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Question Number 148082 by tabata last updated on 25/Jul/21

Commented by tabata last updated on 25/Jul/21

help me msr olaf

$${help}\:{me}\:{msr}\:{olaf} \\ $$

Answered by Olaf_Thorendsen last updated on 25/Jul/21

f(z) = ((z+4)/(z^2 (z^2 +3z+2))), ∣z∣ > 2 f(z) = ((z+4)/(z^2 (z+1)(z+2))) f(z) = (2/z^2 )−(5/(2z)).+(3/(z+1))−(1/(2(z+2))) f(z) = (2/z^2 )−(5/(2z))+(3/z).(1/(1+(1/z)))−(1/(2z)).(1/(1+(2/z))) f(z) = (2/z^2 )−(5/(2z))+(3/z)Σ_(n=0) ^∞ (((−1)^n )/z^n )−(1/(2z)).Σ_(n=0) ^∞ (−1)^n (2^n /z^n ) f(z) = (2/z^2 )−(5/(2z))+Σ_(n=0) ^∞ (−1)^n (3−2^(n−1) )(1/z^(n+1) ) f(z) = (2/z^2 )−(5/(2z))−Σ_(n=1) ^∞ (−1)^n (3−2^(n−2) )(1/z^n ) f(z) = −(1/z^2 )−Σ_(n=3) ^∞ (−1)^n (3−2^(n−2) )(1/z^n ) (if ∣z∣ > 2)

$${f}\left({z}\right)\:=\:\frac{{z}+\mathrm{4}}{{z}^{\mathrm{2}} \left({z}^{\mathrm{2}} +\mathrm{3}{z}+\mathrm{2}\right)},\:\mid{z}\mid\:>\:\mathrm{2} \\ $$$${f}\left({z}\right)\:=\:\frac{{z}+\mathrm{4}}{{z}^{\mathrm{2}} \left({z}+\mathrm{1}\right)\left({z}+\mathrm{2}\right)} \\ $$$${f}\left({z}\right)\:=\:\frac{\mathrm{2}}{{z}^{\mathrm{2}} }−\frac{\mathrm{5}}{\mathrm{2}{z}}.+\frac{\mathrm{3}}{{z}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\left({z}+\mathrm{2}\right)} \\ $$$${f}\left({z}\right)\:=\:\frac{\mathrm{2}}{{z}^{\mathrm{2}} }−\frac{\mathrm{5}}{\mathrm{2}{z}}+\frac{\mathrm{3}}{{z}}.\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{z}}}−\frac{\mathrm{1}}{\mathrm{2}{z}}.\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}}{{z}}} \\ $$$${f}\left({z}\right)\:=\:\frac{\mathrm{2}}{{z}^{\mathrm{2}} }−\frac{\mathrm{5}}{\mathrm{2}{z}}+\frac{\mathrm{3}}{{z}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{z}^{{n}} }−\frac{\mathrm{1}}{\mathrm{2}{z}}.\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{2}^{{n}} }{{z}^{{n}} } \\ $$$${f}\left({z}\right)\:=\:\frac{\mathrm{2}}{{z}^{\mathrm{2}} }−\frac{\mathrm{5}}{\mathrm{2}{z}}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{3}−\mathrm{2}^{{n}−\mathrm{1}} \right)\frac{\mathrm{1}}{{z}^{{n}+\mathrm{1}} } \\ $$$${f}\left({z}\right)\:=\:\frac{\mathrm{2}}{{z}^{\mathrm{2}} }−\frac{\mathrm{5}}{\mathrm{2}{z}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{3}−\mathrm{2}^{{n}−\mathrm{2}} \right)\frac{\mathrm{1}}{{z}^{{n}} } \\ $$$${f}\left({z}\right)\:=\:−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }−\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{3}−\mathrm{2}^{{n}−\mathrm{2}} \right)\frac{\mathrm{1}}{{z}^{{n}} } \\ $$$$\left({if}\:\mid{z}\mid\:>\:\:\mathrm{2}\right) \\ $$

