Show that $f(x) = exp(-1/x^2) quad |x| > 0,$ and $f^(k)(0) = 0 quad$ for$ quad k = 0,1,…$ is not analytic The Next CEO of Stack OverflowCounterexample: For real functions existence of all higher order derivatives doesn't imply analycity.Functions and derivatives with rational and irrational values at the integersnot following a step in ash and novinger example of analytic but does not have primitiveConstructing neighbourhoods for analytic functionsEvery differentiable function is infinitely differentiable?convex, smooth but not analyticIs the function $x|x| + mathrmi y|y|$ analytic?Show that there is a neighborhood around $0$ such that $f(z) neq 0$.Is there any function that is differentiable at z_0 but not analytic at the same point?Confusion about relationship between continuous and real analytic functions
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Show that $f(x) = exp(-1/x^2) quad |x| > 0,$ and $f^(k)(0) = 0 quad$ for$ quad k = 0,1,…$ is not analytic
The Next CEO of Stack OverflowCounterexample: For real functions existence of all higher order derivatives doesn't imply analycity.Functions and derivatives with rational and irrational values at the integersnot following a step in ash and novinger example of analytic but does not have primitiveConstructing neighbourhoods for analytic functionsEvery differentiable function is infinitely differentiable?convex, smooth but not analyticIs the function $x|x| + mathrmi y|y|$ analytic?Show that there is a neighborhood around $0$ such that $f(z) neq 0$.Is there any function that is differentiable at z_0 but not analytic at the same point?Confusion about relationship between continuous and real analytic functions
$begingroup$
In the book of Markushevich, Complex Analysis, at page 10, question 1.5, it is asked to verify that the function
$$f(x) = exp(-1/x^2) quad |x| > 0,$$and
$$f^(k)(0) = 0 quadtextfor quad k = 0,1,...$$
is infinitely differentiable, but not analytic.
It also states that $f$ does not satisfy
$$|f^k(x)| < M k!$$
in a neighbourhood of $0$, but as far as I can see, the function $f^1(x)$ and $f^2$ takes zero values around $0$, and for practical reason, I cannot check the higher derivatives, so I would say that $f$ is analytic around zero, but the book states that it is not.
So, how to show that the function does not satisfy the give inequality above ?
real-analysis complex-analysis
edited Mar 28 at 9:11
Bernard
123k741118
asked Mar 28 at 9:05
onurcanbektasonurcanbektas
3,47911037
$endgroup$
add a comment |
$begingroup$
In the book of Markushevich, Complex Analysis, at page 10, question 1.5, it is asked to verify that the function
$$f(x) = exp(-1/x^2) quad |x| > 0,$$and
$$f^(k)(0) = 0 quadtextfor quad k = 0,1,...$$
is infinitely differentiable, but not analytic.
It also states that $f$ does not satisfy
$$|f^k(x)| < M k!$$
in a neighbourhood of $0$, but as far as I can see, the function $f^1(x)$ and $f^2$ takes zero values around $0$, and for practical reason, I cannot check the higher derivatives, so I would say that $f$ is analytic around zero, but the book states that it is not.
So, how to show that the function does not satisfy the give inequality above ?
real-analysis complex-analysis
edited Mar 28 at 9:11
Bernard
123k741118
asked Mar 28 at 9:05
onurcanbektasonurcanbektas
3,47911037
$endgroup$
$begingroup$
What do you mean by “takes zero values around $0$”?
$endgroup$
– MPW
Mar 28 at 9:15
$begingroup$
The function is not analytic in $0$ as you should see from the result to be proven. This is a well known example.
$endgroup$
– Kezer
Mar 28 at 9:24
add a comment |
$begingroup$
In the book of Markushevich, Complex Analysis, at page 10, question 1.5, it is asked to verify that the function
$$f(x) = exp(-1/x^2) quad |x| > 0,$$and
$$f^(k)(0) = 0 quadtextfor quad k = 0,1,...$$
is infinitely differentiable, but not analytic.
It also states that $f$ does not satisfy
$$|f^k(x)| < M k!$$
in a neighbourhood of $0$, but as far as I can see, the function $f^1(x)$ and $f^2$ takes zero values around $0$, and for practical reason, I cannot check the higher derivatives, so I would say that $f$ is analytic around zero, but the book states that it is not.
So, how to show that the function does not satisfy the give inequality above ?
real-analysis complex-analysis
edited Mar 28 at 9:11
Bernard
123k741118
asked Mar 28 at 9:05
onurcanbektasonurcanbektas
3,47911037
$endgroup$
In the book of Markushevich, Complex Analysis, at page 10, question 1.5, it is asked to verify that the function
$$f(x) = exp(-1/x^2) quad |x| > 0,$$and
$$f^(k)(0) = 0 quadtextfor quad k = 0,1,...$$
is infinitely differentiable, but not analytic.
It also states that $f$ does not satisfy
$$|f^k(x)| < M k!$$
in a neighbourhood of $0$, but as far as I can see, the function $f^1(x)$ and $f^2$ takes zero values around $0$, and for practical reason, I cannot check the higher derivatives, so I would say that $f$ is analytic around zero, but the book states that it is not.
So, how to show that the function does not satisfy the give inequality above ?
real-analysis complex-analysis
real-analysis complex-analysis
edited Mar 28 at 9:11
Bernard
123k741118
asked Mar 28 at 9:05
onurcanbektasonurcanbektas
3,47911037
edited Mar 28 at 9:11
Bernard
123k741118
asked Mar 28 at 9:05
onurcanbektasonurcanbektas
3,47911037
edited Mar 28 at 9:11
Bernard
123k741118
edited Mar 28 at 9:11
Bernard
123k741118
edited Mar 28 at 9:11
Bernard
123k741118
123k741118
asked Mar 28 at 9:05
onurcanbektasonurcanbektas
3,47911037
asked Mar 28 at 9:05
onurcanbektasonurcanbektas
3,47911037
asked Mar 28 at 9:05
onurcanbektasonurcanbektas
3,47911037
3,47911037
$begingroup$
What do you mean by “takes zero values around $0$”?
$endgroup$
– MPW
Mar 28 at 9:15
$begingroup$
The function is not analytic in $0$ as you should see from the result to be proven. This is a well known example.
$endgroup$
– Kezer
Mar 28 at 9:24
add a comment |
$begingroup$
What do you mean by “takes zero values around $0$”?
$endgroup$
– MPW
Mar 28 at 9:15
$begingroup$
The function is not analytic in $0$ as you should see from the result to be proven. This is a well known example.
$endgroup$
– Kezer
Mar 28 at 9:24
$begingroup$
What do you mean by “takes zero values around $0$”?
$endgroup$
– MPW
Mar 28 at 9:15
$begingroup$
What do you mean by “takes zero values around $0$”?
$endgroup$
– MPW
Mar 28 at 9:15
$begingroup$
The function is not analytic in $0$ as you should see from the result to be proven. This is a well known example.
$endgroup$
– Kezer
Mar 28 at 9:24
$begingroup$
The function is not analytic in $0$ as you should see from the result to be proven. This is a well known example.
$endgroup$
– Kezer
Mar 28 at 9:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have to prove two things.
First:
You need to prove that $f$ is infinitelly differentiable. Clearly, the only problem can occur around $0$. To prove that the function is differentiable, you can prove that, for each $k$, the $k$-th derivative of $f$ for nonzero $x$ is equal to $Q(x)cdot e^-frac1x^2$ where $Q$ is some rational function. This can easily be done by induction. It should also be clear that $$lim_xto0 Q(x)cdot e^-frac1x^2 = 0$$ since $$lim_xto 0frace^-frac1x^2x^n = 0$$ for all values of $n$.
Second:
You need to prove $f$ is not infinitelly differentiable. A function is analytic iff its power series is convergent. However, for your $f$, the power series around $0$ is
$$sum_n=0^infty xcdotfracf^(n)(0)n!= sum_n=0^infty xcdot frac0n! = sum_n=0^infty 0 = 0$$
and there is no neighborhood of $0$ at which the power series converges and is equal to $f$. Mind you, the power series converges on $mathbb R$, it just doesn't converge to $f$. For small nonzero values of $x$, the value of $f(x)$ is not equal to $0$.
answered Mar 28 at 9:12
5xum5xum
91.8k394161
$endgroup$
add a comment |
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oldest
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$begingroup$
You have to prove two things.
First:
You need to prove that $f$ is infinitelly differentiable. Clearly, the only problem can occur around $0$. To prove that the function is differentiable, you can prove that, for each $k$, the $k$-th derivative of $f$ for nonzero $x$ is equal to $Q(x)cdot e^-frac1x^2$ where $Q$ is some rational function. This can easily be done by induction. It should also be clear that $$lim_xto0 Q(x)cdot e^-frac1x^2 = 0$$ since $$lim_xto 0frace^-frac1x^2x^n = 0$$ for all values of $n$.
Second:
You need to prove $f$ is not infinitelly differentiable. A function is analytic iff its power series is convergent. However, for your $f$, the power series around $0$ is
$$sum_n=0^infty xcdotfracf^(n)(0)n!= sum_n=0^infty xcdot frac0n! = sum_n=0^infty 0 = 0$$
and there is no neighborhood of $0$ at which the power series converges and is equal to $f$. Mind you, the power series converges on $mathbb R$, it just doesn't converge to $f$. For small nonzero values of $x$, the value of $f(x)$ is not equal to $0$.
answered Mar 28 at 9:12
5xum5xum
91.8k394161
$endgroup$
add a comment |
$begingroup$
You have to prove two things.
First:
You need to prove that $f$ is infinitelly differentiable. Clearly, the only problem can occur around $0$. To prove that the function is differentiable, you can prove that, for each $k$, the $k$-th derivative of $f$ for nonzero $x$ is equal to $Q(x)cdot e^-frac1x^2$ where $Q$ is some rational function. This can easily be done by induction. It should also be clear that $$lim_xto0 Q(x)cdot e^-frac1x^2 = 0$$ since $$lim_xto 0frace^-frac1x^2x^n = 0$$ for all values of $n$.
Second:
You need to prove $f$ is not infinitelly differentiable. A function is analytic iff its power series is convergent. However, for your $f$, the power series around $0$ is
$$sum_n=0^infty xcdotfracf^(n)(0)n!= sum_n=0^infty xcdot frac0n! = sum_n=0^infty 0 = 0$$
and there is no neighborhood of $0$ at which the power series converges and is equal to $f$. Mind you, the power series converges on $mathbb R$, it just doesn't converge to $f$. For small nonzero values of $x$, the value of $f(x)$ is not equal to $0$.
answered Mar 28 at 9:12
5xum5xum
91.8k394161
$endgroup$
add a comment |
$begingroup$
You have to prove two things.
First:
You need to prove that $f$ is infinitelly differentiable. Clearly, the only problem can occur around $0$. To prove that the function is differentiable, you can prove that, for each $k$, the $k$-th derivative of $f$ for nonzero $x$ is equal to $Q(x)cdot e^-frac1x^2$ where $Q$ is some rational function. This can easily be done by induction. It should also be clear that $$lim_xto0 Q(x)cdot e^-frac1x^2 = 0$$ since $$lim_xto 0frace^-frac1x^2x^n = 0$$ for all values of $n$.
Second:
You need to prove $f$ is not infinitelly differentiable. A function is analytic iff its power series is convergent. However, for your $f$, the power series around $0$ is
$$sum_n=0^infty xcdotfracf^(n)(0)n!= sum_n=0^infty xcdot frac0n! = sum_n=0^infty 0 = 0$$
and there is no neighborhood of $0$ at which the power series converges and is equal to $f$. Mind you, the power series converges on $mathbb R$, it just doesn't converge to $f$. For small nonzero values of $x$, the value of $f(x)$ is not equal to $0$.
answered Mar 28 at 9:12
5xum5xum
91.8k394161
$endgroup$
You have to prove two things.
First:
You need to prove that $f$ is infinitelly differentiable. Clearly, the only problem can occur around $0$. To prove that the function is differentiable, you can prove that, for each $k$, the $k$-th derivative of $f$ for nonzero $x$ is equal to $Q(x)cdot e^-frac1x^2$ where $Q$ is some rational function. This can easily be done by induction. It should also be clear that $$lim_xto0 Q(x)cdot e^-frac1x^2 = 0$$ since $$lim_xto 0frace^-frac1x^2x^n = 0$$ for all values of $n$.
Second:
You need to prove $f$ is not infinitelly differentiable. A function is analytic iff its power series is convergent. However, for your $f$, the power series around $0$ is
$$sum_n=0^infty xcdotfracf^(n)(0)n!= sum_n=0^infty xcdot frac0n! = sum_n=0^infty 0 = 0$$
and there is no neighborhood of $0$ at which the power series converges and is equal to $f$. Mind you, the power series converges on $mathbb R$, it just doesn't converge to $f$. For small nonzero values of $x$, the value of $f(x)$ is not equal to $0$.
answered Mar 28 at 9:12
5xum5xum
91.8k394161
edited Mar 28 at 9:18
answered Mar 28 at 9:12
5xum5xum
91.8k394161
answered Mar 28 at 9:12
5xum5xum
91.8k394161
answered Mar 28 at 9:12
5xum5xum
91.8k394161
91.8k394161
add a comment |
add a comment |
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$begingroup$
What do you mean by “takes zero values around $0$”?
$endgroup$
– MPW
Mar 28 at 9:15
$begingroup$
The function is not analytic in $0$ as you should see from the result to be proven. This is a well known example.
$endgroup$
– Kezer
Mar 28 at 9:24