求 n : n*sin(2pi/n)=9pi/5

Solve n sin (2π/n) = 9π/5.

Let f(n) = n sin (2π/n) – 9π/5

f(0) is undefined
f(2) = 2 sin (2π/2) – 9π/5 = – 9π/5 < 0
f(12) = 12 sin (2π/12) – 9π/5 = 6 – 9π/5 > 0
∴There are odd numbers of roots lie between 2 and 12

When 2π/n ≧ π => 0 < n ≦ 2 ,
– 1 ≦ sin (2π/n) ≦ 1
– n ≦ n sin (2π/n) ≦ n
– max {(0, 2] } ≦ n sin (2π/n) ≦ max {(0, 2] }
– 2 ≦ n sin (2π/n) ≦ 2
∵9π/5 > 2
∴n sin (2π/n) – 9π/5 < 0
i.e. f(n) < 0 when 0 < n ≦ 2

f′(n) = sin (2π/n) + n cos (2π/n) × (– 2π/n²)
　　　= sin (2π/n) – (2π/n) cos (2π/n)

When π/2 ≦ 2π/n < π => 2 < n ≦ 4 ,
sin (2π/n) > 0
cos (2π/n) ≦ 0
∴sin (2π/n) – (2π/n) cos (2π/n) > 0
i.e. f′(n) > 0 when 2 < n ≦ 4

When 0 < 2π/n < π/2 => n > 4 ,
tan (2π/n) > 2π/n
[sin (2π/n)]/cos (2π/n) > 2π/n
∵cos (2π/n) > 0
∴sin (2π/n) > (2π/n) cos (2π/n)
sin (2π/n) – (2π/n) cos (2π/n) > 0
i.e. f′(n) > 0 when n > 4

∴f(n) is strictly increasing when n > 2
∴there is only one root lie between 2 and 12

Recall that f(n) is strictly increasing when n > 2
∴there is only one root lie between 2 and +∞

Recall that f(n) < 0 when 0 < n ≦ 2 , also f(0) is undefined
∴the equation n sin (2π/n) – 9π/5 = 0 has only one +ve root

f(– n) = – n sin (2π/(– n)) – 9π/5
　　　= – n sin (– 2π/n) – 9π/5
　　　= – (– n) sin (2π/n) – 9π/5
　　　= n sin (2π/n) – 9π/5
　　　= f(n)
∴f(n) is a even function
Hence the properties of f(n) when n < 0 can be formed by symmetric

Recall that the equation n sin (2π/n) – 9π/5 = 0 has only one +ve root
∴the equation n sin (2π/n) – 9π/5 = 0 has only one –ve root
Hence the equation n sin (2π/n) = 9π/5 has only two roots, where the two roots are only differed by +/– signs

For finding the roots of the equation n sin (2π/n) = 9π/5 , there is not any algebraic method, the only way is find the approximate roots by iteration method. And the fastest iteration method is Newton Method.

The iteration form of the equation n sin (2π/n) = 9π/5 using Newton Method is
f(n_(k + 1))
= n_k – f(n_k)/f′(n_k)
= n_k – [n_k sin (2π/(n_k)) – 9π/5]/[sin (2π/(n_k)) – (2π/(n_k)) cos (2π/(n_k))]
= [n_k sin (2π/(n_k)) – 2π cos (2π/(n_k)) – n_k sin (2π/(n_k)) + 9π/5]/[sin (2π/(n_k)) – (2π/(n_k)) cos (2π/(n_k))]
= [– 2π cos (2π/(n_k)) + 9π/5]/[sin (2π/(n_k)) – (2π/(n_k)) cos (2π/(n_k))]
= [2π cos (2π/(n_k)) – 9π/5]/[(2π/(n_k)) cos (2π/(n_k)) – sin (2π/(n_k))]

Starting with n_0 = 2, we have
n_0 = 2
n_1 = [2π cos (2π/2) – 9π/5]/[(2π/2) cos (2π/2) – sin (2π/2)]
　　= 3.8
n_2 = [2π cos (2π/3.8) – 9π/5]/[(2π/3.8) cos (2π/3.8) – sin (2π/3.8)]
　　= 5.44840
n_3 = [2π cos (2π/5.44840) – 9π/5]/[(2π/5.44840) cos (2π/5.44840) – sin (2π/5.44840)]
　　= 6.95971
n_4 = [2π cos (2π/6.95971) – 9π/5]/[(2π/6.95971) cos (2π/6.95971) – sin (2π/6.95971)]
　　= 7.80571
n_5 = [2π cos (2π/7.80571) – 9π/5]/[(2π/7.80571) cos (2π/7.80571) – sin (2π/7.80571)]
　　= 7.98107
n_6 = [2π cos (2π/7.98107) – 9π/5]/[(2π/7.98107) cos (2π/7.98107) – sin (2π/7.98107)]
　　= 7.98693
n_7 = [2π cos (2π/7.98693) – 9π/5]/[(2π/7.98693) cos (2π/7.98693) – sin (2π/7.98693)]
　　= 7.98693

∴n = 7.9869(corr. to 4 d.p.)
Hence another value of n is – 7.9869(corr. to 4 d.p.)
∴n = ± 7.9869(corr. to 4 d.p.)

對不起，這題無步驟可言。要解決這一題，一定要用numerical method 去解決
最簡單的當然是用bisection：

設a = 1, b = 3*pi, f(n) = n*sin(2pi/n) - 9pi/5
設 mid = (a+b)/2
若 f(mid) < 0 a = mid
若 f(mid) > 0 b = mid
若 f(mid) = 0 mid 就是答案

只要使用Excel 之類的工具，就可很快求出 n = 7.9869 ~ 2.54 pi