8.5-bo'lim: Furye kosinus seriyasi
Ushbu bo'limda biz Furye kosinuslari seriyasini ko'rib chiqamiz. Biz oldingi bo'limda bo'lgani kabi, Furye sinus qatorini ko'rib chiqamiz. Dastlab biz ishlayotgan \(f\left( x \right)\) funksiyasi teng funktsiya deb faraz qilishdan boshlaylik (i.e. \(f\left( { - x} \right) = f\left( x \right)\)) va biz \( - L \le x \le L\) da ushbu funktsiya uchun kosinuslar (ular ham juft) bo'yicha ketma-ket tasvirni yozmoqchimiz. Boshqacha qilib aytganda, biz quyidagilarni qidiramiz:
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \]Ushbu seriya a deb ataladi Furye kosinuslar qatori va shuni esda tutingki, bu holda (Furye sinus seriyasidan farqli o'laroq) biz \(n = 0\) da ketma-ket namoyishni boshlashimiz mumkin, chunki bu atama sinuslardagi kabi nolga teng bo'lmaydi. Bundan tashqari, Furye Sinus seriyasida bo'lgani kabi, kosinuslardagi \(\frac{{n\pi x}}{L}\) argumenti biz keyingi bobda ko'rib chiqiladigan argument bo'lgani uchungina foydalaniladi. Bu erda yagona haqiqiy talab shundaki, biz foydalanadigan funktsiyalar to'plami biz ishlayotgan intervalda ortogonal bo'lishi kerak.
Shuni ham yodda tutingki, biz ushbu seriya aslida \( - L \le x \le L\)-da \(f\left( x \right)\)-ga yaqinlashadi deb taxmin qilamiz. Keyinchalik bo'limi biz ushbu seriyaning konvergentsiyasini batafsilroq ko'rib chiqamiz.
Shunday qilib, \({A_n}\) koeffitsientlari formulasini aniqlash uchun biz \(\left\{ {\cos \left( {\frac{{n\pi x}}{L}} \right)} \right\}_{n\,\, = \,\,0}^\infty \) ning \( - L \le x \le L\) oralig'ida ortogonal to'plam hosil qilishidan foydalanamiz. ko'rsatdi oldingi bo'limda. Ushbu bo'limda biz birozdan keyin kerak bo'ladigan quyidagi formulani ham oldik.
\[\int_{{ - L}}^{L}{{\cos \left( {\frac{{n\pi x}}{L}} \right)\cos \left( {\frac{{m\pi x}}{L}} \right)\,dx}} = \left\{ {\begin{array}{*{20}{l}}{2L}&{{\mbox{if }}n = m = 0}\\L&{{\mbox{if }}n = m \ne 0}\\0&{{\mbox{if }}n \ne m}\end{array}} \right.\]Biz koeffitsientlar formulasini oldingi bo'limda qilganimizdek deyarli bir xil tarzda olamiz. Biz yuqoridagi tasvirdan boshlaymiz va ikkala tomonni \(\cos \left( {\frac{{m\pi x}}{L}} \right)\) ga ko'paytiramiz, bunda \(m\) \(\left\{ {0,1,2,3, \ldots } \right\}\) diapazonidagi sobit butun sondir. Buni qilish,
\[f\left( x \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)} \]Keyinchalik, biz ikkala tomonni \(x = - L\) dan \(x = L\) ga birlashtiramiz va Furye Sinus seriyasi bilan ishlashga muvaffaq bo'lganimizdek, biz yana integral va qatorni almashtira olamiz.
\[\begin{align*}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} & = \int_{{\, - L}}^{L}{{\sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)} }}\,dx\\ & = \sum\limits_{n = 0}^\infty {{A_n}\int_{{\, - L}}^{{\,L}}{{\cos \left( {\frac{{n\,\pi x}}{L}} \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}} \end{align*}\]Endi biz \(n = m\) dan tashqari o'ng tomondagi barcha integrallar nolga teng bo'lishini bilamiz, chunki kosinuslar to'plami \( - L \le x \le L\) oralig'ida ortogonal to'plamni tashkil qiladi. Biroq, biz \(m\) (yoki siz foydalanmoqchi bo'lgan harfga qarab \(n\)) qiymatiga ehtiyot bo'lishimiz kerak. Shunday qilib, barcha integrallarni baholagandan so'ng, biz koeffitsientlar uchun quyidagi formulalar to'plamiga kelamiz.
