Question

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

Answer 1

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.