let-give-f-x-1-2x-2-3x-1-1-find-f-n-x-2-find-f-n-0-3-if-f-x-a-n-x-n-calculate-the-sequence-a-n-

Question Number 33126 by prof Abdo imad last updated on 10/Apr/18

l e t g i v e f (x) = \frac{1}{2 x^{2} - 3 x + 1}

1) f i n d f^{(n)} (x)

2) f i n d f^{(n)} (0)

3) i f f (x) = Σ a_{n} x^{n} c a l c u l a t e t h e s e q u e n c e a_{n}

Commented by prof Abdo imad last updated on 12/Apr/18

1) p o l e s o f f ? 2 x^{2} - 3 x + 1 = 0

Δ = 9 - 4 \times 2 = 1 \Rightarrow x_{1} = \frac{3 + 1}{4} = 1

x_{2} = \frac{3 - 1}{4} = \frac{1}{2} \Rightarrow f (x) = \frac{1}{2 (x - 1) (x - \frac{1}{2})}

= \frac{a}{x - 1} + \frac{b}{x - \frac{1}{2}}

a = \frac{1}{2 . \frac{1}{2}} = 1, b = \frac{1}{2 . (- \frac{1}{2})} = - 1 ⇏

f (x) = \frac{1}{x - 1} - \frac{1}{x - \frac{1}{2}} \Rightarrow

f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{{(x - 1)}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(x - \frac{1}{2})}^{n + 1}} \Rightarrow

2) f^{(n)} (0) = \frac{{(- 1)}^{n} n!}{{(- 1)}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(- \frac{1}{2})}^{n + 1}}

= - (n!) - \frac{{(- 1)}^{n} n!}{{(- 1)}^{n + 1}} 2^{n + 1} = n! 2^{n + 1} - n!

= n! (2^{n + 1} - 1)

3) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

\Rightarrow f (x) = \sum_{n = 0}^{\infty} (2^{n + 1} - 1) x^{n} .