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== 4.4#h(0.6em)Sine and Cosine Series


    
=== 4.4.1:Even Periodic Functions


    You may have noticed by now that an odd function has no cosine terms in the Fourier series and an even function has no sine terms in the Fourier series. This observation is not a coincidence. Let us look at even and odd periodic function in more detail.


    Recall that a function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is odd if #math.equation(block: false, alt: "f open parenthesis minus t close parenthesis equals minus f open parenthesis t close parenthesis")[$f ( − t ) = − f ( t )$]. A function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is even if #math.equation(block: false, alt: "f open parenthesis minus t close parenthesis equals f open parenthesis t close parenthesis")[$f ( − t ) = f ( t )$]. For example, #math.equation(block: false, alt: "cos open parenthesis n t close parenthesis")[$cos ( n t )$] is even and #math.equation(block: false, alt: "sin open parenthesis n t close parenthesis")[$sin ( n t )$] is odd. Similarly the function #math.equation(block: false, alt: "t to the power k")[$t^(k)$] is even if #math.equation(block: false, alt: "k")[$k$] is even and odd when #math.equation(block: false, alt: "k")[$k$] is odd.


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take two functions #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] and #math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$] and define their product #math.equation(block: false, alt: "h open parenthesis t close parenthesis equals f open parenthesis t close parenthesis g open parenthesis t close parenthesis")[$h ( t ) = f ( t ) g ( t )$].


      + Suppose both #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] and #math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$] are odd, is #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$] odd or even?
+ Suppose one is even and one is odd, is #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$] odd or even?
+ Suppose both are even, is #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$] odd or even?
]


    If #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] and #math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$] are both odd, then #math.equation(block: false, alt: "f open parenthesis t close parenthesis plus g open parenthesis t close parenthesis")[$f ( t ) + g ( t )$] is odd. Similarly for even functions. On the other hand, if #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is odd and #math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$] even, then we cannot say anything about the sum #math.equation(block: false, alt: "f open parenthesis t close parenthesis plus g open parenthesis t close parenthesis")[$f ( t ) + g ( t )$]. In fact, the Fourier series of any function is a sum of an odd (the sine terms) and an even (the cosine terms) function.


    In this section we consider odd and even periodic functions. We have previously defined the #math.equation(block: false, alt: "2 L")[$2 L$]-periodic extension of a function defined on the interval #math.equation(block: false, alt: "[ minus L , L ]")[$[ − L , L ]$]. Sometimes we are only interested in the function on the range #math.equation(block: false, alt: "[ 0 , L ]")[$[ 0 , L ]$] and it would be convenient to have an odd (resp. even) function. If the function is odd (resp. even), all the cosine (resp. sine) terms will disappear. What we will do is take the odd (resp. even) extension of the function to #math.equation(block: false, alt: "[ minus L , L ]")[$[ − L , L ]$] and then extend periodically to a #math.equation(block: false, alt: "2 L")[$2 L$]-periodic function.


    Take a function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] defined on #math.equation(block: false, alt: "[ 0 , L ]")[$[ 0 , L ]$]. On #math.equation(block: false, alt: "open parenthesis minus L , L ]")[$( − L , L ]$] define the functions


    #math.equation(block: true, alt: "F sub o d d open parenthesis t close parenthesis equals d e f { f open parenthesis t close parenthesis, i f, 0 less than or equal to t less than or equal to L ,; minus f open parenthesis minus t close parenthesis, i f, minus L less than t less than 0 ,; F sub e v e n open parenthesis t close parenthesis equals d e f { f open parenthesis t close parenthesis, i f, 0 less than or equal to t less than or equal to L ,; f open parenthesis minus t close parenthesis, i f, minus L less than t less than 0 .")[$F_(upright(o) upright(d) upright(d)) ( t ) limits(=)^(upright(d) upright(e) upright(f)) \{ f ( t ) & i f & 0 ≤ t ≤ L , \ − f ( − t ) & i f & − L < t < 0 , \ F_(upright(e) upright(v) upright(e) upright(n)) ( t ) limits(=)^(upright(d) upright(e) upright(f)) \{ f ( t ) & i f & 0 ≤ t ≤ L , \ f ( − t ) & i f & − L < t < 0 .$]


