8.6-bo'lim: Furye qatorlari
Oldingi ikki bo'limda biz alohida Furye sinuslar va Furye kosinuslar qatorlarini ko'rib chiqdik. Endi to'liq Furye qatori ni (Fourier series) o'rganish vaqti keldi. Furye qatorida biz berilgan \( - L \le x \le L\) oraliqda aniqlangan \(f(x)\) funksiyani quyidagi umumiy ko'rinishda yozishga harakat qilamiz:
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \]Demak, to'liq Furye qatori ma'lum ma'noda Furye sinus va Furye kosinus qatorlarining umumiy yig'indisidir. Xuddi oldingi bo'limlardagi kabi, hozircha ushbu qator berilgan oraliqda \(f(x)\) ga haqiqatan ham yaqinlashadimi yoki yo'qmi degan masalaga to'xtalmaymiz. Hozircha u yaqinlashadi deb hisoblaymiz, uning yaqinlashish shartlarini esa keyingi bo'limda batafsil ko'rib chiqamiz.
\({A_n}\) va \({B_n}\) koeffitsiyentlarini aniqlash formulalari ham oldingi bo'limlardagidek usul bilan chiqariladi. Buning uchun \(\left\{ {\cos \left( {\frac{{n\pi x}}{L}} \right)} \right\}_{n = 0}^\infty \) va \(\left\{ {\sin \left( {\frac{{n\pi x}}{L}} \right)} \right\}_{n = 1}^\infty \) funksiyalar sistemasining \( - L \le x \le L\) oraliqda o'zaro ortogonal ekanligidan foydalanamiz. Ushbu ortogonallikdan kelib chiqadigan asosiy integral formulalarini esga olaylik:
\[\begin{align*}& \int_{{ - L}}^{L}{{\cos \left( {\frac{{n\pi x}}{L}} \right)\cos \left( {\frac{{m\pi x}}{L}} \right)\,dx}} = \left\{ {\begin{array}{*{20}{l}}{2L}&{{\mbox{agar }}n = m = 0{\mbox{ bo'lsa}}}\\L&{{\mbox{agar }}n = m \ne 0{\mbox{ bo'lsa}}}\\0&{{\mbox{agar }}n \ne m{\mbox{ bo'lsa}}}\end{array}} \right.\\ & \\ & \int_{{ - L}}^{L}{{\sin \left( {\frac{{n\pi x}}{L}} \right)\sin \left( {\frac{{m\pi x}}{L}} \right)\,dx}} = \left\{ {\begin{array}{*{20}{l}}L&{{\mbox{agar }}n = m{\mbox{ bo'lsa}}}\\0&{{\mbox{agar }}n \ne m{\mbox{ bo'lsa}}}\end{array}} \right.\\ & \\ & \int_{{ - L}}^{L}{{\sin \left( {\frac{{n\pi x}}{L}} \right)\cos \left( {\frac{{m\pi x}}{L}} \right)\,dx}} = 0\end{align*}\]Dastlab qatorning har ikkala tomonini \(\cos \left( {\frac{{m\pi x}}{L}} \right)\) ga ko'paytirib, \( - L\) dan \(L\) gacha integrallaymiz:
\[\int_{{\, - L}}^{L}{{f\left( x \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = \int_{{ - L}}^{L}{{\sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)} \,dx}} + \int_{{ - L}}^{L}{{\sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)} }}\,dx\]Oldingilardagi kabi integral va cheksiz yig'indi amallarining o'rnini almashtiramiz:
\[\int_{{\, - L}}^{L}{{f\left( x \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = \sum\limits_{n = 0}^\infty {{A_n}\int_{{ - L}}^{L}{{\cos \left( {\frac{{n\,\pi x}}{L}} \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}} + \sum\limits_{n = 1}^\infty {{B_n}\int_{{ - L}}^{L}{{\sin \left( {\frac{{n\,\pi x}}{L}} \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)}}\,dx} \]Endi sinus va kosinuslarning ortogonalligidan foydalanamiz. Ikkinchi qatordagi barcha integrallar nolga aylanadi (chunki sinus va kosinus ko'paytmasining integrali doimo nolga teng). Birinchi qatorda esa faqat \(n = m\) bo'lgandagina integral nolga teng bo'lmaydi, qolgan barcha hollarda nolga teng. Natijada:
