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results from limit of a complex sequence
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the limit of a complex sequence An+iBn is zero as n tends to infinity, where An and Bn are two real sequences.
Does this implies that the limit of An and Bn is zero
complex-analysis control-theory
asked Jan 25 at 19:03
i.issai.issa
62
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add a comment |
$begingroup$
the limit of a complex sequence An+iBn is zero as n tends to infinity, where An and Bn are two real sequences.
Does this implies that the limit of An and Bn is zero
complex-analysis control-theory
asked Jan 25 at 19:03
i.issai.issa
62
$endgroup$
$begingroup$
Do you mean $A_n + iB_n$ rather than $A_n+B_n$? Otherwise, no, since you could take $A_n$ to be any sequence at all and take $B_n=-A_n$.
$endgroup$
– MPW
Jan 25 at 19:14
$begingroup$
Sorry, yes i mean An+iBn
$endgroup$
– i.issa
Jan 25 at 19:42
add a comment |
$begingroup$
the limit of a complex sequence An+iBn is zero as n tends to infinity, where An and Bn are two real sequences.
Does this implies that the limit of An and Bn is zero
complex-analysis control-theory
asked Jan 25 at 19:03
i.issai.issa
62
$endgroup$
the limit of a complex sequence An+iBn is zero as n tends to infinity, where An and Bn are two real sequences.
Does this implies that the limit of An and Bn is zero
complex-analysis control-theory
complex-analysis control-theory
asked Jan 25 at 19:03
i.issai.issa
62
asked Jan 25 at 19:03
i.issai.issa
62
edited Jan 25 at 19:41
i.issa
asked Jan 25 at 19:03
i.issai.issa
62
asked Jan 25 at 19:03
i.issai.issa
62
asked Jan 25 at 19:03
i.issai.issa
62
62
$begingroup$
Do you mean $A_n + iB_n$ rather than $A_n+B_n$? Otherwise, no, since you could take $A_n$ to be any sequence at all and take $B_n=-A_n$.
$endgroup$
– MPW
Jan 25 at 19:14
$begingroup$
Sorry, yes i mean An+iBn
$endgroup$
– i.issa
Jan 25 at 19:42
add a comment |
$begingroup$
Do you mean $A_n + iB_n$ rather than $A_n+B_n$? Otherwise, no, since you could take $A_n$ to be any sequence at all and take $B_n=-A_n$.
$endgroup$
– MPW
Jan 25 at 19:14
$begingroup$
Sorry, yes i mean An+iBn
$endgroup$
– i.issa
Jan 25 at 19:42
$begingroup$
Do you mean $A_n + iB_n$ rather than $A_n+B_n$? Otherwise, no, since you could take $A_n$ to be any sequence at all and take $B_n=-A_n$.
$endgroup$
– MPW
Jan 25 at 19:14
$begingroup$
Do you mean $A_n + iB_n$ rather than $A_n+B_n$? Otherwise, no, since you could take $A_n$ to be any sequence at all and take $B_n=-A_n$.
$endgroup$
– MPW
Jan 25 at 19:14
$begingroup$
Sorry, yes i mean An+iBn
$endgroup$
– i.issa
Jan 25 at 19:42
$begingroup$
Sorry, yes i mean An+iBn
$endgroup$
– i.issa
Jan 25 at 19:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have $vert A_n vert le vert A_n +i B_n vert = sqrt{A_n^2 +B_n^2}$.
Saying that $A_n +i B_n$ converges to zero means that $limlimits_{nto infty} vert A_n +i B_n vert=0$. And the inequality above implies that $limlimits_{nto infty} vert A_nvert=0$. Similar proof to be done for $(B_n)$.
