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Dzielenie wielomianów, schemat blokowy

Started by Dreamer, 11 Jul 2011 14:42

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#741 Dreamer

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Posted 23 June 2019 - 11:18

$Permutacja(a,b,c)^{4}^{5}^{6} =\\ \\ a _{i}\cdot ( (a+b+c) ^{4-1}+(b+c)^{4-1}+c ^{4-1} )+\\ a _{i}\cdot ( (a+b+c) ^{5-1}+(b+c)^{5-1}+c ^{5-1} )+\\ a _{i}\cdot ( (a+b+c) ^{6-1}+(b+c)^{6-1}+c ^{6-1} )+\\ a _{i} \cdot b _{i} \cdot ( (b+c)^{4-2}+c ^{4-2})+ \\ a _{i} \cdot b _{i} \cdot ( (b+c)^{5-2}+c ^{5-2})+ \\ a _{i} \cdot b _{i} \cdot ( (b+c)^{6-2}+c ^{6-2})+ \\ a _{i} \cdot b _{i}\cdot c_{i} \cdot (c ^{4-3})+ \\ a _{i} \cdot b _{i}\cdot c_{i} \cdot (c ^{5-3})+ \\ a _{i} \cdot b _{i}\cdot c_{i} \cdot (c ^{6-3})+ \\ b_{j} \cdot ( (b+c)^{4-1}+c ^{4-1})+ \\ b_{j} \cdot ( (b+c)^{4-1}+c ^{5-1})+ \\ b_{j} \cdot ( (b+c)^{4-1}+c ^{6-1})+ \\ b _{j} \cdot c _{j} \cdot (c ^{4-2})+\\ b _{j} \cdot c _{j} \cdot (c ^{5-2})+\\ b _{j} \cdot c _{j} \cdot (c ^{6-2})+\\ c _{j} \cdot (c ^{4-1})\\ c _{j} \cdot (c ^{5-1})\\ c _{j} \cdot (c ^{6-1})$

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#742 Dreamer

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870 posts

Posted 23 June 2019 - 11:18

$Permutacja(a,b,c)(^{4})(^{5})(^{6}) =\\ \\ a _{i}\cdot ( (a+b+c) ^{4-1}+(b+c)^{4-1}+c ^{4-1} )+\\ a _{i}\cdot ( (a+b+c) ^{5-1}+(b+c)^{5-1}+c ^{5-1} )+\\ a _{i}\cdot ( (a+b+c) ^{6-1}+(b+c)^{6-1}+c ^{6-1} )+\\ \\ a _{i} \cdot b _{i} \cdot ( (b+c)^{4-2}+c ^{4-2})+ \\ a _{i} \cdot b _{i} \cdot ( (b+c)^{5-2}+c ^{5-2})+ \\ a _{i} \cdot b _{i} \cdot ( (b+c)^{6-2}+c ^{6-2})+ \\ \\ a _{i} \cdot b _{i}\cdot c_{i} \cdot (c ^{4-3})+ \\ a _{i} \cdot b _{i}\cdot c_{i} \cdot (c ^{5-3})+ \\ a _{i} \cdot b _{i}\cdot c_{i} \cdot (c ^{6-3})+ \\ \\ b_{j} \cdot ( (b+c)^{4-1}+c ^{4-1})+ \\ b_{j} \cdot ( (b+c)^{4-1}+c ^{5-1})+ \\ b_{j} \cdot ( (b+c)^{4-1}+c ^{6-1})+ \\ \\ b _{j} \cdot c _{j} \cdot (c ^{4-2})+\\ b _{j} \cdot c _{j} \cdot (c ^{5-2})+\\ b _{j} \cdot c _{j} \cdot (c ^{6-2})+\\ \\ c _{j} \cdot (c ^{4-1})\\ c _{j} \cdot (c ^{5-1})\\ c _{j} \cdot (c ^{6-1})$

