chapter 103 even and odd functions and half-range ... AWS
CHAPTER 103 EVEN AND ODD FUNCTIONS AND HALF-RANGE FOURIER SERIES EXERCISE 364 Page 1076
1. Determine the Fourier series for the function defined by:
π − 1, − π 〈 x 〈 − 2 π π f(x) = 1, − 〈 x 〈 2 2 π − 1, 2 〈 x 〈 π which is periodic outside of this range of period 2π.
The square wave shown is an even function since it is symmetrical about the f(x) axis.
∞
Hence, the Fourier series is given by:
f(x) = a 0 +
∑a
n
cos nx
n =1
(i.e. the series contains no sine terms) a0 =
π
1
π∫
0
f ( x) d x =
=
an =
2
π
∫
π 0
π {∫ 1
π /2 0
{[ x] π 1
f ( x) cos nx d x =
1d x + ∫
π /2 0
2
π
π π /2
}
−1d x
}
+ [ − x ] π /2 =
{∫
π
π /2 0
1
π
[
1cos nx d x + ∫
1526
(π) + [(– 2π) – (– π)] π π /2
] =0
}
−1cos nx d x
© 2014, John Bird
π /2 π 2 sin nx − sin nx = + π n 0 n π /2
=
− sin(π / 2)n 2 sin(π / 2)n − 0 + 0 − π n n
=
2 2sin(π / 2)n nπ 4 sin = π n 2 π n
When n is even, a n = 0 When n is odd, a n =
and
an =
Hence, a 1 =
4
π
4 for n = 1, 5, 9,... πn −4 for n = 3, 7, 11,... πn
, a3 =
−4 4 , a5 = , and so on 3π 5π
Hence the Fourier series for the above waveform is given by: f(x) =
4 1 1 1 cos x − cos 3 x + cos 5 x − cos 7 x + ... 3 5 7 π
2. Obtain the Fourier series of the function defined by: t +π, −π 〈 t 〈 0 f(t) = t −π, 0 〈 t 〈 π which is periodic of period 2π. Sketch the given function.
The periodic function is shown in the diagram below. Since it is symmetrical about the origin, the ∞
function is odd, and f (t ) = ∑ bn sin nt n =1
1527
© 2014, John Bird
1
π {∫
π
1
f (t ) sin nt d= t π∫π
b= n
−
0 −π
π
(t + π ) sin nt d t + ∫ (t − π ) sin nt d t 0
} π
1 t cos nt sin nt π cos nt t cos nt sin nt π cos nt − + − + − + + π n n2 n −π n n2 n 0 0
=
by integration by parts
π cos nπ π cos nπ 0 − + + n n 1 π −π cos(−nπ ) π cos(−nπ ) = 0 + 0 − − − +0− + n n n π π − 0 + 0 + n
=
1 π π cos(−nπ ) π cos(−nπ ) π cos nπ π cos nπ π 1 2π + − + − = − − − π n n n n n n π n b1 = −
Hence,
2 , 1
b2 = −
2 , 2
b3 = −
2 , 3
b4 = −
2 − = n
2 , and so on 4
i.e.
2 2 2 2 f(t) = − sin t − sin 2t − sin 3t − sin 4t − ... 1 2 3 4
i.e.
1 1 1 f (t ) = −2 sin t + sin 2t + sin 3t + sin 4t + ... 2 3 4
1 − x, − π 〈 x 〈 0 3. Determine the Fourier series defined by f(x) = which is periodic of period 1 + x , 0 〈 x 〈 π 2π. The periodic function is shown in the diagram below. Since it is symmetrical about the f(x) axis, the ∞
f ( x= ) a0 + ∑ an cos nx
function is even, and
n =1
a0 =
1 π 1 f ( x) d x = ∫ − π 2π π
= a= n
1
π
1
π
∫
∫
π 0
f ( x) cos nx d= x π∫π −
π 0
f ( x) d x due to symmetry π
1 x2 1 π2 π (1 + x) d x = x + = π + 1+ − ( 0 ) = π 2 0 π 2 2
π {∫ 1
0 −π
π
}
(1 − x) cos nx d x + ∫ (1 + x) cos nx d x 0
1528
© 2014, John Bird
=
π {∫
=
1 sin nx x sin nx cos nx sin nx x sin nx cos nx − − + + + π n n n 2 −π n n n 2 0
=
1 1 cos(−nπ ) cos nπ 1 + 0 + 0 + − 0 + 0 + 0 − 0 − 2 − 0 − 0 − π n n2 n2 n 2
=
1 1 cos(−nπ ) cos nπ 1 2 + = − ( cos nπ − 1) − 2 + n2 n2 n2 π n2 π n
1
0 −π
}
π
( cos nx − x cos nx ) d x + ∫0 ( cos nx + x cos nx ) d x
π
0
since cos(–nπ) = cos nπ
an = 0
When n is even,
2 4 −1 − 1) =− ( π (1) 2 π
When n = 1,
a1 =
When n = 3,
a3 =
When n = 5,
a5 =
2 4 −1 − 1) =− ( π (3) 2 π (3) 2
2 4 −1 − 1) =− ( π (5) 2 π (5) 2
∞
π
n =1
2
f ( x= ) a0 + ∑ an cos nx = i.e.
