8.4-bo'lim: Furye sinus seriyasi
Ushbu bo'limda biz Furye seriyasini ko'rib chiqishni boshlaymiz. Shuni ta'kidlashimiz kerakki, bu butun sinfni qamrab oladigan mavzu va biz ushbu bo'limda (shuningdek, keyingi ikki bo'limda) nima qilishimiz mavzuga juda qisqacha qarashdan boshqa narsa emas. Bu erda gap, keyingi bobda qisman differensial tenglamalar uchun ba'zi bir asosiy echimlarni amalga oshirishga imkon berish uchun etarli ish qilishdir. Furye seriyasini o'rganishda ko'plab mavzular mavjud bo'lib, biz bu erda to'xtalmaymiz.
Shunday ekan, keling, biz Furye seriyasidan boshlasak ham, boshlaylik. Buning o'rniga, biz ko'rib chiqqan Calculus kursimizga qaytaylik Teylor seriyasi. Teylor Series bilan biz \(f\left( x \right)\) funksiyasining ketma-ket tasvirini yozdik, uning shartlari ba'zi \(x = a\) uchun \(x - a\) vakolatlari edi. Ba'zi shartlar bilan biz buni ko'rsata oldik,
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}{{\left( {x - a} \right)}^n}} \]va qator funktsiyaning o'ziga bog'liq bo'lgan ba'zi \(R\) uchun \(\left| {x - a} \right| < R\) da \(f\left( x \right)\) ga yaqinlashadi.
Buning hech qanday yomon joyi yo'q, lekin bu \(x = a\) da barcha buyurtmalarning hosilalari mavjudligini talab qiladi. Yoki boshqacha qilib aytganda \({f^{\left( n \right)}}\left( a \right)\) \(n = 0,1,2,3, \ldots \) uchun mavjud Shuningdek, ba'zi funksiyalar uchun \(R\) qiymati juda kichik bo'lishi mumkin.
Ushbu ikkita masala (bir nechta boshqa masalalar bilan birga) bu har doim ham funktsiya uchun ketma-ket tasvirni yozishning eng yaxshi usuli emasligini anglatadi. Ko'p hollarda u yaxshi ishlaydi va boshqa turdagi seriyalarga ehtiyoj sezish uchun hech qanday sabab bo'lmaydi. Biroq, boshqa turdagi seriyalar afzalroq yoki talab qilinadigan holatlar mavjud.
Biz keyingi ikki bo'lim davomida funktsiya uchun muqobil seriyali tasvirni yaratamiz. Ushbu bobning qolgan qismidagi yakuniy maqsad funksiyaning sinuslar va kosinuslar nuqtai nazaridan ketma-ket tasvirini yozish bo'ladi.
Biz ishni \(f\left( x \right)\) funksiyasi g‘alati funktsiya (i.e. \(f\left( { - x} \right) = - f\left( x \right)\)). \(f\left( x \right)\) g'alati bo'lgani uchun, biz buning uchun faqat sinuslar bo'yicha ketma-ket tasvirni yozishimiz kerakligi mantiqan to'g'ri keladi (chunki ular ham g'alati funktsiyalardir).
Bu erda biz \(f\left( x \right)\) ni quyidagi ketma-ket tasvir sifatida yozishga harakat qilamiz, a deb ataladi Furye sinus qatori, on \( - L \le x \le L\).
\[\sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \]Bu erda bir nechta masalalarni e'tiborga olish kerak. Birinchidan, shu nuqtada biz ketma-ket ko'rinish \( - L \le x \le L\) da \(f\left( x \right)\) ga yaqinlashishini taxmin qilamiz. U haqiqatan ham birlashadimi yoki yo'qmi, keyinroq ko'rib chiqamiz bo'limi. Biroq, seriya \(f\left( x \right)\) ga yaqinlashadi deb faraz qilsak, Teylor seriyasidan farqli o'laroq, bu tasvir har doim bir xil oraliqda yaqinlashadi va interval funktsiyaga bog'liq emas.
