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I posted a similar question to this, however this question refers to the sum of inverse tangents of any polynomial.

Let $$P(x)=n_1x^{k}+n_2x^{k-1}+n_3x^{k-2}+...+n_{k-1}x+n_{k}$$ where $k$ is a positive integer.

If $P(x)$ has roots: $$\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5,\alpha_6,.........,\alpha_{k-1},\alpha_{k}$$

What is the value of: $$\tan^{-1}(\alpha_1)+\tan^{-1}(\alpha_2)+\tan^{-1}(\alpha_3)+\tan^{-1}(\alpha_4)+.........+\tan^{-1}(\alpha_{k-1})+\tan^{-1}(\alpha_{k})$$

in terms of the coefficients of $P(x)$.

I have found out that for a polynomial of degree 2, the inverse tangents of its roots $\alpha,\beta$ is equal to:

$$tan^{-1}\left(\frac{\alpha+\beta}{1-\alpha\beta}\right)$$

Which leaves us with the sum and product of roots and can then be simplified in terms of its coefficients. However, I am unsure how I can generalise this in continuing onwards to higher degree polynomials?

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  • $\begingroup$ I doubt there is a general formula for higher degree polynomials, since we don't have a relationship between roots and coefficients for degrees higher than $3$. $\endgroup$
    – HappyDay
    Commented Oct 14, 2022 at 10:18
  • $\begingroup$ The pattern continues and you can use induction to prove. Check en.wikipedia.org/wiki/…, and use Vieta to substitute out the elementary symmetric polynomials in the tan identity... $\endgroup$
    – Macavity
    Commented Oct 14, 2022 at 12:57

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