1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

Chapter 1 449

Section 1.6

1. π^1

∫π

−π

(

ln

∣∣

∣∣2cos

(

x

2

)∣∣

∣∣

) 2

dx=

∑∞

n= 1

1

n^2 =

π^2

6.

3. a. Coefficients tend to zero.

b. Coefficients tend to zero, although

∫ 1

− 1

|x|−^1 dxis infinite.

5. The integral must be infinite, because

∑∞

n= 1

a^2 n+b^2 n=∞.

Section 1.7

1. Theequalitytobeprovedis

2sin

( 1

2 y

)( 1

2 +

∑N

n= 1

cos(ny)

)

=sin

((

N+

1

2

)

y

)

.

The left-hand side is transformed as follows:

2sin

( 1

2 y

)( 1

2 +

∑N

n= 1

cos(ny)

)

=sin

( 1

2

y

)

+

∑N

n= 1

2sin

( 1

2

y

)

cos(ny)

=sin

( 1

2 y

)

+

∑N

n= 1

(

sin

((

n+

1

2

)

y

)

−sin

((

n−

1

2

)

y

))

=sin

(

1

2 y

)

+

∑N

n= 1

sin

((

n+^12

)

y

)

−

N∑− 1

n= 0

sin

((

n+^12

)

y

)

=sin

((

N+

1

2

)

y

)

because all other terms cancel.

3.φ( 0 +)=1,φ( 0 −)=−1. See Fig. 1.

5. a.f′(x)=^34 x−^1 /^4 for 0<x<π(andf′is an odd function). Thus,fhas a
   vertical tangent atx=0, although it is continuous there.

b.φ(y)= |y|

3 / 4

2sin(^12 y)

cos

(

1

2 y

)

, −π<y<π