This document discusses higher order derivatives that are useful for N-body simulations. It presents formulas for calculating higher order derivatives of power functions like y=xn, and applies this to derivatives of gravitational force f=mr-3. Specifically: 1) It derives recursive formulas for calculating higher order derivatives of power functions y=xn in terms of previous derivatives. 2) It applies these formulas to calculate derivatives of the gravitational force f=mr-3 in terms of derivatives of r and q=r-3/2. 3) It also describes an alternative approach by Le Guyader (1993) for calculating derivatives of r and q in terms of dot products of r with itself.

1. Higher order derivatives for N-body
simulations
Keigo Nitadori
July 3, 2014
1 Derivatives of power functions
Consider
y = xn
, (1)
with time dependent x and y, and constant n. From
lny =n lnx, (2)
˙y
y
=n
˙x
x
, (3)
we have
0 =n ˙xy − x ˙y,
0 =n ¨xy + (n − 1) ˙x ˙y − x ¨y,
0 =n
...
xy + (2n − 1) ¨x ˙y + (n − 2) ˙x ¨y − x
...
y ,
0 =n
....
x y + (3n − 1)
...
x ˙y + (3n − 3) ¨x ¨y + (n − 3) ˙x
...
y − x
....
y .
(4)
Here, ˙[ ] = d
dt [ ] and we also note [ ](n) = dn
dtn [ ]. From these, higher order derivatives of y are
recursively available in
˙y =[n ˙xy]/x,
¨y =[n ¨xy + (n − 1) ˙x ˙y]/x
...
y =[n
...
xy + (2n − 1) ¨x ˙y + (n − 2) ˙x ¨y]/x,
....
y =[n
....
x y + (3n − 1)
...
x ˙y + (3n − 3) ¨x ¨y + (n − 3) ˙x
...
y ]/x,
(5)
and so on. A general form for k ≥ 1 is
y(k)
=
1
x
k−1
i=0
Ck,ix(k−i)
y(i)
, (6)
1

2. with coefficients
Ck,0 =n
Ck,k = − 1
Ck,i =Ck−1,i−1 + Ck−1,i
=
k − 1
i
n −
k − 1
i − 1
(1 ≤ i ≤ k − 1).
(7)
2 Derivatives of gravitational force
Now we focus on the time derivatives of the gravitational force,
f = m
r
r 3
. (8)
Let
s = (r · r), and q = s−3/2
, (9)
then
f =mqr,
˙f =mq ˙r +
˙q
q
r ,
¨f =mq ¨r + 2
˙q
q
˙r +
¨q
q
r ,
...
f =mq
...
r + 3
˙q
q
¨r + 3
¨q
q
˙r +
...
q
q
r ,
(10)
and
f (k)
=m
k
i=0
k
i
q(i)
r(k−i)
=mq
k
i=0
k
i
q(i)
q
r(k−i)
.
(11)
For the derivatives of q, by using n = −3
2, we have
q =s−3/2
,
˙q
q
= −
3
s
˙s
2
,
¨q
q
= −
3
s
¨s
2
+
5
3
˙s
2
˙q
q
...
q
q
= −
3
s
...
s
2
+
8
3
¨s
2
˙q
q
+
7
3
˙s
2
¨q
q
.
(12)
The coefficients multiplied by 6 are:
2

3. @
@
@k
i
0 1 2 3 4 5
1 3 (2)
2 3 5 (2)
3 3 8 7 (2)
4 3 11 15 9 (2)
5 3 14 26 24 11 (2)
Derivatives of s have simple form in,
s =(r · r),
˙s =2(r · ˙r),
¨s =2(r · ¨r) + 2( ˙r · ˙r),
...
s =2(r ·
...
r ) + 6( ˙r · ¨r),
s(4)
=2(r · r(4)
) + 8( ˙r ·
...
r ) + 6( ¨r · ¨r),
s(5)
=2(r · r(5)
) + 10( ˙r · r(4)
) + 20( ¨r ·
...
r ),
(13)
and a general form is
s(k)
=
k
i=0
k
i
(r(i)
· r(k−i)
)
=



2
(k−1)/2
i=0
k
i
(r(i)
· r(k−i)
) (k is odd)


2
(k−2)/2
i=0
k
i
(r(i)
· r(k−i)
)


+
k
k/2
(r(k/2)
· r(k/2)
) (k is even)
.
(14)
3 Another approach
Le Guyader (1993) took a slightly different approach. For the derivatives of r = r ,
r2
= (r · r), (15)
r ˙r = (r · ˙r), (16)
and after (k − 1) times differentiations,
k−1
i=0
k − 1
i
r(i)
r(k−i)
=
k−1
i=0
k − 1
i
(r(i)
· r(k−i)
). (17)
Finally, we have
r(k)
=
1
r


(r · r(k)
) +
k−1
i=1
k − 1
i
(r(i)
· r(k−i)
) − r(i)
r(k−i)


, (18)
3

4. for k ≥ 2.
For the derivatives of q = r−3, from
r ˙q = −3˙rq, (19)
after differentiating (k − 1) times,
k−1
i=0
k − 1
i
r(i)
q(k−i)
= −3
k−1
i=0
k − 1
i
r(k−i)
q(i)
. (20)
Thus,
q(k)
= −
1
r


3r(k)
q +
k−1
i=1
k − 1
i
3r(k−i)
q(i)
+ r(i)
q(k−i)


, (21)
for k ≥ 2.
References
Le Guyader, C. 1993, A&A, 272, 687. http://adsabs.harvard.edu/abs/1993A&A...
272..687L
4