Question

What is a power series representation for $f\left( x \right) = \ln \left( {1 + x} \right)$ and what is its radius of convergence?

Answer 1

Here, we will first differentiate the function $\ln \left( {1 + x} \right)$ and convert it in the form of $\dfrac{1}{{1 - u}}$ .Then on integrating the expanded function, we will obtain the required power series. After that for the radius of convergence, we will use the ratio test on the power series obtained which states that if $\mathop {\lim }\limits_{n \to \infty } |\dfrac{{{a_{n + 1}}}}{{{a_n}}}|{\text{ }} < 1$ ,then $\sum\limits_{n = 0}^\infty {{a_n}}$ converges.

Complete step by step answer:
Let the given function be written as
$f\left( x \right) = \ln \left( {1 + x} \right){\text{ }} - - - \left( 1 \right)$
Differentiating both sides of the above equation with respect to $x$ we get
$\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {\ln \left( {1 + x} \right)} \right)}}{{dx}}$
Now using the formula
$\dfrac{{d\left( {\ln x} \right)}}{{dx}} = \dfrac{1}{x}$
Therefore, we get
$\Rightarrow {f^1}\left( x \right) = \dfrac{1}{{1 + x}}{\text{ }} - - - \left( 2 \right)$
Now let us consider the right-hand side of the above equation as
$g\left( x \right) = \dfrac{1}{{1 + x}}$

Now, using binomial expansion we know that
${\left( {1 + x} \right)^{ - n}} = 1 - nx + \dfrac{{n\left( {n + 1} \right)}}{{2!}}{x^2} - \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{3!}}{x^3} + .....$
So, ${\left( {1 + x} \right)^{ - 1}} = 1 - \left( 1 \right)x + \dfrac{{1\left( {1 + 1} \right)}}{{2!}}{x^2} - \dfrac{{1\left( {1 + 1} \right)\left( {1 + 2} \right)}}{{3!}}{x^3} + .....$
$\Rightarrow {\left( {1 + x} \right)^{ - 1}} = 1 - x + \dfrac{2}{{2!}}{x^2} - \dfrac{6}{{3!}}{x^3} + .....$
On cancelling the terms, we get
$\Rightarrow {\left( {1 + x} \right)^{ - 1}} = 1 - x + {x^2} - {x^3} + .....$
Now, substituting the value in the equation $\left( 2 \right)$ we get
$\Rightarrow {f^1}\left( x \right) = 1 - x + {x^2} - {x^3} + .....$
Now, integrating both the sides, we have
$\Rightarrow \int {{f^1}\left( x \right)} {\text{ }}dx = \int {\left( {1 - x + {x^2} - {x^3} + .....} \right)} {\text{ }}dx$

We know that
$\int {{f^1}\left( x \right){\text{ }}dx = f\left( x \right)}$
$\Rightarrow \int {{x^n}{\text{ }}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$
Therefore, we get
$f\left( x \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ..... + c$
where c is the constant of integration
Substituting the value in the equation $\left( 1 \right)$ we get
$\ln \left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ..... + c$
$\Rightarrow \ln \left( {1 + x} \right) = \dfrac{{{x^1}}}{1} - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ..... + c$
Now writing the right-hand side in the summation form, we get
$\Rightarrow \ln \left( {1 + x} \right) = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\dfrac{{{x^{n + 1}}}}{{n + 1}}} {\text{ + }}c{\text{ }} - - - \left( 3 \right)$

Now substituting $x = 0$ in the above equation, we have
$\ln \left( {1 + 0} \right) = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\dfrac{{{0^{n + 1}}}}{{n + 1}}} + c$
$\Rightarrow \ln \left( 1 \right) = 0 + c$
We know that $\ln \left( 1 \right) = 0$
Therefore, we get
$0 = 0 + c$
$\Rightarrow c = 0$
Substituting this in equation $\left( 3 \right)$ we get
$\Rightarrow \ln \left( {1 + x} \right) = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\dfrac{{{x^{n + 1}}}}{{n + 1}}} - - - \left( 4 \right)$
which 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 power series.So, for that we will apply the ratio test which states that if $\mathop {\lim }\limits_{n \to \infty } |\dfrac{{{a_{n + 1}}}}{{{a_n}}}|{\text{ }} < 1$ ,then $\sum\limits_{n = 0}^\infty {{a_n}}$ converges

Let $L = \mathop {\lim }\limits_{n \to \infty } |\dfrac{{{a_{n + 1}}}}{{{a_n}}}|{\text{ }}$
From equation $\left( 4 \right)$ we have
${a_n} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$
Therefore,
$\Rightarrow L = \mathop {\lim }\limits_{n \to \infty } |\dfrac{{\dfrac{{{x^{n + 2}}}}{{n + 2}}}}{{\dfrac{{{x^{n + 1}}}}{{n + 1}}}}|{\text{ }}$
$\Rightarrow L = \mathop {\lim }\limits_{n \to \infty } |\dfrac{{x\left( {n + 1} \right)}}{{n + 2}}|{\text{ }}$
$\Rightarrow L = |x|\mathop {\lim }\limits_{n \to \infty } |\dfrac{{n + 1}}{{n + 2}}|{\text{ }}$
As $n$ tends to infinity, $n$ will get closer to $n + 1$ and $n + 2$
and hence $\mathop {\lim }\limits_{n \to \infty } |\dfrac{{n + 1}}{{n + 2}}|{\text{ }} = 1$
Putting this value in the above equation, we get
$L = {\text{ }}|x|$
Now by ratio test, the above limit must be less than one, which means
$|x|{\text{ }} < 1$

Hence, the radius of convergence is equal to one.

Note: Power series are useful in mathematical analysis, where they arise as Taylor series of infinitely differentiable functions. When a power series converges on an interval, then the distance from the centre of convergence to the other end of an interval, where it converges is called the radius of convergence. Also note that the radius of convergence can be determined using other tests as well such as Integral test, Raabe’s test, etc.