Question Number 46425 by maxmathsup by imad last updated on 25/Oct/18
![let p(x)=(x+i)^n −(x−i)^n with i^2 =−1 1) find p(x) at form Σ a_k x^k 2) find the roots of p(x) 3) factorize inside C[x] p(x) 4) factorize inside R[x] the polynom p(x) 5) decompose the fraction F(x)=(1/(p(x)))](https://www.tinkutara.com/question/Q46425.png)
$${let}\:{p}\left({x}\right)=\left({x}+{i}\right)^{{n}} −\left({x}−{i}\right)^{{n}} \:\:\:{with}\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{p}\left({x}\right)\:{at}\:{form}\:\Sigma\:{a}_{{k}} {x}^{{k}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{factorize}\:{inside}\:{C}\left[{x}\right]\:{p}\left({x}\right) \\ $$$$\left.\mathrm{4}\right)\:{factorize}\:{inside}\:{R}\left[{x}\right]\:{the}\:{polynom}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{5}\right)\:{decompose}\:{the}\:{fraction}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{p}\left({x}\right)} \\ $$
Commented by maxmathsup by imad last updated on 18/Nov/18
![1)we have P(x)=Σ_(k=0) ^n C_n ^k i^k x^(n−k) −Σ_(k=0) ^n C_n ^k (−i)^k x^(n−k) =Σ_(k=0) ^n C_n ^k (i^k −(−i)^k )x^(n−k) =Σ_(k=2p ) (...) +Σ_(k=2p+1) (...) =Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) (i^(2p+1) −(−i)^(2p+1) )x^(n−2p−1) =Σ_(p=0) ^([((n−1)/2)]) 2i (−1)^p C_n ^(2p+1) x^(n−2p−1) =2i Σ_(p=0) ^([((n−1)/2)]) (−1)^p C_n ^(2p+1) x^(n−2p−1) .](https://www.tinkutara.com/question/Q48044.png)
$$\left.\mathrm{1}\right){we}\:{have}\:{P}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{i}^{{k}} {x}^{{n}−{k}} \:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−{i}\right)^{{k}} \:{x}^{{n}−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left({i}^{{k}} −\left(−{i}\right)^{{k}} \right){x}^{{n}−{k}} \:=\sum_{{k}=\mathrm{2}{p}\:} \:\:\left(…\right)\:+\sum_{{k}=\mathrm{2}{p}+\mathrm{1}} \left(…\right) \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} {C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \left({i}^{\mathrm{2}{p}+\mathrm{1}} −\left(−{i}\right)^{\mathrm{2}{p}+\mathrm{1}} \right){x}^{{n}−\mathrm{2}{p}−\mathrm{1}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\mathrm{2}{i}\:\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{x}^{{n}−\mathrm{2}{p}−\mathrm{1}} \:=\mathrm{2}{i}\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{x}^{{n}−\mathrm{2}{p}−\mathrm{1}} . \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 18/Nov/18
![2) roots of P(x) P(x)=0 ⇔(x+i)^n =(x−i)^n ⇔(((x+i)/(x−i)))^n =1 ⇒((x+i)/(x−i))=e^((i2kπ)/n) k∈[[1,n−1]] ⇒x+i =(x−i)α_k (α_k =e^((i2k)/n) ) ⇒(1−α_k )x=−i−iα_k ⇒x_k =−i((1+α_k )/(1−α_k )) x_k =−i ((1+e^((i2kπ)/n) )/(1−e^((i2kπ)/n) )) =−i ((1+cos(((2kπ)/n))+i sin(((2kπ)/n)))/(1−cos(((2kπ)/n))−i sin(((2kπ)/n)))) =−i ((2cos^2 (((kπ)/n)) +2i sin(((kπ)/n))cos(((kπ)/n)))/(2sin^2 (((kπ)/n))−2i sin(((kπ)/n))cos(((kπ)/n)))) =−i ((cos(((kπ)/n))e^((ikπ)/n) )/(−isin(((kπ)/n))e^((ikπ)/n) )) =cotan(((kπ)/n)) so the roots of P(x) are x_k =cotan(((kπ)/n)) with k∈[[1,n−1]].](https://www.tinkutara.com/question/Q48046.png)
$$\left.\mathrm{2}\right)\:{roots}\:{of}\:{P}\left({x}\right) \\ $$$${P}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\left({x}+{i}\right)^{{n}} =\left({x}−{i}\right)^{{n}} \:\Leftrightarrow\left(\frac{{x}+{i}}{{x}−{i}}\right)^{{n}} \:=\mathrm{1}\:\:\Rightarrow\frac{{x}+{i}}{{x}−{i}}={e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \:\:\:{k}\in\left[\left[\mathrm{1},{n}−\mathrm{1}\right]\right] \\ $$$$\Rightarrow{x}+{i}\:=\left({x}−{i}\right)\alpha_{{k}} \:\:\:\:\:\:\left(\alpha_{{k}} ={e}^{\frac{{i}\mathrm{2}{k}}{{n}}} \right)\:\Rightarrow\left(\mathrm{1}−\alpha_{{k}} \right){x}=−{i}−{i}\alpha_{{k}} \:\Rightarrow{x}_{{k}} =−{i}\frac{\mathrm{1}+\alpha_{{k}} }{\mathrm{1}−\alpha_{{k}} } \\ $$$${x}_{{k}} =−{i}\:\frac{\mathrm{1}+{e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} }{\mathrm{1}−{e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} }\:=−{i}\:\frac{\mathrm{1}+{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)+{i}\:{sin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)−{i}\:{sin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)} \\ $$$$=−{i}\:\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)\:+\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)} \\ $$$$=−{i}\:\frac{{cos}\left(\frac{{k}\pi}{{n}}\right){e}^{\frac{{ik}\pi}{{n}}} }{−{isin}\left(\frac{{k}\pi}{{n}}\right){e}^{\frac{{ik}\pi}{{n}}} }\:={cotan}\left(\frac{{k}\pi}{{n}}\right)\:\:{so}\:{the}\:{roots}\:{of}\:{P}\left({x}\right)\:{are}\:\:{x}_{{k}} ={cotan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${with}\:{k}\in\left[\left[\mathrm{1},{n}−\mathrm{1}\right]\right]. \\ $$