Ikkinchidan, ketma-ket tasvir sinus kuchlarini o'z ichiga olmaydi (yana buni Teylor seriyasidan farqli o'laroq), aksincha, turli dalillarga ega sinuslarni o'z ichiga oladi.
Nihoyat, \(\frac{{n\pi x}}{L}\) sinuslar argumenti o'zboshimchalik bilan tanlangan va qaysidir ma'noda shunday bo'lgan g'alati tanlov kabi ko'rinishi mumkin. Furye sinus qatori uchun argument bu bo'lishi shart emas, lekin bu erda tanlov uchun bir nechta sabablar mavjud. Birinchidan, bu ba'zi asosiy qisman differensial tenglamalarni echishda yordam berish uchun Furye seriyasidan (umuman, Furye sinus qatori emas) foydalanganda keyingi bobda tabiiy ravishda paydo bo'ladigan argumentdir.
Ushbu argumentdan foydalanishning keyingi sababi shundaki, biz ishlash uchun tanlagan funktsiyalar to'plami, bu holda \(\left\{ {\sin \left( {\frac{{n\pi x}}{L}} \right)} \right\}_{n\,\, = \,\,1}^\infty \), bo'lishi kerak. ortogonal berilgan oraliqda, bu holda \( - L \le x \le L\) va oxirgi bo'limda biz e'tibor bering ko'rsatdi aslida ular shunday. Boshqacha qilib aytganda, biz ishlaydigan funktsiyalarni tanlash va biz ishlayotgan interval qandaydir tarzda bir-biriga bog'langan bo'ladi. Biz boshqacha foydalanishimiz mumkin argument, lekin sinuslar (turli argumentlar bilan) ortogonal ekanligini isbotlashimiz mumkin bo'lgan intervalni tanlashimiz kerak bo'ladi.
Shunday qilib, keling, \(f\left( x \right)\) g'alati funktsiyani hisobga olgan holda, funktsiyani \( - L \le x \le L\) da ifodalash uchun yuqorida keltirilgan shakldagi Furye sinus qatorini topishimiz mumkin deb faraz qilaylik. Bu bizda bo'ladi degan ma'noni anglatadi,
\[f\left( x \right) = \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \]Yuqorida ta'kidlab o'tilganidek, biz buni amalga oshirish mumkinmi yoki yo'qmi va ketma-ket tasvir haqiqatan ham keyingi bo'limda funktsiyaga yaqinlashsa, muhokama qilamiz. Shu nuqtada biz buni amalga oshirish mumkin deb taxmin qilamiz. Endi savol qatordagi \({B_n}\) koeffitsientlarini qanday aniqlash mumkin.
Keling, yuqoridagi qatordan boshlaylik va ikkala tomonni \(\sin \left( {\frac{{m\pi x}}{L}} \right)\) ga ko'paytiramiz, bunda \(m\) \(\left\{ {1,2,3, \ldots } \right\}\) diapazonidagi sobit butun sondir. Boshqacha qilib aytganda, biz ikkala tomonni bu erda ishlayotgan sinuslar to'plamidagi har qanday sinusga ko'paytiramiz. Buni qilish,
\[f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right) = \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)} \]Keling, buning ikkala tomonini \(x = - L\) dan \(x = L\) ga birlashtiraylik.
