x-1-1-1-2-1-3-1-4-

Question Number 45356 by MrW3 last updated on 12/Oct/18

x = \frac{1}{1 + \frac{1}{2 + \frac{1}{3 + \frac{1}{4 + \dots}}}} = ?

Answered by MJS last updated on 12/Oct/18

non - periodic infinite continued fractions

are transcendental numbers, at least I seem

to remember this . we can only approximate

in this case the sequence is

1, \frac{2}{3}, \frac{7}{10}, \frac{30}{43}, \frac{157}{224}, \frac{972}{1393}, \frac{6961}{9976}, \frac{56660}{81201}, \frac{516901}{740785}, \frac{5225670}{7489051}, \dots

the difference between the last two fractions

is \frac{1}{5547776645035} < 10^{- 12}

I get x \approx . 697774657964

Commented by MrW3 last updated on 12/Oct/18

T h a n k y o u f o r t h i s a n s w e r s i r!

Commented by MJS last updated on 12/Oct/18

Sir, may I offer you these as a gift :

these are by your special friend Lambert :

\frac{4}{π} = 1 + \frac{1^{2}}{3 + \frac{2^{2}}{5 + \frac{3^{2}}{7 + \frac{4^{2}}{9 + \frac{5^{2}}{⋱}}}}}

\tan x = \frac{x}{1 - \frac{x^{2}}{3 - \frac{x^{2}}{5 - \frac{x^{2}}{7 - \frac{x^{2}}{⋱}}}}}

these are by Euler :

e - 2 = \frac{1}{1 + \frac{1}{2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{4 + \frac{1}{1 + \frac{1}{1 + \frac{1}{6 + \frac{1}{⋱}}}}}}}}} = \frac{1}{1 + \frac{1}{2 + \frac{2}{3 + \frac{3}{4 + \frac{4}{5 + \frac{5}{6 + \frac{6}{7 + \frac{7}{8 + \frac{8}{⋱}}}}}}}}}

Commented by MrW3 last updated on 13/Oct/18

v e r y i n t e r e s t i n g, t h a n k y o u s i r .

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