# Total area of squares.

We have a square whose length is $1$ unit. Every time we rotate by $\theta$ and scale the square such as you see in the image. Does the total area of squares converge if $\theta$ goes to $0$?

• What have you tried? The hardest part is finding out how the area of the rescaled square relates to the original square. And that's still basic geometry. May 23, 2013 at 3:32
• I'd guess this impossible to answer as currently stated, and we'd have to stipulate conditions on "phi" for us to know when it converges and when it doesn't. Can't the area of the squares come as any real number? Don't plenty of sequences of real numbers diverge? May 23, 2013 at 3:33

Consider a square of side $a$. If the new square rotated by $\theta$ has side of length $b$, we then have $$b \cos(\theta) + b \sin(\theta) = a \implies b = \dfrac{a}{\sqrt2 \sin(\theta+\pi/4)}$$ Hence, the $n^{th}$ square will have a side of length $$\dfrac{a}{2^{n/2}\sin^n(\theta+\pi/4)}$$ and thereby an area of $$\dfrac{a^2}{2^n \sin^{2n}(\theta+\pi/4)}$$ Hence, we want to know for what $\theta$ $$\sum_{n=0}^{\infty} \dfrac{\csc^{2n}(\theta+\pi/4)}{2^n}$$ converges. Can you finish it off from here?
• That means the total area diverges when $\theta$ goes to zero, according to your equation. Thanks. If $\theta>0$ it converges. May 23, 2013 at 3:43
• @newzad Even pictorially, we can see that as $\theta \to 0$, all the squares are the "same" square and thereby the area at each of the infinite squares is the same.
• Yes, I agree with you. But when $\theta$ slightly bigger than $0$, logically it should converge. I was not sure so I asked. May 23, 2013 at 3:56
• @newzad Yes. For $\theta \in (0,\pi/2)$, the series does converge. It diverges at $\theta = 0$ and $\theta=\pi/2$.