by integration by parts
f(x) =
π 2
+1 −
+1−
4
π
cos x −
and so on
4 4 cos 3 x − cos 5 x − ... π (3) 2 π (5) 2
4 1 1 cos x + 2 cos 3 x + 2 cos 5 x + ... 3 5 π
4. In the Fourier series of Problem 3, let x = 0 and deduce a series for π2/8
When x = 0 in the series of Problem 3, f(x) = 1,
π
hence,
1=
i.e.
1=
i.e. and
−
2
π 2
+1−
4 cos 0 cos 0 cos 0 + 2 + 2 + ... π 3 5
+1−
4 1 1 1 1 + 2 + 2 + 2 + ... π 3 5 7
π
4 1 1 1 = − 1 + + + + ... 2 2 2 2 π 3 5 7
π2 8
1 =+
1 1 1 + + + ... 32 52 7 2
5. Show that the Fourier series for the triangular waveform shown is given by: y=
8 1 1 1 sin θ − sin 3θ + sin 5θ − sin 7θ + ... 32 52 72 π2 1529
© 2014, John Bird
The function is periodic of period 2π
The equation of the function between 0 and π/2 is of the straight line form y = mθ + c where gradient, m =
1 2 and intercept, c = 0 = π /2 π
Hence, equation of the line between 0 and π/2 is y =
2θ
π
The equation of the function between π/2 and 3π/2 is of the straight line form y = mθ + c where gradient, m =
2 2 = − −π π
When θ = π, and y = 0 and since y = mθ + c then 0 = −
from which, c = 2. Hence, equation of line between π/2 and 3π/2 is y = −
2θ
π
2
π
π+c
+2
The equation of the function between 3π/2 and 2π is of the straight line form y = mθ + c where gradient, m =
1 2 = π /2 π
When θ = 2π, and y = 0 and since y = mθ + c then 0 =
from which, c = –4. Hence, equation of line between 3π/2 and 2π is y =
2θ
π
2
π
(2π) + c
−4
The triangular wave is an odd function since it is symmetrical about the origin ∞
Hence, the Fourier series is given by:
f(θ) = ∑ ( b n sin nθ)
i.e. a 0 = a n = 0
n =1
bn =
2
π
∫
π 0
f ( x) sin nx d x =
π 2θ 2 π /2 2θ + + 2 sin nθ d θ sin n θ d θ − ∫ ∫ 0 π /2 π π π
= π /2 π π 2 2 θ cos nθ sin nθ 2 θ cos nθ sin nθ 2 cos nθ + − − + − − π π n n 2 0 π n n 2 π /2 n π /2
1530
© 2014, John Bird
=
2 2 (π / 2) cos nπ / 2 sin nπ / 2 + − ( 0 ) − π π n n2
–
2 2 (π ) cos nπ sin nπ (π / 2) cos nπ / 2 sin nπ / 2 − + + −− π π n n2 n n2
–
Thus, b 1 =
4 1 4 π 1 2 2 4 4 4 4 8 −0 + − + 0 − − − = − + + = π2 12 π 2 1 12 π 1 π 2 π π 2 π π 2
b2 =
4 π 4 π π 2 2 2 1 3 4 = + − = − − − − + 0 = b 4 = b 6 = b 8 , and so on π 2 4 π 2 2 4 π 2 2 π π π
b3 =
4 1 4 π 1 2 2 4 4 4 4 8 − + = − − 2 − 2 + 2 − − = − 2 2 − 2 π 3 π 3 3 π 3 π 3 3π π 2 32 3π π 2 32
b5 =
4 1 4 π 1 2 2 4 4 4 4 8 + + = −0 + 2 − 2 + 0 − 2 − − = 2 2 − 2 5 π 5 5 π 5 π 5 5π π 2 52 5π π 2 52 π
It follows that b 7 = ∞
Thus,
y=
∑b
n
n =1
i.e.
