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


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Question Number 55069 by gunawan last updated on 17/Feb/19

$Known analytic function$ $f (z) = \frac{2 (z - 2)}{z (z - 4)}$ $and written as f (z) = \underset{n = 0}{\overset{\propto}{Σ}} a_{n} {(z - 1)}^{n}$ $The value of a_{100} is . . .$
Commented by maxmathsup by imad last updated on 17/Feb/19

$w e h a v e f (z + 1) = \sum_{n = 0}^{\infty} a_{n} z^{n} = g (z) \Rightarrow t a y l o r s e r i e g i v e a_{n} = \frac{g^{(n)} (0)}{n!}$ $b u t g (z) = f (z + 1) = \frac{2 (z - 1)}{(z + 1) (z - 3)} = \frac{a}{z + 1} + \frac{b}{z - 3}$ $a = {l i m}_{z \to - 1} (z + 1) g (z) = \frac{- 4}{- 4} = 1$ $b = {l i m}_{z \to 3} (z - 3) g (z) = \frac{4}{4} = 1 \Rightarrow g (z) = \frac{1}{z + 1} + \frac{1}{z - 3} \Rightarrow g^{(n)} (z) = \frac{{(- 1)}^{n} n!}{{(z + 1)}^{n + 1}} + \frac{{(- 1)}^{n} n!}{{(z - 3)}^{n + 1}}$ $\Rightarrow g^{(n)} (0) = {(- 1)}^{n} n! + \frac{{(- 1)}^{n} n!}{{(- 3)}^{n + 1}} \Rightarrow a_{n} = {(- 1)}^{n} + \frac{{(- 1)}^{n}}{{(- 3)}^{n + 1}}$ $a_{n} = {(- 1)}^{n} - \frac{1}{3^{n + 1}} \Rightarrow a_{100} = 1 - \frac{1}{3^{101}}$
Commented by mr W last updated on 17/Feb/19

$t h a n k s s i r!$
Commented by maxmathsup by imad last updated on 17/Feb/19

$y o u a r e w e l c o m e s i r .$
	Answered by mr W last updated on 17/Feb/19
	$f(z)=((2(z−2))/(z(z−4)))=((2[(z−1)−1])/([(z−1)+1][(z−1)−3])) let t=z−1 f(z)=−((2(t−1))/(3(1+t)(1−(t/3)))) (1/(1+t))=Σ_(k=0) ^∞ (−t)^k =Σ_(k=0) ^∞ (−1)^k t^k (1/(1−(t/3)))=Σ_(k=9) ^∞ ((t/3))^k =Σ_(k=0) ^∞ (t^k /3^k ) f(z)=−(2/3)(t−1)[Σ_(k=0) ^∞ (−1)^k t^k ][Σ_(k=0) ^∞ (t^k /3^k )]=Σ_(n=0) ^∞ a_n t^n =Σ_(n=0) ^∞ a_n (z−1)^n a_(100) =−(2/3){Σ_(k=0) ^(99) (((−1)^(99−k) )/3^k )−Σ_(k=0) ^(100) (((−1)^(100−k) )/3^k )} a_(100) =−(2/3){2Σ_(k=0) ^(99) (((−1)^(99−k) )/3^k )−(1/3^(100) )} a_(100) =−(2/3){2(−(1/3^0 )+(1/3^1 )−(1/3^2 )+...+(1/3^(99) ))−(1/3^(100) )} a_(100) =−(2/3){2(−((1−(−(1/3))^(100) )/(1−(−(1/3)))))−(1/3^(100) )} a_(100) =−(2/3){−(3/2)[1−(1/3^(100) )]−(1/3^(100) )} a_(100) =−(2/3){−(3/2)+(1/(2×3^(100) ))} ⇒a_(100) =1−(1/3^(101) ) in general a_n =(−1)^n −(1/3^(n+1) )$
	$f (z) = \frac{2 (z - 2)}{z (z - 4)} = \frac{2 [(z - 1) - 1]}{[(z - 1) + 1] [(z - 1) - 3]}$ $l e t t = z - 1$ $f (z) = - \frac{2 (t - 1)}{3 (1 + t) (1 - \frac{t}{3})}$ $\frac{1}{1 + t} = \underset{k = 0}{\sum^{\infty}} {(- t)}^{k} = \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} t^{k}$ $\frac{1}{1 - \frac{t}{3}} = \underset{k = 9}{\sum^{\infty}} {(\frac{t}{3})}^{k} = \underset{k = 0}{\sum^{\infty}} \frac{t^{k}}{3^{k}}$ $f (z) = - \frac{2}{3} (t - 1) [\underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} t^{k}] [\underset{k = 0}{\sum^{\infty}} \frac{t^{k}}{3^{k}}] = \underset{n = 0}{\sum^{\infty}} a_{n} t^{n} = \underset{n = 0}{\sum^{\infty}} a_{n} {(z - 1)}^{n}$ $a_{100} = - \frac{2}{3} {\underset{k = 0}{\sum^{99}} \frac{{(- 1)}^{99 - k}}{3^{k}} - \underset{k = 0}{\sum^{100}} \frac{{(- 1)}^{100 - k}}{3^{k}}}$ $a_{100} = - \frac{2}{3} {2 \underset{k = 0}{\sum^{99}} \frac{{(- 1)}^{99 - k}}{3^{k}} - \frac{1}{3^{100}}}$ $a_{100} = - \frac{2}{3} {2 (- \frac{1}{3^{0}} + \frac{1}{3^{1}} - \frac{1}{3^{2}} + . . . + \frac{1}{3^{99}}) - \frac{1}{3^{100}}}$ $a_{100} = - \frac{2}{3} {2 (- \frac{1 - {(- \frac{1}{3})}^{100}}{1 - (- \frac{1}{3})}) - \frac{1}{3^{100}}}$ $a_{100} = - \frac{2}{3} {- \frac{3}{2} [1 - \frac{1}{3^{100}}] - \frac{1}{3^{100}}}$ $a_{100} = - \frac{2}{3} {- \frac{3}{2} + \frac{1}{2 \times 3^{100}}}$ $\Rightarrow a_{100} = 1 - \frac{1}{3^{101}}$ $i n g e n e r a l a_{n} = {(- 1)}^{n} - \frac{1}{3^{n + 1}}$
	Commented by mr W last updated on 17/Feb/19

	$t h e o t h e r m e t h o d i s u s i n g t a i l o r s e r i e s :$ $f (x) = \underset{n = 0}{\sum^{\infty}} \frac{f^{(n)} (a)}{n!} {(x - a)}^{n}$
	Commented by mr W last updated on 17/Feb/19

	$t h i s i s t h e m e t h o d i k n o w, o n l y u s i n g$ $g e o m e t r i c p r o g r e s s i o n .$ $p l e a s e f e e d b a c k i f m y a n s w e r i s c o r r e c t .$
	Commented by gunawan last updated on 17/Feb/19

	$Answer is correct Sir$ $thank you very much$ $i {don}^{'} t know orther solution$
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