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

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

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##1/(1-u) = sum_(n=0)^(N) u^n = 1 + u + u^2 + u^3 + …##

Now, let’s work from ##ln(5-x)## to get to ##1/(1-u)##.

##d/(dx)[ln(5-x)] = -1/(5-x) = -1/5*1/(1-x/5)##

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

1. Differentiated our target.
2. Factored out ##-1/5##.
3. Substituted ##x/5## for ##u##.

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

1. Substitute ##u = x/5##.
2. Multiply by ##-1/5##.
3. Integrate the result.

Since ##int “function”= int”power series of that function”##, we can do this:

##1/(1-x/5) = 1 + x/5 + x^2/25 + x^3/125 + …##

##-1/5*1/(1-x/5) = -1/5 – x/25 – x^2/125 – x^3/625 – …##

##int -1/5*1/(1-x/5)dx = ln(5-x)##

##= int -1/5 – x/25 – x^2/125 – x^3/625 – …dx##

##= mathbf(C) – x/5 – x^2/50 – x^3/375 – x^4/2500 – …##

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 ##1/(1-x)##, we write

##sum_(n=0)^N (x-0)^n = 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 ##lnx##, so the constant ##C## is ##color(green)(ln(5))##. So, we get:

##color(blue)(ln(5-x) = ln(5) – x/5 – x^2/50 – x^3/375 – x^4/2500 – …)##

And then finally, for the radius of convergence, it is ##|x| < 5## because ##ln(5-x)## approaches ##-oo## as ##x->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.

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