find-the-tylor-series-expantion-of-z-2-1-z-1-z-3-

Question Number 155542 by Engr_Jidda last updated on 02/Oct/21

f i n d t h e t y l o r s e r i e s e x p a n t i o n o f \frac{z^{2} - 1}{(z + 1) (z + 3)}

Commented by aliyn last updated on 02/Oct/21

i t h i n k m a c l u r i n s e r i e s b u t n o t t y l o r s e r i e s ?

Commented by aliyn last updated on 02/Oct/21

S o l u t i o n :

f (z) = \frac{z^{2} - 1}{(z + 1) (z + 3)} = \frac{(z + 3) - 4}{(z + 3)} = 1 - 4 {(z + 3)}^{- 1}

f (z) = 1 - 4 {(z + 3)}^{- 1} \Rightarrow f (0) = 1 - 4 {(3)}^{- 1} = - \frac{1}{3}

f^{^{'}} (z) = 4 {(z + 3)}^{- 2} \Rightarrow f^{^{'}} (0) = 4 {(3)}^{- 2} = \frac{4}{9}

f^{^{″}} (z) = - 8 {(z + 3)}^{- 3} \Rightarrow f^{^{″}} (0) = - \frac{8}{27}

f^{^{‴}} (z) = 24 {(z + 3)}^{- 4} \Rightarrow f^{^{‴}} (0) = \frac{24}{81}

f^{(4)} (z) = - 96 {(z + 3)}^{- 5} \Rightarrow f^{(4)} (0) = - \frac{96}{243}

:

:

: f^{n} (z)

∴ f (z) = f (0) + f^{^{'}} (0) (z - a) + \frac{f^{^{″}} (z)}{2!} {(z - a)}^{2} + \frac{f^{^{‴}} (z)}{3!} {(z - a)}^{3} + \frac{f^{(4)} (z)}{4!} {(z - a)}^{4} + \dots \dots + \frac{f^{(n)} (z)}{n!} {(z - a)}^{n}

∴ f (z) = - \frac{1}{3} + \frac{4}{9} z - \frac{8}{27 \times 2!} z^{2} + \frac{24}{81 \times 3!} z^{3} - \frac{96}{243 \times 4!} z^{4} + \dots . \sqrt{}

⟨ M . T ⟩

Commented by Engr_Jidda last updated on 27/Oct/21

t h a n k s

Commented by Engr_Jidda last updated on 27/Oct/21

y e s s i r

Answered by mr W last updated on 02/Oct/21

\frac{z^{2} - 1}{(z + 1) (z + 3)} = \frac{z - 1}{z + 3} = 1 - \frac{4}{3 (1 + \frac{z}{3})}

= 1 - \frac{4}{3} [1 - (\frac{z}{3}) + {(\frac{z}{3})}^{2} - {(\frac{z}{3})}^{3} + \dots]

= - \frac{1}{3} + \frac{4 z}{3^{2}} - \frac{4 z^{2}}{3^{3}} + \frac{4 z^{3}}{3^{4}} - \frac{4 z^{4}}{3^{5}} + \dots

Commented by Engr_Jidda last updated on 27/Oct/21

t h a n k s