\[\int_{{\, - L}}^{L}{{f\left( x \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = \left\{ {\begin{array}{*{20}{c}}{{A_m}\left( {2L} \right)}&{{\mbox{agar }}n = m = 0{\mbox{ bo'lsa}}}\\{{A_m}\left( L \right)}&{{\mbox{agar }}n = m \ne 0{\mbox{ bo'lsa}}}\end{array}} \right.\]Bundan \({A_m}\) koeffitsiyentlarni topsak:
\[\begin{align*}& {A_0} = \frac{1}{{2L}}\int_{{\, - L}}^{L}{{f\left( x \right)\,dx}}\\ & {A_m} = \frac{1}{L}\int_{{\, - L}}^{L}{{f\left( x \right)\cos \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}\hspace{0.25in}\hspace{0.25in}m = 1,2,3, \ldots \end{align*}\]Endi aynan shu jarayonni qaytadan bajaramiz: qatorning har ikkala tomonini \(\sin \left( {\frac{{m\pi x}}{L}} \right)\) ga ko'paytirib, \( - L\) dan \(L\) gacha integrallaymiz va integral bilan yig'indi o'rnini almashtiramiz:
\[\int_{{\, - L}}^{L}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = \sum\limits_{n = 0}^\infty {{A_n}\int_{{ - L}}^{L}{{\cos \left( {\frac{{n\,\pi x}}{L}} \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}} + \sum\limits_{n = 1}^\infty {{B_n}\int_{{ - L}}^{L}{{\sin \left( {\frac{{n\,\pi x}}{L}} \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)}}\,dx} \]Bu holda birinchi qatordagi integral nolga aylanadi, ikkinchi qatorda esa faqat \(n = m\) bo'lgandagina integral \(L\) ga teng bo'lib qoladi:
\[\int_{{\, - L}}^{L}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}} = {B_m}\left( L \right)\]Bundan \({B_m}\) ni aniqlaymiz:
\[{B_m} = \frac{1}{L}\int_{{\, - L}}^{L}{{f\left( x \right)\sin \left( {\frac{{m\,\pi x}}{L}} \right)\,dx}}\hspace{0.25in}\hspace{0.25in}m = 1,2,3, \ldots \]Oldingi ikki bo'limda integral ostidagi funksiya juft bo'lganidan foydalanib, \(0\) dan \(L\) gacha oraliqdagi ikkinchi formulani chiqargan edik. Ammo umumiy holda berilgan \(f(x)\) funksiyaning juft yoki toqligi oldindan ma'lum emas (aksariyat hollarda u juft ham, toq ham bo'lmaydi). Shuning uchun to'liq Furye qatori koeffitsiyentlari uchun ushbu formulalar yagona hisoblanadi.
Misollarni boshlashdan oldin, hisoblashlarda doimo juda ko'p ishlatiladigan muhim munosabatlarni eslatib o'tamiz. Ixtiyoriy butun \(n\) soni uchun:
\[\cos \left( {n\pi } \right) = {\left( { - 1} \right)^n}\hspace{0.25in}\hspace{0.25in}\sin \left( {n\pi } \right) = 0\]Biz ishlaydigan barcha masalalarda \(n\) butun son bo'ladi, shuning uchun bu qiymatlardan qo'shimcha izohsiz to'g'ridan-to'g'ri foydalanamiz.
Shuningdek, sinus toq funksiya (\(\sin(-x) = -\sin(x)\)) va kosinus juft funksiya (\(\cos(-x) = \cos(x)\)) ekanligini unutmang. Integrallarni hisoblashda ushbu xossalardan ham keng foydalanamiz.
