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$f_nto f$ then $phi(x)=Ce^x$?
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Let $f(x)in C_infty(mathbbR)$ and a succession of derivatives $f^(n)(x)$ converges in $C_[a,b]$ to a function $phi(x)$ in each finite interval. Show that $phi(x)=Ce^x$, where $C$ is a constant.
I know that $Ce^xin C_infty$ but I cannot prove $phi(x)=Ce^x$. I think the $max|f_n-f|=max|Ce^x-Ce^x|=0 $ for $n>NinmathbbN$. And I know that $Ce^xin C_[a,b]$, once it is continuous on the $mathbbR$.
Question:
1) How do I prove $phi(x)=Ce^x$? And not for example $sin (x)$?
2) What is intended on this question?
real-analysis functional-analysis
asked Jul 17 at 21:08
Pedro Gomes
1,3192618
add a comment |Â
up vote
5
down vote
favorite
Let $f(x)in C_infty(mathbbR)$ and a succession of derivatives $f^(n)(x)$ converges in $C_[a,b]$ to a function $phi(x)$ in each finite interval. Show that $phi(x)=Ce^x$, where $C$ is a constant.
I know that $Ce^xin C_infty$ but I cannot prove $phi(x)=Ce^x$. I think the $max|f_n-f|=max|Ce^x-Ce^x|=0 $ for $n>NinmathbbN$. And I know that $Ce^xin C_[a,b]$, once it is continuous on the $mathbbR$.
Question:
1) How do I prove $phi(x)=Ce^x$? And not for example $sin (x)$?
2) What is intended on this question?
real-analysis functional-analysis
asked Jul 17 at 21:08
Pedro Gomes
1,3192618
1
One possible way might be to let $g(x) = lim_n to inftyf^(n)(x)$. Try to show that $g'(x) = g(x)$. for any $x in mathbbR$. Then show that $g(x)/e^x$ is constant by showing the derivative is zero.
– Sean Haight
Jul 17 at 21:14
Think like this. $phi$ is the limit of the derivatives of $f$. So, when you derive a high derivative, you still get a little closer to $phi$. But when you take $phi$ and derive it, it will be $phi$ itself (which one has to prove). But the only functions $phi$ that satisfy $phi'=phi$ are of the form $Ce^x$.
– amsmath
Jul 17 at 21:28
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $f(x)in C_infty(mathbbR)$ and a succession of derivatives $f^(n)(x)$ converges in $C_[a,b]$ to a function $phi(x)$ in each finite interval. Show that $phi(x)=Ce^x$, where $C$ is a constant.
I know that $Ce^xin C_infty$ but I cannot prove $phi(x)=Ce^x$. I think the $max|f_n-f|=max|Ce^x-Ce^x|=0 $ for $n>NinmathbbN$. And I know that $Ce^xin C_[a,b]$, once it is continuous on the $mathbbR$.
Question:
1) How do I prove $phi(x)=Ce^x$? And not for example $sin (x)$?
2) What is intended on this question?
real-analysis functional-analysis
asked Jul 17 at 21:08
Pedro Gomes
1,3192618
Let $f(x)in C_infty(mathbbR)$ and a succession of derivatives $f^(n)(x)$ converges in $C_[a,b]$ to a function $phi(x)$ in each finite interval. Show that $phi(x)=Ce^x$, where $C$ is a constant.
I know that $Ce^xin C_infty$ but I cannot prove $phi(x)=Ce^x$. I think the $max|f_n-f|=max|Ce^x-Ce^x|=0 $ for $n>NinmathbbN$. And I know that $Ce^xin C_[a,b]$, once it is continuous on the $mathbbR$.
Question:
1) How do I prove $phi(x)=Ce^x$? And not for example $sin (x)$?
2) What is intended on this question?
real-analysis functional-analysis
asked Jul 17 at 21:08
Pedro Gomes
1,3192618
asked Jul 17 at 21:08
Pedro Gomes
1,3192618
asked Jul 17 at 21:08
Pedro Gomes
1,3192618
asked Jul 17 at 21:08
Pedro Gomes
1,3192618
1,3192618
1
One possible way might be to let $g(x) = lim_n to inftyf^(n)(x)$. Try to show that $g'(x) = g(x)$. for any $x in mathbbR$. Then show that $g(x)/e^x$ is constant by showing the derivative is zero.
