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《数学通报》2014年1月号问题2162及其证明

 王虎应六爻求真 2015-12-23
作者:sqing55
[转载]《数学通报》2014年1月号问题2162及其证明
[转载]《数学通报》2014年1月号问题2162及其证明
              [转载]《数学通报》2014年1月号问题2162及其证明
[转载]《数学通报》2014年1月号问题2162及其证明

[转载]《数学通报》2014年1月号问题2162及其证明
[转载]《数学通报》2014年1月号问题2162及其证明
[转载]《数学通报》2014年1月号问题2162及其证明
[转载]《数学通报》2014年1月号问题2162及其证明
                              
                       

[转载]《数学通报》2014年1月号问题2162及其证明
这是我的摩尔多瓦----摩尔多瓦不等式欣赏
这是我的摩尔多瓦----摩尔多瓦不等式欣赏
 Expressions of the form
 lim_(k->infty)x_0+sqrt(x_1+sqrt(x_2+sqrt(...+x_k)))
(1)

are called nested radicals. Herschfeld (1935) proved that a nested radical of real nonnegative terms converges iff (x_n)^(2^(-n)) is bounded. He also extended this result to arbitrary   powers (which include continued square roots andcontinued fractions as well), a result is known as Herschfeld's convergence theorem.

Nested radicals appear in the computation of pi,

 2/pi=sqrt(1/2)sqrt(1/2+1/2sqrt(1/2))sqrt(1/2+1/2sqrt(1/2+1/2sqrt(1/2)))...
(2)

(Vieta 1593; Wells 1986, p. 50; Beckmann 1989, p. 95), in  trigonometrical values of cosine and sine for argumentsof the form pi/2^n, e.g.,

sin(pi/8) = 1/2sqrt(2-sqrt(2))
(3)
cos(pi/8) = 1/2sqrt(2+sqrt(2))
(4)
sin(pi/(16)) = 1/2sqrt(2-sqrt(2+sqrt(2)))
(5)
cos(pi/(16)) = 1/2sqrt(2+sqrt(2+sqrt(2))).
(6)

Nest radicals also appear in the computation of the  golden ratio

 phi=sqrt(1+sqrt(1+sqrt(1+sqrt(1+...))))
(7)

and plastic constant

 P=RadicalBox[{1, +, RadicalBox[{1, +, RadicalBox[{1, +, ...}, 3]}, 3]}, 3].
(8)

Both of these are special cases of

 x=RadicalBox[{a, +, RadicalBox[{a, +, ...}, n]}, n],
(9)

which can be exponentiated to give

 x^n=a+RadicalBox[{a, +, RadicalBox[{a, +, ...}, n]}, n],
(10)

so solutions are

 x^n=a+x.
(11)

The  silver constant is related to the nested radical expression

 RadicalBox[{7, +, 7, RadicalBox[{7, +, ...}, 3]}, 3].
(12)

There are a number of general formula for nested radicals (Wong and McGuffin). For example,

 x=RadicalBox[{{(, {1, -, q}, )}, {x, ^, n}, +, q, {x, ^, {(, {n, -, 1}, )}}, RadicalBox[{{(, {1, -, q}, )}, {x, ^, n}, +, q, {x, ^, {(, {n, -, 1}, )}}, RadicalBox[..., n]}, n]}, n]
(13)

which gives as special cases

 (b+sqrt(b^2+4a))/2=sqrt(a+bsqrt(a+bsqrt(a+bsqrt(...))))
(14)

(n=2q=1-a/x^2x=b/q),

 x=RadicalBox[{{x, ^, {(, {n, -, 1}, )}}, RadicalBox[{{x, ^, {(, {n, -, 1}, )}}, RadicalBox[{{x, ^, {(, {n, -, 1}, )}}, RadicalBox[..., n]}, n]}, n]}, n]
(15)

(q=1), and

 x=sqrt(xsqrt(xsqrt(xsqrt(xsqrt(...)))))
(16)

(q=1,n=2). Equation (13) also gives rise to

 q^((n^k-1)/(n-1))x^(n^j)=RadicalBox[{{q, ^, {(, {{(, {{n, ^, {(, {k, +, 1}, )}}, -, n}, )}, /, {(, {n, -, 1}, )}}, )}}, {(, {1, -, q}, )}, {x, ^, {(, {n, ^, {(, {j, +, 1}, )}}, )}}, +, ...}, n] ...+RadicalBox[{{q, ^, {(, {{(, {{n, ^, {(, {k, +, 2}, )}}, -, n}, )}, /, {(, {n, -, 1}, )}}, )}}, {(, {1, -, q}, )}, {x, ^, {(, {n, ^, {(, {j, +, 2}, )}}, )}}, +, RadicalBox[..., n]}, n]^_,
(17)

which gives the special case for q=1/2n=2x=1, and k=-1,

 sqrt(2)=sqrt(2/(2^(2^0))+sqrt(2/(2^(2^1))+sqrt(2/(2^(2^2))+sqrt(2/(2^(2^3))+sqrt(2/(2^(2^4))+...))))).
(18)

Equation (◇) can be generalized to

 x^(1/(n-1))=RadicalBox[{x, RadicalBox[{x, RadicalBox[{x, ...}, n]}, n]}, n]
(19)

for integers n>=2, which follows from

1+1/n+1/(n^2)+... = 1/(1-1/n)
(20)
= n/(n-1)
(21)
= 1+1/(n-1)
(22)
1/n+1/(n^2)+1/(n^3)+... = 1/(n-1)
(23)
1/n(1+1/n(1+1/n(1+...))) = 1/(n-1).
(24)

In particular, taking n=3 gives

 sqrt(x)=RadicalBox[{x, RadicalBox[{x, RadicalBox[{x, ...}, 3]}, 3]}, 3].
(25)

(J. R. Fielding, pers. comm., Oct. 8, 2002).

Ramanujan discovered

 x+n+a=sqrt(ax+(n+a)^2+xsqrt(a(x+n)+(n+a)^2+...)) ...+(x+n)sqrt(a(x+2n)+(n+a)^2+(x+2n)sqrt(...))^_^_,
(26)

which gives the special cases

 x+1=sqrt(1+xsqrt(1+(x+1)sqrt(1+(x+2)sqrt(1+...))))
(27)

for a=0n=1 (Ramanujan 1911; Ramanujan 2000, p. 323; Pickover 2002, p. 310), and

 3=sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+5sqrt(...)))))
(28)

for a=0n=1, and x=2. The justification of this process both in general and in the particular example of lnsigma, where sigma is  Somos's quadratic recurrence constant in given by Vijayaraghavan (in Ramanujan 2000, p. 348).

For a nested radical  of the form

 x=sqrt(n+sqrt(n+sqrt(n+...)))
(29)

to be equal a given  real number x, it must be true that

 x=sqrt(n+sqrt(n+sqrt(n+...)))=sqrt(n+x),
(30)

so

 x^2=n+x
(31)

and

 x=1/2(1+sqrt(4n+1)).
(32)

An amusing nested radical follows rewriting the series for e

 e=1+1/(1!)+1/(2!)+1/(3!)+...
(33)

as

 e=1+1+1/2(1+1/3(1+1/4(1+1/5(1+...)))),
(34)

so

 x^(e-2)=sqrt(xRadicalBox[{x, RadicalBox[{x, RadicalBox[{x, ...}, 5]}, 4]}, 3])
(35)
(J. R. Fielding, pers. comm., May 15, 2002).
                                

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