Question

How do you find a power series representation for$\ln (5 - x)$ and what is the radius of convergence?

Answer 1

First of all we must know, what is a power series? So, a power series (in one variable) is an infinite series of the form,
$\dfrac{1}{{1 - u}} = \sum\limits_{n = 0}^N {{u^n}} = 1 + u + {u^2} + {u^3}.......$
So, to find the power series of$\ln (5 - x)$, we are to convert it in the form of $\dfrac{1}{{1 - u}}$, by various operations, and then have to reverse the operations, to get the required result.
A number r is called the radius of convergence of the power series, such that the series converges whenever $\left| {x - c} \right| < r$ and $r$ is $0 < r \leqslant \infty$.

Complete answer:
We know, the general form of power series,
$\dfrac{1}{{1 - u}} = \sum\limits_{n = 0}^N {{u^n}} = 1 + u + {u^2} + {u^3}.......$
Now, we are to convert $\ln (5 - x)$ to the form $\dfrac{1}{{1 - u}}$.
So, $\dfrac{d}{{dx}}\left[ {\ln (5 - x)} \right]$
$= - \dfrac{1}{{5 - x}}$
$= - \dfrac{1}{5} \times \dfrac{1}{{1 - \dfrac{x}{5}}}$
$\therefore \dfrac{d}{{dx}}\left[ {\ln (5 - x)} \right] = - \dfrac{1}{5} \times \dfrac{1}{{1 - \dfrac{x}{5}}}$
Thus, with $u = \dfrac{x}{5}$, we had taken the derivative and then factored out $- \dfrac{1}{5}$.
To get the power series, we have to work in reverse.
We had done this:
Differentiated $\ln (5 - x)$
Factored out$- \dfrac{1}{5}$
Substituted $\dfrac{x}{5}$ for $u$
Now, we are to reverse what we did, starting from the power series itself.
Substitute $u = \dfrac{x}{5}$
Multiply by $- \dfrac{1}{5}$
Integrate the result
Since, $\int {function = \int {{\text{power series of that function}}} }$,
So, we have, $\dfrac{1}{{1 - u}} = 1 + u + {u^2} + {u^3}.......$
$\Rightarrow \dfrac{1}{{1 - \dfrac{x}{5}}} = 1 + \dfrac{x}{5} + \dfrac{{{x^2}}}{{25}} + \dfrac{{{x^3}}}{{125}} + .....$
Multiplying with$- \dfrac{1}{5}$, we get,
$\Rightarrow - \dfrac{1}{5} \cdot \dfrac{1}{{1 - \dfrac{x}{5}}} = - \dfrac{1}{5} - \dfrac{x}{{25}} - \dfrac{{{x^2}}}{{125}} - \dfrac{{{x^3}}}{{625}} - .....$
Integrating, the result, we get,
$\Rightarrow \int { - \dfrac{1}{5}} \cdot \dfrac{1}{{1 - \dfrac{x}{5}}}dx = \int {\left[ { - \dfrac{1}{5} - \dfrac{x}{{25}} - \dfrac{{{x^2}}}{{125}} - \dfrac{{{x^3}}}{{625}} - .....} \right]} dx$
$\Rightarrow \ln (5 - x) = C - \dfrac{x}{5} - \dfrac{{{x^2}}}{{50}} - \dfrac{{{x^3}}}{{375}} - \dfrac{{{x^4}}}{{2500}}......$
Where $C$ is the constant of integration, it is the term for$n = 0$.
For a regular power series derive from$\dfrac{1}{{1 - x}}$, we write,
$\sum\limits_{n - 0}^N {{{(x - 0)}^n}} = \dfrac{1}{{1 - x}}$.
Where the power series is centred around $a = 0$.
We know that the constant must not contain a$x$term (because$x$ is a variable).
The constant cannot be $\ln x$, so the constant $C$ is $\ln (5)$.
So, we get, the required power series is,
$\ln (5 - x) = \ln (5) - \dfrac{x}{5} - \dfrac{{{x^2}}}{{50}} - \dfrac{{{x^3}}}{{375}} - \dfrac{{{x^4}}}{{2500}} - ......$
And, for the radius of convergence, it is,
$\left| x \right| < 5$
Because $\ln (5 - x)$ approaches $- \infty$ as$x \to 5$.
We know that the power series must already converge upon $\ln (5 - x)$ wherever the function exists because it was constructed for the function.

Note:
Power series are useful in mathematical analysis, where they arise as Taylor series of infinitely differentiable functions. Beyond their role in mathematical analysis, power series also occur in combinatorics as generating functions and in electronic engineering. Any polynomial can be easily expressed as a power series around any centre c, although all but finitely many of the coefficients will be zero since a power series has infinitely many terms by definition.