\(m = 0\):\[\int_{{\, - L}}^{{\,L}}{{f\left( x \right)dx}} = {A_{\,0}}\left( {2L} \right)\hspace{0.25in}\,\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,\,\,\,{A_{\,0}} = \frac{1}{{2L}}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\,dx}}\] \(m \ne 0\):
\[\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = {A_{\,m}}\left( L \right)\hspace{0.25in}\,\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,\,\,\,{A_{\,m}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}\]
Hammasini umumlashtirib, juft funktsiyaning Furye kosinus qatori, \( - L \le x \le L\) da \(f\left( x \right)\) quyidagicha berilgan:
\[f\left( x \right)=\sum\limits_{n=0}^{\infty }{{{A}_{n}}\cos \left( \frac{n\,\pi x}{L} \right)} \hspace{0.25in} {{A}_{n}}=\left\{ \begin{array}{*{35}{l}} \displaystyle \frac{1}{2L}\int_{\,-L}^{\,L}{f\left( x \right)\,dx} & \,\,\,\,\,n=0 \\ \displaystyle \frac{1}{L}\int_{\,-L}^{\,L}{f\left( x \right)\cos \left( \frac{n\,\pi x}{L} \right)\,dx} & \,\,\,\,\,n\ne 0 \\ \end{array} \right.\]Nihoyat, misol bilan ishlashdan oldin, shuni ta'kidlaymizki, \(f\left( x \right)\) ham, kosinuslar ham yuqoridagi ikkala integralda juft integraldir va shuning uchun biz \({A_n}\) formulalarini quyidagicha yozishimiz mumkin,
\[{A_n} = \left\{ {\begin{array}{*{20}{l}}{\displaystyle \frac{1}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\,dx}}}&{\,\,\,\,\,n = 0}\\{\displaystyle \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}}&{\,\,\,\,\,n \ne 0}\end{array}} \right.\]Endi bir misolni ko'rib chiqaylik.
Yechim
Bu erda biz aniq bir xil funktsiyaga egamiz va shuning uchun biz qilishimiz kerak bo'lgan narsa koeffitsientlarni hisoblash va ular biroz chalkash bo'lishi mumkin, chunki biz qismlarga bo'lingan integratsiyani ikki marta bajarishimiz kerak. Tasdiqlash uchun haqiqiy integratsiya tafsilotlarini sizga qoldiramiz.
\[{A_0} = \frac{1}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\,dx}} = \frac{1}{L}\int_{{\,0}}^{{\,L}}{{{x^2}\,dx}} = \frac{1}{L}\left( {\frac{{{L^3}}}{3}} \right) = \frac{{{L^2}}}{3}\] \[\begin{align*}{A_n} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{{x^2}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\\ & = \frac{2}{L}\left. {\left( {\frac{L}{{{n^3}{\pi ^3}}}} \right)\left( {2Ln\pi x\cos \left( {\frac{{n\,\pi x}}{L}} \right) + \left( {{n^2}{\pi ^2}{x^2} - 2{L^2}} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \right)} \right|_0^L\\ & = \frac{2}{{{n^3}{\pi ^3}}}\left( {2{L^2}n\pi \cos \left( {n\,\pi } \right) + \left( {{n^2}{\pi ^2}{L^2} - 2{L^2}} \right)\sin \left( {n\,\pi } \right)} \right)\\ & = \frac{{4{L^2}{{\left( { - 1} \right)}^n}}}{{{n^2}{\pi ^2}}}\hspace{0.25in}\hspace{0.25in}n = 1,2,3, \ldots \end{align*}\]Keyin koeffitsientlar,
\[{A_0} = \frac{{{L^2}}}{3}\hspace{0.25in}{A_n} = \frac{{4{L^2}{{\left( { - 1} \right)}^n}}}{{{n^2}{\pi ^2}}},\,\,\,\,n = 1,2,3, \ldots \]Furye kosinus qatori shunday bo'ladi:
\[{x^2} = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} = {A_0} + \sum\limits_{n = 1}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} = \frac{{{L^2}}}{3} + \sum\limits_{n = 1}^\infty {\frac{{4{L^2}{{\left( { - 1} \right)}^n}}}{{{n^2}{\pi ^2}}}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \]E'tibor bering, biz \(n = 0\)-ni bu erda qilganimizdek, ko'pincha seriyadan olib tashlaymiz, chunki u deyarli har doim boshqa koeffitsientlardan farq qiladi va bu bizga koeffitsientlarni seriyaga ulash imkonini beradi.
