مواضيع المحاضرة: EVEN FUNCTION AND ODD FUNCTION
قراءة
عرض

Lecture N o.2-11 by Hussein J. Abd ulHussein

Advanced Calculus II Al Muthanna University, College of Science

1

EVEN FUNCTION AND ODD FUNCTION

(a) Even Function
A function f(x) is said to be even (or symmetric) function if , , f(−x)= f(x)
The graph of such a function is symmetrical with respect to y -axis [f(x)axis ] Here y – axis
is a mirror for the re flection of the curve
The area under such a curve from − π to π is double the area from 0 to π
∴ ∫ f (x)dx = 2∫ f (x)dx π
0 π
− π
A function is called odd (or skew symmetric) function if
f(−x)= −f(x)
Here the area under the cur ve from − π to π is zero.
∫ f (x)dx = 0
π
− π

Expansion of an even function

a0= 1
π∫ f (x)dx = 2
π∫ f (x)dx π
0 π
− π
an= 1
π∫ f (x)cos nx dx = 2
π∫ f (x)cos nx dx
π
0

π
− π

As f(x) and cos nx are bot h even function, therefore, the product of f(x) ∙cos nx is also an
even function
bn= 1
π∫ f (x)sin nx dx = 0 π
− π

As is an odd function so f(x) is also an odd function We need not to calculate bn It saves our lab
our a lot The series of the even function will contain only cosine terms
Expansion of an odd function:
a0= 1
π∫ f (x)dx = 0 π
− π
an= 1
π∫ f (x)cos nx dx = 0
π
− π
[f(x)cos nx is odd function ]

Lecture N o.2-11 by Hussein J. Abd ulHussein

Advanced Calculus II Al Muthanna University, College of Science

2

bn= 1
π∫ f (x)sin nx dx = 2
π∫ f (x)sin nx dx π
0 π
− π , [f(x)cos nx is odd function ]
The series of the odd function will contain only sine terms . The function shown below neither odd
nor even so it contains both sine and cosine terms Fo urier Series
Example 9. Find the Fourier series expansion of the periodic function of period 2π
f(x)= x2.−π ≤ x≤ π
Hence find the sum of the series 1
12− 1
22+ 1
32− 1
42+ ⋯
Solution f(x)= x2.−π ≤ x≤ π
This is an even function ∴ bn= 0
a0= 2
π∫ f (x)dx = 2
π∫ x2dx = 2
π[x3
3]
0
π π
0

π
0
= 2π2
3
an= 2
π∫ f (x)cos nx dx = 2
π∫ x2cos nx dx
π
0

π
0

= 2
π[x2[sin nx
n ]− (2x )[− cos nx
n2 ]+ (2)[− sin nx
n3 ]]0π
= 2
π[ π2sin nπ
n + 2π cos nπ
n2 − 2sin nπ
n3 ]= 4(−1)n
n2
Fourier Series is
f(x)= a0
2 + a1cos x+ a2cos 2x + a3cos 3x + ⋯ + ancos nx + ⋯
x2= π2
3 − 4[cos x
12 − cos 2
22 + cos 3
32 − cos 4
42 + ⋯ ]
On putting x= 0 . we have
0= π2
3 − 4[1
12− 1
22+ 1
32− 1
42+ ⋯ ]
⇒ 1
12− 1
22+ 1
32− 1
42+ ⋯ = π2
3

Lecture N o.2-11 by Hussein J. Abd ulHussein

Advanced Calculus II Al Muthanna University, College of Science

3

Example 10. Obtain a Fourier ex pression for

f(x)= x3.−π < x< π
Solution. f(x)= x3 is odd function
∴ a0= 0 and an= 0

bn= 2
π∫ f (x)sin nx dx = 2
π∫ x3sin nx dx = [∫ uv = uv 1− uv 2+ u′′ v3− u′′′ v4+ ⋯
π
0

π
0
]
= 2
π[x3[− cos nx
n ]− 3x2[− sin nx
n2 ]+ 6x [cos nx
n3 ]− 6[− sin nx
n4 ]] 0π
= 2
π[− π3cos nπ
n + πcos nπ
n3 ]= 2(−1)n[− π2
3 + 6
n3]
∴ x3= 2[− [− π2
1 + 6
13 ]sin x+ [− π2
2 + 6
23]sin 2x − [− π2
3 + 6
33]sin 3x + ⋯ ]
Example 11. Expand the function f(x)= xsin x. as a Fourier series in the interval −π ≤ x≤ π
Hence deduce that 1
1∙3− 1
3∙5+ 1
5∙7− 1
7∙9+ ⋯ = π−2
4
Solution. f(x)= xsin x.

