Answer to Question #220074 in Calculus for Nick

Question

4.Expand log (cosx) about the point.

Answer 1

The given function is $f(x)=log(cosx)$. We have to find the Taylor's series expansion of the given function up to the fourth degree term. So, we will compute up to the fourth derivative of the given function.

The value of the function at $x=\frac{π}3$ is 

$f(\frac{π} 3 )=log(cos\frac{π} 3 )=log(\frac{1} 2)$

The first derivative of $f(x)$ is $f'(x)=\frac{1}{cosx}×−sinx$

The value of the first derivative at x=3π​ is

$f'(\frac{π}{3})=\frac{1}{cos\frac{π}{3}}×−sin\frac{π}{3}=\frac{1}{\frac{1}{2}}×-\frac{\sqrt{3}}{2}=\sqrt{3}$

The second derivative of $f(x)$ is $f''(x) =-sec^2x.$

The value of the second derivative at $x=\frac{π}{3}$ is

$f''(\frac{π}{3})=-sec^2\frac{π}{3}=-\frac{1}{cos^2\frac{π}{3}}×=-\frac{1}{(\frac{1}{2})^2}={4}$

The third derivative of $f(x)$ is $f^{(3)}(x)=-2sec^2xtan^2x$ The value of the third derivative at $x=\frac3π$ ​ is

$f^{(3)}(\frac{π}3)=−2sec^2\frac{π}3tan⁡\frac{π}3=−2×4×\sqrt3=−8\sqrt3$

The fourth derivative of $f(x)$ is $f^{(4)}(x)=4sec^2xtan^2x+2sec^4x$

The value of the fourth derivative at $x=\frac{π}{3}$ Is

$f^{(4)}(\frac{π}{3})=4sec^2\frac{π}{3}tan^2\frac{n}{3}+2sec^4\frac{π}{3}= −4×4×(\sqrt3)^2+2×\frac{1}{(\frac{1}{2})^2}\\=48+2×16=48+32=80$

Now, substituting all these values in the Taylor's expansion series, we get

$f(log(cosx))=log(\frac{1}{2})+(-\sqrt{3})(x-\frac{π}{3})+\frac{(-4)}{2!}(x-\frac{π}{3})^2+\frac{(-8\sqrt3}{3!}(x-\frac{π}{3})^3+\frac{80}{4!}(x-\frac{π}{3})^4$

Simplifying the above equation, we get

$f(log(cosx))=log(\frac{1}{2})+(-\sqrt{3})(x-\frac{π}{3})-2(x-\frac{π}{3})^2-\frac{(4\sqrt3}{3}(x-\frac{π}{3})^3+\frac{10}{3}(x-\frac{π}{3})^4$

The above equation is the Taylor's series expansion up to the fourth degree term of the function $f(x)=log(cosx)$