Endi oldingi bobda qilganimizdek, juft bo'lmagan funksiyaning Furye kosinus qatorini topish uchun nima qilishimiz kerakligini so'raymiz. Furye sinus qatorida bo'lgani kabi, biz ushbu o'zgarishlarni amalga oshirganimizda, biz \( - L \le x \le L\) o'rniga hozir \(0 \le x \le L\) oralig'iga o'tishimiz kerak va biz yana bu nuqtada seriya \(f\left( x \right)\) ga yaqinlashadi deb taxmin qilamiz va bu seriyaning yaqinlashuvi muhokamasini keyingisiga qoldiramiz. bo'limi.
Biz Furye sinus qatorida bo'lgani kabi, bu erda koeffitsientlarni ikki marta topishimiz mumkin edi, ammo buning uchun hech qanday sabab yo'q. Shunday qilib, koeffitsientlar formulalarini olish uchun yuqoridagi barcha ishlarni qayta bajarishimiz mumkin bo'lsa-da, keling, to'g'ridan-to'g'ri koeffitsientlarni topishning ikkinchi usuliga o'tamiz.
Bunday holda, biz buni davom ettirishdan oldin, \( - L \le x \le L\) da \(f\left( x \right)\) funksiyasining teng kengaytmasini aniqlashimiz kerak. Shunday qilib, \(f\left( x \right)\) funktsiyasini hisobga olgan holda, biz funktsiyaning teng kengaytmasini quyidagicha aniqlaymiz:
\[g\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{f\left( x \right)}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{f\left( { - x} \right)}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\]Bu teng funktsiya ekanligini ko'rsatish juda oddiy.
\[g\left( -x \right)=f\left( -\left( -x \right) \right)=f\left( x \right)=g\left( x \right) \hspace{0.25in} \text{for }0<x<L\]va biz \(0 \le x \le L\) da \(g\left( x \right) = f\left( x \right)\) ekanligini ko'rishimiz mumkin va agar \(f\left( x \right)\) allaqachon teng funksiya bo'lsa, biz \( - L \le x \le L\) da \(g\left( x \right) = f\left( x \right)\) ni olamiz.
Keling, ba'zi funktsiyalarni ko'rib chiqaylik va funktsiyalar uchun teng kengaytmalarni chizamiz.
- \(f\left( x \right) = L - x\) on \(0 \le x \le L\)
- \(f\left( x \right) = {x^3}\) on \(0 \le x \le L\)
- \(f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{\frac{L}{2}}&{\,\,\,\,{\mbox{if }}0 \le x \le \frac{L}{2}}\\{x - \frac{L}{2}}&{\,\,\,\,{\mbox{if }}\frac{L}{2} \le x \le L}\end{array}} \right.\)
a \(f\left( x \right) = L - x\) on \(0 \le x \le L\) Yechim
Mana bu funksiyaning teng kengaytmasi.
\[\begin{align*}g\left( x \right) & = \left\{ {\begin{array}{*{20}{l}}{f\left( x \right)}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{f\left( { - x} \right)}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\\ & = \left\{ {\begin{array}{*{20}{l}}{L - x}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{L + x}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\end{align*}\]Bu erda asl funktsiyaning ham, uning tekis kengaytmasining ham grafigi keltirilgan. E'tibor bering, biz "kengaytma" ni bir tekis kengaytmani olishimiz uchun qo'shilayotgan funktsiya qismini aniq qilish uchun chiziqli chiziq bilan qo'yganmiz.
