Dzielenie za pomoc膮 permutacji.
ostatnie wiadomo艣ci | regulamin | latex
| Autor | Wiadomo艣膰 |
Szymon Konieczny post贸w: 11670 |  2023-06-21 15:07:34$\frac{W_{1}+W_{5}x^{4}+W_{12}x^{11}}{(x+a)(x+b)(x+c)}=$ $W_{1}per(a,b,c)^{0}$ $-W_{1}per(a,b,c)^{1}+W_{2}per(a,b,c)^{0}$ $-W_{1}per(a,b,c)^{2}+W_{2}per(a,b,c)^{1}-W_{3}per(a,b,c)^{0}$ $+W_{1}per(a,b,c)^{3}-W_{2}per(a,b,c)^{2}+W_{3}per(a,b,c)^{1}-W_{3}per(a,b,c)^{0}$ $W_{1}per(a,b,c)^{4}-W_{4}per(a,b,c)^{0}$ $W_{1}per(a,b,c)^{5}-W_{4}per(a,b,c)^{1}$ $-W_{1}per(a,b,c)^{6}+W_{4}per(a,b,c)^{2}$ $-W_{1}per(a,b,c)^{7}+W_{4}per(a,b,c)^{3}$ $+W_{1}per(a,b,c)^{8}-W_{4}per(a,b,c)^{4}$ $\frac{+W_{1}per(a,b,c)^{9}-W_{4}per(a,b,c)^{5}}{(x+a)}$ $\frac{-W_{1}per(a,b,c)^{10}+W_{4}per(a,b,c)^{6}}{(x+a)(x+b)}$ $\frac{-W_{1}(c)^{11}+W_{4}(c)^{6}-W_{12}c^{0}}{(x+a)(x+b)(x+c)}$ Wiadomo艣膰 by艂a modyfikowana 2023-06-21 15:09:14 przez Szymon Konieczny |
Szymon Konieczny post贸w: 11670 | 2023-06-21 15:10:06 Pocz膮tki, teraz podstawiamy wz贸r, na permutacj臋. |
Szymon Konieczny post贸w: 11670 | 2023-06-21 15:20:04 $\frac{W_{1}+W_{5}x^{4}+W_{12}x^{11}}{(x+a)(x+b)(x+c)}=$ $W_{1}per(a,b,c)^{0}$ $-W_{1}per(a,b,c)^{1}$ $-W_{1}per(a,b,c)^{2}$ $+W_{1}per(a,b,c)^{3}$ $W_{1}per(a,b,c)^{4}-W_{4}per(a,b,c)^{0}$ $W_{1}per(a,b,c)^{5}-W_{4}per(a,b,c)^{1}$ $-W_{1}per(a,b,c)^{6}+W_{4}per(a,b,c)^{2}$ $-W_{1}per(a,b,c)^{7}+W_{4}per(a,b,c)^{3}$ $+W_{1}per(a,b,c)^{8}-W_{4}per(a,b,c)^{4}$ $\frac{+W_{1}per(a,b,c)^{9}-W_{4}per(a,b,c)^{5}}{(x+a)}$ $\frac{-W_{1}per(a,b,c)^{10}+W_{4}per(a,b,c)^{6}}{(x+a)(x+b)}$ $\frac{-W_{1}(c)^{11}+W_{4}(c)^{6}-W_{12}(c)^{0}}{(x+a)(x+b)(x+c)}$ Wiadomo艣膰 by艂a modyfikowana 2023-06-21 15:23:57 przez Szymon Konieczny |
Szymon Konieczny post贸w: 11670 | 2023-06-21 15:23:06 Za du偶o jak na raz, nie licz臋. |
Szymon Konieczny post贸w: 11670 | 2023-06-21 15:32:30 $per(a,bc)^{3}=(a+b+c)(abc)$ $Per(a,b,c)^{4}=(a+b+c) \cdot Per(a,b,c)^{3}$ $Per(a,b,c)^{5}=(a+b+c) \cdot Per(a,b,c)^{4}$ $Per(a,b,c)^{6}=(abc)(per(a,b,c)^{3})$ $Per(a,b,c)^{7}=(a+b+c) \cdot Per(a,b,c)^{6}$ $Per(a,b,c)^{8}=(a+b+c) \cdot Per(a,b,c)^{7}$ $Per(a,b,c)^{9}=(abc)(per(a,b,c)^{6})$ $Per(a,b,c)^{10}=(a+b+c) \cdot Per(a,b,c)^{9}$ $Per(a,b,c)^{11}=(a+b+c) \cdot per(a,b,c)^{10}$ |
Szymon Konieczny post贸w: 11670 | 2023-06-21 15:44:41 Ol艣ni艂o mnie. To na tym polega probalistyka, na wklejaniu wzoru, do wzoru. |
Szymon Konieczny post贸w: 11670 | 2023-06-21 16:14:57 Wychodzi: $W_{1} \cdot (+Per(a,b,c)^{0}+Per(a,b,c)^{1}-Per(a,b,c)^{2}-Per(a,b,c)^{3}+Per(a,b,c)^{4}+Per(a,b,c)^{5}-Per(a,b,c)^{6}-Per(a,b,c)^{7}+Per(a,b,c)^{8})+$ $W_{4} \cdot (-Per(a,b,c)^{0}-Per(a,b,c){1}+Per(a,b,c)^{2}+Per(a,b,c)^{3}-Per(a,b,c)^{4})+$ $\frac{+W_{1}per(a,b,c)^{9}-W_{4}per(a,b,c)^{5}}{(x+a)}$ $\frac{-W_{1}per(a,b,c)^{10}+W_{4}per(a,b,c)^{6}}{(x+a)(x+b)}$ $\frac{-W_{1}(c)^{11}+W_{4}(c)^{6}-W_{12}c^{0}}{(x+a)(x+b)(x+c)}$ Wiadomo艣膰 by艂a modyfikowana 2023-06-21 16:17:04 przez Szymon Konieczny |
Szymon Konieczny post贸w: 11670 | 2023-06-21 16:21:31 O ja przeros艂a, mnie moja wiedza, na dzisiaj koniec. |
Szymon Konieczny post贸w: 11670 | 2023-06-21 16:33:04 Nikt nie zawi贸d艂. |Po prostu tak mia艂o by膰. Wiecie jak膮 tu widzia艂em piecz臋膰. Nikt, by nie podo艂a艂. |
Szymon Konieczny post贸w: 11670 | 2023-06-21 16:40:41 Jak kto艣 zawi贸d艂, to ja. Mia艂em do艣膰 permutacji i zawiesi艂em obliczenia. A mog艂em co艣 zdzia艂a膰, szybciej |
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