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darrellhaddon@yahoo.com
science forum beginner
Joined: 26 May 2006
Posts: 4
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 Posted: Fri May 26, 2006 1:55 pm Post subject:
Functional equation with delta
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Is it possible to exist continuous R->R function f such that:
f = delta(f o f) (*)
and f(x) = 0 for every real x is false.
If so is f unique?
Notation:
(delta(f))(x) = f(x+1) - f(x)
(f o g)(x) = f(g(x))
Therefore (*) is equivalent to: f(x) = f(f(x+1)) - f(f(x)).
Thank you,
Darrell H |
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Robert B. Israel
science forum Guru
Joined: 24 Mar 2005
Posts: 2151
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| Posted: Fri May 26, 2006 6:34 pm Post subject:
Re: Functional equation with delta
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In article <1148651713.199888.239440@j55g2000cwa.googlegroups.com>,
<darrellhaddon@yahoo.com> wrote:
| Quote: | Is it possible to exist continuous R->R function f such that:
f = delta(f o f) (*)
and f(x) = 0 for every real x is false.
If so is f unique?
Notation:
(delta(f))(x) = f(x+1) - f(x)
(f o g)(x) = f(g(x))
Therefore (*) is equivalent to: f(x) = f(f(x+1)) - f(f(x)).
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One solution is f(x) = -arcsin(sin(pi x))/(2 pi)
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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Patrick Coilland
science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197
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| Posted: Fri May 26, 2006 7:16 pm Post subject:
Re: Functional equation with delta
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Robert Israel nous a récemment amicalement signifié :
| Quote: | In article <1148651713.199888.239440@j55g2000cwa.googlegroups.com>,
darrellhaddon@yahoo.com> wrote:
Is it possible to exist continuous R->R function f such that:
f = delta(f o f) (*)
and f(x) = 0 for every real x is false.
If so is f unique?
Notation:
(delta(f))(x) = f(x+1) - f(x)
(f o g)(x) = f(g(x))
Therefore (*) is equivalent to: f(x) = f(f(x+1)) - f(f(x)).
One solution is f(x) = -arcsin(sin(pi x))/(2 pi)
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I tried things like this and failed :-(
But you're right : it works perfectly !
Bravo !
--
Patrick |
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Patrick Coilland
science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197
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| Posted: Fri May 26, 2006 7:55 pm Post subject:
Re: Functional equation with delta
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darrellhaddon@yahoo.com nous a récemment amicalement signifié :
| Quote: | Is it possible to exist continuous R->R function f such that:
f = delta(f o f) (*)
and f(x) = 0 for every real x is false.
If so is f unique?
Notation:
(delta(f))(x) = f(x+1) - f(x)
(f o g)(x) = f(g(x))
Therefore (*) is equivalent to: f(x) = f(f(x+1)) - f(f(x)).
|
Robert gave a solution.
Here is (based on Robert's idea) a whole family of solutions (to show
that f is not unique) :
Let c be a real in ]0,1/2]
Let u(x) be any function defined on [c,1/2] such that :
u(c) = -c/2
|u(x)| <= c/2 for any x in [c,1/2]
f(x) :
for x in [0,c] : f(x) = -x/2
for x in [c,1/2] : f(x) = u(x)
for x in [-1/2,0] : f(x) = -f(-x)
for x >= 1/2 : f(x) = -f(x-1)
for x <= -1/2 : f(x) = -f(x+1)
Then f(x) is a solution to the problem :
1) f(x) is in [-c/2, c/2]
2) for any x in [-c/2, c/2], f(x) = -x/2
3) f(x+1) = -f(x) for any x in R
4) f(f(x+1)) = -f(x+1)/2 (implied by 1 and 2)
f(f(x)) = -f(x)/2 (implied by 1 and 2)
f(f(x+1)) - f(f(x)) = (f(x)-f(x+1))/2
= (f(x)-(-f(x))/2
= f(x)
The solution of Robert is :
c = 1/2
--
Patrick |
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darrellhaddon@yahoo.com
science forum beginner
Joined: 26 May 2006
Posts: 4
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| Posted: Fri May 26, 2006 9:28 pm Post subject:
Re: Functional equation with delta
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Patrick Coilland wrote:
| Quote: | darrellhaddon@yahoo.com nous a récemment amicalement signifié :
Is it possible to exist continuous R->R function f such that:
f = delta(f o f) (*)
and f(x) = 0 for every real x is false.
If so is f unique?