b \(f\left( x \right) = {x^3}\) on \(0 \le x \le L\) Yechim
Bu funksiyaning teng kengaytmasi,
\[\begin{align*}g\left( x \right) & = \left\{ {\begin{array}{*{20}{l}}{f\left( x \right)}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{f\left( { - x} \right)}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\\ & = \left\{ {\begin{array}{*{20}{l}}{{x^3}}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{ - {x^3}}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\end{align*}\]Funktsiyaning eskizi va juft kengaytmasi:
c \(f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{\frac{L}{2}}&{\,\,\,\,{\mbox{if }}0 \le x \le \frac{L}{2}}\\{x - \frac{L}{2}}&{\,\,\,\,{\mbox{if }}\frac{L}{2} \le x \le L}\end{array}} \right.\) Yechim
Mana bu funksiyaning teng kengaytmasi,
\[\begin{align*}g\left( x \right) & = \left\{ {\begin{array}{*{20}{l}}{f\left( x \right)}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{f\left( { - x} \right)}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\\ & = \left\{ {\begin{array}{*{20}{l}}{x - \frac{L}{2}}&{\,\,\,\,{\mbox{if }}\frac{L}{2} \le x \le L}\\{\frac{L}{2}}&{\,\,\,\,{\mbox{if }}0 \le x \le \frac{L}{2}}\\{\frac{L}{2}}&{\,\,\,\,{\mbox{if }} - \frac{L}{2} \le x \le 0}\\{ - x - \frac{L}{2}}&{\,\,\,\,{\mbox{if }} - L \le x \le - \frac{L}{2}}\end{array}} \right.\end{align*}\]Funktsiyaning eskizi va juft kengaytmasi:
Xo'sh, keling, \(0 \le x \le L\) da istalgan \(f\left( x \right)\) funksiyasining Furye kosinus qatorini topish uchun funktsiyaning teng kengaytmasidan qanday foydalanishimiz haqida o'ylab ko'raylik.
Shunday qilib, \(f\left( x \right)\) funktsiyasini hisobga olsak, biz \(g\left( x \right)\) ga yuqorida ta'riflanganidek teng kengaytma bo'lishiga ruxsat beramiz. Endi \(g\left( x \right)\) \( - L \le x \le L\) da bir tekis funksiyadir va shuning uchun biz uning Furye kosinuslar qatorini yozishimiz mumkin. Bu,
\[g\left( x \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \hspace{0.25in}{A_n} = \left\{ {\begin{array}{*{20}{l}}{\displaystyle \frac{1}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\,dx}}}&{\,\,\,\,\,n = 0}\\{\displaystyle \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}}&{\,\,\,\,\,n \ne 0}\end{array}} \right.\]va konstantalarni hisoblash uchun biz integrallarning ikkinchi shaklidan foydalanamiz.
Endi, chunki biz \(0 \le x \le L\) da bizda \(f\left( x \right) = g\left( x \right)\) borligini bilamiz va shuning uchun \(0 \le x \le L\) da \(f\left( x \right)\) ning Furye kosinus qatori ham berilgan:
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \hspace{0.25in}{A_n} = \left\{ {\begin{array}{*{20}{l}}{\displaystyle \frac{1}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\,dx}}}&{\,\,\,\,\,n = 0}\\{\displaystyle \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}}&{\,\,\,\,\,n \ne 0}\end{array}} \right.\]Keling, bir nechta misollarni ko'rib chiqaylik.
