Matematika/Furje eilutės: Skirtumas tarp puslapio versijų
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24 eilutė:
:<math>=-\frac{1}{2}\left[\frac{\sin((k+n)x)}{k+n}|_{-\pi}^{\pi}-\frac{\sin((k-n)x)}{k-n}|_{-\pi}^{\pi}\right]=0, \; \text{kai} \; k\neq n, \quad (5.1) </math>
:čia <math>\text{d}((k+n)x)=(k+n)\text{d}x, \; \frac{\text{d}((k+n)x)}{k+n}=\text{d}x</math> ir <math>\text{d}((k-n)x)=(k-n)\text{d}x, \; \frac{\text{d}((k-n)x)}{k-n}=\text{d}x</math> bei pasinaudojome trigonometrine formule <math>\sin(A)\cdot\sin(B) = -\frac{1}{2}[\cos(A + B) - \cos (A - B)];</math>
:<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>
: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>
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