1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

1.7 Proof of Convergence 97

=

1

π

∫π

−π

f(z)

(

1

2 +

∑N

n= 1

cos(nz)cos(nx)+sin(nz)sin(nx)

)

dz

(7)

=

1

π

∫π

−π

f(z)

(

1

2 +

∑N

n= 1

cos

(

n(z−x)

)

)

dz. (8)

In this very compact formula forSN(x), we now change the variable of in-
tegration fromztoy=z−x:

SN(x)=π^1

∫π+x

−π+x

f(x+y)

(

1

2 +

∑N

n= 1

cos(ny)

)

dy. (9)

Note that both factors in the integrand are periodic with period 2π.The
interval of integration can be any interval of length 2πwith no change in
the result. (See Exercise 5 of Section 1.) Therefore,

SN(x)=π^1

∫π

−π

f(x+y)

(

1

2 +

∑N

n= 1

cos(ny)

)

dy. (10)

Part 2.Expression forSN(x)−f(x).
Since we must show that the differenceSN(x)−f(x)goes to 0, we need to
havef(x)in a form compatible with that forSN(x).Recallthatxis fixed
(although arbitrary), sof(x)is to be thought of as a number. Lemma 1
suggests the appropriate form,

f(x)=f(x)·

1

π

∫π

−π

(

1

2 +

∑N

n= 1

cos(ny)

)

dy

=π^1

∫π

−π

f(x)

(

1

2 +

∑N

n= 1

cos(ny)

)

dy. (11)

Now, using Eq. (10) to representSN(x),wehave

SN(x)−f(x)=

1

π

∫π

−π

(

f(x+y)−f(x)

)

(

1

2 +

∑N

n= 1

cos(ny)

)

dy. (12)

Part 3.The limit.
ThenextstepistouseLemma2toreplacethesuminEq.(12).Theresult
is

SN(x)−f(x)=π^1

∫π

−π

(

f(x+y)−f(x)

)sin

(

(N+^12 )y

)

2sin(^12 y)

dy. (13)