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Nov 24, 2005, 2:27:44â€¯AM11/24/05

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I want to find a function f : R -> R

satisfying f'(x) = f(f(x)) and f(0) = 1.

I would appreciate any help on this matter.

satisfying f'(x) = f(f(x)) and f(0) = 1.

I would appreciate any help on this matter.

Nov 24, 2005, 3:24:05â€¯AM11/24/05

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--

Ciao,

Graziano

Nov 24, 2005, 3:29:20â€¯AM11/24/05

to

--

Ciao,

Graziano

Nov 24, 2005, 6:25:18â€¯AM11/24/05

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Chers Amis,

relation f '(x) = f( f(x) ) implies f '(f(x) ) =f(f '(x) )=f^[3](x) ,

so f and f ' commute ;

since f(0) = 1 we can' t expect a solution like

f(x) = c*x^a ( c and a constant values to define).

Sorry, no more to tell ya ,

Bon courage, Alain.

Nov 24, 2005, 1:11:05â€¯PM11/24/05

to

A solution of your functional-differential equation is [an appropriate

branch of]

f(x) = (sqrt(3)+i)/2 exp(pi sqrt(3)/6) x^((1 + i sqrt(3))/2)

for x <> 0, but of course that's not real-valued and doesn't satisfy the

initial condition.

Do you have any reason to believe your problem has a solution?

Global existence is not all that common, even for ordinary

differential equations.

An integral form of your equation is

f(x) = 1 + int_0^x f(f(t)) dt

I don't know if iteration of this with a clever starting point

might converge to a solution. But a couple of attempts didn't

give promising results.

Robert Israel isr...@math.ubc.ca

Department of Mathematics http://www.math.ubc.ca/~israel

University of British Columbia Vancouver, BC, Canada

Nov 24, 2005, 2:20:14â€¯PM11/24/05

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A continuuing problem... My Google newsreader show this as f(x) =

f(f(x)), rather than the form f '(x) = f(f(x)) as revealed by later

posts. Somehow, the "prime" gets stripped off. I am trying, with this

post, to solve the stripping problem by leaving a space between the f

and the prime. Hopefully, others will do the same, if it works.

R.G. Vickson

PS: the prime is not stripped off in the "preview post" mode, but it is

omitted in raw reader mode.

Nov 24, 2005, 2:48:29â€¯PM11/24/05

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I'm very curious as to how you obtained your particular solution. Is there some general trick to at least attempt a solution to such equations?

Nov 24, 2005, 2:53:33â€¯PM11/24/05

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In article <18088215.1132861739...@nitrogen.mathforum.org>,

Try f(x) = a x^p and see what equations a and p have to satisfy.

Nov 25, 2005, 2:08:17â€¯AM11/25/05

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"Robert Israel" <isr...@math.ubc.ca> wrote in message

news:dm4vnp$m0t$1...@nntp.itservices.ubc.ca...

> In article <dm3qck$kc9$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:

> >I want to find a function f : R -> R

> >satisfying f'(x) = f(f(x)) and f(0) = 1.

> >I would appreciate any help on this matter.

>

> A solution of your functional-differential equation is [an appropriate

> branch of]

> f(x) = (sqrt(3)+i)/2 exp(pi sqrt(3)/6) x^((1 + i sqrt(3))/2)

> for x <> 0, but of course that's not real-valued and doesn't satisfy the

> initial condition.

>

> Do you have any reason to believe your problem has a solution?

> Global existence is not all that common, even for ordinary

> differential equations.

I don't know if there exists a solution. A non-global, real-valued

solution would also be nice even without the initial condition (but

not the zero solution).

Nov 25, 2005, 3:10:35â€¯PM11/25/05

to

I find that for some insane reason the Google reply box

uses a variable pitch font, which means line endings

that appear tidy there become higgledy-piggledy when

displayed fixed-pitch in preview and published format!

