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  • https://ukrayinska.libretexts.org/%D1%84%D1%96%D0%B7%D0%B8%D0%BA%D0%B8/%D0%9A%D0%B2%D0%B0%D0%BD%D1%82%D0%BE%D0%B2%D0%B0_%D0%BC%D0%B5%D1%85%D0%B0%D0%BD%D1%96%D0%BA%D0%B0/%D0%92%D1%81%D1%82%D1%83%D0%BF%D0%BD%D0%B0_%D0%BA%D0%B2%D0%B0%D0%BD%D1%82%D0%BE%D0%B2%D0%B0_%D0%BC%D0%B5%D1%85%D0%B0%D0%BD%D1%96%D0%BA%D0%B0_(Fitzpatrick)/01%3A_%D0%A2%D0%B5%D0%BE%D1%80%D1%96%D1%8F_%D0%B9%D0%BC%D0%BE%D0%B2%D1%96%D1%80%D0%BD%D0%BE%D1%81%D1%82%D0%B5%D0%B9/1.04%3A_%D0%91%D0%B5%D0%B7%D0%BF%D0%B5%D1%80%D0%B5%D1%80%D0%B2%D0%BD%D0%B8%D0%B9_%D1%80%D0%BE%D0%B7%D0%BF%D0%BE%D0%B4%D1%96%D0%BB_%D0%B9%D0%BC%D0%BE%D0%B2%D1%96%D1%80%D0%BD%D0%BE%D1%81%D1%82%D0%B5%D0%B9
    Більш ранні результати (1.2.4), (1.3.4) та (1.3.11) узагальнюють простим способом, щоб дати:\[\begin{aligned} 1&= \int_{-\infty}^\infty P(u)\,du,\\[0.5ex] \langle u\rangle &= \int_{-\infty}^\infty P(u...Більш ранні результати (1.2.4), (1.3.4) та (1.3.11) узагальнюють простим способом, щоб дати:1=P(u)du,u=P(u)udu,(Δu)2=P(u)(uu)2du=u2u2, відповідно.