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Say, we have two $2^n$-gons: one inscribed in a unit circle and another one circumscribed around the unit circle. After using some basic geometry and limits we arrive at the following result:

$$ \begin{align} \lim_{n\to \infty} \mbox{(perimeter of the inscribed $2^n$-gon)} &= \lim_{n\to \infty} \mbox{(perimeter of the circumscribed $2^n$-gon)} \\ &:= 2\pi \end{align} $$

Yet now (at least according to my professor) we are not allowed to conclude that the perimeter of the unit circle itself equals $2\pi$, because we have only showed it for $2^n$-gons and maybe some other approximations would give us different result.

Now what if we are interested in the areas?

Unit circle and the circumscribed polygon

$$ \begin{align} \lim_{n\to \infty} \mbox{(area of the circumscribed $2^n$-gon)} &= \lim_{n\to \infty} \mbox{($2^n \times $ area of the triangle)} \\ &= \lim_{n\to \infty} (2^n \frac{1 \times s_n}{2}) \\ &= \lim_{n\to \infty} \mbox{($\frac{1}{2} \times$ perimeter of the circumscribed $2^n$-gon)} \\ &= \frac{1}{2} \times 2\pi \mbox{ (using the result above)} \\ &= \pi \end{align} $$

Similarly (but with slightly more geometry involved):

$$\lim_{n\to \infty} \mbox{(area of the inscribed $2^n$-gon)} = \pi $$

My question is: are we now allowed to conclude that the area of the unit circle equals $\pi$? Or do we have a similar problem as the one (with perimeters) described above?

Any help is much appreciated.

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Area is well-behaved, yes, it is OK. Arclength can do strange things. –  André Nicolas Mar 27 '13 at 23:05
    
@André Nicolas, could you please elaborate what is the difference between squeezing the perimeter of the unit circle between two limits of equal value and squeezing the area of the unit circle? Why is that problematic this the former and OK with the latter? Thank you. –  Leo Mar 27 '13 at 23:45
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For the areas, the fact that the circle lies completely within the circumscribed polygons implies that its area is less than the polygons' areas (and similarly for inscribed polygons). There is no such assurance, though, that a shape contained in a second shape has a smaller perimeter than the second shape's perimeter. (It's true in this case, and showing that can actually lead to a formula for the circle's perimeter; but it's not true for all shapes.) –  Greg Martin Mar 28 '13 at 0:03
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Imagine two squares, with square $B$ inside square $A$. Suppose square $A$ has area $1$, and square $B$ has area $0.99$. If $R$ is a region such that $B\subset R\subset A$, then the area of $R$ is between $0.99$ and $1$. Now imagine the same thing, but with perimeters $0.99$ and $1$. The region $R$ could have a lot of zigzags, and perimeter much greater than $1$. In fact the perimeter of $R$ could be infinite. –  André Nicolas Mar 28 '13 at 0:07
    
@André Nicolas and Greg Martin: thank you for the responses. That's a pity one can't accept a comment as an answer. –  Leo Mar 28 '13 at 0:41

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Bringing down comments to be an answer

For the areas, the fact that the circle lies completely within the circumscribed polygons implies that its area is less than the polygons' areas (and similarly for inscribed polygons). There is no such assurance, though, that a shape contained in a second shape has a smaller perimeter than the second shape's perimeter. (It's true in this case, and showing that can actually lead to a formula for the circle's perimeter; but it's not true for all shapes.)

Imagine two squares, with square B inside square A. Suppose square A has area 1, and square B has area 0.99. If R is a region such that B⊂R⊂A, then the area of R is between 0.99 and 1. Now imagine the same thing, but with perimeters 0.99 and 1. The region R could have a lot of zigzags, and perimeter much greater than 1. In fact the perimeter of R could be infinite.

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