Menu Close

decompose-inside-C-x-the-fraction-F-x-1-x-2-1-n-with-n-integr-natural-and-n-1-

Tinku Tara 0 Comments
Pin




Question Number 57922 by maxmathsup by imad last updated on 14/Apr/19
decompose inside C(x) the fraction F(x) =(1/((x^2 +1)^n )) with n integr natural and n≥1
$${decompose}\:{inside}\:{C}\left({x}\right)\:{the}\:{fraction}\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:\:{with}\:{n}\:{integr}\:{natural} \\ $$$${and}\:{n}\geqslant\mathrm{1} \\ $$
Commented by maxmathsup by imad last updated on 30/Apr/19
we have F(x) =(1/((x−i)^n (x+i)^n )) =Σ_(k=1) ^n (a_k /((x−i)^k )) +Σ_(k=1) ^n (b_k /((x+i)^k )) from another side ch. x−i =t give F(x) =G(t) =(1/(t^n (t+2i)^n )) let determine D_(n−1) (0) for u(t) =(1/((t+2i)^n )) ⇒u(t) =Σ_(k=0) ^(n−1) ((u^((k)) (0))/(k!)) t^k +(t^n /(n!))ξ(t^n ) we have u(t) =(t+2i)^(−n) ⇒u^((1)) (t) =(−n)(t+2i)^(−n−1) ⇒ u^((2)) (t)=(−1)^2 n(n+1) (t+2i)^(−n−2) ⇒u^((p)) (t) =(−1)^p n(n+1)...(n+p−1)(t+2i)^(−n−p) ⇒u^((k)) (0) =(−1)^k n(n+1)...(n+k−1)(2i)^(−n−k) ⇒ u(t) =Σ_(k=0) ^(n−1) (((−1)^k n(n+1)....(n+k−1))/(k!(2i)^(n+k) )) t^k +(t^n /(n!)) ξ(t^n ) ⇒ ((u(t))/t^n ) = Σ_(k=0) ^(n−1) (((−1)^k n(n+1)....(n+k−1))/((2i)^(n+k) k! t^(n−k) )) +(1/(n!)) ξ(1) (n−k =p) =Σ_(p=1) ^n (((−1)^(n−p) n(n+1)....(n+n−p))/((n−p)! t^p (2i)^(2n−p) )) +(1/(n!))ξ(1) =Σ_(p=1) ^n (((−1)^(n−p) n(n+1)....(2n−p))/((n−p)! (x−i)^p (2i)^(2n−p) )) +(1/(n!))ξ(1) ⇒ a_k =(((−1)^(n−k) n(n+1)....(2n−k))/((n−k)!(2i)^(2n−k) )) now let find b_k we use ch .x+i =t ⇒ F(x) =H(t) =(1/(t^n (t−2i)^n )) let determine D_(n−1) (0) for v(t) =(t−2i)^(−n) ⇒ v(t) =Σ_(k=0) ^(n−1) ((v^((k)) (0))/(k!)) t^k +t^n ξ(t^n ) but v^((k)) (0) =(((−1)^k n(n+1)....(n+k−1))/((−2i)^(n+k) )) ⇒ ((v(t))/t^n )=Σ_(p=1) ^n (((−1)^(n−p) n(n+1)...(2n−p))/((n−p)!t^p (−2i)^(2n−p) )) +(1/(n!))ξ(1) ⇒ b_k =(((−1)^(n−k) n(n+1)....(2n−k))/((n−k)!(−2i)^(2n−k) )) ⇒ F(x) =Σ_(k=1) ^n (((−1)^(n−k) n(n+1)....(2n−k))/((n−k)!(2i)^(2n−k) (x−i)^k )) +Σ_(k=1) ^n (((−1)^(n−k) n(n+1)...(2n−k))/((n−k)!(−2i)^(2n−k) (x+i)^k )) and we see that b_k =conj(a_k ) so we can search only a_k ... error of typing change ξ(t^n ) by ξ(t)...
