1.3. Heat equation 25

Also,

(1.18)

N−1

x=1

sin2

πjx

N

=

N

2

.

♦

The Nth roots of unity, ζ1, . . . , ζN are the N complex numbers ζ such

that ζN = 1. They are given by

ζk = cos

2kπ

N

+ i sin

2kπ

N

, j = 1, . . . , N.

The roots of unity are spread evenly about the unit circle in C; in particular,

ω1 + ω2 + · · · + ωN = 0,

which implies that

N

j=1

cos

2kπ

N

=

N

j=1

sin

2kπ

N

= 0.

The double angle formula for sine gives

N−1

j=1

sin2

jxπ

N

=

N

j=1

sin2

jxπ

N

=

1

2

N−1

j=0

1 − cos

2jxπ

N

=

N

2

−

1

2

N

j=1

cos

2jxπ

N

.

If x is an integer, the last sum is zero. This gives (1.18).

In particular, if we choose the solution with initial condition

f(x) = 1; f(z) = 0,z = x we can see that

P{Sn∧TA = y | S0 = x} =

2

N

N−1

j=1

φj(x) cos

jπ

N

n

φj(y).

It is interesting to see what happens as n → ∞. For large n, the

sum is very small but it is dominated by the j = 1 and j = N − 1