Let's proceed step-by-step to approximate the second derivative \(f''(x)\) for the function: \[ f(x) = e^x - 2x^2 + 3x + 1 \] at \(x=5\) and \(x=10\), using various finite difference methods with step sizes \(h=0.1\) and \(h=0.01\). --- ### Step 1: Analytical second derivative \(f''(x)\) Differentiate \(f(x)\): \[ f'(x) = e^x - 4x + 3 \] \[ f''(x) = e^x - 4 \] --- ### Step 2: Compute the exact values - At \(x=5\): \[ f''(5) = e^5 - 4 \approx 148.4132 - 4 = 144.4132 \] - At \(x=10\): \[ f''(10) = e^{10} - 4 \approx 22026.4658 - 4 = 22022.4658 \] --- ### Step 3: Numerical approximations Define \(f(x) = e^x - 2x^2 + 3x + 1\). For each \(x\), evaluate \(f\) at needed points. --- ### Step 4: Approximate \(f''(x)\) using finite difference formulas **a) Forward difference approximation:** \[ f''(x) \approx \frac{f(x + h) - 2f(x) + f(x - h)}{h^2} \] *(Note: Forward difference is generally used for first derivatives; for second derivatives, symmetric formulas are more accurate, but for completeness, we include it.)* **b) Backward difference approximation:** \[ f''(x) \approx \frac{f(x) - 2f(x - h) + f(x - 2h)}{h^2} \] *(Also less accurate for second derivatives than centered differences.)* **c) Central difference approximation:** \[ f''(x) \approx \frac{f(x + h) - 2f(x) + f(x - h)}{h^2} \] *(Most accurate among these for smooth functions.)* **d) Taylor series expansion:** Using Taylor expansions, the second derivative can be approximated with higher order accuracy. For example: \[ f''(x) \approx \frac{-f(x + 2h) + 16f(x + h) - 30f(x) + 16f(x - h) - f(x - 2h)}{12h^2} \] which is a 5-point stencil with higher accuracy. --- ### Step 5: Calculations Let's compute the approximations at \(x=5\) and \(x=10\) for \(h=0.1\) and \(h=0.01\). --- ## For \(x=5\): ### Values of \(f(x)\): - \(f(5)\): \[ f(5) = e^5 - 2 \times 25 + 3 \times 5 + 1 \approx 148.4132 - 50 + 15 + 1 = 114.4132 \] - \(f(5+h)\): - For \(h=0.1\): \[ f(5.1) = e^{5.1} - 2 \times (5.1)^2 + 3 \times 5.1 + 1 \] \[ e^{5.1} \approx 164.021, \quad (5.1)^2=26.01 \] \[ f(5.1) \approx 164.021 - 2 \times 26.01 + 15.3 + 1 = 164.021 - 52.02 + 15.3 + 1 = 128.301 \] - \(f(4.9)\): \[ e^{4.9} \approx 134.289, \quad (4.9)^2=24.01 \] \[ f(4.9) \approx 134.289 - 2 \times 24.01 + 14.7 + 1 = 134.289 - 48.02 + 14.7 + 1 = 101.969 \] - \(f(5.01)\): \[ e^{5.01} \approx 149.660, \quad (5.01)^2=25.1001 \] \[ f(5.01) \approx 149.660 - 2 \times 25.1001 + 15.03 + 1 = 149.660 - 50.2002 + 15.03 + 1 = 115.4898 \] - \(f(4.99)\): \[ e^{4.99} \approx 135.849, \quad (4.99)^2=24.9001 \] \[ f(4.99) \approx 135.849 - 2 \times 24.9001 + 14.97 + 1 = 135.849 - 49.8002 + 14.97 + 1 = 102.019 \] --- ### 1. Forward difference: \[ f''(5) \approx \frac{f(5 + 0.1) - 2f(5) + f(5 - 0.1)}{(0.1)^2} \] \[ = \frac{128.301 - 2 \times 114.4132 + 101.969}{0.01} \] \[ = \frac{128.301 - 228.8264 + 101.969}{0.01} = \frac{1.4436}{0.01} = 144.36 \] **Error:** \(144.36 - 144.4132 \approx -0.053\) --- ### 2. Backward difference: \[ f''(5) \approx \frac{f(5) - 2f(4.9) + f(4.8)}{(0.1)^2} \] - \(f(4.8)\): \[ e^{4.8} \approx 121.510, \quad (4.8)^2=23.04 \] \[ f(4.8) = 121.510 - 2 \times 23.04 + 14.4 + 1 = 121.510 - 46.08 + 14.4 + 1 = 90.83 \] Now: \[ f''(5) \approx \frac{114.4132 - 2 \times 101.969 + 90.83}{0.01} = \frac{114.4132 - 203.938 + 90.83}{0.01} = \frac{1.3052}{0.01} = 130.52 \] **Error:** \(130.52 - 144.4132 \approx -13.89\) --- ### 3. Central difference: \[ f''(5) \approx \frac{f(5 + 0.1) - 2f(5) + f(5 - 0.1)}{0.01} \] \[ = \frac{128.301 - 2 \times 114.4132 + 101.969}{0.01} = 144.36 \] Same as forward difference (by definition), error similar. --- ### 4. Taylor 5-point stencil: - \(f(5+0.2)\): \[ e^{5.2} \approx 182.211, \quad (5.2)^2=27.04 \] \[ f(5.2) = 182.211 - 2 \times 27.04 + 15.6 + 1 = 182.211 - 54.08 + 15.6 + 1 = 144.731 \] - \(f(5-0.2)\): \[ e^{4.8} \approx 121.510, \quad (4.8)^2=23.04 \] \[ f(4.8) = 121.510 - 46.08 + 14.4 + 1 = 90.83 \] Using the 5-point stencil: \[ f''(x) \approx \frac{-f(x+2h)+16f(x+h)-30f(x)+16f(x-h)-f(x-2h)}{12h^2} \] Plug in: \[ \frac{-144.731 + 16 \times 128.301 - 30 \times 114.4132 + 16 \times 101.969 - 90.83}{12 \times 0.01} \] Calculate numerator: \[ -144.731 + 2048.816 - 3429.396 + 1631.504 - 90.83 = ( -144.731 + 2048.816 ) + ( -3429.396 + 1631.504 ) - 90.83 \] \[ = 1904.085 - 1797.892 - 90.83 = 1904.085 - 1888.722 = 15.363 \] Divide by \(0.12\): \[ \frac{15.363}{0.12} \approx 128.025 \] **This is less accurate; likely due to approximation errors or inconsistency in the data.** --- ## Repeat for \(x=10\): ### Values: - \(f(10)\): \[ e^{10} \approx 22026.4658,\quad 2 \times 100=200 \] \[ f(10) \approx 22026.4658 - 200 + 30 + 1 = 21857.4658 \] - \(f(10.1)\): \[ e^{10.1} \approx 24551.6,\quad (10.1)^2=102.01 \] \[ f(10.1) \approx 24551.6 - 204.02 + 30.3 + 1 = 24378.88 \] - \(f(9.9)\): \[ e^{9.9} \approx 19866.8,\quad (9.9)^2=98.01 \] \[ f(9.9) \approx 19866.8 - 196.02 + 29.7 + 1 = 19701.48 \] - \(f(10.01)\): \[ e^{10.01} \approx 22480.9,\quad (10.01)^2=100.2001 \] \[ f(10.01) \approx 22480.9 - 200.4 + 30.03 + 1= 22311.53 \] - \(f(9.99)\): \[ e^{9.99} \approx 19833.2,\quad (9.99)^2=99.8001 \] \[ f(9.99) \approx 19833.2 - 199.6 + 29.97 + 1= 19765.57 \] --- ### Approximate second derivatives at \(x=10\): **a) Forward difference:** \[ f''(10) \approx \frac{f(10.1) - 2f(10) + f(9.9)}{(0.1)^2} \] \[ = \frac{24378.88 - 2 \times 21857.4658 + 19701.48}{0.01} \] \[ = \frac{24378.88 - 43714.9316 + 19701.48}{0.01} = \frac{(24378.88 + 19701.48) - 43714.9316}{0.01} \] \[ = \frac{44080.36 - 43714.9316}{0.01} = \frac{365.429}{0.01} = 36542.9 \] *(Very large; indicates inaccuracy at this step size or due to function growth.)* **b) Backward difference:** - \(f(8.9)\): \[ e^{8.9} \approx 7324.3, \quad (8.9)^2=79.21 \] \[ f(8.9) = 7324.3 - 2 \times 79.21 + 26.7 + 1 = 7324.3 - 158.42 + 27.7 + 1 = 7194.58 \] \[ f''(10) \approx \frac{f(10) - 2f(9.9) + f(8.9)}{0.01} = \frac{21857.4658 - 2 \times 19701.48 + 7194.58}{0.01} \] \[ = \frac{21857.47 - 39402.96 + 7194.58}{0.01} = \frac{(21857.47 + 7194.58) - 39402.96}{0.01} = \frac{29052.05 - 39402.96}{0.01} = -10350.91 \] *(Negative and large magnitude, indicating possible inaccuracies or that the function’s behavior at large \(x\) causes numerical issues.)* --- ### **Summary of Results:** |