CHAPTER 102 FOURIER SERIES FOR A NON-PERIODIC FUNCTION
CHAPTER 102 FOURIER SERIES FOR A NON-PERIODIC FUNCTION OVER RANGE 2π EXERCISE 363 Page 1071 1. Show that the Fourier series for the function f(x) = x over the range x = 0 to x = 2π is given by: 1 1 1 sin 2 x + sin 3 x + sin 4 x + ...) 2 3 4
f(x) = π – 2(sin x +
The function f(x) = x is not periodic. The function is shown below in the range 0 to 2π and is then constructed outside of that range so that it is periodic of period 2π (see broken lines) with the resulting saw-tooth waveform
∞
For a Fourier series:
∑( a
f(x) = a 0 +
n
cos nx + b n sin nx)
n =1
a0 =
an =
1 2π 1
π
∫ ∫
2π 0
2π 0
f ( x) d x =
1 2π
∫
f ( x) cos nx d x =
2π 0
2π
xd x =
1
π
∫
2π 0
1 x2 1 4π 2 = − 0 = π 2π 2 0 2π 2
x cos nx d x 2π
1 x sin nx sin nx = −∫ d x n π n 0
by parts (see Chapter 68)
2π
1 x sin nx cos nx = + π n n 2 0 =
1 cos 2nπ 0+ π n2
cos 0 − 0 + 2 = 0 n
1516
© 2014, John Bird
bn =
1
π∫
2π 0
f ( x) sin nx d x =
1
π∫
2π 0
x sin nx d x 2π
1 − x cos nx − cos nx = − ∫ d x n n 0 π
by parts
2π
1 − x cos nx sin nx = + n n 2 0 π =
1 −2π cos 2nπ sin 2nπ sin 0 + − 0 + 2 2 π n n n
=
1 −2π cos 2nπ −2 = cos 2nπ π n n
2 2 2 2 2 When n is odd or even, b n = − . Thus b 1 = 2, b 2 = − , b 3 = − , b 4 = – , b 5 = − , and so n 4 2 3 5
on Thus i.e.
f(x) = x = π – 2 sin x – x = π – 2(sin x +
2 2 2 2 2 sin 2x – sin 3x – sin 4x – sin 5x – sin 6x + … 3 4 5 6 2
1 1 1 1 1 sin 2x + sin 3x + sin 4x + sin 5x + sin 6 x + ... ) 3 4 6 2 5
for values of f(x) between 0 and 2π. For values of f(x) outside the range 0 to 2π the sum of the series is not equal to f(x)
2. Determine the Fourier series for the function defined by: 1 − t , when − π 〈 t 〈 0 f(t) = 1 + t , when 0 〈 t 〈 π Draw a graph of the function within and outside of the given range.
The periodic function is shown in the diagram below
1517
© 2014, John Bird
1 a0 = 2π
π
1 dt f (t )= 2π
∫π −
=
a= n
1
{∫
0 −π
(1 − t ) d t + ∫
π 0
}
t 2 0 t 2 π t − + t + 2 −π 2 0 1 π2 π 2 2π 2π 2 π + + = 1+ 2 π + − ( 0 ) = = 2 2 2π 4π 2 2π
1 (1 + t ) d= t 2π
1 π 2 ( 0 ) − −π − + π 2π 2
π
f (t ) cos nt d= t π∫π −
π {∫ 1
π
0 −π
(1 − t ) cos nt d t + ∫ (1 + t ) cos nt d t 0
}
=
π {∫
=
1 sin nt t sin nt cos nt sin nt t sin nt cos nt by integration by parts − − + + + π n n n 2 −π n n n 2 0
=
1 cos 0 cos(−nπ ) cos nπ cos 0 + 0 + 0 + −0+ 0+ 0 − 0 − 2 − 0 − 0 − π n n2 n2 n 2
=
1 1 cos(−nπ ) cos nπ 1 1 + = − ( 2 cos nπ − 2 ) − 2 + π n n2 n2 n2 π n2
1
0 −π
π
( cos nt − t cos nt ) d t + ∫ 0 ( cos nt + t cos nt ) d t
} π
0
