Forum Navigation:

magazine

FORUMS > Brainteaser Forum < refresh >
Topic Title: arctan(1+ arctan(1+...
Created On Sat Aug 07, 10 07:47 AM
Topic View:

Pages: [ 1 2 3 4 >> Next ]
View thread in raw text format


slops
Member

Posts: 75
Joined: Oct 2007

Sat Aug 07, 10 07:47 AM
User is offline View users profile

Hi all!

What is



-------------------------
Do not stop until you are done.
 
Reply
   
Quote
   
Top
   
Bottom
     



frenchX
Senior Member

Posts: 5911
Joined: Mar 2010

Sat Aug 07, 10 09:14 AM
User is offline

Mapple gives for a=arctan(1+a) with a0=1 the solution a=1.132267725... (It's a bit like as x=tan(x) which doesn't have an analytic solution I fear...).

-------------------------
The important thing is not to stop questioning. Curiosity has its own reason
for existing. Never lose a holy curiosity.

Edited: Sat Aug 07, 10 at 11:15 AM by frenchX
 
Reply
   
Quote
   
Top
   
Bottom
     



slops
Member

Posts: 75
Joined: Oct 2007

Sat Aug 07, 10 11:56 AM
User is offline View users profile

Watch out for AVt

Again it is all a matter of definition!






-------------------------
Do not stop until you are done.
 
Reply
   
Quote
   
Top
   
Bottom
     



Traden4Alpha
Senior Member

Posts: 16882
Joined: Sep 2002

Sat Aug 07, 10 01:08 PM
User is offline

Interesting!

As with the squareroot problem, arctan is multivalued. Arctan(1) = 0.78540 + i*3.14159, where i is any integer. Convention (and numerical calculations) pick n=0. Here we have a infinite number of possible values at each arctan and the possibility that the i^N iterated space of outcomes might or might not fill the real line.

This problem also adds the twist that the units on the angle aren't specified. They could be radians, degrees, or grads. Math usually assumes radians but navigation and surveying rarely use radians. Assumptions about units have caused the loss of billion-dollar spacecraft, so it's best to confirm the units.

-------------------------
"It often happens that a player carries out a deep and complicated calculation, but fails to spot something elementary right at the first move." -- grandmaster Alexander Kotov --inscribed on gift chess sets given by Amaranth hedge fund.
 
Reply
   
Quote
   
Top
   
Bottom
     



AVt
Senior Member

Posts: 1051
Joined: Dec 2001

Sat Aug 07, 10 10:12 PM
User is offline View users profile

Hm - I would read the simple way (principal branch): u(n+1)=arctan(1+u(n)), u(0)=1,
increasing and bounded, thus having a solution and rules on limits then allow to set
up what the French did (where the solution is transcendent).

I see no reason, why to deviate from that (like using identities not valid in all naturals),
what did you had in mind?
 
Reply
   
Quote
   
Top
   
Bottom
     



Paul
Senior Member

Posts: 5300
Joined: Jul 2001

Sat Aug 07, 10 11:17 PM
User is offline View users profile

Quote

Originally posted by: AVt
Hm - I would read the simple way (principal branch): u(n+1)=arctan(1+u(n)), u(0)=1,
increasing and bounded, thus having a solution and rules on limits then allow to set
up what the French did (where the solution is transcendent).

I see no reason, why to deviate from that (like using identities not valid in all naturals),
what did you had in mind?
Plot the multivalued function y=arctan(1+x). And plot y=x. Now do the iteration graphically (i.e the usual thing of going from the arctan(1+x) curve to an axis then to the x then to an axis, and so on). You'll easily see the effect of the choice that you have. Although you have an infinite choice here, as T4A says, you won't find yourself going off into the complex plane unless you force that with your initial condition.

P

 
Reply
   
Quote
   
Top
   
Bottom
     



dunrewpp
Senior Member

Posts: 260
Joined: Apr 2009

Sat Aug 07, 10 11:23 PM
User is offline

Quote

Originally posted by: slops
Hi all!

