HW #06
Due: 10/27/2025
1.
password2.jpg
What is the 10 digit WiFi password? Penalty if you actually compute the integral.


2.
By expanding x4 by the Fourier series, obtain

∞
∑
n=1 
 1
n4
 .
Hint: You can use the result of
∞
∑
n=1 
 1
n2
from #14 lecture
Integrate with respect to
from to .

Examples: Cos[x], Sin[m x], Exp[x], Pi, Infinity, Tan[x] etc..



3.
Solve the following differential equation using the Fourier series and plot the first three non-zero terms over −π < x < π.

u"(x) = f(x),     u(−π)=0,     u(π) = 0,
where
f(x) = ⎧
⎪
⎨
⎪
⎩
−1
    −π < x < 0
1
    0 < x < π.
Solution



    1. Note that x3 cos [(x)/2] √{4−x2} is an odd function so its integral from −2 to 2 is zero. The integration of √{4−x2} from -2 to 2 is the upper half of a circle centered at 0 with a radius 2. Hence it is 2 ×2 ×π/2 = 2π.

    3141592653

    2.
    By expanding x4 by a Fourier series, obtain
    ∞
    ∑
    n=1 
    		 1
    n4
    Solution
    a0 = 1
    π
    		 ⌠
    ⌡     		π

    −π 
    		 x4 dx = 2 π4
    5
    		 ,

    am = 1
    π
    		 ⌠
    ⌡     		π

    −π 
    		 x4 cos m x dx = − 48 (−1)m
    m4
    		 + 8 π2 (−1)m
    m2
    so
    f(x) = π4
    5
    		 + ∞
    ∑
    m=1 
    		 ⎛
    ⎝     		− 48 (−1)m
    m4
    		 + 8 π2 (−1)m
    m2
    		⎞
    ⎠     		cos m x.
    Substitute x = π in the both sides gives
    π4
    =
    π4
    5
    		 + ∞
    ∑
    m=1 
    		 ⎛
    ⎝     		− 48
    m4
    		 + 8 π2
    m2
    		⎞
    ⎠
    		(1)
    =
    π4
    5
    		 − 48 ∞
    ∑
    m=1 
    		 1
    m4
    		 + 8 π2 ∞
    ∑
    m=1 
    		 1
    m2
    		(2)
    Using
    ∞
    ∑
    m=1 
    		 1
    m2
    		 = π2
    6
    		 ,
    the above can be solved for
    ∞
    ∑
    m=1 
    		 1
    m4
    		 = π4
    90
    		 .

    3,

    f(x) = ⎧
    ⎪
    ⎨
    ⎪
    ⎩
    −1
        −π < x < 0
    1
        0 < x < π
    		(3)

    bm
    =
    1
    π
    		 ⌠
    ⌡     		π

    −π 
    		 f(x) sin  mx dx
    =
    1
    π
    		 ⌠
    ⌡     		0

    −π 
    		 (−1) sin  mx dx + 1
    π
    		 ⌠
    ⌡     		π

    0 
    		 (+1) sin  mx dx
    =
    2(1−(−1)m)
    m π
    		 .
    		(4)
    so
    f(x) = ∞
    ∑
    m=1 
    		 2(1−(−1)m)
    m π
    		 sin m x.
    Express the unknown function, u(x), and the known function by Fourier series as
    u(x) = ∞
    ∑
    m=1 
    		 um sin m x,
    then,
    u"(x)
    =
    ∞
    ∑
    m=1 
    		 −m2  um  sin m x
    f(x)
    =
    ∞
    ∑
    m=1 
    		 2(1−(−1)m)
    m π
    		 sin m x.
    		(5)
    from which
    um = − 2(1−(−1)m)
    m3 π
    		 .
    		(6)
    Therefore,
    u(x) = ∞
    ∑
    m=1 
    		 − 2(1−(−1)m)
    m3 π
    		 sin m x.
    The first three non-zeo terms of the Fourier series is
    ~
    u
     
    3 
    		 = − 4 sin(x)
    π
    		 − 4 sin(3 x)
    27 π
    		 − 4 sin(5 x)
    125 π
    		 .
    		(7)
    hw_05_sol_1.jpg


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On 29 Oct 2025, 11:35.