Infinite linear combination of linearly independent reals equaling 0
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Is it possible to find a sequence of real numbers $x_n$ linearly independent over $mathbbQ$ with the property that there exists a sequence of rationals $q_n$ such that $sum q_n x_n = 0,$ but for any sequence of integers $a_n, sum a_nx_n not= 0$?
real-analysis sequences-and-series
asked Aug 25 at 5:34
cats
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up vote
1
down vote
favorite
Is it possible to find a sequence of real numbers $x_n$ linearly independent over $mathbbQ$ with the property that there exists a sequence of rationals $q_n$ such that $sum q_n x_n = 0,$ but for any sequence of integers $a_n, sum a_nx_n not= 0$?
real-analysis sequences-and-series
asked Aug 25 at 5:34
cats
3,443726
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is it possible to find a sequence of real numbers $x_n$ linearly independent over $mathbbQ$ with the property that there exists a sequence of rationals $q_n$ such that $sum q_n x_n = 0,$ but for any sequence of integers $a_n, sum a_nx_n not= 0$?
real-analysis sequences-and-series
asked Aug 25 at 5:34
cats
3,443726
Is it possible to find a sequence of real numbers $x_n$ linearly independent over $mathbbQ$ with the property that there exists a sequence of rationals $q_n$ such that $sum q_n x_n = 0,$ but for any sequence of integers $a_n, sum a_nx_n not= 0$?
real-analysis sequences-and-series
asked Aug 25 at 5:34
cats
3,443726
asked Aug 25 at 5:34
cats
3,443726
asked Aug 25 at 5:34
cats
3,443726
asked Aug 25 at 5:34
cats
3,443726
3,443726
add a comment |Â
add a comment |Â
1 Answer
1
active
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votes
up vote
1
down vote
accepted
Of course; take any transcendental number $0<alpha<1$ and set $x_n:=n(alpha^n+1-alpha^n)$ and $q_n=frac1n$.
answered Aug 25 at 9:37
Servaes
1
Can you elaborate on why this implies there do not exist integers satisfying the same condition?
– cats
Aug 25 at 15:59
It turns out this question is quite easy, though I do not believe the above sequence works since the limit of $x_n$ is $0.$ Any sequence with lim inf not $0$ will do the trick.
– cats
Aug 26 at 5:51
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
accepted
Of course; take any transcendental number $0<alpha<1$ and set $x_n:=n(alpha^n+1-alpha^n)$ and $q_n=frac1n$.
answered Aug 25 at 9:37
Servaes
1
Can you elaborate on why this implies there do not exist integers satisfying the same condition?
– cats
Aug 25 at 15:59
It turns out this question is quite easy, though I do not believe the above sequence works since the limit of $x_n$ is $0.$ Any sequence with lim inf not $0$ will do the trick.
– cats
Aug 26 at 5:51
add a comment |Â
up vote
1
down vote
accepted
Of course; take any transcendental number $0<alpha<1$ and set $x_n:=n(alpha^n+1-alpha^n)$ and $q_n=frac1n$.
answered Aug 25 at 9:37
Servaes
1
Can you elaborate on why this implies there do not exist integers satisfying the same condition?
– cats
Aug 25 at 15:59
It turns out this question is quite easy, though I do not believe the above sequence works since the limit of $x_n$ is $0.$ Any sequence with lim inf not $0$ will do the trick.
– cats
Aug 26 at 5:51
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Of course; take any transcendental number $0<alpha<1$ and set $x_n:=n(alpha^n+1-alpha^n)$ and $q_n=frac1n$.
answered Aug 25 at 9:37
Servaes
1
Of course; take any transcendental number $0<alpha<1$ and set $x_n:=n(alpha^n+1-alpha^n)$ and $q_n=frac1n$.
answered Aug 25 at 9:37
Servaes
1
answered Aug 25 at 9:37
Servaes
1
answered Aug 25 at 9:37
Servaes
1
answered Aug 25 at 9:37
Servaes
1
1
Can you elaborate on why this implies there do not exist integers satisfying the same condition?
– cats
Aug 25 at 15:59
It turns out this question is quite easy, though I do not believe the above sequence works since the limit of $x_n$ is $0.$ Any sequence with lim inf not $0$ will do the trick.
– cats
Aug 26 at 5:51
add a comment |Â
Can you elaborate on why this implies there do not exist integers satisfying the same condition?
– cats
Aug 25 at 15:59
It turns out this question is quite easy, though I do not believe the above sequence works since the limit of $x_n$ is $0.$ Any sequence with lim inf not $0$ will do the trick.
– cats
Aug 26 at 5:51
Can you elaborate on why this implies there do not exist integers satisfying the same condition?
– cats
Aug 25 at 15:59
Can you elaborate on why this implies there do not exist integers satisfying the same condition?
– cats
Aug 25 at 15:59
It turns out this question is quite easy, though I do not believe the above sequence works since the limit of $x_n$ is $0.$ Any sequence with lim inf not $0$ will do the trick.
– cats
Aug 26 at 5:51
It turns out this question is quite easy, though I do not believe the above sequence works since the limit of $x_n$ is $0.$ Any sequence with lim inf not $0$ will do the trick.
– cats
Aug 26 at 5:51
add a comment |Â
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