$ |1+z+z^2 +…+z^n| geq |z|^n $ is valid [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$exists x in N, forall y in N, x ge y$A surprisingly resistant elementary numerical inequalityIs it true that $2max_z | z^n+ (sum_ i=2^n-1 a_iz^i ) + a_1 z+a_0| geq max_z |a_1z+a_0|$?Prove $frac13(x+y+z)^2 geq xy + yz + xz.$If $a_1b_1+…+a_nb_n geq 0$ then is it true that $a_1^2b_1+…+a_n^2b_n geq 0$.Is $a^2+b^2 geq 2ab$ where $a,b in mathbbR$?Proving $lchoose n+nl-d+nchoose d<0$ for $ngeq 6$, $dgeq 3$, $l<d+n-1$Describe all points in the complex plane that verify these inequalities: $| z + 3 i | geq 3 | z − i |$ & $| z - 3/2 i | geq 1$Proving that $sqrt2 |z| geq |x| + |y|$ for $z = x + iy$Proving $sum_i=1^nfrac1p^a_i geq sum_i=1^nfrac1p^b_i$, when $p geq 2$.
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$ |1+z+z^2 +…+z^n| geq |z|^n $ is valid [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$exists x in N, forall y in N, x ge y$A surprisingly resistant elementary numerical inequalityIs it true that $2max_=1 | z^n+ (sum_ i=2^n-1 a_iz^i ) + a_1 z+a_0| geq max_z |a_1z+a_0|$?Prove $frac13(x+y+z)^2 geq xy + yz + xz.$If $a_1b_1+…+a_nb_n geq 0$ then is it true that $a_1^2b_1+…+a_n^2b_n geq 0$.Is $a^2+b^2 geq 2ab$ where $a,b in mathbbR$?Proving $lchoose n+nl-d+nchoose d<0$ for $ngeq 6$, $dgeq 3$, $l<d+n-1$Describe all points in the complex plane that verify these inequalities: $| z + 3 i | geq 3 | z − i |$ & $| z - 3/2 i | geq 1$Proving that $sqrt2 |z| geq |x| + |y|$ for $z = x + iy$Proving $sum_i=1^nfrac1p^a_i geq sum_i=1^nfrac1p^b_i$, when $p geq 2$.
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Let $z=a+ib$ , where $a>0$ & $b $ are integers, then $ |1+z+z^2 +...+z^n| geq |z|^n $ is always holds...
I tried to prove the because this seems to be true... But in vain
complex-analysis inequality complex-numbers
asked Mar 26 at 8:59
user561527user561527
595
$endgroup$
closed as off-topic by uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會 Mar 26 at 13:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會
add a comment |
$begingroup$
Let $z=a+ib$ , where $a>0$ & $b $ are integers, then $ |1+z+z^2 +...+z^n| geq |z|^n $ is always holds...
I tried to prove the because this seems to be true... But in vain
complex-analysis inequality complex-numbers
asked Mar 26 at 8:59
user561527user561527
595
$endgroup$
closed as off-topic by uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會 Mar 26 at 13:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會
$begingroup$
Why does it seem to be true? For which integers did you verify it to be true? Do you have an example of equality?
$endgroup$
– uniquesolution
Mar 26 at 9:05
$begingroup$
I put several integers... That doesn't contradict this inequality
$endgroup$
– user561527
Mar 26 at 9:10
$begingroup$
It‘s definitely true for $n = 2$, I‘ve just verified that.
$endgroup$
– Kezer
Mar 26 at 10:28
add a comment |
$begingroup$
Let $z=a+ib$ , where $a>0$ & $b $ are integers, then $ |1+z+z^2 +...+z^n| geq |z|^n $ is always holds...
I tried to prove the because this seems to be true... But in vain
complex-analysis inequality complex-numbers
asked Mar 26 at 8:59
user561527user561527
595
$endgroup$
Let $z=a+ib$ , where $a>0$ & $b $ are integers, then $ |1+z+z^2 +...+z^n| geq |z|^n $ is always holds...
