In the analytic theory of continued fractions , Euler's continued fraction formula is an identity connecting a certain very general infinite series with an infinite continued fraction . First published in 1748, it was at first regarded as a simple identity connecting a finite sum with a finite continued fraction in such a way that the extension to the infinite case was immediately apparent.[1] convergence problem for infinite continued fractions with complex elements.
Euler derived the formula as
connecting a finite sum of products with a finite continued fraction .
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{\displaystyle a_{0}\left(1+a_{1}\left(1+a_{2}\left(\cdots +a_{n}\right)\cdots \right)\right)=a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+\cdots +a_{0}a_{1}a_{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 induction on n , and is therefore applicable in the limit: if the expression on the left is extended to represent a convergent infinite series , the expression on the right can also be extended to represent a convergent infinite continued fraction .
This is written more compactly using generalized continued fraction notation:
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{\displaystyle a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+\cdots +a_{0}a_{1}a_{2}\cdots a_{n}={\frac {a_{0}}{1+}}\,{\frac {-a_{1}}{1+a_{1}+}}\,{\cfrac {-a_{2}}{1+a_{2}+}}\cdots {\frac {-a_{n}}{1+a_{n}}}.}
Theorem: Let
n
{\displaystyle n}
n
+
1
{\displaystyle n+1}
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{\displaystyle a_{0},a_{1},\ldots ,a_{n}}
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{\displaystyle \sum _{k=0}^{n}\prod _{j=0}^{k}a_{j}={\frac {a_{0}}{1+}}\,{\frac {-a_{1}}{1+a_{1}+}}\cdots {\frac {-a_{n}}{1+a_{n}}}}
and for
n
{\displaystyle n}
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1
,
…
,
b
n
{\displaystyle b_{1},\ldots ,b_{n}}
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1.
{\displaystyle {\frac {-b_{1}}{1+b_{1}+}}\,{\frac {-b_{2}}{1+b_{2}+}}\cdots {\frac {-b_{n}}{1+b_{n}}}\neq -1.}
Proof: We perform a double induction. For
n
=
1
{\displaystyle n=1}
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{\displaystyle {\frac {a_{0}}{1+}}\,{\frac {-a_{1}}{1+a_{1}}}={\frac {a_{0}}{1+{\frac {-a_{1}}{1+a_{1}}}}}={\frac {a_{0}(1+a_{1})}{1}}=a_{0}+a_{0}a_{1}=\sum _{k=0}^{1}\prod _{j=0}^{k}a_{j}}
and
−
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+
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≠
−
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{\displaystyle {\frac {-b_{1}}{1+b_{1}}}\neq -1.}
Now suppose both statements are true for some
n
≥
1
{\displaystyle n\geq 1}
We have
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{\displaystyle {\frac {-b_{1}}{1+b_{1}+}}\,{\frac {-b_{2}}{1+b_{2}+}}\cdots {\frac {-b_{n+1}}{1+b_{n+1}}}={\frac {-b_{1}}{1+b_{1}+x}}}
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{\displaystyle x={\frac {-b_{2}}{1+b_{2}+}}\cdots {\frac {-b_{n+1}}{1+b_{n+1}}}\neq -1}
by applying the induction hypothesis to
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2
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1
{\displaystyle b_{2},\ldots ,b_{n+1}}
But if
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{\displaystyle {\frac {-b_{1}}{1+b_{1}+x}}=-1}
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{\displaystyle b_{1}=1+b_{1}+x}
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{\displaystyle x=-1}
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{\displaystyle {\frac {-b_{1}}{1+b_{1}+}}\,{\frac {-b_{2}}{1+b_{2}+}}\cdots {\frac {-b_{n+1}}{1+b_{n+1}}}\neq -1,}
completing that induction.
