Question

By writing xx as 4+(x−4) and using the Maclaurin expansion for ln(1+t), or otherwise, find the first four non zero terms in the Taylor series expansion for ln⁡(2x) about x=4?

Answer 1

We know that

`d/dx(ln(1 + t)) = 1/(1 + t)`

Now we observe that the series expansion of this is the basic power series, or `sum_(n = 0)^oo (-1)^nt^n`. To get the correct series for `ln(1 + t)`, we must integrate our sum, with respect to `t`.

`sum_(n = 0)^oo (-1)^n t^(n + 1)/((n +1)`

If `1 + t = 2x`, then `t = 2x - 1`. Thus the Maclaurin series for `ln(2x)` will be

`sum_(n = 1)^oo (-1)^(n-1) (2x -1)^(n)/n`

So at `x = 4` the series expansion would be

`sum_(n = 1)^oo (-1)^(n - 1) (2(x - 4) + 4 - 1)^n/n`

`sum_(n = 1)^oo (-1)^(n - 1) (2x - 8 + 4 - 1)^n/n`

`sum_(n = 1)^oo (-1)^(n - 1) (2x- 5)^n/n`