    Extend #math.equation(block: false, alt: "F sub o d d open parenthesis t close parenthesis")[$F_(o d d) ( t )$] and #math.equation(block: false, alt: "F sub e v e n open parenthesis t close parenthesis")[$F_(e v e n) ( t )$] to be #math.equation(block: false, alt: "2 L")[$2 L$]-periodic. Then #math.equation(block: false, alt: "F sub o d d open parenthesis t close parenthesis")[$F_(o d d) ( t )$] is called the #emph[odd periodic extension] of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$], and #math.equation(block: false, alt: "F sub e v e n open parenthesis t close parenthesis")[$F_(e v e n) ( t )$] is called the #emph[even periodic extension] of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]. For the odd extension we generally assume that #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis equals f open parenthesis L close parenthesis equals 0")[$f ( 0 ) = f ( L ) = 0$].


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Check that #math.equation(block: false, alt: "F sub o d d open parenthesis t close parenthesis")[$F_(o d d) ( t )$] is odd and that #math.equation(block: false, alt: "F sub e v e n open parenthesis t close parenthesis")[$F_(e v e n) ( t )$] is even. For #math.equation(block: false, alt: "F sub o d d")[$F_(o d d)$], assume #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis equals f open parenthesis L close parenthesis equals 0")[$f ( 0 ) = f ( L ) = 0$].
]


    #examplebox("Example 1")[][
Take the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t open parenthesis 1 minus t close parenthesis")[$f ( t ) = t ( 1 − t )$] defined on #math.equation(block: false, alt: "[ 0 , 1 ]")[$[ 0 , 1 ]$]. Figure #math.equation(block: false, alt: "1")[$1$] shows the plots of the odd and even extensions of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].


        #figure(figph[Two plots over t from -2 to 2 comparing the odd and even extensions: on the left the humps alternate between 0.25 and -0.25, crossing zero smoothly; on the right all humps are positive, meeting the axis in sharp corners.], alt: "Two plots over t from -2 to 2 comparing the odd and even extensions: on the left the humps alternate between 0.25 and -0.25, crossing zero smoothly; on the right all humps are positive, meeting the axis in sharp corners.", caption: [Figure #math.equation(block: false, alt: "1")[$1$]: Odd and even #math.equation(block: false, alt: "2")[$2$]-periodic extension of #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t open parenthesis 1 minus t close parenthesis")[$f ( t ) = t ( 1 − t )$], #math.equation(block: false, alt: "0 less than or equal to t less than or equal to 1")[$0 ≤ t ≤ 1$].])
]


    
=== Sine and Cosine Series


    Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] be an odd #math.equation(block: false, alt: "2 L")[$2 L$]-periodic function. We write the Fourier series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]. First, we compute the coefficients #math.equation(block: false, alt: "a sub n")[$a_(n)$] (including #math.equation(block: false, alt: "n equals 0")[$n = 0$]) and get


    #math.equation(block: true, alt: "a sub n equals the fraction 1 over L ∫ minus L L f open parenthesis t close parenthesis cos open parenthesis the fraction n π over L t close parenthesis d t equals 0 .")[$a_(n) = frac(1, L) ∫_(− L)^(L) f ( t ) "cos" ( frac(n π, L) t ) d t = 0 .$]


    That is, there are no cosine terms in the Fourier series of an odd function. The integral is zero because #math.equation(block: false, alt: "f open parenthesis t close parenthesis cos open parenthesis the fraction n π over L t close parenthesis")[$f ( t ) cos ( frac(n π, L) t )$] is an odd function (product of an odd and an even function is odd) and the integral of an odd function over a symmetric interval is always zero. The integral of an even function over a symmetric interval #math.equation(block: false, alt: "[ minus L , L ]")[$[ − L , L ]$] is twice the integral of the function over the interval #math.equation(block: false, alt: "[ 0 , L ]")[$[ 0 , L ]$]. The function #math.equation(block: false, alt: "f open parenthesis t close parenthesis sin open parenthesis the fraction n π over L t close parenthesis")[$f ( t ) sin ( frac(n π, L) t )$] is the product of two odd functions and hence is even.


    #math.equation(block: true, alt: "b sub n equals the fraction 1 over L ∫ minus L L f open parenthesis t close parenthesis sin open parenthesis the fraction n π over L t close parenthesis d t equals the fraction 2 over L ∫ 0 L f open parenthesis t close parenthesis sin open parenthesis the fraction n π over L t close parenthesis d t .")[$b_(n) = frac(1, L) ∫_(− L)^(L) f ( t ) sin ( frac(n π, L) t ) d t = frac(2, L) ∫_(0)^(L) f ( t ) sin ( frac(n π, L) t ) d t .$]


    We now write the Fourier series of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] as


    #math.equation(block: true, alt: "∑ n equals 1 infinity b sub n sin open parenthesis the fraction n π over L t close parenthesis .")[$∑_(n = 1)^(∞) b_(n) sin ( frac(n π, L) t ) .$]


    Similarly, if #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is an even #math.equation(block: false, alt: "2 L")[$2 L$]-periodic function. For the same exact reasons as above, we find that #math.equation(block: false, alt: "b sub n equals 0")[$b_(n) = 0$] and


    #math.equation(block: true, alt: "a sub n equals the fraction 2 over L ∫ 0 L f open parenthesis t close parenthesis cos open parenthesis the fraction n π over L t close parenthesis d t .")[$a_(n) = frac(2, L) ∫_(0)^(L) f ( t ) cos ( frac(n π, L) t ) d t .$]


    The formula still works for #math.equation(block: false, alt: "n equals 0")[$n = 0$], in which case it becomes


    #math.equation(block: true, alt: "a sub 0 equals the fraction 2 over L ∫ 0 L f open parenthesis t close parenthesis d t .")[$a_(0) = frac(2, L) ∫_(0)^(L) f ( t ) d t .$]


    The Fourier series is then


    #math.equation(block: true, alt: "the fraction a sub 0 over 2 plus ∑ n equals 1 infinity a sub n cos open parenthesis the fraction n π over L t close parenthesis .")[$frac(a_(0), 2) + ∑_(n = 1)^(∞) a_(n) cos ( frac(n π, L) t ) .$]


    An interesting consequence is that the coefficients of the Fourier series of an odd (or even) function can be computed by just integrating over the half interval #math.equation(block: false, alt: "[ 0 , L ]")[$[ 0 , L ]$]. Therefore, we can compute the Fourier series of the odd (or even) extension of a function by computing certain integrals over the interval where the original function is defined.


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] be a piecewise smooth function defined on #math.equation(block: false, alt: "[ 0 , L ]")[$[ 0 , L ]$]. Then the odd periodic extension of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] has the Fourier series


      #math.equation(block: true, alt: "F sub o d d open parenthesis t close parenthesis equals ∑ n equals 1 infinity b sub n sin open parenthesis the fraction n π over L t close parenthesis ,")[$F_(o d d) ( t ) = ∑_(n = 1)^(∞) b_(n) sin ( frac(n π, L) t ) ,$]


      where


      #math.equation(block: true, alt: "b sub n equals the fraction 2 over L ∫ 0 L f open parenthesis t close parenthesis sin open parenthesis the fraction n π over L t close parenthesis d t .")[$b_(n) = frac(2, L) ∫_(0)^(L) f ( t ) sin ( frac(n π, L) t ) d t .$]


      The even periodic extension of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] has the Fourier series


      #math.equation(block: true, alt: "F sub e v e n open parenthesis t close parenthesis equals the fraction a sub 0 over 2 plus ∑ n equals 1 infinity a sub n cos open parenthesis the fraction n π over L t close parenthesis ,")[$F_(e v e n) ( t ) = frac(a_(0), 2) + ∑_(n = 1)^(∞) a_(n) cos ( frac(n π, L) t ) ,$]


      where


      #math.equation(block: true, alt: "a sub n equals the fraction 2 over L ∫ 0 L f open parenthesis t close parenthesis cos open parenthesis the fraction n π over L t close parenthesis d t .")[$a_(n) = frac(2, L) ∫_(0)^(L) f ( t ) cos ( frac(n π, L) t ) d t .$]
]


    The series #math.equation(block: false, alt: "∑ n equals 1 infinity b sub n sin open parenthesis the fraction n π over L t close parenthesis")[$∑_(n = 1)^(∞) b_(n) sin ( frac(n π, L) t )$] is called the #emph[sine series] of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] and the series #math.equation(block: false, alt: "the fraction a sub 0 over 2 plus ∑ n equals 1 infinity a sub n cos open parenthesis the fraction n π over L t close parenthesis")[$frac(a_(0), 2) + ∑_(n = 1)^(∞) a_(n) cos ( frac(n π, L) t )$] is called the #emph[cosine series] of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]. We often do not actually care what happens outside of #math.equation(block: false, alt: "[ 0 , L ]")[$[ 0 , L ]$]. In this case, we pick whichever series fits our problem better.


