> x
  [1]  0.5  1.5  3.5  6.5 10.5 15.5 21.5 28.5 36.5 45.5 55.5  0.5  1.5  3.5
[15]  6.5 10.5 15.5 21.5 28.5 36.5 45.5 55.5  0.5  1.5  3.5  6.5 10.5 15.5
[29] 21.5 28.5 36.5 45.5 55.5  0.5  1.5  3.5  6.5 10.5 15.5 21.5 28.5 36.5
[43] 45.5 55.5  0.5  1.5  3.5  6.5 10.5 15.5 21.5 28.5 36.5 45.5 55.5  0.5
[57]  1.5  3.5  6.5 10.5 15.5 21.5 28.5 36.5 45.5 55.5  0.5  1.5  3.5  6.5
[71] 10.5 15.5 21.5 28.5 36.5 45.5 55.5  0.5  1.5  3.5  6.5 10.5 15.5 21.5
[85] 28.5 36.5 45.5 55.5  0.5  1.5  3.5  6.5 10.5 15.5 21.5 28.5 36.5 45.5
[99] 55.5  0.5  1.5  3.5  6.5 10.5 15.5 21.5 28.5 36.5 45.5 55.5  0.5  1.5
[113]  3.5  6.5 10.5 15.5 21.5 28.5 36.5 45.5 55.5  0.5  1.5  3.5  6.5 10.5
[127] 15.5 21.5 28.5 36.5 45.5 55.5

> y
  [1] 0 1 5 4 5 4 7 5 6 4 5 1 0 2 6 7 5 3 7 5 5 6 0 2 3 5 6 6 5 4 5 6 4 1 1 5
[37] 6 6 5 6 5 4 3 6 0 1 3 6 4 7 5 5 6 7 9 1 2 2 5 4 6 6 6 7 5 5 0 3 3 4 6 5
[73] 5 5 6 6 4 1 1 4 5 6 7 5 4 6 7 7 0 1 3 6 5 5 5 7 8 6 6 1 2 3 7 7 6 8 8 5
[109] 8 7 2 3 6 3 7 7 6 7 8 8 5 0 0 5 7 6 7 7 7 6 5 7

我遇到一个头疼的问题，就是使用R中的nls函数拟合，方程为
nls(y ~ (a * x)/( b + x), start = list(a = 20, b = 5))
然后我再使用bootstrap来计算a和b的标准误差，结果发现:
直接使用nls计算的a和b的标准误差要大于2000次bootstrap的标准误差，请问是什么原因造成的？

	a	a.sd	b	b.sd
直接拟合	6.5474	0.4734	2.8165	0.4734
bootstrap	6.5431	0.2016	2.8304	0.3955