Matematika/Furje eilutės: Skirtumas tarp puslapio versijų

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199 eilutė:
:''Sprendimas''. Įvedame keitinį <math>x=\frac{\xi l}{\pi}; \; \xi=\frac{\pi x}{l}; \; \text{d}\xi=\frac{\pi}{l}\text{d}x.</math> Funkcija <math>\phi(\xi)</math> apibrėžta atkarpoje <math>[-\pi; \; \pi].</math> Kadangi funkcija <math>f(x)=x^2\;</math> lyginė, tai
:<math>b_n=0;</math>
:<math>a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}\phi(\xi) \text{d}\xi=\frac{2}{\pi}\int_{0}^{\pi}\phi(\xi) \text{d}\xi=\frac{2}{\pi}\int_0^l f(x)\cdot \frac{\pi}{l}\text{d}x=\frac{2}{l}\cdotint_0^l \frac{x^2 \text{d}x=\frac{2}|_0{\pi}\int_0^l \frac{x^3}{3} {d}x=\frac{2}{\pi}\int_0^l \frac{x^3}{3}\cdot \frac{\pi}{l^2}\text{2d}x=l;</math>
:<math>a_n=\frac{2}{l}\int_0^l x\cos\frac{n\pi x}{l}\text{d}x=\frac{2}{l}\left(\frac{l\cdot x\sin\frac{n\pi x}{l}}{n\pi}|_0^l-\frac{l}{n\pi}\int_0^l\sin\frac{n\pi x}{l}\text{d}x \right)=</math>
:<math>=\frac{2}{l}\left(\frac{l^2\sin\frac{n\pi l}{l}}{n\pi}+\frac{l}{n^2\pi^2}\cos\frac{n\pi x}{l}|_0^l \right)=\frac{2}{l}\left(\frac{l^2\sin(n\pi)}{n\pi}+\frac{l}{n^2\pi^2}(\cos\frac{n\pi l}{l}-\cos\frac{n\pi \cdot 0}{l}) \right)=</math>