### 2.33.3 Series expansion : `series`

`series` takes from one to four arguments :

- an expression dependending of a variable (by default `x`),
- an equality variable=value (e.g. x=a) where to compute the series expansion, by default `x=0`,
- an integer n, the order of the series expansion, by default `5`
- a direction `-1, 1` (for unidirectional series expansion) or `0` (for bidirectional series expansion) (by default `0`).

Note that the syntax `...,x,``a``,``n``,...` (instead of `...,x=``a``,``n``,...`) is also accepted.
`series` returns a polynomial in `x-a`, plus a remainder of the form:

`(x-a)```^```n*order_size(x-a)`

where `order_size` is a function such that,

∀ r>0, |
lim
x→ 0
| xr order_size(x) = 0

The order returned by `series` may be smaller than n if cancellations between numerator and denominator occur.

Examples :

- series expansion in the vicinity of `x=0`
  Find an series expansion of x3+sin(x)3/x−sin(x) in the vicinity of `x=0`.
  Input :
  `series(x```^```3+sin(x)```^```3/(x-sin(x)))`
  Output is only a 2nd-order series expansion :
  `6+-27/10*x```^2```+x```^```3*order_size(x)`
  We have lost 3 orders because the valuation of the numerator and denominator is 3. To get a 4-th order expansion, we must therefore take n=7, input:
  `series(x```^```3+sin(x)```^```3/(x-sin(x)),x=0,7)`
  Or :
  `series(x```^```3+sin(x)```^```3/(x-sin(x)),x,0,7)`
  Output is a 4th-order series expansion :
  `6+-27/10*x```^```2+x```^```3+711/1400*x```^```4+ x```^```5*order_size(x)`
- series expansion in the vicinity of `x=a`
  Find a series 4th-order expansion of cos(2x)2 in the vicinity of x=π/6.
  Input:
  `series(cos(2*x)```^```2,x=pi/6, 4)`
  Output :
  `1/4+(-(4*sqrt(3)))/4*(x-pi/6)+(4*3-4)/4*(x-pi/6)```^```2+ 32*sqrt(3)/3/4*(x-pi/6)```^```3+(-16*3+16)/3/4*(x-pi/6)```^```4+ (x-pi/6)```^```5*order_size(x-pi/6)`
- series expansion in the vicinity of `x=+``∞` or `x=-``∞`
  1. Find a 5th-order series expansion of arctan(x) in the vicinity of `x=+``∞`.
     Input :
     `series(atan(x),x=+infinity,5)`
     Output :
     `pi/2-1/x+1/3*(1/x)```^```3+1/-5*(1/x)```^```5+ (1/x)```^```6*order_size(1/x)`
     Note that the expansion variable and the argument of the `order_size` function is h=1/x →x→ + ∞ 0 .
  2. Find a series 2nd-order expansion of (2x−1)e1/x−1 in the vicinity of `x=+``∞`.
     Input :
     `series((2*x-1)*exp(1/(x-1)),x=+infinity,3)`
     Output is only a 1st-order series expansion :
     `2*x+1+2/x+(1/x)```^```2*order_size(1/x)`
     To get a 2nd-order series expansion in 1/x, input:
     `series((2*x-1)*exp(1/(x-1)),x=+infinity,4)`
     Output :
     `2*x+1+2/x+17/6*(1/x)```^```2+(1/x)```^```3*order_size(1/x)`
  3. Find a 2nd-order series expansion of (2x−1)e1/x−1 in the vicinity of `x=-``∞`.
     Input :
     `series((2*x-1)*exp(1/(x-1)),x=-infinity,4)`
     Output :
     `-2*(-x)+1-2*(-1/x)+17/6*(-1/x)```^```2+
     (-1/x)```^```3*order_size(-1/x)`
- unidirectional series expansion.
  The fourth parameter indicates the direction :
  - `1` to do an series expansion in the vicinity of x=a with x>a,
  - `-1` to do an series expansion in the vicinity of x=a with x<a,
  - `0` to do an series expansion in the vicinity of x=a with x ≠ a.
  For example, find a 2nd-order series expansion of (1+x)1/x/x3 in the vicinity of x=0+. Input :
  `series((1+x)```^```(1/x)/x```^```3,x=0,2,1)`
  Output :
  `exp(1)/x```^```3+(-(exp(1)))/2/x```^```2+1/x*order_size(x)`