Numerical check of difference schemes

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

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 $O(h^2)$.