Mixing
What function satisfies $F'(x) = F(2x)$?
$begingroup$
The exponential generating function counting the number of graphs on $n$ labeled vertices satisfies (and is defined by) the equations
$$
F'(x) = F(2x) ; ; ; ; ; F(0) = 1
$$
Is there some closed form or other nice description of this function?
Does it have a name?
Of course, the series itself does not converge for any nonzero $x$, but like the Lambert W function (counting trees) it has combinatorial meaning. And the Lambert W function has a nice description as the inverse of $x e^x$; maybe there is a similar description of $F$?
combinatorics reference-request generating-functions delay-differential-equations
edited Jul 19 '15 at 4:53
Michael Galuza
3,96221536
asked Jul 18 '15 at 7:04
60056005
36.2k751125
$endgroup$
|
show 7 more comments
$begingroup$
The exponential generating function counting the number of graphs on $n$ labeled vertices satisfies (and is defined by) the equations
$$
F'(x) = F(2x) ; ; ; ; ; F(0) = 1
$$
Is there some closed form or other nice description of this function?
Does it have a name?
Of course, the series itself does not converge for any nonzero $x$, but like the Lambert W function (counting trees) it has combinatorial meaning. And the Lambert W function has a nice description as the inverse of $x e^x$; maybe there is a similar description of $F$?
combinatorics reference-request generating-functions delay-differential-equations
edited Jul 19 '15 at 4:53
Michael Galuza
3,96221536
asked Jul 18 '15 at 7:04
60056005
36.2k751125
$endgroup$
2
$begingroup$
Well, if the function is analytic at $0$, then its Maclaurin series is $$sum frac{2^{n(n-1)/2}}{n!},$$from repeated differentiation at $x = 0$, but I'm guessing you knew this.
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:28
1
$begingroup$
Yes I did. I've been making little errors like this all week :-(
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:44
3
$begingroup$
the power series's radius of convergence is zero though.
$endgroup$
– mercio
Jul 18 '15 at 8:20
3
$begingroup$
Hmm.. if we set $H(t)=F(frac{1}{log 2}2^{-t})$ then $H$ satisfies the delay differential equation $H'(t)=-2^{-t}H(t-1)$, which looks like it has many different solutions with $lim_{ttoinfty}H(t)=1$
$endgroup$
– Henning Makholm
Jul 18 '15 at 10:37
1
$begingroup$
@6005: I don't. I just have the heuristic argument that you can choose $H$ freely on, say, the interval $[1,2]$ making sure the $H'(2)=-frac12H(1)$, and then integrate the values of $H(t)$ up to $t=infty$ numerically. Because of the $2^{-t}$ factor it should end up with a horizontal asymptote, which will be nonzero unless you're pretty unlucky. Then scale everything such that the limit is $1$ instead (the equation is linear after all).
$endgroup$
– Henning Makholm
Jul 18 '15 at 20:02
|
show 7 more comments
$begingroup$
The exponential generating function counting the number of graphs on $n$ labeled vertices satisfies (and is defined by) the equations
$$
F'(x) = F(2x) ; ; ; ; ; F(0) = 1
$$
Is there some closed form or other nice description of this function?
Does it have a name?
Of course, the series itself does not converge for any nonzero $x$, but like the Lambert W function (counting trees) it has combinatorial meaning. And the Lambert W function has a nice description as the inverse of $x e^x$; maybe there is a similar description of $F$?
combinatorics reference-request generating-functions delay-differential-equations
edited Jul 19 '15 at 4:53
Michael Galuza
3,96221536
asked Jul 18 '15 at 7:04
60056005
36.2k751125
$endgroup$
The exponential generating function counting the number of graphs on $n$ labeled vertices satisfies (and is defined by) the equations
$$
F'(x) = F(2x) ; ; ; ; ; F(0) = 1
$$
Is there some closed form or other nice description of this function?
Does it have a name?
Of course, the series itself does not converge for any nonzero $x$, but like the Lambert W function (counting trees) it has combinatorial meaning. And the Lambert W function has a nice description as the inverse of $x e^x$; maybe there is a similar description of $F$?
combinatorics reference-request generating-functions delay-differential-equations
combinatorics reference-request generating-functions delay-differential-equations
edited Jul 19 '15 at 4:53
Michael Galuza
3,96221536
asked Jul 18 '15 at 7:04
60056005
36.2k751125
edited Jul 19 '15 at 4:53
Michael Galuza
3,96221536
asked Jul 18 '15 at 7:04
60056005
36.2k751125
edited Jul 19 '15 at 4:53
Michael Galuza
3,96221536
edited Jul 19 '15 at 4:53
Michael Galuza
3,96221536
edited Jul 19 '15 at 4:53
Michael Galuza
3,96221536
3,96221536
asked Jul 18 '15 at 7:04
60056005
36.2k751125
asked Jul 18 '15 at 7:04
60056005
36.2k751125
asked Jul 18 '15 at 7:04
60056005
36.2k751125
36.2k751125
2
$begingroup$
Well, if the function is analytic at $0$, then its Maclaurin series is $$sum frac{2^{n(n-1)/2}}{n!},$$from repeated differentiation at $x = 0$, but I'm guessing you knew this.
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:28
1
$begingroup$
Yes I did. I've been making little errors like this all week :-(
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:44
3
$begingroup$
the power series's radius of convergence is zero though.
$endgroup$
– mercio
Jul 18 '15 at 8:20
3
$begingroup$
Hmm.. if we set $H(t)=F(frac{1}{log 2}2^{-t})$ then $H$ satisfies the delay differential equation $H'(t)=-2^{-t}H(t-1)$, which looks like it has many different solutions with $lim_{ttoinfty}H(t)=1$
$endgroup$
– Henning Makholm
Jul 18 '15 at 10:37
1
$begingroup$
@6005: I don't. I just have the heuristic argument that you can choose $H$ freely on, say, the interval $[1,2]$ making sure the $H'(2)=-frac12H(1)$, and then integrate the values of $H(t)$ up to $t=infty$ numerically. Because of the $2^{-t}$ factor it should end up with a horizontal asymptote, which will be nonzero unless you're pretty unlucky. Then scale everything such that the limit is $1$ instead (the equation is linear after all).
$endgroup$
– Henning Makholm
Jul 18 '15 at 20:02
|
show 7 more comments
2
$begingroup$
Well, if the function is analytic at $0$, then its Maclaurin series is $$sum frac{2^{n(n-1)/2}}{n!},$$from repeated differentiation at $x = 0$, but I'm guessing you knew this.
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:28
1
$begingroup$
Yes I did. I've been making little errors like this all week :-(
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:44
3
$begingroup$
the power series's radius of convergence is zero though.
$endgroup$
– mercio
Jul 18 '15 at 8:20
3
$begingroup$
Hmm.. if we set $H(t)=F(frac{1}{log 2}2^{-t})$ then $H$ satisfies the delay differential equation $H'(t)=-2^{-t}H(t-1)$, which looks like it has many different solutions with $lim_{ttoinfty}H(t)=1$
$endgroup$
– Henning Makholm
Jul 18 '15 at 10:37
1
$begingroup$
@6005: I don't. I just have the heuristic argument that you can choose $H$ freely on, say, the interval $[1,2]$ making sure the $H'(2)=-frac12H(1)$, and then integrate the values of $H(t)$ up to $t=infty$ numerically. Because of the $2^{-t}$ factor it should end up with a horizontal asymptote, which will be nonzero unless you're pretty unlucky. Then scale everything such that the limit is $1$ instead (the equation is linear after all).
$endgroup$
– Henning Makholm
Jul 18 '15 at 20:02
2
2
$begingroup$
Well, if the function is analytic at $0$, then its Maclaurin series is $$sum frac{2^{n(n-1)/2}}{n!},$$from repeated differentiation at $x = 0$, but I'm guessing you knew this.
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:28
$begingroup$
Well, if the function is analytic at $0$, then its Maclaurin series is $$sum frac{2^{n(n-1)/2}}{n!},$$from repeated differentiation at $x = 0$, but I'm guessing you knew this.
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:28
1
1
$begingroup$
Yes I did. I've been making little errors like this all week :-(
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:44
$begingroup$
Yes I did. I've been making little errors like this all week :-(
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:44
3
3
$begingroup$
the power series's radius of convergence is zero though.
$endgroup$
– mercio
Jul 18 '15 at 8:20
$begingroup$
the power series's radius of convergence is zero though.
$endgroup$
– mercio
Jul 18 '15 at 8:20
3
3
$begingroup$
Hmm.. if we set $H(t)=F(frac{1}{log 2}2^{-t})$ then $H$ satisfies the delay differential equation $H'(t)=-2^{-t}H(t-1)$, which looks like it has many different solutions with $lim_{ttoinfty}H(t)=1$
$endgroup$
– Henning Makholm
Jul 18 '15 at 10:37
$begingroup$
Hmm.. if we set $H(t)=F(frac{1}{log 2}2^{-t})$ then $H$ satisfies the delay differential equation $H'(t)=-2^{-t}H(t-1)$, which looks like it has many different solutions with $lim_{ttoinfty}H(t)=1$
$endgroup$
– Henning Makholm
Jul 18 '15 at 10:37
1
1
$begingroup$
@6005: I don't. I just have the heuristic argument that you can choose $H$ freely on, say, the interval $[1,2]$ making sure the $H'(2)=-frac12H(1)$, and then integrate the values of $H(t)$ up to $t=infty$ numerically. Because of the $2^{-t}$ factor it should end up with a horizontal asymptote, which will be nonzero unless you're pretty unlucky. Then scale everything such that the limit is $1$ instead (the equation is linear after all).
$endgroup$
– Henning Makholm
Jul 18 '15 at 20:02
$begingroup$
@6005: I don't. I just have the heuristic argument that you can choose $H$ freely on, say, the interval $[1,2]$ making sure the $H'(2)=-frac12H(1)$, and then integrate the values of $H(t)$ up to $t=infty$ numerically. Because of the $2^{-t}$ factor it should end up with a horizontal asymptote, which will be nonzero unless you're pretty unlucky. Then scale everything such that the limit is $1$ instead (the equation is linear after all).
$endgroup$
– Henning Makholm
Jul 18 '15 at 20:02
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
There are infinitely many differentiable functions $Fcolon [0,infty)tomathbb{R}$ that satisfy
$$
F'(x)=F(2x)qquadtext{and}qquad F(0)=1.
$$
We will follow the argument outlined by @HenningMakholm in the comments. First, consider the substitution $G(x) = Fleft(dfrac{2^x}{log 2}right)$. Plugging this into the above equation gives
$$
G,'(x) = 2^x G(x+1)qquadtext{and}qquad lim_{xto-infty} G(x)=1.
$$
This delay differential equation has a large number of solutions. Roughly speaking, one can choose any function for $G$ on $[0,1]$, and then define
$$
G(x) = 2^{1-x}G,'(x-1)tag*{(1)}
$$
recursively for $x> 1$, and
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
recursively for $x<0$. (See this question for a simpler example of this kind of solution.)
Obstacles and Boundary Conditions
There are two obstacles to this construction:
If $G$ is $C^n$ on the interval $(0,1)$, then $G$ will be $C^{n-1}$ on $(1,2)$ by equation (1), and then $C^{n-2}$ on $(2,3)$, and so forth. Thus, if we want $G$ to be everywhere defined, we must start with a $C^infty$ function on $[0,1]$.
If we start with an arbitrary $C^infty$ function on $[0,1]$ and then switch to definition (1) for $x>1$, we must make sure that $G$ is $C^infty$ at $x=1$, i.e. the the left and right hand derivatives of every order match up at this point.
For the second obstacle, repeatedly differentiating the equation $G(x+1)=2^{-x}G,'(x)$ gives the sequence of equations
$$
G^{(n)}(x+1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i},2^{-x}, G^{(i+1)}(x)
$$
and plugging in $x=0$ yields
$$
G^{(n)}(1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i}, G^{(i+1)}(0).
$$
These are the boundary conditions that $G$ must satisfy on the interval $[0,1]$, but it is not hard to find $C^infty$ functions that satisfy these, e.g. any bump function that satisfies
$$
G^{(n)}(0) = G^{(n)}(1) = 0;;text{ for all }ngeq 1
qquadtext{and}qquad
G(1)=0.
