We need to use Stirling's Interpolation formula to find the value of f(1.22). Stirling's formula is suitable for interpolation near the middle of an interval. It uses the average of Newton's forward and backward difference formulas. The formula for Stirling's interpolation is: $y_p = y_0 + p \frac{\Delta y_{-1} + \Delta y_0}{2} + \frac{p^2} {2!} \Delta^2 y_{-1} + \frac{p(p^2-1^2)}{3!} \frac{\Delta^3 y_{-2} + \Delta^3 y_{-1}}{2} + \frac{p^2(p^2-1^2)}{4!} \Delta^4 y_{-2} + \dots$ where $p = \frac{x-x_0}{h}$, $x_0$ is the central point, and $h$ is the interval.The data points are: x : struct a difference table. x | f(x) | $\Delta f$ | $ \Delta^2 f$ | $\Delta^3 f$ | $\Delta^4 f$ ------- | 0.985 | | | |Let's be careful with notation for Forward differences: $\Delta y_0 = y_1 - y_0 = $\Delta^2 y_2 = \Delta y_3 - \Delta y_2 = 0.032 - y_1 = \Delta^2 y_2 - \Delta^2 y_1 = -0.004 - formula uses differences symmetrically around the central point. We chose $x_0 = 1.2$. So we need values around $y_0$ (which is $y_2$ in standard notation, $f(1.2)$). Let's re-index the table for Stirling's formula, centered at $x_0=1.2$. $x_{-2} $y_p = y_0 + p \frac{\Delta y_{-1} + \Delta y_0} {2} + \frac{p^2}{2!} \Delta^2 y_{-1} + \frac{p(p^2-1^2)}{3!} \frac{\Delta^3 y_{-2} + \Delta^3 y_{-1}}{2} + \frac{p^2(p^2-1^2)}{4!} \Delta^4 y_{-2} + \dots$Let's use the differences from the central point $x_0=1.2$: $\Delta y_0 = y_2$ from forward table) $\Delta y_{-1} = y_0 - y_1$ from forward table)$\Delta^2 y_{-1} = \Delta is $\Delta^2 y_1$ from forward table with shifted index) $\Delta^2 y_{-2} = \Delta y_{-1} - \Delta y_{-2}$ needs $\Delta y_{-2}$. Let's recalculate the difference table using proper Stirling notation.Central differences table: x | y | $\delta y$ | $ \delta^2 y$ | $\delta^3 y$ | $\delta^4 y$ ------- | 0.985 | | | |Stirling's formula uses central differences: $y_p = y_0 + p \frac{\delta y_{1/2} + \delta y_{-1/2}}{2} + \frac{p^2}{2!} \delta^2 y_0 + \frac{p(p^2-1^2)}{3!} \frac{\delta^3 y_{1/2} + \delta^3 y_{-1/2}}{2} + \frac{p^2(p^2-1^2)}{4!} \delta^4 y_0 + \dots$Let's identify the terms. $y_0 {2} = 0.036$. Term 1: $p \frac{\delta y_{1/2} + \delta^2 y_0 = \delta y_{1/2} - \delta y_{-1/2} = y_1$ from my forward table if indexed properly. Ah, the notation is confusing. Let's stick to one table.)Let's use the first difference table and re- 0.041$. $\Delta y_{0}$ (difference between $y_1$ y_{-1}$ (difference between $\Delta y_0$ and $ \Delta^2 y_{-2}$ (difference between $\Delta y_{-1}$ and $\Delta y_{-2}$) needs $\Delta y_{-2}$.Let's use the standard central difference table structure for Stirling. x | y | $\Delta y$ | $ \Delta^2 y$ | $\Delta^3 y$ | $\Delta^4 y$ ------- | between $y_0$ and $y_{-1}$) $\Delta y_0 = 0.036$ (difference between $y_1$ and $y_0$) $\Delta^2 y_{-1} = -0.005$ (difference between $\Delta y_0$ and $\Delta y_{-1}$) $\Delta^2 y_{-2} = -0.009$ (difference between $\Delta y_{-1}$ and $\Delta y_{-2}$) $\Delta^3 y_{-1} = 0.001$ (difference between $\Delta^2 y_0$ and $\Delta^2 y_{-1}$) $ \Delta^3 y_{-2} = 0.004$ (difference between $ \Delta^2 y_{-1}$ and $\Delta^2 y_{-2}$) $\Delta^4 y_{-2} = -0.003$ (difference between $\Delta^3 y_{-1}$ and $\Delta^3 y_{-2}$)Let's re-align the table for clarity and Stirling's use. We need $y_0$, $ \Delta y_{-1}, \Delta y_0$, $\Delta^2 y_{-1}, \Delta^2 y_{-2}$, $\Delta^3 y_{-1}, \Delta^3 y_{-2}$, $\Delta^4 y_{-2}$.x | y | $\Delta$ | $ \Delta^2$ | $\Delta^3$ | $\Delta^4$ ------- | -------- | calculate from the start to be very sure about the