$${we}\:{have}\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{\left({x}−{i}\right)^{{n}} \left({x}+{i}\right)^{{n}} }\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{a}_{{k}} }{\left({x}−{i}\right)^{{k}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{b}_{{k}} }{\left({x}+{i}\right)^{{k}} } \\ $$$${from}\:{another}\:{side}\:\:{ch}.\:{x}−{i}\:={t}\:{give}\:{F}\left({x}\right)\:={G}\left({t}\right)\:=\frac{\mathrm{1}}{{t}^{{n}} \left({t}+\mathrm{2}{i}\right)^{{n}} } \\ $$$${let}\:{determine}\:{D}_{{n}−\mathrm{1}} \left(\mathrm{0}\right)\:{for}\:\:{u}\left({t}\right)\:=\frac{\mathrm{1}}{\left({t}+\mathrm{2}{i}\right)^{{n}} }\:\Rightarrow{u}\left({t}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{{u}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\:{t}^{{k}} \:+\frac{{t}^{{n}} }{{n}!}\xi\left({t}^{{n}} \right) \\ $$$${we}\:{have}\:{u}\left({t}\right)\:=\left({t}+\mathrm{2}{i}\right)^{−{n}} \:\Rightarrow{u}^{\left(\mathrm{1}\right)} \left({t}\right)\:=\left(−{n}\right)\left({t}+\mathrm{2}{i}\right)^{−{n}−\mathrm{1}} \:\Rightarrow \\ $$$${u}^{\left(\mathrm{2}\right)} \left({t}\right)=\left(−\mathrm{1}\right)^{\mathrm{2}} {n}\left({n}+\mathrm{1}\right)\:\left({t}+\mathrm{2}{i}\right)^{−{n}−\mathrm{2}} \:\Rightarrow{u}^{\left({p}\right)} \left({t}\right)\:=\left(−\mathrm{1}\right)^{{p}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{p}−\mathrm{1}\right)\left({t}+\mathrm{2}{i}\right)^{−{n}−{p}} \\ $$$$\Rightarrow{u}^{\left({k}\right)} \left(\mathrm{0}\right)\:=\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−{n}−{k}} \:\Rightarrow \\ $$$${u}\left({t}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)….\left({n}+{k}−\mathrm{1}\right)}{{k}!\left(\mathrm{2}{i}\right)^{{n}+{k}} }\:{t}^{{k}} \:+\frac{{t}^{{n}} }{{n}!}\:\xi\left({t}^{{n}} \right)\:\Rightarrow \\ $$$$\frac{{u}\left({t}\right)}{{t}^{{n}} }\:=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)….\left({n}+{k}−\mathrm{1}\right)}{\left(\mathrm{2}{i}\right)^{{n}+{k}} {k}!\:{t}^{{n}−{k}} }\:+\frac{\mathrm{1}}{{n}!}\:\xi\left(\mathrm{1}\right)\:\left({n}−{k}\:={p}\right) \\ $$$$=\sum_{{p}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−{p}} \:{n}\left({n}+\mathrm{1}\right)….\left({n}+{n}−{p}\right)}{\left({n}−{p}\right)!\:{t}^{{p}} \left(\mathrm{2}{i}\right)^{\mathrm{2}{n}−{p}} }\:+\frac{\mathrm{1}}{{n}!}\xi\left(\mathrm{1}\right) \\ $$$$=\sum_{{p}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−{p}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−{p}\right)}{\left({n}−{p}\right)!\:\left({x}−{i}\right)^{{p}} \left(\mathrm{2}{i}\right)^{\mathrm{2}{n}−{p}} }\:+\frac{\mathrm{1}}{{n}!}\xi\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${a}_{{k}} =\frac{\left(−\mathrm{1}\right)^{{n}−{k}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−{k}\right)}{\left({n}−{k}\right)!