Note that for
x
≠
−
1
{\displaystyle x\neq -1}
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;
{\displaystyle {\frac {1}{1+}}\,{\frac {-a}{1+a+x}}={\frac {1}{1-{\frac {a}{1+a+x}}}}={\frac {1+a+x}{1+x}}=1+{\frac {a}{1+x}};}
if
x
=
−
1
−
a
{\displaystyle x=-1-a}
Using
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=
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{\displaystyle a=a_{1}}
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{\displaystyle x={\frac {-a_{2}}{1+a_{2}+}}\,\cdots {\frac {-a_{n+1}}{1+a_{n+1}}}\neq -1}
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{\displaystyle a_{1},a_{2},\ldots ,a_{n+1}}
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{\displaystyle {\begin{aligned}a_{0}+&a_{0}a_{1}+a_{0}a_{1}a_{2}+\cdots +a_{0}a_{1}a_{2}a_{3}\cdots a_{n+1}\\&=a_{0}+a_{0}(a_{1}+a_{1}a_{2}+\cdots +a_{1}a_{2}a_{3}\cdots a_{n+1})\\&=a_{0}+a_{0}{\big (}{\frac {a_{1}}{1+}}\,{\frac {-a_{2}}{1+a_{2}+}}\,\cdots {\frac {-a_{n+1}}{1+a_{n+1}}}{\big )}\\&=a_{0}{\big (}1+{\frac {a_{1}}{1+}}\,{\frac {-a_{2}}{1+a_{2}+}}\,\cdots {\frac {-a_{n+1}}{1+a_{n+1}}}{\big )}\\&=a_{0}{\big (}{\frac {1}{1+}}\,{\frac {-a_{1}}{1+a_{1}+}}\,{\frac {-a_{2}}{1+a_{2}+}}\,\cdots {\frac {-a_{n+1}}{1+a_{n+1}}}{\big )}\\&={\frac {a_{0}}{1+}}\,{\frac {-a_{1}}{1+a_{1}+}}\,{\frac {-a_{2}}{1+a_{2}+}}\,\cdots {\frac {-a_{n+1}}{1+a_{n+1}}},\end{aligned}}}
completing the other induction.
As an example, the expression
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{\displaystyle a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+a_{0}a_{1}a_{2}a_{3}}
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3
{\displaystyle {\begin{aligned}a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+a_{0}a_{1}a_{2}a_{3}&=a_{0}(a_{1}(a_{2}(a_{3}+1)+1)+1)\\[8pt]&={\cfrac {a_{0}}{\cfrac {1}{a_{1}(a_{2}(a_{3}+1)+1)+1}}}\\[8pt]&={\cfrac {a_{0}}{{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)+1}{a_{1}(a_{2}(a_{3}+1)+1)+1}}-{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)}{a_{1}(a_{2}(a_{3}+1)+1)+1}}}}={\cfrac {a_{0}}{1-{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)}{a_{1}(a_{2}(a_{3}+1)+1)+1}}}}\\[8pt]&={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)+1}{a_{2}(a_{3}+1)+1}}}}}\\[8pt]&={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)}{a_{2}(a_{3}+1)+1}}+{\cfrac {a_{2}(a_{3}+1)+1}{a_{2}(a_{3}+1)+1}}-{\cfrac {a_{2}(a_{3}+1)}{a_{2}(a_{3}+1)+1}}}}}}={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{1+a_{1}-{\cfrac {a_{2}(a_{3}+1)}{a_{2}(a_{3}+1)+1}}}}}}\\[8pt]&={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{1+a_{1}-{\cfrac {a_{2}}{\cfrac {a_{2}(a_{3}+1)+1}{a_{3}+1}}}}}}}\\[8pt]&={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{1+a_{1}-{\cfrac {a_{2}}{{\cfrac {a_{2}(a_{3}+1)}{a_{3}+1}}+{\cfrac {a_{3}+1}{a_{3}+1}}-{\cfrac {a_{3}}{a_{3}+1}}}}}}}}={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{1+a_{1}-{\cfrac {a_{2}}{1+a_{2}-{\cfrac {a_{3}}{1+a_{3}}}}}}}}}\end{aligned}}}
This can be applied to a sequence of any length, and will therefore also apply in the infinite case.
The exponential function
edit
The exponential function e x entire function with a power series expansion that converges uniformly on every bounded domain in the complex plane.
e
x
=
1
+
∑
n
=
1
∞
x
n
n
!
=
1
+
∑
n
=
1
∞
(
∏
i
=
1
n
x
i
)
{\displaystyle e^{x}=1+\sum _{n=1}^{\infty }{\frac {x^{n}}{n!}}=1+\sum _{n=1}^{\infty }\left(\prod _{i=1}^{n}{\frac {x}{i}}\right)\,}
The application of Euler's continued fraction formula is straightforward:
e
x
=
1
1
−
x
1
+
x
−
1
2
x
1
+
1
2
x
−
1
3
x
1
+
1
3
x
−
1
4
x
1
+
1
4
x
−
⋱
.
{\displaystyle e^{x}={\cfrac {1}{1-{\cfrac {x}{1+x-{\cfrac {{\frac {1}{2}}x}{1+{\frac {1}{2}}x-{\cfrac {{\frac {1}{3}}x}{1+{\frac {1}{3}}x-{\cfrac {{\frac {1}{4}}x}{1+{\frac {1}{4}}x-\ddots }}}}}}}}}}.\,}
Applying an equivalence transformation that consists of clearing the fractions this example is simplified to
e
x
=
1
1
−
x
1
+
x
−
x
2
+
x
−
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x
3
+
x
−
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x
4
+
x
−
⋱
{\displaystyle e^{x}={\cfrac {1}{1-{\cfrac {x}{1+x-{\cfrac {x}{2+x-{\cfrac {2x}{3+x-{\cfrac {3x}{4+x-\ddots }}}}}}}}}}\,}
and we can be certain that this continued fraction converges uniformly on every bounded domain in the complex plane because it is equivalent to the power series for e x
The natural logarithm
edit
The Taylor series for the principal branch of the natural logarithm in the neighborhood of 1 is well known:
log
(
1
+
x
)
=
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−
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2
2
+
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3
3
−
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4
+
⋯
=
∑
n
=
1
∞
(
−
1
)
n
+
1
z
n
n
.
{\displaystyle \log(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+\cdots =\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}z^{n}}{n}}.\,}
This series converges when |x | < 1 and can also be expressed as a sum of products:[2]
log
(
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+
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)
=
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+
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x
)
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−
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)
+
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)
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+
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+
⋯
{\displaystyle \log(1+x)=x+(x)\left({\frac {-x}{2}}\right)+(x)\left({\frac {-x}{2}}\right)\left({\frac {-2x}{3}}\right)+(x)\left({\frac {-x}{2}}\right)\left({\frac {-2x}{3}}\right)\left({\frac {-3x}{4}}\right)+\cdots }
Applying Euler's continued fraction formula to this expression shows that
log
(
1
+
x
)
=
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1
−
−
x
2
1
+
−
x
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−
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1
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x
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−
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3
x
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+
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⋱
{\displaystyle \log(1+x)={\cfrac {x}{1-{\cfrac {\frac {-x}{2}}{1+{\frac {-x}{2}}-{\cfrac {\frac {-2x}{3}}{1+{\frac {-2x}{3}}-{\cfrac {\frac {-3x}{4}}{1+{\frac {-3x}{4}}-\ddots }}}}}}}}}
and using an equivalence transformation to clear all the fractions results in
log
(
1
+
x
)
=
x
1
+
x
2
−
x
+
2
2
x
3
−
2
x
+
3
2
x
4
−
3
x
+
⋱
{\displaystyle \log(1+x)={\cfrac {x}{1+{\cfrac {x}{2-x+{\cfrac {2^{2}x}{3-2x+{\cfrac {3^{2}x}{4-3x+\ddots }}}}}}}}}
x | < 1 because it is equivalent to the series from which it was derived.[2]
The trigonometric functions
edit
The Taylor series of the sine function converges over the entire complex plane and can be expressed as the sum of products.
sin
x
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
+
1
)
!
x
2
n
+
1
=
x
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
+
x
9
9
!
−
⋯
=
x
+
(
x
)
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−
x
2
2
⋅
3
)
+
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x
)
(
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2
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)
(
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2
4
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)
+
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x
)
(
−
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2
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3
)
(
−
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2
4
⋅
5
)
(
−
x
2
6
⋅
7
)
+
⋯
{\displaystyle {\begin{aligned}\sin x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-\cdots \\[8pt]&=x+(x)\left({\frac {-x^{2}}{2\cdot 3}}\right)+(x)\left({\frac {-x^{2}}{2\cdot 3}}\right)\left({\frac {-x^{2}}{4\cdot 5}}\right)+(x)\left({\frac {-x^{2}}{2\cdot 3}}\right)\left({\frac {-x^{2}}{4\cdot 5}}\right)\left({\frac {-x^{2}}{6\cdot 7}}\right)+\cdots \end{aligned}}}
Euler's continued fraction formula can then be applied
x
1
−
−
x
2
2
⋅
3
1
+
−
x
2
2
⋅
3
−
−
x
2
4
⋅
5
1
+
−
x
2
4
⋅
5
−
−
x
2
6
⋅
7
1
+
−
x
2
6
⋅
7
−
⋱
{\displaystyle {\cfrac {x}{1-{\cfrac {\frac {-x^{2}}{2\cdot 3}}{1+{\frac {-x^{2}}{2\cdot 3}}-{\cfrac {\frac {-x^{2}}{4\cdot 5}}{1+{\frac {-x^{2}}{4\cdot 5}}-{\cfrac {\frac {-x^{2}}{6\cdot 7}}{1+{\frac {-x^{2}}{6\cdot 7}}-\ddots }}}}}}}}}
An equivalence transformation is used to clear the denominators:
sin
x
=
x
1
+
x
2
2
⋅
3
−
x
2
+
2
⋅
3
x
2
4
⋅
5
−
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2
+
4
⋅
5
x
2
6
⋅
7
−
x
2
+
⋱
.
{\displaystyle \sin x={\cfrac {x}{1+{\cfrac {x^{2}}{2\cdot 3-x^{2}+{\cfrac {2\cdot 3x^{2}}{4\cdot 5-x^{2}+{\cfrac {4\cdot 5x^{2}}{6\cdot 7-x^{2}+\ddots }}}}}}}}.}
The same argument can be applied to the cosine function:
cos
x
=
∑
n
=
0
∞
(
−
1
)
n
(
2
n
)
!
x
2
n
=
1
−
x
2
2
!
+
x
4
4
!
−
x
6
6
!
+
x
8
8
!
−
⋯
=
1
+
−
x
2
2
+
(
−
x
2
2
)
(
−
x
2
3
⋅
4
)
+
(
−
x
2
2
)
(
−
x
2
3
⋅
4
)
(
−
x
2
5
⋅
6
)
+
⋯
=
1
1
−
−
x
2
2
1
+
−
x
2
2
−
−
x
2
3
⋅
4
1
+
−
x
2
3
⋅
4
−
−
x
2
5
⋅
6
1
+
−
x
2
5
⋅
6
−
⋱
{\displaystyle {\begin{aligned}\cos x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+{\frac {x^{8}}{8!}}-\cdots \\[8pt]&=1+{\frac {-x^{2}}{2}}+\left({\frac {-x^{2}}{2}}\right)\left({\frac {-x^{2}}{3\cdot 4}}\right)+\left({\frac {-x^{2}}{2}}\right)\left({\frac {-x^{2}}{3\cdot 4}}\right)\left({\frac {-x^{2}}{5\cdot 6}}\right)+\cdots \\[8pt]&={\cfrac {1}{1-{\cfrac {\frac {-x^{2}}{2}}{1+{\frac {-x^{2}}{2}}-{\cfrac {\frac {-x^{2}}{3\cdot 4}}{1+{\frac {-x^{2}}{3\cdot 4}}-{\cfrac {\frac {-x^{2}}{5\cdot 6}}{1+{\frac {-x^{2}}{5\cdot 6}}-\ddots }}}}}}}}\end{aligned}}}
∴
cos
x
=
1
1
+
x
2
2
−
x
2
+
2
x
2
3
⋅
4
−
x
2
+
3
⋅
4
x
2
5
⋅
6
−
x
2
+
⋱
.
{\displaystyle \therefore \cos x={\cfrac {1}{1+{\cfrac {x^{2}}{2-x^{2}+{\cfrac {2x^{2}}{3\cdot 4-x^{2}+{\cfrac {3\cdot 4x^{2}}{5\cdot 6-x^{2}+\ddots }}}}}}}}.}
The inverse trigonometric functions
edit
The inverse trigonometric functions can be represented as continued fractions.
sin
−
1
x
=
∑
n
=
0
∞
(
2
n
−
1
)
!
!
(
2
n
)
!
!
⋅
x
2
n
+
1
2
n
+
1
=
x
+
(
1
2
)
x
3
3
+
(
1
⋅
3
2
⋅
4
)
x
5
5
+
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
x
7
7
+
⋯
=
x
+
x
(
x
2
2
⋅
3
)
+
x
(
x
2
2
⋅
3
)
(
(
3
x
)
2
4
⋅
5
)
+
x
(
x
2
2
⋅
3
)
(
(
3
x
)
2
4
⋅
5
)
(
(
5
x
)
2
6
⋅
7
)
+
⋯
=
x
1
−
x
2
2
⋅
3
1
+
x
2
2
⋅
3
−
(
3
x
)
2
4
⋅
5
1
+
(
3
x
)
2
4
⋅
5
−
(
5
x
)
2
6
⋅
7
1
+
(
5
x
)
2
6
⋅
7
−
⋱
{\displaystyle {\begin{aligned}\sin ^{-1}x=\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}\cdot {\frac {x^{2n+1}}{2n+1}}&=x+\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}+\cdots \\[8pt]&=x+x\left({\frac {x^{2}}{2\cdot 3}}\right)+x\left({\frac {x^{2}}{2\cdot 3}}\right)\left({\frac {(3x)^{2}}{4\cdot 5}}\right)+x\left({\frac {x^{2}}{2\cdot 3}}\right)\left({\frac {(3x)^{2}}{4\cdot 5}}\right)\left({\frac {(5x)^{2}}{6\cdot 7}}\right)+\cdots \\[8pt]&={\cfrac {x}{1-{\cfrac {\frac {x^{2}}{2\cdot 3}}{1+{\frac {x^{2}}{2\cdot 3}}-{\cfrac {\frac {(3x)^{2}}{4\cdot 5}}{1+{\frac {(3x)^{2}}{4\cdot 5}}-{\cfrac {\frac {(5x)^{2}}{6\cdot 7}}{1+{\frac {(5x)^{2}}{6\cdot 7}}-\ddots }}}}}}}}\end{aligned}}}
An equivalence transformation yields
sin
−
1
x
=
x
1
−
x
2
2
⋅
3
+
x
2
−
2
⋅
3
(
3
x
)
2
4
⋅
5
+
(
3
x
)
2
−
4
⋅
5
(
5
x
2
)
6
⋅
7
+
(
5
x
2
)
−
⋱
.
