This post concludes this series about a curious function:
In the previous post, I derived three of the four parts of this function. Today, I’ll consider the last part ().
The circle that encloses the grey region must have the points and on its circumference; the distance between these points will be , where is the radius of the enclosing circle. Unlike the case of , we no longer have to worry about the origin, which will be safely inside the enclosing circle.
Furthermore, this line segment will be perpendicular to the angle bisector (the dashed line above), and the center of the enclosing circle must be on the angle bisector. Using trigonometry,
,
or
.
We see from this derivation the unfortunate typo in the above Monthly article.
I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.
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One thought on “A natural function with discontinuities (Part 3)”
In the previous post, I derived three of the four parts of this function. Today, I’ll consider the last part (
The circle that encloses the grey region must have the points 
Mean Green Math 
One thought on “A natural function with discontinuities (Part 3)”