To avoid this, I have to prepare replies in notepad and

then paste it into the reply box, which is quite tedious

(but also a very good idea for non-trivial-length posts,

as Google has an alarming habit of failing on posts and

when Internet Explorer is used losing the text!!)

If f'(x) = f(f(x)) can't we just differentiate again,

WRT x, and use the chain rule on the RHS, to obtain

f''(x) = df/df . df/dx = f'(x) and thus conclude that

f'(x) = c.e^x, so that f(x) = c.e^x + d and then the

original condition requires x - 1 = c.(e^x - 1) ?

For any given constant, c, this gives a single point

(for real solutions). So the condition is incompatible

with a curve, and therefore arguably, as single points

can't have a derivative, there is no solution.

Unfortunately the solution above also isn't compatible

with Robert Israel's solution, which leads me to wonder

if my assumption is somehow wrong. But a sanity check

with g(x) = x^2 and h(x) = g(g(x)) = (g(x))^2, so that

h(x) = x^4, shows that the chain rule works with this

example: dh/dx = dh/d(g(x)) . dg/dx = 2g(x).2x = 4x^3.

Nov 25, 2005, 3:11:52â€¯PM11/25/05

to

In article <dm6d8f$1tf3$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:

>

>"Robert Israel" <isr...@math.ubc.ca> wrote in message

>news:dm4vnp$m0t$1...@nntp.itservices.ubc.ca...

>> In article <dm3qck$kc9$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:

>> >I want to find a function f : R -> R

>> >satisfying f'(x) = f(f(x)) and f(0) = 1.

>> >I would appreciate any help on this matter.

>>

>> A solution of your functional-differential equation is [an appropriate

>> branch of]

>> f(x) = (sqrt(3)+i)/2 exp(pi sqrt(3)/6) x^((1 + i sqrt(3))/2)

>> for x <> 0, but of course that's not real-valued and doesn't satisfy the

>> initial condition.

>>

>> Do you have any reason to believe your problem has a solution?

>> Global existence is not all that common, even for ordinary

>> differential equations.

>

>I don't know if there exists a solution. A non-global, real-valued

>solution would also be nice even without the initial condition (but

>not the zero solution).

>

>"Robert Israel" <isr...@math.ubc.ca> wrote in message

>news:dm4vnp$m0t$1...@nntp.itservices.ubc.ca...

>> In article <dm3qck$kc9$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:

>> >I want to find a function f : R -> R

>> >satisfying f'(x) = f(f(x)) and f(0) = 1.

>> >I would appreciate any help on this matter.

>>

>> A solution of your functional-differential equation is [an appropriate

>> branch of]

>> f(x) = (sqrt(3)+i)/2 exp(pi sqrt(3)/6) x^((1 + i sqrt(3))/2)

>> for x <> 0, but of course that's not real-valued and doesn't satisfy the

>> initial condition.

>>

>> Do you have any reason to believe your problem has a solution?

>> Global existence is not all that common, even for ordinary

>> differential equations.

>

>I don't know if there exists a solution. A non-global, real-valued

>solution would also be nice even without the initial condition (but

>not the zero solution).

The integral equation

f(x) = f(a) + int_a^x f(f(t)) dt

suggests considering iteration of the operator T where

Tf(x) = a + int_a^x f(f(t)) dt

Consider this on continuous functions on the interval J = [a-r,a+r]

where r > 0.

Note that if f maps J into J, then so does Tf if

|a| + r <= 1. For convenience, I'll write M = |a| + r.

Moreover, Tf is differentiable in (a-r,a+r)

with (Tf)'(x) = f(f(x)), so in particular |(Tf)'| <= M.

Now note that

|Tf(x) - Tg(x)| <= int_a^x |f(f(t)) - g(g(t))| dt

<= int_a^x (|f(f(t)) - f(g(t))| + |f(g(t)) - g(g(t))|) dt

Suppose f and g map J into J, with

|f'| <= M and |f - g| <= D on this interval. Then

|Tf(x) - Tg(x)| <= int_a^x (M D + D) dt <= r (M + 1) D

We want T to be a contraction on a suitable set of functions,

so assume r (M + 1) < 1. Thus we consider the iteration

f_1(x) = a

f_{n+1} = T(f_n)

and we get estimates

|f_{n+1}(x) - f_n(x)| <= r (M + 1) |f_n(x) - f_{n-1}(x)|

<= ... <= (r (M+1))^(n-1) a r for x in J.

So if |a| + r < 1 and r (|a| + r + 1) < 1 the sequence f_n converges

uniformly to a function f on the interval J, and T(f) = f. Thus

f'(x) = f(f(x)) and f(a) = a. Moreover, f can be written formally

as a power series around x = a:

f(x) = a + a (x - a) + a^2 (x - a)^2/2 + (a^3 + a^4) (x - a)^3/6 + ...

and I think this should converge for |x - a| < r.

Nov 25, 2005, 3:16:04â€¯PM11/25/05

to

In article <1132949435.8...@z14g2000cwz.googlegroups.com>,

<john_r...@sagitta-ps.com> wrote:

>

>C6...@shaw.ca wrote:

>>

>> PiSigma wrote:

>> >

>> > I want to find a function f : R -> R

>> > satisfying f'(x) = f(f(x)) and f(0) = 1.

>> > I would appreciate any help on this matter.

<john_r...@sagitta-ps.com> wrote:

>

>C6...@shaw.ca wrote:

>>

>> PiSigma wrote:

>> >

>> > I want to find a function f : R -> R

>> > satisfying f'(x) = f(f(x)) and f(0) = 1.

>> > I would appreciate any help on this matter.

>If f'(x) = f(f(x)) can't we just differentiate again,

>WRT x, and use the chain rule on the RHS, to obtain

>f''(x) = df/df . df/dx = f'(x)

No. The chain rule says d/dx (f(f(x)) = f'(f(x)) f'(x).

Nov 25, 2005, 7:14:34â€¯PM11/25/05

to

My first reply to this hasn't appeared, and hopefully

it never will because differentiating by parts to give

"f'' = df/df . df/dx" and assuming "df/df" equals 1

is nonsense as this is actually d(f(f(x))/d(f(x)) !

However, I think one can do better by thinking of x

and f(x) as both being parametrized by some other

value whose inverse is t = g(x) and for which f

is a differential operator.

If we let f(x) = dx/dg + d^2(x)/dg^2 then f' = f o f

gives:

df/dx = (d/dg + d^2/dg^2)(dx/dg + d^2(x)/dg^2)

= d^2(x)/dg^2 + 2.d^3(x)/dg^3 + d^4(x)/dg^4

= dx/dg . d/dx(dx/dg + 2.d^2(x)/dg^2 + d^3(x)/dg^3)

Integrating this by parts WRT x as it stands using the

rule "first times integral of second minus integral of

(diff of first times integral of second)" and noticing

that the "diff of first" WRT x, i.e. d/dx(dx/dg), is

d/dg(dx/dx) which is zero, and therefore the second

integral is a constant we obtain a 2nd order ODE in

G := dx/dg

(f =) G + dG/dg = G.(G + 2G.dG/dg + d^2(G)/dg^2)

In principle, this can be solved as G = G(g) with

a couple of constants of integration and then f(x)

and x both expressed parametrically in terms of g.

(Expressing x = x(g) requires another integration,

and hence _another_ constant of integration - where

the hell are they all coming from ?!)

Trying the same trick with just f(x) = dx/dg doesn't

work, and nor does incorporating functions of x as

coefficients of the operator, as this prevents dx/dg

being factored in the expression for df/dx (except

possibly by imposing elaborate conditions on these

functions, which seems fairly pointless as there is

a simple alternative of assuming they are constant

which we have done).