    It is not necessary to start with the full Fourier series to obtain the sine and cosine series. The sine series is really the eigenfunction expansion of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] using eigenfunctions of the eigenvalue problem #math.equation(block: false, alt: "x double prime plus λ x equals 0")[$x^(″) + λ x = 0$] #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$], #math.equation(block: false, alt: "x open parenthesis L close parenthesis equals L")[$x ( L ) = L$]. The cosine series is the eigenfunction expansion of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] using eigenfunctions of the eigenvalue problem #math.equation(block: false, alt: "x double prime plus λ x equals 0")[$x^(″) + λ x = 0$], #math.equation(block: false, alt: "x prime open parenthesis 0 close parenthesis equals 0")[$x^(′) ( 0 ) = 0$], #math.equation(block: false, alt: "x prime open parenthesis L close parenthesis equals L")[$x^(′) ( L ) = L$]. We could have, therefore, gotten the same formulas by defining the inner produ


    #math.equation(block: true, alt: "⟨ f open parenthesis t close parenthesis , g open parenthesis y close parenthesis ⟩ equals ∫ 0 L f open parenthesis t close parenthesis g open parenthesis t close parenthesis d t ,")[$⟨ f ( t ) , g ( y ) ⟩ = ∫_(0)^(L) f ( t ) g ( t ) d t ,$]


    and following the procedure of #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/4%3A_Fourier_series_and_PDEs/4.02%3A_The_trigonometric_series")[Section 4.2]. This point of view is useful, as we commonly use a specific series that arose because our underlying question led to a certain eigenvalue problem. If the eigenvalue problem is not one of the three we covered so far, you can still do an eigenfunction expansion, generalizing the results of this chapter. We will deal with such a generalization in #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/5%3A_Eigenvalue_problems")[Chapter 5].


    #examplebox("Example 2")[][
Find the Fourier series of the even periodic extension of the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t squared")[$f ( t ) = t^(2)$] for #math.equation(block: false, alt: "0 less than or equal to t less than or equal to π")[$0 ≤ t ≤ π$].


      
      #solutionbox[
We want to write #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals the fraction a sub 0 over 2 plus ∑ n equals 1 infinity a sub n cos open parenthesis n t close parenthesis ,")[$f ( t ) = frac(a_(0), 2) + ∑_(n = 1)^(∞) a_(n) cos ( n t ) ,$] where #math.equation(block: true, alt: "a sub 0 equals the fraction 2 over π ∫ 0 π t squared d t equals the fraction 2 π squared over 3 ,")[$a_(0) = frac(2, π) ∫_(0)^(π) t^(2) d t = frac(2 π^(2), 3) ,$] and


        #math.equation(block: true, alt: "a sub n equals the fraction 2 over π ∫ 0 π t squared cos open parenthesis n t close parenthesis d t equals the fraction 2 over π [ t squared the fraction 1 over n sin open parenthesis n t close parenthesis ] 0 π minus the fraction 4 over n π ∫ 0 π t sin open parenthesis n t close parenthesis d t; equals the fraction 4 over n squared π [ t cos open parenthesis n t close parenthesis ] 0 π plus the fraction 4 over n squared π ∫ 0 π cos open parenthesis n t close parenthesis d t equals the fraction 4 open parenthesis minus 1 close parenthesis to the power n over n squared .")[$a_(n) = frac(2, π) ∫_(0)^(π) t^(2) cos ( n t ) d t = frac(2, π) attach([ t^(2) frac(1, n) sin ( n t ) ], t: π, b: 0) − frac(4, n π) ∫_(0)^(π) t sin ( n t ) d t \ = frac(4, n^(2) π) attach([ t cos ( n t ) ], t: π, b: 0) + frac(4, n^(2) π) ∫_(0)^(π) cos ( n t ) d t = frac(4 ( − 1 )^(n), n^(2)) .$]


        Note that we have “detected” the continuity of the extension since the coefficients decay as #math.equation(block: false, alt: "the fraction 1 over n squared")[$frac(1, n^(2))$]. That is, the even extension of #math.equation(block: false, alt: "t squared")[$t^(2)$] has no jump discontinuities. It does have corners, since the derivative, which is an odd function and a sine series, has jumps; it has a Fourier series whose coefficients decay only as #math.equation(block: false, alt: "the fraction 1 over n")[$frac(1, n)$].


