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.