\[\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = \int_{{\, - L}}^{{\,L}}{{\sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)} \,dx}}\]Ayni paytda bizda hal qilish uchun kichik muammo bor. Biz hisob-kitoblardan bilamizki, chekli qatorning integrali (ko‘pincha chekli yig‘indi deb ataladi….) bo‘laklar integrallarining (cheklangan) yig‘indisidan boshqa narsa emas. Boshqacha qilib aytganda, chekli qatorlar uchun biz integral va qatorni almashtira olamiz. Cheksiz qatorlar uchun biz har doim ham buni qila olmaymiz. Cheksiz qatorlarning ayrim integrallari uchun biz integral va qatorni almashtira olmaymiz. Yaxshiyamki, biz uchun bu holatda integral va qatorni o'zaro almashtira olamiz. Buni amalga oshirish va integraldan \({B_n}\) doimiysini faktorlarga ajratish,
\[\begin{align*}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} &= \sum\limits_{n = 1}^\infty {\int_{{\, - L}}^{{\,L}}{{{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}} \\ & = \sum\limits_{n = 1}^\infty {{B_n}\int_{{\, - L}}^{{\,L}}{{\sin \left( {\frac{{n\,\pi x}}{L}} \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}} \end{align*}\]Endi biz oxirgi qismdan eslaymiz isbotladi \(\left\{ {\sin \left( {\frac{{n\pi x}}{L}} \right)} \right\}_{n\,\, = \,\,1}^\infty \) \( - L \le x \le L\) da ortogonal va bu,
\[\int_{{ - L}}^{L}{{\sin \left( {\frac{{n\pi x}}{L}} \right)\sin \left( {\frac{{m\pi x}}{L}} \right)\,dx}} = \left\{ {\begin{array}{*{20}{l}}L&{{\mbox{if }}n = m}\\0&{{\mbox{if }}n \ne m}\end{array}} \right.\]Xo'sh, bu biz uchun nimani anglatadi. Biz ketma-ket \(n\) ning turli qiymatlari bilan ishlaganimizda va integrallarning qiymatini hisoblaganimizda, integrallarning biridan boshqasi nolga teng bo'ladi. Bizda \(n = m\) bo'lganda yagona nolga teng bo'lmagan integral keladi, bu holda integral \(L\) qiymatiga ega bo'ladi. Shunday qilib, seriyadagi yagona nolga teng bo'lmagan atama bizda \(n = m\) bo'lganda keladi va bizning tenglamamiz:
\[\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = {B_{\,m}}L\]Va nihoyat, biz qilishimiz kerak bo'lgan yagona narsa \(L\) ga bo'linish va endi bizda har bir koeffitsient uchun tenglama mavjud.
\[{B_{\,m}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}\hspace{0.25in}m = 1,2,3, \ldots \]Keyin shuni yodda tutingki, biz ikkita g'alati funktsiyani birlashtirganimiz sababli, bu integralning integrali juft va shuning uchun biz buni ham bilamiz,
\[{B_{\,m}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}\hspace{0.25in}m = 1,2,3, \ldots \]Barcha bu ishlarni umumlashtirib, \( - L \le x \le L\) da \(f\left( x \right)\) toq funksiyasining Furye sinus qatori quyidagicha berilgan:
\[\begin{align*}f\left( x \right) & = \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} & {B_{\,n}} & = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \\ & & & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \end{align*}\]Keling, bir misolni tez ko'rib chiqaylik.
Yechim
Avval shuni yodda tutingki, biz ishlayotgan funktsiya aslida g'alati funktsiyadir va shuning uchun bu biz qila oladigan narsadir. Bu erda, albatta, \(f\left( x \right) = x\) uchun koeffitsientlarni hisoblashdan boshqa qiladigan ko'p narsa yo'q.
Mana o'sha ish va biz tekshirish uchun qismlar bo'yicha integratsiyani sizga qoldiramiz. \(n\), \(L\) va \(\pi \) doimiy ekanligini unutmang!
\[\begin{align*}{B_{\,n}} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{x\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{2}{L}\left. {\left( {\frac{L}{{{n^2}{\pi ^2}}}} \right)\left( {L\sin \left( {\frac{{n\,\pi x}}{L}} \right) - n\pi x\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right)} \right|_0^L\\ & = \frac{2}{{{n^2}{\pi ^2}}}\left( {L\sin \left( {n\,\pi } \right) - n\pi L\cos \left( {n\,\pi } \right)} \right)\end{align*}\]Ushbu integrallar, ba'zan, ma'lum bir raqam o'rniga intervalning so'nggi nuqtalari uchun umumiy \(L\) dan foydalansak, biroz chalkash bo'lishi mumkin.
