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- https://ukrayinska.libretexts.org/%D0%A5%D1%96%D0%BC%D1%96%D1%8F/%D0%A4%D1%96%D0%B7%D0%B8%D1%87%D0%BD%D0%B0_%D1%96_%D1%82%D0%B5%D0%BE%D1%80%D0%B5%D1%82%D0%B8%D1%87%D0%BD%D0%B0_%D1%85%D1%96%D0%BC%D1%96%D1%8F/%D0%A4%D1%96%D0%B7%D0%B8%D1%87%D0%BD%D0%B0_%D1%85%D1%96%D0%BC%D1%96%D1%8F_(LibreTexts)/06%3A_%D0%90%D1%82%D0%BE%D0%BC_%D0%B2%D0%BE%D0%B4%D0%BD%D1%8E/6.02%3A_%D0%A5%D0%B2%D0%B8%D0%BB%D1%8C%D0%BE%D0%B2%D1%96_%D1%84%D1%83%D0%BD%D0%BA%D1%86%D1%96%D1%97_%D0%B6%D0%BE%D1%80%D1%81%D1%82%D0%BA%D0%BE%D0%B3%D0%BE_%D1%80%D0%BE%D1%82%D0%B0%D1%82%D0%BE%D1%80%D0%B0_%D0%BD%D0%B0%D0%B7%D0%B8%D0%B2%D0%B0%D1%8E%D1%82%D1%8C%D1%81%D1%8F_%D1%81%D1%84%D0%B5%D1%80%D0%B8%D1%87%D0%BD%D0%B8%D0%BC%D0%B8_%D0%B3%D0%B0%D1%80%D0%BC%D0%BE%D0%BD%D1%96%D0%BA%D0%B0%D0%BC%D0%B8Розв'язками рівняння атома водню Шредінгера є функціями, які є добутком сферичної гармонічної функції та радіальної функції.
- https://ukrayinska.libretexts.org/%D1%84%D1%96%D0%B7%D0%B8%D0%BA%D0%B8/%D0%90%D1%81%D1%82%D1%80%D0%BE%D0%BD%D0%BE%D0%BC%D1%96%D1%8F_%D1%82%D0%B0_%D0%BA%D0%BE%D1%81%D0%BC%D0%BE%D0%BB%D0%BE%D0%B3%D1%96%D1%8F/%D0%9D%D0%B5%D0%B1%D0%B5%D1%81%D0%BD%D0%B0_%D0%BC%D0%B5%D1%85%D0%B0%D0%BD%D1%96%D0%BA%D0%B0_(Tatum)/05%3A_%D0%93%D1%80%D0%B0%D0%B2%D1%96%D1%82%D0%B0%D1%86%D1%96%D0%B9%D0%BD%D0%B5_%D0%BF%D0%BE%D0%BB%D0%B5_%D1%82%D0%B0_%D0%BF%D0%BE%D1%82%D0%B5%D0%BD%D1%86%D1%96%D0%B0%D0%BB/5.11%3A_%D0%9C%D0%BD%D0%BE%D0%B3%D0%BE%D1%87%D0%BB%D0%B5%D0%BD%D0%B8_%D0%9B%D0%B5%D0%B6%D0%B0%D0%BD%D0%B4%D1%80%D0%B0У цьому розділі ми розглянемо достатньо про многочлени Лежандра, щоб бути корисними в наступному розділі.
- https://ukrayinska.libretexts.org/%D1%84%D1%96%D0%B7%D0%B8%D0%BA%D0%B8/%D0%90%D1%81%D1%82%D1%80%D0%BE%D0%BD%D0%BE%D0%BC%D1%96%D1%8F_%D1%82%D0%B0_%D0%BA%D0%BE%D1%81%D0%BC%D0%BE%D0%BB%D0%BE%D0%B3%D1%96%D1%8F/%D0%9D%D0%B5%D0%B1%D0%B5%D1%81%D0%BD%D0%B0_%D0%BC%D0%B5%D1%85%D0%B0%D0%BD%D1%96%D0%BA%D0%B0_(Tatum)/01%3A_%D0%A7%D0%B8%D1%81%D0%B5%D0%BB%D1%8C%D0%BD%D1%96_%D0%BC%D0%B5%D1%82%D0%BE%D0%B4%D0%B8/1.14%3A_%D0%9C%D0%BD%D0%BE%D0%B3%D0%BE%D1%87%D0%BB%D0%B5%D0%BD%D0%B8_%D0%9B%D0%B5%D0%B6%D0%B0%D0%BD%D0%B4%D1%80%D0%B0Ви виявите, що коефіцієнт ofrl - це поліноміальний вираз вx градусіl. (1−2rx+r2)−1/2=P0(x)+P1(x)r+P2(x)r2+P3(x)r3... Коефіці...Ви виявите, що коефіцієнт ofrl - це поліноміальний вираз вx градусіl. \boldsymbol{(1 - 2rx + r^2 )^{-1/2} = P_0 (x) + P_1 (x) r + P_2 (x) r^2 + P_3 (x) r^3 ... \label{1.14.2} \tag{1.14.2}} Коефіцієнти послідовної степеніr - це многочлени Лежандра; коефіцієнтrl, який єPl(x), є поліномом Лежандра порядкуl, і він є поліномом уx включенні членів настільки ж високим, якxl. Pl+1=12ll!dldxl(x2−1)l.
- https://ukrayinska.libretexts.org/%D0%9C%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D0%BA%D0%B0/%D0%94%D0%B8%D1%84%D0%B5%D1%80%D0%B5%D0%BD%D1%86%D1%96%D0%B9%D0%BD%D1%96_%D1%80%D1%96%D0%B2%D0%BD%D1%8F%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%A0%D1%96%D0%B2%D0%BD%D1%8F%D0%BD%D0%BD%D1%8F_%D0%B7_%D1%87%D0%B0%D1%81%D1%82%D0%B8%D0%BD%D0%BD%D0%B8%D0%BC%D0%B8_%D0%BF%D0%BE%D1%85%D1%96%D0%B4%D0%BD%D0%B8%D0%BC%D0%B8_(Walet)/11%3A_%D0%9F%D0%BE%D0%B4%D1%96%D0%BB_%D0%B7%D0%BC%D1%96%D0%BD%D0%BD%D0%B8%D1%85_%D1%83_%D1%82%D1%80%D1%8C%D0%BE%D1%85_%D0%B2%D0%B8%D0%BC%D1%96%D1%80%D0%B0%D1%85/11.02%3A_%D0%92%D0%BB%D0%B0%D1%81%D1%82%D0%B8%D0%B2%D0%BE%D1%81%D1%82%D1%96_%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE%D1%87%D0%BB%D0%B5%D0%BD%D1%96%D0%B2_%D0%9B%D0%B5%D0%B6%D0%B0%D0%BD%D0%B4%D1%80%D0%B0\[\begin{aligned} { \frac{d^{n+1}}{dx^{n+1}}\left[ (x^{2}-1) \frac{d}{dx} (x^{2}-1)^{n} - 2 n x (x^{2}-1)^{n}\right]} &= n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n + 2(n+1) x \frac{d^{n+1}}{dx^{n+1}} (x^2-1...\[\begin{aligned} { \frac{d^{n+1}}{dx^{n+1}}\left[ (x^{2}-1) \frac{d}{dx} (x^{2}-1)^{n} - 2 n x (x^{2}-1)^{n}\right]} &= n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n + 2(n+1) x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n+(x^2-1) \frac{d^{n+2}}{dx^{n+2}} (x^2-1)^n \nonumber\\ &-2n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n - 2n x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n \nonumber\\ &=-n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n + 2 x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n+(x^2-1) \frac{d^{n+2}}{dx^{n+2}} (x^2-1)^n \nonumber\\ &= -\left[…
- https://ukrayinska.libretexts.org/%D0%9C%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D0%BA%D0%B0/%D0%94%D0%B8%D1%84%D0%B5%D1%80%D0%B5%D0%BD%D1%86%D1%96%D0%B9%D0%BD%D1%96_%D1%80%D1%96%D0%B2%D0%BD%D1%8F%D0%BD%D0%BD%D1%8F/%D0%94%D1%80%D1%83%D0%B3%D0%B8%D0%B9_%D0%BA%D1%83%D1%80%D1%81_%D0%B7%D0%B2%D0%B8%D1%87%D0%B0%D0%B9%D0%BD%D0%B8%D1%85_%D0%B4%D0%B8%D1%84%D0%B5%D1%80%D0%B5%D0%BD%D1%86%D1%96%D0%B0%D0%BB%D1%8C%D0%BD%D0%B8%D1%85_%D1%80%D1%96%D0%B2%D0%BD%D1%8F%D0%BD%D1%8C%3A_%D0%B4%D0%B8%D0%BD%D0%B0%D0%BC%D1%96%D1%87%D0%BD%D1%96_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%B8_%D1%82%D0%B0_%D0%BA%D1%80%D0%B0%D0%B9%D0%BE%D0%B2%D1%96_%D0%B7%D0%B0%D0%B4%D0%B0%D1%87%D1%96_(Herman)/07%3A_%D0%A1%D0%BF%D0%B5%D1%86%D1%96%D0%B0%D0%BB%D1%8C%D0%BD%D1%96_%D1%84%D1%83%D0%BD%D0%BA%D1%86%D1%96%D1%97/7.02%3A_%D0%9C%D0%BD%D0%BE%D0%B3%D0%BE%D1%87%D0%BB%D0%B5%D0%BD%D0%B8_%D0%9B%D0%B5%D0%B6%D0%B0%D0%BD%D0%B4%D1%80%D0%B0(x−t)∞∑n=0Pn(x)tn=∞∑n=0nPn(x)tn−1−∞∑n=02nxPn(x)tn+∞∑n=0nPn(x)tn+1. \[\sum_{n=0}^{\infty} n...(x−t)∞∑n=0Pn(x)tn=∞∑n=0nPn(x)tn−1−∞∑n=02nxPn(x)tn+∞∑n=0nPn(x)tn+1. ∞∑n=0nPn(x)tn−1−∞∑n=0(2n+1)xPn(x)tn+∞∑n=0(n+1)Pn(x)tn+1=0 ∫1−1xmPn(x)dx=12nn!∫1−1xmdndxn(x2−1)ndx.