$$
The Limit as $xto-infty$
For negative values of $x$, the function $G(x)$ is defined by the integral equation
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
If we assume that $G(x)geq 0$ for $xin [0,1]$ and, say, that
$$
int_0^1 2^{u-1} G(u),du ;<; frac{G(0)}{10}
$$
then $frac{9}{10}G(0) leq G(x) leq G(0)$ for all $xin[-1,0]$. It follows recursively that
$$
G(0) ;geq; G(x) ;geq; left(frac{9}{10} - frac{1}{log 4}right)G(0) ;>; 0
$$
for all $x<-1$, since
begin{align*}
G(x) ;&=; G(0)-int_{x+1}^1 2^{u-1}G(u),du \
&=; G(0) - int_0^1 2^{u-1}G(u),du - int_{x+1}^0 2^{u-1}G(u),du \
≥ G(0) - frac{G(0)}{10} - int_{-infty}^0 2^{u-1}G(0),du \
&=; left(frac{9}{10} - frac{1}{log 4}right)G(0).
end{align*}
Then $G$ is monotone on $(-infty,0)$, so $displaystylelim_{xto-infty} G(x)$ exists and is positive. Scaling $G$ linearly, we can arrange for this limit to be $1$.
answered Jul 19 '15 at 4:09
Jim BelkJim Belk
37.6k286152
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
There are infinitely many differentiable functions $Fcolon [0,infty)tomathbb{R}$ that satisfy
$$
F'(x)=F(2x)qquadtext{and}qquad F(0)=1.
$$
We will follow the argument outlined by @HenningMakholm in the comments. First, consider the substitution $G(x) = Fleft(dfrac{2^x}{log 2}right)$. Plugging this into the above equation gives
$$
G,'(x) = 2^x G(x+1)qquadtext{and}qquad lim_{xto-infty} G(x)=1.
$$
This delay differential equation has a large number of solutions. Roughly speaking, one can choose any function for $G$ on $[0,1]$, and then define
$$
G(x) = 2^{1-x}G,'(x-1)tag*{(1)}
$$
recursively for $x> 1$, and
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
recursively for $x<0$. (See this question for a simpler example of this kind of solution.)
Obstacles and Boundary Conditions
There are two obstacles to this construction:
If $G$ is $C^n$ on the interval $(0,1)$, then $G$ will be $C^{n-1}$ on $(1,2)$ by equation (1), and then $C^{n-2}$ on $(2,3)$, and so forth. Thus, if we want $G$ to be everywhere defined, we must start with a $C^infty$ function on $[0,1]$.
If we start with an arbitrary $C^infty$ function on $[0,1]$ and then switch to definition (1) for $x>1$, we must make sure that $G$ is $C^infty$ at $x=1$, i.e. the the left and right hand derivatives of every order match up at this point.
For the second obstacle, repeatedly differentiating the equation $G(x+1)=2^{-x}G,'(x)$ gives the sequence of equations
$$
G^{(n)}(x+1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i},2^{-x}, G^{(i+1)}(x)
$$
and plugging in $x=0$ yields
$$
G^{(n)}(1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i}, G^{(i+1)}(0).
$$
These are the boundary conditions that $G$ must satisfy on the interval $[0,1]$, but it is not hard to find $C^infty$ functions that satisfy these, e.g. any bump function that satisfies
$$
G^{(n)}(0) = G^{(n)}(1) = 0;;text{ for all }ngeq 1
qquadtext{and}qquad
G(1)=0.
$$
The Limit as $xto-infty$
For negative values of $x$, the function $G(x)$ is defined by the integral equation
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
If we assume that $G(x)geq 0$ for $xin [0,1]$ and, say, that
$$
int_0^1 2^{u-1} G(u),du ;<; frac{G(0)}{10}
$$
then $frac{9}{10}G(0) leq G(x) leq G(0)$ for all $xin[-1,0]$. It follows recursively that
$$
G(0) ;geq; G(x) ;geq; left(frac{9}{10} - frac{1}{log 4}right)G(0) ;>; 0
$$
for all $x<-1$, since
begin{align*}
G(x) ;&=; G(0)-int_{x+1}^1 2^{u-1}G(u),du \
&=; G(0) - int_0^1 2^{u-1}G(u),du - int_{x+1}^0 2^{u-1}G(u),du \
≥ G(0) - frac{G(0)}{10} - int_{-infty}^0 2^{u-1}G(0),du \
&=; left(frac{9}{10} - frac{1}{log 4}right)G(0).
end{align*}
Then $G$ is monotone on $(-infty,0)$, so $displaystylelim_{xto-infty} G(x)$ exists and is positive. Scaling $G$ linearly, we can arrange for this limit to be $1$.
answered Jul 19 '15 at 4:09
Jim BelkJim Belk
37.6k286152
$endgroup$
add a comment |
$begingroup$
There are infinitely many differentiable functions $Fcolon [0,infty)tomathbb{R}$ that satisfy
$$
F'(x)=F(2x)qquadtext{and}qquad F(0)=1.
$$
We will follow the argument outlined by @HenningMakholm in the comments. First, consider the substitution $G(x) = Fleft(dfrac{2^x}{log 2}right)$. Plugging this into the above equation gives
$$
G,'(x) = 2^x G(x+1)qquadtext{and}qquad lim_{xto-infty} G(x)=1.
$$
This delay differential equation has a large number of solutions. Roughly speaking, one can choose any function for $G$ on $[0,1]$, and then define
$$
G(x) = 2^{1-x}G,'(x-1)tag*{(1)}
$$
recursively for $x> 1$, and
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
recursively for $x<0$. (See this question for a simpler example of this kind of solution.)
Obstacles and Boundary Conditions
There are two obstacles to this construction:
If $G$ is $C^n$ on the interval $(0,1)$, then $G$ will be $C^{n-1}$ on $(1,2)$ by equation (1), and then $C^{n-2}$ on $(2,3)$, and so forth. Thus, if we want $G$ to be everywhere defined, we must start with a $C^infty$ function on $[0,1]$.
If we start with an arbitrary $C^infty$ function on $[0,1]$ and then switch to definition (1) for $x>1$, we must make sure that $G$ is $C^infty$ at $x=1$, i.e. the the left and right hand derivatives of every order match up at this point.
For the second obstacle, repeatedly differentiating the equation $G(x+1)=2^{-x}G,'(x)$ gives the sequence of equations
$$
G^{(n)}(x+1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i},2^{-x}, G^{(i+1)}(x)
$$
and plugging in $x=0$ yields
$$
G^{(n)}(1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i}, G^{(i+1)}(0).
$$
These are the boundary conditions that $G$ must satisfy on the interval $[0,1]$, but it is not hard to find $C^infty$ functions that satisfy these, e.g. any bump function that satisfies
$$
G^{(n)}(0) = G^{(n)}(1) = 0;;text{ for all }ngeq 1
qquadtext{and}qquad
G(1)=0.
$$
The Limit as $xto-infty$
For negative values of $x$, the function $G(x)$ is defined by the integral equation
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
If we assume that $G(x)geq 0$ for $xin [0,1]$ and, say, that
$$
int_0^1 2^{u-1} G(u),du ;<; frac{G(0)}{10}
$$
then $frac{9}{10}G(0) leq G(x) leq G(0)$ for all $xin[-1,0]$. It follows recursively that
$$
G(0) ;geq; G(x) ;geq; left(frac{9}{10} - frac{1}{log 4}right)G(0) ;>; 0
$$
for all $x<-1$, since
begin{align*}
G(x) ;&=; G(0)-int_{x+1}^1 2^{u-1}G(u),du \
&=; G(0) - int_0^1 2^{u-1}G(u),du - int_{x+1}^0 2^{u-1}G(u),du \
≥ G(0) - frac{G(0)}{10} - int_{-infty}^0 2^{u-1}G(0),du \
&=; left(frac{9}{10} - frac{1}{log 4}right)G(0).
end{align*}
Then $G$ is monotone on $(-infty,0)$, so $displaystylelim_{xto-infty} G(x)$ exists and is positive. Scaling $G$ linearly, we can arrange for this limit to be $1$.
answered Jul 19 '15 at 4:09
Jim BelkJim Belk
37.6k286152
$endgroup$
add a comment |
$begingroup$
There are infinitely many differentiable functions $Fcolon [0,infty)tomathbb{R}$ that satisfy
$$
F'(x)=F(2x)qquadtext{and}qquad F(0)=1.
$$
We will follow the argument outlined by @HenningMakholm in the comments. First, consider the substitution $G(x) = Fleft(dfrac{2^x}{log 2}right)$. Plugging this into the above equation gives
$$
G,'(x) = 2^x G(x+1)qquadtext{and}qquad lim_{xto-infty} G(x)=1.
$$
This delay differential equation has a large number of solutions. Roughly speaking, one can choose any function for $G$ on $[0,1]$, and then define
$$
G(x) = 2^{1-x}G,'(x-1)tag*{(1)}
$$
recursively for $x> 1$, and
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
recursively for $x<0$. (See this question for a simpler example of this kind of solution.)
