sin1+sin0.5+sin0.25+sin0.125+sin0.0625+.以后每个sin中的数都是前一个数除以2,

解题思路：
设f0(x)=sinx
f1(x)=sinx+sinx/2
.
fn(x)=sinx+sinx/2+...+sinx/2^n
可以推倒出fn(x)的另一种形态的表达式
要求的问题就是fn(1)=?
fn(x)=(2^n)cos(x/2^1)cos(x/2^2)cos(x/2^3)...cos(x/2^n)sin[x/2^(n-1)]
+[2^(n-1)]cos(x/2^2)cos(x/2^3)...cos(x/2^n)sin[x/2^(n-1)]
+[2^(n-2)]cos(x/2^3)...cos(x/2^n)sin[x/2^(n-1)]
...
+2cos[x/2^(n-1)]
+sin(x/2^n)
=sin(x/2^n){(2^n)cos(x/2^1)cos(x/2^2)cos(x/2^3)...cos[x/2^(n-1)]
+[2^(n-1)]cos(x/2^2)cos(x/2^3)...cos[x/2^(n-1)]
+[2^(n-2)]cos(x/2^3)...cos[x/2^(n-1)]
...
+2cos[x/2^(n-1)]
+1}
fn(x)/sin(x/2^n)=(2^n)cos(x/2^1)cos(x/2^2)cos(x/2^3)...cos[x/2^(n-1)
+[2^(n-1)]cos(x/2^2)cos(x/2^3)...cos[x/2^(n-1)]
+[2^(n-2)]cos(x/2^3)...cos[x/2^(n-1)]
...
+2cos[x/2^(n-1)]
+1
=1+2cos[x/2^(n-1)]{[2^(n-1)cos(x/2^1)cos(x/2^2)cos(x/2^3).cos[x/2^(n-2)]
+[2^(n-2)]cos(x/2^2)cos(x/2^3)...cos[x/2^(n-2)]
+[2^(n-3)]cos(x/2^3)...cos[x/2^(n-2)]
...
+2cos[x/2^(n-2)]
+1}
=1+2cos[x/2^(n-1)]f(n-1)(x)/sin[x/2^(n-1)]
=1+2cos[x/2^(n-1)][fn(x)-sin(x/2^n)]/sin[x/2^(n-1)]
即
fn(x)/sin(x/2^n)
=1+2cos[x/2^(n-1)]fn(x)/sin[x/2^(n-1)]-2cos[x/2^(n-1)]sin(x/2^n)]/sin[x/2^(n-1)]
解得fn(x)={sin[x/2^(n-1)]-2cos[x/2^(n-1)]sin(x/2^n)}/2{cosx/2^n}-cos[x/2^(n-1)]}
fn(1)={sin[1/2^(n-1)]-2cos[1/2^(n-1)]sin(1/2^n)}/2{cos1/2^n}-cos[1/2^(n-1)]}
求n趋近无穷大时的极限limfn(1)
剩余的自己计算吧,累了
求limfn(1)时,把n都变成x,极限大小不会变.但变成x后可以上下求导计算