Yechim
Biz qilishimiz kerak bo'lgan yagona narsa koeffitsientlarni hisoblash, shuning uchun buning uchun ish.
\[{A_0} = \frac{1}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\,dx}} = \frac{1}{L}\int_{{\,0}}^{{\,L}}{{L - x\,dx}} = \frac{L}{2}\] \[\begin{align*}{A_n} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{\left( {L - x} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\\ & = \frac{2}{L}\left. {\left( {\frac{L}{{{n^2}{\pi ^2}}}} \right)\left( {n\pi \left( {L - x} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right) - L\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right)} \right|_0^L\\ & = \frac{2}{L}\left( {\frac{L}{{{n^2}{\pi ^2}}}} \right)\left( { - L\cos \left( {n\,\pi } \right) + L} \right) = \frac{{2L}}{{{n^2}{\pi ^2}}}\left( {1 + {{\left( { - 1} \right)}^{n + 1}}} \right)\hspace{0.25in}\hspace{0.25in}n = 1,2,3, \ldots \end{align*}\]Furye kosinus qatori shunday bo'ladi:
\[f\left( x \right) = \frac{L}{2} + \sum\limits_{n = 1}^\infty {\frac{{2L}}{{{n^2}{\pi ^2}}}\left( {1 + {{\left( { - 1} \right)}^{n + 1}}} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \]E'tibor bering, biz ushbu bo'limdagi birinchi misolda bo'lgani kabi, biz koeffitsientlarni ulashdan oldin \({A_0}\) atamasini olib tashladik.
Keyin toq funksiyaning Furye kosinus qatorini topamiz. E'tibor bering, buni amalga oshirish mumkin, chunki biz haqiqatan ham funktsiyaning teng kengaytmasining Furye kosinus qatorini topmoqdamiz.
Yechim
\({A_0}\) uchun integral etarlicha sodda, ammo qolganlari uchun integral juda tartibsiz bo'ladi, chunki u uchta qismlarga integratsiyani talab qiladi. Haqiqiy integratsiya tafsilotlarini tekshirish uchun sizga qoldiramiz. Mana ish,
\[{A_0} = \frac{1}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\,dx}} = \frac{1}{L}\int_{{\,0}}^{{\,L}}{{{x^3}\,dx}} = \frac{{{L^3}}}{4}\] \[\begin{align*}{A_n} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{{x^3}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\\ & = \frac{2}{L}\left. {\left( {\frac{L}{{{n^4}{\pi ^4}}}} \right)\left( {n\pi x\left( {{n^2}{\pi ^2}{x^2} - 6{L^2}} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right) + \left( {3L{n^2}{\pi ^2}{x^2} - 6{L^3}} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right)} \right|_0^L\\ & = \frac{2}{L}\left( {\frac{L}{{{n^4}{\pi ^4}}}} \right)\left( {n\pi L\left( {{n^2}{\pi ^2}{L^2} - 6{L^2}} \right)\sin \left( {n\,\pi } \right) + \left( {3{L^3}{n^2}{\pi ^2} - 6{L^3}} \right)\cos \left( {n\,\pi } \right) + 6{L^3}} \right)\\ & = \frac{2}{L}\left( {\frac{{3{L^4}}}{{{n^4}{\pi ^4}}}} \right)\left( {2 + \left( {{n^2}{\pi ^2} - 2} \right){{\left( { - 1} \right)}^n}} \right) = \frac{{6{L^3}}}{{{n^4}{\pi ^4}}}\left( {2 + \left( {{n^2}{\pi ^2} - 2} \right){{\left( { - 1} \right)}^n}} \right)\hspace{0.25in}\,\,\,\,\,\,\,\,n = 1,2,3, \ldots \end{align*}\]Bu funktsiya uchun Furye kosinus qatori,
\[f\left( x \right) = \frac{{{L^3}}}{4} + \sum\limits_{n = 1}^\infty {\frac{{6{L^3}}}{{{n^4}{\pi ^4}}}\left( {2 + \left( {{n^2}{\pi ^2} - 2} \right){{\left( { - 1} \right)}^n}} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \]Nihoyat, keling, qismlarga bo'lingan funksiyani tez ko'rib chiqaylik.