{\displaystyle \sin ^{-1}x={\cfrac {x}{1-{\cfrac {x^{2}}{2\cdot 3+x^{2}-{\cfrac {2\cdot 3(3x)^{2}}{4\cdot 5+(3x)^{2}-{\cfrac {4\cdot 5(5x^{2})}{6\cdot 7+(5x^{2})-\ddots }}}}}}}}.}
The continued fraction for the inverse tangent is straightforward:
tan
−
1
x
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
+
1
2
n
+
1
=
x
−
x
3
3
+
x
5
5
−
x
7
7
+
⋯
=
x
+
x
(
−
x
2
3
)
+
x
(
−
x
2
3
)
(
−
3
x
2
5
)
+
x
(
−
x
2
3
)
(
−
3
x
2
5
)
(
−
5
x
2
7
)
+
⋯
=
x
1
−
−
x
2
3
1
+
−
x
2
3
−
−
3
x
2
5
1
+
−
3
x
2
5
−
−
5
x
2
7
1
+
−
5
x
2
7
−
⋱
=
x
1
+
x
2
3
−
x
2
+
(
3
x
)
2
5
−
3
x
2
+
(
5
x
)
2
7
−
5
x
2
+
⋱
.
{\displaystyle {\begin{aligned}\tan ^{-1}x=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{2n+1}}&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots \\[8pt]&=x+x\left({\frac {-x^{2}}{3}}\right)+x\left({\frac {-x^{2}}{3}}\right)\left({\frac {-3x^{2}}{5}}\right)+x\left({\frac {-x^{2}}{3}}\right)\left({\frac {-3x^{2}}{5}}\right)\left({\frac {-5x^{2}}{7}}\right)+\cdots \\[8pt]&={\cfrac {x}{1-{\cfrac {\frac {-x^{2}}{3}}{1+{\frac {-x^{2}}{3}}-{\cfrac {\frac {-3x^{2}}{5}}{1+{\frac {-3x^{2}}{5}}-{\cfrac {\frac {-5x^{2}}{7}}{1+{\frac {-5x^{2}}{7}}-\ddots }}}}}}}}\\[8pt]&={\cfrac {x}{1+{\cfrac {x^{2}}{3-x^{2}+{\cfrac {(3x)^{2}}{5-3x^{2}+{\cfrac {(5x)^{2}}{7-5x^{2}+\ddots }}}}}}}}.\end{aligned}}}
edit
We can use the previous example involving the inverse tangent to construct a continued fraction representation of π . We note that
tan
−
1
(
1
)
=
π
4
,
{\displaystyle \tan ^{-1}(1)={\frac {\pi }{4}},}
And setting x = 1 in the previous result, we obtain immediately
π
=
4
1
+
1
2
2
+
3
2
2
+
5
2
2
+
7
2
2
+
⋱
.
{\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+\ddots }}}}}}}}}}.\,}
The hyperbolic functions
edit
Recalling the relationship between the hyperbolic functions and the trigonometric functions,
sin
i
x
=
i
sinh
x
{\displaystyle \sin ix=i\sinh x}
cos
i
x
=
cosh
x
,
{\displaystyle \cos ix=\cosh x,}
And that
i
2
=
−
1
,
{\displaystyle i^{2}=-1,}
sinh
x
=
x
1
−
x
2
2
⋅
3
+
x
2
−
2
⋅
3
x
2
4
⋅
5
+
x
2
−
4
⋅
5
x
2
6
⋅
7
+
x
2
−
⋱
{\displaystyle \sinh x={\cfrac {x}{1-{\cfrac {x^{2}}{2\cdot 3+x^{2}-{\cfrac {2\cdot 3x^{2}}{4\cdot 5+x^{2}-{\cfrac {4\cdot 5x^{2}}{6\cdot 7+x^{2}-\ddots }}}}}}}}}
cosh
x
=
1
1
−
x
2
2
+
x
2
−
2
x
2
3
⋅
4
+
x
2
−
3
⋅
4
x
2
5
⋅
6
+
x
2
−
⋱
.
