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    Fourier

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    Hasibur Rahman
    (1) A half range Fourier series contains either only sine terms or only cosine terms. This occurs when the function is defined on half an interval and is specified as odd or even. (2) The document provides examples of finding the half range sine and cosine series for the function F(x)=x on the interval 0<x<4. (3) For the half range sine series, the coefficients are found to be bn=(-1)n+1. For the half range cosine series, the coefficients are found to be an=2(−1)n−1.

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    Fourier

    Uploaded by

    Hasibur Rahman
    12 views8 pages
    (1) A half range Fourier series contains either only sine terms or only cosine terms. This occurs when the function is defined on half an interval and is specified as odd or even. (2) The document provides examples of finding the half range sine and cosine series for the function F(x)=x on the interval 0<x<4. (3) For the half range sine series, the coefficients are found to be bn=(-1)n+1. For the half range cosine series, the coefficients are found to be an=2(−1)n−1.

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    (1) A half range Fourier series contains either only sine terms or only cosine terms. This occurs when the function is defined on half an interval and is specified as odd or even. (2) The document provides examples of finding the half range sine and cosine series for the function F(x)=x on the interval 0<x<4. (3) For the half range sine series, the coefficients are found to be bn=(-1)n+1. For the half range cosine series, the coefficients are found to be an=2(−1)n−1.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    12 views8 pages

    Fourier

    Uploaded by

    Hasibur Rahman
    (1) A half range Fourier series contains either only sine terms or only cosine terms. This occurs when the function is defined on half an interval and is specified as odd or even. (2) The document provides examples of finding the half range sine and cosine series for the function F(x)=x on the interval 0<x<4. (3) For the half range sine series, the coefficients are found to be bn=(-1)n+1. For the half range cosine series, the coefficients are found to be an=2(−1)n−1.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 8

    Lecture # 24

    Dr. Bimal Chandra Das

    Engineering Mathematics
    Fourier Series

    18/8/20 1
    Half Range Sine and Cosine Series
    Half Range Sine and Cosine Series : A half range Fourier
    sine or cosine series is a series in which only sine terms or only cosine terms
    are present respectively. When a half range series corresponding
    to a given function is desired
    (i) the function is generally defined in the interval
    (0,l)[which is half of the interval (-l,l)]
    (ii) then the function is specified as odd or even, so that it is clearly
    defined in the other half of the interval, namely ( − l,0).
    In such case, we have
    l
    2 nπ x
    an = 0 and bn = ∫ F(x)sin dx for half range sine series
    l0 l
    l
    2 nπ x
    bn = 0 and an = ∫ F(x)cos dx for half range cosine series.
    l0 l
    18/8/20 2
    Applications
    Problem: Find the Fourier series of
    (i) F(x) = x ; 0 < x < 4 in half range sine series
    (ii ) F(x) = x ; 0 < x < 4 in half range cosine series

    Solution (i): Given that, F(x) = x

    Here, 2l = 4 ∴l = 2
    We have the half range Fourier sine series,
    ∞
    nπ x
    F(x) = ∑ bn sin ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)
    n=1 l
    l
    2
    l
    nπ x 2 nπ x
    where, bn = ∫ F(x) sin dx = ∫ x sin dx
    l0 l 20 2
    18/8/20 3
    Applications
    2
    ⎡ ⎛ 2 nπ x ⎞ ⎛ 2 nπ x ⎞ ⎤
    = ⎢ x ⎜ − cos ⎟ − ∫ 1⋅ ⎜ − cos ⎟ dx ⎥
    ⎣ ⎝ nπ 2 ⎠ ⎝ nπ 2 ⎠ ⎦0
    2
    ⎡ x nπ x 2 nπ x ⎤
    = 2 ⎢ − cos + 2 2 sin ⎥
    ⎣ nπ 2 nπ 2 ⎦0
    ⎡ 2 nπ 2 ⎤ 4
    = 2 ⎢ − cos ⎥ = − cosnπ
    ⎣ nπ 2 ⎦ nπ
    4 4
    = − (−1) = (−1)n+1 [∵sinnπ = 0, cosnπ = (−1)n ]
    n

    nπ nπ

    18/8/20 4
    Applications
    Now, from (1) we have,
    ∞
    4 nπ x
    F(x) = ∑ (−1) sin
    n+1

    n=1 nπ 2
    ∞
    (−1)n+1 nπ x
    = 4∑ sin
    n=1 nπ 2
    4 ⎡ π x 1 2π x 1 3π x ⎤
    = ⎢sin − sin + sin − ⋅⋅⋅⋅⋅⋅⎥
    π⎣ 2 2 2 3 2 ⎦

    Which is the required series.

    18/8/20 5
    Applications
    Problem: Find the Fourier series of
    (i) F(x) = x ; 0 < x < 4 in half range sine series
    (ii ) F(x) = x ; 0 < x < 4 in half range cosine series

    Solution (ii): Given that, F(x) = x


    We have the half range Fourier cosine series with l = 2,
    a0 ∞
    nπ x
    F(x) = + ∑ an cos ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)
    2 n=1 l
    2
    2
    l
    2
    l
    ⎡x ⎤
    2
    where, a0 = ∫ F(x) dx = ∫ x dx = ⎢ ⎥ = 2
    l0 20 ⎣ 2 ⎦0
    l l
    2 nπ x 2 nπ x
    an = ∫ F(x) cos dx = ∫ x cos dx
    l0 l 20 2
    18/8/20 6
    Applications
    2
    ⎡ ⎛ 2 nπ x ⎞ ⎛ 2 nπ x ⎞ ⎤
    = ⎢ x ⎜ sin ⎟ − ∫ 1⋅ ⎜ sin ⎟ dx ⎥
    ⎣ ⎝ nπ 2 ⎠ ⎝ nπ 2 ⎠ ⎦0
    2
    ⎡ x nπ x 2 nπ x ⎤
    = 2 ⎢ − sin + 2 2 cos ⎥
    ⎣ nπ 2 nπ 2 ⎦0

    4 ⎡ nπ 2 ⎤ 4
    = 2 2 ⎢cos − cos0⎥ = 2 2 [cosnπ −1]
    nπ ⎣ 2 ⎦ nπ
    4
    = 2 2 [(−1)n −1] [∵sinnπ = 0, cosnπ = (−1)n ]
    nπ

    18/8/20 7
    Applications
    Now, from (1) we have,
    2 ∞ 4 nπ x
    F(x) = + ∑ 2 2 [(−1) −1]cos
    n

    2 n=1 n π 2
    4 ∞ 1 nπ x
    = 1+ 2 ∑ 2 [(−1) −1]cos
    n

    π n=1 n 2
    4⎡ πx 2 3π x 2 5π x ⎤
    = 1+ 2 ⎢ −2cos +0− 2 cos +0− 2 cos +0⋅⋅⋅⋅⋅⋅⎥
    π ⎣ 2 3 2 5 2 ⎦
    8 ⎡ πx 1 3π x 1 5π x ⎤
    = 1− 2 ⎢cos + 2 cos + 2 cos + ⋅⋅⋅⋅⋅⋅⎥
    π ⎣ 2 3 2 5 2 ⎦

    Which is the required series.


    18/8/20 8

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