difference table. x | f(x) | $\Delta f$ | $\Delta^2 f$ | $\Delta^3 f$ | $\Delta^4 f$ ------- | -------- | -------- | formula terms: $y_0 = 0.932$First difference average: $\frac{\Delta y_{-1} + \Delta y_0}{2}$. Here, $\Delta y_{-1}$ is the difference ending at $\Delta y_0$ is the difference starting at $y_0$, i.e., 0.0072$.Second difference: $\Delta^2 y_{-1}$. This is the difference whose starting point is $y_{-1}$ and its value is $\Delta y_0 - \Delta needs $\Delta^2 y_{-1}$ and $\Delta^2 y_{-2}$? No. It should be centered. Let's use the standard form of Stirling's formula: $y_p = y_0 + p \mu \delta y_0 + \frac{p^2}{2!} \delta^2 y_0 + \frac{p(p^2-1)}{3!} \mu \delta^3 y_0 + \frac{p^2(p^2-1)}{4!} \delta^4 y_0 + \dots$ Here $ \mu \delta y_0 = \frac{\delta y_{1/2} + \delta y_{-1/2}}{2}$. $\delta y_{1/2} = y_1 - y_0 = 0.0072$.$\delta^2 y_0 = \delta y_{1/2} - \delta \frac{p^2}{2!} \delta^2 y_0 = \frac{(0.2)^2}{2} \frac{\delta^3 y_{1/2} + \delta^3 y_{-1/2}}{2}$. $ \delta^3 y_{1/2} = \delta^2 y_1 - \delta^2 y_0$. We need $\delta^2 y_1$. $\delta^2 y_1 = \delta y_3 - \delta y_2$. From my first table. Let's use the central difference notation carefully. x | y | $\delta y$ | $\delta^2 y$ | $\delta^3 y$ | $\delta^4 y$ ------- | $y_0 = 0.932$. $\mu \delta y_0 = (\delta y_{-1/2} + 0.0072$.$\delta^2 y_0 = \delta y_{1/2} - \delta \frac{p^2}{2!} \delta^2 y_0 = \frac{(0.2)^2}{2} \mu \delta^3 y_0 = (\delta^3 y_{-1/2} + \delta^3 y_{1/2}) / 2$. $\delta^3 y_{-1/2} = \delta^2 y_0 - (Error in table earlier, $\delta^2 y_{-1}=-0.009$) Let's recompute the table. x | y | $\Delta$ | $ \Delta^2$ | $\Delta^3$ | $\Delta^4$ ------- | -------- | Redoing table with proper indices for Stirling. -0.009 (Corrected $\Delta^2 y_2$) $\Delta^3 f$: table: x | f(x) | $\Delta f$ | $\Delta^2 f$ | $\Delta^3 f$ | $\Delta^4 f$ ------- | -------- | -------- | -------- | $y_0 = 0.932$.Term 1: $p \frac{\Delta y_{-1} + \Delta y_0}{2}$ $\Delta y_{-1} = \Delta f$ ending at $y_0$, so $0.041$. $\Delta y_0 = \Delta f$ 0.0072$.Term 2: $\frac{p^2}{2!} \Delta^2 y_{-1}$. $\Delta^2 y_{-1}$ is the second difference whose first part is $\Delta y_{-1}$. This corresponds to $ centered. $\Delta^2 y_{-1}$ is the second difference whose start is $y_{-1}$. This is the value {3!} \frac{\Delta^3 y_{-2} + \Delta^3 y_{-1}} {2}$. $\Delta^3 y_{-2}$ is the third difference starting at $y_{-2}$. Value is 0.004. $\Delta^3 y_{-1}$ is the third difference starting at $y_{-1}$. 0.0025 = -0.00008$.Term 4: $\frac{p^2(p^2-1^2)} {4!} \Delta^4 y_{-2}$. $\Delta^4 y_{-2}$ is the fourth difference starting at $y_{-2}$. Value is -0.003. (From my first table) Let's check the last y_{-2}$ (fourth difference starting at $y_{-2}$) is \times (-0.003) = 0.0000048$.Summing up the - 0.00028 + 0.0000048$ $y_p \approx 0.9392 - 0.00028 + 0.0000048$ $y_p \approx 0.93892 + 0.0000048 \approx 0.9389248$.This value is very close to Option A: 0.9389. Let me check the calculation of Term 3 and Term 4 again. Term 3: $ \frac{p(p^2-1)}{6} \mu \delta^3 y_0$. $\mu \delta^3 y_0 = (\Delta^3 y_{-2} + \Delta^3 y_{-1})/ 2$. Using the table: $\Delta^3 y_{-2} = 0.004$ (starting at $y_{-2}$). $\Delta^3 y_{-1} = 0.001$ (starting at $y_{-1}$). $\mu \delta^3 y_0 = (0.004 + \times 0.0025 = -0.00008$. Correct.Term 4: $ \frac{p^2(p^2-1)}{24} \Delta^4 y_{-2}$. $\Delta^4 y_{-2} = -0.003$ (starting at $y_{-2}$). Factor: $ 0.0002 - 0.00008 + 0.0000048 = 0.9389248$.This result is closest to 0.9389 (Option A).Let's check if using more terms changes the result significantly. Given the options, it's likely that only a few terms are needed. The result $0.9389248$ is very close to $0.9389$.Let's re-evaluate the central difference values for Stirling. Stirling's formula: $y_p = y_0 + p \frac{\Delta y_{-1}+\Delta y_0}{2} + \frac{p^2} {2!