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[函数] 找一些整数的非平凡关系

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hbghlyj Post time 2024-2-26 05:11 |Read mode
参见 Wolfram 文档 reference.wolfram.com/language/ref/LatticeReduce.html
从 5 个实数开始
1, ArcTan[1], ArcTan[1/5], ArcTan[1/239], 1

乘以 10^20 并舍入为整数
{100000000000000000000,
78539816339744830962,
19739555984988075837,
418407600207472386,
100000000000000000000}

写出平凡的线性关系
a0 - 100000000000000000000 a4= 0
a1 - 78539816339744830962 a4 = 0
a2 - 19739555984988075837 a4 = 0
a3 - 418407600207472386 a4 = 0

{{1,0,0,0,-100000000000000000000},
{0,1,0,0,-78539816339744830962},
{0,0,1,0,-19739555984988075837},
{0,0,0,1,-418407600207472386}}

执行函数 LatticeReduce 为
{{0,1,-4,1,0},
{-325302,315725,367312,1153518,928458},
{-381213,314234,633857,2221192,-2330529},
{-3210817,4041574,249764,-3042512,306976}}

第1条{0,1,-4,1,0}最优(系数最小)
Pi/4+ArcTan[1/239]-4 ArcTan[1/5]

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 Author| hbghlyj Post time 2024-2-26 05:14
尽管我们进行了舍入,但这等式是准确的。
\[\fracπ4 + \tan^{-1}\frac1{239} - 4\tan^{-1}\frac15=0\]
我想知道函数“LatticeReduce”内部执行了什么?

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 Author| hbghlyj Post time 2024-2-26 05:20
本帖最后由 hbghlyj 于 2024-2-29 18:06 编辑 它继续讲到,
由平行四边形基本区域生成的晶格
Untitled.png
从一个边长很长的平行四边形开始,选择另一个可以生成相同格子但边长更小的平行四边形
具体是如何算的呢
kuing.cjhb.site/forum.php?mod=viewthread&tid=12033

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