Mixing
Definition for real analytic 'function'
$begingroup$
I'm studying analytic functions out of Rudin PMA and the wikipedia article, but I'm not clear on analytic functions. Rudin says a function $f: mathbb{R} rightarrow mathbb{R}$ is analytic at $x = a$ if there exists ${c_n} subset mathbb{R}$ and $R > 0$ such that $f(x) = sum c_{n}(x - a)^n$ for all $|x - a| < R$.
Now, from Rudin I can see that a function $f$ is real analytic on an open set $D subset mathbb{R}$ if it has a power series expansion for each $a in D$. But what do we talk about if we say a function $f$ is ''real analytic''. Like the definition for continuity of a function, I assume this means that a power series expansion may be found for $f$ at every point in its domain. But Rudin doesn't say this explicitly, so I look at Wikipedia.
''The reciprocal of an analytic function that is nowhere zero is analytic''. For the moment, I believe this statement to be false. Consider $f(x) = frac{1}{1-x}$, which is the reciprocal of the analytic function $1-x$. Also $1-x$ is nowhere zero if we don't allow $x = 1$ in the domain. From the basic geometric series formula, we have $f(x) = sum_{n=0}^{infty}x^{n}$ if and only if $|x| < 1$. We can't express $f(x)$ as a power series if $|x| geq 1$. In particular, $2$ belongs tot he domain of $f$ but there is no power series in a neighborhood of $2$. So $f$ is not real analytic at every $x$ in the domain of $f$, meaning $f$ is not real analytic.
I appreciate all feedback. Thanks.
real-analysis
asked Sep 23 '16 at 14:16
user55912user55912
995
$endgroup$
add a comment |
$begingroup$
I'm studying analytic functions out of Rudin PMA and the wikipedia article, but I'm not clear on analytic functions. Rudin says a function $f: mathbb{R} rightarrow mathbb{R}$ is analytic at $x = a$ if there exists ${c_n} subset mathbb{R}$ and $R > 0$ such that $f(x) = sum c_{n}(x - a)^n$ for all $|x - a| < R$.
Now, from Rudin I can see that a function $f$ is real analytic on an open set $D subset mathbb{R}$ if it has a power series expansion for each $a in D$. But what do we talk about if we say a function $f$ is ''real analytic''. Like the definition for continuity of a function, I assume this means that a power series expansion may be found for $f$ at every point in its domain. But Rudin doesn't say this explicitly, so I look at Wikipedia.
''The reciprocal of an analytic function that is nowhere zero is analytic''. For the moment, I believe this statement to be false. Consider $f(x) = frac{1}{1-x}$, which is the reciprocal of the analytic function $1-x$. Also $1-x$ is nowhere zero if we don't allow $x = 1$ in the domain. From the basic geometric series formula, we have $f(x) = sum_{n=0}^{infty}x^{n}$ if and only if $|x| < 1$. We can't express $f(x)$ as a power series if $|x| geq 1$. In particular, $2$ belongs tot he domain of $f$ but there is no power series in a neighborhood of $2$. So $f$ is not real analytic at every $x$ in the domain of $f$, meaning $f$ is not real analytic.
I appreciate all feedback. Thanks.
real-analysis
asked Sep 23 '16 at 14:16
user55912user55912
995
$endgroup$
4
$begingroup$
Yes, you can develop $dfrac{1}{1-x}$ in series around any $x_0$ with $|x_0|>1$. It is said nowhere that it has to be the same series everywhere. The development $dfrac{1}{1-x}=sum_{ngeq0}x^n$ is around $0$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:20
4
$begingroup$
For instance, as you choose $2$, you have, for $x$ in a neighborhood of $2$: $$dfrac{1}{1-x}=dfrac{1}{-1-(x-2)}=dfrac{-1}{1+(x-2)}=-sum_{n=0}^{infty}(-1)^n(x-2)^n$$ Its radius of convergence is $1$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:27
$begingroup$
Okay, well how do we develop a series for some $|x_0| > 1$? Perhaps we look at $f(x) = frac{1}{1-x} = frac{1}{(1 - x_0) - (x - x_0)} = frac{1}{1 - x_0} frac{1}{1 - frac{x - x_0}{1 - x_0}}$ and do some geometric series argument for the right-hand side?
$endgroup$
– user55912
Sep 23 '16 at 14:32
1
$begingroup$
Exactly. An interesting point: the radius of this series is $|x_0-1|$, or exactly the distance from $x_0$ to $1$, the pole closest (actually unique here) to $x_0$. See this.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:35
add a comment |
$begingroup$
I'm studying analytic functions out of Rudin PMA and the wikipedia article, but I'm not clear on analytic functions. Rudin says a function $f: mathbb{R} rightarrow mathbb{R}$ is analytic at $x = a$ if there exists ${c_n} subset mathbb{R}$ and $R > 0$ such that $f(x) = sum c_{n}(x - a)^n$ for all $|x - a| < R$.
Now, from Rudin I can see that a function $f$ is real analytic on an open set $D subset mathbb{R}$ if it has a power series expansion for each $a in D$. But what do we talk about if we say a function $f$ is ''real analytic''. Like the definition for continuity of a function, I assume this means that a power series expansion may be found for $f$ at every point in its domain. But Rudin doesn't say this explicitly, so I look at Wikipedia.
''The reciprocal of an analytic function that is nowhere zero is analytic''. For the moment, I believe this statement to be false. Consider $f(x) = frac{1}{1-x}$, which is the reciprocal of the analytic function $1-x$. Also $1-x$ is nowhere zero if we don't allow $x = 1$ in the domain. From the basic geometric series formula, we have $f(x) = sum_{n=0}^{infty}x^{n}$ if and only if $|x| < 1$. We can't express $f(x)$ as a power series if $|x| geq 1$. In particular, $2$ belongs tot he domain of $f$ but there is no power series in a neighborhood of $2$. So $f$ is not real analytic at every $x$ in the domain of $f$, meaning $f$ is not real analytic.
I appreciate all feedback. Thanks.
real-analysis
asked Sep 23 '16 at 14:16
user55912user55912
995
$endgroup$
I'm studying analytic functions out of Rudin PMA and the wikipedia article, but I'm not clear on analytic functions. Rudin says a function $f: mathbb{R} rightarrow mathbb{R}$ is analytic at $x = a$ if there exists ${c_n} subset mathbb{R}$ and $R > 0$ such that $f(x) = sum c_{n}(x - a)^n$ for all $|x - a| < R$.
Now, from Rudin I can see that a function $f$ is real analytic on an open set $D subset mathbb{R}$ if it has a power series expansion for each $a in D$. But what do we talk about if we say a function $f$ is ''real analytic''. Like the definition for continuity of a function, I assume this means that a power series expansion may be found for $f$ at every point in its domain. But Rudin doesn't say this explicitly, so I look at Wikipedia.
''The reciprocal of an analytic function that is nowhere zero is analytic''. For the moment, I believe this statement to be false. Consider $f(x) = frac{1}{1-x}$, which is the reciprocal of the analytic function $1-x$. Also $1-x$ is nowhere zero if we don't allow $x = 1$ in the domain. From the basic geometric series formula, we have $f(x) = sum_{n=0}^{infty}x^{n}$ if and only if $|x| < 1$. We can't express $f(x)$ as a power series if $|x| geq 1$. In particular, $2$ belongs tot he domain of $f$ but there is no power series in a neighborhood of $2$. So $f$ is not real analytic at every $x$ in the domain of $f$, meaning $f$ is not real analytic.
I appreciate all feedback. Thanks.
real-analysis
real-analysis
asked Sep 23 '16 at 14:16
user55912user55912
995
asked Sep 23 '16 at 14:16
user55912user55912
995
asked Sep 23 '16 at 14:16
user55912user55912
995
asked Sep 23 '16 at 14:16
user55912user55912
995
asked Sep 23 '16 at 14:16
user55912user55912
995
995
4
$begingroup$
Yes, you can develop $dfrac{1}{1-x}$ in series around any $x_0$ with $|x_0|>1$. It is said nowhere that it has to be the same series everywhere. The development $dfrac{1}{1-x}=sum_{ngeq0}x^n$ is around $0$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:20
4
$begingroup$
For instance, as you choose $2$, you have, for $x$ in a neighborhood of $2$: $$dfrac{1}{1-x}=dfrac{1}{-1-(x-2)}=dfrac{-1}{1+(x-2)}=-sum_{n=0}^{infty}(-1)^n(x-2)^n$$ Its radius of convergence is $1$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:27
$begingroup$
Okay, well how do we develop a series for some $|x_0| > 1$? Perhaps we look at $f(x) = frac{1}{1-x} = frac{1}{(1 - x_0) - (x - x_0)} = frac{1}{1 - x_0} frac{1}{1 - frac{x - x_0}{1 - x_0}}$ and do some geometric series argument for the right-hand side?