        Explicitly, the first few terms of the series are


        #math.equation(block: true, alt: "the fraction π squared over 3 minus 4 cos open parenthesis t close parenthesis plus cos open parenthesis 2 t close parenthesis minus the fraction 4 over 9 cos open parenthesis 3 t close parenthesis plus ⋯")[$frac(π^(2), 3) − 4 cos ( t ) + cos ( 2 t ) − frac(4, 9) cos ( 3 t ) + ⋯$]
]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
+ Compute the derivative of the even extension of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] above and verify it has jump discontinuities. Use the actual definition of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$], not its cosine series!
+ Why is it that the derivative of the even extension of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is the odd extension of #math.equation(block: false, alt: "f prime open parenthesis t close parenthesis")[$f^(′) ( t )$]?
]


    
=== Application


    Fourier series ties in to the boundary value problems we studied earlier. Let us see this connection in more detail.


    Suppose we have the boundary value problem for #math.equation(block: false, alt: "0 less than t less than L")[$0 < t < L$].


    #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus λ x open parenthesis t close parenthesis equals f open parenthesis t close parenthesis ,")[$x^(″) ( t ) + λ x ( t ) = f ( t ) ,$]


    for the #emph[Dirichlet boundary conditions] #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0 , x open parenthesis L close parenthesis equals 0")[$x ( 0 ) = 0 , x ( L ) = 0$]. By using the Fredholm alternative (Theorem 4.1.2) we note that as long as #math.equation(block: false, alt: "λ")[$λ$] is #emph[not]an eigenvalue of the underlying homogeneous problem, there exists a unique solution. Note that the eigenfunctions of this eigenvalue problem are the functions #math.equation(block: false, alt: "sin open parenthesis the fraction n π over L t close parenthesis")[$sin ( frac(n π, L) t )$]. Therefore, to find the solution, we first find the Fourier sine series for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]. We write #math.equation(block: false, alt: "x")[$x$] also as a sine series, but with unknown coefficients. We substitute the series for #math.equation(block: false, alt: "x")[$x$] into the equation and solve for the unknown coefficients. If we have the #emph[Neumann boundary conditions] #math.equation(block: false, alt: "x prime open parenthesis 0 close parenthesis equals 0")[$x^(′) ( 0 ) = 0$] and #math.equation(block: false, alt: "x prime open parenthesis L close parenthesis equals 0")[$x^(′) ( L ) = 0$], we do the same procedure using the cosine series.


    Let us see how this method works on examples.


    #examplebox("Example 3")[][
Take the boundary value problem for #math.equation(block: false, alt: "0 less than t less than 1")[$0 < t < 1$],


        #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus 2 x open parenthesis t close parenthesis equals f open parenthesis t close parenthesis ,")[$x^(″) ( t ) + 2 x ( t ) = f ( t ) ,$]


        where #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t")[$f ( t ) = t$] on #math.equation(block: false, alt: "0 less than t less than 1")[$0 < t < 1$], and satisfying the #strong[Dirichlet boundary] conditions #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$] and #math.equation(block: false, alt: "x open parenthesis 1 close parenthesis equals 0")[$x ( 1 ) = 0$]. We write #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] as a sine series


        #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals ∑ n equals 1 infinity c sub n sin open parenthesis n π t close parenthesis .")[$f ( t ) = ∑_(n = 1)^(∞) c_(n) sin ( n π t ) .$]


        Compute


        #math.equation(block: true, alt: "c sub n equals 2 ∫ 0 1 t sin open parenthesis n π t close parenthesis d t equals the fraction 2 open parenthesis minus 1 close parenthesis to the power n plus 1 over n π .")[$c_(n) = 2 ∫_(0)^(1) t sin ( n π t ) d t = frac(2 ( − 1 )^(n + 1), n π) .$]


        We write #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] as


        #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals ∑ n equals 1 infinity b sub n sin open parenthesis n π t close parenthesis .")[$x ( t ) = ∑_(n = 1)^(∞) b_(n) sin ( n π t ) .$]