What is


Either arctan is well-defined or not (at any point in time). If not well-defined, then we don't know what we are talking about (at any point in time). Once it is well-defined, then there is only one answer to the infinitely long expression arctan(1+arctan(1+arctan(1+arctan(1+....)))). The only thing that bothers me is the question: How does the initial value a(0) affect the limit of sequence {a(n)} defined as a(0)=1, a(n+1)=f(1+a(n))?

 
Reply
   
Quote
   
Top
   
Bottom
     



Paul
Senior Member

Posts: 5300
Joined: Jul 2001

Sat Aug 07, 10 11:30 PM
User is offline View users profile

Quote

Originally posted by: dunrewpp
Quote

Originally posted by: slops
Hi all!

What is


Either arctan is well-defined or not (at any point in time). If not well-defined, then we don't know what we are talking about (at any point in time). Once it is well-defined, then there is only one answer to the infinitely long expression arctan(1+arctan(1+arctan(1+arctan(1+....)))). The only thing that bothers me is the question: How does the initial value a(0) affect the limit of sequence {a(n)} defined as a(0)=1, a(n+1)=f(1+a(n))?
A simple plot as I mentioned will show you all you need to know about convergence.

If you don't want to reinvent the wheel then also look at Newton Raphson. (Some on this forum do like reinventing the wheel, including me, and we have found wheels with wonderful shapes, some complex, some chaotic!)

P
 
Reply
   
Quote
   
Top
   
Bottom
     



slops
Member

Posts: 75
Joined: Oct 2007

Sun Aug 08, 10 05:52 AM
User is offline View users profile

There you have an other solution again!

Choose f(x)=arctan(1+atctan(1+x)), in the limit you would have x=arctan(1+atctan(1+x)). That has a totally different solution.




-------------------------
Do not stop until you are done.
 
Reply
   
Quote
   
Top
   
Bottom
     



dunrewpp
Senior Member

Posts: 260
Joined: Apr 2009

Sun Aug 08, 10 06:43 AM
User is offline

Quote

Originally posted by: slops
There you have an other solution again!

Choose f(x)=arctan(1+atctan(1+x)), in the limit you would have x=arctan(1+atctan(1+x)). That has a totally different solution.



First off, I assume that the arctan function is well-defined here. That is, arctan(x)=y if and only if (-pi/2)<y<(pi/2) and tan(y)=x.

Now, it is incorrect to say "x=arctan(1+atctan(1+x))" has a different solution. The equation just quoted has different solutions, one of which agrees with the unique solution of x=arctan(1+x). So, yes, speaking in terms of solution set of an equation, it is a trivial fact that the equation "x=arctan(1+atctan(1+x))" has a different solution set than the equation x=arctan(1+x). The question is: Of all the solutions of "x=arctan(1+atctan(1+x))", which one should we take to be the limit of the sequence b(n+1)=arctan(1+arctan(1+b(n))), b(0)=1? The answer is precisely the one that agrees with the unique solution of x=arctan(1+x).

Here's why:

Let's generalize the original problem.

Let f be a well-defined real-valued function whose domain is the set of real numbers (or other 'appropriate' subset of the reals). Now we look at f(1+f(1+f(1+f(1+f(1+...))))). We can interpret this in two ways amongst many.

One is: a(0)=1, a(n+1)=f(1+a(n)).

Another is: b(0)=1, b(n+1)=f(1+f(1+b(n))).

Now, it is obvious that the sequence {b(n)} is a subsequence of the sequence {a(n)}.

So, if {a(n)} converges to a limit L, then so does the sequence {b(n)}.

More specifically, the sequence {b(n)} defined as b(n+1)=arctan(1+arctan(1+b(n))), b(0)=1 is a subsequence of {a(n)} defined as a(n+1)=arctan(1+a(n)), a(0)=1. Therefore both have the same limit (if at all).
 
