Question

How do you find the power series representation for $\ln \left( {1 - x} \right)$ and what is the radius of convergence?

Answer 1

Here, we will first differentiate the function $\ln \left( {1 - x} \right)$ and expand the differentiated function using the long division method. Then on integrating the expanded function, we will obtain the required power series. For the radius of convergence, we will apply the ratio test on the power series obtained.

Complete step-by-step solution:
Let the given function be written as
$f\left( x \right) = \ln \left( {1 - x} \right)$……………………………$\left( 1 \right)$
Differentiating both sides of the above equation with respect to $x$, we get
$\Rightarrow \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {\ln \left( {1 - x} \right)} \right)}}{{dx}}$
Now using the formula $\dfrac{{d\left( {\ln x} \right)}}{{dx}} = \dfrac{1}{x}$, we get

$\Rightarrow f'\left( x \right) = \dfrac{{ - 1}}{{\left( {1 - x} \right)}}$………………………………..$\left( 2 \right)$
Now, let us consider the RHS of the above equation as
$g\left( x \right) = \dfrac{{ - 1}}{{\left( {1 - x} \right)}}$
Now, when we divide $1$ by $\left( {1 - x} \right)$ using division method, we get
$\dfrac{1}{{\left( {1 - x} \right)}} = 1 + x + {x^2} + {x^3} + ...$
Substituting this in $\left( 2 \right)$ we get
$f'\left( x \right) = - \left( {1 + x + {x^2} + {x^3} + .......} \right)$
Integrating both the sides, we have
$\Rightarrow \int {f'\left( x \right)dx} = \int { - \left( {1 + x + {x^2} + {x^3} + .......} \right)dx}$
We know that $\int {f'\left( x \right)dx} = f\left( x \right)$. So we get
$\Rightarrow f\left( x \right) = \int { - \left( {1 + x + {x^2} + {x^3} + .......} \right)dx}$
On integrating the RHS of the above equation, we get
$\Rightarrow f\left( x \right) = - \left( {x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} + \dfrac{{{x^4}}}{4} + .....} \right) + C$
Putting $\left( 1 \right)$ in the above equation, we get
$\Rightarrow \ln \left( {1 - x} \right) = - \left( {x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} + \dfrac{{{x^4}}}{4} + .....} \right) + C$
$\Rightarrow \ln \left( {1 - x} \right) = - \left( {\dfrac{{{x^1}}}{1} + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} + \dfrac{{{x^4}}}{4} + .....} \right) + C$
Writing the RHS in the contracted form, we get
$\ln \left( {1 - x} \right) = - \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{n}} + C$………………………$\left( 3 \right)$

Substituting $x = 0$ in the above equation, we have
$\Rightarrow \ln \left( {1 - 0} \right) = - \sum\limits_{n = 0}^\infty {\dfrac{{{0^n}}}{n}} + C$
$\Rightarrow \ln \left( 1 \right) = 0 + C$

We know that $\ln \left( 1 \right) = 0$. Substituting this above, we get
$0 = 0 + C$
$\Rightarrow C = 0$
Substituting this in equation $\left( 3 \right)$, we get
$\ln \left( {1 - x} \right) = - \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{n}}$……………………$\left( 4 \right)$
This is the required power series representation for $\ln \left( {1 - x} \right)$.
Now, for the radius of convergence, we have to determine the interval of convergence of the above series. For that, we need to apply the ratio test as below
$L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right|$
From $\left( 4 \right)$ we have ${a_n} = \dfrac{{{x^n}}}{n}$.
Substituting ${a_n} = \dfrac{{{x^n}}}{n}$ in the above equation, we get
$\Rightarrow L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{x^{n + 1}}/n + 1}}{{{x^n}/n}}} \right|$
$\Rightarrow L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{xn}}{{n + 1}}} \right|$
$\Rightarrow L = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{n}{{n + 1}}} \right|$
As $n$ tends to infinity, $n$ will get closer to $n + 1$ and hence $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{n}{{n + 1}}} \right| = 1$. Putting this value in above equation, we get
$\Rightarrow L = \left| x \right|$
Now, by ratio test, the above limit must be less than one, which means
$\left| x \right| < 1$

Hence, the radius of convergence is equal to one.

Note:
When a power series converges on an interval, then the distance from the centre of convergence to the other end of the interval, where it converges, is known as the radius of convergence. Here, we might make a mistake if we forget the negative sign while differentiating the function $\ln \left( {1 - x} \right)$. Also, the radius of convergence can be determined using other tests as well such as the Raabe’s test, the integral test etc. But as the ratio test is the easiest of all, we used it to find the radius of convergence.