2 2 cos nπ 2 cos nπ / 2 − π n n
y=
8 and so on π 72 2
sin nθ =
8
π2
sin θ −
8 8 8 sin 3θ + sin 5θ − sin 7θ + ... π 2 32 π 2 52 π 2 72
8 1 1 1 sin θ − sin 3θ + sin 5θ − sin 7θ + ... π2 32 52 72
1531
© 2014, John Bird
EXERCISE 365 Page 1079
1. Determine the half-range sine series for the function defined by:
π x, 0 〈 x 〈 2 f(x) = 0, π 〈 x 〈 π 2
The periodic function is shown in the diagram below. Since a half-range sine series is required, the ∞
f ( x) = ∑ bn sin nx
function is symmetrical about the origin and
n =1
{
}
π /2
2 π 2 π /2 2 x cos nx sin nx bn = f ( x) sin nx d x =∫ x sin nx d x = − + ∫ π 0 π 0 π n n 2 0
π nπ nπ cos sin 2 2 2 + 2 = − π n n2
Hence,
π π π cos sin 2 2+ 2 b1 = − 2 2 1 1 π
2 1 2 , = 0 + 2 = π 1 π
π cos π 2 2 sin π b2 = − + π 2 22
2 π 2 π = + 0 = , π 4 π 4
π 3π 3π cos sin 2 2 2 + 2 b3 = − π 3 32
by integration by parts
− ( 0 )
2 1 2 , − = 0− 2 = 3 π (3) 2 π
1532
© 2014, John Bird
π cos 2π 2 2 sin 2π b4 = − + π 4 42 ∞
f ( x) = ∑ bn sin nx =
Hence,
n =1
f ( x) =
i.e.
2
π
2 π 2 π − , = − + 0 = π8 π 8
sin x +
and so on
2 π 2 1 2 π sin 2 x − 2 sin 3x − sin 4 x + ... π 4 π 3 π8
π π 2 1 sin x + sin 2 x − sin 3 x − sin 4 x + ... π 4 9 8
2. Obtain (a) the half-range cosine series and (b) the half-range sine series for the function
π 0, 0 〈 t 〈 2 f(t) = 1, π 〈 t 〈 π 2
(a) The periodic function is shown in the diagram below. Since a half-range cosine series is ∞
required, the function is symmetrical about the f(t) axis and
f (t= ) a0 + ∑ an cos nt n =1
a0 =
1
π
π ∫π
/2
1d t =
1
π
[t ] π /2 = π
1 π 1 − = π π 2 2
nπ π sin 2 π 2 sin nt 2 2 1cos nt d t = = an = 0− π ∫π /2 π n π /2 π n
nπ 2sin 2 − = π n
When n is even, an = 0
and
Thus,
3π 5π π 2sin 2sin 2sin 2 1 2 1 2 2 = 2 = 2 = − − − , and so on a1 = − − , a3 = , a5 = π 3 3π π 5 5π π π ∞ 1 2 2 1 2 1 − cos t + cos 3t − cos 5t + ... f (t ) = a0 + ∑ an cos nt = 2 π π 3 π 5 n =1
1533
© 2014, John Bird
1 2 1 1 − cos t − cos 3t + cos 5t − ... f (t ) = 2 π 3 5
i.e.
(b) The periodic function is shown in the diagram below. Since a half-range sine series is required, ∞
f ( x) = ∑ bn sin nx
the function is symmetrical about the origin and
n =1
π
2 π 2 cos nt 2 nπ − = − − bn = 1sin nt d t = cos n π cos ∫ π π /2 π n π /2 nπ 2 Hence,
b1 =−
2 π 2 2 cos π − cos =− ( −1 − 0 ) = , π 2 π π
b2 = −
2 2 2 ( cos 2π − cos π ) = − (1 − −1) = − , 2π 2π π
b3 =−
2 3π
3π cos 3π − cos 2
2 2 =− ( −1 − 0 ) = , 3π 3π
2 2 b4 = 0, − − ( cos 4π − cos 2π ) = (0 − 0) = 4π 4π
b5 =−
2 5π 2 2 ( −1 − 0 ) = , cos 5π − cos =− 5π 2 5π 5π
b6 = −
2 2 2 ( cos 6π − cos 3π ) = − (1 − −1) = − , and so on 6π 6π 3π ∞
Thus,
f (t ) = ∑ bn sin nt = n =1
i.e.
