无理数α∈(0,1)，证明有正整数a1＜a2＜a3＜…使∑(n=1,∞)(-1)^(n+1)／(a1a2…an)=α

jvk0825

請問此題如何求解?

uk702

有意思，和任意的无理数都有一个 n 进制表示正好相反。注意到 ai 是正整数，猜测大概和任一个无理数都有唯一的简单连分式展开相关。

王守恩

   a1,a2,a3,=1,3,24,      A226053     这通项公式应该怎样调整？
1, 3, 24, 815, 2263886, 9073564639850, 176228569027146222763928594, 84205747605016031994416006285857418872429042805656089,

1. Table[Solve[{Floor[-1/x] == x1, Floor[1/(x + x1^-1)] == x2,
   
2. Floor[-1/(x + x1^-1 - x2^-1)] == x3, Floor[1/(x + x1^-1 - x2^-1 + x3^-1)]==x4,
   
3. Floor[-1/(x + x1^-1 - x2^-1 + x3^-1 - x4^-1)] == x5,
   
4. Floor[1/(x + x1^-1 - x2^-1 + x3^-1 - x4^-1 + x5^-1)] == x6,
   
5. Floor[-1/(x + x1^-1 - x2^-1 + x3^-1 - x4^-1 + x5^-1 - x6^-1)] == x7,
   
6. Floor[1/(x + x1^-1 - x2^-1 + x3^-1 - x4^-1 + x5^-1 - x6^-1 + x7^-1)] == x8},
   
7. {x1, x2, x3, x4, x5, x6, x7, x8}], {x, 1/Sqrt[2] - 1, 1/Sqrt[2] - 1}]

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王守恩

主帖是这样一串数:
a1=1,
a2=3,
a3=8,
a4=33,
a5=35,
a6=39201,
a7=39203,
a8=60245508192801,
a9=60245508192803,
a10=218662352649181293830957829984632156775201,
a11=218662352649181293830957829984632156775203,
a12=1045495214949533748385099187796335300703912924044130047385470740995840\
5643210820820334358767544214981903739582932832429216801,
a13=1045495214949533748385099187796335300703912924044130047385470740995840\
5643210820820334358767544214981903739582932832429216803,
......
个位数好像都是1,3,1,3,1,3,...     A091831         Pierce expansion of 1/sqrt(2)

王守恩

无理数α∈(0,1)都可以有解？把\(\sqrt{1/2}\)  换换好像都成功了。
1，算式有点复杂, 思路还是可以看出来的。

1. Table[Solve[{Floor[1/α] == a1,
   
2. Floor[Floor[-1/(α - a1^-1)]*1/a1] == a2,
   
3. Floor[Floor[1/(α - a1^-1 + (a1*a2)^-1)]*1/(a1*a2)] == a3,
   
4. Floor[Floor[-1/(α - a1^-1 + (a1*a2)^-1 - (a1*a2*a3)^-1)]*1/(a1*a2*a3)] == a4,
   
5. Floor[Floor[1/(α - a1 + (a1*a2)^-1 - (a1*a2*a3)^-1
   
6. +(a1*a2*a3*a4)^-1)]*1/(a1*a2*a3*a4)]==a5,
   
7. Floor[Floor[-1/(α - a1 + (a1*a2)^-1 - (a1*a2*a3)^-1
   
8. +(a1*a2*a3*a4)^-1 - (a1*a2*a3*a4*a5)^-1)]*1/(a1*a2*a3*a4*a5)]==a6,
   
9. Floor[Floor[1/(α - a1 + (a1*a2)^-1 - (a1*a2*a3)^-1 + (a1*a2*a3*a4)^-1
   
10. -(a1*a2*a3*a4*a5)^-1+(a1*a2*a3*a4*a5*a6)^-1)]*1/(a1*a2*a3*a4*a5*a6)]==a7},
    
11. {a1, a2, a3, a4, a5, a6, a7, a8}], {α, Sqrt[1/2], Sqrt[1/2]}]

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{{{a1 -> 1, a2 -> 3, a3 -> 8, a4 -> 33, a5 -> 35, a6 -> 39201, a7 -> 39203, a8 -> 60245508192801}}}
2，算式还可以简单？

1. PierceExp[A_, n_] := Join[Array[1 &, Floor[A]],
   
2. First@Transpose@NestList[{Floor[Expand[1 - #[[1]] #[[2]]]^-1],
   
3. Expand[1 - #[[1]] #[[2]]]} &, {Floor[(A - Floor[A])^-1],
   
4. A - Floor[A]}, n - 1]]; PierceExp[N[Sqrt[1/2], 6!], 11]

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{1, 3, 8, 33, 35, 39201, 39203, 60245508192801, 60245508192803,
218662352649181293830957829984632156775201,
218662352649181293830957829984632156775203}

王守恩

欣赏几串数。
\(\frac{1}{1}-\frac{1}{1*2}+\frac{1}{1*2*3}-\frac{1}{1*2*3*4}+\cdots\cdots=\frac{e-1}{e}\)
\(\frac{1}{1}-\frac{1}{1*2}+\frac{1}{1*2*4}-\frac{1}{1*2*4*6}+\cdots\cdots=\sqrt{1/e}\)
\(\frac{1}{1}-\frac{1}{1*3}+\frac{1}{1*3*5}-\frac{1}{1*3*5*7}+\cdots\cdots=\sqrt{\pi/2/e}\)
\(\frac{1}{1}-\frac{1}{1*3}+\frac{1}{1*3*6}-\frac{1}{1*3*6*9}+\cdots\cdots=\sqrt[3]{1/e}\)