Matematika/Furje eilutės: Skirtumas tarp puslapio versijų

:<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}\int_0^l x^2 \text{d}x=\frac{2}{l} \frac{x^3}{3}|_0^l=\frac{2}{l}\cdot \frac{l^3}{3}=\frac{2l^2}{3};</math>

:<math>=-\frac{4}{n\pi}\int_0^{\pi}x\sin(nx)\text{d}x=-\frac{4}{n\pi}\left(-\frac{x\cos(nx)}{n}|_0^{\pi}-\frac{-1}{n}\int_0^{\pi}\cos(nx)\text{d}x\right)=</math>

:<math>=-\frac{4}{n\pi}\left(-\frac{\pi\cos(n\pi)}{n}+\frac{1}{n^2}\sin(nx)|_0^{\pi}\right)=</math>

:<math>=\frac{4}{n^2}\cos(n\pi)=\frac{4}{n^2}\cdot (-1)^n=(-1)^n\frac{4}{n^2};</math>

:Furjė eilutė funkcijos <math>f(x)\;</math> yra tokia

:<math>x^2=\frac{\pi^2}{3}-4\left( \frac{\cos x}{1}-\frac{\cos (2x)}{2^2}+\frac{\cos (3x)}{3^2}-... \right).</math>