f (t ) =
2
π
sin t −
2
π
sin 2t +
2 2 2 sin 3t + 0 + sin 5t − sin 6t + ... 3π 5π 3π
2 1 1 1 sin t − sin 2t + sin 3t + sin 5t − sin 6t + ... 3 5 3 π
3. Find the half-range Fourier sine series for the function f(x) = sin2 x in the range 0 ≤ x ≤ π. Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below. Since a half-range sine series is required, 1534
© 2014, John Bird
∞
the function is symmetrical about the origin and
f ( x) = ∑ bn sin nx n =1
2 π 2 2 π1 1 π sin x sin nx d x = (1 − cos 2 x) sin nx d x = (sin nx − cos 2 x sin nx) d x bn = ∫ ∫ π 0 π 0 2 π ∫0
=
2
π∫
π 0
1 sin nx − ( sin(2 + n) x − sin(2 − n) x ) d x 2 π
1 − cos nx 1 − cos(2 + n) x cos(2 − n) x = − + 2 (2 + n) (2 − n) 0 π n
= 1 − cos nπ cos(2 + n)π cos(2 − n)π −1 1 1 + − − − + π n 2(2 + n) 2(2 − n) n 2(2 + n) 2(2 − n)
Hence, b1 =
1 1 1 1 1 1 8 8 + +1− + 1 − = = π 2(3) 2(1) 2(3) 2(1) π 3 3π
b2 =
1 1 1 1 1 1 1 − + − + = 0 = b4 = b6 = b8 and so on − + π 2 2(4) 2(0) 2 2(4) 2(0)
1 1 1 1 1 1 1 1 2 1 1 10 − 3 − 15 8 + + − + b3 = − = − − 1 = =− π 3 2(5) 2(−1) 3 2(5) 2(−1) π 3 5 π (3)(5) π (3)(5) 1 1 1 1 1 1 1 1 2 1 1 1 42 − 15 − 35 8 + + − + − b5 = − = = − − = π 5 2(7) 2(−3) 5 2(7) 2(−3) π 5 7 3 π (3)(5)(7) π (3)(5)(7) It follows that b7 = −
8 and so on π (5)(7)(9)
∞
Thus,
f ( x) = ∑ bn sin nx = n =1
8 8 8 8 sin x − sin 3 x − sin 5 x − sin 7 x 3π π (3)(5) π (3)(5)(7) π (5)(7)(9) 1535
© 2014, John Bird
8 sin x sin 3 x sin 5 x sin 7 x − − − − ... f ( x) = sin 2 x = π (1)(3) (1)(3)(5) (3)(5)(7) (5)(7)(9)
i.e.
4. Determine the half-range Fourier cosine series in the range x = 0 to x = π for the function defined by:
π 0〈 x〈 x, 2 f(x) = (π − x), π 〈 x 〈 π 2
The periodic function is shown in the diagram below. Since a half-range cosine series is required, ∞
the function is symmetrical about the f(x) axis and
f ( x= ) a0 + ∑ an cos nx n =1
= a0
= an
1
π
{∫
π {∫ 2
π /2 0
π /2 0
xd x + ∫
π π /2
}
(π − x= )d x
x cos nx d x + ∫
π π /2
π /2 π x 2 1 x 2 + − π x π 2 0 2 π /2
=
1 π 2 2 π 2 π 2 π 2 − + π − − π 8 2 2 8
=
1 π 2 π 2 π 2 π 2 1 2π 2 π + − += = π 8 2 2 8 π 8 4
}
(π − x) cos nx d x
π /2 π 2 x sin nx cos nx π sin nπ x sin nx cos nx = + + − − π n n 2 0 n n 2 π /2 n
π nπ nπ sin cos 2 2 2 + 2 = n n2 π
1 cos nπ −0 + 2 +0 −0 − n n2 1536
nπ π nπ nπ sin cos π sin 2 − 2 2 − 2 − 2 n n n
© 2014, John Bird
nπ nπ nπ 2 cos sin sin π π 2 1 cos 2 n nπ π 2 + 2 − − 2= = − − 1 − cos nπ 2 cos 2 2 2 2 2 n n n n πn π n
When n is odd, an = 0
2 2 8 2 ( 2 cos π − 1 − cos 2π ) = ( −2 − 1 − 1) =− =− , 2 4π 4π π (2) π
a2 =
and
a4 =
2 2 4π ) − 1) 0 , ( 2 cos 2π − 1 − cos= ( 2 − 1= 2 16π π (4) 2 2 8 2 2 cos 3π − 1 − cos 6π ) = −2 − 1 − 1) =− =− , ( ( π (6) 2 36π 36π (3) 2 π
a6 =
a8 = 0 , 2 2 8 2 2 cos 5π − 1 − cos10π ) = −2 − 1 − 1) =− =− , and so on ( ( π (10) 2 100π 100π (5) 2 π
a10 =
Thus,
i.e.