Notation:
(delta(f))(x) = f(x+1) - f(x)
(f o g)(x) = f(g(x))
Therefore (*) is equivalent to: f(x) = f(f(x+1)) - f(f(x)).
Robert gave a solution.
Here is (based on Robert's idea) a whole family of solutions (to show
that f is not unique) :
Let c be a real in ]0,1/2]
Let u(x) be any function defined on [c,1/2] such that :
u(c) = -c/2
|u(x)| <= c/2 for any x in [c,1/2]
f(x) :
for x in [0,c] : f(x) = -x/2
for x in [c,1/2] : f(x) = u(x)
for x in [-1/2,0] : f(x) = -f(-x)
for x >= 1/2 : f(x) = -f(x-1)
for x <= -1/2 : f(x) = -f(x+1)
Then f(x) is a solution to the problem :
1) f(x) is in [-c/2, c/2]
2) for any x in [-c/2, c/2], f(x) = -x/2
3) f(x+1) = -f(x) for any x in R
4) f(f(x+1)) = -f(x+1)/2 (implied by 1 and 2)
f(f(x)) = -f(x)/2 (implied by 1 and 2)
f(f(x+1)) - f(f(x)) = (f(x)-f(x+1))/2
= (f(x)-(-f(x))/2
= f(x)
The solution of Robert is :
c = 1/2
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Thanks for the solutions.
Is it possible to exist non-periodic solution or solution with
continuous derivative?
Thank you,
Darrell H |
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darrellhaddon@yahoo.com
science forum beginner
Joined: 26 May 2006
Posts: 4
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| Posted: Fri May 26, 2006 10:40 pm Post subject:
Re: Functional equation with delta
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darrellhaddon@yahoo.com wrote:
| Quote: | Patrick Coilland wrote:
darrellhaddon@yahoo.com nous a récemment amicalement signifié :
Is it possible to exist continuous R->R function f such that:
f = delta(f o f) (*)
and f(x) = 0 for every real x is false.
If so is f unique?
Notation:
(delta(f))(x) = f(x+1) - f(x)
(f o g)(x) = f(g(x))
Therefore (*) is equivalent to: f(x) = f(f(x+1)) - f(f(x)).
Robert gave a solution.
Here is (based on Robert's idea) a whole family of solutions (to show
that f is not unique) :
Let c be a real in ]0,1/2]
Let u(x) be any function defined on [c,1/2] such that :
u(c) = -c/2
|u(x)| <= c/2 for any x in [c,1/2]
f(x) :
for x in [0,c] : f(x) = -x/2
for x in [c,1/2] : f(x) = u(x)
for x in [-1/2,0] : f(x) = -f(-x)
for x >= 1/2 : f(x) = -f(x-1)
for x <= -1/2 : f(x) = -f(x+1)
Then f(x) is a solution to the problem :
1) f(x) is in [-c/2, c/2]
2) for any x in [-c/2, c/2], f(x) = -x/2
3) f(x+1) = -f(x) for any x in R
4) f(f(x+1)) = -f(x+1)/2 (implied by 1 and 2)
f(f(x)) = -f(x)/2 (implied by 1 and 2)
f(f(x+1)) - f(f(x)) = (f(x)-f(x+1))/2
= (f(x)-(-f(x))/2
= f(x)
The solution of Robert is :
c = 1/2
Thanks for the solutions.
Is it possible to exist non-periodic solution or solution with
continuous derivative?
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Maybe this can be useful:
g = f o f =>
f(x) = g(x+1) - g(x) <==> g(x+1) = g(x) + f(x) = (g+f)(x)
In words: g(x+1) is a sum of g(x) and (a iterative root of g)(x).
Darrell H |
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alain verghote
science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 293
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| Posted: Sat May 27, 2006 9:16 am Post subject:
Re: Functional equation with delta
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Dear Friends ,
I do not understand the way the solution f1(x) = -arcsin(sin(pi
x))/(2 pi)
was built : lines a*x +b are not solutions and one value of f1 is
- x/2 .
I've thought about the relation given by darrellhad :g(x+1) = g(x) +
f(x)
but it leads to infinite messy series ,
Alain |
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Patrick Coilland
science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197
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| Posted: Sat May 27, 2006 9:33 am Post subject:
Re: Functional equation with delta
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alainverghote@yahoo.fr nous a récemment amicalement signifié :
| Quote: | Dear Friends ,
I do not understand the way the solution f1(x) = -arcsin(sin(pi
x))/(2 pi)
was built : lines a*x +b are not solutions and one value of f1 is
- x/2 .
