Replace the following function by its taylor polynomial of the given grade, and approximate the error in the given interval:
$f(x) = \arctan(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$
My solution and thoughts
We only need the first three derivatives and the fourth one for the error
$ f'(x) = \frac{1}{1+x^2} \\ f''(x) = \frac{2x}{(x^2+1)^2} \\ f'''(x) = \frac{6x^2-2}{(x^2+1)^3} \\ f^{(iv)}(x) = \frac{24x(x^2-1)}{(x^2+1)^4} $
And by definition we know that
$ T_3(f,x,0) = \sum\limits_{k=0}^3 \frac{f^{(k)}(0)}{k!}x^k $
we get
$ T_3(f,x,0) = x - \frac{x^3}{3} $
We know that
$R_3(x) = \frac{f^{(iv)}(c)}{4!}x^4, |x| \le \frac{1}{10}$
by plugging in the maximal value of $x = c = \frac{1}{10}$ we get for any $c$:
$ \left|f^{(iv)}(c)\right| = \left|\frac{24c(c^2-1)}{(c^2+1)^4}\right| \le \\ \le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4} $
$ \Rightarrow \left|R_3(x)\right| \le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4 \cdot 4!}\cdot\left|x^4\right| \Leftrightarrow \\ \Leftrightarrow \left|R_3(x)\right| \le \frac{10}{101^4} $
Is this solution correct? How can I make it more formally right? (I know I lack some mathematical formalism).