= When n is even,
2 4 ( −1 − 1) =− 2 π (1) π
a1 =
When n = 3,
a3 =
When n = 5,
a5 =
1
π
π
∫π −
2 ( cos nπ − 1) π n2
an = 0
When n = 1,
b= n
since cos(–nπ) = cos nπ
2 4 ( −1 − 1) =− 2 π (3) π (3) 2 2 4 ( −1 − 1) =− 2 π (5) π (5) 2
f (t ) sin nt d= t
π {∫ 1
0 −π
and so on π
(1 − t ) sin nt d t + ∫ (1 + t ) sin nt d t 0
}
=
π {∫
=
1 cos nt t cos nt sin nt cos nt t cos nt sin nt − + − + − − + n n n 2 −π n n n 2 0 π
1
π
( sin nt − t sin nt ) d t + ∫0 ( sin nt + t sin nt ) d t −π 0
} π
0
by integration by parts
cos 0 cos(−nπ ) π cos(−nπ ) + 0 − 0 − − − − 0 − n n n 1 = π cos nπ π cos nπ cos 0 + − − + 0 − − + 0 + 0 n n n =
1 1 cos(−nπ ) π cos(−nπ ) cos nπ π cos nπ 1 + − − + =0 − + π n n n n n n 1518
since cos nπ = cos(–nπ) © 2014, John Bird
∞
Substituting into f(t) = a0 + ∑ ( an cos nt + bn sin nt ) n =1
gives:
f(t) =
i.e.
f(t) =
π 2
π 2
+1 −
+1−
4
cos t −
π
4 4 cos 3t − cos 5t − ... + 0 2 π (3) π (5) 2
4 cos 3t cos 5t cos t + 2 + 2 + ... π 3 5
3. Find the Fourier series for the function f(x) = x + π within the range –π < x < π
The function is shown sketched below
∞
f(x) = a 0 +
∑( a
n
cos nx + b n sin nx)
n =1
1 a0 = 2π
an =
π
∫π −
1 f ( x) d x = 2π
π
1
f ( x) cos nx d x π∫π −
π
1 x2 1 π 2 π2 2 − = x π π + = + − π 2 = π ( x + π ) d x ∫ −π 2π 2 −π 2π 2 2 π
=
1
π
( x + π ) cos nx d x= π∫π −
π
1
( x cos nx + π cos nx) d x π∫π −
π
1 x sin nx sin nx π = by parts (see Chapter 68) d x + sin nx −∫ n n π n −π π
=
=
1 π sin nπ cos nπ π −π sin(−nπ ) cos(−nπ ) π + + sin nπ − + + sin(−nπ ) π n n2 n n n2 n =
bn =
1
π
1 x sin nx cos nx π + + sin nx 2 π n n n −π
1 cos nπ π n 2 1
cos(−nπ ) 0 − = n2
π
f ( x) sin nx d x = ∫ ( x + π ) sin nx d x= π π π∫π −
−
1519
1
π
( x sin nx + π sin nx) d x π∫π −
© 2014, John Bird
π
π cos nπ 1 − x cos nx − cos nx = by parts − ∫ d x − π n n n −π π
1 − x cos nx sin nx π cos nπ = + − π n n2 n −π =
1 −π cos nπ sin nπ π cos nπ + − π n n2 n =
Hence, b 1 =
1 −2π cos nπ π n
2 − cos nπ − ( 0 ) = n
2 2 2 2 2 , b 2 = − , b 3 = , b 4 = – , b 5 = , and so on 1 4 5 2 3
Thus f(x) = x + π = π + 0 + x + π = π + 2(sin x –
i.e.
π cos(−nπ ) sin(−nπ ) π cos(−nπ ) + − − n n2 n
2 2 2 2 2 2 sin x – sin 2x + sin 3x – sin 4x + sin 5x – sin 6x + … 2 3 4 5 6 1 1 1 1 1 1 sin 2x + sin 3x – sin 4x + sin 5x – sin 6 x + ... ) 3 5 6 4 2
4. Determine the Fourier series up to and including the third harmonic for the function defined by: x, when 0 〈 x 〈 π f(x) = 2π − x, when π 〈 x 〈 2π Sketch a graph of the function within and outside of the given range, assuming the period is 2π.
The periodic function is shown in the diagram below
a0 =
1 2π
π
∫π −
{
}
2π x 2 π x 2 2 x π + − ∫0 π 2 π 2 0 1 π 2 4π 2 2 π 2 1 π 2 = (π= )} { − ( 0 ) + 4π 2 − − 2π − = 2π 2 2 2 2π 2
f ( x= )d x
1 2π
π
xd x + ∫
2π
(2π − x)= dx
1520
1 2π
© 2014, John Bird
an =
1
π∫
2π 0
f ( x) cos nx d x =
π {∫ 1
π 0
x cos nx d x + ∫
2π
π
}
(2π − x) cos nx d x
π
2π
=
1 x sin nx cos nx 2π sin nx x sin nx cos nx + + − − π n n 2 0 n n n 2 π
=
1 cos nπ 0 + π n2
=
1 2 cos nπ − 1 − cos 2π n = + cos nπ } { ( cos nπ − 1) π n2 π n2
When n is even,
cos 0 cos 2π n cos nπ − 0 + 2 + 0 − 0 − −0−0− 2 n n n2
2 4 ( −1 − 1) =− 2 π (1) π
a1 =
When n = 3,
a3 =
When n = 5,
a5 =
1
π∫
2π 0
2 4 ( −1 − 1) =− 2 π (3) π (3) 2 2 4 ( −1 − 1) =− 2 π (5) π (5) 2
f ( x) sin = nx d x
π {∫ 1
0 −π
and so on
}
π
x sin nx d x + ∫ (2π − x) sin nx d x 0
π
=
an = 0
When n = 1,
= bn
by integration by parts
2π
1 x cos nx sin nx 2π cos nx x cos nx sin nx − + + − + − 2 π n n 0 n n n 2 π
by integration by parts
2π cos 2nπ 2π cos 2nπ + + 0 − n n 1 π cos nπ = − + 0 − ( 0 + 0 ) + π n 2π cos nπ π cos nπ −− + − 0 n n
=
1 0 {−π cos nπ + 2π cos nπ − π cos nπ } = nπ ∞
Substituting into f(x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
f(x) =
i.e.
f(x) =
π 2
π 2
−
−
4
π
cos x −
4 4 cos 3 x − cos 5 x − ... + 0 π (3) 2 π (5) 2
4 cos 3 x cos 5 x + + ... cos x + 2 2 π 3 5
1521
© 2014, John Bird
5. Expand the function f(θ) = θ 2 in a Fourier series in the range –π < θ < π Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below
1 = a0 2π
π
∫π −
1 f ( x) = dx 2π
{
π
π {∫
1 π 1 = ∫ f (θ ) cos nθ d θ
an
π
}
1 θ ∫ −π θ d= 2π 2
−π
π −π
θ 3 π 1 2π 3 π 2 3 {π 3 − −π= } = = 6π 3 3 −π 6π
θ 2 cos nθ d θ
}
π
1 θ 2 sin nθ 2θ cos nθ 2sin nθ = + − π n n2 n3 −π =
1 2π cos nπ 2π cos(−nπ ) 4 4π cos nπ − 0 − 0 − −= = 0 cos nπ 0 + π π n2 n2 n2 n2 since cos nπ = cos(–nπ)
4 4 4 4 a1 = ( −1) = − = , a2 = (1) 2 , 2 2 2 (2) 2 (1) 1
Hence,
π {∫
1 π 1 = ∫ f (θ ) sin nθ d θ
bn
by integration by parts
π
−π
π −π
θ 2 sin nθ d θ
4 4 a3 = ( −1) = − , 2 (3) 32
a4 =
4 , and so on 42
}
π
1 θ 2 cos nθ 2θ sin nθ 2 cos nθ = − + + π n n2 n3 −π =
1 π 2 cos nπ 2 cos nπ +0+ − π n n3
by integration by parts
2 cos(−nπ ) π 2 cos(−nπ ) +0+ −− = 0 n n3
∞
Substituting into f(θ) = a0 + ∑ ( an cos nθ + bn sin nθ ) n =1
gives:
f(θ) =
i.e.
f(θ) =
π2 3
−
4 4 4 4 cos θ + cos 2θ − cos 3θ + cos 4θ − ... + 0 12 22 32 42
π2
1 1 − 4 cos θ − cos 2θ + cos 3θ − ... 2 2 3 2 3 1522
© 2014, John Bird
6. For the Fourier series obtained in Problem 5, let x = π and deduce the series for
∞
1
∑n n =1
2
When θ = π in Problem 5 above, f(θ) = π 2 Thus,
π2 =
i.e.
π2 = π2 −
i.e.
1 1 1 − 4 cos π − cos 2π + cos 3π − cos 4π + ... 3 22 32 42
π2 3
+
4 4 4 4 + + + + ... 12 22 32 42
π2
1 1 1 1 = 4 + + + + ... 2 2 2 2 3 1 2 3 4
2π 2 1 1 1 1 = 4 + + + + ... 3 12 22 32 42
i.e. and i.e.
π2
2π 2 1 1 1 1 = + + + + ... 3(4) 12 22 32 42
π2 1 1 1 1 + + + + ... = 12 22 32 42 6 ∞
1
∑n
i.e.
n =1
2
=
π2 6
7. Sketch the waveform defined by: 2x 1 + π , when − π 〈 x 〈 0 f(x) = 1 − 2 x , when 0 〈 x 〈 π π Determine the Fourier series in this range.
The periodic function is shown in the diagram below
1523
© 2014, John Bird
1 = a0 2π
π
∫π −
0 π π 1 1 0 2x 2x x2 x 2 f ( x) = dx x 1 + d x + ∫ 0 (1 − ) d= x + + x − 2π ∫ −π π π π −π π 0 2π
=
= an
1 1 π 2 π2 π )} 0 ) {(π − π ) + (π −= ( 0 ) − −π + + π − − ( 0= 2π π π 2π
π
1
f ( x) cos nx d= x π∫π −
π 1 0 2x 2x ∫ −π 1 + cos nx d x + ∫ 0 (1 − ) cos nx d x π π π
π
0
sin nx 2 x sin nx cos nx 1 sin nx 2 x sin nx cos nx by integration by = + + − + + 2 n −π n n 2 0 π n π n π n parts =
1 2 1 2 cos(−nπ ) 2 cos nπ 0 + 0 + 2 − 0 + 0 + + 0 − 0 − 2 π π n π n π n2
=
1 4 2} {2 − 2 cos(−nπ ) − 2 cos nπ += (1 − cos nπ ) 2 2 π n π n2 2
1
4 8 4 8 1 − −1) = = , a3 (= (1 − −1) 2 2 , 2 2 2 2 π (1) π π (3) π (3) 2
a5 =
8 π (5) 2 2
and so on
π 1 0 2x 2x ∫ −π 1 + sin nx d x + ∫ 0 1 − sin nx d x π π π
π
f ( x) sin nx d= x π∫π
b= n
since cos nπ = cos(–nπ)
an = 0
When n is even,
a1 Hence,=
2 1 − 0 − 0 − π n 2
−
0 π cos nx 2 x cos nx sin nx 1 cos nx 2 x cos nx sin nx = − + − + + − − − + by n n n 2 −π n n n 2 0 π π π
integration by parts cos nπ 2π cos nπ + 0 − n + n 1 1 cos(− nπ ) 2π cos(− nπ ) = − − 0 + 0 − − + + 0 + n n π n 1 − − + 0 − 0 n
=
1 1 cos(−nπ ) 2π cos(−nπ ) cos nπ 2π cos nπ 1 0 − − + + = − + n n n n n π n ∞
∞
Substituting into f(x) = a0 + ∑ ( an cos nx + bn sin nx ) = ∑ an cos nx = n 1= n 1
gives:
f(x) =
i.e.
f(x) =
8
π
2
cos x +
8 8 8 cos 3 x + cos 5 x + cos 7 x + ... 2 2 2 2 π (3) π (5) π (7) 2 2
8 1 1 1 cos x + 2 cos 3 x + 2 cos 5 x + 2 cos 7 x + ... 2 3 5 7 π
1524
© 2014, John Bird
8. For the Fourier series of Problem 8, deduce a series for
π2 8
When f(x) = 1 in the series of Problem 8 above, x = 0 hence, i.e.