∞ π 2 2 2 − cos 2 x − f ( x) = a0 + ∑ an cos nx = cos 6 x − cos10 x + ... 4 π (3) 2 π (5) 2 π n =1
π 2 cos 6 x cos10 x f ( x) = − cos 2 x + + + ... 2 2 4 π 3 5
1537
© 2014, John Bird
1. Determine the Fourier series for the function defined by:
π − 1, − π 〈 x 〈 − 2 π π f(x) = 1, − 〈 x 〈 2 2 π − 1, 2 〈 x 〈 π which is periodic outside of this range of period 2π.
The square wave shown is an even function since it is symmetrical about the f(x) axis.
∞
Hence, the Fourier series is given by:
f(x) = a 0 +
∑a
n
cos nx
n =1
(i.e. the series contains no sine terms) a0 =
π
1
π∫
0
f ( x) d x =
=
an =
2
π
∫
π 0
π {∫ 1
π /2 0
{[ x] π 1
f ( x) cos nx d x =
1d x + ∫
π /2 0
2
π
π π /2
}
−1d x
}
+ [ − x ] π /2 =
{∫
π
π /2 0
1
π
[
1cos nx d x + ∫
1526
(π) + [(– 2π) – (– π)] π π /2
] =0
}
−1cos nx d x
© 2014, John Bird
π /2 π 2 sin nx − sin nx = + π n 0 n π /2
=
− sin(π / 2)n 2 sin(π / 2)n − 0 + 0 − π n n
=
2 2sin(π / 2)n nπ 4 sin = π n 2 π n
When n is even, a n = 0 When n is odd, a n =
and
an =
Hence, a 1 =
4
π
4 for n = 1, 5, 9,... πn −4 for n = 3, 7, 11,... πn
, a3 =
−4 4 , a5 = , and so on 3π 5π
Hence the Fourier series for the above waveform is given by: f(x) =
4 1 1 1 cos x − cos 3 x + cos 5 x − cos 7 x + ... 3 5 7 π
2. Obtain the Fourier series of the function defined by: t +π, −π 〈 t 〈 0 f(t) = t −π, 0 〈 t 〈 π which is periodic of period 2π. Sketch the given function.
The periodic function is shown in the diagram below. Since it is symmetrical about the origin, the ∞
function is odd, and f (t ) = ∑ bn sin nt n =1
1527
© 2014, John Bird
1
π {∫
π
1
f (t ) sin nt d= t π∫π
b= n
−
0 −π
π
(t + π ) sin nt d t + ∫ (t − π ) sin nt d t 0
} π
1 t cos nt sin nt π cos nt t cos nt sin nt π cos nt − + − + − + + π n n2 n −π n n2 n 0 0
=
by integration by parts
π cos nπ π cos nπ 0 − + + n n 1 π −π cos(−nπ ) π cos(−nπ ) = 0 + 0 − − − +0− + n n n π π − 0 + 0 + n
=
1 π π cos(−nπ ) π cos(−nπ ) π cos nπ π cos nπ π 1 2π + − + − = − − − π n n n n n n π n b1 = −
Hence,
2 , 1
b2 = −
2 , 2
b3 = −
2 , 3
b4 = −
2 − = n
2 , and so on 4
i.e.
2 2 2 2 f(t) = − sin t − sin 2t − sin 3t − sin 4t − ... 1 2 3 4
i.e.
1 1 1 f (t ) = −2 sin t + sin 2t + sin 3t + sin 4t + ... 2 3 4
1 − x, − π 〈 x 〈 0 3. Determine the Fourier series defined by f(x) = which is periodic of period 1 + x , 0 〈 x 〈 π 2π. The periodic function is shown in the diagram below. Since it is symmetrical about the f(x) axis, the ∞
f ( x= ) a0 + ∑ an cos nx
function is even, and
n =1
a0 =
1 π 1 f ( x) d x = ∫ − π 2π π
= a= n
1
π
1
π
∫
∫
π 0
f ( x) cos nx d= x π∫π −
π 0
f ( x) d x due to symmetry π
1 x2 1 π2 π (1 + x) d x = x + = π + 1+ − ( 0 ) = π 2 0 π 2 2
π {∫ 1
0 −π
π
}
(1 − x) cos nx d x + ∫ (1 + x) cos nx d x 0
1528
© 2014, John Bird
=
π {∫
=
1 sin nx x sin nx cos nx sin nx x sin nx cos nx − − + + + π n n n 2 −π n n n 2 0
=
1 1 cos(−nπ ) cos nπ 1 + 0 + 0 + − 0 + 0 + 0 − 0 − 2 − 0 − 0 − π n n2 n2 n 2
=
1 1 cos(−nπ ) cos nπ 1 2 + = − ( cos nπ − 1) − 2 + n2 n2 n2 π n2 π n
1
0 −π
}
π
( cos nx − x cos nx ) d x + ∫0 ( cos nx + x cos nx ) d x
π
0
since cos(–nπ) = cos nπ
an = 0
When n is even,
2 4 −1 − 1) =− ( π (1) 2 π
When n = 1,
a1 =
When n = 3,
a3 =
When n = 5,
a5 =
2 4 −1 − 1) =− ( π (3) 2 π (3) 2
2 4 −1 − 1) =− ( π (5) 2 π (5) 2
∞
π
n =1
2
f ( x= ) a0 + ∑ an cos nx = i.e.
