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Question Number 155542 by Engr_Jidda last updated on 02/Oct/21
find the tylor series expantion of ((z^2 −1)/((z+1)(z+3)))
$${find}\:{the}\:{tylor}\:{series}\:{expantion}\:{of}\:\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\left({z}+\mathrm{1}\right)\left({z}+\mathrm{3}\right)} \\ $$
Commented by aliyn last updated on 02/Oct/21
i think maclurin series but not tylor series ?
$$\boldsymbol{{i}}\:\boldsymbol{{think}}\:\boldsymbol{{maclurin}}\:\boldsymbol{{series}}\:\boldsymbol{{but}}\:\boldsymbol{{not}}\:\boldsymbol{{tylor}}\:\boldsymbol{{series}}\:? \\ $$
Commented by aliyn last updated on 02/Oct/21
Solution: f(z) = ((z^2 −1)/((z+1)(z+3))) = (((z+3)−4)/((z+3))) = 1 − 4 (z+3)^(−1) f(z) = 1 − 4(z+3)^(−1) ⇒ f(0) = 1 − 4(3)^(−1) = − (1/3) f^′ (z) = 4 (z+3)^(−2) ⇒ f^′ (0) = 4 (3)^(−2) = (4/9) f^(′′) (z) = −8 (z+3)^(−3) ⇒ f^(′′) (0) = − (8/(27)) f^(′′′) (z) = 24 (z+3)^(−4) ⇒ f^(′′′) (0) = ((24)/(81)) f^((4)) (z) = − 96 (z+3)^(−5) ⇒ f^((4)) (0) = − ((96)/(243)) : : : f^n (z) ∴ f (z) = f (0) + f^′ (0) (z−a) + ((f^(′′) (z))/(2!))(z−a)^2 + ((f^(′′′) (z))/(3!))(z−a)^3 + ((f^( (4)) (z))/(4!))(z−a)^4 +......+ ((f^((n)) (z))/(n!))(z−a)^n ∴ f (z) = − (1/3) + (4/9) z − (8/(27×2!)) z^2 + ((24)/(81×3!)) z^3 − ((96)/(243×4!)) z^4 +....(√) ⟨ M . T ⟩
$$\boldsymbol{{Solution}}: \\ $$$$ \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{z}}\right)\:=\:\frac{\boldsymbol{{z}}^{\mathrm{2}} −\mathrm{1}}{\left(\boldsymbol{{z}}+\mathrm{1}\right)\left(\boldsymbol{{z}}+\mathrm{3}\right)}\:=\:\frac{\left(\boldsymbol{{z}}+\mathrm{3}\right)−\mathrm{4}}{\left(\boldsymbol{{z}}+\mathrm{3}\right)}\:=\:\mathrm{1}\:−\:\mathrm{4}\:\left(\boldsymbol{{z}}+\mathrm{3}\right)^{−\mathrm{1}} \\ $$$$ \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{z}}\right)\:=\:\mathrm{1}\:−\:\mathrm{4}\left(\boldsymbol{{z}}+\mathrm{3}\right)^{−\mathrm{1}} \:\:\:\Rightarrow\:\boldsymbol{{f}}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:−\:\mathrm{4}\left(\mathrm{3}\right)^{−\mathrm{1}} \:=\:−\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$$\boldsymbol{{f}}\:^{'} \left(\boldsymbol{{z}}\right)\:\:\:=\:\mathrm{4}\:\left(\boldsymbol{{z}}+\mathrm{3}\right)^{−\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\Rightarrow\:\boldsymbol{{f}}\:^{'} \left(\mathrm{0}\right)\:=\:\mathrm{4}\:\left(\mathrm{3}\right)^{−\mathrm{2}} \:=\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$ \\ $$$$\boldsymbol{{f}}\:^{''} \left(\boldsymbol{{z}}\right)\:\:\:=\:−\mathrm{8}\:\left(\boldsymbol{{z}}+\mathrm{3}\right)^{−\mathrm{3}} \:\:\:\:\:\Rightarrow\:\boldsymbol{{f}}\:^{''} \left(\mathrm{0}\right)\:=\:−\:\frac{\mathrm{8}}{\mathrm{27}} \\ $$$$ \\ $$$$\boldsymbol{{f}}\:^{'''} \left(\boldsymbol{{z}}\right)\:\:=\:\mathrm{24}\:\left(\boldsymbol{{z}}+\mathrm{3}\right)^{−\mathrm{4}} \:\:\:\:\:\:\:\Rightarrow\:\boldsymbol{{f}}\:^{'''} \left(\mathrm{0}\right)\:=\:\frac{\mathrm{24}}{\mathrm{81}} \\ $$$$ \\ $$$$\boldsymbol{{f}}\:^{\left(\mathrm{4}\right)} \left(\boldsymbol{{z}}\right)\:=\:−\:\mathrm{96}\:\left(\boldsymbol{{z}}+\mathrm{3}\right)^{−\mathrm{5}} \:\Rightarrow\:\boldsymbol{{f}}\:^{\left(\mathrm{4}\right)} \left(\mathrm{0}\right)\:=\:−\:\frac{\mathrm{96}}{\mathrm{243}}\: \\ $$$$ \\ $$$$: \\ $$$$: \\ $$$$:\:\boldsymbol{{f}}\:^{\boldsymbol{{n}}} \left(\boldsymbol{{z}}\right)\: \\ $$$$ \\ $$$$\therefore\:\boldsymbol{{f}}\:\left(\boldsymbol{{z}}\right)\:=\:\boldsymbol{{f}}\:\left(\mathrm{0}\right)\:+\:\boldsymbol{{f}}\:^{'} \left(\mathrm{0}\right)\:\left(\boldsymbol{{z}}−\boldsymbol{{a}}\right)\:+\:\frac{\boldsymbol{{f}}\:^{''} \left(\boldsymbol{{z}}\right)}{\mathrm{2}!}\left(\boldsymbol{{z}}−\boldsymbol{{a}}\right)^{\mathrm{2}} \:+\:\frac{\boldsymbol{{f}}\:^{'''} \left(\boldsymbol{{z}}\right)}{\mathrm{3}!}\left(\boldsymbol{{z}}−\boldsymbol{{a}}\right)^{\mathrm{3}} \:+\:\frac{\boldsymbol{{f}}^{\:\left(\mathrm{4}\right)} \left(\boldsymbol{{z}}\right)}{\mathrm{4}!}\left(\boldsymbol{{z}}−\boldsymbol{{a}}\right)^{\mathrm{4}} \:+……+\:\frac{\boldsymbol{{f}}\:^{\left(\boldsymbol{{n}}\right)} \left(\boldsymbol{{z}}\right)}{\boldsymbol{{n}}!}\left(\boldsymbol{{z}}−\boldsymbol{{a}}\right)^{\boldsymbol{{n}}} \\ $$$$ \\ $$$$\therefore\:\boldsymbol{{f}}\:\left(\boldsymbol{{z}}\right)\:=\:−\:\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{4}}{\mathrm{9}}\:\boldsymbol{{z}}\:−\:\frac{\mathrm{8}}{\mathrm{27}×\mathrm{2}!}\:\boldsymbol{{z}}^{\mathrm{2}} +\:\frac{\mathrm{24}}{\mathrm{81}×\mathrm{3}!}\:\boldsymbol{{z}}^{\mathrm{3}} \:−\:\frac{\mathrm{96}}{\mathrm{243}×\mathrm{4}!}\:\boldsymbol{{z}}^{\mathrm{4}} +….\sqrt{} \\ $$$$ \\ $$$$\langle\:\boldsymbol{{M}}\:.\:\boldsymbol{{T}}\:\:\rangle \\ $$$$ \\ $$
Commented by Engr_Jidda last updated on 27/Oct/21
thanks
$${thanks} \\ $$
Commented by Engr_Jidda last updated on 27/Oct/21
yes sir
$${yes}\:{sir} \\ $$
Answered by mr W last updated on 02/Oct/21
((z^2 −1)/((z+1)(z+3)))=((z−1)/(z+3))=1−(4/(3(1+(z/3)))) =1−(4/3)[1−((z/3))+((z/3))^2 −((z/3))^3 +...] =−(1/3)+((4z)/3^2 )−((4z^2 )/3^3 )+((4z^3 )/3^4 )−((4z^4 )/3^5 )+...
$$\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\left({z}+\mathrm{1}\right)\left({z}+\mathrm{3}\right)}=\frac{{z}−\mathrm{1}}{{z}+\mathrm{3}}=\mathrm{1}−\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{1}+\frac{{z}}{\mathrm{3}}\right)} \\ $$$$=\mathrm{1}−\frac{\mathrm{4}}{\mathrm{3}}\left[\mathrm{1}−\left(\frac{{z}}{\mathrm{3}}\right)+\left(\frac{{z}}{\mathrm{3}}\right)^{\mathrm{2}} −\left(\frac{{z}}{\mathrm{3}}\right)^{\mathrm{3}} +…\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{4}{z}}{\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}{z}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} }−\frac{\mathrm{4}{z}^{\mathrm{4}} }{\mathrm{3}^{\mathrm{5}} }+… \\ $$
Commented by Engr_Jidda last updated on 27/Oct/21
thanks
$${thanks} \\ $$

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