Mixing
Writing the limit of a series inside the series
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Let $f(x) = sumlimits_n=0^infty a_n x^n$ be finite or infinite,
where $x$ is a real number and $(a_n)_n$ is an infinite sequence of positive integers.
For any integer $n$, let $ (a_L,n)_L$ be an infinite sequence of integers which equals $a_n$ for any $L$ large enough and for any L, let
$$
f_L(x) = sumlimits_n=0^infty a_L,n x^n.
$$
Is it always true that
$$
lim_L rightarrow infty f_L(x) = f(x)?
$$Is there a famous theorem that can be applied from which the identity follows?
real-analysis sequences-and-series analysis limits
asked Jul 28 at 19:15
QuantumLogarithm
470314
add a comment |Â
up vote
0
down vote
favorite
Let $f(x) = sumlimits_n=0^infty a_n x^n$ be finite or infinite,
where $x$ is a real number and $(a_n)_n$ is an infinite sequence of positive integers.
For any integer $n$, let $ (a_L,n)_L$ be an infinite sequence of integers which equals $a_n$ for any $L$ large enough and for any L, let
$$
f_L(x) = sumlimits_n=0^infty a_L,n x^n.
$$
Is it always true that
$$
lim_L rightarrow infty f_L(x) = f(x)?
$$Is there a famous theorem that can be applied from which the identity follows?
real-analysis sequences-and-series analysis limits
asked Jul 28 at 19:15
QuantumLogarithm
470314
1
We can make so that $f_L(x) = infty$ for any $L$, and any $xneq 0$, but $f(x)$ is convergent. For example, $f(x)=frac11-x$ and $f_L(x)$ has it's tail as $sum_n=k+1^infty n!x^n$.
– Rumpelstiltskin
Jul 28 at 19:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f(x) = sumlimits_n=0^infty a_n x^n$ be finite or infinite,
where $x$ is a real number and $(a_n)_n$ is an infinite sequence of positive integers.
For any integer $n$, let $ (a_L,n)_L$ be an infinite sequence of integers which equals $a_n$ for any $L$ large enough and for any L, let
$$
f_L(x) = sumlimits_n=0^infty a_L,n x^n.
$$
Is it always true that
$$
lim_L rightarrow infty f_L(x) = f(x)?
$$Is there a famous theorem that can be applied from which the identity follows?
real-analysis sequences-and-series analysis limits
asked Jul 28 at 19:15
QuantumLogarithm
470314
Let $f(x) = sumlimits_n=0^infty a_n x^n$ be finite or infinite,
where $x$ is a real number and $(a_n)_n$ is an infinite sequence of positive integers.
For any integer $n$, let $ (a_L,n)_L$ be an infinite sequence of integers which equals $a_n$ for any $L$ large enough and for any L, let
$$
f_L(x) = sumlimits_n=0^infty a_L,n x^n.
$$
Is it always true that
$$
lim_L rightarrow infty f_L(x) = f(x)?
$$Is there a famous theorem that can be applied from which the identity follows?
real-analysis sequences-and-series analysis limits
asked Jul 28 at 19:15
QuantumLogarithm
470314
asked Jul 28 at 19:15
QuantumLogarithm
470314
asked Jul 28 at 19:15
QuantumLogarithm
470314
asked Jul 28 at 19:15
QuantumLogarithm
470314
470314
1
We can make so that $f_L(x) = infty$ for any $L$, and any $xneq 0$, but $f(x)$ is convergent. For example, $f(x)=frac11-x$ and $f_L(x)$ has it's tail as $sum_n=k+1^infty n!x^n$.
– Rumpelstiltskin
Jul 28 at 19:48
add a comment |Â
1
We can make so that $f_L(x) = infty$ for any $L$, and any $xneq 0$, but $f(x)$ is convergent. For example, $f(x)=frac11-x$ and $f_L(x)$ has it's tail as $sum_n=k+1^infty n!x^n$.
– Rumpelstiltskin
Jul 28 at 19:48
1
1
We can make so that $f_L(x) = infty$ for any $L$, and any $xneq 0$, but $f(x)$ is convergent. For example, $f(x)=frac11-x$ and $f_L(x)$ has it's tail as $sum_n=k+1^infty n!x^n$.
– Rumpelstiltskin
Jul 28 at 19:48
We can make so that $f_L(x) = infty$ for any $L$, and any $xneq 0$, but $f(x)$ is convergent. For example, $f(x)=frac11-x$ and $f_L(x)$ has it's tail as $sum_n=k+1^infty n!x^n$.
– Rumpelstiltskin
Jul 28 at 19:48
add a comment |Â
1 Answer
1
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I can interpret "$(a_n,L)$ equals $(a_n)$ for $L$ large enough either as "for every $n$ there exists $k$ such that $a_n,L=a_n$ for $L>k$" or as "there exists $k$ such that for $L>k$ and all $n$, $a_n,L=a_n$.
In the first case, your statement is false. Let $a_n=1$ and $a_L,n=a_n=1$ for $L>n$ and $a_L,n=0$ else. Then $sum a_n x^n=frac11-x=f(x)$ and $sum a_L,n x^n=sum_n=L+1^infty x^n = fracx^L+11-x=f_L(x)$. For $-1<x<1$, $f_L(x)to 0neq f(x)$.
In the second case, your statement is true, as $a_L,n=a_n$ for $L>k$ and for all $n$. So for every $L>k$, $f_L(x)=f(x)$. Then of course $lim_Ltoinfty f_L(x)=f(x)$.
Or do you have another different definition in mind?
answered Jul 28 at 19:52
Kusma
1,087111
Thank you. I meant the first case. Actually, I also know that $a_n$ increases exponentially fast with $n$ and that $a_n,L = a_n$ for any $n< L$, but I don't know if this might make the statement true.
– QuantumLogarithm
Jul 28 at 20:12
If $a_L,n = 1$ for $L > n$ and $a_L,n = 0$ for $n geqslant L$, then $f_L(x) = sum_n=0^L-1 x^n = (1-x^L)/(1-x) to 1/(1-x) = f(x)$, as well, since $lim_L to inftyx^L =0$ $ (|x| < 1)$. It seems your first example does not do the job, or am I missing something.
– RRL
Jul 28 at 21:06
Oh yes, you're right, sorry :( Indeed $n>L$ is not the same as $L>n$... @Adam gave a much better example above.
– Kusma
Jul 28 at 21:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I can interpret "$(a_n,L)$ equals $(a_n)$ for $L$ large enough either as "for every $n$ there exists $k$ such that $a_n,L=a_n$ for $L>k$" or as "there exists $k$ such that for $L>k$ and all $n$, $a_n,L=a_n$.
In the first case, your statement is false. Let $a_n=1$ and $a_L,n=a_n=1$ for $L>n$ and $a_L,n=0$ else. Then $sum a_n x^n=frac11-x=f(x)$ and $sum a_L,n x^n=sum_n=L+1^infty x^n = fracx^L+11-x=f_L(x)$. For $-1<x<1$, $f_L(x)to 0neq f(x)$.
In the second case, your statement is true, as $a_L,n=a_n$ for $L>k$ and for all $n$. So for every $L>k$, $f_L(x)=f(x)$. Then of course $lim_Ltoinfty f_L(x)=f(x)$.
Or do you have another different definition in mind?
answered Jul 28 at 19:52
Kusma
1,087111
Thank you. I meant the first case. Actually, I also know that $a_n$ increases exponentially fast with $n$ and that $a_n,L = a_n$ for any $n< L$, but I don't know if this might make the statement true.
– QuantumLogarithm
Jul 28 at 20:12
If $a_L,n = 1$ for $L > n$ and $a_L,n = 0$ for $n geqslant L$, then $f_L(x) = sum_n=0^L-1 x^n = (1-x^L)/(1-x) to 1/(1-x) = f(x)$, as well, since $lim_L to inftyx^L =0$ $ (|x| < 1)$. It seems your first example does not do the job, or am I missing something.
– RRL
Jul 28 at 21:06
Oh yes, you're right, sorry :( Indeed $n>L$ is not the same as $L>n$... @Adam gave a much better example above.
– Kusma
Jul 28 at 21:53
add a comment |Â
up vote
1
down vote
I can interpret "$(a_n,L)$ equals $(a_n)$ for $L$ large enough either as "for every $n$ there exists $k$ such that $a_n,L=a_n$ for $L>k$" or as "there exists $k$ such that for $L>k$ and all $n$, $a_n,L=a_n$.
In the first case, your statement is false. Let $a_n=1$ and $a_L,n=a_n=1$ for $L>n$ and $a_L,n=0$ else. Then $sum a_n x^n=frac11-x=f(x)$ and $sum a_L,n x^n=sum_n=L+1^infty x^n = fracx^L+11-x=f_L(x)$. For $-1<x<1$, $f_L(x)to 0neq f(x)$.
In the second case, your statement is true, as $a_L,n=a_n$ for $L>k$ and for all $n$. So for every $L>k$, $f_L(x)=f(x)$. Then of course $lim_Ltoinfty f_L(x)=f(x)$.
Or do you have another different definition in mind?
answered Jul 28 at 19:52
Kusma
1,087111
Thank you. I meant the first case. Actually, I also know that $a_n$ increases exponentially fast with $n$ and that $a_n,L = a_n$ for any $n< L$, but I don't know if this might make the statement true.
– QuantumLogarithm
Jul 28 at 20:12
If $a_L,n = 1$ for $L > n$ and $a_L,n = 0$ for $n geqslant L$, then $f_L(x) = sum_n=0^L-1 x^n = (1-x^L)/(1-x) to 1/(1-x) = f(x)$, as well, since $lim_L to inftyx^L =0$ $ (|x| < 1)$. It seems your first example does not do the job, or am I missing something.
– RRL
Jul 28 at 21:06
Oh yes, you're right, sorry :( Indeed $n>L$ is not the same as $L>n$... @Adam gave a much better example above.
– Kusma
Jul 28 at 21:53
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I can interpret "$(a_n,L)$ equals $(a_n)$ for $L$ large enough either as "for every $n$ there exists $k$ such that $a_n,L=a_n$ for $L>k$" or as "there exists $k$ such that for $L>k$ and all $n$, $a_n,L=a_n$.
In the first case, your statement is false. Let $a_n=1$ and $a_L,n=a_n=1$ for $L>n$ and $a_L,n=0$ else. Then $sum a_n x^n=frac11-x=f(x)$ and $sum a_L,n x^n=sum_n=L+1^infty x^n = fracx^L+11-x=f_L(x)$. For $-1<x<1$, $f_L(x)to 0neq f(x)$.
In the second case, your statement is true, as $a_L,n=a_n$ for $L>k$ and for all $n$. So for every $L>k$, $f_L(x)=f(x)$. Then of course $lim_Ltoinfty f_L(x)=f(x)$.
Or do you have another different definition in mind?
answered Jul 28 at 19:52
Kusma
1,087111
I can interpret "$(a_n,L)$ equals $(a_n)$ for $L$ large enough either as "for every $n$ there exists $k$ such that $a_n,L=a_n$ for $L>k$" or as "there exists $k$ such that for $L>k$ and all $n$, $a_n,L=a_n$.
In the first case, your statement is false. Let $a_n=1$ and $a_L,n=a_n=1$ for $L>n$ and $a_L,n=0$ else. Then $sum a_n x^n=frac11-x=f(x)$ and $sum a_L,n x^n=sum_n=L+1^infty x^n = fracx^L+11-x=f_L(x)$. For $-1<x<1$, $f_L(x)to 0neq f(x)$.
In the second case, your statement is true, as $a_L,n=a_n$ for $L>k$ and for all $n$. So for every $L>k$, $f_L(x)=f(x)$. Then of course $lim_Ltoinfty f_L(x)=f(x)$.
