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Całe, życie nad jednym wzorem pracowałem, oto wyniki.
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Szymon K
Użytkownik
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Dodane dnia 20-01-2025 20:07 |
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permutacja.
(a+b+c)
(a(a+b+c)+b(b+c)+c(c))
a(a(a+b+c)+b(b+c)+c(c))+b(b(b+c)+c(c))+cc(c)
a(a(a(a+b+c)+b(b+c)+c(c))+b(b(b+c)+c(c))+cc(c))+b(b(b(b+c)+c(c))+cc(c))+ccc(c)
Ale jak to ugryźć.
Ale jak to ugryźć.
(a+b+c)
(a+b+c)(a)+b(b+c)+c(c)
(a+b+c)(aa+bb+cc)+abc
(a+b+c)(aaa+bbb+ccc)+(a+b+c)(aa+bb+cc)(a+b+c)
(a+b+c)(aaaa+bbbb+cccc)+(a+b+c)(aaa+b +ccc)(a+b+c)+(a+b+c)(aa+bb+cc)(a+b+c)(a+b+c)
(a+b+c)(aaaaa+bbbbb+ccccc)+(a+b+c)(aaaa+bbbb+cccc)(a+b+c)+(a+b+c)(aaa+bb +ccc)(a+b+c)(a+b+c)+(a+b+c)(aa+bb+cc)(a+b+c)(a+b+c)(a+b+c)
per(a,b,c)^{k+3}=Suma (a+b+c)^{k)(a^{n-k{+b^{n-k}+c{n-k})- (a+b+c)^{n-1)(a^{1{+b^{1}+c^{1}
Zgadza się. To pozamiatane.
Tyle lat to katowałem, i dopiero teraz wpadłem na pomysł.
To napisałem, fakultet z dawno zjedzonego tematu. Ciekawe co to zmieni.
W_{1}x^{n}+W_{2}x^{n-1}-...w_{n}x^{0}/(a+x)(b+X)(c+x)=
x^{n-3}(W_{1} +
x^{n-4)(-W_{1}(per(a,b,c)^{1}+W_{2} +
x^{n-4)(-W_{1}(per(a,b,c)^{2}+W_{2}(per(a,b,c)^{1}-W_3})+
x^{n-5)(W_{1(}per(a,b,c)^{3}-W_{2}(per(a,b,c)^{2}+W_3}(per(a,b,c)^{1}-W_{4} +
+...+
x^{2)(+/-W_{1}(per(a,b,c)^{n-4-}+/-W_{2}(per(a,b,c)^{n-5}...-+W_(n-5}(per(a,b,c)^{1}+/-W_{n-4}})+
x^{1} ((+/-W_{1}(per(a,b,c)^{n-3-}+/-W_{2}(per(a,b,c)^{n-4}...-+W_(n-4}(per(a,b,c)^{1}+/-W_{n-3}}))+
((+/-W_{1}(per(a,b,c)^{n-2-}+/-W_{2}(per(a,b,c)^{n-3}...-+W_(n-3}(per(a,b,c)^{1}+/-W_{n-2}})/(x+a)+
((+/-W_{1}(per(a,b,c)^{n-1-}+/-W_{2}(per(a,b,c)^{n-2}...-+W_(n-2}(per(a,b,c)^{1}+/-W_{n-1}})/(x+ a)(x+ +
((+/-W_{1}(c)^{n-}+/-W_{2}(c)^{n-1}...-+W_(n-1}(c)^{1}+/-W_{n}})/(x+a)(x+ (x+ c)
[Dodano 20-01-2025 20:09]
Jak się pozbyć tych emotikonek?
[Dodano 24-01-2025 13:54]
per(a,b,c)^{k+4}= (a+b+c)^{1)(a^{n-1}+b^{n-1}+c{n-1})+per(a,b,c)^{k+3}
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