Assume a square inscribed in a circle. Rotate it by $45^\circ$ and superimpose on the original square. Rotate by $22.5^\circ$ the resulting star like shape and superimpose on the previous shape. Continue this way, each time the rotation is half the previous one and superimpose on the last created shape. The area of the resulting shapes is converging to the area of the circle. Assuming the circle with radius of 1 (area is $\pi$), the sequence for computing the area of the resulting shape may be derived as follows: $$2^{n-1}[1-\tan(\pi /4-\pi /2^n)]$$ This converges to $\pi$, as $n \to \infty$, "based" on the geometrical process described. How is this proven by analytical means? A side note - you may start with any regular polygon and develop a a similar equation that converges to $\pi$ in a slightly different way. It is also interesting that the perimeter of the shapes is not converting to the expected $2\pi$.


2 Answers 2


The limit of the function $$f(x) = \dfrac{1}{2}\dfrac{1-\tan(\pi/4-\pi 2^{-x})}{2^{-x}}$$

as $x\to\infty$ has "indeterminate form" of $0/0$, and so L'Hopital's works fine. Replace $2^{-x}$ with $a(x)$ for convenience, and note $a(x)\to 0$.

$$\,\, \dfrac{1}{2}\lim_{x\to\infty} \dfrac{1-\tan(\pi/4-\pi a(x))}{a(x)} =\,\, \dfrac{1}{2}\lim_{x\to\infty} \dfrac{-\sec^2(\pi/4-\pi a(x))\cdot (-\pi a'(x))}{a'(x)}$$

$$=\dfrac{1}{2}\lim_{x\to\infty} -\sec^2(\pi/4-\pi a(x))\cdot (-\pi) = \dfrac{\pi}{2}\sec^2(\dfrac{\pi}{4}) = \dfrac{\pi}{2}\cdot 2$$

and of course... if $f(x)\to L, f(n)\to L$.


Use the formula of tangent of angle difference, and that $\tan(\pi/4)=1$: $$1-\tan(\pi/4-\pi/2^n)=1-\frac{\tan(\pi/4)-\tan(\pi/2^n)}{1+\tan(\pi/4)\tan(\pi/2^n)}=1-\frac{1-\tan(\pi/2^n)}{1+\tan(\pi/2^n)}=\frac{2\tan(\pi/2^n)}{1+\tan(\pi/2^n)}$$

Then use the fact that $\lim_{x\rightarrow 0}\frac{\tan(x)}{x}=\lim_{x\rightarrow 0}\frac{\sin(x)}{x}\cdot\frac{1}{\cos(x)}=1$



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