1=
π2 8
8 1 1 1 cos 0 + 2 cos 0 + 2 cos 0 + 2 cos 0 + ... 2 3 5 7 π
1 =+
1 1 1 1 + + + + ... 32 52 7 2 92
1525
© 2014, John Bird
f(x) = π – 2(sin x +
The function f(x) = x is not periodic. The function is shown below in the range 0 to 2π and is then constructed outside of that range so that it is periodic of period 2π (see broken lines) with the resulting saw-tooth waveform
∞
For a Fourier series:
∑( a
f(x) = a 0 +
n
cos nx + b n sin nx)
n =1
a0 =
an =
1 2π 1
π
∫ ∫
2π 0
2π 0
f ( x) d x =
1 2π
∫
f ( x) cos nx d x =
2π 0
2π
xd x =
1
π
∫
2π 0
1 x2 1 4π 2 = − 0 = π 2π 2 0 2π 2
x cos nx d x 2π
1 x sin nx sin nx = −∫ d x n π n 0
by parts (see Chapter 68)
2π
1 x sin nx cos nx = + π n n 2 0 =
1 cos 2nπ 0+ π n2
cos 0 − 0 + 2 = 0 n
1516
© 2014, John Bird
bn =
1
π∫
2π 0
f ( x) sin nx d x =
1
π∫
2π 0
x sin nx d x 2π
1 − x cos nx − cos nx = − ∫ d x n n 0 π
by parts
2π
1 − x cos nx sin nx = + n n 2 0 π =
1 −2π cos 2nπ sin 2nπ sin 0 + − 0 + 2 2 π n n n
=
1 −2π cos 2nπ −2 = cos 2nπ π n n
2 2 2 2 2 When n is odd or even, b n = − . Thus b 1 = 2, b 2 = − , b 3 = − , b 4 = – , b 5 = − , and so n 4 2 3 5
on Thus i.e.
f(x) = x = π – 2 sin x – x = π – 2(sin x +
2 2 2 2 2 sin 2x – sin 3x – sin 4x – sin 5x – sin 6x + … 3 4 5 6 2
1 1 1 1 1 sin 2x + sin 3x + sin 4x + sin 5x + sin 6 x + ... ) 3 4 6 2 5
for values of f(x) between 0 and 2π. For values of f(x) outside the range 0 to 2π the sum of the series is not equal to f(x)
2. Determine the Fourier series for the function defined by: 1 − t , when − π 〈 t 〈 0 f(t) = 1 + t , when 0 〈 t 〈 π Draw a graph of the function within and outside of the given range.
The periodic function is shown in the diagram below
1517
© 2014, John Bird
1 a0 = 2π
π
1 dt f (t )= 2π
∫π −
=
a= n
1
{∫
0 −π
(1 − t ) d t + ∫
π 0
}
t 2 0 t 2 π t − + t + 2 −π 2 0 1 π2 π 2 2π 2π 2 π + + = 1+ 2 π + − ( 0 ) = = 2 2 2π 4π 2 2π
1 (1 + t ) d= t 2π
1 π 2 ( 0 ) − −π − + π 2π 2
π
f (t ) cos nt d= t π∫π −
π {∫ 1
π
0 −π
(1 − t ) cos nt d t + ∫ (1 + t ) cos nt d t 0
}
=
π {∫
=
1 sin nt t sin nt cos nt sin nt t sin nt cos nt by integration by parts − − + + + π n n n 2 −π n n n 2 0
=
1 cos 0 cos(−nπ ) cos nπ cos 0 + 0 + 0 + −0+ 0+ 0 − 0 − 2 − 0 − 0 − π n n2 n2 n 2
=
1 1 cos(−nπ ) cos nπ 1 1 + = − ( 2 cos nπ − 2 ) − 2 + π n n2 n2 n2 π n2
1
0 −π
π
( cos nt − t cos nt ) d t + ∫ 0 ( cos nt + t cos nt ) d t
} π
0
= When n is even,
2 4 ( −1 − 1) =− 2 π (1) π
a1 =
When n = 3,
a3 =
When n = 5,
a5 =
1
π
π
∫π −
2 ( cos nπ − 1) π n2
an = 0
When n = 1,
b= n
since cos(–nπ) = cos nπ
2 4 ( −1 − 1) =− 2 π (3) π (3) 2 2 4 ( −1 − 1) =− 2 π (5) π (5) 2
f (t ) sin nt d= t
π {∫ 1
0 −π
and so on π
(1 − t ) sin nt d t + ∫ (1 + t ) sin nt d t 0
}
=
π {∫
=
1 cos nt t cos nt sin nt cos nt t cos nt sin nt − + − + − − + n n n 2 −π n n n 2 0 π
1
π
( sin nt − t sin nt ) d t + ∫0 ( sin nt + t sin nt ) d t −π 0
} π
0
by integration by parts
cos 0 cos(−nπ ) π cos(−nπ ) + 0 − 0 − − − − 0 − n n n 1 = π cos nπ π cos nπ cos 0 + − − + 0 − − + 0 + 0 n n n =
1 1 cos(−nπ ) π cos(−nπ ) cos nπ π cos nπ 1 + − − + =0 − + π n n n n n n 1518
since cos nπ = cos(–nπ) © 2014, John Bird
∞
Substituting into f(t) = a0 + ∑ ( an cos nt + bn sin nt ) n =1
gives:
f(t) =
i.e.
