f(x) = tan 2.a = 0

The given function is $f\left(x\right)={\tan }^{-1}\frac{x}{2}$, a=0.

(a) To find: First four terms of Taylor's series/Maclaurin series centered at a.

(b) To Write: Power series using summation notation.

(c) To Determine: Interval of convergence.

Taylor's series for a function f is given by,

$f\left(x\right)=f\left(0\right)+\frac{1}{1!}f\text{'}\left(0\right)\left(x-0\right)+\frac{1}{2!}f\text{'}\text{'}\left(0\right){\left(x-0\right)}^2+\frac{1}{3!}f\text{'}\text{'}\text{'}\left(0\right){\left(x-0\right)}^3+...$                    (i)

$f\left(x\right)={\tan }^{-1}\frac{x}{2}$.

$a_0=f\left(0\right)=0$

$f\text{'}\left(x\right)=\frac{1}{2}\frac{1}{1+{\displaystyle \frac{x^2}{4}}}=\frac{2}{4+x^2}$,

$f\text{'}\left(0\right)=\frac{4}{4}=1$

$a_1=f\text{'}\left(0\right)=1$

$f\text{'}\text{'}\left(x\right)=\frac{-2\left(2x\right)}{{\left(4+x^2\right)}^2}=\frac{-4x}{{\left(4+x^2\right)}^2}$

$f\text{'}\text{'}\left(0\right)=0$,

$a_2=\frac{1}{2!}f\text{'}\text{'}\left(0\right)=0$

$f\text{'}\text{'}\text{'}\left(x\right)=\frac{{\left(4+x^2\right)}^2\times \left(-4\right)-\left(-4x\right)2\left(4+x^2\right)\times 2x}{{\left(4+x^2\right)}^4}=\frac{-4\left(4+x^2\right)+16x}{{\left(4+x^2\right)}^3}$

$f\text{'}\text{'}\text{'}\left(0\right)=\frac{-16}{64}=\frac{-1}{4}$

$a_3=\frac{1}{3!}f\text{'}\text{'}\text{'}\left(0\right)=\frac{-1}{4\times 3!}=\frac{-1}{24}$,

$f^{\left(1v\right)}\left(x\right)=\frac{{\left(4+x^2\right)}^3\left(\left(-8x\right)+16\right)-\left(-4\left(4+x^2\right)+16x\right)3{\left(4+x^2\right)}^2\times 2x}{{\left(4+x^2\right)}^6}=-\frac{-48x\left(x^2-4\right)}{{\left(4+x^2\right)}^4}$

$f^{\left(iv\right)}\left(0\right)=0$

$a_4=\frac{1}{4!}{f}^{\left(iv\right)}\left(0\right)=0$,

$f^{\left(v\right)}\left(x\right)=\frac{48\left(5x^4-40x^2+16\right)}{{\left(x^2+4\right)}^5}$

$f^v\left(0\right)=\frac{48\times 16}{4^5}=\frac{3\times 4^2\times 4^2}{4^5}=\frac{3}{4}$.

$a_5=\frac{1}{5!}{f}^{\left(v\right)}\left(0\right)=\frac{1}{5\times 4\times 3\times 2\times 1}\times \frac{3}{4}=\frac{1}{160}$.

$f^{vi}\left(x\right)=\frac{-480x\left(3x^4-40x^2+48\right)}{{\left(x^2+4\right)}^6}$

$f^{\left(vi\right)}\left(0\right)=0$.

$a_6=\frac{1}{6!}{f}^{\left(vi\right)}\left(0\right)=0$.

$f^{\left(vii\right)}\left(x\right)=\frac{1440\left(7x^6-140x^4+336x^2-64\right)}{{\left(x^2+4\right)}^7}$

$f^{\left(vii\right)}\left(0\right)=\frac{1440\times \left(-64\right)}{4^7}=-\frac{3^2\times 4^2\times 10\times 4^3}{4^7}=-\frac{90}{4^2}$

$a_7=\frac{1}{7!}{f}^{\left(vii\right)}\left(0\right)=\frac{-90}{4^2}\times \frac{1}{7\times 6\times 5\times 4\times 3\times 2\times 1}=\frac{-3\times 3\times 2\times 5}{4^2\times 7\times 6\times 5\times 4\times 3\times 2\times 1}=\frac{-1}{4\times 2\times 7\times 4^2}=-\frac{1}{896}$.

Thus, From (i),

$f\left(x\right)=\frac{x}{2}-\frac{x^3}{24}+\frac{x^5}{160}-\frac{x^7}{896}$+...

these are the first four terms of Taylor's expansion of the function  $f\left(x\right)={\tan }^{-1}\frac{x}{2}$.