Question

How do you find the Maclaurin series for tanx?

Answer 1

Hint: In this question, we need to find the Maclaurins series for tanx. For this, we will find some derivative of tanx i.e if f(x) = tanx. We will try to find f(0), f'(0), f''(0), f'''(0), . . . . . . . . After that we will use the formula of Maclaurins series to evaluate our answer. Maclaurins series is given by $ f\left( x \right)=f\left( 0 \right)+f'\left( 0 \right)x+\dfrac{f''\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{f'''\left( 0 \right)}{3!}{{x}^{3}}+\cdots \ \cdots $ .

Complete step by step answer:
Here we are given the function as tanx for which we need to find the Maclaurin’s series. Let us suppose that f(x) = tanx.
We know, Maclaurins series for a function f(x) is given by:
 $ f\left( x \right)=f\left( 0 \right)+f'\left( 0 \right)x+\dfrac{f''\left( 0 \right)}{2!}{{x}^{2}}+\dfrac{f'''\left( 0 \right)}{3!}{{x}^{3}}+\cdots \ \cdots $ .
So let us evaluate the values of f(0), f'(0), f''(0), f'''(0), . . . . . . . . .
For this, we need to evaluate f'(x), f''(x), f'''(x).
We have f(x) = tanx.
Putting x = 0 we get f(0) = tan0.
We know that tan0 is 0 so we get $ f\left( 0 \right)=0\cdots \cdots \cdots \left( 1 \right) $ .
Differentiating f(x) with respect to x we get $ f'\left( x \right)=\dfrac{d}{dx}\tan x $ .
We know that, derivative of tanx is $ {{\sec }^{2}}x $ so we have $ f'\left( x \right)={{\sec }^{2}}x $ .
We know that $ {{\sec }^{2}}x=1+{{\tan }^{2}}x $ so we get $ f'\left( x \right)=1+{{\tan }^{2}}x $ .
Putting x = 0 we get $ f'\left( 0 \right)=1+{{\tan }^{2}}0 $ .
Using tan0 = 0 we get $ f'\left( 0 \right)=1+0\Rightarrow f'\left( 0 \right)=1\cdots \cdots \cdots \left( 2 \right) $ .
Differentiating f'(x) with respect to x we get $ f''\left( x \right)=\dfrac{d}{dx}\left( 1+{{\tan }^{2}}x \right) $ .
We know that $ \dfrac{d\left( 1 \right)}{dx}=0 $ so let us find derivative of $ {{\tan }^{2}}x $ . Using chain rule, derivatives of $ {{\tan }^{2}}x $ will be given by $ 2{{\tan }^{2}}x\cdot \dfrac{d}{dx}\tan x=2\tan x{{\sec }^{2}}x $ .
Therefore we have $ f''\left( x \right)=0+2\tan x{{\sec }^{2}}x\Rightarrow f''\left( x \right)=2\tan x{{\sec }^{2}}x $ .
Using $ {{\sec }^{2}}x=1+{{\tan }^{2}}x $ we get $ f''\left( x \right)=2\tan x\left( 1+{{\tan }^{2}}x \right) $ .
Putting x = 0 we get $ f''\left( x0 \right)=2\tan 0\left( 1+{{\tan }^{2}}0 \right) $ .
Again using tan0 = 0 we get $ f''\left( 0 \right)=2\left( 0 \right)\left( 1+0 \right)=0\Rightarrow f''\left( 0 \right)=0\cdots \cdots \cdots \left( 3 \right) $ .
Differentiating f''(x) with respect to x we get $ f'''\left( x \right)=\dfrac{d}{dx}\left( 2\tan x\left( 1+{{\tan }^{2}}x \right) \right) $ .
For this, let us use the product rule. We know that, for any two functions u and v product rule is given as $ \dfrac{duv}{dx}=uv'+u'v $ where u' and v' are derivatives of u and v respectively, we get,
\[f'''\left( x \right)=2\left( \tan x\dfrac{d}{dx}\left( 1+{{\tan }^{2}}x \right)+\dfrac{d}{dx}\left( \tan x \right)\cdot \left( 1+{{\tan }^{2}}x \right) \right)\].
Using previous results for derivatives of \[\left( 1+{{\tan }^{2}}x \right)\] and tanx we get,
\[f'''\left( x \right)=2\left( \tan x\left( 2\tan x\left( 1+{{\tan }^{2}}x \right) \right)+\left( 1+{{\tan }^{2}}x \right)\left( 1+{{\tan }^{2}}x \right) \right)\].
Simplifying we get, \[f'''\left( x \right)=2\left( 2{{\tan }^{2}}x\left( 1+{{\tan }^{2}}x \right)+{{\left( 1+{{\tan }^{2}}x \right)}^{2}} \right)\].
Putting x = 0 we get \[f'''\left( 0 \right)=2\left( 2{{\tan }^{2}}0\left( 1+{{\tan }^{2}}0 \right)+{{\left( 1+{{\tan }^{2}}0 \right)}^{2}} \right)\].
Again using tan0 = 0 we get,
\[\begin{align}
  & f'''\left( 0 \right)=2\left( 2\times 0\left( 1+0 \right)+{{\left( 1+0 \right)}^{2}} \right) \\
 & \Rightarrow f'''\left( 0 \right)=2\left( 0+{{1}^{2}} \right) \\
 & \Rightarrow f'''\left( 0 \right)=2\left( 1 \right) \\
 & \Rightarrow f'''\left( 0 \right)=2\cdots \cdots \cdots \left( 4 \right) \\
\end{align}\]
Putting all these values in the Maclaurins series we get,
 $ \tan x=0+\left( 1 \right)x+\dfrac{0}{2!}{{x}^{2}}+\dfrac{2}{3!}{{x}^{3}}+\cdots \ \cdots $ .
Simplifying we get $ \tan x=x+\dfrac{2}{3!}{{x}^{3}}+\cdots \ \cdots $ .
Expanding 3! as $ 3\times 2 $ we get $ \tan x=x+\dfrac{2}{3\times 2}{{x}^{3}}+\cdots \ \cdots \Rightarrow \tan x=x+\dfrac{{{x}^{3}}}{3}+\cdots \ \cdots $ .
Therefore, the Maclaurins series for tanx is given as $ \tan x=x+\dfrac{{{x}^{3}}}{3}+\cdots \ \cdots $ .

Note:
 Students should take care while finding all the derivatives. They should note that all even values will be equal to 0, so we have Maclaurin’s series in odd order only. They can find more functions to increase expansion.