"one value of f1 is - x/2" is wrong : |
f1 = -x/2 only for x in [-1/2, 1/2]
We have :
P1: f1(x) = -x/2 for x in [-1/2, 1/2]
P2: f1(x+1) = -f1(x) for any x in R
For any x, f1(x) is in [-1/4, 1/4]
==> P3: f1(f1(x)) = -f1(x)/2 (using P1)
So :
f1(f1(x+1)) = -f1(x+1)/2 (using P3)
= f1(x)/2 (using P2)
f1(f1(x)) = -f1(x)/2 (using P3)
==>
f1(f1(x+1)) - f1(f1(x)) = f1(x)
Q.E.D.
--
Patrick |
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alain verghote
science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 293
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| Posted: Sat May 27, 2006 9:57 am Post subject:
Re: Functional equation with delta
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OK , for your explanation
I 've also seen yours based on Robert's ,
Does it still work with
f1(f1(x+a)) +b*f1(f1(x)) = f1(x) ; a, b any constant
Alain |
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Patrick Coilland
science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197
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| Posted: Sat May 27, 2006 10:24 am Post subject:
Re: Functional equation with delta
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alainverghote@yahoo.fr nous a récemment amicalement signifié :
| Quote: | OK , for your explanation
I 've also seen yours based on Robert's ,
Does it still work with
f1(f1(x+a)) +b*f1(f1(x)) = f1(x) ; a, b any constant
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If b different from -1 :
f1(x) = x/(b+1) - a/(b+1)^3 is a solution
--
Patrick |
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Patrick Coilland
science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197
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| Posted: Sat May 27, 2006 3:12 pm Post subject:
Re: Functional equation with delta
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darrellhaddon@yahoo.com nous a récemment amicalement signifié :
| Quote: |
f(f(x+1)) - f(f(x)) = f(x)
Is it possible to exist non-periodic solution or solution with
continuous derivative?
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It is possible to find solution with continuous derivative.
Let us modify a little bit my previous solution :
Let c be a real in ]0,1/2[
Let u(x) be any function defined on [c,1/2] such that :
u(c) = -c/2
|u(x)| <= c/2 for any x in [c,1/2]
u'(x) exists and :
u'(c) = -1/2
u'(1/2) = 0
f(x) periodic with period 2 and defined on [-1/2, 3/2] :
for x in [0,c] : f(x) = -x/2
for x in [c,1/2] : f(x) = u(x)
for x in [-1/2,0] : f(x) = -f(-x)
for x in [1/2,3/2] : f(x) = -f(x-1)
Then f(x) is a solution to the problem, continuous and with continuous
derivative.
With this method, it is possible to build C0, C1, C2, .... solutions.
But not analytic one. In fact, it does not exist any analytic solution
periodic with period 2.
I think it exists also non periodic solutions.
--
Patrick |
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Patrick Coilland
science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197
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| Posted: Sat May 27, 2006 3:18 pm Post subject:
Re: Functional equation with delta
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Patrick Coilland nous a récemment amicalement signifié :
| Quote: | |u(x)| <= c/2 for any x in [c,1/2]
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Correction :
|u(x)| <= c for any x in [c,1/2]
else it would be impossible to have u(c) = -c/2 and u'(c) = -1/2
--
Patrick |
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alain verghote
science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 293
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| Posted: Sat May 27, 2006 3:31 pm Post subject:
Re: Functional equation with delta
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Dear friends ,
well ,
f(x) = -arcsin(sin(pi x))/(2 pi) , how do you arrive at
this proved right solution which steps , analogy ...
Of course it is very particular ;since with b = -1
f1(f1(x+c)) -f1(f1(x)) = f1(x) ; c any constant
we do not get lines solutions a*x+b
Alain |
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Patrick Coilland
science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197
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| Posted: Sat May 27, 2006 3:43 pm Post subject:
Re: Functional equation with delta
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alainverghote@yahoo.fr nous a récemment amicalement signifié :
| Quote: | Dear friends ,
well ,
f(x) = -arcsin(sin(pi x))/(2 pi) , how do you arrive at
this proved right solution which steps , analogy ...
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I don't know how Robert did.
But :
f(f(x+1)) - f(f(x)) = f(x) lead to look for periodic functions.
Then, it is easy to show that f(x) = -x/2 on at least a non zero
interval around 0.
And hence the simpliest solution given by Robert.