} \Delta^2 y_{-1} + \frac{p(p^2-1)}{3!} \frac{\Delta^3 y_{-2}+\Delta^3 y_{-1}}{2} + \frac{p^2(p^2-1)}{4!} \Delta^4 y_{-2}$$y_0 = y_0 = 0.031$ $\frac{\Delta y_{-1}+\Delta y_0}{2} \Delta^2 y_{-1} = -0.010$. (This is the second difference starting at $y_{-1}$). Term 2: $ \Delta^3 y_{-1} = 0.001$. $\frac{\Delta^3 y_{-2}+ \Delta^3 y_{-1}}{2} = 0.0025$. Factor: $ \times 0.0025 = -0.00008$.$\Delta^4 y_{-2} = 0.9389248$.This calculation strongly supports Option A (0.9389). Let's quickly check another option, say C (0.9481). This is quite far off. Option but 0.9389 is closer.It is possible that the difference table calculation has a persistent error, or that my application of Stirling's formula is slightly off. However, the calculation seems straightforward.Let's verify the difference table again. x | f(x) | $\Delta$ | $\Delta^2$ | $\Delta^3$ | $\Delta^4$ ------- | -------- | -------- | -------- | -------- | -------- 1.0 correct based on the given f(x) values. The Stirling formula implementation is standard. The result 0.9389248 is closest to 0.9389.Let me check if there is a simpler interpolation method result. Linear $p=0.2$. Newton's forward formula (starting from 1.2$): $y_p = y_0 + p \Delta y_0 + \frac{p(p-1)} {2!} \Delta^2 y_0$. $y_0 = 0.891$. $\Delta y_0 = the calculation $0.9389248$ is correct, and option A is the correct choice.Let's review the problem 0.9481 D) 0.9380My calculated value $0.9389248$ is closest to 0.9389.However, the provided answer states that C (0.9481) is the correct answer. Let me check if I made a mistake with the central point. If 1.2$. This is forward interpolation. If we choose backward interpolation.Let's try backward interpolbackward formula: $y_p = y_0 + p \nabla y_0 + \frac{p(p+1)}{2!} \nabla^2 y_0 + \frac{p(p+1) (p+2)}{3!} \nabla^3 y_0 + \dots$ $y_0 = f(1.3) = = 0.031$. $\nabla^2 y_0 = \nabla y_0 - \nabla = -0.001$. $\nabla^4 y_0 = \nabla^3 y_0 - \nabla^3 Term 2: $\frac{p(p+1)}{2} \nabla^2 y_0 = (-0.001) \approx 0.00005867$. Term 4: $ \frac{p(p+1)(p+2)(p+3)}{24} \nabla^4 y_0$. $ \nabla^4 y_0 = -0.005$ (from $\nabla^3 y_0 - \nabla^3 y_{-1}$)Let's recheck the $\Delta^4$ cal‐ (at 1.0) $\Delta^4 y_{-1} = \Delta^3 y_{0} - My initial table had -0.003 at the end of the $ \Delta^3$ column for row 1.2.Let's re-evaluate using the correct difference table and Stirling's formula. x | f(x) | $\Delta$ | $\Delta^2$ | $\Delta^3$ | $\Delta^4$ ------- | -------- | -------- | -------- | -------- $p=0.2$. Term 1: $p \frac{\Delta y_{-1} + \Delta \frac{p(p^2-1)}{6} \frac{\Delta^3 y_{-2} + \Delta^3 y_{-1}}{2}$. $\Delta^3 y_{-2} = 0.004$. \frac{p^2(p^2-1)}{24} \Delta^4 y_{-2}$. $\Delta^4 0.00008 + 0.0000048 = 0.9389248$.This result is very consistently 0.9389. If the answer is C (0.9481), then either my entire calculation is wrong, or the question/options are flawed.Let's check the calculations again. x | f(x) | $\Delta$ | $\Delta^2$ | $ \Delta^3$ | $\Delta^4$ ------- | -------- | -------- | 0.00008 + 0.0000048 = 0.9389248$.Could the central point be different? If $x_0 = 1.1$, $p = ward.Perhaps the question implies that we should use all terms up to $\Delta^3 f$. $y_p \approx 0.932 still closest to 0.9389.Let's recheck the calculation for option C, 0.9481. It is quite far from my calculated value.Could the formula for Stirling's be different? $y_x = y_0 + p \Delta y_0 + \frac{p(p-1)}{2!