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Known-analytic-function-f-z-2-z-2-z-z-4-and-written-as-f-z-n-0-a-n-z-1-n-The-value-of-a-100-is-

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Question Number 55069 by gunawan last updated on 17/Feb/19
Known analytic function f(z)=((2(z−2))/(z(z−4))) and written as f(z)=Σ_(n=0) ^(∝) a_n (z−1)^n The value of a_(100) is...
$$\mathrm{Known}\:\mathrm{analytic}\:\mathrm{function} \\ $$$${f}\left({z}\right)=\frac{\mathrm{2}\left({z}−\mathrm{2}\right)}{{z}\left({z}−\mathrm{4}\right)} \\ $$$$\mathrm{and}\:\mathrm{written}\:\mathrm{as}\:{f}\left({z}\right)=\underset{{n}=\mathrm{0}} {\overset{\propto} {\Sigma}}\:{a}_{{n}} \left({z}−\mathrm{1}\right)^{{n}} \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}_{\mathrm{100}} \:\mathrm{is}… \\ $$
Commented by maxmathsup by imad last updated on 17/Feb/19
we have f(z+1) =Σ_(n=0) ^∞ a_n z^n =g(z) ⇒taylor serie give a_n =((g^((n)) (0))/(n!)) but g(z)=f(z+1) =((2(z−1))/((z+1)(z−3))) =(a/(z+1)) +(b/(z−3)) a =lim_(z→−1) (z+1)g(z) =((−4)/(−4)) =1 b =lim_(z→3) (z−3)g(z) =(4/4) =1 ⇒g(z)=(1/(z+1)) +(1/(z−3)) ⇒g^((n)) (z)=(((−1)^n n!)/((z+1)^(n+1) )) +(((−1)^n n!)/((z−3)^(n+1) )) ⇒g^((n)) (0) =(−1)^n n! +(((−1)^n n!)/((−3)^(n+1) )) ⇒a_n =(−1)^n +(((−1)^n )/((−3)^(n+1) )) a_n =(−1)^n −(1/3^(n+1) ) ⇒a_(100) =1−(1/3^(101) )
$${we}\:{have}\:{f}\left({z}+\mathrm{1}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} {a}_{{n}} {z}^{{n}} \:={g}\left({z}\right)\:\:\Rightarrow{taylor}\:{serie}\:{give}\:{a}_{{n}} =\frac{{g}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!} \\ $$$${but}\:{g}\left({z}\right)={f}\left({z}+\mathrm{1}\right)\:=\frac{\mathrm{2}\left({z}−\mathrm{1}\right)}{\left({z}+\mathrm{1}\right)\left({z}−\mathrm{3}\right)}\:=\frac{{a}}{{z}+\mathrm{1}}\:+\frac{{b}}{{z}−\mathrm{3}} \\ $$$${a}\:={lim}_{{z}\rightarrow−\mathrm{1}} \left({z}+\mathrm{1}\right){g}\left({z}\right)\:=\frac{−\mathrm{4}}{−\mathrm{4}}\:=\mathrm{1} \\ $$$${b}\:={lim}_{{z}\rightarrow\mathrm{3}} \left({z}−\mathrm{3}\right){g}\left({z}\right)\:=\frac{\mathrm{4}}{\mathrm{4}}\:=\mathrm{1}\:\Rightarrow{g}\left({z}\right)=\frac{\mathrm{1}}{{z}+\mathrm{1}}\:+\frac{\mathrm{1}}{{z}−\mathrm{3}}\:\Rightarrow{g}^{\left({n}\right)} \left({z}\right)=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({z}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:+\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({z}−\mathrm{3}\right)^{{n}+\mathrm{1}} } \\ $$$$\Rightarrow{g}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\left(−\mathrm{1}\right)^{{n}} {n}!\:+\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left(−\mathrm{3}\right)^{{n}+\mathrm{1}} }\:\Rightarrow{a}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \:+\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(−\mathrm{3}\right)^{{n}+\mathrm{1}} } \\ $$$${a}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \:−\frac{\mathrm{1}}{\mathrm{3}^{{n}+\mathrm{1}} }\:\Rightarrow{a}_{\mathrm{100}} \:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{101}} } \\ $$
Commented by mr W last updated on 17/Feb/19
thanks sir!
$${thanks}\:{sir}! \\ $$
Commented by maxmathsup by imad last updated on 17/Feb/19
you are welcome sir.
