Question

How do you do the taylor expansion for \[f\left( x \right)=\log \left( x+1 \right)\] at x = 0?

Answer 1

Hint: Consider the formula for the taylor expansion of a function f (x) given as: - \[f\left( x \right)=f\left( a \right)+\dfrac{\left( x-a \right)}{1!}f'\left( a \right)+\dfrac{{{\left( x-a \right)}^{2}}}{2!}f''\left( a \right)+\dfrac{{{\left( x-a \right)}^{3}}}{3!}f'''\left( a \right)+......\] and substitute the value of ‘a’ equal to 0. Here, f’, f’’, f’’’,……… represents the first derivative, second derivative, third derivative……… respectively of the function f (x). Use the formula given as: - \[m\left( m-1 \right)\left( m-2 \right).......1=m!\] to simplify the expression and get the required answer.

Complete step-by-step answer:
Here, we have been provided with the function \[f\left( x \right)=\log \left( x+1 \right)\] and we are asked to write its expanded form at x = 0. We have to use the taylor series or expansion formula for the answer.
Now, we know that taylor 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 given that we have to find the expansion at x = 0, that means the value of ‘a’ is 0. So, substituting the value of a = 0 in the above formula we get,
\[\Rightarrow f\left( x \right)=f\left( 0 \right)+\dfrac{x}{1!}f'\left( 0 \right)+\dfrac{{{x}^{2}}}{2!}f''\left( 0 \right)+\dfrac{{{x}^{3}}}{3!}f'''\left( 0 \right)+......\] - (1)
Now, we need to find the derivatives of \[f\left( x \right)=\log \left( x+1 \right)\] at x = 0. So, let us find them one – by – one.
\[\begin{align}
  & \Rightarrow f\left( x \right)=\log \left( x+1 \right)\Rightarrow f\left( 0 \right)=\log 1=0 \\
 & \Rightarrow f'\left( x \right)=\dfrac{1}{x+1}\Rightarrow f'\left( 0 \right)=\dfrac{1}{0+1}=1 \\
 & \Rightarrow f''\left( x \right)=\dfrac{-1}{{{\left( x+1 \right)}^{2}}}\Rightarrow f''\left( x \right)=\dfrac{-1}{{{\left( 0+1 \right)}^{2}}}=-1=-\left( 1! \right) \\
 & \Rightarrow f'''\left( x \right)=\dfrac{+2}{{{\left( x+1 \right)}^{3}}}\Rightarrow f'''\left( 0 \right)=\dfrac{2}{{{\left( 0+1 \right)}^{3}}}=2=2\times 1=2! \\
 & \Rightarrow f''''\left( x \right)=\dfrac{-2\times 3}{{{\left( x+1 \right)}^{4}}}\Rightarrow f''''\left( 0 \right)=\dfrac{-2\times 3}{{{\left( 0+1 \right)}^{4}}}=-2\times 3=-\left( 3! \right) \\
\end{align}\]
On observing the above pattern we get,
\[\begin{align}
  & \Rightarrow f'''''\left( 0 \right)=4! \\
 & \Rightarrow f''''''\left( 0 \right)=-\left( 5! \right) \\
\end{align}\]
And this will go on up to infinite. So, substituting all the obtained values in the taylor expansion formula, we get,
\[\Rightarrow f\left( x \right)=0+\dfrac{x}{1!}\times 1+\dfrac{{{x}^{2}}}{2!}\times \left[ -\left( -1! \right) \right]+\dfrac{{{x}^{3}}}{3!}\times 2!+\dfrac{{{x}^{4}}}{4!}\times \left[ -\left( 3! \right) \right]+\dfrac{{{x}^{5}}}{5!}\times 4!+\dfrac{{{x}^{6}}}{6!}\times \left[ -\left( 5! \right) \right]+......\]
\[\Rightarrow f\left( x \right)=x-\dfrac{{{x}^{2}}}{2!}\times 1!+\dfrac{{{x}^{3}}}{3!}\times 2!-\dfrac{{{x}^{4}}}{4!}\times 3!+\dfrac{{{x}^{5}}}{5!}\times 4!-\dfrac{{{x}^{6}}}{6!}\times 5!+......\]
The above expression can be simplified as: -
\[\Rightarrow f\left( x \right)=x-\dfrac{{{x}^{2}}}{2\times 1!}\times 1!+\dfrac{{{x}^{3}}}{3\times 2!}\times 2!-\dfrac{{{x}^{4}}}{4\times 3!}\times 3!+\dfrac{{{x}^{5}}}{5\times 4!}\times 4!-\dfrac{{{x}^{6}}}{6\times 5!}\times 5!+......\]
Cancelling the common factors we get,
\[\Rightarrow f\left( x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{5}}}{5}-\dfrac{{{x}^{6}}}{6}+......\]
Hence, the above expression represents the Taylor expansion of \[f\left( x \right)=\log \left( x+1 \right)\] at x = 0.

Note: One may note that equation (1) 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. You must remember both the formulas to solve the question. 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. So, we obtain a general pattern for the series to get the answer.