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HamCall (October 1991)
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1989-03-07
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TROUBLE
This slide show consists of various graphs of the same function, seen over
different domains.
Frequently people are under the impression that if "enough points" of a function
are plotted, and then joined, the resultant graph accurately portrays the
original function. This is the principle used by all computer graphing
packages. People are even questioning the wisdom of teaching curve sketching in
calculus, because of such packages. The following slide show is designed to
display some of the pitfalls of graphing by plotting points, and then joining
them. In fact calculus is essential in deciding the accuracy of any graph.
The function used exclusively throughout this slide show is
1.2
f(x) = sin[4000 π (x + 1) ] ,
which was created for this purpose by Richard Thompson. Notice that f(x) is
continuous everywhere, and that as x goes to 0, f(x) goes to 0. Also notice
that f(x) is neither even nor odd.
In all the following slides exactly the same process is followed:
1) An interval, containing the origin, is selected and divided into equally
spaced x values.
2) f(x) is calculated at each of these x values.
3) These points are plotted and joined.
A. [-.1, .1] with about 160 sample points.
The first slide displays f(x) in the interval [-.1, .1] using 158 sample
points.
From now on the number of sample points is increased by 1 until the total
points sampled is 162. Surprise.
None of these graphs accurately portray f(x). In fact, using an elementary
argument involving calculus, it is easy to show that f(x) has 485 relative
extrema between 0 and .1 . No wonder that plotting only 160 points may be
inaccurate!
B. [-.1, .1002] to [-.1, .0998] in steps of -.00005
The first slide displays f(x) in the interval -.1 < x < .1002. The
vertical line is the y-axis. Notice where the function crosses the y-axis.
From now on the left hand end point is left untouched, and the right hand
end point is reduced by .00005, so the new interval is -.1 < x < .10015.
Surprise. This process is repeated until the final domain is -.1 < x < .0998.
None of these graphs accurately portray f(x).
C. [-.0845, .0845] to [-.082, .082] in steps of -.0005
The first slide displays f(x) in the interval -.0845 < x < .0845. This is
a smaller domain than the (A) set of slides.
Now the left and right hand end points are reduced simultaneously by .0005,
so the new interval is -.0840 < x < .0840. Surprise. This process is repeated,
until the final domain is -.082 < x < .082.
None of these graphs accurately portray f(x).
D. [-.0006, .0006] to [-.003, .003] in steps of .0012
The first slide displays f(x) in the interval -.0006 < x < .0006. This is
a much smaller domain than either the (A) or (B) sets of slides. What you are
looking at is an accurate graph of f(x). How do we know? Because, using
calculus, we can show that the first two maxima which are to the right of the y-
axis occur at x = .000104 and x = .000521, and the first minima is at .000312.
This is clearly the case in this graph.
Now the left and right hand end points are increased by .0012, so the new
interval is -.0018 < x < .0018, and the region containing the three relative
extrema mentioned above is indicated by a box. Finally, this process is
repeated, and the final domain is -.003 < x < .003. These graphs portray f(x)
accurately. It is easily shown that, between 0 and .003, f(x) has 14 relative
extrema in agreement with the last slide.
When viewing the slides, the following keys are operational:
HOME takes you to the first slide in the sequence you selected
END takes you to the last slide in the sequence you selected
UP ARROW takes you to the previous slide in the sequence you selected
F9 immediately quits the program
These keys do NOT operate like that while you are reading this document.
When you have finished reading this document, press Q to quit.