Can I use Nest iteration variable as the variable of Limit?












2












$begingroup$


Can anybody tell me what is wrong with this: 



Limit[Nest[1/(1 + #) &, x0, n], n -> ∞]





nest







share|improve this question











$endgroup$








  • 2




    $begingroup$
    The 3rd argument of Nest must be numeric
    $endgroup$
    – Coolwater
    Nov 18 '18 at 14:02






  • 2




    $begingroup$
    This FixedPoint[1/(1 + #) &, #] & /@ Range[0., 4, 1] might help.
    $endgroup$
    – Αλέξανδρος Ζεγγ
    Nov 18 '18 at 14:17






  • 1




    $begingroup$
    Or for an exact value x /. Solve[{1/(1 + x) == x, x > 0}, x][[1]]
    $endgroup$
    – Bob Hanlon
    Nov 18 '18 at 14:35
















2












$begingroup$


Can anybody tell me what is wrong with this: 



Limit[Nest[1/(1 + #) &, x0, n], n -> ∞]





nest







share|improve this question











$endgroup$








  • 2




    $begingroup$
    The 3rd argument of Nest must be numeric
    $endgroup$
    – Coolwater
    Nov 18 '18 at 14:02






  • 2




    $begingroup$
    This FixedPoint[1/(1 + #) &, #] & /@ Range[0., 4, 1] might help.
    $endgroup$
    – Αλέξανδρος Ζεγγ
    Nov 18 '18 at 14:17






  • 1




    $begingroup$
    Or for an exact value x /. Solve[{1/(1 + x) == x, x > 0}, x][[1]]
    $endgroup$
    – Bob Hanlon
    Nov 18 '18 at 14:35














2












2








2


1



$begingroup$


Can anybody tell me what is wrong with this: 



Limit[Nest[1/(1 + #) &, x0, n], n -> ∞]





nest







share|improve this question











$endgroup$




Can anybody tell me what is wrong with this: 



Limit[Nest[1/(1 + #) &, x0, n], n -> ∞]





nest




nest






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 18 '18 at 14:20









Αλέξανδρος Ζεγγ

4,2341929




4,2341929










asked Nov 18 '18 at 13:53









pablopablo

162




162








  • 2




    $begingroup$
    The 3rd argument of Nest must be numeric
    $endgroup$
    – Coolwater
    Nov 18 '18 at 14:02






  • 2




    $begingroup$
    This FixedPoint[1/(1 + #) &, #] & /@ Range[0., 4, 1] might help.
    $endgroup$
    – Αλέξανδρος Ζεγγ
    Nov 18 '18 at 14:17






  • 1




    $begingroup$
    Or for an exact value x /. Solve[{1/(1 + x) == x, x > 0}, x][[1]]
    $endgroup$
    – Bob Hanlon
    Nov 18 '18 at 14:35














  • 2




    $begingroup$
    The 3rd argument of Nest must be numeric
    $endgroup$
    – Coolwater
    Nov 18 '18 at 14:02






  • 2




    $begingroup$
    This FixedPoint[1/(1 + #) &, #] & /@ Range[0., 4, 1] might help.
    $endgroup$
    – Αλέξανδρος Ζεγγ
    Nov 18 '18 at 14:17






  • 1




    $begingroup$
    Or for an exact value x /. Solve[{1/(1 + x) == x, x > 0}, x][[1]]
    $endgroup$
    – Bob Hanlon
    Nov 18 '18 at 14:35








2




2




$begingroup$
The 3rd argument of Nest must be numeric
$endgroup$
– Coolwater
Nov 18 '18 at 14:02




$begingroup$
The 3rd argument of Nest must be numeric
$endgroup$
– Coolwater
Nov 18 '18 at 14:02




2




2




$begingroup$
This FixedPoint[1/(1 + #) &, #] & /@ Range[0., 4, 1] might help.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 18 '18 at 14:17




$begingroup$
This FixedPoint[1/(1 + #) &, #] & /@ Range[0., 4, 1] might help.
$endgroup$
– Αλέξανδρος Ζεγγ
Nov 18 '18 at 14:17




1




1




$begingroup$
Or for an exact value x /. Solve[{1/(1 + x) == x, x > 0}, x][[1]]
$endgroup$
– Bob Hanlon
Nov 18 '18 at 14:35




$begingroup$
Or for an exact value x /. Solve[{1/(1 + x) == x, x > 0}, x][[1]]
$endgroup$
– Bob Hanlon
Nov 18 '18 at 14:35










1 Answer
1






active

oldest

votes


















6












$begingroup$

Nest is a functional programming construct whereas Limit works primarily with mathematical expressions. It simply has no way to work with Nest[...].