f(t) =
π 2
π 2
+1 −
+1−
4
cos t −
π
4 4 cos 3t − cos 5t − ... + 0 2 π (3) π (5) 2
4 cos 3t cos 5t cos t + 2 + 2 + ... π 3 5
3. Find the Fourier series for the function f(x) = x + π within the range –π < x < π
The function is shown sketched below
∞
f(x) = a 0 +
∑( a
n
cos nx + b n sin nx)
n =1
1 a0 = 2π
an =
π
∫π −
1 f ( x) d x = 2π
π
1
f ( x) cos nx d x π∫π −
π
1 x2 1 π 2 π2 2 − = x π π + = + − π 2 = π ( x + π ) d x ∫ −π 2π 2 −π 2π 2 2 π
=
1
π
( x + π ) cos nx d x= π∫π −
π
1
( x cos nx + π cos nx) d x π∫π −
π
1 x sin nx sin nx π = by parts (see Chapter 68) d x + sin nx −∫ n n π n −π π
=
=
1 π sin nπ cos nπ π −π sin(−nπ ) cos(−nπ ) π + + sin nπ − + + sin(−nπ ) π n n2 n n n2 n =
bn =
1
π
1 x sin nx cos nx π + + sin nx 2 π n n n −π
1 cos nπ π n 2 1
cos(−nπ ) 0 − = n2
π
f ( x) sin nx d x = ∫ ( x + π ) sin nx d x= π π π∫π −
−
1519
1
π
( x sin nx + π sin nx) d x π∫π −
© 2014, John Bird
π
π cos nπ 1 − x cos nx − cos nx = by parts − ∫ d x − π n n n −π π
1 − x cos nx sin nx π cos nπ = + − π n n2 n −π =
1 −π cos nπ sin nπ π cos nπ + − π n n2 n =
Hence, b 1 =
1 −2π cos nπ π n
2 − cos nπ − ( 0 ) = n
2 2 2 2 2 , b 2 = − , b 3 = , b 4 = – , b 5 = , and so on 1 4 5 2 3
Thus f(x) = x + π = π + 0 + x + π = π + 2(sin x –
i.e.
π cos(−nπ ) sin(−nπ ) π cos(−nπ ) + − − n n2 n
2 2 2 2 2 2 sin x – sin 2x + sin 3x – sin 4x + sin 5x – sin 6x + … 2 3 4 5 6 1 1 1 1 1 1 sin 2x + sin 3x – sin 4x + sin 5x – sin 6 x + ... ) 3 5 6 4 2
4. Determine the Fourier series up to and including the third harmonic for the function defined by: x, when 0 〈 x 〈 π f(x) = 2π − x, when π 〈 x 〈 2π Sketch a graph of the function within and outside of the given range, assuming the period is 2π.
The periodic function is shown in the diagram below
a0 =
1 2π
π
∫π −
{
}
2π x 2 π x 2 2 x π + − ∫0 π 2 π 2 0 1 π 2 4π 2 2 π 2 1 π 2 = (π= )} { − ( 0 ) + 4π 2 − − 2π − = 2π 2 2 2 2π 2
f ( x= )d x
1 2π
π
xd x + ∫
2π
(2π − x)= dx
1520
1 2π
© 2014, John Bird
an =
1
π∫
2π 0
f ( x) cos nx d x =
π {∫ 1
π 0
x cos nx d x + ∫
2π
π
}
(2π − x) cos nx d x
π
2π
=
1 x sin nx cos nx 2π sin nx x sin nx cos nx + + − − π n n 2 0 n n n 2 π
=
1 cos nπ 0 + π n2
=
1 2 cos nπ − 1 − cos 2π n = + cos nπ } { ( cos nπ − 1) π n2 π n2
When n is even,
cos 0 cos 2π n cos nπ − 0 + 2 + 0 − 0 − −0−0− 2 n n n2
2 4 ( −1 − 1) =− 2 π (1) π
a1 =
When n = 3,
a3 =
When n = 5,
a5 =
1
π∫
2π 0
2 4 ( −1 − 1) =− 2 π (3) π (3) 2 2 4 ( −1 − 1) =− 2 π (5) π (5) 2
f ( x) sin = nx d x
π {∫ 1
0 −π
and so on
}
π
x sin nx d x + ∫ (2π − x) sin nx d x 0
π
=
an = 0
When n = 1,
= bn
by integration by parts
2π
1 x cos nx sin nx 2π cos nx x cos nx sin nx − + + − + − 2 π n n 0 n n n 2 π
by integration by parts
2π cos 2nπ 2π cos 2nπ + + 0 − n n 1 π cos nπ = − + 0 − ( 0 + 0 ) + π n 2π cos nπ π cos nπ −− + − 0 n n
=
1 0 {−π cos nπ + 2π cos nπ − π cos nπ } = nπ ∞
Substituting into f(x) = a0 + ∑ ( an cos nx + bn sin nx ) n =1
gives:
f(x) =
i.e.
f(x) =
π 2
π 2
−
−
4
π
cos x −
4 4 cos 3 x − cos 5 x − ... + 0 π (3) 2 π (5) 2
4 cos 3 x cos 5 x + + ... cos x + 2 2 π 3 5
1521
© 2014, John Bird
5. Expand the function f(θ) = θ 2 in a Fourier series in the range –π < θ < π Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below
1 = a0 2π
π
∫π −
1 f ( x) = dx 2π
{
π
π {∫
1 π 1 = ∫ f (θ ) cos nθ d θ
an
π
}
1 θ ∫ −π θ d= 2π 2
−π
π −π
θ 3 π 1 2π 3 π 2 3 {π 3 − −π= } = = 6π 3 3 −π 6π
θ 2 cos nθ d θ
}
π
1 θ 2 sin nθ 2θ cos nθ 2sin nθ = + − π n n2 n3 −π =
1 2π cos nπ 2π cos(−nπ ) 4 4π cos nπ − 0 − 0 − −= = 0 cos nπ 0 + π π n2 n2 n2 n2 since cos nπ = cos(–nπ)
4 4 4 4 a1 = ( −1) = − = , a2 = (1) 2 , 2 2 2 (2) 2 (1) 1
Hence,
π {∫
1 π 1 = ∫ f (θ ) sin nθ d θ
bn
by integration by parts
π
−π
π −π
θ 2 sin nθ d θ
4 4 a3 = ( −1) = − , 2 (3) 32
a4 =
4 , and so on 42
}
π
1 θ 2 cos nθ 2θ sin nθ 2 cos nθ = − + + π n n2 n3 −π =
1 π 2 cos nπ 2 cos nπ +0+ − π n n3
by integration by parts
2 cos(−nπ ) π 2 cos(−nπ ) +0+ −− = 0 n n3
∞
Substituting into f(θ) = a0 + ∑ ( an cos nθ + bn sin nθ ) n =1
gives:
f(θ) =
i.e.
f(θ) =
π2 3
−
4 4 4 4 cos θ + cos 2θ − cos 3θ + cos 4θ − ... + 0 12 22 32 42
π2
1 1 − 4 cos θ − cos 2θ + cos 3θ − ... 2 2 3 2 3 1522
© 2014, John Bird
6. For the Fourier series obtained in Problem 5, let x = π and deduce the series for
∞
1
∑n n =1
2
When θ = π in Problem 5 above, f(θ) = π 2 Thus,
π2 =
i.e.