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Answered by mr W last updated on 17/Feb/19
f(z)=((2(z−2))/(z(z−4)))=((2[(z−1)−1])/([(z−1)+1][(z−1)−3])) let t=z−1 f(z)=−((2(t−1))/(3(1+t)(1−(t/3)))) (1/(1+t))=Σ_(k=0) ^∞ (−t)^k =Σ_(k=0) ^∞ (−1)^k t^k (1/(1−(t/3)))=Σ_(k=9) ^∞ ((t/3))^k =Σ_(k=0) ^∞ (t^k /3^k ) f(z)=−(2/3)(t−1)[Σ_(k=0) ^∞ (−1)^k t^k ][Σ_(k=0) ^∞ (t^k /3^k )]=Σ_(n=0) ^∞ a_n t^n =Σ_(n=0) ^∞ a_n (z−1)^n a_(100) =−(2/3){Σ_(k=0) ^(99) (((−1)^(99−k) )/3^k )−Σ_(k=0) ^(100) (((−1)^(100−k) )/3^k )} a_(100) =−(2/3){2Σ_(k=0) ^(99) (((−1)^(99−k) )/3^k )−(1/3^(100) )} a_(100) =−(2/3){2(−(1/3^0 )+(1/3^1 )−(1/3^2 )+...+(1/3^(99) ))−(1/3^(100) )} a_(100) =−(2/3){2(−((1−(−(1/3))^(100) )/(1−(−(1/3)))))−(1/3^(100) )} a_(100) =−(2/3){−(3/2)[1−(1/3^(100) )]−(1/3^(100) )} a_(100) =−(2/3){−(3/2)+(1/(2×3^(100) ))} ⇒a_(100) =1−(1/3^(101) ) in general a_n =(−1)^n −(1/3^(n+1) )
$${f}\left({z}\right)=\frac{\mathrm{2}\left({z}−\mathrm{2}\right)}{{z}\left({z}−\mathrm{4}\right)}=\frac{\mathrm{2}\left[\left({z}−\mathrm{1}\right)−\mathrm{1}\right]}{\left[\left({z}−\mathrm{1}\right)+\mathrm{1}\right]\left[\left({z}−\mathrm{1}\right)−\mathrm{3}\right]} \\ $$$${let}\:{t}={z}−\mathrm{1} \\ $$$${f}\left({z}\right)=−\frac{\mathrm{2}\left({t}−\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−\frac{{t}}{\mathrm{3}}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{t}}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{t}\right)^{{k}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} {t}^{{k}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\frac{{t}}{\mathrm{3}}}=\underset{{k}=\mathrm{9}} {\overset{\infty} {\sum}}\left(\frac{{t}}{\mathrm{3}}\right)^{{k}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{t}^{{k}} }{\mathrm{3}^{{k}} } \\ $$$${f}\left({z}\right)=−\frac{\mathrm{2}}{\mathrm{3}}\left({t}−\mathrm{1}\right)\left[\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} {t}^{{k}} \right]\left[\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{t}^{{k}} }{\mathrm{3}^{{k}} }\right]=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {t}^{{n}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} \left({z}−\mathrm{1}\right)^{{n}} \\ $$$${a}_{\mathrm{100}} =−\frac{\mathrm{2}}{\mathrm{3}}\left\{\underset{{k}=\mathrm{0}} {\overset{\mathrm{99}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{99}−{k}} }{\mathrm{3}^{{k}} }−\underset{{k}=\mathrm{0}} {\overset{\mathrm{100}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{100}−{k}} }{\mathrm{3}^{{k}} }\right\} \\ $$$${a}_{\mathrm{100}} =−\frac{\mathrm{2}}{\mathrm{3}}\left\{\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\mathrm{99}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{99}−{k}} }{\mathrm{3}^{{k}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{100}} }\right\} \\ $$$${a}_{\mathrm{100}} =−\frac{\mathrm{2}}{\mathrm{3}}\left\{\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{0}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{99}} }\right)−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{100}} }\right\} \\ $$$${a}_{\mathrm{100}} =−\frac{\mathrm{2}}{\mathrm{3}}\left\{\mathrm{2}\left(−\frac{\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{100}} }{\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)}\right)−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{100}} }\right\} \\ $$$${a}_{\mathrm{100}} =−\frac{\mathrm{2}}{\mathrm{3}}\left\{−\frac{\mathrm{3}}{\mathrm{2}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{100}} }\right]−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{100}} }\right\} \\ $$$${a}_{\mathrm{100}} =−\frac{\mathrm{2}}{\mathrm{3}}\left\{−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}^{\mathrm{100}} }\right\} \\ $$$$\Rightarrow{a}_{\mathrm{100}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{101}} } \\ $$$$ \\ $$$${in}\:{general}\:{a}_{{n}} =\left(−\mathrm{1}\right)^{{n}} −\frac{\mathrm{1}}{\mathrm{3}^{{n}+\mathrm{1}} } \\ $$
Commented by mr W last updated on 17/Feb/19
the other method is using tailor series: f(x)=Σ_(n=0) ^∞ ((f^((n)) (a))/(n!))(x−a)^n
$${the}\:{other}\:{method}\:{is}\:{using}\:{tailor}\:{series}: \\ $$$${f}\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{f}^{\left({n}\right)} \left({a}\right)}{{n}!}\left({x}−{a}\right)^{{n}} \\ $$
Commented by mr W last updated on 17/Feb/19
this is the method i know, only using geometric progression. please feedback if my answer is correct.
$${this}\:{is}\:{the}\:{method}\:{i}\:{know},\:{only}\:{using} \\ $$$${geometric}\:{progression}. \\ $$$${please}\:{feedback}\:{if}\:{my}\:{answer}\:{is}\:{correct}. \\ $$
Commented by gunawan last updated on 17/Feb/19
Answer is correct Sir thank you very much i don′t know orther solution
$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{Sir} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$$$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{orther}\:\mathrm{solution} \\ $$

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