--- title: "Numerical check of difference schemes" subject: "Applied Computational Fluid Dynamics" chapter: 3 key: "semester_7/applied_computational_fluid_dynamics:finite-difference-derivative-numerical" source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/finite-difference-derivative-numerical site: BCT Engineering --- # Numerical check of difference schemes Applied Computational Fluid Dynamics · Chapter 3: Discretization Methods - Asked in: year not given, 6 marks - Video: [Numerical Differentiation | Forward, Backward and Central Difference | Numerical Computation](https://www.youtube.com/watch?v=5zwxZ439nTA) — The Infinite Math - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/finite-difference-derivative-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-3 ## Question For the function $f(x)=\sin x$ (x in radians), evaluate $f'(x)$ at $x=1.0$ using forward, backward and central differences with $h=0.1$ and then $h=0.05$. Compare with the exact value, find the percentage errors and verify the order of accuracy of each scheme. ## Short answer Exact cos(1)=0.5403. With h=0.1: forward 0.4974, backward 0.5814, central 0.5394. With h=0.05: forward 0.5190, backward 0.5611, central 0.5401. Central error reduces by ~4 (second order), forward/backward by ~2 (first order). ## Answer Exact: $f'(1)=\cos 1=0.54030$. Values needed: $\sin0.9=0.783327$, $\sin0.95=0.813416$, $\sin1.0=0.841471$, $\sin1.05=0.867423$, $\sin1.1=0.891207$. #### h = 0.1 - Forward: $\dfrac{\sin1.1-\sin1.0}{0.1}=\dfrac{0.891207-0.841471}{0.1}=0.49736$ - Backward: $\dfrac{0.841471-0.783327}{0.1}=0.58144$ - Central: $\dfrac{0.891207-0.783327}{0.2}=0.53940$ #### h = 0.05 - Forward: $\dfrac{0.867423-0.841471}{0.05}=0.51904$ - Backward: $\dfrac{0.841471-0.813416}{0.05}=0.56111$ - Central: $\dfrac{0.867423-0.813416}{0.1}=0.54008$ #### Errors | Scheme | $h$ | Value | Error | % error | |---|---|---|---|---| | Forward | 0.1 | 0.49736 | -0.04294 | 7.95 | | Forward | 0.05 | 0.51904 | -0.02126 | 3.93 | | Backward | 0.1 | 0.58144 | +0.04114 | 7.61 | | Backward | 0.05 | 0.56111 | +0.02081 | 3.85 | | Central | 0.1 | 0.53940 | -0.00090 | 0.17 | | Central | 0.05 | 0.54008 | -0.00023 | 0.04 | #### Order of accuracy Ratio of errors when $h$ is halved: - Forward: $0.04294/0.02126=2.02\approx2^1$, so first order. - Backward: $0.04114/0.02081=1.98\approx2^1$, so first order. - Central: $0.00090/0.000225=4.0=2^2$, so second order. Here error means (numerical $-$ exact). The signs agree with theory: the forward formula equals $f'+\tfrac h2f''$, and since $f''=-\sin x<0$ its error is negative; the backward error has the opposite sign; the central error is far smaller. **Answer:** Central difference is best (0.5394 and 0.5401 against exact 0.5403); forward and backward are $O(h)$ and central is $O(h^2)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For sin x at x = 1 with h = 0.1, which scheme gives the closest value to cos 1 = 0.5403? A. Forward, 0.4974 B. Backward, 0.5814 C. Central, 0.5394 D. All equal Answer: C. Central difference gives 0.5394 (0.17% error) compared with about 8% for forward and backward. (Key point: Best scheme) 2. How is the order of accuracy verified? A. By comparing signs of errors B. By the ratio of errors when h is halved C. By checking the exact value only D. By using larger h Answer: B. A ratio of about 2 means first order, and about 4 means second order. (Key point: Order verification) 3. For the central scheme the error ratio on halving h was about: A. 2 B. 1 C. 4 D. 8 Answer: C. 0.00090/0.000225 = 4, confirming second order. (Key point: Central error ratio) 4. The forward error is negative because: A. The forward formula equals f' + (h/2)f'' and f'' = -sin x < 0 B. The step is too large C. Round-off dominates D. cos 1 is negative Answer: A. Forward estimate = f' + (h/2) f'', and f'' is negative for sin x at x = 1, so error is negative. 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