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a-b-c-C-x-0-a-y-0-b-z-0-c-x-n-1-x-n-y-n-y-n-1-y-n-z-n-z-n-1-z-n-x-n-x-n-y-n-z-n-n-1-




Question Number 4795 by 123456 last updated on 13/Mar/16
a,b,c∈C  x_0 =a  y_0 =b  z_0 =c  x_(n+1) =x_n −y_n   y_(n+1) =y_n −z_n   z_(n+1) =z_n −x_n   x_n +y_n +z_n =?,n≥1
$${a},{b},{c}\in\mathbb{C} \\ $$$${x}_{\mathrm{0}} ={a} \\ $$$${y}_{\mathrm{0}} ={b} \\ $$$${z}_{\mathrm{0}} ={c} \\ $$$${x}_{{n}+\mathrm{1}} ={x}_{{n}} −{y}_{{n}} \\ $$$${y}_{{n}+\mathrm{1}} ={y}_{{n}} −{z}_{{n}} \\ $$$${z}_{{n}+\mathrm{1}} ={z}_{{n}} −{x}_{{n}} \\ $$$${x}_{{n}} +{y}_{{n}} +{z}_{{n}} =?,{n}\geqslant\mathrm{1} \\ $$
Answered by Yozzii last updated on 13/Mar/16
Adding the three recurrence equations  we obtain   x_(n+1) +y_(n+1) +z_(n+1) =x_n −y_n +y_n −z_n +z_n −x_n =0.  ⇒x_n +y_n +z_n =0 for all n∈N. (∗)  Corollary: Since x_0 =a, y_0 =b, z_0 =c  ⇒ a+b+c=0 from (∗) where a,b,c∈C.  Indeed, ∃a,b,c∈C such that a+b+c=0.  E.g a=−1,b=3, c=−2 or a=1,b=e^(2πi/3) ,c=e^(−2πi/3) .
$${Adding}\:{the}\:{three}\:{recurrence}\:{equations} \\ $$$${we}\:{obtain}\: \\ $$$${x}_{{n}+\mathrm{1}} +{y}_{{n}+\mathrm{1}} +{z}_{{n}+\mathrm{1}} ={x}_{{n}} −{y}_{{n}} +{y}_{{n}} −{z}_{{n}} +{z}_{{n}} −{x}_{{n}} =\mathrm{0}. \\ $$$$\Rightarrow{x}_{{n}} +{y}_{{n}} +{z}_{{n}} =\mathrm{0}\:{for}\:{all}\:{n}\in\mathbb{N}.\:\left(\ast\right) \\ $$$${Corollary}:\:{Since}\:{x}_{\mathrm{0}} ={a},\:{y}_{\mathrm{0}} ={b},\:{z}_{\mathrm{0}} ={c} \\ $$$$\Rightarrow\:{a}+{b}+{c}=\mathrm{0}\:{from}\:\left(\ast\right)\:{where}\:{a},{b},{c}\in\mathbb{C}. \\ $$$${Indeed},\:\exists{a},{b},{c}\in\mathbb{C}\:{such}\:{that}\:{a}+{b}+{c}=\mathrm{0}. \\ $$$${E}.{g}\:{a}=−\mathrm{1},{b}=\mathrm{3},\:{c}=−\mathrm{2}\:{or}\:{a}=\mathrm{1},{b}={e}^{\mathrm{2}\pi{i}/\mathrm{3}} ,{c}={e}^{−\mathrm{2}\pi{i}/\mathrm{3}} . \\ $$
Commented by prakash jain last updated on 13/Mar/16
How do you conclude a+b+c=0.  Only relation for x_(n+1)  is given so   a+b+c need not be equal to zero.  x_n +y_n +z_n =0 ∀(a,b,c)∈C^3  if n≥1
$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{conclude}\:{a}+{b}+{c}=\mathrm{0}. \\ $$$$\mathrm{Only}\:\mathrm{relation}\:\mathrm{for}\:{x}_{{n}+\mathrm{1}} \:\mathrm{is}\:\mathrm{given}\:\mathrm{so}\: \\ $$$${a}+{b}+{c}\:\mathrm{need}\:\mathrm{not}\:\mathrm{be}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{zero}. \\ $$$${x}_{{n}} +{y}_{{n}} +{z}_{{n}} =\mathrm{0}\:\forall\left({a},{b},{c}\right)\in\mathbb{C}^{\mathrm{3}} \:\mathrm{if}\:{n}\geqslant\mathrm{1} \\ $$
Commented by Yozzii last updated on 13/Mar/16
Sorry. Error made.
$${Sorry}.\:{Error}\:{made}. \\ $$$$ \\ $$

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