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13.5.2  Series expansion

The taylor or series command finds Taylor expansions.

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).

Example

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

or:

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

or (be careful with the order of the arguments):

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

or:

series(sin(x),x,2,1)
     
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 consider

  
x3+sin(x)3
x−sin(x)
.
taylor(x^3+sin(x)^3/(x-sin(x)),x=0,5)
     
6−
27
10
 x2+x3+
711
1400
 x4+x6 order_size⎛
⎝
x⎞
⎠
          

which is only a 2nd degree expansion. Indeed the numerator and denominator valuation is 3, hence you lose 3 orders. To get order 4, you should use n=7.

taylor(x^3+sin(x)^3/(x-sin(x)),x=0,7)
     
6−
27
10
 x2+x3+
711
1400
 x4−
737
14000
 x6+x8 order_size⎛
⎝
x⎞
⎠
          

a fourth degree expansion.

Examples

Find a 4th-order expansion of cos(2x)2 in the vicinity of x=π/6.

taylor(cos(2*x)^2,x=pi/6, 4)
     
1
4
−√
3
 ⎛
⎜
⎜
⎝
x−
π 
6
⎞
⎟
⎟
⎠
+2 ⎛
⎜
⎜
⎝
x−
π 
6
⎞
⎟
⎟
⎠
2



 
+
8
3
 √
3
 ⎛
⎜
⎜
⎝
x−
π 
6
⎞
⎟
⎟
⎠
3



 
−
8
3
 ⎛
⎜
⎜
⎝
x−
π 
6
⎞
⎟
⎟
⎠
4



 
+⎛
⎜
⎜
⎝
x−
π 
6
⎞
⎟
⎟
⎠
5



 
 order_size⎛
⎜
⎜
⎝
x−
π 
6
⎞
⎟
⎟
⎠
          

Find a 5th-order series expansion of arctan(x) in the vicinity of x=+∞.

series(atan(x),x=+infinity,5)
     
π 
2
−
1
x
+
⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
3



 
3
−
⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
5



 
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 → 0 as x→+∞.

Find a 2nd-order expansion of (2x−1)e1/x−1 in the vicinity of x=+∞.

series((2*x-1)*exp(1/(x-1)),x=+infinity,3)

Output (only a 1st-order series expansion):

     
2 ⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
−1



 
+1+
2
x
+
17
6
 ⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
2



 
+⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
3



 
 order_size⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
          

Note that this is only a 1st-order expansion. To get a 2nd-order series expansion in 1/x:

series((2*x-1)*exp(1/(x-1)),x=+infinity,4)
     
2 ⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
−1



 
+1+
2
x
+
17
6
 ⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
2



 
+
47
12
 ⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
3



 
+⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
4



 
 order_size⎛
⎜
⎜
⎝
1
x
⎞
⎟
⎟
⎠
          

Find a 2nd-order series expansion of (2x−1)e1/x−1 in the vicinity of x=-∞.

series((2*x-1)*exp(1/(x-1)),x=-infinity,4)
     
−2 ⎛
⎜
⎜
⎝
−
1
x
⎞
⎟
⎟
⎠
−1



 
+1+
2
x
+
17
6
 ⎛
⎜
⎜
⎝
−
1
x
⎞
⎟
⎟
⎠
2



 
−
47
12
 ⎛
⎜
⎜
⎝
−
1
x
⎞
⎟
⎟
⎠
3



 
+⎛
⎜
⎜
⎝
−
1
x
⎞
⎟
⎟
⎠
4



 
 order_size⎛
⎜
⎜
⎝
−
1
x
⎞
⎟
⎟
⎠
          

Find a 2nd-order series expansion of (1+x)1/x/x3 in the vicinity of x=0+.

series((1+x)^(1/x)/x^3,x=0,2,1)

(Note that this is a one-sided series expansion, since dir=1.)

     
e x−3−
e
2
 x−2+x−1 order_size⎛
⎝
x⎞
⎠
          

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