But one could adjust the constant coefficients to

try and end up with an ODE in G which can be more

readily solved, e.g. whose terms can be collected

entirely into exact derivatives.

Cheers

John R Ramsden (jhnr...@yahoo.com.uk)

* Remove m from com to reply

* From address is defunct

Nov 26, 2005, 9:43:26â€¯PM11/26/05

to

john_r...@sagitta-ps.com wrote:

>

> [..] noticing that the "diff of first" WRT x, i.e. d/dx(dx/dg),

> is d/dg(dx/dx) which is zero,

No this isn't right either, because x and g are not independent.

In fact d/dx(dx/dg) = d/dx(1 / dg/dx) = - d^2(g)/dx^2 / (dg/dx)^2

However, one can make a lot more progress by assuming:

f(x) = (ax + b).dx/dg

and I'll post the details to tomorrow (unless someone else

gets time to nip in first ..)

Nov 27, 2005, 4:09:46â€¯AM11/27/05

to

More generally, suppose f maps the disk U = {z: |z - a| <= r} in the

complex

plane into itself, and satisfies f(b) = c where b and c are in U. This

leads to

the operator

Tf(z) = c + int_b^z f(f(t)) dt

and then Tf also maps U into itself if |c - a + a (a - b)| <= (1 - |a|

- |a-b| - r) r.

If f is the image of such a function under T, |f'(z)| <= |a| + r on U,

and then

T is a strict contraction on such functions if (r + |b-a|) | (|a| + r +

1) < 1.

If so, the sequence f_n with f_1 = c and f_{n+1} = T f_n converges

uniformly

on U to a function f that is analytic in the interior of U, satisfying

your equation

with f(b) = c. The region of analyticity could turn out to be larger.

For example, you could take a = b = 0, c = 1/4, r = 1/2. The Taylor

series of

the function f, which by the above analysis is guaranteed to converge

for

|z| < 1/2, is approximately

1/4+ .3388226478 x + .06310286283 x^2 + .01058266737 x^3 +.001876688787

x^4 +.0003615747008 x^5 + .00007425636819 x^6 + .00001596727294 x^7

+ .000003555019431 x^8 + .0000008135397704 x^9 + ...

This approximation maps the disk of radius approximately 3.5 centred at

0 into itself,

and the series appears to converge in that disk.

Nov 27, 2005, 1:47:26â€¯PM11/27/05

to

Robert Israel wrote:

> Robert Israel wrote:

> > In article <dm6d8f$1tf3$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:

> > >

> > >"Robert Israel" <isr...@math.ubc.ca> wrote in message

> > >news:dm4vnp$m0t$1...@nntp.itservices.ubc.ca...

> > >> In article <dm3qck$kc9$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:

> > >> >I want to find a function f : R -> R

> > >> >satisfying f'(x) = f(f(x)) and f(0) = 1.

> > >> >I would appreciate any help on this matter.

> > >>

> > >> A solution of your functional-differential equation is [an appropriate

> > >> branch of]

> > >> f(x) = (sqrt(3)+i)/2 exp(pi sqrt(3)/6) x^((1 + i sqrt(3))/2)

> > >> for x <> 0, but of course that's not real-valued and doesn't satisfy the

> > >> initial condition.

> > >>

> > >> Do you have any reason to believe your problem has a solution?

> > >> Global existence is not all that common, even for ordinary

> > >> differential equations.

> > >

> > >I don't know if there exists a solution. A non-global, real-valued

> > >solution would also be nice even without the initial condition (but

> > >not the zero solution).

Still more generally, suppose U is a convex open proper subset of the

complex plane

containing b and c, and let S be the set of analytic functions f: U ->

closure(U) such

that f(b) = c. Suppose T (as defined above) maps S into itself. Now S

is convex and compact in the topology of uniform convergence on compact

subsets of U, and the Tychonoff fixed point theorem says that T has a

fixed point.