Yechim
Bu erda har bir koeffitsient uchun integrallarni ajratishimiz kerak. Mana koeffitsientlar.
\[\begin{align*}{A_0} & = \frac{1}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\,dx}} = \frac{1}{L}\left[ {\int_{{\,0}}^{{\,\frac{L}{2}}}{{f\left( x \right)\,dx}} + \int_{{\,\frac{L}{2}}}^{{\,L}}{{f\left( x \right)\,dx}}} \right]\\ & = \frac{1}{L}\left[ {\int_{{\,0}}^{{\,\frac{L}{2}}}{{\frac{L}{2}\,dx}} + \int_{{\,\frac{L}{2}}}^{{\,L}}{{x - \frac{L}{2}\,dx}}} \right] = \frac{1}{L}\left[ {\frac{{{L^2}}}{4} + \frac{{{L^2}}}{8}} \right] = \frac{1}{L}\left[ {\frac{{3{L^2}}}{8}} \right] = \frac{{3L}}{8}\end{align*}\]Qolgan koeffitsientlar uchun biz bajarishimiz kerak bo'lgan integraldir.
\[\begin{align*}{A_n} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} & = \frac{2}{L}\left[ {\int_{{\,0}}^{{\,\frac{L}{2}}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} + \int_{{\,\frac{L}{2}}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}} \right]\\ & = \frac{2}{L}\left[ {\int_{{\,0}}^{{\,\frac{L}{2}}}{{\frac{L}{2}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} + \int_{{\,\frac{L}{2}}}^{{\,L}}{{\left( {x - \frac{L}{2}} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}} \right]\end{align*}\]Hayotni biroz osonlashtirish uchun keling, bularning har birini alohida bajaramiz.
\[\int_{{\,0}}^{{\,\frac{L}{2}}}{{\frac{L}{2}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{L}{2}\left. {\left( {\frac{L}{{n\pi }}} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \right|_0^{\frac{L}{2}} = \frac{L}{2}\left( {\frac{L}{{n\pi }}} \right)\sin \left( {\frac{{n\,\pi }}{2}} \right) = \frac{{{L^2}}}{{2n\pi }}\sin \left( {\frac{{n\,\pi }}{2}} \right)\] \[\begin{align*}\int_{{\,\frac{L}{2}}}^{{\,L}}{{\left( {x - \frac{L}{2}} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} & = \left. {\frac{L}{{n\pi }}\left( {\frac{L}{{n\pi }}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\, + \left( {x - \frac{L}{2}} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \right)} \right|_{\frac{L}{2}}^L\\ & = \frac{L}{{n\pi }}\left( {\frac{L}{{n\pi }}\cos \left( {n\,\pi } \right)\, + \frac{L}{2}\sin \left( {n\,\pi } \right) - \frac{L}{{n\pi }}\cos \left( {\frac{{n\,\pi }}{2}} \right)} \right)\\ & = \frac{{{L^2}}}{{{n^2}{\pi ^2}}}\left( {{{\left( { - 1} \right)}^n} - \cos \left( {\frac{{n\,\pi }}{2}} \right)} \right)\end{align*}\]Bularni birlashtirib, beradi,
\[\begin{align*}{A_n} & = \frac{2}{L}\left[ {\frac{{{L^2}}}{{2n\pi }}\sin \left( {\frac{{n\,\pi }}{2}} \right) + \frac{{{L^2}}}{{{n^2}{\pi ^2}}}\left( {{{\left( { - 1} \right)}^n} - \cos \left( {\frac{{n\,\pi }}{2}} \right)} \right)} \right]\\ & = \frac{{2L}}{{{n^2}{\pi ^2}}}\left[ {{{\left( { - 1} \right)}^n} - \cos \left( {\frac{{n\,\pi }}{2}} \right) + \frac{{n\pi }}{2}\sin \left( {\frac{{n\,\pi }}{2}} \right)} \right]\end{align*}\]Shunday qilib, barcha ishlardan so'ng, Furye kosinus seriyasi,
\[f\left( x \right) = \frac{{3L}}{8} + \sum\limits_{n = 1}^\infty {\frac{{2L}}{{{n^2}{\pi ^2}}}\left[ {{{\left( { - 1} \right)}^n} - \cos \left( {\frac{{n\,\pi }}{2}} \right) + \frac{{n\pi }}{2}\sin \left( {\frac{{n\,\pi }}{2}} \right)} \right]\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \]E'tibor bering, biz Furye sinus qatorida ko'rganimizdek, ko'pgina koeffitsientlar bilan shug'ullanish juda chalkash bo'ladi.