Reply
   
Quote
   
Top
   
Bottom
     



Paul
Senior Member

Posts: 5300
Joined: Jul 2001

Sun Aug 08, 10 11:40 AM
User is offline View users profile

What is

?

P

Edited: Sun Aug 08, 10 at 11:41 AM by Paul
 
Reply
   
Quote
   
Top
   
Bottom
     



frenchX
Senior Member

Posts: 5911
Joined: Mar 2010

Sun Aug 08, 10 12:56 PM
User is offline

The solution of this kind of problem strongly depends on the initial condition.
Example on Mapple the solution of x=tan(-1+x) with x0=0.1 gives x=2.13... while with x0=5 it gives x=5.53.... As Traden and others said the tan function is multivaluated modulo 3.14 and so each intricated tangent would be defined with a modulo so you have a kind of chaotic behaviour (maybe with some periodic cycles)

Example :

x0=4 x=2.132
x0=4.5 x=-3.43
x0=5 x=5.53

etc... in this case the best method to find the roots is to use the graphic method and you will find a set of discrete values. For info you have a quite similar equation for the energy levels in a quantum well which gives an infinite number of discrete levels (fortunately or we will be stucked in the fundamental state for ever )


-------------------------
The important thing is not to stop questioning. Curiosity has its own reason
for existing. Never lose a holy curiosity.

Edited: Sun Aug 08, 10 at 12:59 PM by frenchX
 
Reply
   
Quote
   
Top
   
Bottom
     



Traden4Alpha
Senior Member

Posts: 16882
Joined: Sep 2002

Sun Aug 08, 10 01:15 PM
User is offline

Quote

Originally posted by: Paul
What is

?

P
Indeed! The unstable solutions in which a(n) = a(0) of Paul's iterated function are the in-limit convergence values of the original "1+ arctan...." iteration. Using FrenchX result, we find that this will be 1.132267725... + i*pi.


Quote

Originally posted by: dunrewpp
First off, I assume that the arctan function is well-defined here. That is, arctan(x)=y if and only if (-pi/2)<y<(pi/2) and tan(y)=x.
That's nice that you state the restriction on arctan(x), but it's a dangerous definition.

Arbitrary decisions on the "conventional" definitions of an inverse can lead to incorrect results. For instance, systems involving mixed trigonometric functions may have no solutions in the conventional restricted ranges of the inverses. For example, "conventional" restrictions on arctan(x) = y limit the range to (-pi/2)<y<(pi/2) but "conventional" restrictions on arccos(x) = y limit the range of this function to (0)<y<(pi). Solutions to equations such as -tan(y) = cos(y) don't fall inside these arbitrary ranges. Other equations might involve solutions in integers of the cycle counts on inverses (e.g., cos(x/10) = sin(x)/10 ). But without a mathematics of mutlivalued inverses, we are powerless to cope with such equations.

Why is a multivalued inverse not "well-defined"? Shouldn't we have types of math that can handle multivalued mappings often found with inverses such as squareroot or inverse trigonometric functions?

-------------------------
"It often happens that a player carries out a deep and complicated calculation, but fails to spot something elementary right at the first move." -- grandmaster Alexander Kotov --inscribed on gift chess sets given by Amaranth hedge fund.

Edited: Sun Aug 08, 10 at 01:27 PM by Traden4Alpha
 
Reply
   
Quote
   
Top
   
Bottom
     



dunrewpp
Senior Member

Posts: 260
Joined: Apr 2009

Sun Aug 08, 10 01:19 PM
User is offline

Quote

Originally posted by: Paul
What is

?

P


Here's my answer:

Set a(0)=A (where A can be anything in the case of converging sequences, although I have no proof for this assertion), and also set a(n+1)=tan(-1+a(n)). Now, either the sequence {a(n)} converges to a limit L, or it does not converge at all. If {a(n)} converges to a limit L, then all other sequences 'extracted' from your infinitely embedded tan thingy will converge to the same limit L. If the sequence {a(n)} converges to a limit L, then L would be a solution to x=tan(-1+x). Setting u=-1+x, we get u+1=tan(u). This equation obviously has infinitely many solutions.