by integration by parts
f(x) =
π 2
+1 −
+1−
4
π
cos x −
and so on
4 4 cos 3 x − cos 5 x − ... π (3) 2 π (5) 2
4 1 1 cos x + 2 cos 3 x + 2 cos 5 x + ... 3 5 π
4. In the Fourier series of Problem 3, let x = 0 and deduce a series for π2/8
When x = 0 in the series of Problem 3, f(x) = 1,
π
hence,
1=
i.e.
1=
i.e. and
−
2
π 2
+1−
4 cos 0 cos 0 cos 0 + 2 + 2 + ... π 3 5
+1−
4 1 1 1 1 + 2 + 2 + 2 + ... π 3 5 7
π
4 1 1 1 = − 1 + + + + ... 2 2 2 2 π 3 5 7
π2 8
1 =+
1 1 1 + + + ... 32 52 7 2
5. Show that the Fourier series for the triangular waveform shown is given by: y=
8 1 1 1 sin θ − sin 3θ + sin 5θ − sin 7θ + ... 32 52 72 π2 1529
© 2014, John Bird
The function is periodic of period 2π
The equation of the function between 0 and π/2 is of the straight line form y = mθ + c where gradient, m =
1 2 and intercept, c = 0 = π /2 π
Hence, equation of the line between 0 and π/2 is y =
2θ
π
The equation of the function between π/2 and 3π/2 is of the straight line form y = mθ + c where gradient, m =
2 2 = − −π π
When θ = π, and y = 0 and since y = mθ + c then 0 = −
from which, c = 2. Hence, equation of line between π/2 and 3π/2 is y = −
2θ
π
2
π
π+c
+2
The equation of the function between 3π/2 and 2π is of the straight line form y = mθ + c where gradient, m =
1 2 = π /2 π
When θ = 2π, and y = 0 and since y = mθ + c then 0 =
from which, c = –4. Hence, equation of line between 3π/2 and 2π is y =
2θ
π
2
π
(2π) + c
−4
The triangular wave is an odd function since it is symmetrical about the origin ∞
Hence, the Fourier series is given by:
f(θ) = ∑ ( b n sin nθ)
i.e. a 0 = a n = 0
n =1
bn =
2
π
∫
π 0
f ( x) sin nx d x =
π 2θ 2 π /2 2θ + + 2 sin nθ d θ sin n θ d θ − ∫ ∫ 0 π /2 π π π
= π /2 π π 2 2 θ cos nθ sin nθ 2 θ cos nθ sin nθ 2 cos nθ + − − + − − π π n n 2 0 π n n 2 π /2 n π /2
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© 2014, John Bird
=
2 2 (π / 2) cos nπ / 2 sin nπ / 2 + − ( 0 ) − π π n n2
–
2 2 (π ) cos nπ sin nπ (π / 2) cos nπ / 2 sin nπ / 2 − + + −− π π n n2 n n2
–
Thus, b 1 =
4 1 4 π 1 2 2 4 4 4 4 8 −0 + − + 0 − − − = − + + = π2 12 π 2 1 12 π 1 π 2 π π 2 π π 2
b2 =
4 π 4 π π 2 2 2 1 3 4 = + − = − − − − + 0 = b 4 = b 6 = b 8 , and so on π 2 4 π 2 2 4 π 2 2 π π π
b3 =
4 1 4 π 1 2 2 4 4 4 4 8 − + = − − 2 − 2 + 2 − − = − 2 2 − 2 π 3 π 3 3 π 3 π 3 3π π 2 32 3π π 2 32
b5 =
4 1 4 π 1 2 2 4 4 4 4 8 + + = −0 + 2 − 2 + 0 − 2 − − = 2 2 − 2 5 π 5 5 π 5 π 5 5π π 2 52 5π π 2 52 π
It follows that b 7 = ∞
Thus,
y=
∑b
n
n =1
i.e.
