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,

$$\forall r>0, \quad \lim_{x\rightarrow 0} x^r \mbox{order\_size}(x) = 0$$

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 $${\frac{{x^3+\sin(x)^3}}{{x-\sin(x)}}}$$ 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 = ${\frac{{\pi}}{{6}}}$.
  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=+$\infty$ or x=-$\infty$
  1. Find a 5th-order series expansion of arctan(x) in the vicinity of x=+$\infty$.
     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 = $${\frac{{1}}{{x}}}$$ $$\rightarrow_{{x\rightarrow + \infty}}^{}$$ 0.
  2. Find a series 2nd-order expansion of (2x - 1)e^{${\frac{}{}}$}1x-1 in the vicinity of x=+$\infty$.
     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)e^{${\frac{}{}}$}1x-1) in the vicinity of x=-$\infty$.
     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 $\neq$ a.
  For example, find a 2nd-order series expansion of  ${\frac{{(1+x)^\frac{1}{x}}}{{x^3}}}$  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)