96 Chapter 1 Fourier Series and Integrals
Proof: Let the pointxbe chosen; it is to remain fixed. To begin with, we
assume thatfiscontinuousatx, so the sum of the series should bef(x).
Another way to say this is thatNlim→∞SN(x)−f(x)=^0 ,
whereSNis the partial sum of the Fourier series off,SN(x)=a 0 +∑N
n= 1ancos(nx)+bnsin(nx). (2)Of course, thea’s andb’s are the Fourier coefficients off,a 0 = 21 π∫π−πf(z)dz,an=^1
π∫π−πf(z)cos(nz)dz,bn=1
2 π∫π−πf(z)sin(nz)dz.(3)
The integrals havezas their variable of integration, but that does not affect
their value.
Part 1.Transformation ofSN(x).
In order to show a relationship betweenSN(x)andf, we replace the co-
efficients in Eq. (2) by the integrals that define them and use elementary
algebra on the results:SN(x)= 21 π∫π−πf(z)dx+∑N
n= 1[
1
π∫π−πf(z)cos(nz)dzcos(nx)+
1
π∫π−πf(z)sin(nz)dzsin(nx)]
(4)
=
1
2 π∫π−πf(z)dx+∑N
n= 1[ 1
π∫π−πf(z)cos(nz)cos(nx)dz+^1
π∫π−πf(z)sin(nz)sin(nx)dz]
(5)
= 21 π∫π−πf(z)dx+∑N
n= 1[
1
π∫π−πf(z)(
cos(nz)cos(nx)+sin(nz)sin(nx))
dz