Solving functional differential equation $f'(x)=2f(2x)-f(x)$
Clash Royale CLAN TAG#URR8PPP
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Show that there is at least a nonzero function $f$, differentiable on $[0,+infty)$, satisfying $$f'(x)=2f(2x)-f(x) qquad forall x>0 $$
$$M_n:=int_0^inftyx^nf(x)dx<infty qquad forall nin
mathbbN $$
My best idea so far is to assume that the solution is a power series, i.e.
$$ f(x)=sum_n=0^inftya_nx^nqquad forall x>0$$
Then the equation becomes
$$ sum_n=0^inftyna_nx^n-1=2sum_n=0^inftya_n2^nx^n-sum_n=0^inftya_nx^n$$
equating all the coefficients of the same degree I get
$$na_n=(2^n-1)a_n-1qquad forall ngeq 1$$
So setting $a_0=1$, I get
$$a_n=frac1n!prod_k=1^n(2^k-1) qquad forall n$$
But does the power series actually converge? Using Hadamard's formula, and that $2^k-1geq 2^k-1$,
$$ |a_n|^1/ngeqfrac1(n!)^1/nleft[2^n(n-1)/2right]^1/nsimfracen(2pi n)^1/2n2^(n-1)/2to infty$$
so the radius of converge of the series is $0$, so it doesn't actually define a solution on $[0,+infty)$.
differential-equations functional-equations
asked Dec 6 '17 at 0:48
Lorenzo Quarisa
2,734316
add a comment |Â
up vote
6
down vote
favorite
Show that there is at least a nonzero function $f$, differentiable on $[0,+infty)$, satisfying $$f'(x)=2f(2x)-f(x) qquad forall x>0 $$
$$M_n:=int_0^inftyx^nf(x)dx<infty qquad forall nin
mathbbN $$
My best idea so far is to assume that the solution is a power series, i.e.
$$ f(x)=sum_n=0^inftya_nx^nqquad forall x>0$$
Then the equation becomes
$$ sum_n=0^inftyna_nx^n-1=2sum_n=0^inftya_n2^nx^n-sum_n=0^inftya_nx^n$$
equating all the coefficients of the same degree I get
$$na_n=(2^n-1)a_n-1qquad forall ngeq 1$$
So setting $a_0=1$, I get
$$a_n=frac1n!prod_k=1^n(2^k-1) qquad forall n$$
But does the power series actually converge? Using Hadamard's formula, and that $2^k-1geq 2^k-1$,
$$ |a_n|^1/ngeqfrac1(n!)^1/nleft[2^n(n-1)/2right]^1/nsimfracen(2pi n)^1/2n2^(n-1)/2to infty$$
so the radius of converge of the series is $0$, so it doesn't actually define a solution on $[0,+infty)$.
differential-equations functional-equations
asked Dec 6 '17 at 0:48
Lorenzo Quarisa
2,734316
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Show that there is at least a nonzero function $f$, differentiable on $[0,+infty)$, satisfying $$f'(x)=2f(2x)-f(x) qquad forall x>0 $$
$$M_n:=int_0^inftyx^nf(x)dx<infty qquad forall nin
mathbbN $$
My best idea so far is to assume that the solution is a power series, i.e.
$$ f(x)=sum_n=0^inftya_nx^nqquad forall x>0$$
Then the equation becomes
$$ sum_n=0^inftyna_nx^n-1=2sum_n=0^inftya_n2^nx^n-sum_n=0^inftya_nx^n$$
equating all the coefficients of the same degree I get
$$na_n=(2^n-1)a_n-1qquad forall ngeq 1$$
So setting $a_0=1$, I get
$$a_n=frac1n!prod_k=1^n(2^k-1) qquad forall n$$
But does the power series actually converge? Using Hadamard's formula, and that $2^k-1geq 2^k-1$,
$$ |a_n|^1/ngeqfrac1(n!)^1/nleft[2^n(n-1)/2right]^1/nsimfracen(2pi n)^1/2n2^(n-1)/2to infty$$
so the radius of converge of the series is $0$, so it doesn't actually define a solution on $[0,+infty)$.
differential-equations functional-equations
asked Dec 6 '17 at 0:48
Lorenzo Quarisa
2,734316
Show that there is at least a nonzero function $f$, differentiable on $[0,+infty)$, satisfying $$f'(x)=2f(2x)-f(x) qquad forall x>0 $$
$$M_n:=int_0^inftyx^nf(x)dx<infty qquad forall nin
mathbbN $$
My best idea so far is to assume that the solution is a power series, i.e.
$$ f(x)=sum_n=0^inftya_nx^nqquad forall x>0$$
Then the equation becomes
$$ sum_n=0^inftyna_nx^n-1=2sum_n=0^inftya_n2^nx^n-sum_n=0^inftya_nx^n$$
equating all the coefficients of the same degree I get
$$na_n=(2^n-1)a_n-1qquad forall ngeq 1$$
So setting $a_0=1$, I get
$$a_n=frac1n!prod_k=1^n(2^k-1) qquad forall n$$
But does the power series actually converge? Using Hadamard's formula, and that $2^k-1geq 2^k-1$,
$$ |a_n|^1/ngeqfrac1(n!)^1/nleft[2^n(n-1)/2right]^1/nsimfracen(2pi n)^1/2n2^(n-1)/2to infty$$
so the radius of converge of the series is $0$, so it doesn't actually define a solution on $[0,+infty)$.
differential-equations functional-equations
differential-equations functional-equations
asked Dec 6 '17 at 0:48
Lorenzo Quarisa
2,734316
asked Dec 6 '17 at 0:48
Lorenzo Quarisa
2,734316
asked Dec 6 '17 at 0:48
Lorenzo Quarisa
2,734316
asked Dec 6 '17 at 0:48
Lorenzo Quarisa
2,734316
asked Dec 6 '17 at 0:48
Lorenzo Quarisa
2,734316
2,734316
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
Well, it looks like your best idea (which by itself was not bad at all) fails precisely for the reasons you so comprehensively described. So what shall we do?
Guess what?
We turn it the other way around. Let's look for a function in the form of
$$f(x)=sum_n=0^infty a_ne^-2^nx,quad xgeqslant0$$
You might wonder what's with $2^ncdot x$ up there in the exponent. Well, it's simple: I was going to write $e^-nx$ when I realized we won't be needing most of those terms, so let it be this way.
Now, the equation becomes
$$-sum_n=0^infty2^na_ne^-2^nx=2sum_n=0^inftya_ne^-2^n+1x-sum_n=0^inftya_ne^-2^nx$$
or
$$sum_n=0^infty2^na_ne^-2^nx = -2sum_n=1^inftya_n-1e^-2^nx+sum_n=0^inftya_ne^-2^nx$$
which gives
$$(2^n-1)a_n=-2a_n-1,quad ngeqslant1$$
So continuing in your footsteps, I set $a_0=1$ and get
$$a_n=frac2^nprod_k=1^n(2^k-1), quad ngeqslant1$$
which makes my series converge pretty fast, for the same reason why your series fails to do so.
Now that's our solution.
answered Dec 6 '17 at 12:06
Ivan Neretin
8,40721533
Thanks. That leaves the question of the convergence of the integrals $M_n$. I need to exchange a series and integral sign so for that I need the uniform convergence of the series $sum_n=0^inftya_n x^n e^-2^nx$. My idea would be using the fact that $a_nx^n e^-2^nxleq a_nx^n$, and that the series $sum_n=0^inftya_nx^n$ converges uniformly on $[0,+infty)$ because $(a_n)^1/nto 0$. Is this correct? Another question: couldn't a solution also be of the form $sum_n=0^inftya_nx^-n$? Did you jump to inverse exponentials just to make sure it would converge quickly enough?
– Lorenzo Quarisa
Dec 6 '17 at 19:32
No. True, $sum_n=0^inftya_nx^n$ does converge for all $x$, but it definitely does not converge uniformly on $(0,+infty)$. Then again, how the combination $a_nx^n$ came about in the first place?
– Ivan Neretin
Dec 6 '17 at 19:39
$x^-n$ behave badly at 0. But your guess is also true: yes, that's to make sure it converges quickly.
– Ivan Neretin
Dec 6 '17 at 19:40
I see, convergence of power series only implies uniform convergence on compact subsets of the convergence disk, and $[0,+infty)$ is not compact so that doesn't work. Actually I just need the fact that $x^ne^-2^nx$ is bounded (uniformly in $x$ and $n$), so I can bound the series with $Csum_n=0^inftya_n$ which clearly converges and doesn't depend on $x$, so the first series must converge uniformly.
– Lorenzo Quarisa
Dec 6 '17 at 19:52
1
You don't have the combination $x^ne^-2^nx$. The $n$ in the definition of your $M_n$ and another $n$ in the definition of my series are two totally different, unrelated things.
– Ivan Neretin
Dec 6 '17 at 20:05
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Well, it looks like your best idea (which by itself was not bad at all) fails precisely for the reasons you so comprehensively described. So what shall we do?
Guess what?
We turn it the other way around. Let's look for a function in the form of
$$f(x)=sum_n=0^infty a_ne^-2^nx,quad xgeqslant0$$
You might wonder what's with $2^ncdot x$ up there in the exponent. Well, it's simple: I was going to write $e^-nx$ when I realized we won't be needing most of those terms, so let it be this way.
Now, the equation becomes
$$-sum_n=0^infty2^na_ne^-2^nx=2sum_n=0^inftya_ne^-2^n+1x-sum_n=0^inftya_ne^-2^nx$$
or
$$sum_n=0^infty2^na_ne^-2^nx = -2sum_n=1^inftya_n-1e^-2^nx+sum_n=0^inftya_ne^-2^nx$$
which gives
$$(2^n-1)a_n=-2a_n-1,quad ngeqslant1$$
So continuing in your footsteps, I set $a_0=1$ and get
$$a_n=frac2^nprod_k=1^n(2^k-1), quad ngeqslant1$$
which makes my series converge pretty fast, for the same reason why your series fails to do so.
Now that's our solution.
answered Dec 6 '17 at 12:06
Ivan Neretin
8,40721533
Thanks. That leaves the question of the convergence of the integrals $M_n$. I need to exchange a series and integral sign so for that I need the uniform convergence of the series $sum_n=0^inftya_n x^n e^-2^nx$. My idea would be using the fact that $a_nx^n e^-2^nxleq a_nx^n$, and that the series $sum_n=0^inftya_nx^n$ converges uniformly on $[0,+infty)$ because $(a_n)^1/nto 0$. Is this correct? Another question: couldn't a solution also be of the form $sum_n=0^inftya_nx^-n$? Did you jump to inverse exponentials just to make sure it would converge quickly enough?
– Lorenzo Quarisa
Dec 6 '17 at 19:32
No. True, $sum_n=0^inftya_nx^n$ does converge for all $x$, but it definitely does not converge uniformly on $(0,+infty)$. Then again, how the combination $a_nx^n$ came about in the first place?
– Ivan Neretin
Dec 6 '17 at 19:39
$x^-n$ behave badly at 0. But your guess is also true: yes, that's to make sure it converges quickly.
– Ivan Neretin
Dec 6 '17 at 19:40
I see, convergence of power series only implies uniform convergence on compact subsets of the convergence disk, and $[0,+infty)$ is not compact so that doesn't work. Actually I just need the fact that $x^ne^-2^nx$ is bounded (uniformly in $x$ and $n$), so I can bound the series with $Csum_n=0^inftya_n$ which clearly converges and doesn't depend on $x$, so the first series must converge uniformly.
– Lorenzo Quarisa
Dec 6 '17 at 19:52
1
You don't have the combination $x^ne^-2^nx$. The $n$ in the definition of your $M_n$ and another $n$ in the definition of my series are two totally different, unrelated things.
– Ivan Neretin
Dec 6 '17 at 20:05
 |Â
show 1 more comment
up vote
7
down vote
accepted
Well, it looks like your best idea (which by itself was not bad at all) fails precisely for the reasons you so comprehensively described. So what shall we do?
Guess what?
We turn it the other way around. Let's look for a function in the form of
$$f(x)=sum_n=0^infty a_ne^-2^nx,quad xgeqslant0$$
You might wonder what's with $2^ncdot x$ up there in the exponent. Well, it's simple: I was going to write $e^-nx$ when I realized we won't be needing most of those terms, so let it be this way.
Now, the equation becomes
$$-sum_n=0^infty2^na_ne^-2^nx=2sum_n=0^inftya_ne^-2^n+1x-sum_n=0^inftya_ne^-2^nx$$
or
$$sum_n=0^infty2^na_ne^-2^nx = -2sum_n=1^inftya_n-1e^-2^nx+sum_n=0^inftya_ne^-2^nx$$
which gives
$$(2^n-1)a_n=-2a_n-1,quad ngeqslant1$$
So continuing in your footsteps, I set $a_0=1$ and get
$$a_n=frac2^nprod_k=1^n(2^k-1), quad ngeqslant1$$
which makes my series converge pretty fast, for the same reason why your series fails to do so.
Now that's our solution.
answered Dec 6 '17 at 12:06
Ivan Neretin
8,40721533
Thanks. That leaves the question of the convergence of the integrals $M_n$. I need to exchange a series and integral sign so for that I need the uniform convergence of the series $sum_n=0^inftya_n x^n e^-2^nx$. My idea would be using the fact that $a_nx^n e^-2^nxleq a_nx^n$, and that the series $sum_n=0^inftya_nx^n$ converges uniformly on $[0,+infty)$ because $(a_n)^1/nto 0$. Is this correct? Another question: couldn't a solution also be of the form $sum_n=0^inftya_nx^-n$? Did you jump to inverse exponentials just to make sure it would converge quickly enough?
– Lorenzo Quarisa
Dec 6 '17 at 19:32
No. True, $sum_n=0^inftya_nx^n$ does converge for all $x$, but it definitely does not converge uniformly on $(0,+infty)$. Then again, how the combination $a_nx^n$ came about in the first place?
– Ivan Neretin
Dec 6 '17 at 19:39
$x^-n$ behave badly at 0. But your guess is also true: yes, that's to make sure it converges quickly.
– Ivan Neretin
Dec 6 '17 at 19:40
I see, convergence of power series only implies uniform convergence on compact subsets of the convergence disk, and $[0,+infty)$ is not compact so that doesn't work. Actually I just need the fact that $x^ne^-2^nx$ is bounded (uniformly in $x$ and $n$), so I can bound the series with $Csum_n=0^inftya_n$ which clearly converges and doesn't depend on $x$, so the first series must converge uniformly.
– Lorenzo Quarisa
Dec 6 '17 at 19:52
1
You don't have the combination $x^ne^-2^nx$. The $n$ in the definition of your $M_n$ and another $n$ in the definition of my series are two totally different, unrelated things.
– Ivan Neretin
Dec 6 '17 at 20:05
 |Â
show 1 more comment
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Well, it looks like your best idea (which by itself was not bad at all) fails precisely for the reasons you so comprehensively described. So what shall we do?
Guess what?
We turn it the other way around. Let's look for a function in the form of
$$f(x)=sum_n=0^infty a_ne^-2^nx,quad xgeqslant0$$
You might wonder what's with $2^ncdot x$ up there in the exponent. Well, it's simple: I was going to write $e^-nx$ when I realized we won't be needing most of those terms, so let it be this way.
Now, the equation becomes
$$-sum_n=0^infty2^na_ne^-2^nx=2sum_n=0^inftya_ne^-2^n+1x-sum_n=0^inftya_ne^-2^nx$$
or
$$sum_n=0^infty2^na_ne^-2^nx = -2sum_n=1^inftya_n-1e^-2^nx+sum_n=0^inftya_ne^-2^nx$$
which gives
$$(2^n-1)a_n=-2a_n-1,quad ngeqslant1$$
So continuing in your footsteps, I set $a_0=1$ and get
$$a_n=frac2^nprod_k=1^n(2^k-1), quad ngeqslant1$$
which makes my series converge pretty fast, for the same reason why your series fails to do so.
Now that's our solution.
answered Dec 6 '17 at 12:06
Ivan Neretin
8,40721533
Well, it looks like your best idea (which by itself was not bad at all) fails precisely for the reasons you so comprehensively described. So what shall we do?
Guess what?
We turn it the other way around. Let's look for a function in the form of
$$f(x)=sum_n=0^infty a_ne^-2^nx,quad xgeqslant0$$
You might wonder what's with $2^ncdot x$ up there in the exponent. Well, it's simple: I was going to write $e^-nx$ when I realized we won't be needing most of those terms, so let it be this way.
Now, the equation becomes
$$-sum_n=0^infty2^na_ne^-2^nx=2sum_n=0^inftya_ne^-2^n+1x-sum_n=0^inftya_ne^-2^nx$$
or
$$sum_n=0^infty2^na_ne^-2^nx = -2sum_n=1^inftya_n-1e^-2^nx+sum_n=0^inftya_ne^-2^nx$$
which gives
$$(2^n-1)a_n=-2a_n-1,quad ngeqslant1$$
So continuing in your footsteps, I set $a_0=1$ and get
$$a_n=frac2^nprod_k=1^n(2^k-1), quad ngeqslant1$$
which makes my series converge pretty fast, for the same reason why your series fails to do so.
Now that's our solution.
answered Dec 6 '17 at 12:06
Ivan Neretin
8,40721533
answered Dec 6 '17 at 12:06
Ivan Neretin
8,40721533
answered Dec 6 '17 at 12:06
Ivan Neretin
8,40721533
answered Dec 6 '17 at 12:06
Ivan Neretin
8,40721533
8,40721533
Thanks. That leaves the question of the convergence of the integrals $M_n$. I need to exchange a series and integral sign so for that I need the uniform convergence of the series $sum_n=0^inftya_n x^n e^-2^nx$. My idea would be using the fact that $a_nx^n e^-2^nxleq a_nx^n$, and that the series $sum_n=0^inftya_nx^n$ converges uniformly on $[0,+infty)$ because $(a_n)^1/nto 0$. Is this correct? Another question: couldn't a solution also be of the form $sum_n=0^inftya_nx^-n$? Did you jump to inverse exponentials just to make sure it would converge quickly enough?
– Lorenzo Quarisa
Dec 6 '17 at 19:32
No. True, $sum_n=0^inftya_nx^n$ does converge for all $x$, but it definitely does not converge uniformly on $(0,+infty)$. Then again, how the combination $a_nx^n$ came about in the first place?
– Ivan Neretin
Dec 6 '17 at 19:39
$x^-n$ behave badly at 0. But your guess is also true: yes, that's to make sure it converges quickly.
– Ivan Neretin
Dec 6 '17 at 19:40
I see, convergence of power series only implies uniform convergence on compact subsets of the convergence disk, and $[0,+infty)$ is not compact so that doesn't work. Actually I just need the fact that $x^ne^-2^nx$ is bounded (uniformly in $x$ and $n$), so I can bound the series with $Csum_n=0^inftya_n$ which clearly converges and doesn't depend on $x$, so the first series must converge uniformly.
– Lorenzo Quarisa
Dec 6 '17 at 19:52
1
You don't have the combination $x^ne^-2^nx$. The $n$ in the definition of your $M_n$ and another $n$ in the definition of my series are two totally different, unrelated things.
– Ivan Neretin
Dec 6 '17 at 20:05
 |Â
show 1 more comment
Thanks. That leaves the question of the convergence of the integrals $M_n$. I need to exchange a series and integral sign so for that I need the uniform convergence of the series $sum_n=0^inftya_n x^n e^-2^nx$. My idea would be using the fact that $a_nx^n e^-2^nxleq a_nx^n$, and that the series $sum_n=0^inftya_nx^n$ converges uniformly on $[0,+infty)$ because $(a_n)^1/nto 0$. Is this correct? Another question: couldn't a solution also be of the form $sum_n=0^inftya_nx^-n$? Did you jump to inverse exponentials just to make sure it would converge quickly enough?
– Lorenzo Quarisa
Dec 6 '17 at 19:32
No. True, $sum_n=0^inftya_nx^n$ does converge for all $x$, but it definitely does not converge uniformly on $(0,+infty)$. Then again, how the combination $a_nx^n$ came about in the first place?
– Ivan Neretin
Dec 6 '17 at 19:39
$x^-n$ behave badly at 0. But your guess is also true: yes, that's to make sure it converges quickly.
– Ivan Neretin
Dec 6 '17 at 19:40
I see, convergence of power series only implies uniform convergence on compact subsets of the convergence disk, and $[0,+infty)$ is not compact so that doesn't work. Actually I just need the fact that $x^ne^-2^nx$ is bounded (uniformly in $x$ and $n$), so I can bound the series with $Csum_n=0^inftya_n$ which clearly converges and doesn't depend on $x$, so the first series must converge uniformly.
– Lorenzo Quarisa
Dec 6 '17 at 19:52
1
You don't have the combination $x^ne^-2^nx$. The $n$ in the definition of your $M_n$ and another $n$ in the definition of my series are two totally different, unrelated things.
– Ivan Neretin
Dec 6 '17 at 20:05
Thanks. That leaves the question of the convergence of the integrals $M_n$. I need to exchange a series and integral sign so for that I need the uniform convergence of the series $sum_n=0^inftya_n x^n e^-2^nx$. My idea would be using the fact that $a_nx^n e^-2^nxleq a_nx^n$, and that the series $sum_n=0^inftya_nx^n$ converges uniformly on $[0,+infty)$ because $(a_n)^1/nto 0$. Is this correct? Another question: couldn't a solution also be of the form $sum_n=0^inftya_nx^-n$? Did you jump to inverse exponentials just to make sure it would converge quickly enough?
– Lorenzo Quarisa
Dec 6 '17 at 19:32
Thanks. That leaves the question of the convergence of the integrals $M_n$. I need to exchange a series and integral sign so for that I need the uniform convergence of the series $sum_n=0^inftya_n x^n e^-2^nx$. My idea would be using the fact that $a_nx^n e^-2^nxleq a_nx^n$, and that the series $sum_n=0^inftya_nx^n$ converges uniformly on $[0,+infty)$ because $(a_n)^1/nto 0$. Is this correct? Another question: couldn't a solution also be of the form $sum_n=0^inftya_nx^-n$? Did you jump to inverse exponentials just to make sure it would converge quickly enough?
– Lorenzo Quarisa
Dec 6 '17 at 19:32
No. True, $sum_n=0^inftya_nx^n$ does converge for all $x$, but it definitely does not converge uniformly on $(0,+infty)$. Then again, how the combination $a_nx^n$ came about in the first place?
– Ivan Neretin
Dec 6 '17 at 19:39
No. True, $sum_n=0^inftya_nx^n$ does converge for all $x$, but it definitely does not converge uniformly on $(0,+infty)$. Then again, how the combination $a_nx^n$ came about in the first place?
– Ivan Neretin
Dec 6 '17 at 19:39
$x^-n$ behave badly at 0. But your guess is also true: yes, that's to make sure it converges quickly.
– Ivan Neretin
Dec 6 '17 at 19:40
$x^-n$ behave badly at 0. But your guess is also true: yes, that's to make sure it converges quickly.
– Ivan Neretin
Dec 6 '17 at 19:40
I see, convergence of power series only implies uniform convergence on compact subsets of the convergence disk, and $[0,+infty)$ is not compact so that doesn't work. Actually I just need the fact that $x^ne^-2^nx$ is bounded (uniformly in $x$ and $n$), so I can bound the series with $Csum_n=0^inftya_n$ which clearly converges and doesn't depend on $x$, so the first series must converge uniformly.
– Lorenzo Quarisa
Dec 6 '17 at 19:52
I see, convergence of power series only implies uniform convergence on compact subsets of the convergence disk, and $[0,+infty)$ is not compact so that doesn't work. Actually I just need the fact that $x^ne^-2^nx$ is bounded (uniformly in $x$ and $n$), so I can bound the series with $Csum_n=0^inftya_n$ which clearly converges and doesn't depend on $x$, so the first series must converge uniformly.
– Lorenzo Quarisa
Dec 6 '17 at 19:52
1
1
You don't have the combination $x^ne^-2^nx$. The $n$ in the definition of your $M_n$ and another $n$ in the definition of my series are two totally different, unrelated things.
– Ivan Neretin
Dec 6 '17 at 20:05
You don't have the combination $x^ne^-2^nx$. The $n$ in the definition of your $M_n$ and another $n$ in the definition of my series are two totally different, unrelated things.
– Ivan Neretin
Dec 6 '17 at 20:05
 |Â
show 1 more comment
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