Question

How do you find the maclaurin series expansion of `1/(x+3)`?

Answer 1

`1/(x+3) = 1/3 - x/9 + x^2/27 - x^3/81 + x^4/243 - ...`

Explanation:

We seek the Maclaurin series of

`f(x) = 1/(x+3)`

Which we can write as:

`f(x) = 1/(3(x/3+1))`
`\ \ \ \ \ \ \ = 1/3(1+x/3)^(-1)`

As such we can directly apply the Binomial Theorem

`(1+x)^n = 1 + nx + (n(n-1))/(2!)x^2 + (n(n-1)(n-2))/(3!)x^3 + ...`

So, we get:

`f(x) = 1/3{ 1 + (-1)(x/3) + ((-1)(-2))/(2!)(x/3)^2`
`\ \ \ \ \ \ \ \ \ \ \ + ((-1)(-2)(-3))/(3!)(x/3)^3`
`\ \ \ \ \ \ \ \ \ \ \ + ((-1)(-2)(-3)(-4))/(4!)(x/3)^4 + ... }`

`\ \ \ \ \ \ \ = 1/3{ 1 - (x/3) + (x/3)^2 - (x/3)^3 + (x/3)^4 - ... }`

`\ \ \ \ \ \ \ = 1/3{ 1 - x/3 + x^2/9 - x^3/27 + x^4/81 - ... }`

`\ \ \ \ \ \ \ = 1/3 - x/9 + x^2/27 - x^3/81 + x^4/243 - ...`

Answer 2

`1/(x+3) = 1/3-x/9+x^2/27-x^3/81+... = sum_(n=0)^oo (-1)^n x^n/3^(n+1)`

with radius of convergence `3`.

Explanation:

Another way of finding the Maclaurin series is basically to write it out term by term as follows...

If the Maclaurin series for `1/(x+3)` is `sum_(n=0)^oo a_n x^n`, then `1 = (3+x)sum_(n=0)^oo a_n x^n`

So examining each power of `x` in turn, we can start writing:

`1 = (3+x)(1/3...`

since we require `a_0 = 1/3` in order that the constant term (i.e. the term in `x^0`) is `1` when multiplied by `(3+x)`.

Note that this will result in an unwanted term `x * 1/3`. So to cancel that out, the next term must be `-x/9` to give `-x/3` when multiplied by `3`. So our product looks like this:

`1 = (3+x)(1/3-x/9...`

This will result in an unwanted term `-x^2/9`. To cancel that out, the next term must be `x^2/27`...

`1 = (3+x)(1/3-x/9+x^2/27...`

Repeating, it soon becomes clear that the series we want is a geometric series with common ratio `-x/3`, looking like this:

`1 = (3+x)(1/3-x/9+x^2/27-x^3/81+...)`

So:

`1/(x+3) = 1/3-x/9+x^2/27-x^3/81+... = sum_(n=0)^oo (-1)^n x^n/3^(n+1)`

As a bonus, note that such a geometric series will converge if and only if the absolute value of the common ratio is less than `1`, i.e.

`abs(-x/3) < 1`

and hence if and only if:

`abs(x) < 3`

In other words, the radius of convergence is `3`.

Answer 3

`1/(x+3) = sum_(n=0)^oo (-1)^n x^n/3^(n+1)`

Explanation:

Yet another way to solve it uses the geometric series:

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

Then:

`1/(x+3) = 1/3(1/(1-(-x/3)))`

`1/(x+3) = 1/3sum_(n=0)^oo (-x/3)^n`

`1/(x+3) = sum_(n=0)^oo (-1)^n x^n/3^(n+1)`