        We plug in to obtain


        #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus 2 x open parenthesis t close parenthesis equals ∑ n equals 1 infinity minus b sub n n squared π squared sin open parenthesis n π t close parenthesis ⏟ x double prime plus 2 ∑ n equals 1 infinity b sub n sin open parenthesis n π t close parenthesis ⏟ x; equals ∑ n equals 1 infinity b sub n open parenthesis 2 minus n squared π squared close parenthesis sin open parenthesis n π t close parenthesis; equals f open parenthesis t close parenthesis equals ∑ n equals 1 infinity the fraction 2 open parenthesis minus 1 close parenthesis to the power n plus 1 over n π sin open parenthesis n π t close parenthesis .")[$x^(″) ( t ) + 2 x ( t ) = limits(underbrace(∑_(n = 1)^(∞) − b_(n) n^(2) π^(2) sin ( n π t )))_(x^(″)) + 2 limits(underbrace(∑_(n = 1)^(∞) b_(n) sin ( n π t )))_(x) \ = ∑_(n = 1)^(∞) b_(n) ( 2 − n^(2) π^(2) ) sin ( n π t ) \ = f ( t ) = ∑_(n = 1)^(∞) frac(2 attach(( − 1 ), t: n + 1), n π) sin ( n π t ) .$]


        Therefore, #math.equation(block: true, alt: "b sub n open parenthesis 2 minus n squared π squared close parenthesis equals the fraction 2 open parenthesis minus 1 close parenthesis to the power n plus 1 over n π .")[$b_(n) ( 2 − n^(2) π^(2) ) = frac(2 ( − 1 )^(n + 1), n π) .$] or #math.equation(block: true, alt: "b sub n equals the fraction 2 open parenthesis minus 1 close parenthesis to the power n plus 1 over n π open parenthesis 2 minus n squared π squared close parenthesis .")[$b_(n) = frac(2 ( − 1 )^(n + 1), n π ( 2 − n^(2) π^(2) )) .$]


        That #math.equation(block: false, alt: "2 minus n squared π squared")[$2 − n^(2) π^(2)$] is not zero for any #math.equation(block: false, alt: "n")[$n$], and that we can solve for #math.equation(block: false, alt: "b sub n")[$b_(n)$], is precisely because #math.equation(block: false, alt: "2")[$2$] is not an eigenvale of the problem. We have thus obtained a Fourier series for the solution


        #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals ∑ n equals 1 infinity the fraction 2 open parenthesis minus 1 close parenthesis to the power n plus 1 over n π open parenthesis 2 minus n squared π squared close parenthesis sin open parenthesis n π t close parenthesis .")[$x ( t ) = ∑_(n = 1)^(∞) frac(2 ( − 1 )^(n + 1), n π ( 2 − n^(2) π^(2) )) sin ( n π t ) .$]


        See Figure #math.equation(block: false, alt: "2")[$2$] for a graph of the solution. Notice that because the eigenfunctions satisfy the boundary conditions, and #math.equation(block: false, alt: "x")[$x$] is written in terms of the boundary conditions, then #math.equation(block: false, alt: "x")[$x$] satisfies the boundary conditions.


        #figure(figph[Plot of the solution over x from 0 to 1: the curve leaves 0, sags into a single smooth trough with a minimum of about -0.081 near x equal to 0.55, then returns to 0 at x equal to 1.], alt: "Plot of the solution over x from 0 to 1: the curve leaves 0, sags into a single smooth trough with a minimum of about -0.081 near x equal to 0.55, then returns to 0 at x equal to 1.", caption: [Figure #math.equation(block: false, alt: "2")[$2$]: Plot of the solution of #math.equation(block: false, alt: "x double prime plus 2 x equals t")[$x^(″) + 2 x = t$], #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$], #math.equation(block: false, alt: "x open parenthesis 1 close parenthesis equals 0")[$x ( 1 ) = 0$].])
]


    #examplebox("Example 4")[][
Similarly we handle the Neumann conditions. Take the boundary value problem for #math.equation(block: false, alt: "0 less than t less than 1")[$0 < t < 1$],


        #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus 2 x open parenthesis t close parenthesis equals f open parenthesis t close parenthesis ,")[$x^(″) ( t ) + 2 x ( t ) = f ( t ) ,$]


        where again #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t")[$f ( t ) = t$] on #math.equation(block: false, alt: "0 less than t less than 1")[$0 < t < 1$], but now satisfying the Neumann boundary conditions #math.equation(block: false, alt: "x prime open parenthesis 0 close parenthesis equals 0")[$x^(′) ( 0 ) = 0$] and #math.equation(block: false, alt: "x prime open parenthesis 1 close parenthesis equals 0")[$x^(′) ( 1 ) = 0$]. We write #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] as a cosine series


