Editing Simple continued fraction (section)

==Non-simple continued fraction==
{{main|Continued fraction (non-simple)}}

A non-simple continued fraction is an expression of the form

:<math>x = b_0 + \cfrac{a_1}{b_1 + \cfrac{a_2}{b_2 + \cfrac{a_3}{b_3 + \cfrac{a_4}{b_4 + \ddots\,}}}}</math>

where the ''a''<sub>''n''</sub> (''n'' &gt; 0) are the partial numerators, the ''b''<sub>''n''</sub> are the partial denominators, and the leading term ''b''<sub>0</sub> is called the ''integer'' part of the continued fraction.

To illustrate the use of non-simple continued fractions, consider the following example. The sequence of partial denominators of the simple continued fraction of {{pi}} does not show any obvious pattern:

:<math>\pi=[3;7,15,1,292,1,1,1,2,1,3,1,\ldots]</math>

or

:<math>\pi=3+\cfrac{1}{7+\cfrac{1}{15+\cfrac{1}{1+\cfrac{1}{292+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{1+\ddots}}}}}}}}}}}</math>

However, several non-simple continued fractions for {{pi}} have a perfectly regular structure, such as:

:<math>
\pi=\cfrac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\cfrac{7^2}{2+\cfrac{9^2}{2+\ddots}}}}}}
=\cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ddots}}}}}
=3+\cfrac{1^2}{6+\cfrac{3^2}{6+\cfrac{5^2}{6+\cfrac{7^2}{6+\cfrac{9^2}{6+\ddots}}}}}
</math>

:<math>\displaystyle \pi=2+\cfrac{2}{1+\cfrac{1}{1/2+\cfrac{1}{1/3+\cfrac{1}{1/4+\ddots}}}}=2+\cfrac{2}{1+\cfrac{1\cdot2}{1+\cfrac{2\cdot3}{1+\cfrac{3\cdot4}{1+\ddots}}}}</math>

:<math> \displaystyle \pi=2+\cfrac{4}{3+\cfrac{1\cdot3}{4+\cfrac{3\cdot5}{4+\cfrac{5\cdot7}{4+\ddots}}}}</math>

The first two of these are special cases of the [[Inverse trigonometric functions#Variant: Continued fractions for arctangent|arctangent]] function with {{pi}} = 4 arctan (1) and the fourth and fifth one can be derived using the [[Wallis product]].{{sfn|Bunder|Tonien|2017}}{{sfn|Scheinerman|Pickett|Coleman|2008}}

:<math>
\pi=3+\cfrac{1}{6+\cfrac{1^3+2^3}{6\cdot1^2+1^2\cfrac{1^3+2^3+3^3+4^3}{6\cdot2^2+2^2\cfrac{1^3+2^3 +3^3+4^3+5^3+6^3}{6\cdot3^2+3^2\cfrac{1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3}{6\cdot4^2+\ddots}}}}}
</math>

The continued fraction of <math>\pi</math> above consisting of cubes uses the Nilakantha series and an exploit from Leonhard Euler.{{sfn|Foster|2015}}