I will suppose `c0`

is `1`

, else we divide with its square root, which should exist in the ground field for the coefficients. (Feel free to change the following code, so that `c0`

is also a general variable.)

The following does the job for me:

```
c = [1,] + list( var( 'c1,c2,c3,c4,c5,c6' ) )
print "c is", c
var( 'x' );
T = taylor( sqrt( 1 + sum( [ c[k]*x^k for k in [1..6] ] ) ), x, 0, 6 )
# print T # for a long story
print "T has the following coefficients:"
for coeff, deg in T.coefficients( x ):
print "Degree %s: %s" % ( deg, coeff )
```

Results:

```
c is [1, c1, c2, c3, c4, c5, c6]
x
T has the following coefficients:
Degree 0: 1
Degree 1: 1/2*c1
Degree 2: -1/8*c1^2 + 1/2*c2
Degree 3: 1/16*c1^3 - 1/4*c1*c2 + 1/2*c3
Degree 4: -5/128*c1^4 + 3/16*c1^2*c2 - 1/8*c2^2 - 1/4*c1*c3 + 1/2*c4
Degree 5: 7/256*c1^5 - 5/32*c1^3*c2 + 3/16*c1*c2^2 + 1/16*(3*c1^2 - 4*c2)*c3 - 1/4*c1*c4 + 1/2*c5
Degree 6: -21/1024*c1^6 + 35/256*c1^4*c2 - 15/64*c1^2*c2^2 + 1/16*c2^3 - 1/32*(5*c1^3 - 12*c1*c2)*c3 - 1/8*c3^2 + 1/16*(3*c1^2 - 4*c2)*c4 - 1/4*c1*c5 + 1/2*c6
```