Question

What is the Maclaurin series for `ln(5-x)`?

Answer 1

`ln(5-x) = 4 - sum_{n=1}^{\infty}((-4)^n/n(\sum_{r=1}^{\infty}{::}_nC_r(-4)^{-r}))x^n)`.

Explanation:

`ln(5-x)=ln(1-(x-4))`

Then, `ln(1-x) = -(sum_{n=1}^{\infty}x^n/n)`,
`ln(1-(x-4)) = -(sum_{n=1}^{\infty}(x-4)^n/n)`,
`ln(5-x) = -sum_{n=1}^{\infty}sum_{r=1}^{n}(({::}_nC_r (-4)^{n-r})/n x^r )`,
`ln(5-x) = -(-4 + sum_{n=1}^{\infty}((-4)^n/n(\sum_{r=1}^{\infty}{::}_nC_r(-4)^{-r}))x^n)`,
`ln(5-x) = 4 - sum_{n=1}^{\infty}((-4)^n/n(\sum_{r=1}^{\infty}{::}_nC_r(-4)^{-r}))x^n)`.

Expansion of `ln(1-x)` holds for `abs(x)<1`. Then our expansion for `ln(5-x)` holds for `abs(x-4)<1` or `3<x<5`.

Answer 2

`ln(5-x) = ln(5) - x/5-x^2/50-x^3/375-x^4/2500-x^5/15625- ...`

And in sigma notation, this is

`ln(5-x) = ln5 -sum_(n=1)^oo (x^n)/(n5^n)`

Explanation:

The power series for `ln(1-x)` is often quoted in examination formula books etc, and is as follows:

`ln(1-x) = -x-x^2/2-x^3/3-x^4/4-x^5/5 + ...`
`" " = -sum_(n=1)^oo x^n/n`

So for the series we require, we have:

`ln(5-x) = ln(5(1-x/5))`
`" " = ln(5) + ln (1-x/5)`

So then if we substitute `x` for `x/5` in the initial expansion we get:

`ln(5-x) = ln(5) + {-(x/5)-(x/5)^2/2-(x/5)^3/3-(x/5)^4/4-(x/5)^5/5} + ...`

`" " = ln(5) - {x/5+(x^2/25)/2+(x^3/125)/3+(x4/625)/4+(x^5/3125)/5} + ...`

`" " = ln(5) - x/5-x^2/50-x^3/375-x^4/2500-x^5/15625- ...`

And in sigma notation, this is

`ln(5-x) = ln5 -sum_(n=1)^oo (x^n)/(n5^n)`