1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

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62 Chapter 1 Fourier Series and Integrals


We can now summarize our results. In order for the proposed equality

f(x)=a 0 +

∑∞

n= 1

(

ancos(nx)+bnsin(nx)

)

(2)

to hold, thea’s andb’s must be chosen according to the formulas


a 0 = 21 π

∫π

−π

f(x)dx, (3)

an=

1

π

∫π

−π

f(x)cos(nx)dx, (4)

bn=^1
π

∫π

−π

f(x)sin(nx)dx. (5)

When the coefficients are chosen this way, the right-hand side of Eq. (1) is
called theFourier seriesoff.Thea’s andb’s are calledFourier coefficients.We
have not yet answered question (b) about equality, so we write


f(x)∼a 0 +

∑∞

n= 1

(

ancos(nx)+bnsin(nx)

)

to indicate that the Fourier seriescorrespondstof(x). See the CD for an ani-
mated example.


Example.
Suppose thatf(x)is periodic with period 2πand is given by the formula
f(x)=xin the interval−π<x<π(see Fig. 2). According to our formulas,


a 0 =^1
2 π

∫π

−π

f(x)dx=^1
2 π

∫π

−π

xdx= 0 ,

an=π^1

∫π

−π

f(x)cos(nx)dx=π^1

∫π

−π

xcos(nx)dx

=

1

π

[cos(nx)
n^2 +

xsin(nx)
n

]∣∣

∣∣

π

−π

= 0 ,

bn=π^1

∫π

−π

f(x)sin(nx)dx=π^1

∫π

−π

xsin(nx)dx

=

1

π

[sin(nx)
n^2 −

xcos(nx)
n

]∣∣

∣∣

π

−π
=π^1 (−^2 π)ncosnπ=^2 n(− 1 )n+^1.
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