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连分数理论
阅读量:7005 次
发布时间:2019-06-28

本文共 1936 字,大约阅读时间需要 6 分钟。

1、

== The original formula ==

[[Euler]] derived the formula as
connecting a finite sum of products with a finite continued fraction.

\[

a_0 + a_0a_1 + a_0a_1a_2 + \cdots + a_0a_1a_2\cdots a_n =
\cfrac{a_0}{1 - \cfrac{a_1}{1 + a_1 - \cfrac{a_2}{1 + a_2 - \cfrac{\ddots}{\ddots
\cfrac{a_{n-1}}{1 + a_{n-1} - \cfrac{a_n}{1 + a_n}}}}}}\,
\]

The identity is easily established by [[mathematical induction|induction]] on ''n'', and is therefore applicable in the limit: if the expression on the left is extended to represent a [[convergent series|convergent infinite series]], the expression on the right can also be extended to represent a convergent infinite continued fraction.

2、

==Derivation==

Let $f_0, f_1, f_2, \dots$ be a sequence of analytic functions so that
\[f_{i-1} - f_i = k_i\,z\,f_{i+1}\]
for all $i > 0$, where each $k_i$ is a constant.

Then

\[\frac{f_{i-1}}{f_i} = 1 + k_i z \frac{f_{i+1}}{
{f_i}}, \,\] and so \[\frac{f_i}{f_{i-1}} = \frac{1}{1 + k_i z \frac{f_{i+1}}{
{f_i}}}\]

Setting $g_i = f_i / f_{i-1}$,

\[g_i = \frac{1}{1 + k_i z g_{i+1}},\]
So
\[g_1 = \frac{f_1}{f_0} = \cfrac{1}{1 + k_1 z g_2} = \cfrac{1}{1 + \cfrac{k_1 z}{1 + k_2 z g_3}}
= \cfrac{1}{1 + \cfrac{k_1 z}{1 + \cfrac{k_2 z}{1 + k_3 z g_4}}} = \dots\]

Repeating this ad infinitum produces the continued fraction expression

\[\frac{f_1}{f_0} = \cfrac{1}{1 + \cfrac{k_1 z}{1 + \cfrac{k_2 z}{1 + \cfrac{k_3 z}{1 + {}\ddots}}}}\]

In Gauss's continued fraction, the functions $f_i$ are hypergeometric functions of the form ${}_0F_1$, ${}_1F_1$, and ${}_2F_1$, and the equations $f_{i-1} - f_i = k_i z f_{i+1}$ arise as identities between functions where the parameters differ by integer amounts. These identities can be proven in several ways, for example by expanding out the series and comparing coefficients, or by taking the derivative in several ways and eliminating it from the equations generated.

转载于:https://www.cnblogs.com/Eufisky/p/7821460.html

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