# Obtain the Maclaurin’s series expansion for ${\log _e}\left( {1 + x} \right)$.

Answer

Verified

364.5k+ views

Hint: Find the 1st, 2nd, 3rd… order derivatives at x=0 and substitute in general form of Maclaurin’s series expansion.

Let us suppose the given function as \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\]

According to Maclaurin’s series expansion for any function,

\[f\left( x \right) = f\left( 0 \right) + \dfrac{x}{{1!}}f'\left( 0 \right) + \dfrac{{{x^2}}}{{2!}}f''\left( 0 \right) + \dfrac{{{x^3}}}{{3!}}f'''\left( 0 \right) + ......{\text{ }} \to {\text{(1)}}\]

Now, let us find \[f'\left( x \right)\] by differentiating \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\] with respect to $x$, we get

\[{\text{ }}f'\left( x \right) = \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {{{\log }_e}\left( {1 + x} \right)} \right)}}{{dx}} = \left( {\dfrac{1}{{1 + x}}} \right)\left( {\dfrac{{d\left( {1 + x} \right)}}{{dx}}} \right) = \dfrac{1}{{1 + x}}\]

For \[f''\left( x \right)\], differentiate \[f'\left( x \right)\] with respect to $x$, we get

\[f''\left( x \right) = \dfrac{{d\left( {f'\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {\dfrac{1}{{1 + x}}} \right)}}{{dx}} = \dfrac{{d\left[ {{{\left( {1 + x} \right)}^{ - 1}}} \right]}}{{dx}} = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\left( {\dfrac{{d\left( {1 + x} \right)}}{{dx}}} \right) = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\]

Similarly, \[f'''\left( x \right) = \dfrac{{d\left( {f''\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( { - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}} \right)}}{{dx}} = - \dfrac{{d\left( {{{\left( {1 + x} \right)}^{ - 2}}} \right)}}{{dx}} = \dfrac{2}{{{{\left( {1 + x} \right)}^3}}}\]

Now, put \[x = 0\] in the expressions of \[f\left( x \right),{\text{ }}f'\left( x \right),{\text{ }}f''\left( x \right),f'''\left( x \right)\], etc.

Therefore, \[f\left( 0 \right) = {\log _e}\left( {1 + 0} \right) = {\log _e}\left( 1 \right) = 0\]

\[f'\left( 0 \right) = \dfrac{1}{{1 + 0}} = 1\], \[f''\left( 0 \right) = - \dfrac{1}{{{{\left( {1 + 0} \right)}^2}}} = - 1\], \[f'''\left( 0 \right) = \dfrac{2}{{{{\left( {1 + 0} \right)}^3}}} = 2\], etc.

Now substitute all the above values in equation (1), we get

\[

f\left( x \right) = 0 + \dfrac{x}{1} \times 1 + \dfrac{{{x^2}}}{2} \times \left( { - 1} \right) + \dfrac{{{x^3}}}{6} \times 2 + ....... \\

\Rightarrow f\left( x \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ....... \\

\]

The above equation represents the Maclaurin’s series expansion for \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\].

Note: These types of problems are solved by finding the first derivative, second derivative, third derivative and so on of the given function and then finally substituting the value of the variable as 0. In this particular question, we have calculated only up to the third derivative just in order to avoid unnecessary computations. If we observe carefully the next term can be predicted easily.

Let us suppose the given function as \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\]

According to Maclaurin’s series expansion for any function,

\[f\left( x \right) = f\left( 0 \right) + \dfrac{x}{{1!}}f'\left( 0 \right) + \dfrac{{{x^2}}}{{2!}}f''\left( 0 \right) + \dfrac{{{x^3}}}{{3!}}f'''\left( 0 \right) + ......{\text{ }} \to {\text{(1)}}\]

Now, let us find \[f'\left( x \right)\] by differentiating \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\] with respect to $x$, we get

\[{\text{ }}f'\left( x \right) = \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {{{\log }_e}\left( {1 + x} \right)} \right)}}{{dx}} = \left( {\dfrac{1}{{1 + x}}} \right)\left( {\dfrac{{d\left( {1 + x} \right)}}{{dx}}} \right) = \dfrac{1}{{1 + x}}\]

For \[f''\left( x \right)\], differentiate \[f'\left( x \right)\] with respect to $x$, we get

\[f''\left( x \right) = \dfrac{{d\left( {f'\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {\dfrac{1}{{1 + x}}} \right)}}{{dx}} = \dfrac{{d\left[ {{{\left( {1 + x} \right)}^{ - 1}}} \right]}}{{dx}} = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\left( {\dfrac{{d\left( {1 + x} \right)}}{{dx}}} \right) = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\]

Similarly, \[f'''\left( x \right) = \dfrac{{d\left( {f''\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( { - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}} \right)}}{{dx}} = - \dfrac{{d\left( {{{\left( {1 + x} \right)}^{ - 2}}} \right)}}{{dx}} = \dfrac{2}{{{{\left( {1 + x} \right)}^3}}}\]

Now, put \[x = 0\] in the expressions of \[f\left( x \right),{\text{ }}f'\left( x \right),{\text{ }}f''\left( x \right),f'''\left( x \right)\], etc.

Therefore, \[f\left( 0 \right) = {\log _e}\left( {1 + 0} \right) = {\log _e}\left( 1 \right) = 0\]

\[f'\left( 0 \right) = \dfrac{1}{{1 + 0}} = 1\], \[f''\left( 0 \right) = - \dfrac{1}{{{{\left( {1 + 0} \right)}^2}}} = - 1\], \[f'''\left( 0 \right) = \dfrac{2}{{{{\left( {1 + 0} \right)}^3}}} = 2\], etc.

Now substitute all the above values in equation (1), we get

\[

f\left( x \right) = 0 + \dfrac{x}{1} \times 1 + \dfrac{{{x^2}}}{2} \times \left( { - 1} \right) + \dfrac{{{x^3}}}{6} \times 2 + ....... \\

\Rightarrow f\left( x \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ....... \\

\]

The above equation represents the Maclaurin’s series expansion for \[f\left( x \right) = {\log _e}\left( {1 + x} \right)\].

Note: These types of problems are solved by finding the first derivative, second derivative, third derivative and so on of the given function and then finally substituting the value of the variable as 0. In this particular question, we have calculated only up to the third derivative just in order to avoid unnecessary computations. If we observe carefully the next term can be predicted easily.

Last updated date: 30th Sep 2023

•

Total views: 364.5k

•

Views today: 5.64k

Recently Updated Pages

What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How many millions make a billion class 6 maths CBSE

Which are the Top 10 Largest Countries of the World?

How many crores make 10 million class 7 maths CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE