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"questionTitle": "Piecewise function Fourier series",
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"questionContent": "Find $a_0$, $a_n$ and $b_n$ for the Fourier series of $f(x)$, which is assumed to have period $4$.\n\n  \n\n$$\nf(x)= \\begin{cases}0, & -2 \\leq x<-1 \\\\\\ \\frac{2 k}{3}, & -1 \\leq x<1 \\\\\\ -\\frac{k}{2}, & 1 \\leq x<2.\\end{cases}\n$$\n",
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"publishedPartAnswerContent": "$$\na_0=\\frac{5k}{12}\n$$\n\n  \n\n  \n\n$$\n\\begin{align*}\na_n &= \\dfrac{11k}{6n \\pi} \\sin \\left( \\dfrac{n \\pi}{2} \\right) \\\\[1em]\n &= \\dfrac{11k}{6n \\pi} \\frac{1-(-1)^n}{2} (-1)^{^{\\frac{n+3}{2}}} \\\\[1em]\n&= (-1)^{n+1}\\frac{11k}{6(2n-1)\\pi}\n\\end{align*}\n$$\n\n  \n\n  \n\n$$\n\\begin{align*}\nb_n &= \\frac{k}{2n \\pi} \\left( \\cos \\left( \\frac{n \\pi}{2} \\right) - \\cos(n \\pi) \\right) \\\\[1em]\nb_n &= \\frac{k}{2n \\pi} \\left( \\frac{1+(-1)^n}{2}(-1)^{\\frac{n}{2}} - (-1)^n \\right)\n\\end{align*}\n\n$$\n",
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"content": "Recall the Fourier series equations for period $2L$:\n\n  \n\n$$\na_0 = \\frac{1}{L} \\int_{-L}^L {f(x)} \\, \\text{d}x\n\n$$\n\n  \n\n$$\na_n = \\frac{1}{L} \\int_{-L}^L {f(x)} \\cos\\left(\\frac{n \\pi x}{L}\\right) \\, \\text{d}x\n$$\n\n  \n\n$$\nb_n = \\frac{1}{L} \\int_{-L}^L {f(x)} \\sin\\left(\\frac{n \\pi x}{L}\\right) \\, \\text{d}x\n$$\n\n***\n\n$$\n2L=4\n$$\n\n$$\nL=2\n$$\n\n***\n\n### **Finding $a_0$:**\n\n$$\na_0 = \\frac{1}{2} \\int_{-2}^2 {f(x)} \\, \\text{d}x\n$$\n\n  \n\n$$\na_0 = \\frac{1}{2} \\left( \\int_{-2}^{-1} 0 \\, \\text{d}x + \\int_{-1}^1 \\frac{2k}{3} \\, \\text{d}x + \\int_1^2 -\\frac{k}{2} \\, \\text{d}x \\right)\n\n$$\n\n  \n\n$$\na_0=\\frac{5k}{12}\n$$\n\n***\n\n### **Finding $a_n$:**\n\n$$\na_n = \\frac{1}{2} \\int_{-2}^2 {f(x)} \\cos \\left( \\frac{n \\pi x}{L} \\right) \\, \\text{d}x\n\n$$\n\n  \n\n$$\na_n = \\frac{1}{2} \\left( \\int_{-2}^{-1} 0 \\, \\text{d}x + \\int_{-1}^1 \\frac{2k}{3} \\cos \\left( \\frac{n \\pi x}{2} \\right) \\, \\text{d}x + \\int_1^2 -\\frac{k}{2} \\cos \\left( \\frac{n \\pi x}{2} \\right) \\, \\text{d}x \\right)\n\n$$\n\n  \n\nAfter evaluating the integrals, the following is obtained:\n\n  \n\n$$\na_n=\\frac{k}{3}\\left(\\frac{2}{n \\pi} \\sin \\left(\\frac{n \\pi}{2}\\right)-\\frac{2}{n \\pi} \\sin \\left(-\\frac{n \\pi}{2}\\right)\\right)-\\frac{k}{4}\\left(\\frac{2}{n \\pi} \\sin (n \\pi)-\\frac{2}{n \\pi} \\sin \\left(\\frac{n \\pi}{2}\\right)\\right)\n$$\n\n  \n\n* The second $\\sin$ term can be written as: $-\\frac{2}{n \\pi} \\sin \\left(-\\frac{n \\pi}{2}\\right)=\\frac{2}{n \\pi} \\sin \\left(\\frac{n \\pi}{2}\\right)$\n* The third $\\sin$ term is always zero.\n\n  \n\n$$\na_n=\\frac{k}{3}\\left(\\frac{4}{n \\pi} \\sin \\left(\\frac{n \\pi}{2}\\right)\\right)-\\frac{k}{4}\\left(-\\frac{2}{n \\pi} \\sin \\left(\\frac{n \\pi}{2}\\right)\\right)\n$$\n\n  \n\nSome manipulation results in:\n\n  \n\n$$\na_n = \\dfrac{11k}{6n \\pi} \\sin \\left( \\dfrac{n \\pi}{2} \\right)\n$$\n\n**Further simplification of $a_n$:**\n\nThe $\\sin \\left( \\dfrac{n \\pi}{2} \\right)$ term can be simplified by considering the pattern with increasing $n$:\n\n  \n\n$$\n\\sin\\left(\\frac{n\\pi}{2}\\right) = \\begin{cases}\n0 & \\text{if $n=0$} \\\\\n1 & \\text{if $n=1$} \\\\\n0 & \\text{if $n=2$} \\\\\n-1 & \\text{if $n=3$} \\\\\n0 & \\text{if $n=4$} \\\\\n\\vdots & \\vdots\n\\end{cases}\n$$\n\nThis can be achieved as follows:\n\n$$\n\\sin \\left( \\dfrac{n \\pi}{2} \\right)=\\frac{1-(-1)^n}{2} (-1)^{^{\\frac{n+3}{2}}}\n$$\n\n  \n\n$$\na_n = \\dfrac{11k}{6n \\pi} \\frac{1-(-1)^n}{2} (-1)^{^{\\frac{n+3}{2}}}\n$$\n\n$$\n\n\n$$\n\n  \n\nHowever, since $\\sin \\left( \\dfrac{n \\pi}{2} \\right)$ is zero for all even $n$, the expression can alternatively be written as:\n\n  \n\n$$\na_n=(-1)^{n+1}\\frac{11k}{6(2n-1)\\pi} \n$$\n\n***\n\n### **Finding $b_n$:**\n\n$$\nb_n = \\frac{1}{2} \\int_{-2}^2 {f(x)} \\sin\\left(\\frac{n \\pi x}{L}\\right) \\, \\text{d}x\n$$\n\n  \n\n$$\nb_n = \\frac{1}{2} \\left( \\int_{-2}^{-1} 0 \\, \\text{d}x + \\int_{-1}^1 \\frac{2k}{3} \\sin \\left( \\frac{n \\pi x}{2} \\right) \\, \\text{d}x + \\int_1^2 -\\frac{k}{2} \\sin \\left( \\frac{n \\pi x}{2} \\right) \\, \\text{d}x \\right)\n$$\n\n  \n\nAfter evaluating the integrals, the following is obtained:\n\n   \n\n$$\nb_n=\\frac{k}{3}\\left(-\\frac{2}{n \\pi} \\cos \\left(\\frac{n \\pi}{2}\\right)+\\frac{2}{n \\pi} \\cos \\left(\\frac{n \\pi}{2}\\right)\\right)-\\frac{k}{4}\\left(-\\frac{2}{n \\pi} \\cos (n \\pi)+\\frac{2}{n \\pi} \\cos \\left(\\frac{n \\pi}{2}\\right)\\right)\n$$\n\n(note that the second $\\cos$ term has positive argument, because $\\cos$ is an even function.)\n\n  \n\nSimplifying this expression yields the answer. Note that $\\cos(n\\pi)$ has been replaced with $(-1)^n$.\n\n  \n\n$$\nb_n = \\frac{k}{2n \\pi} \\left( \\cos \\left( \\frac{n \\pi}{2} \\right) - \\cos(n \\pi) \\right)\n$$\n\n  \n\n**Further simplification of** $b_n$**:**\n\n* The $\\cos(n\\pi)$ term can be replaced by $(-1)^n$.\n* The $\\cos\\left(\\frac{n\\pi}{2}\\right)$ term can be simplified by considering the pattern with increasing $n$:\n\n   \n\n$$\n\\cos\\left(\\frac{n\\pi}{2}\\right) = \\begin{cases}\n1 & \\text{if $n=0$} \\\\\n0 & \\text{if $n=1$} \\\\\n-1 & \\text{if $n=2$} \\\\\n0 & \\text{if $n=3$} \\\\\n1 & \\text{if $n=4$} \\\\\n\\vdots & \\vdots\n\\end{cases}\n$$\n\nThis can be achieved as follows:\n\n$$\n\\cos\\left(\\frac{n\\pi}{2}\\right) =\\frac{1+(-1)^n}{2}(-1)^{\\frac{n}{2}}\n$$\n\nFinally, this yields:\n\n  \n\n$$\nb_n = \\frac{k}{2n \\pi} \\left( \\frac{1+(-1)^n}{2}(-1)^{\\frac{n}{2}} - (-1)^n \\right)\n$$\n\n  \n\n(Note that this expression may seem less concise that simply including the cos(n\\*pi/2) but is much more desirable and efficient in a numerical algorithm).\n"
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