let-f-x-x-3-cos-2x-calculate-f-n-x-and-f-n-0-

Question Number 90960 by mathmax by abdo last updated on 27/Apr/20

l e t f (x) = x^{3} c o s (2 x)

c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (0)

Commented by mathmax by abdo last updated on 27/Apr/20

l e i b n i z g i v e f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(x^{3})}^{(k)} {(c o s (2 x))}^{(n - k)}

f i r s t l e t f i n d {(c o s (2 x))}^{(p)}

{(c o s (2 x))}^{(1)} = - 2 s i n (2 x) = 2 c o s (2 x + \frac{π}{2})

{(c o s (2 x))}^{(2)} = 4 c o s (2 x + 2 \frac{π}{2}) l e t s u p p o s e

{(c o s (2 x))}^{(p)} = 2^{p} c o s (2 x + \frac{p π}{2}) \Rightarrow {(c o s (2 x))}^{(p + 1)} = - 2^{p + 1} s i n (2 x + \frac{p π}{2})

= 2^{p + 1} c o s (2 x + \frac{(p + 1) π}{2}) \Rightarrow

f^{(n)} (x) = C_{n}^{0} x^{3} 2^{n} c o s (2 x + \frac{n π}{2}) + C_{n}^{1} 3 x^{2} 2^{n - 1} c o s (2 x + \frac{(n - 1) π}{2})

+ C_{n}^{2} (6 x) 2^{n - 2} c o s (2 x + \frac{(n - 2) π}{2}) + C_{n}^{3} 6 2^{n - 3} c o s (2 x + \frac{(n - 3) π}{2})

f^{(n)} (x) = 2^{n} x^{3} c o s (2 x + \frac{n π}{2}) + 3 n 2^{n - 1} x^{2} c o s (2 x + \frac{(n - 1) π}{2})

+ 3 n (n - 1) 2^{n - 2} x c o s (2 x + \frac{(n - 2) π}{2}) + n (n - 1) (n - 2) 2^{n - 3} c o s (2 x + \frac{(n - 3) π}{2})

Commented by mathmax by abdo last updated on 27/Apr/20

f^{(n)} (0) = n (n - 1) (n - 2) 2^{n - 3} c o s (\frac{(n - 3) π}{2})

Answered by MWSuSon last updated on 27/Apr/20

f^{(n)} (x) = [2^{n} \cos (2 x + \frac{n π}{2}) x^{3} + n 2^{n - 1} \cos (2 x + \frac{(n - 1) π}{2}) 3 x^{2}

+ \frac{n (n - 1) 2^{n - 2} \cos (2 x + \frac{(n - 2) π}{2}) 6 x}{2} + \frac{n (n - 1) (n - 2) 2^{n - 3} \cos (2 x + \frac{(n - 3) π}{2}) 6}{6} .

f^{(n)} (0) = [n (n - 1) (n - 2) 2^{n - 3} \cos (\frac{(n - 3) π}{2})]

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