Question

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

Answer 1

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

`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.

We had done this:

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.