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grapefruit
Euclid


Joined: 23 Mar 2014
Posts: 29

PostPosted: Fri Apr 25, 2014 4:00 pm    Post subject: problema Reply with quote

Daca x1,x2....x_n-1 sunt radacinile diferite de 1 ale ecuatiei x^n-1=0
Sa se arate ca 1/(1-x1)+1/(1-x2)+....+1/(1-x_n-1)=(n-1)/2
Problema apare in capitolul derivate.
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grapefruit
Euclid


Joined: 23 Mar 2014
Posts: 29

PostPosted: Sat May 10, 2014 8:54 pm    Post subject: Reply with quote

cineva?
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Adriana Nistor
Thales


Joined: 07 Aug 2008
Posts: 116
Location: Bucuresti

PostPosted: Sun May 11, 2014 12:44 pm    Post subject: Reply with quote

Eu. Smile)
Tinand cont ca x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1) si ca x^n-1=(x-1)(x-x_1)...(x-x_{n-1}) , avem ca x^{n-1}+x^{n-2}+...+x+1=(x-x_1)...(x-x_{n-1}) (1).
Derivand relatia (1) avem ca:
(n-1)x^{n-2}+(n-2)x^{n-3}+...+2x+1= (x-x_2)(x-x_3)...(x-x_{n-1})+(x-x_1)(x-x_3)(x-x_4)..(x-x_{n-1})+...+(x-x_1)(x-x_2)..(x-x_{n-2})
Dar MD(membrul drept) se poate rescrie asa:
(x-x_2)(x-x_3)...(x-x_{n-1})+(x-x_1)(x-x_3)(x-x_4)..(x-x_{n-1})+...+(x-x_1)(x-x_2)..(x-x_{n-2})=\frac{(x-x_1)...(x-x_{n-1})}{x-x_1}+\frac{(x-x_1)...(x-x_{n-1})}{x-x_2}+...+\frac{(x-x_1)...(x-x_{n-1})}{x-x_{n-1}}
Asadar, avem ca:
(n-1)x^{n-2}+(n-2)x^{n-3}+...+2x+1=\frac{(x-x_1)...(x-x_{n-1})}{x-x_1}+\frac{(x-x_1)...(x-x_{n-1})}{x-x_2}+...+\frac{(x-x_1)...(x-x_{n-1})}{x-x_{n-1}} (2)
Evaluam relatia (1) in x=1:
n=(1-x_1)(1-x_2)..(1-x_{n-1}) (3)
Evaluam relatia (2) in x=1:
(n-1)+(n-2)+..+2+1= \frac{(1-x_1)(1-x_2)..(1-x_{n-1})}{1-x_1}+\frac{(1-x_1)(1-x_2)..(1-x_{n-1})}{1-x_2}+...+\frac{(1-x_1)(1-x_2)..(1-x_{n-1})}{1-x_{n-1}
Folosind si relatia (3), relatia de mai sus devine:
\frac{n(n-1)}{2}=\frac{n}{1-x_1}+\frac{n}{1-x_2}+...+\frac{n}{1-x_{n-1}}
Asadar :
\frac{n-1}{2}=\frac{1}{1-x_1}+\frac{1}{1-x_2}+..+\frac{1}{1-x_{n-1}}
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grapefruit
Euclid


Joined: 23 Mar 2014
Posts: 29

PostPosted: Mon May 12, 2014 3:09 pm    Post subject: Reply with quote

Nu prea imi dau seama cum ati derivat acel produs lung de la inceput..
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Adriana Nistor
Thales


Joined: 07 Aug 2008
Posts: 116
Location: Bucuresti

PostPosted: Mon May 12, 2014 5:58 pm    Post subject: Reply with quote

Presupun ca problema de intelegere este la derivarea membrului drept al relatiei (1).
Am folosit ca daca f_1, f_2,...,f_k sunt functii de variabila x, atunci produsul lor derivat este:
(f_1* f_2*...*f_k)\prime= f_1\prime* f_2 *f_3*...*f_k+f_1* f_2\prime* f_3*...*f_k+...+f_1* f_2*...*f_{k-1}*f_k\prime

La noi, in cazul concret, k=n-1, f_1=x-x_1,...f_{n-1}=x-x_{n-1}
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grapefruit
Euclid


Joined: 23 Mar 2014
Posts: 29

PostPosted: Mon May 12, 2014 5:59 pm    Post subject: Reply with quote

grapefruit wrote:
Nu prea imi dau seama cum ati derivat acel produs lung de la inceput..


Edit :Am observat ,ati luat primul factor ca fiind un f si restu la g si ati derivat ca f*g..
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Adriana Nistor
Thales


Joined: 07 Aug 2008
Posts: 116
Location: Bucuresti

PostPosted: Mon May 12, 2014 5:59 pm    Post subject: Reply with quote

Asa este.
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