Question

What is the taylor series expansion for the tangent function $\left( \tan x \right)$?

Answer 1

Assume the given tangent function as $f\left( x \right)=\tan x$. Consider the formula for the taylor expansion of a function f (x) given as: - $f\left( x \right)=f\left( a \right)+\dfrac{\left( x-a \right)}{1!}f'\left( a \right)+\dfrac{{{\left( x-a \right)}^{2}}}{2!}f''\left( a \right)+\dfrac{{{\left( x-a \right)}^{3}}}{3!}f'''\left( a \right)+......$ where ‘a’ denotes the point around which the expansion is to be found. Here, f’, f’’, f’’’……… represents the first derivative, second derivative, third derivative……… respectively of the function f (x).

Complete step by step answer:
Here we have been provided with the tangent function $\left( \tan x \right)$ and we are asked to write its taylor series expansion expression. Let us assume the given function as $f\left( x \right)$, so we have,
$\Rightarrow f\left( x \right)=\tan x$
Now, we know that taylor series expansion of any function f (x) is given by the formula: - $f\left( x \right)=f\left( a \right)+\dfrac{\left( x-a \right)}{1!}f'\left( a \right)+\dfrac{{{\left( x-a \right)}^{2}}}{2!}f''\left( a \right)+\dfrac{{{\left( x-a \right)}^{3}}}{3!}f'''\left( a \right)+......$.
Here, ‘a’ denotes the value of x at which the expansion of the function is to be found. In the above question it is not given that at which point we have to find the expansion, so we will consider a = 0.
Now, we need to find the derivatives of $\tan x$ at x = 0. So, let us find them one – by – one.

& \Rightarrow f\left( x \right)=\tan x\Rightarrow f\left( 0 \right)=0 \\\ & \Rightarrow f'\left( x \right)={{\sec }^{2}}x\Rightarrow f'\left( 0 \right)=1 \\\ & \Rightarrow f''\left( x \right)=2\sec x\times \sec x\tan x=2{{\sec }^{2}}x\tan x\Rightarrow f''\left( 0 \right)=0 \\\ & \Rightarrow f'''\left( x \right)=2\left[ 2{{\sec }^{2}}x{{\tan }^{2}}x+{{\sec }^{4}}x \right]\Rightarrow f'''\left( 0 \right)=2 \\\ \end{aligned}$$ As we will go on with this process we will get the following values of higher derivatives: $$\begin{aligned} & \Rightarrow f''''\left( 0 \right)=0 \\\ & \Rightarrow f'''''\left( 0 \right)=16 \\\ \end{aligned}$$ We will not go further after five terms, so substituting the above obtained values in the expansion formula we get, $$\Rightarrow \tan x=0+\dfrac{x}{1!}\times 1+\dfrac{{{x}^{2}}}{2!}\times 0+\dfrac{{{x}^{3}}}{3!}\times 2+\dfrac{{{x}^{4}}}{4!}\times 0+\dfrac{{{x}^{5}}}{5!}\times 16+......$$ $$\therefore \tan x=x+\dfrac{1}{3}{{x}^{2}}+\dfrac{2}{15}{{x}^{5}}+......$$ Hence the above expression represents the taylor expansion of $\tan x$ at x = 0. **Note:** You may note that taylor expansion formula at for a = 0 has a particular name of its own. It is called the Mclaurin series, i.e., Taylor series at x = 0. It is a special case of taylor series. Note that generally we stop finding the derivatives after the $${{5}^{th}}$$ derivative because the series is continued till infinity and we cannot go on finding the derivatives till infinity.