π2 = π2 −
i.e.
1 1 1 − 4 cos π − cos 2π + cos 3π − cos 4π + ... 3 22 32 42
π2 3
+
4 4 4 4 + + + + ... 12 22 32 42
π2
1 1 1 1 = 4 + + + + ... 2 2 2 2 3 1 2 3 4
2π 2 1 1 1 1 = 4 + + + + ... 3 12 22 32 42
i.e. and i.e.
π2
2π 2 1 1 1 1 = + + + + ... 3(4) 12 22 32 42
π2 1 1 1 1 + + + + ... = 12 22 32 42 6 ∞
1
∑n
i.e.
n =1
2
=
π2 6
7. Sketch the waveform defined by: 2x 1 + π , when − π 〈 x 〈 0 f(x) = 1 − 2 x , when 0 〈 x 〈 π π Determine the Fourier series in this range.
The periodic function is shown in the diagram below
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© 2014, John Bird
1 = a0 2π
π
∫π −
0 π π 1 1 0 2x 2x x2 x 2 f ( x) = dx x 1 + d x + ∫ 0 (1 − ) d= x + + x − 2π ∫ −π π π π −π π 0 2π
=
= an
1 1 π 2 π2 π )} 0 ) {(π − π ) + (π −= ( 0 ) − −π + + π − − ( 0= 2π π π 2π
π
1
f ( x) cos nx d= x π∫π −
π 1 0 2x 2x ∫ −π 1 + cos nx d x + ∫ 0 (1 − ) cos nx d x π π π
π
0
sin nx 2 x sin nx cos nx 1 sin nx 2 x sin nx cos nx by integration by = + + − + + 2 n −π n n 2 0 π n π n π n parts =
1 2 1 2 cos(−nπ ) 2 cos nπ 0 + 0 + 2 − 0 + 0 + + 0 − 0 − 2 π π n π n π n2
=
1 4 2} {2 − 2 cos(−nπ ) − 2 cos nπ += (1 − cos nπ ) 2 2 π n π n2 2
1
4 8 4 8 1 − −1) = = , a3 (= (1 − −1) 2 2 , 2 2 2 2 π (1) π π (3) π (3) 2
a5 =
8 π (5) 2 2
and so on
π 1 0 2x 2x ∫ −π 1 + sin nx d x + ∫ 0 1 − sin nx d x π π π
π
f ( x) sin nx d= x π∫π
b= n
since cos nπ = cos(–nπ)
an = 0
When n is even,
a1 Hence,=
2 1 − 0 − 0 − π n 2
−
0 π cos nx 2 x cos nx sin nx 1 cos nx 2 x cos nx sin nx = − + − + + − − − + by n n n 2 −π n n n 2 0 π π π
integration by parts cos nπ 2π cos nπ + 0 − n + n 1 1 cos(− nπ ) 2π cos(− nπ ) = − − 0 + 0 − − + + 0 + n n π n 1 − − + 0 − 0 n
=
1 1 cos(−nπ ) 2π cos(−nπ ) cos nπ 2π cos nπ 1 0 − − + + = − + n n n n n π n ∞
∞
Substituting into f(x) = a0 + ∑ ( an cos nx + bn sin nx ) = ∑ an cos nx = n 1= n 1
gives:
f(x) =
i.e.
f(x) =
8
π
2
cos x +
8 8 8 cos 3 x + cos 5 x + cos 7 x + ... 2 2 2 2 π (3) π (5) π (7) 2 2
8 1 1 1 cos x + 2 cos 3 x + 2 cos 5 x + 2 cos 7 x + ... 2 3 5 7 π
1524
© 2014, John Bird
8. For the Fourier series of Problem 8, deduce a series for
π2 8
When f(x) = 1 in the series of Problem 8 above, x = 0 hence, i.e.
1=
π2 8
8 1 1 1 cos 0 + 2 cos 0 + 2 cos 0 + 2 cos 0 + ... 2 3 5 7 π
1 =+
1 1 1 1 + + + + ... 32 52 7 2 92
1525
© 2014, John Bird