This can instead be handled by converting the nested expression into a solved recurrence.



recval = 

 RSolveValue[{f[n] == 1/(1 + f[n - 1]), f[0] == x0}, f[n], n]



(* Out[591]= ((2/(1 + Sqrt[5]))^

   n (-2^(1 - n) (1 - Sqrt[5])^n + 

     2^(1 - n) (1 + Sqrt[5])^n + (1/2 (1 - Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 - Sqrt[5]))^n x0 - (1/2 (1 + Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 + Sqrt[5]))^n x0))/(1 + Sqrt[

   5] - ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 

   Sqrt[5] ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 2 x0 - 

   2 ((1 - Sqrt[5])/(1 + Sqrt[5]))^n x0) *)


Since we are only really interested in integer n I will use DiscreteLimit on this.



dlim = DiscreteLimit[recval, n -> Infinity]



(* Out[592]= (2 + (-1 + Sqrt[5]) x0)/(1 + Sqrt[5] + 2 x0) *)


This is actually independent of initial value:



FullSimplify[dlim]



(* Out[610]= 1/2 (-1 + Sqrt[5]) *)


Another well known way to deduce candidate values for the limit is to solve the fixed point equation.



Solve[x == 1/(1 + x), x]



(* Out[594]= {{x -> 1/2 (-1 - Sqrt[5])}, {x -> 1/2 (-1 + Sqrt[5])}} *)





share|improve this answer









$endgroup$













  • $begingroup$
    +1 You get the same result with Limit[recval, n -> Infinity] // FullSimplify. Also, it is useful to include FullSimplify in definition of recval
    $endgroup$
    – Bob Hanlon
    Nov 18 '18 at 16:16











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









6












$begingroup$

Nest is a functional programming construct whereas Limit works primarily with mathematical expressions. It simply has no way to work with Nest[...].



This can instead be handled by converting the nested expression into a solved recurrence.



recval = 

 RSolveValue[{f[n] == 1/(1 + f[n - 1]), f[0] == x0}, f[n], n]



(* Out[591]= ((2/(1 + Sqrt[5]))^

   n (-2^(1 - n) (1 - Sqrt[5])^n + 

     2^(1 - n) (1 + Sqrt[5])^n + (1/2 (1 - Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 - Sqrt[5]))^n x0 - (1/2 (1 + Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 + Sqrt[5]))^n x0))/(1 + Sqrt[

   5] - ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 

   Sqrt[5] ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 2 x0 - 

   2 ((1 - Sqrt[5])/(1 + Sqrt[5]))^n x0) *)


Since we are only really interested in integer n I will use DiscreteLimit on this.



dlim = DiscreteLimit[recval, n -> Infinity]



(* Out[592]= (2 + (-1 + Sqrt[5]) x0)/(1 + Sqrt[5] + 2 x0) *)


This is actually independent of initial value:



FullSimplify[dlim]



(* Out[610]= 1/2 (-1 + Sqrt[5]) *)


Another well known way to deduce candidate values for the limit is to solve the fixed point equation.



Solve[x == 1/(1 + x), x]



(* Out[594]= {{x -> 1/2 (-1 - Sqrt[5])}, {x -> 1/2 (-1 + Sqrt[5])}} *)





share|improve this answer









$endgroup$













  • $begingroup$
    +1 You get the same result with Limit[recval, n -> Infinity] // FullSimplify. Also, it is useful to include FullSimplify in definition of recval
    $endgroup$
    – Bob Hanlon
    Nov 18 '18 at 16:16
















6












$begingroup$

Nest is a functional programming construct whereas Limit works primarily with mathematical expressions. It simply has no way to work with Nest[...].



This can instead be handled by converting the nested expression into a solved recurrence.



recval = 

 RSolveValue[{f[n] == 1/(1 + f[n - 1]), f[0] == x0}, f[n], n]



(* Out[591]= ((2/(1 + Sqrt[5]))^

   n (-2^(1 - n) (1 - Sqrt[5])^n + 

     2^(1 - n) (1 + Sqrt[5])^n + (1/2 (1 - Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 - Sqrt[5]))^n x0 - (1/2 (1 + Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 + Sqrt[5]))^n x0))/(1 + Sqrt[

   5] - ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 

   Sqrt[5] ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 2 x0 - 

   2 ((1 - Sqrt[5])/(1 + Sqrt[5]))^n x0) *)


Since we are only really interested in integer n I will use DiscreteLimit on this.



dlim = DiscreteLimit[recval, n -> Infinity]



(* Out[592]= (2 + (-1 + Sqrt[5]) x0)/(1 + Sqrt[5] + 2 x0) *)


This is actually independent of initial value:



FullSimplify[dlim]



(* Out[610]= 1/2 (-1 + Sqrt[5]) *)


Another well known way to deduce candidate values for the limit is to solve the fixed point equation.