a0= 2
π∫ f (x)dx = 2
π∫ xsin x dx
π
0

π
0
(Here xsin x is an even function )
= 2
π[x(− cos x)− (1)(− sin x)] 0π = 2
π(π)= 2
an= 2
π∫ f (x)cos nx dx = 2
π∫ xsin xcos nx dx
π
0

π
0

= 1
π∫ x{sin (n+ 1)x− sin (n− 1)x} dx
π
0

= 1
π∫ xsin (n+ 1)xdx − 1
π∫ x sin (n− 1)x dx
π
0

π
0

Lecture N o.2-11 by Hussein J. Abd ulHussein

Advanced Calculus II Al Muthanna University, College of Science

4

= 1
π [x (− cos (n+ 1)x
n+ 1 )
0
π
− (1){− sin (n+ 1)x
(n+ 1)2 }]
0
π

− 1
π [x (− cos (n− 1)x
n− 1 )
0
π
− (1){− sin (n− 1)x
(n− 1)2 }]
0
π

= 1
π[−π(−1)n+1
n+ 1 + 0]− 1
π[−π(−1)n−1
n− 1 − 0]

= − (−1)n+1

n+ 1 + (−1)n−1
n− 1 = (−1)n+1 [ 1
n+ 1+ 1
n− 1]= 2(−1)n+1
n2− 1
a1= 2
π∫ xsin xcos xdx
π
0
= 1
π∫ x sin2x dx
π
0

= 1
π[x(− cos 2x
2 )− (1)(− sin 2x
4 )]
0
π
= [− π
2]= − 1
2
bn= 0 [As xsin xsin nx is an odd function ]
Hence
f(x)= 1− 1
2cos x+ 2∑ (−1)n+1
n2− 1

n=2
cos nx = 1− 1
2cos x+ 2∑ (−1)n+1
(n− 1)(n+ 1)

n=2
cos nx
xsin x= 1+ 2[− 1
4cos x− 1
1∙3cos 2x + 1
2∙4cos 3x − 1
3∙5cos 4x + ⋯ ]
Putting x= π
2 . we get π
2 = 1+ 2{1
1∙3− 1
3∙5+ 1
5∙7− ⋯ }
or π
4 = 1
2+ 1
1∙3− 1
3∙5+ 1
5∙7− ⋯ ⇒ π
4− 1
2+ 1
1∙3− 1
3∙5+ 1
5∙7− ⋯

⇒ π− 2

4 = 1
1∙3− 1
3∙5+ 1
5∙7− ⋯

Lecture N o.2-11 by Hussein J. Abd ulHussein

Advanced Calculus II Al Muthanna University, College of Science

5

Example 12. Find the Fourier Series expansion for the function

f(x)= xcos x.−π < x< π
Solution Since xcos x. is an odd function therefore, a0= an= 0
Let xcos x= ∑ bn sin bx . where
bn= 2
π∫ xcos x sin nx dx
π
0
= 1
π∫ x {sin (n+ 1)x+ sin (n− 1)x}dx
π
0

= 1
π∫ x sin (n+ 1)xdx
π
0
+ 1
π∫ x+ sin (n− 1)xdx
π
0

= 1
π [x (− cos (n+ 1)x
n+ 1 )+ sin (n+ 1)x
(n+ 1)2 ]
0
π
+ 1
π[x (−xcos (n− 1)x
n− 1 )+ sin (n− 1)x
(n− 1)2 ]
0
π

= 1
π[π {cos (n+ 1)π
n+ 1 − cos (n− 1)π
n− 1 }]

⇒ bn= {− (−1)n+1

n+ 1 + (−1)n−1
n− 1 }.n≠ 1
bn= −(−1)n+1[ 1
n+ 1+ 1
n− 1]
= − { 1
n+ 1+ 1
n− 1}= −2n
n2− 1 . If n is odd ;n≠ 1
But bn= { 1
n+ 1+ 1
n− 1}= 2n
n2− 1 . If n is even ;n ≠ 1
If n= 1.then b1= 2
π∫ xcos x sin nx dx
π
0
= 1
π∫ xsin 2x dx
π
0

∴ cos x= b1sin x+ b2sin 2x + b3sin x+ ⋯

= − 1
2sin x+ 4sin 2x
22− 1 − 6sin 3x
32− 1 + ⋯


رفعت المحاضرة من قبل: Cruz Maldanado
المشاهدات: لقد قام عضو واحد فقط و 21 زائراً بقراءة هذه المحاضرة






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