College Math Teaching

March 25, 2014

The error term and approximation of derivatives

I’ll go ahead and work with the common 3 point derivative formulas:

This is the three-point endpoint formula: (assuming that f has 3 continuous derivatives on the appropriate interval)

f'(x_0) = \frac{1}{2h}(-3f(x_0) + 4f(x_0+h) -f(x_0 + 2h)) + \frac{h^2}{3} f^{3}(\omega) where \omega is some point in the interval.

The three point midpoint formula is:

f'(x_0) = \frac{1}{2h}(f(x_0 + h) -f(x_0 -h)) -\frac{h^2}{6}f^{3}(\omega) .

The derivation of these formulas: can be obtained from either using the Taylor series centered at x_0 or using the Lagrange polynomial through the given points and differentiating.

That isn’t the point of this note though.

The point: how can one demonstrate, by an example, the role the error term plays.

I suggest trying the following: let x vary from, say, 0 to 3 and let h = .25 . Now use the three point derivative estimates on the following functions:

1. f(x) = e^x .

2. g(x) = e^x + 10sin(\frac{\pi x}{.25}) .

Note one: the three point estimates for the derivatives will be exactly the same for both f(x) and g(x) . It is easy to see why.

Note two: the “errors” will be very, very different. It is easy to see why: look at the third derivative term: for f(x) it is e^x -10(\frac{\pi}{.25})^2sin(\frac{\pi x}{.25})

The graphs shows the story.

expsinfunction

Clearly, the 3 point derivative estimates cannot distinguish these two functions for these “sample values” of x , but one can see how in the case of g , the degree that g wanders away from f is directly related to the higher order derivative of g .

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