Endi \(n\) butun son ekanligidan foydalanib, biz \(\sin \left( {n\pi } \right) = 0\) va \(\cos \left( {n\pi } \right) = {\left( { - 1} \right)^n}\) ekanligini bilamiz. Shuning uchun bizda,
\[{B_{\,n}} = \frac{2}{{{n^2}{\pi ^2}}}\left( { - n\pi L{{\left( { - 1} \right)}^n}} \right) = \frac{{{{\left( { - 1} \right)}^{n + 1}}2L}}{{n\pi }}\hspace{0.25in}\hspace{0.25in}n = 1,2,3 \ldots \]Furye sinus qatori shunday bo'ladi:
\[x = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}2L}}{{n\pi }}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} = \frac{{2L}}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \]Shu o‘rinda shuni ta’kidlashimiz kerakki, biz bu yerda o‘z ishimizning barchasini bo‘lmasa ham, ko‘p qismini yakuniy nuqtalar uchun ma’lum raqamlarga ega intervallar o‘rniga umumiy intervalda (\( - L \le x \le L\) yoki \(0 \le x \le L\)) bajaramiz. Buning bir nechta sabablari bor. Birinchidan, u har doim yoqimli bo'lgan ushbu shaklning istalgan oralig'ida ishlaydigan ancha umumiy formulani beradi. Ikkinchidan, biz keyingi bobda bunday ish bilan shug'ullanganimizda, u ham umumiy intervallarda bo'ladi, shuning uchun biz ularga endi ko'nikishimiz mumkin.
Endi, toq funktsiyaning Furye sinus qatorini topish yaxshi va yaxshi, lekin agar biz biron sababga ko'ra g'alati bo'lmagan funksiya uchun Furye sinus qatorini topmoqchi bo'lsak-chi? Buni qanday qilishni ko'rish uchun biz o'zgartirish kiritishimiz kerak. Yuqoridagi ish \( - L \le x \le L\) intervalida bajarildi. Agar funktsiya g'alati bo'lmasa, biz \(0 \le x \le L\) oralig'ida ishlaymiz. Buning sababi birozdan keyin ma'lum bo'ladi.
Shunday qilib, endi biz \(f\left( x \right)\) uchun \(0 \le x \le L\) oralig'ida ketma-ket tasvirni topishga harakat qilamiz, bu shaklda,
\[\sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \]yoki boshqacha aytganda,
\[f\left( x \right) = \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \]\( - L \le x \le L\)-da Furye sinus seriyasi bilan qilganimizdek, seriyalar aslida \(f\left( x \right)\)-ga yaqinlashadi deb taxmin qilamiz va biz seriyaning yaqinlashuvini keyinroq muhokama qilishni to'xtatamiz. bo'limi.
Koeffitsientlar uchun formulalar yaratishning ikkita usuli mavjud, \({B_n}\), garchi biz birozdan keyin ularning haqiqatan ham bir xil ekanligini, faqat turli nuqtai nazardan ko'rib chiqilishini ko'ramiz.