        #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals the fraction c sub 0 over 2 plus ∑ n equals 1 infinity c sub n cos open parenthesis n π t close parenthesis ,")[$f ( t ) = frac(c_(0), 2) + ∑_(n = 1)^(∞) c_(n) cos ( n π t ) ,$]


        where


        #math.equation(block: true, alt: "c sub 0 equals 2 ∫ 0 1 t d t equals 1")[$c_(0) = 2 ∫_(0)^(1) t d t = 1$]


        and


        #math.equation(block: true, alt: "the fraction minus 4 over π squared n squared i f n o d d ,; 0 i f n e v e n .")[$frac(− 4, π^(2) n^(2)) upright(i) upright(f) " " italic(n) " " upright(o) upright(d) upright(d) , \ 0 upright(i) upright(f) " " italic(n) " " upright(e) upright(v) upright(e) upright(n) .$]


        We also write #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] as a cosine series


        #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals the fraction a sub 0 over 2 plus ∑ n equals 1 infinity a sub n cos open parenthesis n π t close parenthesis .")[$x ( t ) = frac(a_(0), 2) + ∑_(n = 1)^(∞) a_(n) cos ( n π t ) .$]


        We plug in to obtain


        #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus 2 x open parenthesis t close parenthesis equals ∑ n equals 1 infinity [ minus a sub n n squared π squared cos open parenthesis n π t close parenthesis ] plus a sub 0 plus 2 ∑ n equals 1 infinity [ a sub n cos open parenthesis n π t close parenthesis ]; equals a sub 0 plus ∑ n equals 1 infinity a sub n open parenthesis 2 minus n squared π squared close parenthesis cos open parenthesis n π t close parenthesis; equals f open parenthesis t close parenthesis equals the fraction 1 over 2 plus ∑ n equals 1 n o d d infinity the fraction minus 4 over π squared n squared cos open parenthesis n π t close parenthesis .")[$x^(″) ( t ) + 2 x ( t ) = ∑_(n = 1)^(∞) [ − a_(n) n^(2) π^(2) cos ( n π t ) ] + a_(0) + 2 ∑_(n = 1)^(∞) [ a_(n) cos ( n π t ) ] \ = a_(0) + ∑_(n = 1)^(∞) a_(n) ( 2 − n^(2) π^(2) ) cos ( n π t ) \ = f ( t ) = frac(1, 2) + ∑_(limits(n = 1)_(n " " o d d))^(∞) frac(− 4, π^(2) n^(2)) cos ( n π t ) .$]


        Therefore, #math.equation(block: false, alt: "a equals the fraction 1 over 2")[$a = frac(1, 2)$] and #math.equation(block: false, alt: "a sub n equals 0")[$a_(n) = 0$] for #math.equation(block: false, alt: "n")[$n$] even (#math.equation(block: false, alt: "n greater than or equal to 2")[$n ≥ 2$]) and for #math.equation(block: false, alt: "n")[$n$] odd we have


        #math.equation(block: true, alt: "a sub n open parenthesis 2 minus n squared π squared close parenthesis equals the fraction minus 4 over π squared n squared ,")[$a_(n) ( 2 − n^(2) π^(2) ) = frac(− 4, π^(2) n^(2)) ,$]


        or


        #math.equation(block: true, alt: "a sub n equals the fraction minus 4 over n squared π squared open parenthesis 2 minus n squared π squared close parenthesis .")[$a_(n) = frac(− 4, n^(2) π^(2) ( 2 − n^(2) π^(2) )) .$]


        The Fourier series for the solution #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] is


        #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals the fraction 1 over 4 plus ∑ n equals 1 n o d d infinity the fraction minus 4 over n squared π squared open parenthesis 2 minus n squared π squared close parenthesis cos open parenthesis n π t close parenthesis .")[$x ( t ) = frac(1, 4) + ∑_(limits(n = 1)_(n " " o d d))^(∞) frac(− 4, n^(2) π^(2) ( 2 − n^(2) π^(2) )) cos ( n π t ) .$]
]