|

楼主 |
发表于 2018-12-1 20:59
|
显示全部楼层
本帖最后由 愚工688 于 2018-12-1 13:00 编辑
在9楼最后,我写道:当然再大一些到1600亿附近的偶数,使用 μ=0.162的修正系数进行计算,则出现正的相对误差是很可能的。
下面验证一下:
G(160000000000) = 229574132;
inf( 160000000000 )≈ 229559235.1 , Δ≈-0.0000649,infS( 160000000000 )= 172169426.33 , k(m)= 1.33333
G(160000000002) = 367315420;
inf( 160000000002 )≈ 367295743.5 , Δ≈-0.0000536,infS( 160000000002 )= 172169426.33 , k(m)= 2.13334
G(160000000004) = 187842530;
inf( 160000000004 )≈ 187821192.4 , Δ≈-0.0001136,infS( 160000000004 )= 172169426.34 , k(m)= 1.09091
G(160000000006) = 233415788;
inf( 160000000006 )≈ 233400374.2 , Δ≈-0.00006604,infS( 160000000006 )= 172169426.34 , k(m)= 1.35564
G(160000000008) = 364844031;
inf( 160000000008 )≈ 364820136.6 , Δ≈-0.00006549,infS( 160000000008 )= 172169426.34 , k(m)= 2.11896
G(160000000010) = 229594896;
inf( 160000000010 )≈ 229576603.6 , Δ≈-0.00007968,infS( 160000000010 )= 172169426.34 , k(m)= 1.33343
G(160000000012) = 172177687;
inf( 160000000012 )≈ 172169426.4 , Δ≈-0.00004798,infS( 160000000012 )= 172169426.35 , k(m)= 1
G(160000000014) = 344517826;
inf( 160000000014 )≈ 344485068.8 , Δ≈-0.00009508,infS( 160000000014 )= 172169426.35 , k(m)= 2.00085
G(160000000016) = 173984721;
inf( 160000000016 )≈ 173963873.9 , Δ≈-0.0001198,infS( 160000000016 )= 172169426.35 , k(m)= 1.01042
G(160000000018) = 172437293;
inf( 160000000018 )≈ 172421542.3 , Δ≈-0.00009134,infS( 160000000018 )= 172169426.35 , k(m)= 1.00146
G(160000000020) = 608929641;
inf( 160000000020 )≈ 608894599.4 , Δ≈-0.00005755,infS( 160000000020 )= 172169426.35 , k(m)= 3.5366
G(160000000022) = 172220877;
inf( 160000000022 )≈ 172211531.9 , Δ≈-0.00005426,infS( 160000000022 )= 172169426.36 , k(m)= 1.00024
计算式:
inf( 160000000000 ) = 1/(1+ .162 )*( 160000000000 /2 -2)*p(m) ≈ 229559235.1
inf( 160000000002 ) = 1/(1+ .162 )*( 160000000002 /2 -2)*p(m) ≈ 367295743.5
inf( 160000000004 ) = 1/(1+ .162 )*( 160000000004 /2 -2)*p(m) ≈ 187821192.4
inf( 160000000006 ) = 1/(1+ .162 )*( 160000000006 /2 -2)*p(m) ≈ 233400374.2
inf( 160000000008 ) = 1/(1+ .162 )*( 160000000008 /2 -2)*p(m) ≈ 364820136.6
inf( 160000000010 ) = 1/(1+ .162 )*( 160000000010 /2 -2)*p(m) ≈ 229576603.6
inf( 160000000012 ) = 1/(1+ .162 )*( 160000000012 /2 -2)*p(m) ≈ 172169426.4
inf( 160000000014 ) = 1/(1+ .162 )*( 160000000014 /2 -2)*p(m) ≈ 344485068.8
inf( 160000000016 ) = 1/(1+ .162 )*( 160000000016 /2 -2)*p(m) ≈ 173963873.9
inf( 160000000018 ) = 1/(1+ .162 )*( 160000000018 /2 -2)*p(m) ≈ 172421542.3
inf( 160000000020 ) = 1/(1+ .162 )*( 160000000020 /2 -2)*p(m) ≈ 608894599.4
inf( 160000000022 ) = 1/(1+ .162 )*( 160000000022 /2 -2)*p(m) ≈ 172211531.9
验证结果是连续的12个偶数中没有出现正的相对误差。但是相对误差的绝对值都已经很小,最大的也不到0.00012 。也许是计算的偶数样本比较少,因此没有遇上出现出现正的相对误差的偶数。
当然如果偶数继续增大到1700亿,那么出现正的相对误差则是必然的。
后面会继续验证,看看预测的怎么样。
|
|