1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

92 Chapter 1 Fourier Series and Integrals

already have an expression for the middle integral. The last one can be found
by replacingfwithgin Eq. (1):

∫a

−a

g^2 (x)dx=a

[

2 A^20 +

∑N

1

A^2 n+B^2 n

]

. (4)

Now we have a formula forENin terms of the variablesA 0 ,An,Bn:

EN=

∫a

−a

f^2 (x)dx− 2 a

[

2 A 0 a 0 +

∑N

1

Anan+Bnbn

]

+a

[

2 A^20 +

∑N

1

A^2 n+B^2 n

]

. (5)

The errorENtakes its minimum value when all of the partial derivatives
with respect to the variables are zero. We must then solve the equations

∂EN

∂A 0

=− 4 aa 0 + 4 aA 0 = 0 ,

∂EN
∂An=−^2 aan+^2 aAn=^0 ,
∂EN
∂Bn=−^2 abn+^2 aBn=^0.
These equations require thatA 0 =a 0 ,An=an,Bn=bn.Thusgshould be
chosen to be thetruncatedFourier series off,

g(x)=a 0 +

∑N

n= 1

ancos

(nπx

a

)

+bnsin

(nπx

a

)

in order to minimizeEN.
Now that we know which choice ofA’s andB’s minimizesEN,wecancom-
pute that minimum value. After some algebra, we see that

min(EN)=

∫a

−a

f^2 (x)dx−a

[

2 a^20 +

∑N

1

a^2 n+b^2 n

]

. (6)

Eventhisminimumerrormustbegreaterthanorequaltozero,andthuswe
have theBessel inequality

1

a

∫a

−a

f^2 (x)dx≥ 2 a^20 +

∑N

1

a^2 n+b^2 n. (7)