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Nov 25, 2006, 6:02:43 AM11/25/06

to

$VersionNumber

5.2

5.2

Limit[Nest[Sqrt[5 + #1] & , 5, n], n -> Infinity]

Nest::intnm : "Non - negative machine - size integer expected at

position 3 in ` ` ".

Limit[Nest[Sqrt[5 + #1] & , 5, n], n -> Infinity]

$VersionNumber

4.0

Needs["Calculus`Limit`"]

Limit[Nest[Sqrt[5 + #]&, 5, n], n -> Infinity]

(1/2)*(1 + Sqrt[21])

N[%]

2.79129

(*Check*)

Table[N[Nest[Sqrt[5 + #1] & , 5, n], 20], {n, 2, 10}]

{2.8569700138728056542,2.8030287215568815242,2.7933901842665806726,2.\

7916644111115112098,2.7913553000489764255,2.7912999301488502752,2.\

7912900118312411350,2.7912882351758732577,2.7912879169257823550}

$VersionNumber

5.2

Needs["Calculus`Limit`"]

Limit::obslt: Functionality previously provided by Limit.m is

superseded by the kernel Limit function. The package Calculus`Limit` is

obsolete.

Nov 26, 2006, 4:24:55 AM11/26/06

to

I contacted Wolfram tech support about this bug. They replied that 4.0

was in error, and that 5.2 is working as it should. (!) I don't buy

their argument. I think they should fix it, to make it possible to

compute the exact limit in this case. Nevertheless, here is their

reply:

was in error, and that 5.2 is working as it should. (!) I don't buy

their argument. I think they should fix it, to make it possible to

compute the exact limit in this case. Nevertheless, here is their

reply:

I believe that Mathematica is behaving correctly in

this example.

Nest and NestList allow you to apply functions a fixed number of times.

In

this sense, the behavior of Ver. 5.X is correct while the behavior of

the

package "Limit" for Ver. 4 is inappropriate.

Often you may want to apply functions until the result no longer

changes.

You can do this using FixedPoint.

In[2]:=

FixedPoint[N[Sqrt[5 + #1] &, 40], N[5, 40]]

Out[2]=

2.791287847477920003294023596864004244492

Note that, the following won't terminate.

In[3]:=

FixedPoint[Sqrt[5 + #1] &, 5]

Out[3]=

$Aborted

because "SameTest" is done symbolically. You may do

In[4]:=

FixedPoint[Sqrt[5 + #1] &, 5.0000000000000000]

Out[4]=

2.791287

though it will take a lot more time if you add more "0"'s to the right

of

5.0000000000000000

This is again because of symbolic "SameTest."

If you do want to symbolically compute the fixed point of the function,

then, as you pointed out, you have to manually solve the equation:

In[5]:=

Solve[x == Sqrt[5 + x]]

Out[5]=

1 + Sqrt[21]

{{x -> ------------}}

2

Nov 27, 2006, 4:26:41 AM11/27/06

to

I have done work on Limit over the past few years so I can comment from

that perspective. While I might have worded the response a bit

differently, I side with Tech Support on this. Let me give a few reasons.

Limit is designed to work on functions defined on a continuum, and known

in a closed form. The function in question satisfies neither requirement.

While it might be possible to find a closed form function that agrees with

it at integers, it is not something that will be done automatically

(indeed, such a function might or might not be what was "intended"). I'll

note that when I tried I was not able to get such a function for this

example using RSolve.

It is true that the Calculus`Limit code had some tricks built in for

handling examples such as this. Neat. (Or, if you are of the right age and

from the English-speaking world, groovy). What it did not have was

consistency or maintainability. The decision to scrap it was not exactly a

difficult call; it was riddled with bugs and simply not fixable.

