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

Deriving a difference quotient

mark

Part 1

Show that

$f''(x) \approx \frac{2f(x+h) - 3f(x)+f(x-2h)}{3h^2}.$

What is the order of the approximation expressed in $O$ notation?

Part 2

Apply part 1 to the data set

$\begin{array}{c|cccc} x&0 & 0.2 & 0.3 & 0.4 \\ \hline y&1 & 1.22 & 1.27 & 1.28 \\ \end{array}$

mark

nathan

Part 1

First let’s write out the Taylor expansion for $2f(x+h)$ and $f(x-2h)$ .

$f(x-2h) \approx f(x) - 2f'(x)h + 2f''(x)h^{2} + O(h^{3})$

$2f(x+h) \approx 2f(x) + 2f'(x)h + f''(x)h^2 + O(h^3)$

If we add these two equations together we get:

$f(x-2h) - 3f(x) + 2f(x+h) \approx 3f''(x)h^2 + 2O(h^3).$

Now divide both sides by $3h^2$ .

$f''(x) \approx \frac{2f(x+h) - 3f(x) + f(x-2h)}{3h^2} - O(h)$

Part 2

Our value of $h$ will be $\Delta x$ since we are restricted by these four data points. So, $h=0.1$ . The question does not specify which $x$ value that we need to approximate the second derivative for, but as it turns out, $f''(0.2)$ is the only one allowed given the data.

$f''(0.2) \approx \frac{2f(0.3) - 3f(0.2) + f(0)}{0.03}$

$\approx \frac{2(1.27) - 3(1.22) + 1}{0.03}$

$f''(0.2) \approx -4$