Given-f-x-sin-2-x-find-the-expansion-of-f-x-up-to-the-n-th-term-

Question Number 98953 by Ar Brandon last updated on 17/Jun/20

G iven f (x) = \sin^{2} x find the expansion of f (x)

up to the n^{th} term .

Answered by mr W last updated on 17/Jun/20

f (x) = \sin^{2} x = \frac{1}{2} (1 - \cos 2 x)

= \frac{1}{2} - \frac{1}{2} (1 - \frac{{(2 x)}^{2}}{2!} + \frac{{(2 x)}^{4}}{4!} - \dots)

= \frac{1}{2} - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} 2^{2 n - 1} x^{2 n}}{(2 n)!}

Answered by mathmax by abdo last updated on 17/Jun/20

f (x) = \sum_{k = 0}^{n} \frac{f^{(k)} (0)}{k!} x^{k} + \dots . we have f (x) = \sin^{2} x \Rightarrow f^{^{'}} (x) = 2 sinx cosx \Rightarrow

f^{(k)} (x) = 2 {(sinxcosx)}^{(k - 1)} = 2 \sum_{p = 0}^{k - 1} C_{k - 1}^{p} \sin^{(p)} (x) \cos^{(k - p)} (x)

f^{(k)} (x) = 2 \sum_{p = 0}^{k - 1} \sin (x + \frac{p π}{2}) \cos (x + \frac{(k - p) π}{2}) \Rightarrow

f^{(k)} (0) = 2 \sum_{p = 0}^{k - 1} \sin (\frac{p π}{2}) \cos (\frac{(k - p) π}{2}) \Rightarrow

\sin^{2} x = 2 \sum_{k = 1}^{n} \frac{1}{k!} (\sum_{p = 0}^{k - 1} \sin (\frac{p π}{2}) \cos (\frac{(k - p) π}{2})) x^{k} + \dots .