Birinchi usul - \(f\left( x \right)\) g'alati emasligiga e'tibor bermaslik va biz yuqorida qilganimizdek davom etamiz, faqat bu safar biz bundan foydalanamiz. isbotladi oldingi bo'limda \(\left\{ {\sin \left( {\frac{{n\pi x}}{L}} \right)} \right\}_{n\,\, = \,\,1}^\infty \) shuningdek \(0 \le x \le L\) da ortogonal to'plamni tashkil qiladi va bu,
\[\int_{0}^{L}{{\sin \left( {\frac{{n\pi x}}{L}} \right)\sin \left( {\frac{{m\pi x}}{L}} \right)\,dx}} = \left\{ {\begin{array}{*{20}{l}}{\frac{L}{2}}&{{\mbox{if }}n = m}\\0&{{\mbox{if }}n \ne m}\end{array}} \right.\]Shunday qilib, agar biz buni qilsak, biz qilishimiz kerak bo'lgan narsa seriyamizning ikkala tomonini \(\sin \left( {\frac{{m\pi x}}{L}} \right)\) ga ko'paytirish, 0 dan \(L\) ga integrallash va integral va qatorlarni almashish,
\[\int_{0}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = \sum\limits_{n = 1}^\infty {{B_n}\int_{0}^{{\,L}}{{\sin \left( {\frac{{n\,\pi x}}{L}} \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}} \]Endi integralni ulab, biz erishamiz,
\[\int_{0}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = {B_m}\left( {\frac{L}{2}} \right)\]Koeffitsientni yechib, biz erishamiz,
\[{B_m} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,m = 1,2,3, \ldots \]E'tibor bering, bu koeffitsientlarning ikkinchi shakli bilan bir xil bo'lib, biz yuqorida \(f\left( x \right)\) g'alati va \( - L \le x \le L\) oralig'ida ishlagan deb hisoblaymiz. Biz aslida bir xil koeffitsientlarga erishganimiz ajablanarli emas, chunki biz koeffitsientlarni yaratishning ikkinchi usulini ko'rib chiqqandan keyin ko'ramiz.
Koeffitsientlarni hosil qilishning ikkinchi usulini ko'rib chiqishdan oldin yana bir kontseptsiyani qisqacha ko'rib chiqishimiz kerak. \(f\left( x \right)\) funksiyasini hisobga olib, biz ni aniqlaymiz g'alati kengaytma yangi funktsiya bo'lishi uchun \(f\left( x \right)\),
\[g\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{f\left( x \right)}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{ - f\left( { - x} \right)}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\]Bu g'alati funktsiya ekanligini ko'rish juda oson.
\[g\left( { - x} \right) = - f\left( { - \left( { - x} \right)} \right) = - f\left( x \right) = - g\left( x \right)\hspace{0.25in}\hspace{0.25in}{\mbox{for }}0 < x < L\]va biz \(0 \le x \le L\) da bizda \(g\left( x \right) = f\left( x \right)\) borligini ham bilishimiz mumkin. Shuni ham yodda tutingki, agar \(f\left( x \right)\) allaqachon g'alati funktsiya bo'lsa, biz \( - L \le x \le L\) da \(g\left( x \right) = f\left( x \right)\) ni olamiz.
Davom etishdan oldin bir nechta g'alati kengaytmalarni ko'rib chiqaylik.
- \(f\left( x \right) = L - x\) on \(0 \le x \le L\)
- \(f\left( x \right) = 1 + {x^2}\) on \(0 \le x \le L\)
- \(f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{\frac{L}{2}}&{\,\,\,\,{\mbox{if }}0 \le x \le \frac{L}{2}}\\{x - \frac{L}{2}}&{\,\,\,\,{\mbox{if }}\frac{L}{2} \le x \le L}\end{array}} \right.\)
Munozarani ko'rsatish
Bular bilan g'alati kengaytmani aniqlab, keyin uni eskiz qilishdan ko'ra ko'p narsa yo'q.
a \(f\left( x \right) = L - x\) on \(0 \le x \le L\) Yechim
Mana bu funktsiyaning g'alati kengaytmasi.
\[\begin{align*}g\left( x \right) & = \left\{ {\begin{array}{*{20}{l}}{f\left( x \right)}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{ - f\left( { - x} \right)}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\\ & = \left\{ {\begin{array}{*{20}{l}}{L - x}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{ - L - x}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\end{align*}\]Quyida funksiya va uning toq kengaytmasining grafigi keltirilgan. E'tibor bering, biz "kengaytma" ni to'q kengaytmani olishimizga imkon berish uchun qo'shilayotgan funktsiya qismini aniq qilish uchun chiziqli chiziq bilan qo'yganmiz.