Or do you have another different definition in mind?
answered Jul 28 at 19:52
Kusma
1,087111
answered Jul 28 at 19:52
Kusma
1,087111
answered Jul 28 at 19:52
Kusma
1,087111
answered Jul 28 at 19:52
Kusma
1,087111
1,087111
Thank you. I meant the first case. Actually, I also know that $a_n$ increases exponentially fast with $n$ and that $a_n,L = a_n$ for any $n< L$, but I don't know if this might make the statement true.
– QuantumLogarithm
Jul 28 at 20:12
If $a_L,n = 1$ for $L > n$ and $a_L,n = 0$ for $n geqslant L$, then $f_L(x) = sum_n=0^L-1 x^n = (1-x^L)/(1-x) to 1/(1-x) = f(x)$, as well, since $lim_L to inftyx^L =0$ $ (|x| < 1)$. It seems your first example does not do the job, or am I missing something.
– RRL
Jul 28 at 21:06
Oh yes, you're right, sorry :( Indeed $n>L$ is not the same as $L>n$... @Adam gave a much better example above.
– Kusma
Jul 28 at 21:53
add a comment |Â
Thank you. I meant the first case. Actually, I also know that $a_n$ increases exponentially fast with $n$ and that $a_n,L = a_n$ for any $n< L$, but I don't know if this might make the statement true.
– QuantumLogarithm
Jul 28 at 20:12
If $a_L,n = 1$ for $L > n$ and $a_L,n = 0$ for $n geqslant L$, then $f_L(x) = sum_n=0^L-1 x^n = (1-x^L)/(1-x) to 1/(1-x) = f(x)$, as well, since $lim_L to inftyx^L =0$ $ (|x| < 1)$. It seems your first example does not do the job, or am I missing something.
– RRL
Jul 28 at 21:06
Oh yes, you're right, sorry :( Indeed $n>L$ is not the same as $L>n$... @Adam gave a much better example above.
– Kusma
Jul 28 at 21:53
Thank you. I meant the first case. Actually, I also know that $a_n$ increases exponentially fast with $n$ and that $a_n,L = a_n$ for any $n< L$, but I don't know if this might make the statement true.
– QuantumLogarithm
Jul 28 at 20:12
Thank you. I meant the first case. Actually, I also know that $a_n$ increases exponentially fast with $n$ and that $a_n,L = a_n$ for any $n< L$, but I don't know if this might make the statement true.
– QuantumLogarithm
Jul 28 at 20:12
If $a_L,n = 1$ for $L > n$ and $a_L,n = 0$ for $n geqslant L$, then $f_L(x) = sum_n=0^L-1 x^n = (1-x^L)/(1-x) to 1/(1-x) = f(x)$, as well, since $lim_L to inftyx^L =0$ $ (|x| < 1)$. It seems your first example does not do the job, or am I missing something.
– RRL
Jul 28 at 21:06
If $a_L,n = 1$ for $L > n$ and $a_L,n = 0$ for $n geqslant L$, then $f_L(x) = sum_n=0^L-1 x^n = (1-x^L)/(1-x) to 1/(1-x) = f(x)$, as well, since $lim_L to inftyx^L =0$ $ (|x| < 1)$. It seems your first example does not do the job, or am I missing something.
– RRL
Jul 28 at 21:06
Oh yes, you're right, sorry :( Indeed $n>L$ is not the same as $L>n$... @Adam gave a much better example above.
– Kusma
Jul 28 at 21:53
Oh yes, you're right, sorry :( Indeed $n>L$ is not the same as $L>n$... @Adam gave a much better example above.
– Kusma
Jul 28 at 21:53
add a comment |Â
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1
We can make so that $f_L(x) = infty$ for any $L$, and any $xneq 0$, but $f(x)$ is convergent. For example, $f(x)=frac11-x$ and $f_L(x)$ has it's tail as $sum_n=k+1^infty n!x^n$.
– Rumpelstiltskin
Jul 28 at 19:48