Solving an difficult inequality

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The discussion focuses on solving a difficult inequality related to the convergence of a function defined as f(x) = x/8 + arctg(x) using Newton's method. The user is trying to establish convergence within a specific interval by manipulating the inequality |xk - xk+1| < 2xk. They derived a new inequality involving the tangent function and found that y = x is less than y = tan{x(x²+17)/8(x²+1)} for x values less than approximately 1.829. The user expresses frustration with Mathematica's handling of the arctg expression and acknowledges the importance of rearranging the inequality for a solution. Ultimately, they seek guidance on finding the solution to the inequality.
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Ill take it from the top :smile: please excuse me for any misspellings.

In Numerical analysis, I have to show that a function only converges towards a solution in a certain interval.
The function is

f(x)=x/8+arctg(x)

Im using Newtons method, i.e.

xk+1 = xk - f(x)/f'(x)

And i know it will converge if I am in an interval that satisfies

|xk-xk+1| < 2xk

I then use that

f(x)/f'(x) < 2xk

giving me

1+8*arctg(x)/x
--------------- < 2
1+8/1+x^2

My problem is now how to solve this inequality, I'm trying with mathematica, but i can't make it work :cry:
Its not important how I solve it, i just need a solution, and guidelines to finding it...
Anyone up for the task?
 
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I rearranged your inequality to get,

x < tan{x(x²+17)/8(x²+1)}

I plotted y = tan{x(x²+17)/8(x²+1)} and y = x using graphmatica and I got y = x < y = tan{x(x²+17)/8(x²+1)} for x < 1.829 (approx.)
 

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Thanks!
Sometimes I hate myself :-(
The problem for mathematica is obviously the arctg expression, so y the h*** didnt i think of reaaranging it...
i will go do that now... thanks man!
 
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