b \(f\left( x \right) = 1 + {x^2}\) on \(0 \le x \le L\) Yechim
Birinchidan, bu aniq bir tekis funktsiya ekanligini unutmang. Biroq, bu biz uning g'alati kengaytmasini aniqlay olmaymiz degani emas. Ushbu funktsiyaning g'alati kengaytmasi:
\[\begin{align*}g\left( x \right) & = \left\{ {\begin{array}{*{20}{l}}{f\left( x \right)}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{ - f\left( { - x} \right)}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\\ & = \left\{ {\begin{array}{*{20}{l}}{1 + {x^2}}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{ - 1 - {x^2}}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right.\end{align*}\]Asl funktsiyaning eskizi va uning g'alati kengaytmasi,
c \(\displaystyle f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{\frac{L}{2}}&{\,\,\,\,{\mbox{if }}0 \le x \le \frac{L}{2}}\\{x - \frac{L}{2}}&{\,\,\,\,{\mbox{if }}\frac{L}{2} \le x \le L}\end{array}} \right.\) Yechim
Keling, avval ushbu funktsiya uchun g'alati kengaytmani yozamiz.
\[g\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{f\left( x \right)}&{\,\,\,\,{\mbox{if }}0 \le x \le L}\\{ - f\left( { - x} \right)}&{\,\,\,\,{\mbox{if }} - L \le x \le 0}\end{array}} \right. = \left\{ {\begin{array}{*{20}{l}}{x - \frac{L}{2}}&{\,\,\,\,{\mbox{if }}\frac{L}{2} \le x \le L}\\{\frac{L}{2}}&{\,\,\,\,{\mbox{if }}0 \le x \le \frac{L}{2}}\\{ - \frac{L}{2}}&{\,\,\,\,{\mbox{if }} - \frac{L}{2} \le x \le 0}\\{x + \frac{L}{2}}&{\,\,\,\,{\mbox{if }} - L \le x \le - \frac{L}{2}}\end{array}} \right.\]Asl funktsiyaning eskizi va uning g'alati kengaytmasi:
G'alati kengaytmaning ta'rifi (va bir nechta misollar) bilan biz endi \(0 \le x \le L\) da \(f\left( x \right)\) funktsiyasi uchun Furye sinus qatorining koeffitsientlari formulalarini olishning ikkinchi usulini ko'rib chiqishimiz mumkin. Birinchidan, bunday funktsiyani hisobga olgan holda, yuqoridagi kabi uning g'alati kengaytmasini aniqlang. Shu nuqtada, \(g\left( x \right)\) g'alati funktsiya bo'lganligi sababli, biz \( - L \le x \le L\) da \(g\left( x \right)\) uchun Furye sinus seriyasi (va hali \(f\left( x \right)\) EMAS) ekanligini bilamiz,
\[g\left( x \right) = \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \,\hspace{0.25in}\hspace{0.25in}{B_{\,n}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{g\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \]Biroq, biz \(g\left( x \right) = f\left( x \right)\)-ni \(0 \le x \le L\)-da bilganimiz sababli, biz \(0 \le x \le L\)-da ekanmiz, bizda ham borligini ko'rishimiz mumkin,
\[f\left( x \right) = \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \,\hspace{0.25in}\hspace{0.25in}{B_{\,n}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,n = 1,2,3, \ldots \]Shunday qilib, formulaga qanday kelganimizdan qat'i nazar, koeffitsientlar uchun aynan bir xil formula va ikkinchi usul, ular nima uchun bu erda bir xil ekanligini asoslab beradi, biz ularni Furye sinus qatori uchun g'alati funktsiya uchun olganimizda bo'lgani kabi.
Keling, 2-misolda ko'rib chiqqan har bir funksiya uchun Furye sinus qatorini topamiz.
