Question

请帮助我 联合间隔时间！

该行＆＃34; For...＆＃34;输出根存在的间隔（根：2.94和5,52）。我必须考虑一个评论：

如果在区间{x * [i]，x [i + 1]}和{x [i + 1]中，x ** [i + 1]}可以是等式的根，则范围{ x * [i]，x ** [i + 1]}必须至少有一个根。

 X = {-2, 6}
    spx = {-2, -1.90577, -1.81153, -1.59327, -1.375, -1.35785, -1.3407,    -1.24655, -1.22941, -1.11811, -0.934054, -0.80167, -0.75, -0.625,-0.5, -0.25, -0.0981238, 0.303752, 0.651876, 0.94833, 1, 1.5, 1.75,2.11731, 2.5, 2.5625, 2.625, 3.3125, 3.75, 4, 4.00964, 4.01928,4.25964, 4.36731, 4.5, 4.75, 5, 5.25, 5.5, 5.75, 6}
    spfw = {33.3632, 43.263, 51.6709, 55.5421, 57.1266, 57.2511, 57.3756,58.059, 58.0778, 58.1995,56.846,55.1903,54.5739,53.0828,51.1542,48.9959,48.0325,42.2533, 36.408,30.7952,30.1551,28.6446,23.138,19.4168,6.47053,5.90328,5.32951,-0.513959, -0.750527, -6.38895, -6.39157, -6.39418,-6.36456, -6.09357, -6.28599, -5.25369, -4.19539, -2.18625, -0.133803,2.90414, 6.171}
    spfn = {33.3632, 40.2933, 46.5882, 51.9781, 55.5583, 55.5708, 55.5762, 55.4604, 55.4393, 55.0045, 530116, 51.1309, 50.4546, 48.1226, 45.6012, 43.402, 42.066, 37.5522, 32.6864, 28.1979, 28.0685,25.7067, 17.5943,13.5547, -2.97428, -3.21054, -3.36422, -5.05466, -5.1301, -6.4392,-6.76879, -6.48231, -7.20196, -7.00719, -7.53373, -6.00246, -4.41058,-2.8187, -1.16621, 2.35765, 6.04694}

    For[i = 1, i < Length@spfn, i++,
     If[! (((0 < spfn[[i]]) && (0 < spfn[[i + 1]])) || 
       ((spfw[[i]] < 0) && (spfw[[i + 1]] < 0))),
       Print["1) exists root on: {", spx[[i]], ";", spx[[i + 1]], "}"]]]

所以结果包括5个区间：

1) exists root on: {2.11731;2.5}
1) exists root on: {2.5;2.5625}
1) exists root on: {2.5625;2.625}
1) exists root on: {2.625;3.3125}
1) exists root on: {5.5;5.75}

由于第一个根是2.94，它必须进入4个第一个区间，最后一个区间为5.52。因此，在考虑该注释后，该行应输出两个间隔。我尝试使用IntervalUnion，但它不起作用:( 请帮我在这一行编写这句话。

Answer 1

修改后的答案。有关## &[]的解释，请参阅this link。

spxpairs = Interval /@ Partition[spx, 2, 1];
spfwpairs = Interval /@ Partition[spfw, 2, 1];
spfnpairs = Interval /@ Partition[spfn, 2, 1];

ans = Apply[IntervalUnion,
   If[Not[0 < Min@#1 || Max@#2 < 0], #3, ## &[]] &
    @@@ Transpose[{spfnpairs, spfwpairs, spxpairs}]];

Column[Prepend[List @@ ans, "Roots exist on :"], Spacings -> 1]

根源存在于：

{2.11731,3.3125}

{5.5,5.75}

Answer 2

是的，有一种更简洁的方式......一旦你有了你的间隔清单（如克里斯的答案），你就是这样做：

 IntervalUnion @@ intervals

（* Interval [{2.11731,3.3125}，{5.5,5.75}] *）

如何联合间隔？

2 个答案: