Find-the-maximum-value-of-i-1-n-sin-5-i-with-i-1-n-sin-i-0-

Question Number 106941 by mr W last updated on 08/Aug/20

F i n d t h e m a x i m u m v a l u e o f \underset{i = 1}{\sum^{n}} \sin^{5} θ_{i}

w i t h \underset{i = 1}{\sum^{n}} \sin θ_{i} = 0 .

Answered by mr W last updated on 08/Aug/20

l e t x_{i} = \sin θ_{i}

- 1 ⩽ x_{i} ⩽ 1

l e t x_{1} = x_{2} = \dots = x_{m} = 1,

\Rightarrow x_{m + 1} = x_{m + 2} = \dots = x_{n} = - \frac{m}{n - m}

S w = \underset{i = 1}{\sum^{n}} \sin^{5} θ_{i} = m \times 1^{5} + (n - m) \times {(- \frac{m}{n - m})}^{5}

S = m - \frac{m^{5}}{{(n - m)}^{4}} = m [1 - {(\frac{m}{n - m})}^{4}]

\frac{d S}{d m} = 1 - \frac{5 m^{4}}{{(n - m)}^{4}} - \frac{4 m^{5}}{{(n - m)}^{5}} = 0

n^{3} - 5 n^{2} m + 10 {n m}^{2} - 10 m^{3} = 0

10 {(\frac{m}{n})}^{3} - 10 {(\frac{m}{n})}^{2} + 5 (\frac{m}{n}) - 1 = 0

\Rightarrow \frac{m}{n} \approx 0.3773

\Rightarrow m \approx 0.3773 n

S_{m a x} \approx 0.3773 n [1 - {(\frac{0.3773}{1 - 0.3773})}^{4}] \approx 0.3265 n

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