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-

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 \dots

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) = \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}} + \dots + \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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