Question

Obtain Taylor's series expansion for $\log \left( \cos x \right)$ about the point $x=\dfrac{\pi }{3}$ upto the fourth degree term.

Answer 1

Hint: We will look at the Taylor's series expansion for a function $f\left( x \right)$ about a point $x=a$. Then we will find the derivative of the given function. As Taylor's series expansion has terms with higher-order derivatives, we will compute them for the given function. We will then substitute the value $x=\dfrac{\pi }{3}$ in the derivatives obtained. We will put the obtained results in Taylor's series expansion.

Complete step-by-step solution
The Taylor's series expansion a function $f\left( x \right)$ about a point $x=a$ is given by
$f\left( x \right)=f\left( a \right)+{f}'\left( a \right)\left( x-a \right)+\dfrac{{f}''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\dfrac{{{f}^{(3)}}\left( a \right)}{3!}{{\left( x-a \right)}^{3}}+\cdots $
The given function is $f\left( x \right)=\log \left( \cos x \right)$. We have to find the Taylor's series expansion of the given function upto the fourth degree term. So, we will compute upto the fourth derivative of the given function.
The value of the function at $x=\dfrac{\pi }{3}$ is $f\left( \dfrac{\pi }{3} \right)=\log \left( \cos \dfrac{\pi }{3} \right)=\log \left( \dfrac{1}{2} \right)$.
The first derivative of $f\left( x \right)$ is ${f}'\left( x \right)=\dfrac{1}{\cos x}\times -\sin x$. The value of the first derivative at $x=\dfrac{\pi }{3}$ is ${f}'\left( \dfrac{\pi }{3} \right)=\dfrac{1}{\cos \dfrac{\pi }{3}}\times -\sin \dfrac{\pi }{3}=\dfrac{1}{\left( \dfrac{1}{2} \right)}\times -\dfrac{\sqrt{3}}{2}=-\sqrt{3}$.
The second derivative of $f\left( x \right)$ is ${f}''\left( x \right)=-{{\sec }^{2}}x$. The value of the second derivative at $x=\dfrac{\pi }{3}$ is ${f}''\left( \dfrac{\pi }{3} \right)=-{{\sec }^{2}}\dfrac{\pi }{3}=-\dfrac{1}{{{\cos }^{2}}\dfrac{\pi }{3}}=-\dfrac{1}{{{\left( \dfrac{1}{2} \right)}^{2}}}=-4$.
The third derivative of $f\left( x \right)$ is ${{f}^{(3)}}\left( x \right)=-2{{\sec }^{2}}x\tan x$. The value of the second derivative at $x=\dfrac{\pi }{3}$ is ${{f}^{(3)}}\left( \dfrac{\pi }{3} \right)=-2{{\sec }^{2}}\dfrac{\pi }{3}\tan \dfrac{\pi }{3}=-2\times 4\times \sqrt{3}=-8\sqrt{3}$.
The fourth derivative of $f\left( x \right)$ is ${{f}^{(4)}}\left( x \right)=4{{\sec }^{2}}x{{\tan }^{2}}x+2{{\sec }^{4}}x$. The value of the second derivative at $x=\dfrac{\pi }{3}$ is
 $\begin{align}
  & {{f}^{(4)}}\left( \dfrac{\pi }{3} \right)=4{{\sec }^{2}}\dfrac{\pi }{3}{{\tan }^{2}}\dfrac{\pi }{3}+2{{\sec }^{4}}\dfrac{\pi }{3} \\
 & =4\times 4\times {{\sqrt{3}}^{2}}+2\times \dfrac{1}{{{\left( \dfrac{1}{2} \right)}^{2}}}=48+2\times 16=48+32=80
\end{align}$
Now, substituting all these values in the Taylor's expansion series, we get
$f\left( \log \left( \cos x \right) \right)=\log \left( \dfrac{1}{2} \right)+\left( -\sqrt{3} \right)\left( x-\dfrac{\pi }{3} \right)+\dfrac{\left( -4 \right)}{2!}{{\left( x-\dfrac{\pi }{3} \right)}^{2}}+\dfrac{\left( -8\sqrt{3} \right)}{3!}{{\left( x-\dfrac{\pi }{3} \right)}^{3}}+\dfrac{80}{4!}{{\left( x-\dfrac{\pi }{3} \right)}^{4}}$
Simplifying the above equation, we get
$f\left( \log \left( \cos x \right) \right)=\log \left( \dfrac{1}{2} \right)-\sqrt{3}\left( x-\dfrac{\pi }{3} \right)-2{{\left( x-\dfrac{\pi }{3} \right)}^{2}}-\dfrac{4\sqrt{3}}{3}{{\left( x-\dfrac{\pi }{3} \right)}^{3}}+\dfrac{10}{3}{{\left( x-\dfrac{\pi }{3} \right)}^{4}}$
The above equation is the Taylor's series expansion up to the fourth degree term of the function $f\left( x \right)=\log \left( \cos x \right)$.

Note: In this type of question, it is necessary that we are familiar with the derivatives of standard functions. It is also important that we know the values of trigonometric functions for standard angles. This will make the calculations a little bit easier. It is useful to calculate every derivative separately so that we can avoid making errors in the calculations.