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Question-96293




Question Number 96293 by 06122004 last updated on 31/May/20
Answered by mathmax by abdo last updated on 31/May/20
let P(x) =(x−(1/(cosa)))(x−(1/(cos(2a))))....(x−(1/(cos(na))))=Π_(k=1) ^n  (x−(1/(cos(ka))))  so (1/(cos(ka))) are roots of this polynom  we have ((P^′ (x))/(P(x))) =Σ_(k=1) ^n  (1/(x−cos(ka)))  x=0 ⇒Σ_(k=1) ^n  (1/(cos(ka))) =−((P^′ (0))/(P(0)))  P(0) =(((−1)^n )/(cosa.cos(2a)....cos(na)))   rest to find P^′ (0) ....be continued...
$$\mathrm{let}\:\mathrm{P}\left(\mathrm{x}\right)\:=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cosa}}\right)\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{2a}\right)}\right)….\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{na}\right)}\right)=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{ka}\right)}\right) \\ $$$$\mathrm{so}\:\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{ka}\right)}\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{this}\:\mathrm{polynom} \\ $$$$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{P}^{'} \left(\mathrm{x}\right)}{\mathrm{P}\left(\mathrm{x}\right)}\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{cos}\left(\mathrm{ka}\right)} \\ $$$$\mathrm{x}=\mathrm{0}\:\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{ka}\right)}\:=−\frac{\mathrm{P}^{'} \left(\mathrm{0}\right)}{\mathrm{P}\left(\mathrm{0}\right)} \\ $$$$\mathrm{P}\left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{cosa}.\mathrm{cos}\left(\mathrm{2a}\right)….\mathrm{cos}\left(\mathrm{na}\right)}\:\:\:\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{P}^{'} \left(\mathrm{0}\right)\:….\mathrm{be}\:\mathrm{continued}… \\ $$

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