Let's do a little bit of case analyses.

Case 1: (-pi/2)<(-1+A)<(pi/4).

Then a(1)<1. So, a(2)=tan(-1+a(1))<0. As a(n) gets to be more and more negative, it will go round until tan(-1+a(n)) becomes positive, and as a(n) becomes more and more positive, it will go around until tan(-1+a(n)) becomes negative again. This swinging between negative and positive values will continue for ever. This generates an interesting dynamics. A question to ponder over is this: How often does a(n) change sign?

Case 2: Any situation for A. Same story as Case 1.

So the sequence {a(n)} does not converge.

I think I have a conjecture. Here it is: Let F be a well-defined real-valued function with a suitable domain. Define a sequence {a(n,A)} as follows: a(n+1,A)=F(c+a(n,A)), a(0)=A, where c is a known fixed number. Now, if the sequence {a(n,A)} converges to a limit L for a certain value A, then for any value of A, the sequence {a(n,A)} converges to the same limit L. That is, L is independent of A.






Edited: Sun Aug 08, 10 at 01:20 PM by dunrewpp
 
Reply
   
Quote
   
Top
   
Bottom
     



frenchX
Senior Member

Posts: 5911
Joined: Mar 2010

Sun Aug 08, 10 01:24 PM
User is offline

Quote

Originally posted by: Traden4Alpha<br
Why is a multivalued inverse not "well-defined"? Shouldn't we have types of math that can handle multivalued mappings often found with inverses such as squareroot or inverse trigonometric functions?


From Wolfram

"The discontinuities of multivalued functions in the complex plane are commonly handled through the adoption of branch cuts, but use of Riemann surfaces is another possibility. " but I have to say that multivalued functions in the complex plane, branch cuts and Rieman surfaces are really pain in the ass ...



-------------------------
The important thing is not to stop questioning. Curiosity has its own reason
for existing. Never lose a holy curiosity.
 
Reply
   
Quote
   
Top
   
Bottom
     



Traden4Alpha
Senior Member

Posts: 16882
Joined: Sep 2002

Sun Aug 08, 10 01:28 PM
User is offline

Quote

Originally posted by: frenchX
Quote

Originally posted by: Traden4Alpha<br
Why is a multivalued inverse not "well-defined"? Shouldn't we have types of math that can handle multivalued mappings often found with inverses such as squareroot or inverse trigonometric functions?


From Wolfram

"The discontinuities of multivalued functions in the complex plane are commonly handled through the adoption of branch cuts, but use of Riemann surfaces is another possibility. " but I have to say that multivalued functions in the complex plane, branch cuts and Rieman surfaces are really pain in the ass ...
LOL!

No one guaranteed that math (or physics) had to be easy.



-------------------------
"It often happens that a player carries out a deep and complicated calculation, but fails to spot something elementary right at the first move." -- grandmaster Alexander Kotov --inscribed on gift chess sets given by Amaranth hedge fund.
 
Reply
   
Quote
   
Top
   
Bottom
     



AVt
Senior Member

Posts: 1051
Joined: Dec 2001

Sun Aug 08, 10 01:38 PM
User is offline View users profile

You may look at it as asking for fixed points of a fct (unique or not), playing with
branches is another variant. If contracting (no, it is not in the usual metric), then
any sequence will do.
 
Reply
   
Quote
   
Top
   
Bottom
     



dunrewpp
Senior Member

Posts: 260
Joined: Apr 2009

Sun Aug 08, 10 02:03 PM
User is offline

Quote

Originally posted by: Traden4Alpha

That's nice that you state the restriction on arctan(x), but it's a dangerous definition.


This is a gem! How do you come up with these things, Traden4Alpha? How could a decent definition be "dangerous"? A definition is a definition. All that matters is whether or not it is done appropriately.