Endi misollarni qadamma-qadam ko'rib chiqaylik.
Yechim
Koeffitsiyentlarni formulalar bo'yicha ketma-ket hisoblaymiz:
\[{A_0} = \frac{1}{{2L}}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\,dx}} = \frac{1}{{2L}}\int_{{\, - L}}^{{\,L}}{{(L - x)\,dx}} = \frac{1}{2L} \left[ Lx - \frac{x^2}{2} \right]_{-L}^L = \frac{1}{2L} \left[ \left(L^2 - \frac{L^2}{2}\right) - \left(-L^2 - \frac{L^2}{2}\right) \right] = L\]Endi \({A_n}\) koeffitsiyentlarni hisoblaymiz (\(n = 1, 2, 3, \ldots\)):
\[\begin{align*}{A_n} &= \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{\left( {L - x} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\\ & = \frac{1}{L}\left. {\left( {\frac{L}{{{n^2}{\pi ^2}}}} \right)\left( {n\pi \left( {L - x} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right) - L\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right)} \right|_{ - L}^L\\ & = \frac{1}{L}\left( {\frac{L}{{{n^2}{\pi ^2}}}} \right)\left( { - 2n\pi L\sin \left( { - n\,\pi } \right)} \right) = 0\hspace{0.25in}\hspace{0.25in}n = 1,2,3, \ldots \end{align*}\]Endi \({B_n}\) koeffitsiyentlarini bo'laklab integrallash orqali hisoblaymiz:
\[\begin{align*}{B_{\,n}} &= \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{\left( {L - x} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\\ & = \frac{1}{L}\left. {\left( { - \frac{L}{{{n^2}{\pi ^2}}}} \right)\left[ {L\sin \left( {\frac{{n\,\pi x}}{L}} \right) - n\pi \left( {x - L} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right]} \right|_{ - L}^L\\ & = \frac{1}{L}\left[ {\frac{{{L^2}}}{{{n^2}{\pi ^2}}}\left( {2n\pi \cos \left( {n\pi } \right) - 2\sin \left( {n\pi } \right)} \right)} \right] = \frac{{2L{{\left( { - 1} \right)}^n}}}{{n\pi }}\hspace{0.25in}\hspace{0.25in}n = 1,2,3, \ldots \end{align*}\]E'tibor bering, ushbu misolda \({A_0} \ne 0\) bo'ldi, lekin barcha \(n \ge 1\) lar uchun \({A_n} = 0\) chiqdi. Bunday holatlar amaliyotda tez-tez uchrab turadi.
Demak, izlanayotgan Furye qatori quyidagicha bo'ladi:
\[\begin{align*}f\left( x \right) &= \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \\ & = {A_0} + \sum\limits_{n = 1}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \\ & = L + \sum\limits_{n = 1}^\infty {\frac{{2L{{\left( { - 1} \right)}^n}}}{{n\pi }}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \end{align*}\]Oldingi misolda ko'rganimizdek, ba'zan \({A_0} \ne 0\) va barcha \(n \ge 1\) uchun \({A_n} = 0\) bo'lishi mumkin. Bu holat \(f(x)\) funksiyaning xususiyatiga bog'liq bo'ladi.
Endi navbatdagi masalaga o'taylik.
Yechim
Funksiya bo'lakli berilgani sababli, har bir koeffitsiyentni topish uchun integrallarni ikki qismga ajratib hisoblaymiz:
\[\begin{align*}{A_0} & = \frac{1}{{2L}}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\,dx}} = \frac{1}{{2L}}\left[ {\int_{{\, - L}}^{{\,0}}{{f\left( x \right)\,dx}} + \int_{{\,0}}^{{\,L}}{{f\left( x \right)\,dx}}} \right]\\ & = \frac{1}{{2L}}\left[ {\int_{{\, - L}}^{{\,0}}{{L\,dx}} + \int_{{\,0}}^{{\,L}}{{2x\,dx}}} \right] = \frac{1}{{2L}}\left[ {{L^2} + {L^2}} \right] = L\end{align*}\]Endi \({A_n}\) koeffitsiyentini yozamiz:
\[\begin{align*}{A_n} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} & = \frac{1}{L}\left[ {\int_{{\, - L}}^{{\,0}}{{L\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} + \int_{{\,0}}^{{\,L}}{{2x\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}} \right]\end{align*}\]Ushbu ikkita integralni alohida-alohida hisoblaymiz:
\[\int_{{\, - L}}^{{\,0}}{{L\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \left. {\left( {\frac{{{L^2}}}{{n\pi }}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \right)} \right|_{ - L}^0 = \frac{{{L^2}}}{{n\pi }}\sin \left( {n\pi } \right) = 0\] \[\begin{align*}\int_{{\,0}}^{{\,L}}{{2x\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} & = \left. {\left( {\frac{{2L}}{{{n^2}{\pi ^2}}}} \right)\left( {L\cos \left( {\frac{{n\,\pi x}}{L}} \right)\, + n\pi x\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,} \right)} \right|_0^L\\ & = \left( {\frac{{2L}}{{{n^2}{\pi ^2}}}} \right)\left( {L\cos \left( {n\pi } \right)\, + n\pi L\sin \left( {n\pi } \right) - L\cos \left( 0 \right)} \right)\\ & = \left( {\frac{{2{L^2}}}{{{n^2}{\pi ^2}}}} \right)\left( {{{\left( { - 1} \right)}^n} - 1} \right)\end{align*}\]Ikkala natijani qo'shib, \(\frac{1}{L}\) ga ko'paytirsak, \({A_n}\) hosil bo'ladi:
\[{A_n} = \frac{{2L}}{{{n^2}{\pi ^2}}}\left( {{{\left( { - 1} \right)}^n} - 1} \right),\hspace{0.25in}\hspace{0.25in}n = 1,2,3, \ldots \]Endi xuddi shunday usulda \({B_n}\) koeffitsiyentlarni hisoblaymiz:
\[\begin{align*}{B_n} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} & = \frac{1}{L}\left[ {\int_{{\, - L}}^{{\,0}}{{L\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} + \int_{{\,0}}^{{\,L}}{{2x\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}} \right]\end{align*}\]Ushbu integrallarni ham alohida hisoblaymiz:
\[\int_{{\, - L}}^{{\,0}}{{L\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \left. {\left( { - \frac{{{L^2}}}{{n\pi }}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \right)} \right|_{ - L}^0 = - \frac{{{L^2}}}{{n\pi }}\left( {1 - \cos \left( { - n\pi } \right)} \right) = \frac{{{L^2}}}{{n\pi }}\left( {{{\left( { - 1} \right)}^n} - 1} \right)\] \[\begin{align*}\int_{{\,0}}^{{\,L}}{{2x\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} & = \left. {\left( {\frac{{2L}}{{{n^2}{\pi ^2}}}} \right)\left( {L\sin \left( {\frac{{n\,\pi x}}{L}} \right)\, - n\pi x\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,} \right)} \right|_0^L\\ & = \left( {\frac{{2L}}{{{n^2}{\pi ^2}}}} \right)\left( {L\sin \left( {n\pi } \right)\, - n\pi L\cos \left( {n\pi } \right)} \right) = - \frac{{2{L^2}}}{{n\pi }}{\left( { - 1} \right)^n}\end{align*}\]Ushbu ikki ifodani birlashtirib \({B_n}\) ni topamiz:
\[\begin{align*}{B_n} & = \frac{1}{L}\left[ {\frac{{{L^2}}}{{n\pi }}\left( {{{\left( { - 1} \right)}^n} - 1} \right) - \frac{{2{L^2}}}{{n\pi }}{{\left( { - 1} \right)}^n}} \right]\\ & = \frac{1}{L}\left[ { - \frac{{{L^2}}}{{n\pi }}{{\left( { - 1} \right)}^n} - \frac{{{L^2}}}{{n\pi }}} \right] = - \frac{L}{{n\pi }}\left( {{{\left( { - 1} \right)}^n} + 1} \right),\hspace{0.25in}\hspace{0.25in}n = 1,2,3, \ldots \end{align*}\]Endi barcha topilgan koeffitsiyentlarni umumiy Furye qatoriga qo'yamiz:
\[\begin{align*}f\left( x \right) & = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \\ & = {A_0} + \sum\limits_{n = 1}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \\ & = L + \sum\limits_{n = 1}^\infty {\left[ {\frac{{2L}}{{{n^2}{\pi ^2}}}\left( {{{\left( { - 1} \right)}^n} - 1} \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right) - \frac{L}{{n\pi }}\left( {{{\left( { - 1} \right)}^n} + 1} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \right]} \end{align*}\]Yuqoridagi misoldan ko'rinib turibdiki, bo'lakli funksiyalar uchun integrallarni hisoblash ancha ko'p mehnat talab qiladi.
Keyingi ikki misol to'liq Furye qatori bilan Furye sinus va kosinus qatorlari o'rtasidagi chuqur bog'liqlikni ko'rsatish uchun keltirilgan.
Yechim
Dastlab \({A_n}\) koeffitsiyent integrallarini tekshiramiz:
\[{A_0} = \frac{1}{{2L}}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\,dx}} = \frac{1}{{2L}}\int_{{\, - L}}^{{\,L}}{{x\,dx}} = 0\] \[{A_n} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{x\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = 0\]Har ikkala holatda ham simmetrik \(\left[ { - L,L} \right]\) oraliqda toq funksiya integrallanmoqda (\(x\) toq, kosinus juft, ularning ko'paytmasi esa toq funksiyadir). Simmetrik oraliqda toq funksiyaning aniq integrali doimo nolga teng bo'ladi!
Endi \({B_n}\) uchun integralni hisoblaymiz:
\[{B_{\,n}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{x\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{x\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\]Bu yerda biz juft funksiyani integrallayapmiz (\(x\) toq va sinus toq, toq funksiyalarning ko'paytmasi esa juft bo'ladi). Shu sababli integralni \(0\) dan \(L\) gacha oraliqqa keltirib, \(2\) ga ko'paytirish mumkin. Biz aynan shu integralni oldingi Furye sinuslar qatori bo'limidagi 1-misolda batafsil hisoblagan edik. O'sha natijadan foydalansak:
\[{B_n} = \frac{{{{\left( { - 1} \right)}^{n + 1}}2L}}{{n\pi }}\hspace{0.25in}\hspace{0.25in}n = 1,2,3, \ldots \]U holda izlanayotgan Furye qatori quyidagicha bo'ladi:
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}2L}}{{n\pi }}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} \]Agar siz ortga qaytib, Furye sinuslar qatori bo'limidagi 1-misolga qarasangiz, o'sha misolda ham \( - L \le x \le L\) oraliqda \(f\left( x \right) = x\) ning Furye sinus qatorini topgan edik. Eng muhimi: o'sha misolda olingan javob bu yerdagi javob bilan aynan bir xildir.
Agar chuqurroq o'ylab ko'rsak, bu ajablanarli emas. Har ikkala holatda ham biz \( - L \le x \le L\) simmetrik oraliqda toq funksiyani ko'rib chiqdik. Toq funksiya bo'lgani uchun Furye qatoridagi kosinuslar koeffitsiyenti \({A_n}\) simmetrik oraliqda toq funksiyani integrallashni talab qiladi va natijada nolga aylanadi. Demak, simmetrik oraliqdagi toq funksiyaning Furye qatori aslida Furye sinuslar qatorining xususiy holatidir.
