Dérét Fourier: Béda antarrépisi

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Baris ka-33:
 
Lamun ''x''(''t'') ngalambangkeun hiji fungsi tina variabel bébas ''t'' mangka ieu fungsi biasana dianggap minangka [[fungsi périodik]] kalayan périoda 2π, dina kalimah sejen bisa dinyatakeun yén ''x''(''t''+2π) = ''x''(''t''), pikeun sakabéh angka ril ''t''. Pikeun nuliskeun éta fungsi minangka pajumlahan [[dérét (matematika)|dérét]] fungsi sinusioda anu tanpa wates réana, urang kudu ngagunakeun pajumlahan fungsi-fungsi [[sinus]] jeung [[kosinus]] anu tanpa wates dina interval [-π,π], saperti anu dilakukeun ku Fourier (tingali kutipan di luhur).
 
{{tarjamahkeun|en}}
===Rumus Fourier pikeun fungsi périodik 2&pi ku cara ngagunakeun fungsi-fungsi sinus jeung kosinus===
 
Baris ka-48:
:<math>x(t) = \frac{a_0}{2} +\sum_{n=1}^{\infty}[a_n \cos(nt) + b_n \sin(nt)]</math>
 
mangrupakeun '''dérét Fourier''' pikeun ''x'' dina interval [-&pi;,&pi;]. Dérét Fourier henteu kudu konvergén (ngurucut), so there may not be equality in the formula above. It is one of the main questions in [[Harmonic analysis]] to decide when equality holds. If a function is [[square-integrable]] on the interval [-&pi;,&pi;], then it can be represented in that interval by the previous formula.
 
=== ExampleConto: ahiji simpledérét Fourier seriesbasajan ===
[[ImageGambar:Periodic identity.png|thumb|right|PlotGambar ofhiji afungsi periodic identity functionpériodik - ahiji [[sawtoothgelombang huntu waveragaji]].]]
[[ImageGambar:Periodic identity function.gif|thumb|right|400px|AnimatedGambar plotanimasi oflima thedérét firstFourier fiveparsial successivemunggaran partial Fouriernu seriespaturut-turut.]]
 
Ayeuna urang ngagunakeun rumus di luhur pikeun néangan dérét Fourier tina hiji fungsi nu basajan nyaéta fungsi huntu ragaji (saperti dijelaskeun dina gambar sabeulah katuhu)''':'''
We now use the formulae above to give a Fourier series expansion of a very simple function. Consider a sawtooth function (as depicted in the figure)''':'''
 
:<math>f(x(t) = xt, \quad \mathrm{for} \quad -\pi < xt < \pi,</math>
:<math>f(x(t + 2\pi) = fx(xt), \quad \mathrm{for} \quad -\infty < xt < \infty.</math>
 
InDina thishal caseieu, thekoéfisien Fourier coefficientsditangtukeun areku givencara bykieu:
 
:<math>\begin{align}
a_n &{} = \frac{1}{\pi}\int_{-\pi}^{\pi}xt \cos(nxnt)\,dxdt = 0. \\
b_n &{}= \frac{1}{\pi}\int_{-\pi}^{\pi} xt \sin(nxnt)\, dxdt = 2\frac{(-1)^{n+1}}{n}.\end{align}</math>
 
And thereforeSahingga''':'''
 
{{NumBlk|:
|<math>
\begin{align}
f(x(t) &= \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[a_n\cos\left(nxnt\right)+b_n\sin\left(nxnt\right)\right] \\
&=2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \sin(nxnt), \quad \mathrm{for} \quad -\infty < xt < \infty .
\end{align}
</math>
|Eq.1}}
 
[[ImageGambar:Fourier heat in a plate.png|thumb|right|HeatDistribusi distributionpanas indina ahiji metalpelat platemétal, usingngagunakeun métoda Fourier's method.]]
 
OneUrang noticesbisa thatningali theyén dérét Fourier seriestina expansionfungsi ofurang ournu functionawal lookskatémbong muchjauh lessleuwih simplebasajan thanbatan the formularumus ''fx''(''xt'')=''xt'',. andSanajan soloba itconto ispanerapanana, notdi immediatelydieu apparentngan whyditémbongkeun oneconto wouldtina needitungan thisnu Fourierdumasar series.kana Whilemotivasi thereForier arengajawab manypersamaan applicationspanas. Contona, we cite Fourier's motivation of solving the heat equation.{{tarjamahkeun|Sunda}} For example, consider a metal plate in the shape of a square whose side measures &pi; meters, with coordinates <math>(x,y) \in [0,\pi] \times [0,\pi]</math>. If there is no heat source within the plate, and if three of the four sides are held at 0 degrees [[celsius]], while the fourth side, given by ''y''=&pi;, is maintained at the temperature gradient ''T''(''x'',&pi;) = ''x'' degrees celsius, for ''x'' in (0,&pi;), then one can show that the stationary heat distribution (or the heat distribution after a long period of time has elapsed) is given by
 
:<math>T(x,y) = 2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sin(nx) {\sinh(ny) \over \sinh(n\pi)}.</math>