$endgroup$
– user55912
Sep 23 '16 at 14:32
1
$begingroup$
Exactly. An interesting point: the radius of this series is $|x_0-1|$, or exactly the distance from $x_0$ to $1$, the pole closest (actually unique here) to $x_0$. See this.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:35
add a comment |
4
$begingroup$
Yes, you can develop $dfrac{1}{1-x}$ in series around any $x_0$ with $|x_0|>1$. It is said nowhere that it has to be the same series everywhere. The development $dfrac{1}{1-x}=sum_{ngeq0}x^n$ is around $0$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:20
4
$begingroup$
For instance, as you choose $2$, you have, for $x$ in a neighborhood of $2$: $$dfrac{1}{1-x}=dfrac{1}{-1-(x-2)}=dfrac{-1}{1+(x-2)}=-sum_{n=0}^{infty}(-1)^n(x-2)^n$$ Its radius of convergence is $1$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:27
$begingroup$
Okay, well how do we develop a series for some $|x_0| > 1$? Perhaps we look at $f(x) = frac{1}{1-x} = frac{1}{(1 - x_0) - (x - x_0)} = frac{1}{1 - x_0} frac{1}{1 - frac{x - x_0}{1 - x_0}}$ and do some geometric series argument for the right-hand side?
$endgroup$
– user55912
Sep 23 '16 at 14:32
1
$begingroup$
Exactly. An interesting point: the radius of this series is $|x_0-1|$, or exactly the distance from $x_0$ to $1$, the pole closest (actually unique here) to $x_0$. See this.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:35
4
4
$begingroup$
Yes, you can develop $dfrac{1}{1-x}$ in series around any $x_0$ with $|x_0|>1$. It is said nowhere that it has to be the same series everywhere. The development $dfrac{1}{1-x}=sum_{ngeq0}x^n$ is around $0$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:20
$begingroup$
Yes, you can develop $dfrac{1}{1-x}$ in series around any $x_0$ with $|x_0|>1$. It is said nowhere that it has to be the same series everywhere. The development $dfrac{1}{1-x}=sum_{ngeq0}x^n$ is around $0$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:20
4
4
$begingroup$
For instance, as you choose $2$, you have, for $x$ in a neighborhood of $2$: $$dfrac{1}{1-x}=dfrac{1}{-1-(x-2)}=dfrac{-1}{1+(x-2)}=-sum_{n=0}^{infty}(-1)^n(x-2)^n$$ Its radius of convergence is $1$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:27
$begingroup$
For instance, as you choose $2$, you have, for $x$ in a neighborhood of $2$: $$dfrac{1}{1-x}=dfrac{1}{-1-(x-2)}=dfrac{-1}{1+(x-2)}=-sum_{n=0}^{infty}(-1)^n(x-2)^n$$ Its radius of convergence is $1$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:27
$begingroup$
Okay, well how do we develop a series for some $|x_0| > 1$? Perhaps we look at $f(x) = frac{1}{1-x} = frac{1}{(1 - x_0) - (x - x_0)} = frac{1}{1 - x_0} frac{1}{1 - frac{x - x_0}{1 - x_0}}$ and do some geometric series argument for the right-hand side?
$endgroup$
– user55912
Sep 23 '16 at 14:32
$begingroup$
Okay, well how do we develop a series for some $|x_0| > 1$? Perhaps we look at $f(x) = frac{1}{1-x} = frac{1}{(1 - x_0) - (x - x_0)} = frac{1}{1 - x_0} frac{1}{1 - frac{x - x_0}{1 - x_0}}$ and do some geometric series argument for the right-hand side?
$endgroup$
– user55912
Sep 23 '16 at 14:32
1
1
$begingroup$
Exactly. An interesting point: the radius of this series is $|x_0-1|$, or exactly the distance from $x_0$ to $1$, the pole closest (actually unique here) to $x_0$. See this.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:35
$begingroup$
Exactly. An interesting point: the radius of this series is $|x_0-1|$, or exactly the distance from $x_0$ to $1$, the pole closest (actually unique here) to $x_0$. See this.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:35
add a comment |
1 Answer
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$begingroup$
Basically for a (real) function $f$ to be analytic, it has a power series on some neighborhood. So to quote what you have your power series is $sum c_n(x-a)^n$ and neighborhood $x in (-a + R, a + R).$ You might think, well surely this ought to be true if the function is smooth right? No, look at this classical example
$$f(x) = left{begin{matrix}
e^{-1/x} & x > 0\
0& x leq 0
end{matrix}right.$$
Now $forall n geq 0$, $f^n(0) = 0$. Hence this is not real analytic as it cannot be equal to its own Taylor series.
answered Apr 13 '18 at 6:56
IAmNoOneIAmNoOne
2,63541221
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add a comment |
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$begingroup$
Basically for a (real) function $f$ to be analytic, it has a power series on some neighborhood. So to quote what you have your power series is $sum c_n(x-a)^n$ and neighborhood $x in (-a + R, a + R).$ You might think, well surely this ought to be true if the function is smooth right? No, look at this classical example
$$f(x) = left{begin{matrix}
e^{-1/x} & x > 0\
0& x leq 0
end{matrix}right.$$
Now $forall n geq 0$, $f^n(0) = 0$. Hence this is not real analytic as it cannot be equal to its own Taylor series.
answered Apr 13 '18 at 6:56
IAmNoOneIAmNoOne
2,63541221
$endgroup$
add a comment |
$begingroup$
Basically for a (real) function $f$ to be analytic, it has a power series on some neighborhood. So to quote what you have your power series is $sum c_n(x-a)^n$ and neighborhood $x in (-a + R, a + R).$ You might think, well surely this ought to be true if the function is smooth right? No, look at this classical example
$$f(x) = left{begin{matrix}
e^{-1/x} & x > 0\
0& x leq 0
end{matrix}right.$$
Now $forall n geq 0$, $f^n(0) = 0$. Hence this is not real analytic as it cannot be equal to its own Taylor series.
answered Apr 13 '18 at 6:56
IAmNoOneIAmNoOne
2,63541221
$endgroup$
add a comment |
$begingroup$
Basically for a (real) function $f$ to be analytic, it has a power series on some neighborhood. So to quote what you have your power series is $sum c_n(x-a)^n$ and neighborhood $x in (-a + R, a + R).$ You might think, well surely this ought to be true if the function is smooth right? No, look at this classical example
$$f(x) = left{begin{matrix}
e^{-1/x} & x > 0\
0& x leq 0
end{matrix}right.$$
Now $forall n geq 0$, $f^n(0) = 0$. Hence this is not real analytic as it cannot be equal to its own Taylor series.
answered Apr 13 '18 at 6:56
IAmNoOneIAmNoOne
2,63541221
$endgroup$
Basically for a (real) function $f$ to be analytic, it has a power series on some neighborhood. So to quote what you have your power series is $sum c_n(x-a)^n$ and neighborhood $x in (-a + R, a + R).$ You might think, well surely this ought to be true if the function is smooth right? No, look at this classical example
$$f(x) = left{begin{matrix}
e^{-1/x} & x > 0\
0& x leq 0
end{matrix}right.$$
Now $forall n geq 0$, $f^n(0) = 0$. Hence this is not real analytic as it cannot be equal to its own Taylor series.
answered Apr 13 '18 at 6:56
IAmNoOneIAmNoOne
2,63541221
edited Apr 13 '18 at 19:45
answered Apr 13 '18 at 6:56
IAmNoOneIAmNoOne
2,63541221
answered Apr 13 '18 at 6:56
IAmNoOneIAmNoOne
2,63541221
answered Apr 13 '18 at 6:56
IAmNoOneIAmNoOne
2,63541221
2,63541221
add a comment |
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$begingroup$
Yes, you can develop $dfrac{1}{1-x}$ in series around any $x_0$ with $|x_0|>1$. It is said nowhere that it has to be the same series everywhere. The development $dfrac{1}{1-x}=sum_{ngeq0}x^n$ is around $0$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:20
4
$begingroup$
For instance, as you choose $2$, you have, for $x$ in a neighborhood of $2$: $$dfrac{1}{1-x}=dfrac{1}{-1-(x-2)}=dfrac{-1}{1+(x-2)}=-sum_{n=0}^{infty}(-1)^n(x-2)^n$$ Its radius of convergence is $1$.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:27
$begingroup$
Okay, well how do we develop a series for some $|x_0| > 1$? Perhaps we look at $f(x) = frac{1}{1-x} = frac{1}{(1 - x_0) - (x - x_0)} = frac{1}{1 - x_0} frac{1}{1 - frac{x - x_0}{1 - x_0}}$ and do some geometric series argument for the right-hand side?
$endgroup$
– user55912
Sep 23 '16 at 14:32
1
$begingroup$
Exactly. An interesting point: the radius of this series is $|x_0-1|$, or exactly the distance from $x_0$ to $1$, the pole closest (actually unique here) to $x_0$. See this.
$endgroup$
– Jean-Claude Arbaut
Sep 23 '16 at 14:35