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Given: $p(x) = x^4 - 5773x^3 - 46464x^2 - 5773x + 46$

What is the sum of all arctan of all the roots of $p(x)$?

2 Answers 2

19

Let $p_1,p_2,p_3,p_4$ be the roots of $p(x)=0$

So, we need $\arctan p_1+\arctan p_2+\arctan p_3+\arctan p_4$

Using Vieta's Formulae,

$\sum p_i=p_1+p_2+p_3+p_4=\frac{5773}1$

$\sum p_ip_j=p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4=\frac{46464}1$

$\sum p_ip_jp_k=p_1p_2p_3+p_1p_2p_4+p_1p_3p_4+p_2p_3p_4=\frac{5773}1$

$p_1p_2p_3p_4=\frac{46}1$

We know, $\tan(A+B+C+D)$

$=\frac{\tan A+\tan B+\tan C+\tan D -(\tan A\tan B\tan C+\tan A\tan B\tan D+\tan A\tan C\tan D+\tan B\tan C\tan D)}{1- (\tan A\tan B+\tan A\tan C+\tan A\tan D+\tan B\tan C+\tan B\tan D+\tan C\tan D)+ \tan A\tan B\tan C\tan D}$

If we put $p_1=\tan A,p_2=\tan B$ etc.,

$\tan(A+B+C+D)=\frac{p_1+p_2+p_3+p_4 -(p_1p_2p_3+p_1p_2p_4+p_1p_3p_4+p_2p_3p_4)}{1- (p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4)+ p_1p_2p_3p_4}$

$=\frac{5773-5773}{1-46464+46}=0$

So, $\tan(A+B+C+D)=0\implies A+B+C+D=\arctan (0)$

$\implies \arctan p_1+\arctan p_2+\arctan p_3+\arctan p_4=\arctan (0)$

The general value of $\arctan (0)$ is $n\pi$ where $n$ is any integer, the special value being $0$(putting $n=0$)

  • 0
    @labbhattacharjee: +1 thank you for the effort2012-12-25
13

Nothing additional, but for the OP a more elementary approach.

$(x-a_1)(x-a_2)(x-a_3)(x-a_4) = x^4 - c_1x^3 + c_2x^2 - c_3x + a_1a_2a_3a_4.$ Where $c_1 = \Sigma a_i$, $c_2 = \Sigma_{i and $c_3 = \Sigma_{i . Let $t_i=\tan^{-1}a_i.$ You want to compute $\Sigma t_i.$ Here, the trick is knowing the $\tan(x+y)$ expansion $\tan(x+y) = \frac{\tan x + \tan y}{1- \tan x \tan y}.$
Apply twice for $x=t_1+t_2$ and $y=t_3+t_4$. $\tan(t_1+t_2+t_3+t_4) = \frac{\tan(t_1+t_2) + \tan(t_3+t_4)}{1+\tan(t_1+t_2)\tan(t_3+t_4)}=\frac{\frac{a_1 + a_2}{1- a_1a_2}+\frac{a_3 + a_4}{1- a_3a_4}}{1-\frac{a_1 + a_2}{1- a_1a_2}\frac{a_3 + a_4}{1- a_3a_4}} = \frac{(a_1+a_2)(1-a_3a_4)+(a_3+a_4)(1-a_1a_2)}{(1-a_1a_2)(1-a_3a_4)-(a_1+a_2)(a_3+a_4)}=\frac{c_1-c_3}{1+c_4-c_2}.$ For this polynomial $c_1=c_3$, thefore $\tan\Sigma t_i =0,$ so is $\Sigma t_i =0$