let-f-x-x-7-arctan-2x-1-calculate-f-4-0-and-f-7-0-2-calculate-f-5-1-

Question Number 130534 by mathmax by abdo last updated on 26/Jan/21

let f (x) = x^{7} \arctan (2 x)

1) calculate f^{(4)} (0) and f^{(7)} (0)

2) calculate f^{(5)} (1)

Answered by mathmax by abdo last updated on 30/Jan/21

f (x) = x^{7} \arctan (2 x) \Rightarrow f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(x^{7})}^{(k)} {(\arctan (2 x))}^{(n - k)}

= x^{7} {(\arctan (2 x))}^{(n)} + C_{n}^{1} {7 x}^{6} {(\arctan (2 x))}^{(n - 1)} + 7 . {6 C}_{n}^{2} x^{5} {(\arctan (2 x))}^{(n - 2)}

+ 7.6.5 C_{n}^{3} x^{4} {(\arctan (2 x))}^{(n - 3)} + 7.6.5.4 C_{n}^{4} x^{3} {(\arctan (2 x))}^{(n - 4)}

+ 7.6.5.4.3 C_{n}^{5} x^{2} {(\arctan (2 x))}^{(n - 5)} + 7.6.5.4.3 . {2 C}_{n}^{6} x {(\arctan (2 x))}^{(n - 6)}

+ 7.6.5.4.3.2.1 C_{n}^{7} {(\arctan (2 x))}^{(n - 7)}

we have {(\arctan (2 x))}^{(1)} = \frac{2}{1 + {4 x}^{2}} = \frac{2}{4 (x^{2} + \frac{1}{4})}

= \frac{1}{2 (x - \frac{i}{2}) (x + \frac{i}{2})} = \frac{1}{2 \times \frac{2 i}{2}} (\frac{1}{x - \frac{i}{2}} - \frac{1}{x + \frac{i}{2}}) = \frac{1}{2 i} (\frac{1}{x - \frac{i}{2}} - \frac{1}{x + \frac{i}{2}}) \Rightarrow

(\arctan {(2 x)}^{(m)} = \frac{1}{2 i} {\frac{{(- 1)}^{m - 1} (m - 1)!}{{(x - \frac{i}{2})}^{m}} - \frac{{(- 1)}^{m - 1} (m - 1)!}{{(x + \frac{i}{2})}^{m}}}

= \frac{{(- 1)}^{m - 1} (m - 1)!}{2 i} {\frac{2 iIm {(x + \frac{i}{2})}^{m}}{{(x^{2} + \frac{1}{4})}^{m}}}

= {(- 1)}^{m - 1} (m - 1)! \times \frac{Im {(x + \frac{i}{2})}^{m}}{{(x^{2} + \frac{1}{4})}^{m}} \dots . be continued \dots