Shuni esda tutingki, biz ushbu shaklning istalgan oralig'i uchun umumiy formulani olish uchun to'g'ri so'nggi nuqta uchun aniq raqamlar o'rniga yana umumiy intervallar ustida ishlamoqdamiz va bu keyingi bobda yana bir turdagi ishdir.
Shuningdek, biz integratsiya tafsilotlarini tekshirish uchun yana sizga qoldiramiz. Aksariyat hollarda bu integralda ko'rsatilgan barcha konstantalar (\(n\), \(L\), \(\pi \) va boshqalar) bilan murakkab bo'lgan qismlar bilan juda oddiy integratsiyani o'z ichiga oladi.
Yechim
Bu erda koeffitsientlarni hisoblashdan boshqa qiladigan ko'p narsa yo'q, shuning uchun ular mana,
\[\begin{align*}{B_{\,n}} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{\left( {L - x} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\\ & = \frac{2}{L}\left. {\left( { - \frac{L}{{{n^2}{\pi ^2}}}} \right)\left[ {L\sin \left( {\frac{{n\,\pi x}}{L}} \right) - n\pi \left( {x - L} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right]} \right|_0^L\\ & = \frac{2}{L}\left[ {\frac{{{L^2}}}{{{n^2}{\pi ^2}}}\left( {n\pi - \sin \left( {n\pi } \right)} \right)} \right] = \frac{{2L}}{{n\pi }}\end{align*}\]Soddalashtirish jarayonida \(n\) butun son ekanligini unutmang.
Shunday qilib, koeffitsientlar bilan biz ushbu funktsiya uchun quyidagi Furye sinus qatorini olamiz.
\[f\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{{2L}}{{n\pi }}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \,\]Keyingi misolda shuni ta'kidlash qiziqki, biz ushbu bo'limni faqat toq funktsiyalarni ko'rib chiqishni boshlaganimizda, endi \(0 \le x \le L\) da juft funktsiyaning Furye sinus qatorini topamiz. Shuni esda tutingki, biz haqiqatan ham ushbu funktsiyaning g'alati kengaytmasining Furye sinus qatorini topmoqdamiz va shuning uchun biz yaxshimiz.
Yechim
Bunday holda koeffitsientlar biroz chalkash bo'lishi mumkin, chunki integrallar qismlarga ikki marta integratsiyani o'z ichiga oladi. Mana koeffitsientlar uchun ish.
\[\begin{align*}{B_{\,n}} & = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{\left( {1 + {x^2}} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\\ & = \frac{2}{L}\left( {\frac{L}{{{n^3}{\pi ^3}}}} \right)\left[ {\left( {2{L^2} - {n^2}{\pi ^2}\left( {1 + {x^2}} \right)} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right) + 2Ln\pi x\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \right]_0^L\\ & = \frac{2}{L}\left( {\frac{L}{{{n^3}{\pi ^3}}}} \right)\left[ {\left( {2{L^2} - {n^2}{\pi ^2}\left( {1 + {L^2}} \right)} \right)\cos \left( {n\pi } \right) + 2{L^2}n\pi \sin \left( {n\pi } \right) - \left( {2{L^2} - {n^2}{\pi ^2}} \right)} \right]\\ & = \frac{2}{{{n^3}{\pi ^3}}}\left[ {\left( {2{L^2} - {n^2}{\pi ^2}\left( {1 + {L^2}} \right)} \right){{\left( { - 1} \right)}^n} - 2{L^2} + {n^2}{\pi ^2}} \right]\end{align*}\]Yuqorida ta'kidlab o'tilganidek, koeffitsientlar eng yoqimli emas, lekin ular mavjud. Bu funktsiya uchun Furye sinus qatori,
\[f\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{2}{{{n^3}{\pi ^3}}}\left[ {\left( {2{L^2} - {n^2}{\pi ^2}\left( {1 + {L^2}} \right)} \right){{\left( { - 1} \right)}^n} - 2{L^2} + {n^2}{\pi ^2}} \right]\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \,\]Ushbu bo'limning oxirgi misolida biz qismlarga bo'lingan funktsiyaning Furye sinus qatorini topamiz va integrallarni biroz murakkablashtirishi mumkin, ammo ular vaqti-vaqti bilan paydo bo'ladi va shuning uchun biz ular bilan shug'ullanishimiz kerak.