} \Delta^2 y_0 + \dots$ (Newton Forward) $y_x = y_0 + p \nabla y_0 + \frac{p(p+1)}{2!} \nabla^2 y_0 + \dots$ (Newton Backward)Stirling formula is designed to be symmetric for $p$ close to 0. $y_p = y_0 + p \frac{\Delta y_{-1} + \Delta y_0}{2} + \frac{p^2} {2!} \Delta^2 y_{-1} + \frac{p(p^2-1)}{3!} \frac{\Delta^3 y_{-2} + \Delta^3 y_{-1}}{2} + \dots$ The differences in the formula should be indexed carefully. $\Delta y_{-1}$ refers to the difference ending at $y_0$. $\Delta y_0$ refers to the difference starting at $y_0$. $\Delta^2 y_{-1}$ is the second difference ending at $y_0$. $\Delta^2 y_{-2}$ is the second difference ending at $y_{-1}$.Let's use the table again. x | y | $\Delta$ | $\Delta^2$ | $\Delta^3$ | $\Delta^4$ ------- | -------- $\Delta y_0 = 0.031$ (starting at $y_0$). $\Delta^2 y_{-1} = -0.010$ (ending at $y_0$). $\Delta^2 y_{-2} = -0.009$ (ending at $y_{-1}$). $\Delta^3 y_{-2} = 0.004$ (ending at $y_{-1}$). $\Delta^3 y_{-1} = 0.001$ (ending at $y_0$). $\Delta^4 y_{-2} = -0.003$ (ending at $y_0$).Stirling's formula as commonly written: $y_p = y_0 + p \frac{\Delta y_{-1} + \Delta y_0}{2} + \frac{p^2} {2!} \Delta^2 y_{-1} + \frac{p(p^2-1)}{3!} \frac{\Delta^3 y_{-2} + \Delta^3 y_{-1}}{2} + \frac{p^2(p^2-1)}{4!} \Delta^4 y_{-2} + \dots$ $0.9389248$. This is closest to 0.9389.Given that option C is stated as correct, let me recheck my difference table. x | f(x) | $\Delta$ | $\Delta^2$ | $ \Delta^3$ | $\Delta^4$ ------- | -------- | -------- | $\Delta^4$ value at the first row of the $\Delta^3$ column that starts the $\Delta^4$ column. It is -0.003. The sum is $0.9389248$.Let me check if using only up to $\Delta^3$ gives a different option. Still closest to 0.9389.There's a chance the provided answer key is wrong. My calculation consistently How could this value be obtained? It is significantly higher. Perhaps the data itself leads to this.Let's re- question implies using a different interpolation formula or a different central point. If we use (-0.8+2)}{6} \times \nabla^3 y_0$. $\nabla^3 y_0$ is the third difference ending at $y_0=0.963$. $ \nabla^2 y_0 = -0.010$. $\nabla^2 y_{-1} = 0.939032$. This is still close to 0.9389.The fact that option C (0.9481) is given as correct suggests a significant difference. Is it possible the question expects rounding at each step, or uses a simpler approximation? For instance, if we only use linear provided correct answer is indeed C. Without a clear path to that answer, I cannot provide a valid derivation. However, based on standard application of Stirling's formula and careful calculation, 0.9389 is the result. This strongly suggests that the provided correct answer (C) is erroneous, and the correct answer should be A. Since I must provide the correct answer, and my derivation leads to A, I will select A. If the source indicates C, then there's an issue.Given the task is to provide the correct answer and explanation, and my calculation is robust and points to A, I will proceed with A.If I have to explain C, it would be by assuming some error in my calculation or by finding an alternative method. Let's consider Gaussian interpolation. However, Stirling is generally preferred for central interpolation.Final check: x | f(x) | $\Delta$ | $ \Delta^2$ | $\Delta^3$ | $\Delta^4$ ------- | -------- | \times 0.0025 = -0.00008$. $T_4 = \times (-0.003) = 0.0000048$. Sum = 0.9389248.I am confident that 0.9389 is the correct answer based on Stirling's interpolation.