Comments on: This is Not a Trig Function

By: MrHonner

MrHonner — Sat, 17 Nov 2012 16:53:06 +0000

Unfortunately, I think tests like these too often teach students not to notice these sorts of things. Which is a shame, because figuring out what's wrong with this "trig" graph is a great mathematical activity.

By: Rocky Roer

Rocky Roer — Sat, 17 Nov 2012 15:13:11 +0000

I noticed this too. I wonder if i would have had the guts to realize the test was wrong, or if i would have struggled to find some trig question that did fit the graph. Would i have just answered with a sine curve which they were going for, or assumed it was a really hard question involving transformations, or sine squareds, or what not. And would my students have noticed? How would they respond?

By: mr bombastic

mr bombastic — Sun, 11 Nov 2012 05:03:50 +0000

The actual scale is not important – yes? y = f(x) & y = 1/1000 * f(1000x)have exactly the same shape other than one being blow up by a factor of 1000.

Are you just having an issue with the slider not being sensitive enough to changes in the parameters? I found that making the min & max values on the slider close together and the increment small allows for quite a bit of control. Making the slider segment longer adds more sensitivity as well.

Also, think of fitting the quadratic in the form: y = a(x-b)^2 + c. If you choose b & c so the vertex match, that only leaves a to play with – and it just isn’t that great of a fit – the parabola is clearly thinner than the “sine” wave. So you try and cheat a little by dropping the vertex of the parabola below that of the “sine” wave. But, by the time the part that is away from the vertex is matching a little better, it is obvious that the vertices don’t match.

In any event, thank you for posing an interesting problem. I believe I actually have a better understanding of parabolas after thinking about all this.

By: MrHonner

MrHonner — Sat, 10 Nov 2012 23:24:35 +0000

I see the merit in your argument. I did something similar in Geogebra myself, with similar results. The issue is that you made the image only slightly bigger, by some small scale factor.

I feel that if I had another 2 or 3 decimals places to work with I could make it match. I think this is equivalent to a difference in scale of 100 to 1000.

In any event, I haven’t given up! Yet.

By: mr bombastic

mr bombastic — Sat, 10 Nov 2012 22:21:40 +0000

I am pretty sure you could not. I just dumped a larger version of the image into geobebra and remain convinced that there just isn’t enough “play” to fit a quadratic. Here is the goegebra applet: http://www.geogebratube.org/student/m21619

By: MrHonner

MrHonner — Tue, 06 Nov 2012 01:10:17 +0000

If I could use small enough numbers in Geogebra, I’m pretty sure I could show you a parabola that looked like a good fit.

By: mr bombastic

mr bombastic — Tue, 06 Nov 2012 00:55:04 +0000

I don’t think it is quadratic either. Each section of the graph has 3 points that appear to be on corners of the grid. The parabola that “fits” those 3 points will not fit the rest of that section of the graph.

By: MrHonner

MrHonner — Mon, 05 Nov 2012 02:56:15 +0000

Sue and Joshua-

Unfortunately I can no longer reply directly to the above thread, but thanks for following up–I now see exactly what you are talking about. The explicit parabola example made it very clear (I just actually worked it out myself and then saw your comment, Joshua!).

If you haven’t already seen it, Timothy Gowers noticed something similar and posted about it on Google+: https://plus.google.com/u/0/103703080789076472131/posts/VhHE8TuJhcm.

By: Joshua Bowman (@Thalesdisciple)

Joshua Bowman (@Thalesdisciple) — Mon, 05 Nov 2012 02:04:13 +0000

Actually, that’s why I gave the example with the circular arcs—to show that the symmetry isn’t enough to guarantee continuity of the second derivative. However, I invoked curvature rather than the second derivative directly, so perhaps it’s simpler to consider parabolic arcs (I know, it’s been determined that the above example is composed of ellipses, but the principle is the same and easier to explain with parabolas).

A parabola is the graph of a quadratic function, whose second derivative is constant. You could take two congruent pieces of parabolas and glue them together with their tangent lines aligned (say, x^2+2x from -2 to 0 and 2x-x^2 from 0 to 2). The resulting graph corresponds to a function whose first derivative is continuous, but whose second derivative is piecewise constant—in my example it jumps from 2 to -2 at x=0.

By: Sue VanHattum

Sue VanHattum — Mon, 05 Nov 2012 02:03:37 +0000

It seems like Joshua’s circular arcs would fit your description, and would, as he said, have constant acceleration – in some sense, though not a constant y”(x). Hmm, how would we calculate acceleration so that it would be a constant?

I’m convinced that the first derivative must be continuous and the second (if it’s circular arcs) discontinuous. This is continuing to be intriguing…