Yechim
Bu erda koeffitsientlar uchun integral.
\[\begin{align*}{B_{\,n}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} &= \frac{2}{L}\left[ {\int_{{\,0}}^{{\,\frac{L}{2}}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} + \int_{{\,\frac{L}{2}}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}} \right]\\ & = \frac{2}{L}\left[ {\int_{{\,0}}^{{\,\frac{L}{2}}}{{\frac{L}{2}\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} + \int_{{\,\frac{L}{2}}}^{{\,L}}{{\left( {x - \frac{L}{2}} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}} \right]\end{align*}\]E'tibor bering, biz asl funktsiyaning bo'lak-bo'lak tabiati tufayli integralni ajratishimiz kerak. Keling, ikkita integralni alohida bajaramiz
\[\int_{{\,0}}^{{\,\frac{L}{2}}}{{\frac{L}{2}\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = - \left. {\left( {\frac{L}{2}} \right)\left( {\frac{L}{{n\pi }}} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right|_0^{\frac{L}{2}} = \frac{{{L^2}}}{{2n\pi }}\left( {1 - \cos \left( {\frac{{n\,\pi }}{2}} \right)} \right)\] \[\begin{align*}\int_{{\,\frac{L}{2}}}^{{\,L}}{{\left( {x - \frac{L}{2}} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} & = \left. {\frac{L}{{{n^2}{\pi ^2}}}\left[ {L\sin \left( {\frac{{n\,\pi x}}{L}} \right) - n\pi \left( {x - \frac{L}{2}} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right]} \right|_{\frac{L}{2}}^L\\ & = \frac{L}{{{n^2}{\pi ^2}}}\left[ {L\sin \left( {n\pi } \right) - \frac{{n\pi L}}{2}\cos \left( {n\pi } \right) - L\sin \left( {\frac{{n\,\pi }}{2}} \right)} \right]\\ & = - \frac{{{L^2}}}{{{n^2}{\pi ^2}}}\left[ {\frac{{n\pi {{\left( { - 1} \right)}^n}}}{2} + \sin \left( {\frac{{n\,\pi }}{2}} \right)} \right]\end{align*}\]Bularning barchasini birlashtirganda,
\[\begin{align*}{B_{\,n}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} & = \frac{2}{L}\left( {\frac{{{L^2}}}{{2n\pi }}} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 1}} - \cos \left( {\frac{{n\,\pi }}{2}} \right) + \frac{2}{{n\pi }}\sin \left( {\frac{{n\,\pi }}{2}} \right)} \right]\\ & = \frac{L}{{n\pi }}\left[ {1 + {{\left( { - 1} \right)}^{n + 1}} - \cos \left( {\frac{{n\,\pi }}{2}} \right) - \frac{2}{{n\pi }}\sin \left( {\frac{{n\,\pi }}{2}} \right)} \right]\end{align*}\]Demak, bu funksiya uchun Furye sinus qatori,
\[f\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{L}{{n\pi }}\left[ {1 + {{\left( { - 1} \right)}^{n + 1}} - \cos \left( {\frac{{n\,\pi }}{2}} \right) - \frac{2}{{n\pi }}\sin \left( {\frac{{n\,\pi }}{2}} \right)} \right]\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \,\]Oldingi ikkita misol ko'rsatganidek, bular uchun koeffitsientlar juda chalkash bo'lishi mumkin, lekin bu ko'pincha shunday bo'ladi va shuning uchun bizni haddan tashqari hayajonlantirishga yo'l qo'ymasligimiz kerak.