How do you solve $Arctan(a^2) - arctan a - 0.22 = 0$ ?

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Answer 1 · 2016-08-11T23:21:51

$a = -0.187135$

$f(a) = arctan(a^2) - arctan( a) - 0.22 = 0$

Given $a_0$ we can expand $f(a)$ in Taylor series, at the $a_0$ vicinity until the linear term giving

$f(a) = f(a_0) + f'(a_0)(a-a_0) + O(a,a_0)$

If we need $f(a_1) = 0$ then supposing that $abs(a_1-a_0) < epsilon$ and that $O(a_1,a_0)$ is small enough, then

$a_1 = a_0 - f(a_0)/(f'(a_0))$ or

$a_{k+1} = a_k - f(a_k)/(f'(a_k))$

In the present case we have

$f´(a) =(2 a)/(1 +a^4) -1/(1 + a^2)$

Applying iteratively this process we obtain

$a_0 = 0.$
$a_1=-0.22$
$a_2 = -0.187754$
$a_3 =-0.187135$
$a_4 =-0.187135$

obtaining a result within $6$ sd.

How do you solve Arctan(a^2) - arctan a - 0.22 = 0Arctan(a^2) - arctan a - 0.22 = 0?