This fixed point will be an analytic function on U that satisfies f'(z)

= f(f(z)) and f(b) = c.

Nov 28, 2005, 3:07:17â€¯PM11/28/05

to

Robert Israel has given some local existence results. In fact if I

haven't made an error it can be shown there is a global solution f:

R->R but it is strictly decreasing and has a fixed point in (-1,0).

On the other hand it's easy to show there is no global solution f: R->R

with f(0) > 0. Such a solution would have to have f(x) > 0 for all x >

0 - otherwise let x = inf{x' > 0 | f(x') = 0}. Note that x > 0 and f(x)

= 0 but f'(x) = f(0) > 0, which is impossible.

It quickly follows that f^(n)(x) > 0 for all x > 0 so that f(x)/x^n ->

infinity as x->infinity, for all n. In particular, we can find a > 1

with f(a) > a. A straightforward induction using the

functional-differential equation shows then that:

f^(n)(a) >= a^(n(n-1)/2+1)

so that f^(n)(a)/n! -> infinity as n->infinity. This contradicts the

remainder form of Taylor's theorem:

f(a+h) = f(a) + f'(a)h/1! + ... + f^(n)(a)h^n/n! +

f^(n+1)(a+th)t^(n+1)/(n+1)!

>= f(a) + f'(a)h/1! + ... + f^(n)(a)h^n/n!

(as f^(n+1) > 0)

because the left hand side is a fixed number and the right hand side

tends to infinity as n->infinity.

Michael

Nov 28, 2005, 5:32:20â€¯PM11/28/05

to

In article <1133208437....@g47g2000cwa.googlegroups.com>,

michaeld <mich...@cantab.net> wrote:

>PiSigma wrote:

>> I want to find a function f : R -> R

>> satisfying f'(x) = f(f(x)) and f(0) = 1.

>> I would appreciate any help on this matter.

michaeld <mich...@cantab.net> wrote:

>PiSigma wrote:

>> I want to find a function f : R -> R

>> satisfying f'(x) = f(f(x)) and f(0) = 1.

>> I would appreciate any help on this matter.

>Robert Israel has given some local existence results. In fact if I

>haven't made an error it can be shown there is a global solution f:

>R->R but it is strictly decreasing and has a fixed point in (-1,0).

I'd be interested in seeing your proof of this.

Nov 29, 2005, 11:23:17â€¯AM11/29/05

to

Robert Israel wrote:

> In article <1133208437....@g47g2000cwa.googlegroups.com>,

> michaeld <mich...@cantab.net> wrote:

> >PiSigma wrote:

> >> I want to find a function f : R -> R

> >> satisfying f'(x) = f(f(x)) and f(0) = 1.

> >> I would appreciate any help on this matter.

>

> >Robert Israel has given some local existence results. In fact if I

> >haven't made an error it can be shown there is a global solution f:

> >R->R but it is strictly decreasing and has a fixed point in (-1,0).

>

> I'd be interested in seeing your proof of this.

> In article <1133208437....@g47g2000cwa.googlegroups.com>,

> michaeld <mich...@cantab.net> wrote:

> >PiSigma wrote:

> >> I want to find a function f : R -> R

> >> satisfying f'(x) = f(f(x)) and f(0) = 1.

> >> I would appreciate any help on this matter.

>

> >Robert Israel has given some local existence results. In fact if I

> >haven't made an error it can be shown there is a global solution f:

> >R->R but it is strictly decreasing and has a fixed point in (-1,0).

>

> I'd be interested in seeing your proof of this.

OK, well in brief outline:

The formal substitution y = f(x) gives dy/dx = f(y), which suggests the

equation:

integral[-1/2 ... f(x)] dy/f(y) = x + 1/2

Now start off with a local (strictly decreasing) solution f: (-1,0) ->

(-1,0) with a fixed point at -1/2 say - this can be constructed as you

outlined using fixed point theorems. The integrand on the left hand

side of the above makes sense for -1/2 < y < 0 (since f(y) has already

been defined) so this can be used to define the f(x) (in the integral's

limit) upto a point a < -1 such that f(a) = 0. So we now have a

solution f: [a,0) -> (-1,0] with f(a) = 0.

Doing something similar but integrating from -1/2 down to a gives us an

extension to f: [a,infinity) -> (a,0) with f(x) -> a as x->infinity.

Finally integrating from -1/2 to infinity extends f to R. I think then

a simple check shows that f'(x) = f(f(x)). I haven't really thought

about whether this is likely to be analytic - though very naively you'd

probably expect so, because f is extended by inverting integrals of

functions that are (locally) analytic.

Maybe I'll be able to give a better explanation later.

Michael

Nov 29, 2005, 9:12:08â€¯PM11/29/05

to

michaeld wrote:

[...]

OK, I'll try and write this out a bit better.

Start with the analytic solution f: [-1,0] -> (-1,0) to f' = f o f with

f(-1/2) = -1/2. This is the solution given by the Taylor series you

wrote down when the fixed point is set to -1/2. You can show using

fixed point theorems that the radius of convergence is > 1/2 so that

the power series converges on [-1,0].

This function is strictly decreasing. Let a = f(-1) so that -1 < a < 0.

Define now:

F(t) = -1 + integral[a ... t] dy/f(y)

This makes sense say for a <= t <= 0 because f is defined and

continuous on [a,0].

Then, defined on [a,0], we have that F is continuous, strictly

decreasing and F(a) = -1. Let b = F(0). Then F: [a,0] -> [b,-1] is a

bijection; in fact it is differentiable with negative derivative, F'(t)

= 1/f(t) (where left/right differentiability is understood at the

endpoints).

Therefore F has an inverse, which is a bijection [b,-1] -> [a,0] that

is differentiable with negative derivative (again understanding

left/right differentiability at the endpoints). Define this inverse to

be f, so that F(f(x)) = x for x in [b,-1]. This is OK because before

now we only defined f on [-1,0] so we haven't defined anything twice

(except for at -1 but both definitions give us the value a).

We now have f: [b,0] -> [-1,0]. To check it is differentiable at x =

-1... well it's clearly continuous here. The right derivative exists

and equals f(f(-1)) = f(a) (because of the original definition of f on

[-1,0].) The left derivative exists and equals 1/F'(a) which is also

f(a). Hence we are differentiable at x = -1. We're clearly

differenentiable everywhere else on [b,0] (again with left/right

derivatives being understood as appropriate).

Therefore f: [b,0] -> [-1,0] is differentiable, with negative

derivative, and with f(b) = 0. And we have that:

x = -1 + integral[a ... f(x)] dy/f(y)

for x in [b,-1].

Since f is differentiable we can differentiate the above equation and

use FTC to deduce that 1 = f'(x)/f(f(x)) so that f'(x) = f(f(x)) for x

in [b,-1]. Therefore this holds on [b,0].

So we have a strictly decreasing f: [b,0] -> [-1,0] satisfying f'(x) =

f(f(x)) and f(b) = 0.

Next look at:

G(t) = b - integral[t ... 0] dy/f(y)

for b < t <= 0. This defines a differentiable function with G(0) = b

and G(t) -> infinity as t->b (since f(t) tends to 0 linearly as t -> b

so its integral diverges). By inverting this we can extend f to

[b,infinity] and this gives us a differentiable f: [b,infinity] ->

[b,0] satisfying f' = f o f.

Finally the extension to R is obtained by looking at:

H(t) = b + integral[0 ... t] dy/f(y)

for 0 <= t < infinity. H maps [0,infinity) to (-infinity,b]

surjectively (as f(t)->b as t->infinity) so we invert this to get a

function from (-infinity,b] to [0,infinity) and we can check again that

f' = f o f everywhere.