Obstacles and Boundary Conditions
There are two obstacles to this construction:
If $G$ is $C^n$ on the interval $(0,1)$, then $G$ will be $C^{n-1}$ on $(1,2)$ by equation (1), and then $C^{n-2}$ on $(2,3)$, and so forth. Thus, if we want $G$ to be everywhere defined, we must start with a $C^infty$ function on $[0,1]$.
If we start with an arbitrary $C^infty$ function on $[0,1]$ and then switch to definition (1) for $x>1$, we must make sure that $G$ is $C^infty$ at $x=1$, i.e. the the left and right hand derivatives of every order match up at this point.
For the second obstacle, repeatedly differentiating the equation $G(x+1)=2^{-x}G,'(x)$ gives the sequence of equations
$$
G^{(n)}(x+1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i},2^{-x}, G^{(i+1)}(x)
$$
and plugging in $x=0$ yields
$$
G^{(n)}(1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i}, G^{(i+1)}(0).
$$
These are the boundary conditions that $G$ must satisfy on the interval $[0,1]$, but it is not hard to find $C^infty$ functions that satisfy these, e.g. any bump function that satisfies
$$
G^{(n)}(0) = G^{(n)}(1) = 0;;text{ for all }ngeq 1
qquadtext{and}qquad
G(1)=0.
$$
The Limit as $xto-infty$
For negative values of $x$, the function $G(x)$ is defined by the integral equation
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
If we assume that $G(x)geq 0$ for $xin [0,1]$ and, say, that
$$
int_0^1 2^{u-1} G(u),du ;<; frac{G(0)}{10}
$$
then $frac{9}{10}G(0) leq G(x) leq G(0)$ for all $xin[-1,0]$. It follows recursively that
$$
G(0) ;geq; G(x) ;geq; left(frac{9}{10} - frac{1}{log 4}right)G(0) ;>; 0
$$
for all $x<-1$, since
begin{align*}
G(x) ;&=; G(0)-int_{x+1}^1 2^{u-1}G(u),du \
&=; G(0) - int_0^1 2^{u-1}G(u),du - int_{x+1}^0 2^{u-1}G(u),du \
≥ G(0) - frac{G(0)}{10} - int_{-infty}^0 2^{u-1}G(0),du \
&=; left(frac{9}{10} - frac{1}{log 4}right)G(0).
end{align*}
Then $G$ is monotone on $(-infty,0)$, so $displaystylelim_{xto-infty} G(x)$ exists and is positive. Scaling $G$ linearly, we can arrange for this limit to be $1$.
answered Jul 19 '15 at 4:09
Jim BelkJim Belk
37.6k286152
$endgroup$
There are infinitely many differentiable functions $Fcolon [0,infty)tomathbb{R}$ that satisfy
$$
F'(x)=F(2x)qquadtext{and}qquad F(0)=1.
$$
We will follow the argument outlined by @HenningMakholm in the comments. First, consider the substitution $G(x) = Fleft(dfrac{2^x}{log 2}right)$. Plugging this into the above equation gives
$$
G,'(x) = 2^x G(x+1)qquadtext{and}qquad lim_{xto-infty} G(x)=1.
$$
This delay differential equation has a large number of solutions. Roughly speaking, one can choose any function for $G$ on $[0,1]$, and then define
$$
G(x) = 2^{1-x}G,'(x-1)tag*{(1)}
$$
recursively for $x> 1$, and
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
recursively for $x<0$. (See this question for a simpler example of this kind of solution.)
Obstacles and Boundary Conditions
There are two obstacles to this construction:
If $G$ is $C^n$ on the interval $(0,1)$, then $G$ will be $C^{n-1}$ on $(1,2)$ by equation (1), and then $C^{n-2}$ on $(2,3)$, and so forth. Thus, if we want $G$ to be everywhere defined, we must start with a $C^infty$ function on $[0,1]$.