– Sean Haight
Jul 17 at 21:14
Think like this. $phi$ is the limit of the derivatives of $f$. So, when you derive a high derivative, you still get a little closer to $phi$. But when you take $phi$ and derive it, it will be $phi$ itself (which one has to prove). But the only functions $phi$ that satisfy $phi'=phi$ are of the form $Ce^x$.
– amsmath
Jul 17 at 21:28
add a comment |Â
1
One possible way might be to let $g(x) = lim_n to inftyf^(n)(x)$. Try to show that $g'(x) = g(x)$. for any $x in mathbbR$. Then show that $g(x)/e^x$ is constant by showing the derivative is zero.
– Sean Haight
Jul 17 at 21:14
Think like this. $phi$ is the limit of the derivatives of $f$. So, when you derive a high derivative, you still get a little closer to $phi$. But when you take $phi$ and derive it, it will be $phi$ itself (which one has to prove). But the only functions $phi$ that satisfy $phi'=phi$ are of the form $Ce^x$.
– amsmath
Jul 17 at 21:28
1
1
One possible way might be to let $g(x) = lim_n to inftyf^(n)(x)$. Try to show that $g'(x) = g(x)$. for any $x in mathbbR$. Then show that $g(x)/e^x$ is constant by showing the derivative is zero.
– Sean Haight
Jul 17 at 21:14
One possible way might be to let $g(x) = lim_n to inftyf^(n)(x)$. Try to show that $g'(x) = g(x)$. for any $x in mathbbR$. Then show that $g(x)/e^x$ is constant by showing the derivative is zero.
– Sean Haight
Jul 17 at 21:14
Think like this. $phi$ is the limit of the derivatives of $f$. So, when you derive a high derivative, you still get a little closer to $phi$. But when you take $phi$ and derive it, it will be $phi$ itself (which one has to prove). But the only functions $phi$ that satisfy $phi'=phi$ are of the form $Ce^x$.
– amsmath
Jul 17 at 21:28
Think like this. $phi$ is the limit of the derivatives of $f$. So, when you derive a high derivative, you still get a little closer to $phi$. But when you take $phi$ and derive it, it will be $phi$ itself (which one has to prove). But the only functions $phi$ that satisfy $phi'=phi$ are of the form $Ce^x$.
– amsmath
Jul 17 at 21:28
add a comment |Â
2 Answers
2
active
oldest
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up vote
2
down vote
accepted
Recall this theorem:
Let $I = [a,b]$ and $f_n : I to mathbbR$ a sequence of differentiable functions such that the sequence of derivatives $(f_n')_n$ converges uniformly to a function $g : I to mathbbR$. Also $exists x_0 in I$ such that the sequence $(f_n(x_0))_n$ converges. Then $(f_n)_n$ converges uniformly to a differentiable function $f : I to mathbbR$ with $f' = g$.
This theorem can be extended for functions $f_n : mathbbR to mathbbR$ by restricting the domain to $[-R, R]$, say. We only lose the uniform convergence of $(f_n)_n$ to $f$ on $mathbbR$, but pointwise convergence remains.
Since $f_n' to phi$ and $f_n to phi$ uniformly, we conclude that $(f_n)_n$ converges pointwise to a differentiable function $psi$ such that $psi' = phi$. But we already know that $f_n to phi$ uniformly so necessarily $psi = phi$.
Therefore
$$phi' = phi$$
and the only functions satisfying this are $phi(x) = Ce^x$.
Additional details:
The assumptions from the exercise imply that $f_n to phi$ and $f_n' to phi$ in $C^infty(mathbbR)$, that means uniform convergence over $mathbbR$. Then in particular for any $R>0$ we have $f_n|_[-R, R] to phi|_[-R, R]$ and $f_n'|_[-R, R] to phi|_[-R, R]$ uniformly on $[-R, R]$. In particular there exists $x_0 in [-R, R]$ such that $f_n|_[-R, R](x_0) to phi|_[-R, R](x_0)$. The theorem implies that there exists a differentiable function $psi : [-R, R] to mathbbR$ such that $f_n|_[-R, R](x_0) to psi$ uniformly and $psi' = phi|_[-R, R]$. But already $f_n|_[-R, R] to phi|_[-R, R]$ uniformly so $psi = phi|_[-R, R]$. Hence we conclude $phi'|_[-R, R] = psi' = phi|_[-R, R]$.
Since $R > 0$ was arbitrary, we conclude $phi' = phi$. Now this implies $psi(x) = Ce^x$. See for example here, the simplest argument is differentiating $phi(x)e^-x$:
$$fracddxphi(x)e^-x = phi'(x)e^-x - phi(x)e^-x =phi(x)e^-x - phi(x)e^-x = 0 $$
So $phi(x)e^-x equiv C$ or $phi(x) =Ce^-x$.
answered Jul 17 at 23:03
mechanodroid
22.2k52041
Thanks for your answer! I understood the theorem but I fail to see the connection that you made to the exercise, like $phi(x)=Ce^x$. Could you develop a little bit more! Thanks in advance!
– Pedro Gomes
Jul 18 at 20:48
1
@PedroGomes Have a look now, I added some details.
– mechanodroid
Jul 18 at 21:10
add a comment |Â
up vote
3
down vote
Let $xin mathbbR$. We know that $f^(n)to phi$ uniformly on $[0,x]$. Thus $phi$ is continuous on $[0,x]$, and
$$lim_nto inftyint_0^xf^(n)(t)dt= int_0^xphi(t)dt $$
where swapping integral and limit is allowed by uniform convergence on a compact domain. On the other hand, by fundamental theorem of calculus,
$$lim_nto infty int_0^xf^(n)(t)dt=lim_nto inftyleft[ f^(n-1)(x)-f^(n-1)(0)right] =phi(x)-phi(0)$$
Hence
$$int_0^x phi(t)dt=phi(x)-phi(0) $$
where $phi:mathbbRto mathbbR$ is a continuous function. Now you just need to solve the above equation for $phi$.
answered Jul 17 at 21:25
Lorenzo Quarisa
2,319314
How do I solve the equation for $phi$? Is it supposed to give me $Ce^x$?Thanks in advance!
– Pedro Gomes
Jul 18 at 20:59
Since $phi$ is continuous, by fundamental theorem of calculus the left-hand side of the last equality is differentiable, hence the right-hand side is too and differentiating term by term you get $phi'(x)=phi(x)$ for all $x$. Can you solve this differential equation?
– Lorenzo Quarisa
Jul 18 at 21:35
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Recall this theorem:
Let $I = [a,b]$ and $f_n : I to mathbbR$ a sequence of differentiable functions such that the sequence of derivatives $(f_n')_n$ converges uniformly to a function $g : I to mathbbR$. Also $exists x_0 in I$ such that the sequence $(f_n(x_0))_n$ converges. Then $(f_n)_n$ converges uniformly to a differentiable function $f : I to mathbbR$ with $f' = g$.
This theorem can be extended for functions $f_n : mathbbR to mathbbR$ by restricting the domain to $[-R, R]$, say. We only lose the uniform convergence of $(f_n)_n$ to $f$ on $mathbbR$, but pointwise convergence remains.
Since $f_n' to phi$ and $f_n to phi$ uniformly, we conclude that $(f_n)_n$ converges pointwise to a differentiable function $psi$ such that $psi' = phi$. But we already know that $f_n to phi$ uniformly so necessarily $psi = phi$.
Therefore
$$phi' = phi$$
and the only functions satisfying this are $phi(x) = Ce^x$.
Additional details:
The assumptions from the exercise imply that $f_n to phi$ and $f_n' to phi$ in $C^infty(mathbbR)$, that means uniform convergence over $mathbbR$. Then in particular for any $R>0$ we have $f_n|_[-R, R] to phi|_[-R, R]$ and $f_n'|_[-R, R] to phi|_[-R, R]$ uniformly on $[-R, R]$. In particular there exists $x_0 in [-R, R]$ such that $f_n|_[-R, R](x_0) to phi|_[-R, R](x_0)$. The theorem implies that there exists a differentiable function $psi : [-R, R] to mathbbR$ such that $f_n|_[-R, R](x_0) to psi$ uniformly and $psi' = phi|_[-R, R]$. But already $f_n|_[-R, R] to phi|_[-R, R]$ uniformly so $psi = phi|_[-R, R]$. Hence we conclude $phi'|_[-R, R] = psi' = phi|_[-R, R]$.
Since $R > 0$ was arbitrary, we conclude $phi' = phi$. Now this implies $psi(x) = Ce^x$. See for example here, the simplest argument is differentiating $phi(x)e^-x$:
$$fracddxphi(x)e^-x = phi'(x)e^-x - phi(x)e^-x =phi(x)e^-x - phi(x)e^-x = 0 $$
So $phi(x)e^-x equiv C$ or $phi(x) =Ce^-x$.
answered Jul 17 at 23:03
mechanodroid
22.2k52041
Thanks for your answer! I understood the theorem but I fail to see the connection that you made to the exercise, like $phi(x)=Ce^x$. Could you develop a little bit more! Thanks in advance!
– Pedro Gomes
Jul 18 at 20:48
1
@PedroGomes Have a look now, I added some details.
– mechanodroid
Jul 18 at 21:10
add a comment |Â
up vote
2
down vote
accepted
Recall this theorem:
Let $I = [a,b]$ and $f_n : I to mathbbR$ a sequence of differentiable functions such that the sequence of derivatives $(f_n')_n$ converges uniformly to a function $g : I to mathbbR$. Also $exists x_0 in I$ such that the sequence $(f_n(x_0))_n$ converges. Then $(f_n)_n$ converges uniformly to a differentiable function $f : I to mathbbR$ with $f' = g$.
This theorem can be extended for functions $f_n : mathbbR to mathbbR$ by restricting the domain to $[-R, R]$, say. We only lose the uniform convergence of $(f_n)_n$ to $f$ on $mathbbR$, but pointwise convergence remains.
Since $f_n' to phi$ and $f_n to phi$ uniformly, we conclude that $(f_n)_n$ converges pointwise to a differentiable function $psi$ such that $psi' = phi$. But we already know that $f_n to phi$ uniformly so necessarily $psi = phi$.
Therefore
$$phi' = phi$$
and the only functions satisfying this are $phi(x) = Ce^x$.
Additional details:
The assumptions from the exercise imply that $f_n to phi$ and $f_n' to phi$ in $C^infty(mathbbR)$, that means uniform convergence over $mathbbR$. Then in particular for any $R>0$ we have $f_n|_[-R, R] to phi|_[-R, R]$ and $f_n'|_[-R, R] to phi|_[-R, R]$ uniformly on $[-R, R]$. In particular there exists $x_0 in [-R, R]$ such that $f_n|_[-R, R](x_0) to phi|_[-R, R](x_0)$. The theorem implies that there exists a differentiable function $psi : [-R, R] to mathbbR$ such that $f_n|_[-R, R](x_0) to psi$ uniformly and $psi' = phi|_[-R, R]$. But already $f_n|_[-R, R] to phi|_[-R, R]$ uniformly so $psi = phi|_[-R, R]$. Hence we conclude $phi'|_[-R, R] = psi' = phi|_[-R, R]$.
Since $R > 0$ was arbitrary, we conclude $phi' = phi$. Now this implies $psi(x) = Ce^x$. See for example here, the simplest argument is differentiating $phi(x)e^-x$:
$$fracddxphi(x)e^-x = phi'(x)e^-x - phi(x)e^-x =phi(x)e^-x - phi(x)e^-x = 0 $$
So $phi(x)e^-x equiv C$ or $phi(x) =Ce^-x$.
answered Jul 17 at 23:03
mechanodroid
22.2k52041
Thanks for your answer! I understood the theorem but I fail to see the connection that you made to the exercise, like $phi(x)=Ce^x$. Could you develop a little bit more! Thanks in advance!
– Pedro Gomes
Jul 18 at 20:48
1
@PedroGomes Have a look now, I added some details.
– mechanodroid
Jul 18 at 21:10
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Recall this theorem:
Let $I = [a,b]$ and $f_n : I to mathbbR$ a sequence of differentiable functions such that the sequence of derivatives $(f_n')_n$ converges uniformly to a function $g : I to mathbbR$. Also $exists x_0 in I$ such that the sequence $(f_n(x_0))_n$ converges. Then $(f_n)_n$ converges uniformly to a differentiable function $f : I to mathbbR$ with $f' = g$.
This theorem can be extended for functions $f_n : mathbbR to mathbbR$ by restricting the domain to $[-R, R]$, say. We only lose the uniform convergence of $(f_n)_n$ to $f$ on $mathbbR$, but pointwise convergence remains.
Since $f_n' to phi$ and $f_n to phi$ uniformly, we conclude that $(f_n)_n$ converges pointwise to a differentiable function $psi$ such that $psi' = phi$. But we already know that $f_n to phi$ uniformly so necessarily $psi = phi$.
Therefore
$$phi' = phi$$
and the only functions satisfying this are $phi(x) = Ce^x$.
Additional details:
The assumptions from the exercise imply that $f_n to phi$ and $f_n' to phi$ in $C^infty(mathbbR)$, that means uniform convergence over $mathbbR$. Then in particular for any $R>0$ we have $f_n|_[-R, R] to phi|_[-R, R]$ and $f_n'|_[-R, R] to phi|_[-R, R]$ uniformly on $[-R, R]$. In particular there exists $x_0 in [-R, R]$ such that $f_n|_[-R, R](x_0) to phi|_[-R, R](x_0)$. The theorem implies that there exists a differentiable function $psi : [-R, R] to mathbbR$ such that $f_n|_[-R, R](x_0) to psi$ uniformly and $psi' = phi|_[-R, R]$. But already $f_n|_[-R, R] to phi|_[-R, R]$ uniformly so $psi = phi|_[-R, R]$. Hence we conclude $phi'|_[-R, R] = psi' = phi|_[-R, R]$.
Since $R > 0$ was arbitrary, we conclude $phi' = phi$. Now this implies $psi(x) = Ce^x$. See for example here, the simplest argument is differentiating $phi(x)e^-x$:
$$fracddxphi(x)e^-x = phi'(x)e^-x - phi(x)e^-x =phi(x)e^-x - phi(x)e^-x = 0 $$
So $phi(x)e^-x equiv C$ or $phi(x) =Ce^-x$.
answered Jul 17 at 23:03
mechanodroid
22.2k52041
Recall this theorem:
Let $I = [a,b]$ and $f_n : I to mathbbR$ a sequence of differentiable functions such that the sequence of derivatives $(f_n')_n$ converges uniformly to a function $g : I to mathbbR$. Also $exists x_0 in I$ such that the sequence $(f_n(x_0))_n$ converges. Then $(f_n)_n$ converges uniformly to a differentiable function $f : I to mathbbR$ with $f' = g$.
This theorem can be extended for functions $f_n : mathbbR to mathbbR$ by restricting the domain to $[-R, R]$, say. We only lose the uniform convergence of $(f_n)_n$ to $f$ on $mathbbR$, but pointwise convergence remains.
Since $f_n' to phi$ and $f_n to phi$ uniformly, we conclude that $(f_n)_n$ converges pointwise to a differentiable function $psi$ such that $psi' = phi$. But we already know that $f_n to phi$ uniformly so necessarily $psi = phi$.
Therefore
$$phi' = phi$$
and the only functions satisfying this are $phi(x) = Ce^x$.
Additional details:
The assumptions from the exercise imply that $f_n to phi$ and $f_n' to phi$ in $C^infty(mathbbR)$, that means uniform convergence over $mathbbR$. Then in particular for any $R>0$ we have $f_n|_[-R, R] to phi|_[-R, R]$ and $f_n'|_[-R, R] to phi|_[-R, R]$ uniformly on $[-R, R]$. In particular there exists $x_0 in [-R, R]$ such that $f_n|_[-R, R](x_0) to phi|_[-R, R](x_0)$. The theorem implies that there exists a differentiable function $psi : [-R, R] to mathbbR$ such that $f_n|_[-R, R](x_0) to psi$ uniformly and $psi' = phi|_[-R, R]$. But already $f_n|_[-R, R] to phi|_[-R, R]$ uniformly so $psi = phi|_[-R, R]$. Hence we conclude $phi'|_[-R, R] = psi' = phi|_[-R, R]$.
Since $R > 0$ was arbitrary, we conclude $phi' = phi$. Now this implies $psi(x) = Ce^x$. See for example here, the simplest argument is differentiating $phi(x)e^-x$:
$$fracddxphi(x)e^-x = phi'(x)e^-x - phi(x)e^-x =phi(x)e^-x - phi(x)e^-x = 0 $$
So $phi(x)e^-x equiv C$ or $phi(x) =Ce^-x$.
answered Jul 17 at 23:03
mechanodroid
22.2k52041
edited Jul 18 at 21:10
answered Jul 17 at 23:03
mechanodroid
22.2k52041
answered Jul 17 at 23:03
mechanodroid
22.2k52041
answered Jul 17 at 23:03
mechanodroid
22.2k52041
22.2k52041
Thanks for your answer! I understood the theorem but I fail to see the connection that you made to the exercise, like $phi(x)=Ce^x$. Could you develop a little bit more! Thanks in advance!
– Pedro Gomes
Jul 18 at 20:48
1
@PedroGomes Have a look now, I added some details.
– mechanodroid
Jul 18 at 21:10
add a comment |Â
Thanks for your answer! I understood the theorem but I fail to see the connection that you made to the exercise, like $phi(x)=Ce^x$. Could you develop a little bit more! Thanks in advance!
– Pedro Gomes
Jul 18 at 20:48
1
@PedroGomes Have a look now, I added some details.
– mechanodroid
Jul 18 at 21:10
Thanks for your answer! I understood the theorem but I fail to see the connection that you made to the exercise, like $phi(x)=Ce^x$. Could you develop a little bit more! Thanks in advance!
– Pedro Gomes
Jul 18 at 20:48
Thanks for your answer! I understood the theorem but I fail to see the connection that you made to the exercise, like $phi(x)=Ce^x$. Could you develop a little bit more! Thanks in advance!
– Pedro Gomes
Jul 18 at 20:48
1
1
@PedroGomes Have a look now, I added some details.
– mechanodroid
Jul 18 at 21:10
@PedroGomes Have a look now, I added some details.
– mechanodroid
Jul 18 at 21:10
add a comment |Â
up vote
3
down vote
Let $xin mathbbR$. We know that $f^(n)to phi$ uniformly on $[0,x]$. Thus $phi$ is continuous on $[0,x]$, and
$$lim_nto inftyint_0^xf^(n)(t)dt= int_0^xphi(t)dt $$
where swapping integral and limit is allowed by uniform convergence on a compact domain. On the other hand, by fundamental theorem of calculus,
$$lim_nto infty int_0^xf^(n)(t)dt=lim_nto inftyleft[ f^(n-1)(x)-f^(n-1)(0)right] =phi(x)-phi(0)$$
Hence
$$int_0^x phi(t)dt=phi(x)-phi(0) $$
where $phi:mathbbRto mathbbR$ is a continuous function. Now you just need to solve the above equation for $phi$.
answered Jul 17 at 21:25
Lorenzo Quarisa
2,319314
How do I solve the equation for $phi$? Is it supposed to give me $Ce^x$?Thanks in advance!
– Pedro Gomes
Jul 18 at 20:59
Since $phi$ is continuous, by fundamental theorem of calculus the left-hand side of the last equality is differentiable, hence the right-hand side is too and differentiating term by term you get $phi'(x)=phi(x)$ for all $x$. Can you solve this differential equation?
– Lorenzo Quarisa
Jul 18 at 21:35
add a comment |Â
up vote
3
down vote
Let $xin mathbbR$. We know that $f^(n)to phi$ uniformly on $[0,x]$. Thus $phi$ is continuous on $[0,x]$, and
$$lim_nto inftyint_0^xf^(n)(t)dt= int_0^xphi(t)dt $$
where swapping integral and limit is allowed by uniform convergence on a compact domain. On the other hand, by fundamental theorem of calculus,
$$lim_nto infty int_0^xf^(n)(t)dt=lim_nto inftyleft[ f^(n-1)(x)-f^(n-1)(0)right] =phi(x)-phi(0)$$
Hence
$$int_0^x phi(t)dt=phi(x)-phi(0) $$
where $phi:mathbbRto mathbbR$ is a continuous function. Now you just need to solve the above equation for $phi$.
answered Jul 17 at 21:25
Lorenzo Quarisa
2,319314
How do I solve the equation for $phi$? Is it supposed to give me $Ce^x$?Thanks in advance!
– Pedro Gomes
Jul 18 at 20:59
Since $phi$ is continuous, by fundamental theorem of calculus the left-hand side of the last equality is differentiable, hence the right-hand side is too and differentiating term by term you get $phi'(x)=phi(x)$ for all $x$. Can you solve this differential equation?
– Lorenzo Quarisa
Jul 18 at 21:35
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $xin mathbbR$. We know that $f^(n)to phi$ uniformly on $[0,x]$. Thus $phi$ is continuous on $[0,x]$, and
$$lim_nto inftyint_0^xf^(n)(t)dt= int_0^xphi(t)dt $$
where swapping integral and limit is allowed by uniform convergence on a compact domain. On the other hand, by fundamental theorem of calculus,
$$lim_nto infty int_0^xf^(n)(t)dt=lim_nto inftyleft[ f^(n-1)(x)-f^(n-1)(0)right] =phi(x)-phi(0)$$
Hence
$$int_0^x phi(t)dt=phi(x)-phi(0) $$
where $phi:mathbbRto mathbbR$ is a continuous function. Now you just need to solve the above equation for $phi$.
answered Jul 17 at 21:25
Lorenzo Quarisa
2,319314
Let $xin mathbbR$. We know that $f^(n)to phi$ uniformly on $[0,x]$. Thus $phi$ is continuous on $[0,x]$, and
$$lim_nto inftyint_0^xf^(n)(t)dt= int_0^xphi(t)dt $$
where swapping integral and limit is allowed by uniform convergence on a compact domain. On the other hand, by fundamental theorem of calculus,
$$lim_nto infty int_0^xf^(n)(t)dt=lim_nto inftyleft[ f^(n-1)(x)-f^(n-1)(0)right] =phi(x)-phi(0)$$
Hence
$$int_0^x phi(t)dt=phi(x)-phi(0) $$
where $phi:mathbbRto mathbbR$ is a continuous function. Now you just need to solve the above equation for $phi$.
answered Jul 17 at 21:25
Lorenzo Quarisa
2,319314
answered Jul 17 at 21:25
Lorenzo Quarisa
2,319314
answered Jul 17 at 21:25
Lorenzo Quarisa
2,319314
answered Jul 17 at 21:25
Lorenzo Quarisa
2,319314
2,319314
How do I solve the equation for $phi$? Is it supposed to give me $Ce^x$?Thanks in advance!
– Pedro Gomes
Jul 18 at 20:59
Since $phi$ is continuous, by fundamental theorem of calculus the left-hand side of the last equality is differentiable, hence the right-hand side is too and differentiating term by term you get $phi'(x)=phi(x)$ for all $x$. Can you solve this differential equation?
– Lorenzo Quarisa
Jul 18 at 21:35
add a comment |Â
How do I solve the equation for $phi$? Is it supposed to give me $Ce^x$?Thanks in advance!
– Pedro Gomes
Jul 18 at 20:59
Since $phi$ is continuous, by fundamental theorem of calculus the left-hand side of the last equality is differentiable, hence the right-hand side is too and differentiating term by term you get $phi'(x)=phi(x)$ for all $x$. Can you solve this differential equation?
– Lorenzo Quarisa
Jul 18 at 21:35
How do I solve the equation for $phi$? Is it supposed to give me $Ce^x$?Thanks in advance!
– Pedro Gomes
Jul 18 at 20:59
How do I solve the equation for $phi$? Is it supposed to give me $Ce^x$?Thanks in advance!
– Pedro Gomes
Jul 18 at 20:59
Since $phi$ is continuous, by fundamental theorem of calculus the left-hand side of the last equality is differentiable, hence the right-hand side is too and differentiating term by term you get $phi'(x)=phi(x)$ for all $x$. Can you solve this differential equation?
– Lorenzo Quarisa
Jul 18 at 21:35
Since $phi$ is continuous, by fundamental theorem of calculus the left-hand side of the last equality is differentiable, hence the right-hand side is too and differentiating term by term you get $phi'(x)=phi(x)$ for all $x$. Can you solve this differential equation?
– Lorenzo Quarisa
Jul 18 at 21:35
add a comment |Â
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1
One possible way might be to let $g(x) = lim_n to inftyf^(n)(x)$. Try to show that $g'(x) = g(x)$. for any $x in mathbbR$. Then show that $g(x)/e^x$ is constant by showing the derivative is zero.
– Sean Haight
Jul 17 at 21:14
Think like this. $phi$ is the limit of the derivatives of $f$. So, when you derive a high derivative, you still get a little closer to $phi$. But when you take $phi$ and derive it, it will be $phi$ itself (which one has to prove). But the only functions $phi$ that satisfy $phi'=phi$ are of the form $Ce^x$.
– amsmath
Jul 17 at 21:28