Question

Expand \[\log \left( {1 + x} \right)\] in ascending powers of \[x\] up to the term containing \[{x^4}\].

Answer 1

Hint: We will take \[f\left( x \right) = \log \left( {1 + x} \right)\]. Then we will do successive differentiation of \[f\left( x \right)\] to find the values of \[{f'}\left( x \right)\], \[{f^{''}}\left( x \right)\], \[{f^{'''}}\left( x \right)\] and \[{f^{''''}}\left( x \right)\]. Putting x equals to zero, we will then find the values of \[{f'}\left( 0 \right)\], \[{f^{''}}\left( 0 \right)\], \[{f^{'''}}\left( 0 \right)\] and\[{f^{''''}}\left( 0 \right)\]. Putting all these values in Maclaurin’s theorem we will get the required expansion of \[\log \left( {1 + x} \right)\].

Complete step by step answer:
Let \[f\left( x \right) = \log \left( {1 + x} \right)\]
Putting \[x = 0\], we get
\[ \Rightarrow f\left( 0 \right) = \log \left( {1 + 0} \right)\]
\[ \Rightarrow f\left( 0 \right) = \log \left( 1 \right)\]
As \[\log \left( 1 \right) = 0\], we get
\[ \Rightarrow f\left( 0 \right) = 0\]
Now, on differentiating \[f\left( x \right)\] we get
\[ \Rightarrow {f'}\left( x \right) = \dfrac{{dy}}{{dx}}\left( {\log \left( {1 + x} \right)} \right)\]
On simplifying right-hand side of above equation, we get
\[ \Rightarrow {f'}\left( x \right) = \dfrac{1}{{1 + x}}\]
Putting \[x = 0\], we get
\[ \Rightarrow {f'}\left( 0 \right) = \dfrac{1}{{1 + 0}}\]
On simplification, we get
\[ \Rightarrow {f'}\left( 0 \right) = 1\]
Now, differentiating \[{f'}\left( x \right)\], we get
\[ \Rightarrow {f^{''}}\left( x \right) = \dfrac{{dy}}{{dx}}\left( {\dfrac{1}{{1 + x}}} \right)\]
On simplifying right-hand side of above equation, we get
\[ \Rightarrow {f^{''}}\left( x \right) = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}\]
Putting \[x = 0\], we get
\[ \Rightarrow {f^{''}}\left( 0 \right) = - \dfrac{1}{{{{\left( {1 + 0} \right)}^2}}}\]
On simplification, we get
\[ \Rightarrow {f^{''}}\left( 0 \right) = - 1\]
In the same way, differentiating \[{f^{''}}\left( x \right)\], we get
\[ \Rightarrow {f^{'''}}\left( x \right) = \dfrac{{dy}}{{dx}}\left( { - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}} \right)\]
On simplifying right-hand side of above equation, we get
\[ \Rightarrow {f^{'''}}\left( x \right) = \dfrac{2}{{{{\left( {1 + x} \right)}^3}}}\]
Putting \[x = 0\], we get
\[ \Rightarrow {f^{'''}}\left( 0 \right) = \dfrac{2}{{{{\left( {1 + 0} \right)}^3}}}\]
On simplification, we get
\[ \Rightarrow {f^{'''}}\left( 0 \right) = 2\]
Now, differentiating \[{f^{'''}}\left( x \right)\], we get
\[ \Rightarrow {f^{''''}}\left( x \right) = \dfrac{{dy}}{{dx}}\left( {\dfrac{2}{{{{\left( {1 + x} \right)}^3}}}} \right)\]
On simplifying right-hand side of above equation, we get
\[ \Rightarrow {f^{''''}}\left( x \right) = - \dfrac{6}{{{{\left( {1 + x} \right)}^4}}}\]
Putting \[x = 0\], we get
\[ \Rightarrow {f^{''''}}\left( 0 \right) = - \dfrac{6}{{{{\left( {1 + 0} \right)}^4}}}\]
On simplification, we get
\[ \Rightarrow {f^{''''}}\left( 0 \right) = - 6\]
Therefore, we get the values as \[f\left( 0 \right) = 0\], \[{f'}\left( 0 \right) = 1\], \[{f^{''}}\left( 0 \right) = - 1\], \[{f^{'''}}\left( 0 \right) = 2\] and \[{f^{''''}}\left( 0 \right) = - 6\].
From Maclaurin’s theorem, we know
\[f\left( x \right) = f\left( 0 \right) + x{f'}\left( 0 \right) + \dfrac{{{x^2}}}{{2!}}{f^{''}}\left( 0 \right) + \dfrac{{{x^3}}}{{3!}}{f^{'''}}\left( 0 \right) + \dfrac{{{x^4}}}{{4!}}{f^{''''}}\left( 0 \right) + ...\]
Putting the values of \[f\left( x \right)\], \[f\left( 0 \right)\], \[{f'}\left( 0 \right)\], \[{f^{''}}\left( 0 \right)\], \[{f^{'''}}\left( 0 \right)\] and \[{f^{''''}}\left( 0 \right)\], we get
\[ \Rightarrow \log \left( {1 + x} \right) = 0 + x \times \left( 1 \right) + \dfrac{{{x^2}}}{{2!}} \times \left( { - 1} \right) + \dfrac{{{x^3}}}{{3!}} \times \left( 2 \right) + \dfrac{{{x^4}}}{{4!}} \times \left( { - 6} \right) + ...\]
On simplification, we get
\[ \Rightarrow \log \left( {1 + x} \right) = x - \dfrac{{{x^2}}}{{2!}} + \dfrac{{2{x^3}}}{{3!}} - \dfrac{{6{x^4}}}{{4!}} + ...\]
On further simplification we get
\[ \Rightarrow \log \left( {1 + x} \right) = x - \dfrac{{{x^2}}}{{2 \times 1}} + \dfrac{{2{x^3}}}{{3 \times 2 \times 1}} - \dfrac{{6{x^4}}}{{4 \times 3 \times 2 \times 1}} + ...\]
After calculation, we get
\[ \Rightarrow \log \left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ...\]
Therefore, expanding \[\log \left( {1 + x} \right)\] in ascending powers of \[x\] up to the term containing \[{x^4}\], we get
\[\log \left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ...\]

Note:
Here we have differentiated \[f\left( x \right)\] four times because we required expansion up to \[{x^4}\]. For example, if we require expansion up to \[{x^6}\] then we have to differentiate \[f\left( x \right)\] six times. So, we have to differentiate the given \[f\left( x \right)\] up to the power of \[x\] required.