Bottom line is that the Limit function was not designed or intended to

handle functions defined on only a sequence of points, or to handle

functions not given in explicit closed form. The Calculus`Limit function

had limited (if you will) capabilities in this respect. It is correct to

say that this particular aspect of the package function was not taken on

by the built-in Limit. For examples such as this one gets results exactly

as indicated in the TS response: use Solve to find an exact representation

of the fixed point for the iteration.

Daniel Lichtblau

Wolfram Research

Nov 27, 2006, 4:39:10 AM11/27/06

to

dimitris schrieb:

...

> $VersionNumber

> 5.2

>

> Needs["Calculus`Limit`"]

> Limit::obslt: Functionality previously provided by Limit.m is

> superseded by the kernel Limit function. The package Calculus`Limit` is

> obsolete.

>

...

> $VersionNumber

> 5.2

>

> Needs["Calculus`Limit`"]

> Limit::obslt: Functionality previously provided by Limit.m is

> superseded by the kernel Limit function. The package Calculus`Limit` is

> obsolete.

>

Hi Dimitris,

if you've got the old Limit.m somewhere on your disks, you can Get[] it:

In[1]:=

$VersionNumber

Out[1]=

5.1

In[2]:=

Limit[Nest[Sqrt[5 + #1] & , 5, n], n -> Infinity]

Nest::intnm: ....

Out[2]=

Limit[Nest[Sqrt[5 + #1] & , 5, n], n -> Infinity]

In[3]:=

Get["D:\\Programme\\Wolfram \

Research\\Mathematica\\4.0\\Addons\\StandardPackages\\Calculus\\Limit.m"]

In[4]:=

Limit[Nest[Sqrt[5 + #1] & , 5, n], n -> Infinity]

Out[4]=

(1/2)*(1 + Sqrt[21])

P²

Nov 27, 2006, 4:56:21 AM11/27/06

to

For anyone cares have adopted this limit for the Mathematica Guidebook

for Symbolics of M. Trott where it is used version 4.0.

Dimitris

Nov 28, 2006, 6:13:42 AM11/28/06

to

da...@wolfram.com wrote:

> For examples such as this one gets results exactly

> as indicated in the TS response: use Solve to find an exact representation

> of the fixed point for the iteration.

>

>

> Daniel Lichtblau

> Wolfram Research

> For examples such as this one gets results exactly

> as indicated in the TS response: use Solve to find an exact representation

> of the fixed point for the iteration.

>

>

> Daniel Lichtblau

> Wolfram Research

Yes, you can use Solve[x == Sqrt[5 + x]] to get the exact solution to

this particular problem.

But how is the user supposed to know what equation to solve?

It is unfortunate that Mathematica is unable to give an exact solution

to the limit of expressions like:

Sqrt[10]

Sqrt[5+Sqrt[10]]

Sqrt[5+ Sqrt[5+Sqrt[10]]]

...

Nov 28, 2006, 6:30:57 AM11/28/06

to

Dear Daniel,

Thanks a lot for yor response.

Your explanation is clear. I got the point!

I really appreciate your assistance

Best Regards

Dimitris

P.S. Thanks also to everyone else that replied me.

Nov 29, 2006, 2:59:43 AM11/29/06

to math...@smc.vnet.net

The user should learn some (fairly elementary) mathematics.

The limit of a nested expression of the form: Nest[f,...], if it

exists, is a fixed point of the function f. A fixed point of f is a

root of f[x]-x==0.

The limit of a nested expression of the form: Nest[f,...], if it

exists, is a fixed point of the function f. A fixed point of f is a

root of f[x]-x==0.

Andrzej Kozlowski

Tokyo, Japan

On 28 Nov 2006, at 20:03, bobba...@frii.com wrote:

da...@wolfram.com wrote:

For examples such as this one gets results exactly

as indicated in the TS response: use Solve to find an exact representation

of the fixed point for the iteration.

Daniel Lichtblau

Wolfram Research

Yes, you can use Solve[x == Sqrt[5 + x]] to get the exact solution to

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