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What is up with limsup ? Is Wikipedia wrong about their smooth non-Analytic argument?
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In this article on Non-analytic smooth function seems to offer the following argument:
$|F^(n)(x_o)| geq e^-sqrtnn^n + O(q^n) $ implies $ displaystyle limsup_n rightarrow infty left( fracF^(n)(x_o)n! right)^1/n = infty$
where $q = 2^m$ for some positive integer $m$. My initial thought was the implication followed by comparison since
$$ limsup_n rightarrow infty left(frace^-sqrtnn^n + O(q^n)n!right)^1/n = infty qquad (textfalse)$$
But, I believe the limit above actually converges. In particular, this example calculation in wolfram alpha and similar calculations show the $O(q^n)$-type term is irrelevant and the limit of $left(e^-sqrtnn^n/n! right)^1/n rightarrow e$ as $n rightarrow infty$. Since the limsup matches the limit when it exists we cannot conclude the limsup of $left(frace^-sqrtnn^n + O(q^n)n!right)^1/n$ diverges. Therefore, there is no comparison to be made and I don't see how we are able to say anything about $limsup_n rightarrow infty left( fracF^(n)(x_o)n! right)^1/n$.
Question: is Wikipedia wrong here? Show me the error of my ways please.
real-analysis examples-counterexamples limsup-and-liminf
asked Jul 20 at 19:38
James S. Cook
12.8k22668
add a comment |Â
up vote
4
down vote
favorite
In this article on Non-analytic smooth function seems to offer the following argument:
$|F^(n)(x_o)| geq e^-sqrtnn^n + O(q^n) $ implies $ displaystyle limsup_n rightarrow infty left( fracF^(n)(x_o)n! right)^1/n = infty$
where $q = 2^m$ for some positive integer $m$. My initial thought was the implication followed by comparison since
$$ limsup_n rightarrow infty left(frace^-sqrtnn^n + O(q^n)n!right)^1/n = infty qquad (textfalse)$$
But, I believe the limit above actually converges. In particular, this example calculation in wolfram alpha and similar calculations show the $O(q^n)$-type term is irrelevant and the limit of $left(e^-sqrtnn^n/n! right)^1/n rightarrow e$ as $n rightarrow infty$. Since the limsup matches the limit when it exists we cannot conclude the limsup of $left(frace^-sqrtnn^n + O(q^n)n!right)^1/n$ diverges. Therefore, there is no comparison to be made and I don't see how we are able to say anything about $limsup_n rightarrow infty left( fracF^(n)(x_o)n! right)^1/n$.
Question: is Wikipedia wrong here? Show me the error of my ways please.
real-analysis examples-counterexamples limsup-and-liminf
asked Jul 20 at 19:38
James S. Cook
12.8k22668
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In this article on Non-analytic smooth function seems to offer the following argument:
$|F^(n)(x_o)| geq e^-sqrtnn^n + O(q^n) $ implies $ displaystyle limsup_n rightarrow infty left( fracF^(n)(x_o)n! right)^1/n = infty$
where $q = 2^m$ for some positive integer $m$. My initial thought was the implication followed by comparison since
$$ limsup_n rightarrow infty left(frace^-sqrtnn^n + O(q^n)n!right)^1/n = infty qquad (textfalse)$$
But, I believe the limit above actually converges. In particular, this example calculation in wolfram alpha and similar calculations show the $O(q^n)$-type term is irrelevant and the limit of $left(e^-sqrtnn^n/n! right)^1/n rightarrow e$ as $n rightarrow infty$. Since the limsup matches the limit when it exists we cannot conclude the limsup of $left(frace^-sqrtnn^n + O(q^n)n!right)^1/n$ diverges. Therefore, there is no comparison to be made and I don't see how we are able to say anything about $limsup_n rightarrow infty left( fracF^(n)(x_o)n! right)^1/n$.
Question: is Wikipedia wrong here? Show me the error of my ways please.
real-analysis examples-counterexamples limsup-and-liminf
asked Jul 20 at 19:38
James S. Cook
12.8k22668
In this article on Non-analytic smooth function seems to offer the following argument:
$|F^(n)(x_o)| geq e^-sqrtnn^n + O(q^n) $ implies $ displaystyle limsup_n rightarrow infty left( fracF^(n)(x_o)n! right)^1/n = infty$
where $q = 2^m$ for some positive integer $m$. My initial thought was the implication followed by comparison since
$$ limsup_n rightarrow infty left(frace^-sqrtnn^n + O(q^n)n!right)^1/n = infty qquad (textfalse)$$
But, I believe the limit above actually converges. In particular, this example calculation in wolfram alpha and similar calculations show the $O(q^n)$-type term is irrelevant and the limit of $left(e^-sqrtnn^n/n! right)^1/n rightarrow e$ as $n rightarrow infty$. Since the limsup matches the limit when it exists we cannot conclude the limsup of $left(frace^-sqrtnn^n + O(q^n)n!right)^1/n$ diverges. Therefore, there is no comparison to be made and I don't see how we are able to say anything about $limsup_n rightarrow infty left( fracF^(n)(x_o)n! right)^1/n$.
Question: is Wikipedia wrong here? Show me the error of my ways please.
real-analysis examples-counterexamples limsup-and-liminf
asked Jul 20 at 19:38
James S. Cook
12.8k22668
edited Jul 20 at 20:44
asked Jul 20 at 19:38
James S. Cook
12.8k22668
asked Jul 20 at 19:38
James S. Cook
12.8k22668
asked Jul 20 at 19:38
James S. Cook
12.8k22668
12.8k22668
add a comment |Â
add a comment |Â
1 Answer
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That's a mistake in the wikipedia article. The given estimate using the term for $k = n$ only shows
$$limsup_n to infty :biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant e,.$$
To fix it, replace $cos (kx)$ with $cos (k^rx)$ for your favourite $r > 1$ in the definition of $F$.
However, the function $F$ as defined in the wikipedia article is indeed nowhere analytic, as one can see using the term for $k = n^2$ for the estimate. Then we obtain
$$F^(n)(x_0) geqslant e^-nn^2n - Ccdot q^n$$
and that yields
$$biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant ncdot bigl(1 + o(1)bigr),.$$
answered Jul 20 at 20:16
Daniel Fischer♦
171k16154274
Perhaps $r=3$ is a good choice. I checked, wolframalpha.com/input/… and found the modification of $n^n$ to $n^3n$ suffices to make it blow-up. I think the introduction of $r$ as you indicate does not spoil anything else, so, I am cautiously optimistic you have tamed this beast. Many thanks.
– James S. Cook
Jul 20 at 20:50
Every $r > 1$ works, the dominant term becomes $e^-sqrtnn^rn$, and $$biggl(fracn^rne^sqrtnn!biggr)^1/n sim en^r-1,.$$
– Daniel Fischer♦
Jul 20 at 20:57
Thanks, I think I can fix the $r$ for my application, but this freedom is helpful. It means we have a family of such functions parametrized by $r$. Neat.
– James S. Cook
Jul 20 at 20:58
I'm embarrassed to admit that I've been fooled by wikipedia @JamesS.Cook. While their argument is wrong, the function given there works without modification. See my edited answer.
– Daniel Fischer♦
Jul 20 at 22:01
but how can we set $k=n^2$ since $k$ is supposed to be $2^m$ for some $m$. I guess we can choose $m$ such that $2^m approxeq n^2$ for the argument?
– James S. Cook
Jul 21 at 1:34
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
That's a mistake in the wikipedia article. The given estimate using the term for $k = n$ only shows
$$limsup_n to infty :biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant e,.$$
To fix it, replace $cos (kx)$ with $cos (k^rx)$ for your favourite $r > 1$ in the definition of $F$.
However, the function $F$ as defined in the wikipedia article is indeed nowhere analytic, as one can see using the term for $k = n^2$ for the estimate. Then we obtain
$$F^(n)(x_0) geqslant e^-nn^2n - Ccdot q^n$$
and that yields
$$biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant ncdot bigl(1 + o(1)bigr),.$$
answered Jul 20 at 20:16
Daniel Fischer♦
171k16154274
Perhaps $r=3$ is a good choice. I checked, wolframalpha.com/input/… and found the modification of $n^n$ to $n^3n$ suffices to make it blow-up. I think the introduction of $r$ as you indicate does not spoil anything else, so, I am cautiously optimistic you have tamed this beast. Many thanks.
– James S. Cook
Jul 20 at 20:50
Every $r > 1$ works, the dominant term becomes $e^-sqrtnn^rn$, and $$biggl(fracn^rne^sqrtnn!biggr)^1/n sim en^r-1,.$$
– Daniel Fischer♦
Jul 20 at 20:57
Thanks, I think I can fix the $r$ for my application, but this freedom is helpful. It means we have a family of such functions parametrized by $r$. Neat.
– James S. Cook
Jul 20 at 20:58
I'm embarrassed to admit that I've been fooled by wikipedia @JamesS.Cook. While their argument is wrong, the function given there works without modification. See my edited answer.
– Daniel Fischer♦
Jul 20 at 22:01
but how can we set $k=n^2$ since $k$ is supposed to be $2^m$ for some $m$. I guess we can choose $m$ such that $2^m approxeq n^2$ for the argument?
– James S. Cook
Jul 21 at 1:34
 |Â
show 4 more comments
up vote
4
down vote
accepted
That's a mistake in the wikipedia article. The given estimate using the term for $k = n$ only shows
$$limsup_n to infty :biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant e,.$$
To fix it, replace $cos (kx)$ with $cos (k^rx)$ for your favourite $r > 1$ in the definition of $F$.
However, the function $F$ as defined in the wikipedia article is indeed nowhere analytic, as one can see using the term for $k = n^2$ for the estimate. Then we obtain
$$F^(n)(x_0) geqslant e^-nn^2n - Ccdot q^n$$
and that yields
$$biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant ncdot bigl(1 + o(1)bigr),.$$
answered Jul 20 at 20:16
Daniel Fischer♦
171k16154274
Perhaps $r=3$ is a good choice. I checked, wolframalpha.com/input/… and found the modification of $n^n$ to $n^3n$ suffices to make it blow-up. I think the introduction of $r$ as you indicate does not spoil anything else, so, I am cautiously optimistic you have tamed this beast. Many thanks.
– James S. Cook
Jul 20 at 20:50
Every $r > 1$ works, the dominant term becomes $e^-sqrtnn^rn$, and $$biggl(fracn^rne^sqrtnn!biggr)^1/n sim en^r-1,.$$
– Daniel Fischer♦
Jul 20 at 20:57
Thanks, I think I can fix the $r$ for my application, but this freedom is helpful. It means we have a family of such functions parametrized by $r$. Neat.
– James S. Cook
Jul 20 at 20:58
I'm embarrassed to admit that I've been fooled by wikipedia @JamesS.Cook. While their argument is wrong, the function given there works without modification. See my edited answer.
– Daniel Fischer♦
Jul 20 at 22:01
but how can we set $k=n^2$ since $k$ is supposed to be $2^m$ for some $m$. I guess we can choose $m$ such that $2^m approxeq n^2$ for the argument?
– James S. Cook
Jul 21 at 1:34
 |Â
show 4 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
That's a mistake in the wikipedia article. The given estimate using the term for $k = n$ only shows
$$limsup_n to infty :biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant e,.$$
To fix it, replace $cos (kx)$ with $cos (k^rx)$ for your favourite $r > 1$ in the definition of $F$.
However, the function $F$ as defined in the wikipedia article is indeed nowhere analytic, as one can see using the term for $k = n^2$ for the estimate. Then we obtain
$$F^(n)(x_0) geqslant e^-nn^2n - Ccdot q^n$$
and that yields
$$biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant ncdot bigl(1 + o(1)bigr),.$$
answered Jul 20 at 20:16
Daniel Fischer♦
171k16154274
That's a mistake in the wikipedia article. The given estimate using the term for $k = n$ only shows
$$limsup_n to infty :biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant e,.$$
To fix it, replace $cos (kx)$ with $cos (k^rx)$ for your favourite $r > 1$ in the definition of $F$.
However, the function $F$ as defined in the wikipedia article is indeed nowhere analytic, as one can see using the term for $k = n^2$ for the estimate. Then we obtain
$$F^(n)(x_0) geqslant e^-nn^2n - Ccdot q^n$$
and that yields
$$biggl(fraclvert F^(n)(x_0)rvertn!biggr)^frac1n geqslant ncdot bigl(1 + o(1)bigr),.$$
answered Jul 20 at 20:16
Daniel Fischer♦
171k16154274
edited Jul 20 at 21:57
answered Jul 20 at 20:16
Daniel Fischer♦
171k16154274
answered Jul 20 at 20:16
Daniel Fischer♦
171k16154274
answered Jul 20 at 20:16
Daniel Fischer♦
171k16154274
171k16154274
Perhaps $r=3$ is a good choice. I checked, wolframalpha.com/input/… and found the modification of $n^n$ to $n^3n$ suffices to make it blow-up. I think the introduction of $r$ as you indicate does not spoil anything else, so, I am cautiously optimistic you have tamed this beast. Many thanks.
– James S. Cook
Jul 20 at 20:50
Every $r > 1$ works, the dominant term becomes $e^-sqrtnn^rn$, and $$biggl(fracn^rne^sqrtnn!biggr)^1/n sim en^r-1,.$$
– Daniel Fischer♦
Jul 20 at 20:57
Thanks, I think I can fix the $r$ for my application, but this freedom is helpful. It means we have a family of such functions parametrized by $r$. Neat.
– James S. Cook
Jul 20 at 20:58
I'm embarrassed to admit that I've been fooled by wikipedia @JamesS.Cook. While their argument is wrong, the function given there works without modification. See my edited answer.
– Daniel Fischer♦
Jul 20 at 22:01
but how can we set $k=n^2$ since $k$ is supposed to be $2^m$ for some $m$. I guess we can choose $m$ such that $2^m approxeq n^2$ for the argument?
– James S. Cook
Jul 21 at 1:34
 |Â
show 4 more comments
Perhaps $r=3$ is a good choice. I checked, wolframalpha.com/input/… and found the modification of $n^n$ to $n^3n$ suffices to make it blow-up. I think the introduction of $r$ as you indicate does not spoil anything else, so, I am cautiously optimistic you have tamed this beast. Many thanks.
– James S. Cook
Jul 20 at 20:50
Every $r > 1$ works, the dominant term becomes $e^-sqrtnn^rn$, and $$biggl(fracn^rne^sqrtnn!biggr)^1/n sim en^r-1,.$$
– Daniel Fischer♦
Jul 20 at 20:57
Thanks, I think I can fix the $r$ for my application, but this freedom is helpful. It means we have a family of such functions parametrized by $r$. Neat.
– James S. Cook
Jul 20 at 20:58
I'm embarrassed to admit that I've been fooled by wikipedia @JamesS.Cook. While their argument is wrong, the function given there works without modification. See my edited answer.
– Daniel Fischer♦
Jul 20 at 22:01
but how can we set $k=n^2$ since $k$ is supposed to be $2^m$ for some $m$. I guess we can choose $m$ such that $2^m approxeq n^2$ for the argument?
– James S. Cook
Jul 21 at 1:34
Perhaps $r=3$ is a good choice. I checked, wolframalpha.com/input/… and found the modification of $n^n$ to $n^3n$ suffices to make it blow-up. I think the introduction of $r$ as you indicate does not spoil anything else, so, I am cautiously optimistic you have tamed this beast. Many thanks.
– James S. Cook
Jul 20 at 20:50
Perhaps $r=3$ is a good choice. I checked, wolframalpha.com/input/… and found the modification of $n^n$ to $n^3n$ suffices to make it blow-up. I think the introduction of $r$ as you indicate does not spoil anything else, so, I am cautiously optimistic you have tamed this beast. Many thanks.
– James S. Cook
Jul 20 at 20:50
Every $r > 1$ works, the dominant term becomes $e^-sqrtnn^rn$, and $$biggl(fracn^rne^sqrtnn!biggr)^1/n sim en^r-1,.$$
– Daniel Fischer♦
Jul 20 at 20:57
Every $r > 1$ works, the dominant term becomes $e^-sqrtnn^rn$, and $$biggl(fracn^rne^sqrtnn!biggr)^1/n sim en^r-1,.$$
– Daniel Fischer♦
Jul 20 at 20:57
Thanks, I think I can fix the $r$ for my application, but this freedom is helpful. It means we have a family of such functions parametrized by $r$. Neat.
– James S. Cook
Jul 20 at 20:58
Thanks, I think I can fix the $r$ for my application, but this freedom is helpful. It means we have a family of such functions parametrized by $r$. Neat.
– James S. Cook
Jul 20 at 20:58
I'm embarrassed to admit that I've been fooled by wikipedia @JamesS.Cook. While their argument is wrong, the function given there works without modification. See my edited answer.
– Daniel Fischer♦
Jul 20 at 22:01
I'm embarrassed to admit that I've been fooled by wikipedia @JamesS.Cook. While their argument is wrong, the function given there works without modification. See my edited answer.
– Daniel Fischer♦
Jul 20 at 22:01
but how can we set $k=n^2$ since $k$ is supposed to be $2^m$ for some $m$. I guess we can choose $m$ such that $2^m approxeq n^2$ for the argument?
– James S. Cook
Jul 21 at 1:34
but how can we set $k=n^2$ since $k$ is supposed to be $2^m$ for some $m$. I guess we can choose $m$ such that $2^m approxeq n^2$ for the argument?
– James S. Cook
Jul 21 at 1:34
 |Â
show 4 more comments
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