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Question Number 31546 by prof Abdo imad last updated on 10/Mar/18

l e t c o n s i d e r t h e n u m r t i c a l f u n c t i o n

f (x) = \frac{1}{x^{2} + x + 1} c a l c u l a t e f^{(n)} (x) t h e n g i v e

f^{(n)} (0) .

Answered by prof Abdo imad last updated on 10/Apr/18

l e t d e c o m p o s e f (x) i n s i d e C (x)

r o o t s o f x^{2} + x + 1 = 0

Δ = 1 - 4 = - 3 = {(i \sqrt{3})}^{2} \Rightarrow x_{1} = \frac{- 1 + i \sqrt{3}}{2} = j

x_{2} = \frac{- 1 - i \sqrt{3}}{2} = \bar{j} s o

f (x) = \frac{1}{(x - j) (x - \bar{j})} = \frac{a}{x - j} + \frac{b}{x - \bar{j}}

a = \frac{1}{j - \bar{j}} = \frac{1}{i \sqrt{3}}, b = \frac{1}{\bar{j} - j} = \frac{1}{- i \sqrt{3}} \Rightarrow

f (x) = \frac{1}{i \sqrt{3}} (\frac{1}{x - j} - \frac{1}{x - \bar{j}}) \Rightarrow

f^{(n)} (x) = \frac{1}{i \sqrt{3}} ({(\frac{1}{x - j})}^{n} - {(\frac{1}{x - \bar{j}})}^{n})

= \frac{1}{i \sqrt{3}} (\frac{{(- 1)}^{n} n!}{{(x - j)}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(x - \bar{j})}^{n + 1}})

= \frac{{(- 1)}^{n} n!}{i \sqrt{3}} (\frac{1}{{(x - j)}^{n + 1}} - \frac{1}{{(x - \bar{j})}^{n + 1}}) \Rightarrow

f^{(n)} (0) = \frac{{(- 1)}^{n} n!}{i \sqrt{3}} (\frac{{(- 1)}^{n + 1}}{j^{n + 1}} - \frac{{(- 1)}^{n + 1}}{\overset{-^{n + 1}}{j}})

= \frac{- n!}{i \sqrt{3}} (\frac{- j^{n + 1} + \overset{- n + 1}{j}}{1}) = \frac{n!}{i \sqrt{3}} (j^{n + 1} - \overset{-^{n + 1}}{j})

= \frac{n!}{i \sqrt{3}} (2 i I m (j^{n + 1}) = \frac{2 n!}{\sqrt{3}} I m (e^{2 i (n + 1) \frac{π}{3}}) \Rightarrow

f^{(n)} (0) = \frac{2 (n!)}{\sqrt{3}} s i n (\frac{2 (n + 1) π}{3}) .