2 2 cos nπ 2 cos nπ / 2 − π n n
y=
8 and so on π 72 2
sin nθ =
8
π2
sin θ −
8 8 8 sin 3θ + sin 5θ − sin 7θ + ... π 2 32 π 2 52 π 2 72
8 1 1 1 sin θ − sin 3θ + sin 5θ − sin 7θ + ... π2 32 52 72
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© 2014, John Bird
EXERCISE 365 Page 1079
1. Determine the half-range sine series for the function defined by:
π x, 0 〈 x 〈 2 f(x) = 0, π 〈 x 〈 π 2
The periodic function is shown in the diagram below. Since a half-range sine series is required, the ∞
f ( x) = ∑ bn sin nx
function is symmetrical about the origin and
n =1
{
}
π /2
2 π 2 π /2 2 x cos nx sin nx bn = f ( x) sin nx d x =∫ x sin nx d x = − + ∫ π 0 π 0 π n n 2 0
π nπ nπ cos sin 2 2 2 + 2 = − π n n2
Hence,
π π π cos sin 2 2+ 2 b1 = − 2 2 1 1 π
2 1 2 , = 0 + 2 = π 1 π
π cos π 2 2 sin π b2 = − + π 2 22
2 π 2 π = + 0 = , π 4 π 4
π 3π 3π cos sin 2 2 2 + 2 b3 = − π 3 32
by integration by parts
− ( 0 )
2 1 2 , − = 0− 2 = 3 π (3) 2 π
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© 2014, John Bird
π cos 2π 2 2 sin 2π b4 = − + π 4 42 ∞
f ( x) = ∑ bn sin nx =
Hence,
n =1
f ( x) =
i.e.
2
π
2 π 2 π − , = − + 0 = π8 π 8
sin x +
and so on
2 π 2 1 2 π sin 2 x − 2 sin 3x − sin 4 x + ... π 4 π 3 π8
π π 2 1 sin x + sin 2 x − sin 3 x − sin 4 x + ... π 4 9 8
2. Obtain (a) the half-range cosine series and (b) the half-range sine series for the function
π 0, 0 〈 t 〈 2 f(t) = 1, π 〈 t 〈 π 2
(a) The periodic function is shown in the diagram below. Since a half-range cosine series is ∞
required, the function is symmetrical about the f(t) axis and
f (t= ) a0 + ∑ an cos nt n =1
a0 =
1
π
π ∫π
/2
1d t =
1
π
[t ] π /2 = π
1 π 1 − = π π 2 2
nπ π sin 2 π 2 sin nt 2 2 1cos nt d t = = an = 0− π ∫π /2 π n π /2 π n
nπ 2sin 2 − = π n
When n is even, an = 0
and
Thus,
3π 5π π 2sin 2sin 2sin 2 1 2 1 2 2 = 2 = 2 = − − − , and so on a1 = − − , a3 = , a5 = π 3 3π π 5 5π π π ∞ 1 2 2 1 2 1 − cos t + cos 3t − cos 5t + ... f (t ) = a0 + ∑ an cos nt = 2 π π 3 π 5 n =1
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1 2 1 1 − cos t − cos 3t + cos 5t − ... f (t ) = 2 π 3 5
i.e.
(b) The periodic function is shown in the diagram below. Since a half-range sine series is required, ∞
f ( x) = ∑ bn sin nx
the function is symmetrical about the origin and
n =1
π
2 π 2 cos nt 2 nπ − = − − bn = 1sin nt d t = cos n π cos ∫ π π /2 π n π /2 nπ 2 Hence,
b1 =−
2 π 2 2 cos π − cos =− ( −1 − 0 ) = , π 2 π π
b2 = −
2 2 2 ( cos 2π − cos π ) = − (1 − −1) = − , 2π 2π π
b3 =−
2 3π
3π cos 3π − cos 2
2 2 =− ( −1 − 0 ) = , 3π 3π
2 2 b4 = 0, − − ( cos 4π − cos 2π ) = (0 − 0) = 4π 4π
b5 =−
2 5π 2 2 ( −1 − 0 ) = , cos 5π − cos =− 5π 2 5π 5π
b6 = −
2 2 2 ( cos 6π − cos 3π ) = − (1 − −1) = − , and so on 6π 6π 3π ∞
Thus,
f (t ) = ∑ bn sin nt = n =1
i.e.
f (t ) =
2
π
sin t −
2
π
sin 2t +
2 2 2 sin 3t + 0 + sin 5t − sin 6t + ... 3π 5π 3π
2 1 1 1 sin t − sin 2t + sin 3t + sin 5t − sin 6t + ... 3 5 3 π
3. Find the half-range Fourier sine series for the function f(x) = sin2 x in the range 0 ≤ x ≤ π. Sketch the function within and outside of the given range.
