Trap rule error in periodic functions
Any function that has all derivatives at its beginning, t=0, equal to all derivatives at its
end, t=T, can be expanded as a periodic function.
A periodic function can always be expanded as

f (t )   c j e
2 i
jt
T
(1)
j 0
Note that cj is complex and that

f (t  T )   c j e
2 i j
t T
T
j 0

  c j e 2 i j e 2 i j t / T   f (t )
(2)
j 0
Since the first exponential is always 1. The integral from 0 to T of f is
T
T



e2 i j  1 
2 i j  t / T 
(3)
f
(
t
)
dt

c
e
dx

c
T




  c0T
j
j  0, j
0
2

i
j
j 0
j

0


0
The value of the integral is given by the c0 term alone. All other terms in the expansion integrate
to zero.
1d 0 

Let h  T / N ; tk  h  k 
 so that the midpoint trap rule approximation to this integral is
2 

k 1d 0 / 2
T

k N
N
k  N 2 i j k
T
T  k  N 2 i j N
T 
2 i j  tk / T 
 ij
f
(
t
)
dt

c
h
e

c
1

c
e

c
T

c
e
e N




j 
0
j
0
j
0
N k 1
N j 1 k 1
N j 1
j 0
k 1
k 1
(4)
Note the interchange of summation orders. Let
T

f (t )dt  c0T 
0
T
N

k N
j 1
k 1
 c j e ij  z j k
zj  e
2 i
j
N
, so that
(5)
The last term can be written as a sum of zk and we can use the familiar relation for z1
N

N 1

N 1
zj
1
k
k
k
  z j   z j   z j   z jk 
1  z j k 0
1 zj
k 0
k N
k 0
N 1
So that
z
k 0
k
j

Note that z j N  e
z jN 1
1 zj
2 i
jN
N
(7)
 e2 ij  1 so that the numerator in (7) is always zero. The denominator is
also zero for j=mN for which zmN  e
T

0
(6)
2 i
mN
N
 1 . In this case the last sum in (5) is N so that

f (t )dt  c0T  T  cmN e  imN
(8)
m 1
The terms that enter are not uniform. If for some accidental reason c10 = 0, 10 point
integration will seem exact up to c20. Changing N from 10 to 20 will give a false accuracy
reading. The correct change is from 10 to 11 or 13, Do not simply double the points.