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#743 Dreamer

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870 posts

Posted 23 June 2019 - 11:19

$Permutacja(a,b,c,d)(^{4})(^{5})(^{6}) =\\ \\ a _{i}\cdot ( (a+b+c+d) ^{4-1}+(b+c+d)^{4-1}+(c+d )^{4-1}+d ^{4-1} )+\\ a _{i}\cdot ( (a+b+c+d) ^{5-1}+(b+c+d)^{5-1}+(c+d )^{5-1}+d ^{5-1} )+\\ a_{i}\cdot ( (a+b+c+d) ^{4-1}+(b+c+d)^{6-1}+(c+d )^{6-1}+d ^{6-1} )+\\ \\ a _{i}\cdot b_{i} \cdot ( (b+c+d)^{4-2}+(c+d) ^{4-2})+d^{4-2})+ \\ a _{i}\cdot b_{i} \cdot ( (b+c+d)^{5-2}+(c+d) ^{5-2})+d^{5-2})+ \\ a _{i}\cdot b_{i} \cdot ( (b+c+d)^{6-2}+(c+d) ^{6-2})+d^{6-2})+ \\ \\ a_{i} \cdot b _{i}\cdot c_{i} \cdot (c+d) ^{4-3} + (d ^{4-3})+ \\ a_{i} \cdot b _{i}\cdot c_{i} \cdot (c+d) ^{5-3} +(d ^{5-3})+ \\ a_{i} \cdot b_{i}\cdot c_{i} \cdot (c+d) ^{6-3} +(d ^{6-3})+ \\ \\ a_{i} \cdot b _{i}\cdot c_{i} \cdot d \cdot d ^{5-4} + \\ a_{i} \cdot b_{i}\cdot c_{i} \cdot d \cdot d ^{6-4} +\\ \\ b_{j} \cdot ( (b+c+d)^{4-1}+c ^{4-1})+ \\ b_{j} \cdot ( (b+c+d)^{5-1}+c ^{5-1})+ \\ b_{j} \cdot ( (b+c+d)^{5-1}+c ^{5-1})+ \\ \\ b_{j} \cdot c_{j} \cdot ((c +d)^{42})+d^{4-2})+\\ b_{j} \cdot c_{j} \cdot ((c +d)^{5-2})+d^{5-2})+\\ b_{j} \cdot c_{j} \cdot ((c +d)^{6-2})+d^{6-2})+\\ \\ b_{j} \cdot c_{j} \cdot d _{j} \cdot (d ^{4-3})+\\ b_{j} \cdot c_{j} \cdot d _{j} \cdot (d ^{5-3})+\\ b_{j} \cdot c_{j} \cdot d_{j} \cdot (d ^{6-3})+\\ c_{l} \cdot ((c +d)^{4-1})+d^{4-1})+\\ c_{l} \cdot ((c +d)^{5-1})+d^{5-1})+\\ c_{l} \cdot ((c +d)^{6-1})+d^{6-1})+\\ c_{l} \cdot d _{l} \cdot (d ^{4-2})+\\ c_{l} \cdot d _{l} \cdot (d ^{5-2})+\\ c_{l} \cdot d_{l} \cdot (d ^{6-2})+\\ d_{m} \cdot ( (d)^{4-1}+\\ d_{m} \cdot ( (d)^{5-1}+\\ d_{m} \cdot ( (d)^{5-1}+\\$

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#744 Dreamer

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870 posts

Posted 23 June 2019 - 11:21

$Permutacja ^{n}= \sum_{k=1 j=1}^{n} a _{j} ^{n-k} \cdot b _{j} \cdot c _{j} \cdot ...\cdot k_{j} \sum_{n}^{k} (a+b+c+d+...+k) ^{n-j} +(b+c+d+...+k) ^{n-j} +...+k ^{n-j}$

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#745 Dreamer

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870 posts

Posted 23 June 2019 - 11:23

Najpierw pierwsza suma:

$Permutacja ^{n}= \sum_{k=1 j=1}^{n} a _{j} ^{n-k} \cdot b _{j} \cdot c _{j} \cdot ...\cdot k_{j} +b _{i}+c _{i}+...k _{i} +...k _{i}$

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#746 Dreamer

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870 posts

Posted 23 June 2019 - 11:23

Mamy dane pierwiastki:

$a \cdot b \cdot c \cdot d \cdot ... \cdot k$

Przy każdym pierwiastku mamy indeks dolny, naszą zmienną potrzebną w drugiej sumie.: