veceriraby

Answered

2022-01-29

Denote by $V}_{(x,y)$ the vertex at ordered pair (x,y) in the Cartesian coordinate system. Denote by $A}_{(x,y)$ the measure of angle $\mathrm{\angle}{V}_{(x,y)}{V}_{(0,0)}{V}_{(1,0)}$ . Let

$\theta =\sum _{0\le i,j\le 5\text{}\text{}(i,j)\ne (0,0)}{A}_{(i,j)}$

Find$\theta \text{}\left(\text{mod}\text{}2\pi \right)$

I understand the question and everything, but I am slightly overwhelmed about how to find an organized approach to computing the sum of angles. I got the value down to the sum

$\sum _{0\le i,j\le 5\text{}\text{}(i,j)\ne (0,0)}\mathrm{arctan}\frac{i}{j}$

But I don't know how to evaluate this.$\mathrm{arctan}(x+y)$ formula fails. I think there is a different approach I am not aware of.

Is there a nice way to generalize to

$\sum _{0\le i,j\le k\text{}\text{}(i,j)\ne (0,0)}{A}_{(i,j)}$

Find

I understand the question and everything, but I am slightly overwhelmed about how to find an organized approach to computing the sum of angles. I got the value down to the sum

But I don't know how to evaluate this.

Is there a nice way to generalize to

Answer & Explanation

Jason Duke

Expert

2022-01-30Added 11 answers

I got it now, Note that if we represent $A}_{m,n}+{A}_{n,m$ in terms of complex numbers we get $arg(m+ni)(n+mi)=\frac{\pi}{2}$ . There are $k(k+1)$ of these numbers (ignnoring diagonals so we get $\left(k\frac{k+1}{2}\right)\left(\frac{\pi}{2}\right)$ . Adding back the diagonal (ignoring (0,0)) we get $k\frac{\pi}{4}$

$\left(k\frac{k+1}{2}\right)\left(\frac{\pi}{2}\right)+k\frac{\pi}{4}=\frac{k(k+2)\pi}{4}$

Which is the general solution. Substituting k=5 gives$35\frac{\pi}{4}\equiv 3\frac{\pi}{4}\left(\text{mod}\text{}2\pi \right)$

Which is the general solution. Substituting k=5 gives

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