The figure shows a non-overlapping trace on a $4\times4$ grid which visits all points exactly once. Imagine the same type of trace on an $n\times n$ grid, where $n$ can be arbitrarily large. Using the fact that $\sum_{k=1}^\infty {1\over k}$ diverges to infinity, show that the sum of all acute angles of the trace also diverges to infinity when $n$ tends to infinity, despite the angles tending to 0 the closer the path gets to the grid’s diagonal.
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Let’s prove that
$\sum_{k=1}^{\infty}{\frac{1}{k}}$diverges. For each$n,$how many terms in the series are between$\frac{1}{2^n}$(inclusive) and$\frac{1}{2^{n+1}}$(exclusive)? By bounding each term$\frac{1}{k}$in the original series with the largest$\frac{1}{2^n}$such that$\frac{1}{2^n} \leq \frac{1}{k},$compute this new sum. Hence argue why$\sum_{k=1}^{\infty}{\frac{1}{k}}$diverges. -
(Source: Wikipedia) Using this diagram, express $\sin \theta, \cos \theta$and$\tan \theta$in terms of$x.$Hence show that$\cos(\arcsin x) = \sqrt{1-x^2}.$Find a similar expression for$\cos(\arctan x).$
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To show the sum of all the angles diverges, it suffices to show the sum of a subset of the angles diverges.
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Express the
$i^{th}$largest angle$\theta_i$that touches the top of the square in terms of$i$. -
The angle
$\theta_i$can be expressed as the difference between two angles. -
The larger angle is
$\frac{\pi}{4}$, and the tangent of the smaller angle can be expressed as the ratio between the sides of the triangle along the grid. -
Remember that the function
$\tan x$is concave in the region$(0,\frac{\pi}{4}).$ -
For positive integer
$i,$$0 < \frac{i}{i+1} < 1,$so$\arctan \frac{i}{i+1}$will lie in the region$(0,\frac{\pi}{4}).$ -
By drawing the graphs
$y = ({\frac{\pi}{4}})^{-1}x$and$y = \tan x,$you can see that$\tan x$lies below the line in the region$(0,\frac{\pi}{4}).$ -
This means that
$\arctan \frac{i}{i+1} \leq \frac{\pi}{4}\frac{i}{i+1}.$
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The full sum
$S$is greater than the sum of only the angles that touch the top of the square. Call this subsequence of angles$\theta_i$. By considering the right-angled triangles with sides$i$and$i+1, \theta_i$can be derived as:$$ \begin{aligned} \theta_i &= \frac{\pi}{4} - \arctan{\frac{i}{i+1}} \\ &\geq \frac{\pi}{4} - \frac{\pi}{4}\frac{i}{i+1} \\ &= \frac{\pi}{4}(\frac{1}{i+1}) \end{aligned} $$where the inequality is due to the concavity of$\tan x$in the region$(0, \frac{\pi}{4})$and$\tan \frac{\pi}{4}=1.$We then have that:$S \geq \sum_{i=0}^\infty \theta_i \geq \frac{\pi}{4}\sum_{i=0}^{\infty}\frac{1}{i+1},$and the latter diverges. Hence$S$diverges.