{\displaystyle \cosh x={\cfrac {1}{1-{\cfrac {x^{2}}{2+x^{2}-{\cfrac {2x^{2}}{3\cdot 4+x^{2}-{\cfrac {3\cdot 4x^{2}}{5\cdot 6+x^{2}-\ddots }}}}}}}}.}
The inverse hyperbolic functions
edit
The inverse hyperbolic functions are related to the inverse trigonometric functions similar to how the hyperbolic functions are related to the trigonometric functions,
sin
−
1
i
x
=
i
sinh
−
1
x
{\displaystyle \sin ^{-1}ix=i\sinh ^{-1}x}
tan
−
1
i
x
=
i
tanh
−
1
x
,
{\displaystyle \tan ^{-1}ix=i\tanh ^{-1}x,}
And these continued fractions are easily derived:
sinh
−
1
x
=
x
1
+
x
2
2
⋅
3
−
x
2
+
2
⋅
3
(
3
x
)
2
4
⋅
5
−
(
3
x
)
2
+
4
⋅
5
(
5
x
2
)
6
⋅
7
−
(
5
x
2
)
+
⋱
{\displaystyle \sinh ^{-1}x={\cfrac {x}{1+{\cfrac {x^{2}}{2\cdot 3-x^{2}+{\cfrac {2\cdot 3(3x)^{2}}{4\cdot 5-(3x)^{2}+{\cfrac {4\cdot 5(5x^{2})}{6\cdot 7-(5x^{2})+\ddots }}}}}}}}}
tanh
−
1
x
=
x
1
−
x
2
3
+
x
2
−
(
3
x
)
2
5
+
3
x
2
−
(
5
x
)
2
7
+
5
x
2
−
⋱
.
{\displaystyle \tanh ^{-1}x={\cfrac {x}{1-{\cfrac {x^{2}}{3+x^{2}-{\cfrac {(3x)^{2}}{5+3x^{2}-{\cfrac {(5x)^{2}}{7+5x^{2}-\ddots }}}}}}}}.}
^ Leonhard Euler (1748), "18", Introductio in analysin infinitorum , vol. I^ a b This series converges for |x | < 1, by Abel's test (applied to the series for log(1 − x )).
![{\displaystyle {\begin{aligned}a_{0}+a_{0}a_{1}+a_{0}a_{1}a_{2}+a_{0}a_{1}a_{2}a_{3}&=a_{0}(a_{1}(a_{2}(a_{3}+1)+1)+1)\\[8pt]&={\cfrac {a_{0}}{\cfrac {1}{a_{1}(a_{2}(a_{3}+1)+1)+1}}}\\[8pt]&={\cfrac {a_{0}}{{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)+1}{a_{1}(a_{2}(a_{3}+1)+1)+1}}-{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)}{a_{1}(a_{2}(a_{3}+1)+1)+1}}}}={\cfrac {a_{0}}{1-{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)}{a_{1}(a_{2}(a_{3}+1)+1)+1}}}}\\[8pt]&={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)+1}{a_{2}(a_{3}+1)+1}}}}}\\[8pt]&={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{{\cfrac {a_{1}(a_{2}(a_{3}+1)+1)}{a_{2}(a_{3}+1)+1}}+{\cfrac {a_{2}(a_{3}+1)+1}{a_{2}(a_{3}+1)+1}}-{\cfrac {a_{2}(a_{3}+1)}{a_{2}(a_{3}+1)+1}}}}}}={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{1+a_{1}-{\cfrac {a_{2}(a_{3}+1)}{a_{2}(a_{3}+1)+1}}}}}}\\[8pt]&={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{1+a_{1}-{\cfrac {a_{2}}{\cfrac {a_{2}(a_{3}+1)+1}{a_{3}+1}}}}}}}\\[8pt]&={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{1+a_{1}-{\cfrac {a_{2}}{{\cfrac {a_{2}(a_{3}+1)}{a_{3}+1}}+{\cfrac {a_{3}+1}{a_{3}+1}}-{\cfrac {a_{3}}{a_{3}+1}}}}}}}}={\cfrac {a_{0}}{1-{\cfrac {a_{1}}{1+a_{1}-{\cfrac {a_{2}}{1+a_{2}-{\cfrac {a_{3}}{1+a_{3}}}}}}}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2220d0ee296b162bc6d1d865431c25db083f18de)