If we start with an arbitrary $C^infty$ function on $[0,1]$ and then switch to definition (1) for $x>1$, we must make sure that $G$ is $C^infty$ at $x=1$, i.e. the the left and right hand derivatives of every order match up at this point.
For the second obstacle, repeatedly differentiating the equation $G(x+1)=2^{-x}G,'(x)$ gives the sequence of equations
$$
G^{(n)}(x+1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i},2^{-x}, G^{(i+1)}(x)
$$
and plugging in $x=0$ yields
$$
G^{(n)}(1) ;=; sum_{i=0}^n binom{n}{i}(-log 2)^{n-i}, G^{(i+1)}(0).
$$
These are the boundary conditions that $G$ must satisfy on the interval $[0,1]$, but it is not hard to find $C^infty$ functions that satisfy these, e.g. any bump function that satisfies
$$
G^{(n)}(0) = G^{(n)}(1) = 0;;text{ for all }ngeq 1
qquadtext{and}qquad
G(1)=0.
$$
The Limit as $xto-infty$
For negative values of $x$, the function $G(x)$ is defined by the integral equation
$$
G(x) = G(0) - int_{x+1}^1 2^{u-1} G(u),dutag*{(2)}
$$
If we assume that $G(x)geq 0$ for $xin [0,1]$ and, say, that
$$
int_0^1 2^{u-1} G(u),du ;<; frac{G(0)}{10}
$$
then $frac{9}{10}G(0) leq G(x) leq G(0)$ for all $xin[-1,0]$. It follows recursively that
$$
G(0) ;geq; G(x) ;geq; left(frac{9}{10} - frac{1}{log 4}right)G(0) ;>; 0
$$
for all $x<-1$, since
begin{align*}
G(x) ;&=; G(0)-int_{x+1}^1 2^{u-1}G(u),du \
&=; G(0) - int_0^1 2^{u-1}G(u),du - int_{x+1}^0 2^{u-1}G(u),du \
≥ G(0) - frac{G(0)}{10} - int_{-infty}^0 2^{u-1}G(0),du \
&=; left(frac{9}{10} - frac{1}{log 4}right)G(0).
end{align*}
Then $G$ is monotone on $(-infty,0)$, so $displaystylelim_{xto-infty} G(x)$ exists and is positive. Scaling $G$ linearly, we can arrange for this limit to be $1$.
answered Jul 19 '15 at 4:09
Jim BelkJim Belk
37.6k286152
edited Jan 12 at 12:26
answered Jul 19 '15 at 4:09
Jim BelkJim Belk
37.6k286152
answered Jul 19 '15 at 4:09
Jim BelkJim Belk
37.6k286152
answered Jul 19 '15 at 4:09
Jim BelkJim Belk
37.6k286152
37.6k286152
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2
$begingroup$
Well, if the function is analytic at $0$, then its Maclaurin series is $$sum frac{2^{n(n-1)/2}}{n!},$$from repeated differentiation at $x = 0$, but I'm guessing you knew this.
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:28
1
$begingroup$
Yes I did. I've been making little errors like this all week :-(
$endgroup$
– Theo Bendit
Jul 18 '15 at 7:44
3
$begingroup$
the power series's radius of convergence is zero though.
$endgroup$
– mercio
Jul 18 '15 at 8:20
3
$begingroup$
Hmm.. if we set $H(t)=F(frac{1}{log 2}2^{-t})$ then $H$ satisfies the delay differential equation $H'(t)=-2^{-t}H(t-1)$, which looks like it has many different solutions with $lim_{ttoinfty}H(t)=1$
$endgroup$
– Henning Makholm
Jul 18 '15 at 10:37
1
$begingroup$
@6005: I don't. I just have the heuristic argument that you can choose $H$ freely on, say, the interval $[1,2]$ making sure the $H'(2)=-frac12H(1)$, and then integrate the values of $H(t)$ up to $t=infty$ numerically. Because of the $2^{-t}$ factor it should end up with a horizontal asymptote, which will be nonzero unless you're pretty unlucky. Then scale everything such that the limit is $1$ instead (the equation is linear after all).
$endgroup$
– Henning Makholm
Jul 18 '15 at 20:02