I can't see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278...(1+3/4)/(2+5/6)~~0.6176 (1 +(3+(7)/(8))/(4+(9)/(10)))/(2+(5+(11)/(12))/(6+(13)/(14)))~~0.6175 Going all the way up to 62 gives a result of 0.627841944566, so it seems to converge. Is it possible to find a value for this? Will it have a closed form solution?

Mariyah Bell 2022-10-25 Answered
How can this expression be calculated? 1 + 3 4 2 + 5 6
I can't see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278...
1 + 3 4 2 + 5 6 0.6176
1 + 3 + 7 8 4 + 9 10 2 + 5 + 11 12 6 + 13 14 0.6175
Going all the way up to 62 gives a result of 0.627841944566, so it seems to converge.
Is it possible to find a value for this? Will it have a closed form solution?
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Answers (1)

toliwask
Answered 2022-10-26 Author has 15 answers
Define
f m ( n ) = { n + f m ( 2 n + 1 ) f m ( 2 n + 2 ) if  n < m , n otherwise.
Then
f 0 ( 0 ) = 0 , f 1 ( 0 ) = 0 + 1 2 , f 2 ( 0 ) = 0 + 1 + 3 4 2 , f 3 ( 0 ) = 0 + 1 + 3 4 2 + 5 6 ,
and you're looking for the value of lim m f m ( 0 )
One can show via reverse induction over n = m , , 1 , 0 that f m ( n ) [ n , n + 1 ]. So in the limit, defining f ( n ) = lim m f m ( n ), we have f ( n ) [ n , n + 1 ]
Using interval arithmetic we can then obtain rigorous bounds on f ( n ). Define the interval-valued function
[ f ] m ( n ) = { n + [ f ] m ( 2 n + 1 ) [ f ] m ( 2 n + 2 ) if  n < m , [ n , n + 1 ] otherwise,
where the usual interval arithmetic rules apply,
x + [ a , b ] = [ x + a , x + b ] , [ a 1 , b 1 ] [ a 2 , b 2 ] = [ a 1 b 2 , a 2 b 1 ]
(because all our intervals are positive, except [ f ] 0 ( 0 ) which never appears in a denominator). It should be possible to show that f k ( n ) [ f ] m ( n ) for all k m, and so f ( n ) [ f ] m ( n ). Assuming that's true,
[ f ] 1023 ( 0 ) = [ 0.62784196682396 542323 , 0.62784196682396 734620 ]
narrows down the desired number f ( 0 ) to 14 significant digits.
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