![{\displaystyle {\begin{aligned}\sin x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-\cdots \\[8pt]&=x+(x)\left({\frac {-x^{2}}{2\cdot 3}}\right)+(x)\left({\frac {-x^{2}}{2\cdot 3}}\right)\left({\frac {-x^{2}}{4\cdot 5}}\right)+(x)\left({\frac {-x^{2}}{2\cdot 3}}\right)\left({\frac {-x^{2}}{4\cdot 5}}\right)\left({\frac {-x^{2}}{6\cdot 7}}\right)+\cdots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc847450258bfb8bba32de1c626c47468155a663)
![{\displaystyle {\begin{aligned}\cos x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+{\frac {x^{8}}{8!}}-\cdots \\[8pt]&=1+{\frac {-x^{2}}{2}}+\left({\frac {-x^{2}}{2}}\right)\left({\frac {-x^{2}}{3\cdot 4}}\right)+\left({\frac {-x^{2}}{2}}\right)\left({\frac {-x^{2}}{3\cdot 4}}\right)\left({\frac {-x^{2}}{5\cdot 6}}\right)+\cdots \\[8pt]&={\cfrac {1}{1-{\cfrac {\frac {-x^{2}}{2}}{1+{\frac {-x^{2}}{2}}-{\cfrac {\frac {-x^{2}}{3\cdot 4}}{1+{\frac {-x^{2}}{3\cdot 4}}-{\cfrac {\frac {-x^{2}}{5\cdot 6}}{1+{\frac {-x^{2}}{5\cdot 6}}-\ddots }}}}}}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfba04d28890375b61217b61171786d3860d958a)
![{\displaystyle {\begin{aligned}\sin ^{-1}x=\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}\cdot {\frac {x^{2n+1}}{2n+1}}&=x+\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}+\cdots \\[8pt]&=x+x\left({\frac {x^{2}}{2\cdot 3}}\right)+x\left({\frac {x^{2}}{2\cdot 3}}\right)\left({\frac {(3x)^{2}}{4\cdot 5}}\right)+x\left({\frac {x^{2}}{2\cdot 3}}\right)\left({\frac {(3x)^{2}}{4\cdot 5}}\right)\left({\frac {(5x)^{2}}{6\cdot 7}}\right)+\cdots \\[8pt]&={\cfrac {x}{1-{\cfrac {\frac {x^{2}}{2\cdot 3}}{1+{\frac {x^{2}}{2\cdot 3}}-{\cfrac {\frac {(3x)^{2}}{4\cdot 5}}{1+{\frac {(3x)^{2}}{4\cdot 5}}-{\cfrac {\frac {(5x)^{2}}{6\cdot 7}}{1+{\frac {(5x)^{2}}{6\cdot 7}}-\ddots }}}}}}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e48943c46be1565da33ae0668dce771c7e1586fb)
![{\displaystyle {\begin{aligned}\tan ^{-1}x=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{2n+1}}&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots \\[8pt]&=x+x\left({\frac {-x^{2}}{3}}\right)+x\left({\frac {-x^{2}}{3}}\right)\left({\frac {-3x^{2}}{5}}\right)+x\left({\frac {-x^{2}}{3}}\right)\left({\frac {-3x^{2}}{5}}\right)\left({\frac {-5x^{2}}{7}}\right)+\cdots \\[8pt]&={\cfrac {x}{1-{\cfrac {\frac {-x^{2}}{3}}{1+{\frac {-x^{2}}{3}}-{\cfrac {\frac {-3x^{2}}{5}}{1+{\frac {-3x^{2}}{5}}-{\cfrac {\frac {-5x^{2}}{7}}{1+{\frac {-5x^{2}}{7}}-\ddots }}}}}}}}\\[8pt]&={\cfrac {x}{1+{\cfrac {x^{2}}{3-x^{2}+{\cfrac {(3x)^{2}}{5-3x^{2}+{\cfrac {(5x)^{2}}{7-5x^{2}+\ddots }}}}}}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d12abf4c05581843e491d5c3609e1a43b17638b1)