The periodic function is shown in the diagram below. Since a half-range sine series is required, 1534
© 2014, John Bird
∞
the function is symmetrical about the origin and
f ( x) = ∑ bn sin nx n =1
2 π 2 2 π1 1 π sin x sin nx d x = (1 − cos 2 x) sin nx d x = (sin nx − cos 2 x sin nx) d x bn = ∫ ∫ π 0 π 0 2 π ∫0
=
2
π∫
π 0
1 sin nx − ( sin(2 + n) x − sin(2 − n) x ) d x 2 π
1 − cos nx 1 − cos(2 + n) x cos(2 − n) x = − + 2 (2 + n) (2 − n) 0 π n
= 1 − cos nπ cos(2 + n)π cos(2 − n)π −1 1 1 + − − − + π n 2(2 + n) 2(2 − n) n 2(2 + n) 2(2 − n)
Hence, b1 =
1 1 1 1 1 1 8 8 + +1− + 1 − = = π 2(3) 2(1) 2(3) 2(1) π 3 3π
b2 =
1 1 1 1 1 1 1 − + − + = 0 = b4 = b6 = b8 and so on − + π 2 2(4) 2(0) 2 2(4) 2(0)
1 1 1 1 1 1 1 1 2 1 1 10 − 3 − 15 8 + + − + b3 = − = − − 1 = =− π 3 2(5) 2(−1) 3 2(5) 2(−1) π 3 5 π (3)(5) π (3)(5) 1 1 1 1 1 1 1 1 2 1 1 1 42 − 15 − 35 8 + + − + − b5 = − = = − − = π 5 2(7) 2(−3) 5 2(7) 2(−3) π 5 7 3 π (3)(5)(7) π (3)(5)(7) It follows that b7 = −
8 and so on π (5)(7)(9)
∞
Thus,
f ( x) = ∑ bn sin nx = n =1
8 8 8 8 sin x − sin 3 x − sin 5 x − sin 7 x 3π π (3)(5) π (3)(5)(7) π (5)(7)(9) 1535
© 2014, John Bird
8 sin x sin 3 x sin 5 x sin 7 x − − − − ... f ( x) = sin 2 x = π (1)(3) (1)(3)(5) (3)(5)(7) (5)(7)(9)
i.e.
4. Determine the half-range Fourier cosine series in the range x = 0 to x = π for the function defined by:
π 0〈 x〈 x, 2 f(x) = (π − x), π 〈 x 〈 π 2
The periodic function is shown in the diagram below. Since a half-range cosine series is required, ∞
the function is symmetrical about the f(x) axis and
f ( x= ) a0 + ∑ an cos nx n =1
= a0
= an
1
π
{∫
π {∫ 2
π /2 0
π /2 0
xd x + ∫
π π /2
}
(π − x= )d x
x cos nx d x + ∫
π π /2
π /2 π x 2 1 x 2 + − π x π 2 0 2 π /2
=
1 π 2 2 π 2 π 2 π 2 − + π − − π 8 2 2 8
=
1 π 2 π 2 π 2 π 2 1 2π 2 π + − += = π 8 2 2 8 π 8 4
}
(π − x) cos nx d x
π /2 π 2 x sin nx cos nx π sin nπ x sin nx cos nx = + + − − π n n 2 0 n n 2 π /2 n
π nπ nπ sin cos 2 2 2 + 2 = n n2 π
1 cos nπ −0 + 2 +0 −0 − n n2 1536
nπ π nπ nπ sin cos π sin 2 − 2 2 − 2 − 2 n n n
© 2014, John Bird
nπ nπ nπ 2 cos sin sin π π 2 1 cos 2 n nπ π 2 + 2 − − 2= = − − 1 − cos nπ 2 cos 2 2 2 2 2 n n n n πn π n
When n is odd, an = 0
2 2 8 2 ( 2 cos π − 1 − cos 2π ) = ( −2 − 1 − 1) =− =− , 2 4π 4π π (2) π
a2 =
and
a4 =
2 2 4π ) − 1) 0 , ( 2 cos 2π − 1 − cos= ( 2 − 1= 2 16π π (4) 2 2 8 2 2 cos 3π − 1 − cos 6π ) = −2 − 1 − 1) =− =− , ( ( π (6) 2 36π 36π (3) 2 π
a6 =
a8 = 0 , 2 2 8 2 2 cos 5π − 1 − cos10π ) = −2 − 1 − 1) =− =− , and so on ( ( π (10) 2 100π 100π (5) 2 π
a10 =
Thus,
i.e.
∞ π 2 2 2 − cos 2 x − f ( x) = a0 + ∑ an cos nx = cos 6 x − cos10 x + ... 4 π (3) 2 π (5) 2 π n =1
π 2 cos 6 x cos10 x f ( x) = − cos 2 x + + + ... 2 2 4 π 3 5
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