w******o
发帖数: 442 | 1 xn = (1+cos(pi/(2^n+1)))/2
ask limit:
L = x1*x2*........*xn
when n goes to infinite.
or x1=1/2, xn=(1+sqrt(xn-1))/2
L/2 = x1*x2*........*xn
These two Ls are same. | f******k
发帖数: 297 | 2 i guess you mean x_n=(1+cos(pi/2^(n+1)))/2. in that case
since x_n=cos(pi/2^(n+2))^2 you can obtain the product easily
by multiplying sin(pi/2^(n+2))^2 in the end.
【在 w******o 的大作中提到】
: xn = (1+cos(pi/(2^n+1)))/2
: ask limit:
: L = x1*x2*........*xn
:
: when n goes to infinite.
: or x1=1/2, xn=(1+sqrt(xn-1))/2
: L/2 = x1*x2*........*xn
: These two Ls are same.
| w******o
发帖数: 442 | 3 yes, you are right.Thank you very much.
【在 f******k 的大作中提到】
: i guess you mean x_n=(1+cos(pi/2^(n+1)))/2. in that case
: since x_n=cos(pi/2^(n+2))^2 you can obtain the product easily
: by multiplying sin(pi/2^(n+2))^2 in the end.
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