Oscillating Infinite Continued Fractions

Number Theory Level 3

What is the equation for oscillating infinite continued fractions of type [ 1 ; m , n , m , n , m , n , . . . ] [1; m, n, m, n, m, n, ...] as shown below?

1 + 1 m + 1 n + 1 m + 1 n + . . . \Large 1 + \frac{1}{m + \frac{1}{n + \frac{1}{m + \frac{1}{n + ...}}}}

m n ( m n + 4 ) − m n + 2 m 2 m \frac{\sqrt{mn(mn + 4)} - mn + 2m}{2m} 5 n 2 + 3 m − 1 + 3 m n − n 2 n m + 1 \frac{\sqrt{5n^2 + 3m - 1} + 3mn - n}{2nm + 1} 3 m n ( n − 1 ) + 2 m − 2 n + 2 m + n − 1 \frac{\sqrt{3mn(n - 1)} + 2m - 2n + 2}{m + n - 1} There is no such equation m n + 2 m + 2 n + 1 + 2 m + 1 2 n + 1 \frac{\sqrt{mn + 2m + 2n + 1} + 2m + 1}{2n + 1} 9 n ( 3 m n + 7 ) − 4 m n 2 − 2 3 n 2 \frac{\sqrt{9n(3mn + 7)} - 4mn^2 - 2}{3n^2}

Jonathan Pappas by Jonathan Pappas , 19, USA

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3 solutions

Dwaipayan Shikari
Nov 22, 2020

Let's take,

1 m + 1 n + 1 m + 1 ⋯ = a \frac{1}{m+\frac{1}{n+\frac{1}{m+\frac{1}{\cdots}}}} = a

So we can say that a = 1 m + 1 n + a \textrm{ So we can say that } a = \frac{1}{m+\frac{1}{n+a}}

So , n + a m n + a m + 1 = a \frac{n+a}{mn+am+1} = a

m a 2 + m n a − n = 0 ma^2 +mna-n =0

a = − m n + ( m 2 n 2 + 4 m n ) 2 m a = \frac{-mn +(\sqrt{m^2n^2 +4mn})}{2m}

Answer is 1 + a = m n ( m n + 4 ) − m n + 2 m 2 m 1+a =\boxed{\frac{\sqrt{mn(mn+4)}-mn+2m}{2m}}

0 Helpful 0 Interesting 1 Brilliant 0 Confused
Chew-Seong Cheong
Nov 18, 2020

Let

1 + x = 1 + 1 m + 1 n + 1 m + 1 n + ⋯ x = 1 m + 1 n + 1 m + 1 n + ⋯ = 1 m + 1 n + x = n + x m x + m n + 1 m x 2 + m n x + x = n + x m x 2 + m n x − n = 0 ⟹ x = m 2 n 2 + 4 m n − m n 2 m ⟹ 1 + x = m n ( m n + 4 ) − m n + 2 m 2 m \large \begin{aligned} 1 + x & = 1 + \frac 1{m+\frac 1{n+\frac 1{m+\frac 1{n + \cdots}}}} \\ x & = \frac 1{m+\frac 1{n+\frac 1{m+\frac 1{n + \cdots}}}} \\ & = \frac 1{m+\frac 1{n+x}} \\ & = \frac {n+x}{mx+mn+1} \\ mx^2 + mnx + x & = n+x \\ mx^2 + mnx - n & = 0 \\ \implies x & = \frac {\sqrt{m^2n^2+4mn}-mn}{2m} \\ \implies 1+x & = \boxed{\frac {\sqrt{mn(mn+4)}-mn+2m}{2m}} \end{aligned}

0 Helpful 0 Interesting 1 Brilliant 0 Confused
Jonathan Pappas
Nov 18, 2020

I started out by finding a pattern for [ 1 ; 1 , n , 1 , n , 1 , n , . . . ] [1;1,n,1,n,1,n,...]

  • 1 + 1 1 + 1 1 + 1 1 + 1 1 + . . . = 1.618033988749895 = 5 + 1 2 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + ...}}}} = 1.618033988749895 = \frac{\sqrt{5} + 1}{2}
  • 1 + 1 1 + 1 2 + 1 1 + 1 2 + . . . = 1.7320508075688772 = 12 − 0 2 1 + \frac{1}{1 + \frac{1}{2 + \frac{1}{1 + \frac{1}{2 + ...}}}} = 1.7320508075688772 = \frac{\sqrt{12} - 0}{2}
  • 1 + 1 1 + 1 3 + 1 1 + 1 3 + . . . = 1.79128784747792 = 21 − 1 2 1 + \frac{1}{1 + \frac{1}{3 + \frac{1}{1 + \frac{1}{3 + ...}}}} = 1.79128784747792 = \frac{\sqrt{21} - 1}{2}
  • 1 + 1 1 + 1 4 + 1 1 + 1 4 + . . . = 1.8284271247461903 = 32 − 2 2 1 + \frac{1}{1 + \frac{1}{4 + \frac{1}{1 + \frac{1}{4 + ...}}}} = 1.8284271247461903 = \frac{\sqrt{32} - 2}{2}
  • 1 + 1 1 + 1 5 + 1 1 + 1 5 + . . . = 1.8541019662496847 = 45 − 3 2 1 + \frac{1}{1 + \frac{1}{5 + \frac{1}{1 + \frac{1}{5 + ...}}}} =1.8541019662496847 = \frac{\sqrt{45} - 3}{2}

etc.

I found this equal to n ( n + 4 ) − n + 2 2 \frac{\sqrt{n(n + 4)} - n + 2}{2}

Then, when looking at other patterns, I found a larger pattern:

  • [ 1 ; 1 , n , 1 , n , 1 , n , . . . ] = 1 n ( 1 n + 4 ) − 1 n + 2 2 [1;1,n,1,n,1,n,...] = \frac{\sqrt{1n(1n + 4)} - 1n + 2}{2}

  • [ 1 ; 2 , n , 2 , n , 2 , n , . . . ] = 2 n ( 2 n + 4 ) − 2 n + 4 4 [1;2,n,2,n,2,n,...] = \frac{\sqrt{2n(2n + 4)} - 2n + 4}{4}

  • [ 1 ; 3 , n , 3 , n , 3 , n , . . . ] = 3 n ( 3 n + 4 ) − 3 n + 6 6 [1;3,n,3,n,3,n,...] = \frac{\sqrt{3n(3n + 4)} - 3n + 6}{6}

  • [ 1 ; 4 , n , 4 , n , 4 , n , . . . ] = 4 n ( 4 n + 4 ) − 4 n + 8 8 [1;4,n,4,n,4,n,...] = \frac{\sqrt{4n(4n + 4)} - 4n + 8}{8}

  • [ 1 ; 5 , n , 5 , n , 5 , n , . . . ] = 5 n ( 5 n + 4 ) − 5 n + 10 10 [1;5,n,5,n,5,n,...] = \frac{\sqrt{5n(5n + 4)} - 5n + 10}{10}

  • [ 1 ; 6 , n , 6 , n , 6 , n , . . . ] = 6 n ( 6 n + 4 ) − 6 n + 12 12 [1;6,n,6,n,6,n,...] = \frac{\sqrt{6n(6n + 4)} - 6n + 12}{12}

etc.

From this pattern, you can ascertain:

  • [ 1 ; m , n , m , n , m , n , . . . ] = m n ( m n + 4 ) − m n + 2 m 2 m [1;m,n,m,n,m,n,...] = \frac{\sqrt{mn(mn + 4)} - mn + 2m}{2m}
0 Helpful 0 Interesting 1 Brilliant 1 Confused

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