The good question now is darrel's one : is there any non periodic
solution ?
| Quote: |
Of course it is very particular ;since with b = -1
f1(f1(x+c)) -f1(f1(x)) = f1(x) ; c any constant
we do not get lines solutions a*x+b
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The simpliest solution here is
f(x) = -c*arcsin(sin(pi x/c))/(2 pi)
--
Patrick |
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Patrick Coilland
science forum Guru Wannabe
Joined: 29 Jan 2006
Posts: 197
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| Posted: Sun May 28, 2006 4:23 pm Post subject:
Re: Functional equation with delta
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darrellhaddon@yahoo.com nous a récemment amicalement signifié :
| Quote: | Is it possible to exist continuous R->R function f such that:
f = delta(f o f) (*)
and f(x) = 0 for every real x is false.
If so is f unique?
Notation:
(delta(f))(x) = f(x+1) - f(x)
(f o g)(x) = f(g(x))
Therefore (*) is equivalent to: f(x) = f(f(x+1)) - f(f(x)).
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Prop P : f(x) = f(f(x+1)) - f(f(x)).
I take interest in the case f non periodic and I suggest a method for
building a family of solutions.
The construction given below shows that it exist an infinity of non
periodic solutions. All the solutions built in this way are null
on ]-inf, 2], and stricty increasing for x >=2. They are continuous, non
periodic, non analytic. I think it is easy to have C1 solutions
(continuous derivative).
1) Construction of f(x) for x <= 5
==================================
Let u(x) be an increasing continuous bijection from [2,3] in [0,2] with
u(2) = 0 and u(3) = 2. Then v(x) = x + u(x) is an increasing bijection
from [2,3] in [2,5] with v(2) = 2 and v(3) = 5. v^[-1](x) exists for x
in [2,5]
Let a_n be the following sequence :
a_0 = 3, a_1 = 4, n > 1 : a_n = 2 + v^[-1](a_(n-2))
a_n is an increasing sequence whose limit is 5.
Let g_n(x), defined on [a_n, a_(n+1)], be the following sequence of
functions :
g_0(x) = x - 1
n > 0 : g_n(x) = g_(n-1)^[-1]( v(x - 2) )
g_n(x) is increasing on [a_n, a_(n+1)] and
g_n(a_n) = a_(n-1) for every n > 0
g_n(a_(n+1)) = a_n for every n >= 0
*** definition of f(x) on ]-inf, 5] ***
For x <= 2 : f(x) = 0
For x in [2,3] : f(x) = u(x)
For x in [a_n, a_(n+1)] : f(x) = g_n(x) for every n >= 0
For x = 5 : f(5) = 5
2) Construction of f(x) for x in ]5,6]
=======================================
We can define g(x) = f(f(x-1)) + f(x-1). This is completely well defined
on ]5,6].
We shall now consider that g(x) is f(f(x)) and build f(x) (there exist
an infinity of such f) with the classical method :
Choose any real "c" in ]6, 10[
Let a_n be the following sequence :
b_0 = c, b_1 = 6, n > 1 : b_n = g^[-1](b_(n-2))
b_n is a decreasing sequence whose limit is 5.
Let w(x) be any increasing continuous bijection from [b_2,b_1] in
[b_1,b_0] with w(b_2) = b_1 and w(b_1) = b_0.
Let h_n(x), defined on [b_(n+2), b_(n+1)], be the following sequence of
functions :
h_0(x) = w(x)
h_(n+1)(x) = h_n^[-1](g(x))
h_n(x) is increasing on [b_(n+2), b_(n+1)] and :
h_n(b_(n+2)) = b_(n+1)
h_n(b_(n+1)) = b_n
*** definition of f(x) in ]5,6] ***
For x in [b_(n+2), b_(n+1)], f(x) = h_n(x) for every n >=0
3) Construction of f(x) for x in ]n,n+1] for any n >=6
======================================================
We can prolongate g(x) = g(x-1) + f(x-1)
And then we can easyly prolongate f(x) (as the unique function such that
:
f(x) is such that already defined on ]-inf, n]
f(f(x)) = g(x) on ]n, n+1]
This is the classical method for construction of "square roots" of
increasing continuous fonctions g(x) with g(x)>x
I don't give any details but it is easy to give if anyone is interested.
4) The end
==========
I gave no details on the fact the solution :
- is continuous
- respect f(x) = f(f(x+1)) - f(f(x)) for any x in R
But, once again, I can give these details if anyone is interested.
--
Patrick |
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