Biroq, ixtiyoriy funksiyani \(0 \le x \le L\) oraliqda Furye sinus qatoriga yoyganimizda natijalar bunday bir xil bo'lishini kutmaslik kerak. Buni yuqoridagi 1-misolni Furye sinus qatori bo'limidagi 3-misol bilan solishtirish orqali ko'rishingiz mumkin. Har ikkala misolda ham \(f(x) = x - L\) qatori topilgan bo'lsa-da, mutlaqo turli natijalar olingan.
Nima uchun turli javoblar chiqdi? Sababi shundaki, \(0 \le x \le L\) oraliqda Furye sinus qatorini topganimizda biz aslida berilgan funksiyani \( - L \le x \le L\) oraliqqa toq davom ettiramiz (odd extension), so'ngra natijani \(0 \le x \le L\) ga toraytiramiz. To'liq Furye qatorida esa biz toq davom ettirishdan emas, balki butun \( - L \le x \le L\) dagi funksiyaning o'zidan to'g'ridan-to'g'ri foydalanamiz. Funksiyalar har xil bo'lgani uchun qatorlar ham tabiiyki turlicha chiqadi.
Xuddi shuningdek: agar toq funksiyaning Furye qatori Furye sinus qatori bilan bir xil bo'lsa, demak juft funksiyaning Furye qatori ham Furye kosinus qatorining xususiy holati bo'lishi kerak. Buni quyidagi misol orqali tekshirib ko'raylik.
Yechim
\({A_n}\) koeffitsiyent integrallarini hisoblaymiz. Bu holda funksiya ham, kosinus ham juft bo'lgani sababli, ularning ko'paytmasi juft bo'ladi va integralni soddalashtirish mumkin:
\[{A_0} = \frac{1}{{2L}}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\,dx}} = \frac{1}{{2L}}\int_{{\, - L}}^{{\,L}}{{{x^2}\,dx}} = \frac{1}{L}\int_{{\,0}}^{{\,L}}{{{x^2}\,dx}} = \frac{1}{L} \left. \frac{x^3}{3} \right|_0^L = \frac{L^2}{3}\] \[{A_n} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{{x^2}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{{x^2}\cos \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\]Ushbu ikkala integral ham oldingi Furye kosinuslar qatori bo'limidagi 1-misolda bajarilgan edi. O'sha natijalarga ko'ra:
\[{A_0} = \frac{{{L^2}}}{3}\hspace{0.25in}\hspace{0.25in}{A_n} = \frac{{4{L^2}{{\left( { - 1} \right)}^n}}}{{{n^2}{\pi ^2}}},\,\,\,\,\,n = 1,2,3, \ldots \]Endi \({B_n}\) uchun integralni qaraymiz:
\[{B_{\,n}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = \frac{1}{L}\int_{{\, - L}}^{{\,L}}{{{x^2}\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}} = 0\]Bu yerda \(x^2\) juft funksiya, sinus esa toq funksiya bo'lgani sababli ularning ko'paytmasi toq funksiya bo'ladi. Simmetrik \( - L \le x \le L\) oraliqda toq funksiyaning aniq integrali nolga tengdir!
Shunday qilib, Furye qatori quyidagi ko'rinishni oladi:
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {{A_n}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\,\pi x}}{L}} \right)} = \frac{{{L^2}}}{3} + \sum\limits_{n = 1}^\infty {\frac{{4{L^2}{{\left( { - 1} \right)}^n}}}{{{n^2}{\pi ^2}}}\cos \left( {\frac{{n\,\pi x}}{L}} \right)} \]Misolni boshlashdan oldin aytib o'tganimizdek, bu natija Furye kosinuslar qatori bo'limidagi 1-misol natijasi bilan to'liq mos keladi. Demak, juft funksiyaning Furye qatori Furye kosinuslar qatorining aynan o'zidir.