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Suggestion for notation of a sequence
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What could be a good notation for a sequence $(x_n)$ chosen from a vector space $displaystyle V= bigcup_{nin mathbb N} f_n(V_n)$ where all $V_n$ are vector spaces and $f_n$ are linear maps.
real-analysis sequences-and-series notation
edited Jan 22 at 6:49
El borito
666216
asked Jan 22 at 6:04
Math LoverMath Lover
1,004315
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add a comment |
$begingroup$
What could be a good notation for a sequence $(x_n)$ chosen from a vector space $displaystyle V= bigcup_{nin mathbb N} f_n(V_n)$ where all $V_n$ are vector spaces and $f_n$ are linear maps.
real-analysis sequences-and-series notation
edited Jan 22 at 6:49
El borito
666216
asked Jan 22 at 6:04
Math LoverMath Lover
1,004315
$endgroup$
add a comment |
$begingroup$
What could be a good notation for a sequence $(x_n)$ chosen from a vector space $displaystyle V= bigcup_{nin mathbb N} f_n(V_n)$ where all $V_n$ are vector spaces and $f_n$ are linear maps.
real-analysis sequences-and-series notation
edited Jan 22 at 6:49
El borito
666216
asked Jan 22 at 6:04
Math LoverMath Lover
1,004315
$endgroup$
What could be a good notation for a sequence $(x_n)$ chosen from a vector space $displaystyle V= bigcup_{nin mathbb N} f_n(V_n)$ where all $V_n$ are vector spaces and $f_n$ are linear maps.
real-analysis sequences-and-series notation
real-analysis sequences-and-series notation
edited Jan 22 at 6:49
El borito
666216
asked Jan 22 at 6:04
Math LoverMath Lover
1,004315
edited Jan 22 at 6:49
El borito
666216
asked Jan 22 at 6:04
Math LoverMath Lover
1,004315
edited Jan 22 at 6:49
El borito
666216
edited Jan 22 at 6:49
El borito
666216
edited Jan 22 at 6:49
El borito
666216
666216
asked Jan 22 at 6:04
Math LoverMath Lover
1,004315
asked Jan 22 at 6:04
Math LoverMath Lover
1,004315
asked Jan 22 at 6:04
Math LoverMath Lover
1,004315
1,004315
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Using $n$ for both the index in $x_n$ and to index the union which defined $V$ doesn't seem like a good idea. Here's why: if $x_n inbigcup_{m in Bbb N} f_mbig(V_mbig)$, there is $m(n) in Bbb N$ such that $x_n in f_{m(n)}big(V_{m(n)}big)$. This means that there is $v_{m(n)} in V_{m(n)}$ such that $x_n = f_{m(n)}big(v_{m(n)}big)$. You want to be able to distinguish things. So $$(x_n)_{n in Bbb N} = big(,f_{m(n)}big(v_{m(n)}big)big)_{n in Bbb N},$$and the function $mcolon Bbb N to Bbb N$ depends on the sequence itself.
edited Jan 22 at 7:53
El borito
666216
answered Jan 22 at 6:18
Ivo TerekIvo Terek
46.4k954142
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
Using $n$ for both the index in $x_n$ and to index the union which defined $V$ doesn't seem like a good idea. Here's why: if $x_n inbigcup_{m in Bbb N} f_mbig(V_mbig)$, there is $m(n) in Bbb N$ such that $x_n in f_{m(n)}big(V_{m(n)}big)$. This means that there is $v_{m(n)} in V_{m(n)}$ such that $x_n = f_{m(n)}big(v_{m(n)}big)$. You want to be able to distinguish things. So $$(x_n)_{n in Bbb N} = big(,f_{m(n)}big(v_{m(n)}big)big)_{n in Bbb N},$$and the function $mcolon Bbb N to Bbb N$ depends on the sequence itself.
edited Jan 22 at 7:53
El borito
666216
answered Jan 22 at 6:18
Ivo TerekIvo Terek
46.4k954142
$endgroup$
add a comment |
$begingroup$
Using $n$ for both the index in $x_n$ and to index the union which defined $V$ doesn't seem like a good idea. Here's why: if $x_n inbigcup_{m in Bbb N} f_mbig(V_mbig)$, there is $m(n) in Bbb N$ such that $x_n in f_{m(n)}big(V_{m(n)}big)$. This means that there is $v_{m(n)} in V_{m(n)}$ such that $x_n = f_{m(n)}big(v_{m(n)}big)$. You want to be able to distinguish things. So $$(x_n)_{n in Bbb N} = big(,f_{m(n)}big(v_{m(n)}big)big)_{n in Bbb N},$$and the function $mcolon Bbb N to Bbb N$ depends on the sequence itself.
edited Jan 22 at 7:53
El borito
666216
answered Jan 22 at 6:18
Ivo TerekIvo Terek
46.4k954142
$endgroup$
add a comment |
$begingroup$
Using $n$ for both the index in $x_n$ and to index the union which defined $V$ doesn't seem like a good idea. Here's why: if $x_n inbigcup_{m in Bbb N} f_mbig(V_mbig)$, there is $m(n) in Bbb N$ such that $x_n in f_{m(n)}big(V_{m(n)}big)$. This means that there is $v_{m(n)} in V_{m(n)}$ such that $x_n = f_{m(n)}big(v_{m(n)}big)$. You want to be able to distinguish things. So $$(x_n)_{n in Bbb N} = big(,f_{m(n)}big(v_{m(n)}big)big)_{n in Bbb N},$$and the function $mcolon Bbb N to Bbb N$ depends on the sequence itself.
edited Jan 22 at 7:53
El borito
666216
answered Jan 22 at 6:18
Ivo TerekIvo Terek
46.4k954142
$endgroup$
Using $n$ for both the index in $x_n$ and to index the union which defined $V$ doesn't seem like a good idea. Here's why: if $x_n inbigcup_{m in Bbb N} f_mbig(V_mbig)$, there is $m(n) in Bbb N$ such that $x_n in f_{m(n)}big(V_{m(n)}big)$. This means that there is $v_{m(n)} in V_{m(n)}$ such that $x_n = f_{m(n)}big(v_{m(n)}big)$. You want to be able to distinguish things. So $$(x_n)_{n in Bbb N} = big(,f_{m(n)}big(v_{m(n)}big)big)_{n in Bbb N},$$and the function $mcolon Bbb N to Bbb N$ depends on the sequence itself.
edited Jan 22 at 7:53
El borito
666216
answered Jan 22 at 6:18
Ivo TerekIvo Terek
46.4k954142
edited Jan 22 at 7:53
El borito
666216
edited Jan 22 at 7:53
El borito
666216
edited Jan 22 at 7:53
El borito
666216
666216
answered Jan 22 at 6:18
Ivo TerekIvo Terek
46.4k954142
answered Jan 22 at 6:18
Ivo TerekIvo Terek
46.4k954142
answered Jan 22 at 6:18
Ivo TerekIvo Terek
46.4k954142
46.4k954142
add a comment |
add a comment |
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