Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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...

$$\mathrm{let}\:\mathrm{P}\left(\mathrm{x}\right)\:=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cosa}}\right)\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{2a}\right)}\right)....\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{na}\right)}\right)=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{ka}\right)}\right) \\ $$$$\mathrm{so}\:\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{ka}\right)}\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{this}\:\mathrm{polynom} \\ $$$$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{P}^{'} \left(\mathrm{x}\right)}{\mathrm{P}\left(\mathrm{x}\right)}\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{cos}\left(\mathrm{ka}\right)} \\ $$$$\mathrm{x}=\mathrm{0}\:\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{ka}\right)}\:=−\frac{\mathrm{P}^{'} \left(\mathrm{0}\right)}{\mathrm{P}\left(\mathrm{0}\right)} \\ $$$$\mathrm{P}\left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{cosa}.\mathrm{cos}\left(\mathrm{2a}\right)....\mathrm{cos}\left(\mathrm{na}\right)}\:\:\:\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{P}^{'} \left(\mathrm{0}\right)\:....\mathrm{be}\:\mathrm{continued}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com