Commented by tabata last updated on 25/Jul/21

put sir (3/(z+1)) is converg ajust ∣z∣<1 how Σ_(n=0) ^∞ (−1)^n ((1/z))^(n )

$${put}\:{sir}\:\frac{\mathrm{3}}{{z}+\mathrm{1}}\:{is}\:{converg}\:{ajust}\:\mid{z}\mid<\mathrm{1}\:{how} \\ $$$$ \\ $$$$\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{{z}}\right)^{{n}\:} \\ $$

Commented by tabata last updated on 25/Jul/21

i think =Σ_(n=0) ^∞ (−1)^n (z)^n

$${i}\:{think}\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left({z}\right)^{{n}} \\ $$

Commented by qaz last updated on 25/Jul/21

∣z∣>2 ⇒(1/(∣z∣))<(2/(∣z∣))<1

$$\mid\mathrm{z}\mid>\mathrm{2}\:\:\:\Rightarrow\frac{\mathrm{1}}{\mid\mathrm{z}\mid}<\frac{\mathrm{2}}{\mid\mathrm{z}\mid}<\mathrm{1} \\ $$

Commented by mathmax by abdo last updated on 25/Jul/21

f(z)=((z+4)/(z^2 (z^2 +3z+2)))=((z+4)/z^2 )((1/(z+1))−(1/(z+2))) =(1/z^2 )(((z+4)/(z+1))−((z+4)/(z+2)))=(1/z^2 )(1+(3/(z+1))−1−(2/(z+2))) =(3/(z^2 (z+1)))−(2/(z^2 (z+2))) we have ∣z∣>2 ⇒∣(1/z)∣<(1/2) and ∣(2/z)∣<1 f(z)=(3/(z^3 (1+(1/z))))−(2/(z^3 (1+(2/z)))) =(3/z^3 )Σ_(n=0) ^∞ (((−1)^n )/z^n )−(2/z^3 )Σ_(n=0) ^(∞ ) (−1)^n ((2/z))^n =3Σ_(n=0) ^∞ (((−1)^n )/z^(n+3) )−2 Σ_(n=0) ^∞ (((−2)^n )/z^(n+3) ) =Σ_(n=0) ^∞ (−1)^n +(−2)^(n+1) )×(1/z^(n+3) )

$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{z}+\mathrm{4}}{\mathrm{z}^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} +\mathrm{3z}+\mathrm{2}\right)}=\frac{\mathrm{z}+\mathrm{4}}{\mathrm{z}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{z}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{z}+\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} }\left(\frac{\mathrm{z}+\mathrm{4}}{\mathrm{z}+\mathrm{1}}−\frac{\mathrm{z}+\mathrm{4}}{\mathrm{z}+\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{z}+\mathrm{1}}−\mathrm{1}−\frac{\mathrm{2}}{\mathrm{z}+\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{z}^{\mathrm{2}} \left(\mathrm{z}+\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{z}^{\mathrm{2}} \left(\mathrm{z}+\mathrm{2}\right)}\:\:\mathrm{we}\:\mathrm{have}\:\mid\mathrm{z}\mid>\mathrm{2}\:\Rightarrow\mid\frac{\mathrm{1}}{\mathrm{z}}\mid<\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\mid\frac{\mathrm{2}}{\mathrm{z}}\mid<\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{3}}{\mathrm{z}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{z}}\right)}−\frac{\mathrm{2}}{\mathrm{z}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{z}}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{z}^{\mathrm{3}} }\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{z}^{\mathrm{n}} }−\frac{\mathrm{2}}{\mathrm{z}^{\mathrm{3}} }\sum_{\mathrm{n}=\mathrm{0}} ^{\infty\:} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\frac{\mathrm{2}}{\mathrm{z}}\right)^{\mathrm{n}} \\ $$$$=\mathrm{3}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{z}^{\mathrm{n}+\mathrm{3}} }−\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{2}\right)^{\mathrm{n}} }{\mathrm{z}^{\mathrm{n}+\mathrm{3}} } \\ $$$$\left.=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} +\left(−\mathrm{2}\right)^{\mathrm{n}+\mathrm{1}} \right)×\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{n}+\mathrm{3}} } \\ $$

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