^x dx \] Recurrence relation: - \[ I_0 = 1 - \frac{1}{e} \] - \[ I_n = 1 - nI_{n-1}, \quad n = 1, 2, \ldots \] --- ## **Step 1: Compute \(I_n\) (exact and recurrence) for \(n = 0\) to \(30\)** ### **A. Exact Value Calculation** The exact value for \(I_n\) can be calculated directly from the definition: \[ I_n = \frac{1}{e} \int_0^1 x^n e^x dx \] This can be computed using integration by parts or numerically (e.g., using Python/Matlab or a calculator). #### **Integration by parts** Let \(u = x^n\), \(dv = e^x dx\), then \(du = n x^{n-1} dx\), \(v = e^x\): \[ \int x^n e^x dx = x^n e^x - \int n x^{n-1} e^x dx \] Applying limits (from 0 to 1): \[ \int_0^1 x^n e^x dx = [x^n e^x]_0^1 - n \int_0^1 x^{n-1} e^x dx \] \[ = [1^n e^1 - 0^n e^0] - n \int_0^1 x^{n-1} e^x dx \] \[ = [e - 0] - n \int_0^1 x^{n-1} e^x dx \] Thus, \[ \int_0^1 x^n e^x dx = e - n \int_0^1 x^{n-1} e^x dx \] Now, divide both sides by \(e\): \[ I_n = \frac{1}{e} \left(e - n \int_0^1 x^{n-1} e^x dx\right) = 1 - n I_{n-1} \] ### **B. Recurrence Value Calculation** From above, just use: - \(I_0 = 1 - \frac{1}{e}\) - \(I_n = 1 - nI_{n-1}\) ### **C. Relative Error Calculation** \[ \text{Relative Error} = \frac{|I_{n,\text{exact}} - I_{n,\text{recurrence}}|}{|I_{n,\text{exact}}|} \] --- ### **Python Code (for computation)** Here is Python code to calculate and display the results: ```python import numpy as np from scipy.integrate import quad e = np.exp(1) # Exact Integral def exact_integral(n): integrand = lambda x: x**n * np.exp(x) result, _ = quad(integrand, 0, 1) return result / e # Recurrence I_rec = [1 - 1/e] for n in range(1, 31): I_rec.append(1 - n * I_rec[-1]) # Compute exact values and errors I_exact = [] errors = [] for n in range(31): exact = exact_integral(n) I_exact.append(exact) rel_error = abs(exact - I_rec[n]) / abs(exact) errors.append(rel_error) # Print results (rounded to 7 significant figures for values, 2 decimal places for error) print(" n Exact Value Recurrence Value Relative Error") for n in range(31): print(f"{n:2d} {I_exact[n]:.7g} {I_rec[n]:.7g} {errors[n]:.2e}") ``` #### **Sample Output Table (First Few n):** | n | Exact Value | Recurrence Value | Relative Error | |---|-------------|------------------|---------------| | 0 | 0.6321206 | 0.6321206 | 0.00e+00 | | 1 | 0.3678794 | 0.3678794 | 0.00e+00 | | 2 | 0.2642411 | 0.2642411 | 0.00e+00 | | ... | ... | ... | ... | | 30 | ... | ... | ... | **(You will need to run the code to get all 31 rows.)** --- ## **Step 2: Plotting** 1. **Plot 1:** n (x-axis) vs. I_exact and I_rec (y-axis), both curves on one graph. 2. **Plot 2:** n (x-axis) vs. relative error (y-axis). --- ## **Step 3: Discussion** ### a. **How error might influence engineering decisions if high precision is required** - If the relative error is significant, engineering predictions (e.g., settlement, creep) may be inaccurate, leading to unsafe or over-conservative designs. - For high-precision requirements, such as safety-critical structures, small errors can accumulate and become unacceptable. ### b. **Suitability for real-time field calculations** - The recurrence method is computationally fast (just arithmetic), so it's suitable for real-time calculations as long as the error is within acceptable limits. - For large n, if errors grow, it may become unsuitable. ### c. **At what values of n the recurrence remains accurate** - Typically, recurrence is accurate for small to moderate n. - For large n (e.g., n > 15-20), the error may grow (as can be seen from the error plot). - The value of n where relative error exceeds a threshold (e.g., 1%) marks the limit of acceptable accuracy. --- ## **Summary Table (Template for Report)** | n | I_exact (7 sig. fig.) | I_rec (7 sig. fig.) | Rel. Error (%) | |---|-----------------------|---------------------|----------------| | 0 | 0.6321206 | 0.6321206 | 0.00 | | 1 | 0.3678794 | 0.3678794 | 0.00 | | ... | ... | ... | ... | | 30 | ... | ... | ... | --- ## **Final Answer Summary** 1. **Table:** For \( n = 0 \) to \( 30 \), present the exact, recurrence, and error values to required precision. 2. **Plot:** Show both value curves and error curve. 3. **Discussion:** Address points (a)-(c) as above. --- If you need the full table or actual plots, let me know if you want a specific tool (e.g., Python, Excel) and I can provide the code or process!

Certainly! Let's proceed step-by-step, providing detailed calculations, explanations, and final numerical results for each part. --- ## **Problem Restatement** Given the integral: \[ I_n = \frac{1}{e} \int_0^1 x^n e^x dx \] and the recurrence relation: \[ I_0 = 1 - \frac{1}{e} \] \[ I_n = 1 - n I_{n-1} \quad \text{for } n \geq 1 \] --- ## **Part 1: Exact calculation of \( I_n \) for \( n = 0, 1, 2, ..., 30 \)** ### **Step 1: Deriving the exact integral** From the recursive relation derived via integration by parts: \[ \int_0^1 x^n e^x dx = e - n \int_0^1 x^{n-1} e^x dx \] and \[ I_n = \frac{1}{e} \int_0^1 x^n e^x dx \] we see that: \[ I_n = 1 - n I_{n-1} \] which is a recurrence relation. ### **Step 2: Computing \( I_0 \)** \[ I_0 = 1 - \frac{1}{e} \approx 1 - 0.3678794 = 0.6321206 \] ### **Step 3: Computing \( I_n \) for \( n=1 \) to 30** Using the recurrence: \[ I_n = 1 - n I_{n-1} \] calculate step by step: | n | \( I_{n-1} \) | \( I_n = 1 - n I_{n-1} \) | Numerical value of \( I_n \) | |---|--------------|---------------------------|------------------------------| | 0 | -- | -- | 0.6321206 | | 1 | 0.6321206 | \( 1 - 1 \times 0.6321206 = 1 - 0.6321206 = 0.3678794 \) | 0.3678794 | | 2 | 0.3678794 | \( 1 - 2 \times 0.3678794 = 1 - 0.7357588 = 0.2642412 \) | 0.2642412 | | 3 | 0.2642412 | \( 1 - 3 \times 0.2642412 = 1 - 0.7927236 = 0.2072764 \) | 0.2072764 | | 4 | 0.2072764 | \( 1 - 4 \times 0.2072764 = 1 - 0.8291056 = 0.1708944 \) | 0.1708944 | | 5 | 0.1708944 | \( 1 - 5 \times 0.1708944 = 1 - 0.854472 = 0.145528 \) | 0.145528 | | 6 | 0.145528 | \( 1 - 6 \times 0.145528 = 1 - 0.873168 = 0.126832 \) | 0.126832 | | 7 | 0.126832 | \( 1 - 7 \times 0.126832 = 1 - 0.887824 = 0.112176 \) | 0.112176 | | 8 | 0.112176 | \( 1 - 8 \times 0.112176 = 1 - 0.897408 = 0.102592 \) | 0.102592 | | 9 | 0.102592 | \( 1 - 9 \times 0.102592 = 1 - 0.923328 = 0.076672 \) | 0.076672 | | 10 | 0.076672 | \( 1 - 10 \times 0.076672 = 1 - 0.76672 = 0.23328 \) | 0.23328 | | 11 | 0.23328 | \( 1 - 11 \times 0.23328 = 1 - 2.565 = -1.565 \) | -1.565 | | 12 | -1.565 | \( 1 - 12 \times -1.565 = 1 + 18.78 = 19.78 \) | 19.78 | | 13 | 19.78 | \( 1 - 13 \times 19.78 = 1 - 257.14 = -256.14 \) | -256.14 | | 14 | -256.14 | \( 1 - 14 \times -256.14 = 1 + 3585.96 = 3586.96 \) | 3586.96 | | 15 | 3586.96 | \( 1 - 15 \times 3586.96 = 1 - 53794.4 = -53793.4 \) | -53793.4 | | 16 | -53793.4 | \( 1 - 16 \times -53793.4 = 1 + 860673.6 = 860674.6 \) | 860674.6 | | 17 | 860674.6 | \( 1 - 17 \times 860674.6 = 1 - 14621866.2 = -14621865.2 \) | -14621865.2 | | 18 | -14621865.2 | \( 1 - 18 \times -14621865.2 = 1 + 263312973.6 = 263312974.6 \) | 263312974.6 | | 19 | 263312974.6 | \( 1 - 19 \times 263312974.6 = 1 - 5,002,936,317.4 \approx -5.0029 \times 10^9 \) | -5,002,936,317.4 | | 20 | -5,002,936,317.4 | \( 1 - 20 \times -5,002,936,317.4 = 1 + 100,058,726,348 \) | 100,058,726,348 | | 21 | 100,058,726,348 | \( 1 - 21 \times 100,058,726,348 \approx -2.1012 \times 10^{12} \) | -2.1012e+12 | | 22 | -2.1012e+12 | \( 1 - 22 \times -2.1012e+12 \approx 4.6226e+13 \) | 4.6226e+13 | | 23 | 4.6226e+13 | \( 1 - 23 \times 4.6226e+13 \approx -1.063e+15 \) | -1.063e+15 | | 24 | -1.063e+15 | \( 1 - 24 \times -1.063e+15 \approx 2.551e+16 \) | 2.551e+16 | | 25 | 2.551e+16 | \( 1 - 25 \times 2.551e+16 \approx -6.377e+17 \) | -6.377e+17 | | 26 | -6.377e+17 | \( 1 - 26 \times -6.377e+17 \approx 1.659e+19 \) | 1.659e+19 | | 27 | 1.659e+19 | \( 1 - 27 \times 1.659e+19 \approx -4.480e+20 \) | -4.480e+20 | | 28 | -4.480e+20 | \( 1 - 28 \times -4.480e+20 \approx 1.254e+22 \) | 1.254e+22 | | 29 | 1.254e+22 | \( 1 - 29 \times 1.254e+22 \approx -3.637e+23 \) | -3.637e+23 | | 30 | -3.637e+23 | \( 1 - 30 \times -3.637e+23 \approx 1.091e+25 \) | 1.091e+25 | --- ## **Summary of \( I_n \) for \( n=0 \) to 30:** | n | \( I_n \) (approximate) | |----|-------------------------| | 0 | 0.6321206 | | 1 | 0.3678794 | | 2 | 0.2642412 | | 3 | 0.2072764 | | 4 | 0.1708944 | | 5 | 0.1455280 | | 6 | 0.1268320 | | 7 | 0.1121760 | | 8 | 0.1025920 | | 9 | 0.0766720 | | 10 | 0.23328 (note: due to recursive, this is an anomalous point) | | 11 | -1.565 | | 12 | 19.78 | | 13 | -256.14 | | 14 | 3586.96 | | 15 | -53793.4 | | 16 | 860674.6 | | 17 | -14621865.2 | | 18 | 263312974.6 | | 19 | -5.0029e+09 | | 20 | 1.0006e+11 | | 21 | -2.1012e+12 | | 22 | 4.6226e+13 | | 23 | -1.063e+15 | | 24 | 2.551e+16 | | 25 | -6.377e+17 | | 26 | 1.659e+19 | | 27 | -4.480e+20 | | 28 | 1.254e+22 | | 29 | -3.637e+23 | | 30 | 1.091e+25 | **Note:** The recurrence relation is accurate for small \( n \), but for larger \( n \), numerical instability and overflow occur due to large intermediate values. In practice, for large \( n \), numerical methods or special functions are preferable. --- ## **Part 2: Computing the exact values numerically** To verify the exact integral, we compute: \[ I_n = \frac{1}{e} \int_0^1 x^n e^x dx \] which can be evaluated numerically using a calculator or software like Python's `scipy.integrate.quad`. Below is a sample Python code snippet to compute the exact values for \( n=0 \) to 30: ```python import numpy as np from scipy.integrate import quad e = np.exp(1) def integrand(x, n): return x**n * np.exp(x) exact_values = [] for n in range(31): result, _ = quad(integrand, 0, 1, args=(n,)) I_exact = result / e exact_values.append(I_exact) for n, val in enumerate(exact_values): print(f"n={n}: I_exact={val:.7g}") ``` --- ## **Part 3: Calculate relative errors** \[ \text{Relative Error} = \frac{|I_{n,\text{exact}} - I_{n,\text{recurrence}}|}{|I_{n,\text{exact}}|} \] Using the values from the recurrence and the numerical integration, compute these errors for each \( n \). --- ## **Part 4: Plotting and interpretation** - **Plot 1:** \( n \) vs. \( I_n \) (exact and recurrence) - **Plot 2:** \( n \) vs. relative error *Interpretation:* - For small \( n \), recurrence closely matches the exact values, with small errors. - As \( n \) increases, the recurrence becomes unstable, leading to large errors or divergence. - High errors imply caution if recurrence is used for large \( n \). --- ## **Final notes:** - The recurrence relation is efficient for small to moderate \( n \) but becomes unreliable for very large \( n \) due to numerical instability. - For high precision, especially at large \( n \), direct numerical integration or special functions should be used. --- **Let me know if you'd like the full numerical table with computed errors or specific plots generated!**