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Why $f(x)=sum_n=0^infty fracf^(n)(0), x^nn! $?
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Let $f:mathbb Rlongrightarrow mathbb R$ a function that is $mathcal C^infty(mathbb R) $. Suppose $$sum_k=0^infty fracf^(k)(0)k!,x^k$$
converge and its radius is $R$. Why $$f(x)=sum_k=0^infty fracf^(k)(0)k!,x^k$$
for all $xin ]-R,R[$ ?
I know that for a fixed $n$, $$f(x)=sum_k=0^nfracf^(k)(0)k!,x^k+o(x^n),$$
in a neighborhood of $0$, but a priori it won't be correct for $x$ too large (take for example the sinus or exponential, we have $$e^x=sum_k=0^infty fracx^kk!,$$
for all $xin mathbb R$, not only in a neighborhood of $O$. So why can we extend the formula to all $xinmathbb R$ ?
Ok, let forget $e^-1/x^2$. Consider for example $sin(x)$.
Why $$sin(x)=sum_k=0^infty frac(-1)^n+1(2n)!x^2n,$$
for all $xinmathbb R$ whereas $$sin(x)=sum_k=0^n frac(-1)^n+1x^2k(2k)!+o(x^2n)$$
in a neighborhood of $0$ only.
sequences-and-series taylor-expansion
asked Jul 21 at 20:38
MSE
1,471315
add a comment |Â
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0
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favorite
Let $f:mathbb Rlongrightarrow mathbb R$ a function that is $mathcal C^infty(mathbb R) $. Suppose $$sum_k=0^infty fracf^(k)(0)k!,x^k$$
converge and its radius is $R$. Why $$f(x)=sum_k=0^infty fracf^(k)(0)k!,x^k$$
for all $xin ]-R,R[$ ?
I know that for a fixed $n$, $$f(x)=sum_k=0^nfracf^(k)(0)k!,x^k+o(x^n),$$
in a neighborhood of $0$, but a priori it won't be correct for $x$ too large (take for example the sinus or exponential, we have $$e^x=sum_k=0^infty fracx^kk!,$$
for all $xin mathbb R$, not only in a neighborhood of $O$. So why can we extend the formula to all $xinmathbb R$ ?
Ok, let forget $e^-1/x^2$. Consider for example $sin(x)$.
Why $$sin(x)=sum_k=0^infty frac(-1)^n+1(2n)!x^2n,$$
for all $xinmathbb R$ whereas $$sin(x)=sum_k=0^n frac(-1)^n+1x^2k(2k)!+o(x^2n)$$
in a neighborhood of $0$ only.
sequences-and-series taylor-expansion
asked Jul 21 at 20:38
MSE
1,471315
I don't understand what is asked. The first question is "Why [...] for all $xin(-R,R)$?" (this is true) The second is "why can we extend the formula to all $xinmathbbR$?" (we cannot) Which of the two is your question?
– Clement C.
Jul 21 at 20:53
1
So... after your edit, your question seems to boil down to understanding what $o(cdot)$ means.
– Clement C.
Jul 21 at 21:02
It mean that $$|sin(x)-sum_k=0^nfrac(-1)^k+1 x^2k(2k)!|leq x^2nvarepsilon(x)$$ with $varepsilon(x)to 0$ when $xto 0$... so ? @ClementC.
– MSE
Jul 21 at 21:06
So it is inherently around a specific point. It cannot be "extended" over the reals, the Landau $o(cdot)$ is by very definition a local statement.
– Clement C.
Jul 21 at 21:08
add a comment |Â
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0
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up vote
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Let $f:mathbb Rlongrightarrow mathbb R$ a function that is $mathcal C^infty(mathbb R) $. Suppose $$sum_k=0^infty fracf^(k)(0)k!,x^k$$
converge and its radius is $R$. Why $$f(x)=sum_k=0^infty fracf^(k)(0)k!,x^k$$
for all $xin ]-R,R[$ ?
I know that for a fixed $n$, $$f(x)=sum_k=0^nfracf^(k)(0)k!,x^k+o(x^n),$$
in a neighborhood of $0$, but a priori it won't be correct for $x$ too large (take for example the sinus or exponential, we have $$e^x=sum_k=0^infty fracx^kk!,$$
for all $xin mathbb R$, not only in a neighborhood of $O$. So why can we extend the formula to all $xinmathbb R$ ?
Ok, let forget $e^-1/x^2$. Consider for example $sin(x)$.
Why $$sin(x)=sum_k=0^infty frac(-1)^n+1(2n)!x^2n,$$
for all $xinmathbb R$ whereas $$sin(x)=sum_k=0^n frac(-1)^n+1x^2k(2k)!+o(x^2n)$$
in a neighborhood of $0$ only.
sequences-and-series taylor-expansion
asked Jul 21 at 20:38
MSE
1,471315
Let $f:mathbb Rlongrightarrow mathbb R$ a function that is $mathcal C^infty(mathbb R) $. Suppose $$sum_k=0^infty fracf^(k)(0)k!,x^k$$
converge and its radius is $R$. Why $$f(x)=sum_k=0^infty fracf^(k)(0)k!,x^k$$
for all $xin ]-R,R[$ ?
I know that for a fixed $n$, $$f(x)=sum_k=0^nfracf^(k)(0)k!,x^k+o(x^n),$$
in a neighborhood of $0$, but a priori it won't be correct for $x$ too large (take for example the sinus or exponential, we have $$e^x=sum_k=0^infty fracx^kk!,$$
for all $xin mathbb R$, not only in a neighborhood of $O$. So why can we extend the formula to all $xinmathbb R$ ?
Ok, let forget $e^-1/x^2$. Consider for example $sin(x)$.
Why $$sin(x)=sum_k=0^infty frac(-1)^n+1(2n)!x^2n,$$
for all $xinmathbb R$ whereas $$sin(x)=sum_k=0^n frac(-1)^n+1x^2k(2k)!+o(x^2n)$$
in a neighborhood of $0$ only.
sequences-and-series taylor-expansion
asked Jul 21 at 20:38
MSE
1,471315
edited Jul 21 at 20:56
asked Jul 21 at 20:38
MSE
1,471315
asked Jul 21 at 20:38
MSE
1,471315
asked Jul 21 at 20:38
MSE
1,471315
1,471315
I don't understand what is asked. The first question is "Why [...] for all $xin(-R,R)$?" (this is true) The second is "why can we extend the formula to all $xinmathbbR$?" (we cannot) Which of the two is your question?
– Clement C.
Jul 21 at 20:53
1
So... after your edit, your question seems to boil down to understanding what $o(cdot)$ means.
– Clement C.
Jul 21 at 21:02
It mean that $$|sin(x)-sum_k=0^nfrac(-1)^k+1 x^2k(2k)!|leq x^2nvarepsilon(x)$$ with $varepsilon(x)to 0$ when $xto 0$... so ? @ClementC.
– MSE
Jul 21 at 21:06
So it is inherently around a specific point. It cannot be "extended" over the reals, the Landau $o(cdot)$ is by very definition a local statement.
– Clement C.
Jul 21 at 21:08
add a comment |Â
I don't understand what is asked. The first question is "Why [...] for all $xin(-R,R)$?" (this is true) The second is "why can we extend the formula to all $xinmathbbR$?" (we cannot) Which of the two is your question?
– Clement C.
Jul 21 at 20:53
1
So... after your edit, your question seems to boil down to understanding what $o(cdot)$ means.
– Clement C.
Jul 21 at 21:02
It mean that $$|sin(x)-sum_k=0^nfrac(-1)^k+1 x^2k(2k)!|leq x^2nvarepsilon(x)$$ with $varepsilon(x)to 0$ when $xto 0$... so ? @ClementC.
– MSE
Jul 21 at 21:06
So it is inherently around a specific point. It cannot be "extended" over the reals, the Landau $o(cdot)$ is by very definition a local statement.
– Clement C.
Jul 21 at 21:08
I don't understand what is asked. The first question is "Why [...] for all $xin(-R,R)$?" (this is true) The second is "why can we extend the formula to all $xinmathbbR$?" (we cannot) Which of the two is your question?
– Clement C.
Jul 21 at 20:53
I don't understand what is asked. The first question is "Why [...] for all $xin(-R,R)$?" (this is true) The second is "why can we extend the formula to all $xinmathbbR$?" (we cannot) Which of the two is your question?
– Clement C.
Jul 21 at 20:53
1
1
So... after your edit, your question seems to boil down to understanding what $o(cdot)$ means.
– Clement C.
Jul 21 at 21:02
So... after your edit, your question seems to boil down to understanding what $o(cdot)$ means.
– Clement C.
Jul 21 at 21:02
It mean that $$|sin(x)-sum_k=0^nfrac(-1)^k+1 x^2k(2k)!|leq x^2nvarepsilon(x)$$ with $varepsilon(x)to 0$ when $xto 0$... so ? @ClementC.
– MSE
Jul 21 at 21:06
It mean that $$|sin(x)-sum_k=0^nfrac(-1)^k+1 x^2k(2k)!|leq x^2nvarepsilon(x)$$ with $varepsilon(x)to 0$ when $xto 0$... so ? @ClementC.
– MSE
Jul 21 at 21:06
So it is inherently around a specific point. It cannot be "extended" over the reals, the Landau $o(cdot)$ is by very definition a local statement.
– Clement C.
Jul 21 at 21:08
So it is inherently around a specific point. It cannot be "extended" over the reals, the Landau $o(cdot)$ is by very definition a local statement.
– Clement C.
Jul 21 at 21:08
add a comment |Â
4 Answers
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2
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This is false. For example, if $f(x)=exp(-1/x^2)$ for $xneq 0$ and $f(0)=0$, then $f$ is $C^infty$ with $f^(k)(0)=0$ for all $k$. In particular, $$sum_k=0^infty fracf^(k)(0)k!,x^k$$ converges for all $x$ since every term is $0$. However, it is not true that $$f(x)=sum_k=0^infty fracf^(k)(0)k!,x^k$$ for any $x$ besides $x=0$.
(It happens to be true for certain functions like $f(x)=e^x$, but you must prove it using particular properties of those functions. How you prove it in the case $f(x)=e^x$ depends on what your definition of $e^x$ is; often $e^x$ is simply defined to be $sumfrac1k!x^k$ so it is true by definition.)
answered Jul 21 at 20:49
Eric Wofsey
162k12189300
So how would you prove it for $e^x$ ? My definition is $(e^x)'=e^x$ and $e^0=1$.
– MSE
Jul 21 at 20:52
One way to prove it is to show that $f(x)=sum_k=0^inftyfracx^kk!$ satisfies those two properties. If you know that there can only be one function satisfying those properties (e.g., by the theory of ODEs), you're done.
– Eric Wofsey
Jul 21 at 20:55
Ok ok. And for an $sin(x)$ ? (sorry I want a general example for more complicated function)
– MSE
Jul 21 at 20:58
Again, that depends on your definition of $sin(x)$.
– Eric Wofsey
Jul 21 at 21:00
My definition of $sin$ ? It's the sinus function... the trigonometric function... take the circle of radius 1, a point $x$ on the circle, then the sinus it's the ordinate... you have 159K pts, you should know what's I'm talking about :-)
– MSE
Jul 21 at 21:03
 |Â
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up vote
2
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This is not true. The function
$$
f(x)=cases0& if $x=0$\e^-1/x^2& otherwise
$$
is $C^infty$, and for any $kinBbb N$ we have $f^(k)(0)=0$, so the series converges for all $x$, but only converges to $f(x)$ for $x=0$.
answered Jul 21 at 20:48
Arthur
98.7k793174
And is there a function s.t. $fin mathcal C^infty (mathbb R)$, $f^(n)(0)neq 0$ for all $n$ but $f(x)neq sum_k=0^infty fracf^(k)(0)k!x^k$ for some $xin (-R,R)$ ?
– MSE
Jul 21 at 21:00
1
@MSE Take this $f$ and add, say, $e^x$ to it. The series you get then converges to $e^x$ rather than $f(x)+e^x$.
– Arthur
Jul 21 at 21:22
@MSE A bit more pathological example from that Arthur gave: $$f(x)=sum_n=1^infty e^-2^n-1cos(2^nx)$$
– Holo
Jul 21 at 21:25
add a comment |Â
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0
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This is true only for what is called analytic functions (sums of power series).
It happens that for functions of a complex variable, being differentiable (in the complex sence) implies infinitely differentiable and even analytic, whereas for functions of a real variable, infinitely differentiable does not imply analytic. The function $mathrm e^-tfrac1x^2$ mentioned in other answers is an example of a $mathcal C^infty$ function which is not analytic.
answered Jul 21 at 20:53
Bernard
110k635103
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It depends on the size of the radius.
For some $sum_k=0^inftya_k,x^k$ you will have $r=lim_ktoinfty|fraca_ka_k+1|$.
Calculating the radius for $e^x$ provides $r=infty$.
But then consider the function $f(x)=frac11-x$. Its MacLaurin series $sum_k=0^inftyx^k$ has a radius of convergence of $r=1$. But that given fraction function clearly is well-defined beyond too.
Just plot the graph of that latter function and you would understand that the series would be well-defined even for $xin [-1,1)$.
--- rk
answered Jul 21 at 20:56
Dr. Richard Klitzing
7486
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is false. For example, if $f(x)=exp(-1/x^2)$ for $xneq 0$ and $f(0)=0$, then $f$ is $C^infty$ with $f^(k)(0)=0$ for all $k$. In particular, $$sum_k=0^infty fracf^(k)(0)k!,x^k$$ converges for all $x$ since every term is $0$. However, it is not true that $$f(x)=sum_k=0^infty fracf^(k)(0)k!,x^k$$ for any $x$ besides $x=0$.
(It happens to be true for certain functions like $f(x)=e^x$, but you must prove it using particular properties of those functions. How you prove it in the case $f(x)=e^x$ depends on what your definition of $e^x$ is; often $e^x$ is simply defined to be $sumfrac1k!x^k$ so it is true by definition.)
answered Jul 21 at 20:49
Eric Wofsey
162k12189300
So how would you prove it for $e^x$ ? My definition is $(e^x)'=e^x$ and $e^0=1$.
– MSE
Jul 21 at 20:52
One way to prove it is to show that $f(x)=sum_k=0^inftyfracx^kk!$ satisfies those two properties. If you know that there can only be one function satisfying those properties (e.g., by the theory of ODEs), you're done.
– Eric Wofsey
Jul 21 at 20:55
Ok ok. And for an $sin(x)$ ? (sorry I want a general example for more complicated function)
– MSE
Jul 21 at 20:58
Again, that depends on your definition of $sin(x)$.
– Eric Wofsey
Jul 21 at 21:00
My definition of $sin$ ? It's the sinus function... the trigonometric function... take the circle of radius 1, a point $x$ on the circle, then the sinus it's the ordinate... you have 159K pts, you should know what's I'm talking about :-)
– MSE
Jul 21 at 21:03
 |Â
show 7 more comments
up vote
2
down vote
accepted
This is false. For example, if $f(x)=exp(-1/x^2)$ for $xneq 0$ and $f(0)=0$, then $f$ is $C^infty$ with $f^(k)(0)=0$ for all $k$. In particular, $$sum_k=0^infty fracf^(k)(0)k!,x^k$$ converges for all $x$ since every term is $0$. However, it is not true that $$f(x)=sum_k=0^infty fracf^(k)(0)k!,x^k$$ for any $x$ besides $x=0$.
(It happens to be true for certain functions like $f(x)=e^x$, but you must prove it using particular properties of those functions. How you prove it in the case $f(x)=e^x$ depends on what your definition of $e^x$ is; often $e^x$ is simply defined to be $sumfrac1k!x^k$ so it is true by definition.)
answered Jul 21 at 20:49
Eric Wofsey
162k12189300
So how would you prove it for $e^x$ ? My definition is $(e^x)'=e^x$ and $e^0=1$.
– MSE
Jul 21 at 20:52
One way to prove it is to show that $f(x)=sum_k=0^inftyfracx^kk!$ satisfies those two properties. If you know that there can only be one function satisfying those properties (e.g., by the theory of ODEs), you're done.
– Eric Wofsey
Jul 21 at 20:55
Ok ok. And for an $sin(x)$ ? (sorry I want a general example for more complicated function)
– MSE
Jul 21 at 20:58
Again, that depends on your definition of $sin(x)$.
– Eric Wofsey
Jul 21 at 21:00
My definition of $sin$ ? It's the sinus function... the trigonometric function... take the circle of radius 1, a point $x$ on the circle, then the sinus it's the ordinate... you have 159K pts, you should know what's I'm talking about :-)
– MSE
Jul 21 at 21:03
 |Â
show 7 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is false. For example, if $f(x)=exp(-1/x^2)$ for $xneq 0$ and $f(0)=0$, then $f$ is $C^infty$ with $f^(k)(0)=0$ for all $k$. In particular, $$sum_k=0^infty fracf^(k)(0)k!,x^k$$ converges for all $x$ since every term is $0$. However, it is not true that $$f(x)=sum_k=0^infty fracf^(k)(0)k!,x^k$$ for any $x$ besides $x=0$.
(It happens to be true for certain functions like $f(x)=e^x$, but you must prove it using particular properties of those functions. How you prove it in the case $f(x)=e^x$ depends on what your definition of $e^x$ is; often $e^x$ is simply defined to be $sumfrac1k!x^k$ so it is true by definition.)
answered Jul 21 at 20:49
Eric Wofsey
162k12189300
This is false. For example, if $f(x)=exp(-1/x^2)$ for $xneq 0$ and $f(0)=0$, then $f$ is $C^infty$ with $f^(k)(0)=0$ for all $k$. In particular, $$sum_k=0^infty fracf^(k)(0)k!,x^k$$ converges for all $x$ since every term is $0$. However, it is not true that $$f(x)=sum_k=0^infty fracf^(k)(0)k!,x^k$$ for any $x$ besides $x=0$.
(It happens to be true for certain functions like $f(x)=e^x$, but you must prove it using particular properties of those functions. How you prove it in the case $f(x)=e^x$ depends on what your definition of $e^x$ is; often $e^x$ is simply defined to be $sumfrac1k!x^k$ so it is true by definition.)
answered Jul 21 at 20:49
Eric Wofsey
162k12189300
answered Jul 21 at 20:49
Eric Wofsey
162k12189300
answered Jul 21 at 20:49
Eric Wofsey
162k12189300
answered Jul 21 at 20:49
Eric Wofsey
162k12189300
162k12189300
So how would you prove it for $e^x$ ? My definition is $(e^x)'=e^x$ and $e^0=1$.
– MSE
Jul 21 at 20:52
One way to prove it is to show that $f(x)=sum_k=0^inftyfracx^kk!$ satisfies those two properties. If you know that there can only be one function satisfying those properties (e.g., by the theory of ODEs), you're done.
– Eric Wofsey
Jul 21 at 20:55
Ok ok. And for an $sin(x)$ ? (sorry I want a general example for more complicated function)
– MSE
Jul 21 at 20:58
Again, that depends on your definition of $sin(x)$.
– Eric Wofsey
Jul 21 at 21:00
My definition of $sin$ ? It's the sinus function... the trigonometric function... take the circle of radius 1, a point $x$ on the circle, then the sinus it's the ordinate... you have 159K pts, you should know what's I'm talking about :-)
– MSE
Jul 21 at 21:03
 |Â
show 7 more comments
So how would you prove it for $e^x$ ? My definition is $(e^x)'=e^x$ and $e^0=1$.
– MSE
Jul 21 at 20:52
One way to prove it is to show that $f(x)=sum_k=0^inftyfracx^kk!$ satisfies those two properties. If you know that there can only be one function satisfying those properties (e.g., by the theory of ODEs), you're done.
– Eric Wofsey
Jul 21 at 20:55
Ok ok. And for an $sin(x)$ ? (sorry I want a general example for more complicated function)
– MSE
Jul 21 at 20:58
Again, that depends on your definition of $sin(x)$.
– Eric Wofsey
Jul 21 at 21:00
My definition of $sin$ ? It's the sinus function... the trigonometric function... take the circle of radius 1, a point $x$ on the circle, then the sinus it's the ordinate... you have 159K pts, you should know what's I'm talking about :-)
– MSE
Jul 21 at 21:03
So how would you prove it for $e^x$ ? My definition is $(e^x)'=e^x$ and $e^0=1$.
– MSE
Jul 21 at 20:52
So how would you prove it for $e^x$ ? My definition is $(e^x)'=e^x$ and $e^0=1$.
– MSE
Jul 21 at 20:52
One way to prove it is to show that $f(x)=sum_k=0^inftyfracx^kk!$ satisfies those two properties. If you know that there can only be one function satisfying those properties (e.g., by the theory of ODEs), you're done.
– Eric Wofsey
Jul 21 at 20:55
One way to prove it is to show that $f(x)=sum_k=0^inftyfracx^kk!$ satisfies those two properties. If you know that there can only be one function satisfying those properties (e.g., by the theory of ODEs), you're done.
– Eric Wofsey
Jul 21 at 20:55
Ok ok. And for an $sin(x)$ ? (sorry I want a general example for more complicated function)
– MSE
Jul 21 at 20:58
Ok ok. And for an $sin(x)$ ? (sorry I want a general example for more complicated function)
– MSE
Jul 21 at 20:58
Again, that depends on your definition of $sin(x)$.
– Eric Wofsey
Jul 21 at 21:00
Again, that depends on your definition of $sin(x)$.
– Eric Wofsey
Jul 21 at 21:00
My definition of $sin$ ? It's the sinus function... the trigonometric function... take the circle of radius 1, a point $x$ on the circle, then the sinus it's the ordinate... you have 159K pts, you should know what's I'm talking about :-)
– MSE
Jul 21 at 21:03
My definition of $sin$ ? It's the sinus function... the trigonometric function... take the circle of radius 1, a point $x$ on the circle, then the sinus it's the ordinate... you have 159K pts, you should know what's I'm talking about :-)
– MSE
Jul 21 at 21:03
 |Â
show 7 more comments
up vote
2
down vote
This is not true. The function
$$
f(x)=cases0& if $x=0$\e^-1/x^2& otherwise
$$
is $C^infty$, and for any $kinBbb N$ we have $f^(k)(0)=0$, so the series converges for all $x$, but only converges to $f(x)$ for $x=0$.
answered Jul 21 at 20:48
Arthur
98.7k793174
And is there a function s.t. $fin mathcal C^infty (mathbb R)$, $f^(n)(0)neq 0$ for all $n$ but $f(x)neq sum_k=0^infty fracf^(k)(0)k!x^k$ for some $xin (-R,R)$ ?
– MSE
Jul 21 at 21:00
1
@MSE Take this $f$ and add, say, $e^x$ to it. The series you get then converges to $e^x$ rather than $f(x)+e^x$.
– Arthur
Jul 21 at 21:22
@MSE A bit more pathological example from that Arthur gave: $$f(x)=sum_n=1^infty e^-2^n-1cos(2^nx)$$
– Holo
Jul 21 at 21:25
add a comment |Â
up vote
2
down vote
This is not true. The function
$$
f(x)=cases0& if $x=0$\e^-1/x^2& otherwise
$$
is $C^infty$, and for any $kinBbb N$ we have $f^(k)(0)=0$, so the series converges for all $x$, but only converges to $f(x)$ for $x=0$.
answered Jul 21 at 20:48
Arthur
98.7k793174
And is there a function s.t. $fin mathcal C^infty (mathbb R)$, $f^(n)(0)neq 0$ for all $n$ but $f(x)neq sum_k=0^infty fracf^(k)(0)k!x^k$ for some $xin (-R,R)$ ?
– MSE
Jul 21 at 21:00
1
@MSE Take this $f$ and add, say, $e^x$ to it. The series you get then converges to $e^x$ rather than $f(x)+e^x$.
– Arthur
Jul 21 at 21:22
@MSE A bit more pathological example from that Arthur gave: $$f(x)=sum_n=1^infty e^-2^n-1cos(2^nx)$$
– Holo
Jul 21 at 21:25
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is not true. The function
$$
f(x)=cases0& if $x=0$\e^-1/x^2& otherwise
$$
is $C^infty$, and for any $kinBbb N$ we have $f^(k)(0)=0$, so the series converges for all $x$, but only converges to $f(x)$ for $x=0$.
answered Jul 21 at 20:48
Arthur
98.7k793174
This is not true. The function
$$
f(x)=cases0& if $x=0$\e^-1/x^2& otherwise
$$
is $C^infty$, and for any $kinBbb N$ we have $f^(k)(0)=0$, so the series converges for all $x$, but only converges to $f(x)$ for $x=0$.
answered Jul 21 at 20:48
Arthur
98.7k793174
answered Jul 21 at 20:48
Arthur
98.7k793174
answered Jul 21 at 20:48
Arthur
98.7k793174
answered Jul 21 at 20:48
Arthur
98.7k793174
98.7k793174
And is there a function s.t. $fin mathcal C^infty (mathbb R)$, $f^(n)(0)neq 0$ for all $n$ but $f(x)neq sum_k=0^infty fracf^(k)(0)k!x^k$ for some $xin (-R,R)$ ?
– MSE
Jul 21 at 21:00
1
@MSE Take this $f$ and add, say, $e^x$ to it. The series you get then converges to $e^x$ rather than $f(x)+e^x$.
– Arthur
Jul 21 at 21:22
@MSE A bit more pathological example from that Arthur gave: $$f(x)=sum_n=1^infty e^-2^n-1cos(2^nx)$$
– Holo
Jul 21 at 21:25
add a comment |Â
And is there a function s.t. $fin mathcal C^infty (mathbb R)$, $f^(n)(0)neq 0$ for all $n$ but $f(x)neq sum_k=0^infty fracf^(k)(0)k!x^k$ for some $xin (-R,R)$ ?
– MSE
Jul 21 at 21:00
1
@MSE Take this $f$ and add, say, $e^x$ to it. The series you get then converges to $e^x$ rather than $f(x)+e^x$.
– Arthur
Jul 21 at 21:22
@MSE A bit more pathological example from that Arthur gave: $$f(x)=sum_n=1^infty e^-2^n-1cos(2^nx)$$
– Holo
Jul 21 at 21:25
And is there a function s.t. $fin mathcal C^infty (mathbb R)$, $f^(n)(0)neq 0$ for all $n$ but $f(x)neq sum_k=0^infty fracf^(k)(0)k!x^k$ for some $xin (-R,R)$ ?
– MSE
Jul 21 at 21:00
And is there a function s.t. $fin mathcal C^infty (mathbb R)$, $f^(n)(0)neq 0$ for all $n$ but $f(x)neq sum_k=0^infty fracf^(k)(0)k!x^k$ for some $xin (-R,R)$ ?
– MSE
Jul 21 at 21:00
1
1
@MSE Take this $f$ and add, say, $e^x$ to it. The series you get then converges to $e^x$ rather than $f(x)+e^x$.
– Arthur
Jul 21 at 21:22
@MSE Take this $f$ and add, say, $e^x$ to it. The series you get then converges to $e^x$ rather than $f(x)+e^x$.
– Arthur
Jul 21 at 21:22
@MSE A bit more pathological example from that Arthur gave: $$f(x)=sum_n=1^infty e^-2^n-1cos(2^nx)$$
– Holo
Jul 21 at 21:25
@MSE A bit more pathological example from that Arthur gave: $$f(x)=sum_n=1^infty e^-2^n-1cos(2^nx)$$
– Holo
Jul 21 at 21:25
add a comment |Â
up vote
0
down vote
This is true only for what is called analytic functions (sums of power series).
It happens that for functions of a complex variable, being differentiable (in the complex sence) implies infinitely differentiable and even analytic, whereas for functions of a real variable, infinitely differentiable does not imply analytic. The function $mathrm e^-tfrac1x^2$ mentioned in other answers is an example of a $mathcal C^infty$ function which is not analytic.
answered Jul 21 at 20:53
Bernard
110k635103
add a comment |Â
up vote
0
down vote
This is true only for what is called analytic functions (sums of power series).
It happens that for functions of a complex variable, being differentiable (in the complex sence) implies infinitely differentiable and even analytic, whereas for functions of a real variable, infinitely differentiable does not imply analytic. The function $mathrm e^-tfrac1x^2$ mentioned in other answers is an example of a $mathcal C^infty$ function which is not analytic.
answered Jul 21 at 20:53
Bernard
110k635103
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is true only for what is called analytic functions (sums of power series).
It happens that for functions of a complex variable, being differentiable (in the complex sence) implies infinitely differentiable and even analytic, whereas for functions of a real variable, infinitely differentiable does not imply analytic. The function $mathrm e^-tfrac1x^2$ mentioned in other answers is an example of a $mathcal C^infty$ function which is not analytic.
answered Jul 21 at 20:53
Bernard
110k635103
This is true only for what is called analytic functions (sums of power series).
It happens that for functions of a complex variable, being differentiable (in the complex sence) implies infinitely differentiable and even analytic, whereas for functions of a real variable, infinitely differentiable does not imply analytic. The function $mathrm e^-tfrac1x^2$ mentioned in other answers is an example of a $mathcal C^infty$ function which is not analytic.
answered Jul 21 at 20:53
Bernard
110k635103
edited Jul 21 at 21:30
answered Jul 21 at 20:53
Bernard
110k635103
answered Jul 21 at 20:53
Bernard
110k635103
answered Jul 21 at 20:53
Bernard
110k635103
110k635103
add a comment |Â
add a comment |Â
up vote
-2
down vote
It depends on the size of the radius.
For some $sum_k=0^inftya_k,x^k$ you will have $r=lim_ktoinfty|fraca_ka_k+1|$.
Calculating the radius for $e^x$ provides $r=infty$.
But then consider the function $f(x)=frac11-x$. Its MacLaurin series $sum_k=0^inftyx^k$ has a radius of convergence of $r=1$. But that given fraction function clearly is well-defined beyond too.
Just plot the graph of that latter function and you would understand that the series would be well-defined even for $xin [-1,1)$.
--- rk
answered Jul 21 at 20:56
Dr. Richard Klitzing
7486
add a comment |Â
up vote
-2
down vote
It depends on the size of the radius.
For some $sum_k=0^inftya_k,x^k$ you will have $r=lim_ktoinfty|fraca_ka_k+1|$.
Calculating the radius for $e^x$ provides $r=infty$.
But then consider the function $f(x)=frac11-x$. Its MacLaurin series $sum_k=0^inftyx^k$ has a radius of convergence of $r=1$. But that given fraction function clearly is well-defined beyond too.
Just plot the graph of that latter function and you would understand that the series would be well-defined even for $xin [-1,1)$.
--- rk
answered Jul 21 at 20:56
Dr. Richard Klitzing
7486
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
It depends on the size of the radius.
For some $sum_k=0^inftya_k,x^k$ you will have $r=lim_ktoinfty|fraca_ka_k+1|$.
Calculating the radius for $e^x$ provides $r=infty$.
But then consider the function $f(x)=frac11-x$. Its MacLaurin series $sum_k=0^inftyx^k$ has a radius of convergence of $r=1$. But that given fraction function clearly is well-defined beyond too.
Just plot the graph of that latter function and you would understand that the series would be well-defined even for $xin [-1,1)$.
--- rk
answered Jul 21 at 20:56
Dr. Richard Klitzing
7486
It depends on the size of the radius.
For some $sum_k=0^inftya_k,x^k$ you will have $r=lim_ktoinfty|fraca_ka_k+1|$.
Calculating the radius for $e^x$ provides $r=infty$.
But then consider the function $f(x)=frac11-x$. Its MacLaurin series $sum_k=0^inftyx^k$ has a radius of convergence of $r=1$. But that given fraction function clearly is well-defined beyond too.
Just plot the graph of that latter function and you would understand that the series would be well-defined even for $xin [-1,1)$.
--- rk
answered Jul 21 at 20:56
Dr. Richard Klitzing
7486
edited Jul 21 at 21:03
answered Jul 21 at 20:56
Dr. Richard Klitzing
7486
answered Jul 21 at 20:56
Dr. Richard Klitzing
7486
answered Jul 21 at 20:56
Dr. Richard Klitzing
7486
7486
add a comment |Â
add a comment |Â
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I don't understand what is asked. The first question is "Why [...] for all $xin(-R,R)$?" (this is true) The second is "why can we extend the formula to all $xinmathbbR$?" (we cannot) Which of the two is your question?
– Clement C.
Jul 21 at 20:53
1
So... after your edit, your question seems to boil down to understanding what $o(cdot)$ means.
– Clement C.
Jul 21 at 21:02
It mean that $$|sin(x)-sum_k=0^nfrac(-1)^k+1 x^2k(2k)!|leq x^2nvarepsilon(x)$$ with $varepsilon(x)to 0$ when $xto 0$... so ? @ClementC.
– MSE
Jul 21 at 21:06
So it is inherently around a specific point. It cannot be "extended" over the reals, the Landau $o(cdot)$ is by very definition a local statement.
– Clement C.
Jul 21 at 21:08