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

Oct 24, 2015

We can start from the power series that you were taught during the semester:

$\frac{1}{1 - u} = {\sum}_{n = 0}^{N} {u}^{n} = 1 + u + {u}^{2} + {u}^{3} + \ldots$

Now, let's work from $\ln \left(5 - x\right)$ to get to $\frac{1}{1 - u}$.

$\frac{d}{\mathrm{dx}} \left[\ln \left(5 - x\right)\right] = - \frac{1}{5 - x} = - \frac{1}{5} \cdot \frac{1}{1 - \frac{x}{5}}$

Thus, with $u = \frac{x}{5}$, we had just taken the derivative and then factored out $- \frac{1}{5}$. To get the power series, we have to work backwards.

1. Differentiated our target.
2. Factored out $- \frac{1}{5}$.
3. Substituted $\frac{x}{5}$ for $u$.

Now, we just reverse what we did, starting from the power series itself.

1. Substitute $u = \frac{x}{5}$.
2. Multiply by $- \frac{1}{5}$.
3. Integrate the result.

Since $\int \text{function"= int"power series of that function}$, we can do this:

$\frac{1}{1 - \frac{x}{5}} = 1 + \frac{x}{5} + {x}^{2} / 25 + {x}^{3} / 125 + \ldots$

$- \frac{1}{5} \cdot \frac{1}{1 - \frac{x}{5}} = - \frac{1}{5} - \frac{x}{25} - {x}^{2} / 125 - {x}^{3} / 625 - \ldots$

$\int - \frac{1}{5} \cdot \frac{1}{1 - \frac{x}{5}} \mathrm{dx} = \ln \left(5 - x\right)$

$= \int - \frac{1}{5} - \frac{x}{25} - {x}^{2} / 125 - {x}^{3} / 625 - \ldots \mathrm{dx}$

$= \setminus m a t h b f \left(C\right) - \frac{x}{5} - {x}^{2} / 50 - {x}^{3} / 375 - {x}^{4} / 2500 - \ldots$

Notice how we still have to figure out the constant $C$ because we performed the indefinite integral. $C$ is the term for $n = 0$.

For a regular power series derived from $\frac{1}{1 - x}$, we write

${\sum}_{n = 0}^{N} {\left(x - 0\right)}^{n} = \frac{1}{1 - x}$.
where the power series is centered around $a = 0$ since it's really the Maclaurin series (meaning, the Taylor series centered around $a = 0$).

We know that the constant must not contain an $x$ term (because $x$ is a variable). The constant cannot be $\ln x$, so the constant $C$ is $\textcolor{g r e e n}{\ln \left(5\right)}$. So, we get:

$\textcolor{b l u e}{\ln \left(5 - x\right) = \ln \left(5\right) - \frac{x}{5} - {x}^{2} / 50 - {x}^{3} / 375 - {x}^{4} / 2500 - \ldots}$

And then finally, for the radius of convergence, it is $| x | < 5$ because $\ln \left(5 - x\right)$ approaches $- \infty$ as $x \to 5$. We know that the power series must already converge upon $\ln \left(5 - x\right)$ wherever the function exists because it was constructed for the function.