An archive of Mark's Spring 2018 Numerical Analysis course.

Symmetric difference quotient and data

mark

Suppose f is known to satisfy the data

\begin{array}{c|cccccc} x & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline y & 0 & 0.04 & 0.16 & 0.35 & 0.6 & 0.84 \\ \end{array}
  • Use the symmetric difference quotient for f' to estimate f'(0.4).
  • Use the symmetric difference quotient for f'' to estimate f''(0.4).
mark
mark
Cornelius

The symmetric difference quotient for f'(x) is:

f'(x) \approx \frac{f(x+h)-f(x-h)}{2h}.

Let us approximate f(0.4) using h=0.2.

f'(x) \approx \frac{f(0.6)-f(0.2)}{2(0.2)} = \frac{0.35-0.04}{0.4}.
Cornelius

f''(x) \approx \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}

Once again we are going to pick h = 0.2.

So f''(0.4) \approx \frac{f(0.6) + 2f(0.4) + f(0.2)}{0.2^2} = \frac{0.35-(2)(0.16)+0.04}{0.2^2}

opernie

way to take the low hanging fruit

MatheMagician

This forum is getting savage.