Question

How do you find the first five terms of the Taylor series for $f\left( x \right)={{x}^{8}}+{{x}^{4}}+3$ at $x=1$ ?

Answer 1

Taylor series is given as $f\left( x \right)+\dfrac{{f}'\left( x \right)}{1!}x+\dfrac{{f}''\left( x \right)}{2!}{{x}^{2}}+\dfrac{{f}'''\left( x \right)}{3!}{{x}^{3}}+........\infty$ where, ${f}'\left( x \right),{f}''\left( x \right),{f}'''\left( x \right)$ are the first, second and third derivatives of the function$f\left( x \right)$. Here, $1!,2!,3!$ are the factorials of 1,2 and 3 respectively. Therefore, we shall calculate a few terms and then observe the pattern occurring to find our final solution.

Complete step by step solution:
We have to find the first five terms of this series for $x=1$.
Let us differentiate the function $f\left( x \right)={{x}^{8}}+{{x}^{4}}+3$ now. We shall use the property of differentiation given as $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ several times to differentiate the function.
$\Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( {{x}^{8}}+{{x}^{4}}+3 \right)$
$\Rightarrow {f}'\left( x \right)=8{{x}^{7}}+4{{x}^{3}}$ …………….. (1)
Again differentiating the function with respect to x, we get:
$\Rightarrow {f}''\left( x \right)=\dfrac{d}{dx}{f}'\left( x \right)$
$\begin{aligned} & \Rightarrow {f}''\left( x \right)=\dfrac{d}{dx}\left( 8{{x}^{7}}+4{{x}^{3}} \right) \\\ & \Rightarrow {f}''\left( x \right)=56{{x}^{6}}+12{{x}^{2}} \\\ \end{aligned}$
$\Rightarrow {f}''\left( x \right)=56{{x}^{6}}+12{{x}^{2}}$ …………………… (2)
Third time differentiating the function with respect to x, we get
$\begin{aligned} & \Rightarrow \dfrac{d}{dx}{f}''\left( x \right)=\dfrac{d}{dx}\left( 56{{x}^{6}}+12{{x}^{2}} \right) \\\ & \Rightarrow {f}'''\left( x \right)=\dfrac{d}{dx}\left( 56{{x}^{6}}+12{{x}^{2}} \right) \\\ \end{aligned}$
$\Rightarrow {f}'''\left( x \right)=336{{x}^{5}}+24x$
$\Rightarrow {f}'''\left( x \right)=336{{x}^{5}}+24x$ ………………. (3)
Fourth time differentiating the function with respect to x, we get
$\begin{aligned} & \Rightarrow \dfrac{d}{dx}{f}'''\left( x \right)=\dfrac{d}{dx}\left( 336{{x}^{5}}+24x \right) \\\ & \Rightarrow f''''\left( x \right)=\dfrac{d}{dx}\left( 336{{x}^{5}}+24x \right) \\\ \end{aligned}$
$\Rightarrow f''''\left( x \right)=1680{{x}^{4}}+24$ ………………. (4)
From (1), (2), (3) and (4), we substitute the value of x equals to 1 to calculate the value of the first, second and third derivative of function at point $x=1$.
$\Rightarrow {f}'\left( 1 \right)=8{{\left( 1 \right)}^{7}}+4{{\left( 1 \right)}^{3}}$
$\Rightarrow {f}'\left( 1 \right)=12$
Now, ${f}''\left( 1 \right)=56{{\left( 1 \right)}^{6}}+12{{\left( 1 \right)}^{2}}$
$\Rightarrow {f}''\left( 1 \right)=68$
And, ${f}'''\left( 1 \right)=336{{\left( 1 \right)}^{5}}+24\left( 1 \right)$
$\Rightarrow {f}'''\left( 1 \right)=360$
Also, $f''''\left( 1 \right)=1680{{\left( 1 \right)}^{4}}+24$
$\Rightarrow f''''\left( x \right)=1704$
Also, $f\left( 1 \right)={{1}^{8}}+{{1}^{4}}+3$
$\Rightarrow f\left( 1 \right)=55$
Hence, the value of the first, second, third and fourth derivative of function at point $x=1$is 12, 68, 360 and 1704 respectively.
The Taylor series is given as $f\left( x \right)+\dfrac{{f}'\left( x \right)}{1!}x+\dfrac{{f}''\left( x \right)}{2!}{{x}^{2}}+\dfrac{{f}'''\left( x \right)}{3!}{{x}^{3}}+........\infty$.
Now, we shall calculate the first five terms of this series at $x=1$ one-by-one.
First term: $f\left( 1 \right)=55$
Second term: $\dfrac{{f}'\left( x \right)}{1!}x=\dfrac{{f}'\left( 1 \right)}{1!}1$
$\Rightarrow \dfrac{{f}'\left( 1 \right)}{1!}1=12$
Third term: $\dfrac{{f}''\left( x \right)}{2!}{{x}^{2}}=\dfrac{{f}''\left( 1 \right)}{2!}{{1}^{2}}$
$\Rightarrow \dfrac{{f}''\left( 1 \right)}{2!}{{1}^{2}}=\dfrac{68}{2}=34$
Fourth term: $\dfrac{{f}'''\left( x \right)}{3!}{{x}^{3}}=\dfrac{{f}'''\left( 1 \right)}{3!}{{1}^{3}}$
$\Rightarrow \dfrac{{f}'''\left( 1 \right)}{3!}{{1}^{3}}=\dfrac{360}{6}=60$
Fifth term: $\dfrac{f''''\left( x \right)}{4!}{{x}^{4}}=\dfrac{f''''\left( 1 \right)}{4!}{{1}^{4}}$
$\Rightarrow \dfrac{f''''\left( 1 \right)}{4!}{{1}^{4}}=\dfrac{1704}{24}=71$
Therefore, the first five terms of the Taylor series for $f\left( x \right)={{x}^{8}}+{{x}^{4}}+3$ at $x=1$ are 55, 12, 34, 60, 71.

Note:
Maclaurin series is a special case of Taylor series centered at $x=0$. It is given as $f\left( 0 \right)+\dfrac{{f}'\left( 0 \right)}{1!}x+\dfrac{{f}''\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{{f}'''\left( 0 \right)}{3!}{{x}^{3}}+........\infty$ where, ${f}'\left( 0 \right),{f}''\left( 0 \right),{f}'''\left( 0 \right)$ are the first, second and third derivatives of the function$f\left( x \right)$ at $x=0$. Likewise, $1!,2!,3!$ are the factorials of 1,2 and 3 respectively.