The notion of uniqueness is indelibly interwoven with the use of symbols in language. A symbol to function effectively as a referent for something else must of necessity refer to something unique at any one point in time. Mathematics has even enshrined the notion of uniqueness by defining the idea of "function". Among normal human beings, the relation from a hand to the 'hand-owning' person is a functional relationship. As soon as you know a hand, you can be certain that there is a unique individual to whom the hand belongs. Conversely, the relation from a person to a hand belonging to the person is not a functional relationship. Knowing the person does not necessarily identify the hand uniquely. Same goes with a function like arctan. Same goes with every word I use or you use -- at any one point in time.

It is in fact more dangerous when symbols are not clearly and uniquely defined or well-defined. Here is an example. Let A(x)=B(x) denote an equation, and let R[A(x)=B(x)] denote the root (or solution) of the equation A(x)=B(x). For example, R[3x-2=x+4]=R[8-x=x+2]=3.

Having defined the idea of root of an equation, let's use it. We will now prove that the equation x(x-3)=x has a unique root.

Proof: Let A and B be two roots of x(x-3)=x. So R[x(x-3)=x]=A and R[x(x-3)=x]=B. Since A and B are equal to the same thing; i.e., R[x(x-3)=x], then they must be equal themselves, so A=B. We just proved that any two roots of x(x-3)=x must be identical. Hence x(x-3)=x has a unique root.

This is an impeccable proof!

Obviously x(x-3)=x does not have a unique root. It has two roots; they are 0 and 4. So, what's wrong with our definition (and symbol or notation) R[A(x)=B(x)]? Answer: The symbol or definition or notation assumes uniqueness without having proven so in the first place! That is why the proof, while absolutely impeccable, gave such a nonsensical result!

You like multivalued thingies? So, let me ask you this: How many different meanings does the word "dangerous" have? Is the word multivalued? If not, then your word "dangerous" is dangerous! No? If your word is multivalued, which meaning am I supposed to understand? As soon as you are done specifying the meaning of the world "dangerous" by using other words, are those words multivalued themselves? If so, will you have specified the meaning of the world "dangerous"? ETC. ETC.




Edited: Sun Aug 08, 10 at 02:08 PM by dunrewpp
 
Reply
   
Quote
   
Top
   
Bottom
     



AVt
Senior Member

Posts: 1051
Joined: Dec 2001

Sun Aug 08, 10 02:45 PM
User is offline View users profile

F(x) = f(f(x)) is contracting for f(x) = arctan(x+1) on the Reals, 0.6 seems to be a Lipschitz constant
(if I made no error) and that's enough for uniqueness (so the starting value does not matter).

@dunrewpp: occasionally there is no deeper need to argue ...
 
Reply
   
Quote
   
Top
   
Bottom
     



Paul
Senior Member

Posts: 5300
Joined: Jul 2001

Sun Aug 08, 10 05:05 PM
User is offline View users profile

Context is everything.

If you are five years old and your teacher asks "The area of a square field is four square metres, how long is a side?" You answer "two."

If you are 10 and the teacher says "What is the square root of 16?", you answer "Plus or minus four."

If you are 18 and your high school exam involves lots of square roots and arctans then unless told otherwise you consider all branches, and accept or reject depending on context. Anything less and you will lose marks.

If a test asks about the convergence of "square root of (1 plus square root of (one plus...))" and it's one question out of thirty and the test lasts one hour then you give the simplest answer.

If the test is an hour long and says "The square root of (1 plus square root of (one plus...)), discuss." Then you fill as many pages as you can with as many branches of maths as possible.

And if you are discussing any of these on an internet forum with at least some interest in mathematics then the sky is the limit! The world wide web is your oyster!

P

 
Reply
   
Quote
   
Top
   
Bottom
     

Pages: [ 1 2 3 4 >> Next ]
View thread in raw text format
FORUMS > Brainteaser Forum < refresh >

Forum Navigation:

© All material, including contents and design, copyright Wilmott Electronic Media Limited - FuseTalk 4.01 © 1999-2014 FuseTalk Inc. Terms & Conditions