Matematika/Furje eilutės: Skirtumas tarp puslapio versijų
Ištrintas turinys Pridėtas turinys
26 eilutė:
:<math>\int_{-\pi}^{\pi}\sin(kx)\cos(nx)\text{d}x=\frac{1}{2}\int_{-\pi}^{\pi}[\sin((k+n)x)+\sin((k-n)x)]\text{d}x=-\frac{1}{2}\Big(\frac{\cos((k+n)x)}{k+n}+\frac{\cos((k-n)x)}{k-n}\Big)|_{-\pi}^\pi=0. \quad (5.2)</math>
:Kai (5.2) integrale <math>k=n=p,</math> tai toks integralas irgi lygus nuliui, nes
:<math>\int_{-\pi}^{\pi}\sin(px)\cos(px)\text{d}x=\int_{-\pi}^{\pi}\sin(px)\cos(px)\frac{\text{d}(\sin(px))}{p\cos(px)}=\frac{1}{p}\int_{-\pi}^{\pi}\sin(px)\text{d}(\sin(px))=\frac{1}{2p}\sin^2(px)|_{-\pi}^{\pi}=0;</math> čia <math>\text{d}(\sin(px))=p\cos(px) \; dx.</math>
:Pagaliau,
:<math>\int_{-\pi}^{\pi} \cos^2(kx) \; \text{d}x=\frac{1}{2}\int_{-\pi}^{\pi}(1+\cos(2k x))\text{d}x=\frac{1}{2}\left(x+\frac{1}{2k}\sin(2kx)\right)|_{-\pi}^{\pi}=\frac{1}{2}(\pi-(-\pi))=\pi, \quad (6)</math>
| |||