\left(\mathrm{2}{i}\right)^{\mathrm{2}{n}−{k}} }\:\:\:\:{now}\:{let}\:{find}\:{b}_{{k}} \:\:{we}\:{use}\:{ch}\:.{x}+{i}\:={t}\:\Rightarrow \\ $$$${F}\left({x}\right)\:={H}\left({t}\right)\:=\frac{\mathrm{1}}{{t}^{{n}} \left({t}−\mathrm{2}{i}\right)^{{n}} }\:\:{let}\:{determine}\:{D}_{{n}−\mathrm{1}} \left(\mathrm{0}\right)\:{for}\:{v}\left({t}\right)\:=\left({t}−\mathrm{2}{i}\right)^{−{n}} \:\Rightarrow \\ $$$${v}\left({t}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{{v}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\:{t}^{{k}} \:\:\:+{t}^{{n}} \:\xi\left({t}^{{n}} \right)\:\:\:\:{but}\:\:{v}^{\left({k}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)….\left({n}+{k}−\mathrm{1}\right)}{\left(−\mathrm{2}{i}\right)^{{n}+{k}} }\:\Rightarrow \\ $$$$\frac{{v}\left({t}\right)}{{t}^{{n}} }=\sum_{{p}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−{p}} {n}\left({n}+\mathrm{1}\right)…\left(\mathrm{2}{n}−{p}\right)}{\left({n}−{p}\right)!{t}^{{p}} \left(−\mathrm{2}{i}\right)^{\mathrm{2}{n}−{p}} }\:+\frac{\mathrm{1}}{{n}!}\xi\left(\mathrm{1}\right)\:\:\Rightarrow \\ $$$${b}_{{k}} =\frac{\left(−\mathrm{1}\right)^{{n}−{k}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−{k}\right)}{\left({n}−{k}\right)!\left(−\mathrm{2}{i}\right)^{\mathrm{2}{n}−{k}} }\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−{k}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−{k}\right)}{\left({n}−{k}\right)!\left(\mathrm{2}{i}\right)^{\mathrm{2}{n}−{k}} \left({x}−{i}\right)^{{k}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−{k}} {n}\left({n}+\mathrm{1}\right)…\left(\mathrm{2}{n}−{k}\right)}{\left({n}−{k}\right)!\left(−\mathrm{2}{i}\right)^{\mathrm{2}{n}−{k}} \left({x}+{i}\right)^{{k}} } \\ $$$${and}\:{we}\:{see}\:{that}\:{b}_{{k}} ={conj}\left({a}_{{k}} \right)\:{so}\:{we}\:{can}\:{search}\:{only}\:{a}_{{k}} … \\ $$$${error}\:{of}\:{typing}\:\:{change}\:\xi\left({t}^{{n}} \right)\:{by}\:\xi\left({t}\right)… \\ $$
Commented by maxmathsup by imad last updated on 30/Apr/19
F(x) =Σ_(k=1) ^n (((−1)^(n−k) n(n+1)...(2n−k−1))/((2i)^ (n−k)! (x−i)^k )) +Σ_(k=1) ^n (((−1)^(n−k) n(n+1)...(2n−k−1))/((−2i)^(2n−k) (n−k)!(x+i)^k ))
$${F}\left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−{k}} \:{n}\left({n}+\mathrm{1}\right)…\left(\mathrm{2}{n}−{k}−\mathrm{1}\right)}{\left(\mathrm{2}{i}\right)^{} \:\:\left({n}−{k}\right)!\:\left({x}−{i}\right)^{{k}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{n}−{k}} {n}\left({n}+\mathrm{1}\right)…\left(\mathrm{2}{n}−{k}−\mathrm{1}\right)}{\left(−\mathrm{2}\boldsymbol{{i}}\right)^{\mathrm{2}\boldsymbol{{n}}−\boldsymbol{{k}}} \left(\boldsymbol{{n}}−\boldsymbol{{k}}\right)!\left({x}+{i}\right)^{{k}} } \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *