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Stone-Weierstrass theorem means every function has a power serie expansion
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I know that every function doesn't have a power series expansion.
Yet what I don't understand is that for every $C^infty$ functions there is a sequence of polynomial $(P_n)$ such that $P_n$ converges uniformly to $f$. That's to say :
$$forall x in [a,b], f(x) = lim_n to infty sum_k = 0^infty a_k,nx^k$$
But then because it converges uniformly why can't I say that :
$$forall x in [a,b], f(x) = sum_k = 0^infty lim_n to infty a_k,nx^k$$
And so $f$ has a power series expansion with coefficients: $lim_n to infty a_k,nx^k$.
calculus real-analysis power-series
edited 3 hours ago
Bernard
113k636105
asked 3 hours ago
auhasard
929
add a comment |Â
up vote
5
down vote
favorite
I know that every function doesn't have a power series expansion.
Yet what I don't understand is that for every $C^infty$ functions there is a sequence of polynomial $(P_n)$ such that $P_n$ converges uniformly to $f$. That's to say :
$$forall x in [a,b], f(x) = lim_n to infty sum_k = 0^infty a_k,nx^k$$
But then because it converges uniformly why can't I say that :
$$forall x in [a,b], f(x) = sum_k = 0^infty lim_n to infty a_k,nx^k$$
And so $f$ has a power series expansion with coefficients: $lim_n to infty a_k,nx^k$.
calculus real-analysis power-series
edited 3 hours ago
Bernard
113k636105
asked 3 hours ago
auhasard
929
3
Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
– Kavi Rama Murthy
3 hours ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I know that every function doesn't have a power series expansion.
Yet what I don't understand is that for every $C^infty$ functions there is a sequence of polynomial $(P_n)$ such that $P_n$ converges uniformly to $f$. That's to say :
$$forall x in [a,b], f(x) = lim_n to infty sum_k = 0^infty a_k,nx^k$$
But then because it converges uniformly why can't I say that :
$$forall x in [a,b], f(x) = sum_k = 0^infty lim_n to infty a_k,nx^k$$
And so $f$ has a power series expansion with coefficients: $lim_n to infty a_k,nx^k$.
calculus real-analysis power-series
edited 3 hours ago
Bernard
113k636105
asked 3 hours ago
auhasard
929
I know that every function doesn't have a power series expansion.
Yet what I don't understand is that for every $C^infty$ functions there is a sequence of polynomial $(P_n)$ such that $P_n$ converges uniformly to $f$. That's to say :
$$forall x in [a,b], f(x) = lim_n to infty sum_k = 0^infty a_k,nx^k$$
But then because it converges uniformly why can't I say that :
$$forall x in [a,b], f(x) = sum_k = 0^infty lim_n to infty a_k,nx^k$$
And so $f$ has a power series expansion with coefficients: $lim_n to infty a_k,nx^k$.
calculus real-analysis power-series
calculus real-analysis power-series
edited 3 hours ago
Bernard
113k636105
asked 3 hours ago
auhasard
929
edited 3 hours ago
Bernard
113k636105
asked 3 hours ago
auhasard
929
edited 3 hours ago
Bernard
113k636105
edited 3 hours ago
Bernard
113k636105
edited 3 hours ago
Bernard
113k636105
113k636105
asked 3 hours ago
auhasard
929
asked 3 hours ago
auhasard
929
asked 3 hours ago
auhasard
929
929
3
Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
– Kavi Rama Murthy
3 hours ago
add a comment |Â
3
Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
– Kavi Rama Murthy
3 hours ago
3
3
Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
– Kavi Rama Murthy
3 hours ago
Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
– Kavi Rama Murthy
3 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
$$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.
answered 3 hours ago
5xum
85.2k388154
Yes, but because I have the uniform convergence the switch of limits should work?
– auhasard
3 hours ago
@auhasard It should? Why?
– 5xum
3 hours ago
add a comment |Â
up vote
0
down vote
What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?
answered 21 mins ago
hartkp
68134
add a comment |Â
up vote
0
down vote
Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.
answered 18 mins ago
Ethan Bolker
37.2k54299
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.
answered 3 hours ago
5xum
85.2k388154
Yes, but because I have the uniform convergence the switch of limits should work?
– auhasard
3 hours ago
@auhasard It should? Why?
– 5xum
3 hours ago
add a comment |Â
up vote
3
down vote
$$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.
answered 3 hours ago
5xum
85.2k388154
Yes, but because I have the uniform convergence the switch of limits should work?
– auhasard
3 hours ago
@auhasard It should? Why?
– 5xum
3 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.
answered 3 hours ago
5xum
85.2k388154
$$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.
answered 3 hours ago
5xum
85.2k388154
answered 3 hours ago
5xum
85.2k388154
answered 3 hours ago
5xum
85.2k388154
answered 3 hours ago
5xum
85.2k388154
85.2k388154
Yes, but because I have the uniform convergence the switch of limits should work?
– auhasard
3 hours ago
@auhasard It should? Why?
– 5xum
3 hours ago
add a comment |Â
Yes, but because I have the uniform convergence the switch of limits should work?
– auhasard
3 hours ago
@auhasard It should? Why?
– 5xum
3 hours ago
Yes, but because I have the uniform convergence the switch of limits should work?
– auhasard
3 hours ago
Yes, but because I have the uniform convergence the switch of limits should work?
– auhasard
3 hours ago
@auhasard It should? Why?
– 5xum
3 hours ago
@auhasard It should? Why?
– 5xum
3 hours ago
add a comment |Â
up vote
0
down vote
What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?
answered 21 mins ago
hartkp
68134
add a comment |Â
up vote
0
down vote
What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?
answered 21 mins ago
hartkp
68134
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?
answered 21 mins ago
hartkp
68134
What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?
answered 21 mins ago
hartkp
68134
answered 21 mins ago
hartkp
68134
answered 21 mins ago
hartkp
68134
answered 21 mins ago
hartkp
68134
68134
add a comment |Â
add a comment |Â
up vote
0
down vote
Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.
answered 18 mins ago
Ethan Bolker
37.2k54299
add a comment |Â
up vote
0
down vote
Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.
answered 18 mins ago
Ethan Bolker
37.2k54299
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.
answered 18 mins ago
Ethan Bolker
37.2k54299
Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.
answered 18 mins ago
Ethan Bolker
37.2k54299
answered 18 mins ago
Ethan Bolker
37.2k54299
answered 18 mins ago
Ethan Bolker
37.2k54299
answered 18 mins ago
Ethan Bolker
37.2k54299
37.2k54299
add a comment |Â
add a comment |Â
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3
Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
– Kavi Rama Murthy
3 hours ago