Let's solve the problem step-by-step: --- ## **Given:** \[ y = e^{-x^2} = \sum_{i=0}^{\infty} \frac{(-1)^i x^{2i}}{i!} \] We are to use \( x = 2 \). --- ### **a. Set up a table for the summation vs \( i \) for \( x = 2 \):** Let's write out the first few terms for \( x = 2 \): \[ y = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \frac{x^8}{4!} - \ldots \] Substitute \( x = 2 \): - \( x^2 = 4 \) - \( x^4 = 16 \) - \( x^6 = 64 \) - \( x^8 = 256 \) Let's compute the terms for \( i = 0 \) to \( 4 \): | \(i\) | Term formula | Value | Cumulative sum | |-------|------------------------------------|------------------|-------------------| | 0 | \( \frac{(-1)^0 2^{0}}{0!} \) | \( 1 \) | 1 | | 1 | \( \frac{(-1)^1 2^{2}}{1!} \) | \( -4 \) | 1 - 4 = **-3** | | 2 | \( \frac{(-1)^2 2^{4}}{2!} \) | \( 16/2 = 8 \) | -3 + 8 = **5** | | 3 | \( \frac{(-1)^3 2^{6}}{3!} \) | \( -64/6 ≈ -10.6667 \) | 5 - 10.6667 = **-5.6667** | | 4 | \( \frac{(-1)^4 2^{8}}{4!} \) | \( 256/24 ≈ 10.6667 \) | -5.6667 + 10.6667 = **5** | So, the table: | \(i\) | Term | Value | Cumulative Sum | |-------|-----------------------------|--------------|----------------| | 0 | \( 1 \) | 1 | 1 | | 1 | \( -4 \) | -4 | -3 | | 2 | \( 8 \) | 8 | 5 | | 3 | \( -10.6667 \) | -10.6667 | -5.6667 | | 4 | \( 10.6667 \) | 10.6667 | 5 | --- ### **b. Compare with the actual value using \( e^{-x^2} \) where \( x = 2 \):** \[ y = e^{-2^2} = e^{-4} \] Calculate \( e^{-4} \): \[ e^{-4} \approx 0.0183 \] --- ### **c. Discussion** - With only the first five terms, the partial sums oscillate and do **not** converge quickly for \( x = 2 \). - The actual value \( e^{-4} \approx 0.0183 \) is **much smaller** than the partial sums calculated above. - You would need many more terms for the partial sum to approach \( 0.0183 \). --- ### **Final Answer** #### **a. Table (for \( x = 2 \)):** | \(i\) | Term Formula | Value | Cumulative Sum | |-------|-----------------------------------|--------------|----------------| | 0 | \( 1 \) | 1 | 1 | | 1 | \( -4 \) | -4 | -3 | | 2 | \( 8 \) | 8 | 5 | | 3 | \( -10.6667 \) | -10.6667 | -5.6667 | | 4 | \( 10.6667 \) | 10.6667 | 5 | #### **b. Comparison** - **Series partial sum after 5 terms:** \( 5 \) - **Actual value \( e^{-4} \):** \( 0.0183 \) - **The series converges very slowly for \( x = 2 \); you need many terms for an accurate result.** --- If you want, you can continue the series in Excel using `=(-1)^i * (2)^(2*i) / FACT(i)` to see how many terms it takes to get close to `0.0183`.

Let's proceed step-by-step to analyze the series and compare it with the exact value: --- ### Step 1: Understand the series The series expansion for \( e^{-x^2} \) is given by: \[ e^{-x^2} = \sum_{i=0}^{\infty} \frac{(-1)^i x^{2i}}{i!} \] We are asked to evaluate this series for \( x = 2 \). --- ### Step 2: Compute the first few terms Calculate each term explicitly: - For \( i=0 \): \[ \frac{(-1)^0 \times 2^{0}}{0!} = \frac{1 \times 1}{1} = 1 \] - For \( i=1 \): \[ \frac{(-1)^1 \times 2^{2}}{1!} = \frac{-1 \times 4}{1} = -4 \] - For \( i=2 \): \[ \frac{(-1)^2 \times 2^{4}}{2!} = \frac{1 \times 16}{2} = 8 \] - For \( i=3 \): \[ \frac{(-1)^3 \times 2^{6}}{3!} = \frac{-1 \times 64}{6} \approx -10.6667 \] - For \( i=4 \): \[ \frac{(-1)^4 \times 2^{8}}{4!} = \frac{1 \times 256}{24} \approx 10.6667 \] Now, sum these terms cumulatively: - Sum after \( i=0 \): \( 1 \) - Sum after \( i=1 \): \( 1 - 4 = -3 \) - Sum after \( i=2 \): \( -3 + 8 = 5 \) - Sum after \( i=3 \): \( 5 - 10.6667 \approx -5.6667 \) - Sum after \( i=4 \): \( -5.6667 + 10.6667 = 5 \) --- ### Step 3: Compare with the exact value The exact value of \( e^{-x^2} \) at \( x=2 \): \[ e^{-4} \approx 0.0183 \] ### Step 4: Observation - The partial sums after just 5 terms are around 5, which is vastly larger than \( 0.0183 \). - This indicates that the series converges very slowly at \( x=2 \) and takes many more terms for the partial sum to approach the true value. - As the number of terms increases, the sum will oscillate and gradually approach \( 0.0183 \). ### **Final Summary:** - **Setup of the first few terms:** - \( i=0 \): \( 1 \) - \( i=1 \): \( -4 \) - \( i=2 \): \( 8 \) - \( i=3 \): \( -10.6667 \) - \( i=4 \): \( 10.6667 \) - **Partial sum after these terms:** approximately **5** - **Actual value \( e^{-4} \):** approximately **0.0183** - **Conclusion:** To get an accurate approximation of \( e^{-4} \) from this series, especially at \( x=2 \), many more terms are required because the series converges slowly for larger \( x \). --- **Note:** In practice, for computational accuracy, use software like Excel with the `FACT()` function to add more terms until the partial sum stabilizes near 0.0183.

The series for \( e^{-x^2} \) converges slowly for larger \( x \), requiring many terms for accurate approximation. As \( x = 2 \), initial partial sums significantly overestimate the true value \( e^{-4} \approx 0.0183 \), highlighting the slow convergence of the series in this region.