This proves that the answer to your question is positive.
answered Jan 25 at 20:03
mathcounterexamples.netmathcounterexamples.net
27k22158
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add a comment |
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It might help to visualize a complex number as a point in the plane. So if you have a sequence of points in the plane which converge towards the origin, will the x component converge towards the y axis and the y component converge towards the x axis? Clearly the answer is yes.
answered Jan 25 at 19:50
Jagol95Jagol95
2637
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
You have $vert A_n vert le vert A_n +i B_n vert = sqrt{A_n^2 +B_n^2}$.
Saying that $A_n +i B_n$ converges to zero means that $limlimits_{nto infty} vert A_n +i B_n vert=0$. And the inequality above implies that $limlimits_{nto infty} vert A_nvert=0$. Similar proof to be done for $(B_n)$.
This proves that the answer to your question is positive.
answered Jan 25 at 20:03
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
You have $vert A_n vert le vert A_n +i B_n vert = sqrt{A_n^2 +B_n^2}$.
Saying that $A_n +i B_n$ converges to zero means that $limlimits_{nto infty} vert A_n +i B_n vert=0$. And the inequality above implies that $limlimits_{nto infty} vert A_nvert=0$. Similar proof to be done for $(B_n)$.
This proves that the answer to your question is positive.
answered Jan 25 at 20:03
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
You have $vert A_n vert le vert A_n +i B_n vert = sqrt{A_n^2 +B_n^2}$.
Saying that $A_n +i B_n$ converges to zero means that $limlimits_{nto infty} vert A_n +i B_n vert=0$. And the inequality above implies that $limlimits_{nto infty} vert A_nvert=0$. Similar proof to be done for $(B_n)$.
This proves that the answer to your question is positive.
answered Jan 25 at 20:03
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
You have $vert A_n vert le vert A_n +i B_n vert = sqrt{A_n^2 +B_n^2}$.
Saying that $A_n +i B_n$ converges to zero means that $limlimits_{nto infty} vert A_n +i B_n vert=0$. And the inequality above implies that $limlimits_{nto infty} vert A_nvert=0$. Similar proof to be done for $(B_n)$.
This proves that the answer to your question is positive.
answered Jan 25 at 20:03
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 25 at 20:03
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 25 at 20:03
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 25 at 20:03
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
add a comment |
add a comment |
$begingroup$
It might help to visualize a complex number as a point in the plane. So if you have a sequence of points in the plane which converge towards the origin, will the x component converge towards the y axis and the y component converge towards the x axis? Clearly the answer is yes.
answered Jan 25 at 19:50
Jagol95Jagol95
2637
$endgroup$
add a comment |
$begingroup$
It might help to visualize a complex number as a point in the plane. So if you have a sequence of points in the plane which converge towards the origin, will the x component converge towards the y axis and the y component converge towards the x axis? Clearly the answer is yes.
answered Jan 25 at 19:50
Jagol95Jagol95
2637
$endgroup$
add a comment |
$begingroup$
It might help to visualize a complex number as a point in the plane. So if you have a sequence of points in the plane which converge towards the origin, will the x component converge towards the y axis and the y component converge towards the x axis? Clearly the answer is yes.
answered Jan 25 at 19:50
Jagol95Jagol95
2637
$endgroup$
It might help to visualize a complex number as a point in the plane. So if you have a sequence of points in the plane which converge towards the origin, will the x component converge towards the y axis and the y component converge towards the x axis? Clearly the answer is yes.
answered Jan 25 at 19:50
Jagol95Jagol95
2637
answered Jan 25 at 19:50
Jagol95Jagol95
2637
answered Jan 25 at 19:50
Jagol95Jagol95
2637
answered Jan 25 at 19:50
Jagol95Jagol95
2637
2637
add a comment |
add a comment |
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$begingroup$
Do you mean $A_n + iB_n$ rather than $A_n+B_n$? Otherwise, no, since you could take $A_n$ to be any sequence at all and take $B_n=-A_n$.
$endgroup$
– MPW
Jan 25 at 19:14
$begingroup$
Sorry, yes i mean An+iBn
$endgroup$
– i.issa
Jan 25 at 19:42