Solve[x == 1/(1 + x), x]



(* Out[594]= {{x -> 1/2 (-1 - Sqrt[5])}, {x -> 1/2 (-1 + Sqrt[5])}} *)





share|improve this answer









$endgroup$













  • $begingroup$
    +1 You get the same result with Limit[recval, n -> Infinity] // FullSimplify. Also, it is useful to include FullSimplify in definition of recval
    $endgroup$
    – Bob Hanlon
    Nov 18 '18 at 16:16














6












6








6





$begingroup$

Nest is a functional programming construct whereas Limit works primarily with mathematical expressions. It simply has no way to work with Nest[...].



This can instead be handled by converting the nested expression into a solved recurrence.



recval = 

 RSolveValue[{f[n] == 1/(1 + f[n - 1]), f[0] == x0}, f[n], n]



(* Out[591]= ((2/(1 + Sqrt[5]))^

   n (-2^(1 - n) (1 - Sqrt[5])^n + 

     2^(1 - n) (1 + Sqrt[5])^n + (1/2 (1 - Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 - Sqrt[5]))^n x0 - (1/2 (1 + Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 + Sqrt[5]))^n x0))/(1 + Sqrt[

   5] - ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 

   Sqrt[5] ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 2 x0 - 

   2 ((1 - Sqrt[5])/(1 + Sqrt[5]))^n x0) *)


Since we are only really interested in integer n I will use DiscreteLimit on this.



dlim = DiscreteLimit[recval, n -> Infinity]



(* Out[592]= (2 + (-1 + Sqrt[5]) x0)/(1 + Sqrt[5] + 2 x0) *)


This is actually independent of initial value:



FullSimplify[dlim]



(* Out[610]= 1/2 (-1 + Sqrt[5]) *)


Another well known way to deduce candidate values for the limit is to solve the fixed point equation.



Solve[x == 1/(1 + x), x]



(* Out[594]= {{x -> 1/2 (-1 - Sqrt[5])}, {x -> 1/2 (-1 + Sqrt[5])}} *)





share|improve this answer









$endgroup$



Nest is a functional programming construct whereas Limit works primarily with mathematical expressions. It simply has no way to work with Nest[...].



This can instead be handled by converting the nested expression into a solved recurrence.



recval = 

 RSolveValue[{f[n] == 1/(1 + f[n - 1]), f[0] == x0}, f[n], n]



(* Out[591]= ((2/(1 + Sqrt[5]))^

   n (-2^(1 - n) (1 - Sqrt[5])^n + 

     2^(1 - n) (1 + Sqrt[5])^n + (1/2 (1 - Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 - Sqrt[5]))^n x0 - (1/2 (1 + Sqrt[5]))^n x0 + 

     Sqrt[5] (1/2 (1 + Sqrt[5]))^n x0))/(1 + Sqrt[

   5] - ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 

   Sqrt[5] ((1 - Sqrt[5])/(1 + Sqrt[5]))^n + 2 x0 - 

   2 ((1 - Sqrt[5])/(1 + Sqrt[5]))^n x0) *)


Since we are only really interested in integer n I will use DiscreteLimit on this.



dlim = DiscreteLimit[recval, n -> Infinity]



(* Out[592]= (2 + (-1 + Sqrt[5]) x0)/(1 + Sqrt[5] + 2 x0) *)


This is actually independent of initial value:



FullSimplify[dlim]



(* Out[610]= 1/2 (-1 + Sqrt[5]) *)


Another well known way to deduce candidate values for the limit is to solve the fixed point equation.



Solve[x == 1/(1 + x), x]



(* Out[594]= {{x -> 1/2 (-1 - Sqrt[5])}, {x -> 1/2 (-1 + Sqrt[5])}} *)






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 18 '18 at 14:44









Daniel LichtblauDaniel Lichtblau

46.9k276163




46.9k276163












  • $begingroup$
    +1 You get the same result with Limit[recval, n -> Infinity] // FullSimplify. Also, it is useful to include FullSimplify in definition of recval
    $endgroup$
    – Bob Hanlon
    Nov 18 '18 at 16:16


















  • $begingroup$
    +1 You get the same result with Limit[recval, n -> Infinity] // FullSimplify. Also, it is useful to include FullSimplify in definition of recval
    $endgroup$
    – Bob Hanlon
    Nov 18 '18 at 16:16
















$begingroup$
+1 You get the same result with Limit[recval, n -> Infinity] // FullSimplify. Also, it is useful to include FullSimplify in definition of recval
$endgroup$
– Bob Hanlon
Nov 18 '18 at 16:16




$begingroup$
+1 You get the same result with Limit[recval, n -> Infinity] // FullSimplify. Also, it is useful to include FullSimplify in definition of recval
$endgroup$
– Bob Hanlon
Nov 18 '18 at 16:16


















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