So we get a function f: R->R that is differentiable with f' = f o f,

f(-1/2) = -1/2, f(b) = 0, f(x) -> b as x->infinity and f(x) ~ -bx as

x->-infinity for some b < -1.

Does that look right now?

Btw, I changed my mind about the function being analytic. It's probably

locally analytic or something but I personally very much doubt it

extends to an analytic function f: C->C.

Michael

Nov 30, 2005, 2:13:52â€¯AM11/30/05

to

In article <1133316728.3...@g44g2000cwa.googlegroups.com>,

michaeld <mich...@cantab.net> wrote:

michaeld <mich...@cantab.net> wrote:

>Btw, I changed my mind about the function being analytic. It's probably

>locally analytic or something but I personally very much doubt it

>extends to an analytic function f: C->C.

So f should be analytic on a neighbourhood of R, but probably not entire.

Indeed, from calculating the first 20 Taylor coefficients of the

series of f around -1/2, it looks to me like the radius of convergence

should be somewhere near 4.

Nov 30, 2005, 3:33:06â€¯PM11/30/05

to

> Find a function satisfying f '(x) = f(f(x))

If y =f (x) , y' = f (y) .So if f is given we can solve for y = f (x).

E.g., if f (u) = u, y = c e^x and if f(u) = sqrt(1+u^2), then y =

sinh(x + c) etc.

But if the functional dependence f is not explicitly given, your

problem is indeterminate or underspecified...However, I like your

question, nee a trap.

Dec 1, 2005, 10:04:52â€¯AM12/1/05

to

"Narasimham" <math...@hotmail.com> wrote in message

news:1133382786.2...@g47g2000cwa.googlegroups.com...

> > Find a function satisfying f '(x) = f(f(x))

>

> If y =f (x) , y' = f (y) .So if f is given we can solve for y = f (x).

> E.g., if f (u) = u, y = c e^x and if f(u) = sqrt(1+u^2), then y =

> sinh(x + c) etc.

If f is given there is no unknown function...

>

> But if the functional dependence f is not explicitly given, your

> problem is indeterminate or underspecified...However, I like your

Robert Israel and michaeld have given local solutions, so I don't

see in what sense the problem is indeterminate or underspecified.

> question, nee a trap.

>

??

Dec 1, 2005, 12:25:01â€¯PM12/1/05

to

The solution f: R->R I gave is global on R.... it's just that

(according to various heuristic arguments and Robert's numerical

evidence) it does not extend to the complex plane C.

It also violates the initial condition f(0) = 1 you gave, but as I

noted earlier there is no global f: R->R satisfying f' = f o f and f(0)

> 0.

As for questions of uniqueness.... I'm not completely sure whether the

solution f: R->R is the unique solution with f(-1/2) = -1/2. I _think_

I can show that it is, but I have to check a few details. On the other

hand it's very easy to see that the local solutions with positive fixed

points are (locally) unique. For example, there is a unique solution

f:(0,1) -> (0,1) with f(1/2) = 1/2.

Michael

Dec 2, 2005, 10:52:29â€¯PM12/2/05

to

Shall be reverting to it later, doubt if it is adequately well-posed.

For time being took liberty to post your problem into Mathematica

group, hope it is in order...

Among those discussed, Andrzej Kozlowski 's numerical solution for a

"chained" function EQ = { f ' [x] == f [ Cos[x] ], f [0] == 1} works

out alright.

Regards

Narasimham

Dec 7, 2005, 4:59:14â€¯PM12/7/05

to

This is from a thread that has probably evaporated from most news spools

by now. The question was about solutions to the equation

f' ( x ) = f ( f ( x ) )

where, during the thread, the initial condition f ( -1/2 ) = -1/2 was

added. (Equivalently, f(x) = h(x+1/2) - 1/2 where h now satisfies

h'= h o h - 1/2 and h(0)=0.)

by now. The question was about solutions to the equation

f' ( x ) = f ( f ( x ) )

where, during the thread, the initial condition f ( -1/2 ) = -1/2 was

added. (Equivalently, f(x) = h(x+1/2) - 1/2 where h now satisfies

h'= h o h - 1/2 and h(0)=0.)

