Guessing and reasoning about  first and then checking it

(I did this problem in Aug. but finished it thirdly on computer until Oct. 5, 2009)

 

 

Chen Tian

 

 

 

Every secondary student knows  is the function of a circle with radius of unit 1 centered at the origin of  coordinate system.

But why is it true?

 

We know is the distance from point (x,y) to the origin  by Pythagorean theorem. So  will represent that the distance from any possible point (x,y)  to  equals1, which symbolizes the definition of a unit circle, and because, we transform   into .

 

Let’s indeed use graphing calculator to graph the function  to make sure our reasoning was right.

 

 

 

Now let’s guess what the graph of the function =1 looks like.

 

Case (1) 0≤x, y≤1

, ,  

so when x goes from o to , y goes down from 1 to , and then x will go from  to 1, y will go from down to 0.

So the curve in the first quadrant will still look like an arc, BUT,  is smaller than  when x takes the same value, so , namely, for the same y. Therefore, the curve will look a little bit “polygonal” than the “smooth” circumference of a circle, and the “turning point” of the “trend” of the curve is (,), which is approximately (0.793701, 0.793701).

 

Case (2):1

Thus, 1. So 0  y0. The greater x goes, the smaller y goes.

 

Case (3): X0

Thus, 0. So 1  y1. The smaller x goes, the greater y goes.

 

Based on our analysis, we graph function =1, shown in the following figure.

 

 

 

Aha, our guess was right!

 

 

Let’s keep moving to =1.

 

Since ,cannot be negative, so both of them have to be less than or equal to 1, and so 0≤|x|,|y|≤1 is the only case.

 

We only consider 0≤x, y≤1 here, since the reasoning for the curve in the other quadrants would be extremely similar.

*. Similarly as case (1) for the function =1, we guess the curve in the first quadrant will look a little bit sharper at the “turning point”  Ň0.840896, 0.840896than that of .

 

The whole curve would look like a “square with round corners”. Let’s check it up by the graph:

 

 

 

 

For, it’s easy to guess that the whole shape of the curve would like look that of , but the “turning point” (,)Ň(0.870551, 0.870551) will make it “sharper” than  at (). Here’s its graph:

 

 

 

Now we make the conjecture that the graph of   will be either in a “polygonal shape” when n is even or in a “pulse shape” when n is odd. And when n even gets greater the graph will approach to a real square; when n odd gets greater the graph in the first quadrant will become a sharper and sharper “peak”.

 

Let’s try  and  as examples.

 

 

 

They work also for our above conjecture.

 

 

One thing I think is important to mention is if we change the right-hand side of the function to a “non-one” number m, i.e., , the curve would not be as “big” or “small” as students may expect, since the “turning point” will be , not a linear or quadratic change.

 

For example, some students may simply predict that if is a “unit square” then 36=62  will be a bigger “square” with side of length  for the same reason like increasing the radius of the circle. But they were wrong because they ignored the fact we mentioned above that the side of the “square” is determined by , not by for the “genuine circle”. In other word, for a circle , but for any  even it’s not the case as some student might predict. We show the graph of   in the following figure to check the side of the “square” is Ň 2(1.01456)=2.02912, not .

 

 

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