**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 .

LetÕs
indeed use graphing calculator to graph the function to make sure our reasoning was right.

Now
letÕs guess what the graph of the function =1
looks like.

Case
(1) 0²x, y²1

, ,

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 ÒpolygonalÓ than the ÒsmoothÓ
circumference of a circle, and the Òturning pointÓ of the ÒtrendÓ of the curve
is (,), which is approximately (0.793701, 0.793701).

Case
(2):＞1

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

Case
(3): X＜0

Thus, ＜0. So ＞1 y＞1.
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!

LetÕs
keep moving to =1.

Since
,cannot
be negative, so both of them have to be less than or equal to 1, and so 0²|x|,|y|²1
is the only case.

We only consider 0²x, y²1 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
Òturning pointÓ（ ， ）Å（0.840896, 0.840896）than that of .

The
whole curve would look like a Òsquare with round cornersÓ. LetÕs check it up by
the graph:

For,
itÕs easy to guess that the whole shape of the curve would like look that of ,
but the Òturning pointÓ (,)Å(0.870551, 0.870551) will make it ÒsharperÓ than
at (). HereÕs its graph:

Now
we make the conjecture that the graph of
will be either in a Òpolygonal
shapeÓ when n is even or in a Òpulse shapeÓ 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 ÒpeakÓ.

LetÕs
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 Ònon-oneÓ number m, i.e., ,
the curve would not be as ÒbigÓ or ÒsmallÓ as students may expect, since the
Òturning pointÓ will be ,
not a linear or quadratic change.

For
example, some students may simply predict that if is
a Òunit squareÓ then 36=6^{2 }will be a bigger ÒsquareÓ 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 ÒsquareÓ is determined by , not by
for the Ògenuine circleÓ. In other word, for a circle ,
but for any even itÕs not the case as some student
might predict. We show the graph of in the following figure to check
the side of the ÒsquareÓ is Å 2(1.01456)=2.02912, not .

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