In article <dmjjfg$gkj$1...@nntp.itservices.ubc.ca>,

Robert Israel <isr...@math.ubc.ca> wrote:

>So f should be analytic on a neighbourhood of R, but probably not entire.

>Indeed, from calculating the first 20 Taylor coefficients of the

>series of f around -1/2, it looks to me like the radius of convergence

>should be somewhere near 4.

I extended the calculation to the first 50+ coefficients. (The

coefficient of x^n is an integer divided by n! 2^(-1 + n(n-1) ).

The numerators grow to several hundred digits by n=50.)

As Robert suggests, the numerators are of comparable magnitude

after multiplication by 4^n -- but not quite! In the range he

checked, all but the first few of these products are smaller (in

magnitude) than 0.5; but the 23rd is around 0.57; the 31st is 1.06; the

41st is nearly 2.0, the 44th is nearly 4.0, and the 53rd is nearly 8.0.

So I don't have quite so much confidence that the series

converges on a disc that large!

I also checked for divergence of the series on the circle of radius 4.

It seems that the values of the partial sums are growing fastest

when x is near 4 exp(+-2i) , which is interesting (but it's hard

to say much from limited data).

So I am led to ask how one might obtain better information about

the function. Here are a couple of things I tried, which prompt

more questions.

1. What's a good way to solve such an equation numerically?

Maple balked at the request for a numeric solution. I tried to

think of some clever way to expand the function in terms of, say,

a basis of orthogonal polynomials, but there didn't seem to be

a good way to use the defining equation to rewrite the integrals

in any useful way. Some good numerical approximations might give

support to the conjecture that the function is analytic along

the circle, except at those two points. To discover that the

critical points are (numerically) exactly at exp(2i), or exp(2/3 pi)

or whatever might give some clues to the behaviour of the function.

2. Since the defining equation requires computing composites, I

thought of extending the problem in the following way. Suppose

we look for a commuting family of operators H_t on (a neighborhood

of the origin in) the complex plane. In the present case I mean

by that a set of functions H_t(x) such that

H_s( H_t(x) ) = H_t( H_s(x) )

The thinking is that both sides should represent H_{st} (x),

so that in particular H_1(x) will be the identity map. We can

compute the power series necessary to guarantee as well that

H_{-1/2} = h; more generally H_t(x) is the unique function in

the family for which (H_t)'(x) = t. Here are the first few terms

of the expansion:

H_t(x) = t*x + 1/6*t*(t-1)*x^2 + 1/36*(t-1)*t*(3*t+1)*x^3

+ 1/5184*(155*t^2-61*t-85)*t*(t-1)*x^4 + ...

The point of introducing the H's is that it is no longer necessary

to refer to composites; the composite of two H's is another H.

Thus the ODE which mentions composites is equivalent to

dH/dx (x,-1/2) = H (x, 1/4) - 1/2

So if, say, someone knows how to solve delay-differential equations,

e.g. dH/dx(x,t) = H(x,t+3/4) - 1/2, then that person could solve

the original equation (using whatever interpretation of "solve" one

prefers). Not being a differential equations guy myself, I didn't

know how to go any further. (I was hoping to see some nice patterns

involving dH/dx and dH/dt after finding the power series above,

but nothing obvous leaps out.)

One of the reasons I thought to pursue this a bit more was that

in Oct 2000 we had a thread about a differential equation

(x^2 y')' = -x^2 y^2,

which, to my surprise, had an _essential_ singularity some distance away

from the point where we were computing Taylor series. I wonder whether

this equations has the same feature? The old thread had message-ID

<8t3fsc$ms1$1...@news1.math.niu.edu>.

dave

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