### 5.34.2 Taylor expansion : taylor

taylor takes from one to four arguments :

- an expression depending of a variable (by default x),
- an equality variable=value (e.g. x=a) where to compute the Taylor 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,n,a,... (instead of ...,x=a,n,...) is also accepted.
taylor 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

For regular series expansion, order_size is a bounded function, but for non regular series expansion, it might tend slowly to infinity, for example like a power of ln(x).
Input :

taylor(sin(x),x=1,2)

Or (be careful with the order of the arguments !) :

taylor(sin(x),x,2,1)

Output :

sin(1)+cos(1)*(x-1)+(-(1/2*sin(1)))*(x-1)`^`2+ (x-1)`^`3*order_size(x-1)

Remark
The order returned by taylor may be smaller than n if cancellations between numerator and denominator occur, for example

taylor( |
x3+sin(x)3
x−sin(x)
| )

Input :

taylor(x`^`3+sin(x)`^`3/(x-sin(x)))

The output is only a 2nd-order series expansion :

6+-27/10*x`^2`+x`^`3*order_size(x)

Indeed the numerator and denominator valuation is 3, hence we lose 3 orders. To get order 4, we should use n=7.
Input :

taylor(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)