I tried to prove the because this seems to be true... But in vain
complex-analysis inequality complex-numbers
complex-analysis inequality complex-numbers
asked Mar 26 at 8:59
user561527user561527
595
asked Mar 26 at 8:59
user561527user561527
595
asked Mar 26 at 8:59
user561527user561527
595
asked Mar 26 at 8:59
user561527user561527
595
asked Mar 26 at 8:59
user561527user561527
595
595
closed as off-topic by uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會 Mar 26 at 13:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會
closed as off-topic by uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會 Mar 26 at 13:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – uniquesolution, MisterRiemann, Robert Wolfe, YiFan, GNUSupporter 8964民主女神 地下教會
$begingroup$
Why does it seem to be true? For which integers did you verify it to be true? Do you have an example of equality?
$endgroup$
– uniquesolution
Mar 26 at 9:05
$begingroup$
I put several integers... That doesn't contradict this inequality
$endgroup$
– user561527
Mar 26 at 9:10
$begingroup$
It‘s definitely true for $n = 2$, I‘ve just verified that.
$endgroup$
– Kezer
Mar 26 at 10:28
add a comment |
$begingroup$
Why does it seem to be true? For which integers did you verify it to be true? Do you have an example of equality?
$endgroup$
– uniquesolution
Mar 26 at 9:05
$begingroup$
I put several integers... That doesn't contradict this inequality
$endgroup$
– user561527
Mar 26 at 9:10
$begingroup$
It‘s definitely true for $n = 2$, I‘ve just verified that.
$endgroup$
– Kezer
Mar 26 at 10:28
$begingroup$
Why does it seem to be true? For which integers did you verify it to be true? Do you have an example of equality?
$endgroup$
– uniquesolution
Mar 26 at 9:05
$begingroup$
Why does it seem to be true? For which integers did you verify it to be true? Do you have an example of equality?
$endgroup$
– uniquesolution
Mar 26 at 9:05
$begingroup$
I put several integers... That doesn't contradict this inequality
$endgroup$
– user561527
Mar 26 at 9:10
$begingroup$
I put several integers... That doesn't contradict this inequality
$endgroup$
– user561527
Mar 26 at 9:10
$begingroup$
It‘s definitely true for $n = 2$, I‘ve just verified that.
$endgroup$
– Kezer
Mar 26 at 10:28
$begingroup$
It‘s definitely true for $n = 2$, I‘ve just verified that.
$endgroup$
– Kezer
Mar 26 at 10:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The inequality is true. For $n=0,1$ the inequality is clear, hence let $n geq 2$. The case $z=1$ is also clear, so let $z neq 1$.
Note that $1+z+dots + z^n = fracz^n+1-1z-1$. So the given inequality is equivalent to
$$ left|z - frac1z^n right| geq |z-1| iff left|z - frac1z^n right|^2 geq |z-1|^2.$$
Now adhere to $|z|^2 = z overlinez$. So equivalently we have to prove
$$ z overlinez - left(fraczoverlinez^n + fracoverlinezz^n right) + frac1z^n overlinez^n geq zoverlinez - (z + overlinez) + 1.$$ This is equivalent to
$$2 operatornameRe(z) + frac1 - left(z^n+1 + overlinez^n+1 right) geq 1 iff 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 1. $$ But the estimation $$ 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 2 operatornameRe(z) + frac1 - frac2 overset?geq 1 $$ almost finishes. Note that $operatornameRe(z) geq 1$ by assumption. As $z neq 1$ by assumption, $frac2 leq 1$, so again, we're done.
Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.
answered Mar 26 at 11:04
KezerKezer
1,405621
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The inequality is true. For $n=0,1$ the inequality is clear, hence let $n geq 2$. The case $z=1$ is also clear, so let $z neq 1$.
Note that $1+z+dots + z^n = fracz^n+1-1z-1$. So the given inequality is equivalent to
$$ left|z - frac1z^n right| geq |z-1| iff left|z - frac1z^n right|^2 geq |z-1|^2.$$
Now adhere to $|z|^2 = z overlinez$. So equivalently we have to prove
$$ z overlinez - left(fraczoverlinez^n + fracoverlinezz^n right) + frac1z^n overlinez^n geq zoverlinez - (z + overlinez) + 1.$$ This is equivalent to
$$2 operatornameRe(z) + frac1 - left(z^n+1 + overlinez^n+1 right) geq 1 iff 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 1. $$ But the estimation $$ 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 2 operatornameRe(z) + frac1 - frac2 overset?geq 1 $$ almost finishes. Note that $operatornameRe(z) geq 1$ by assumption. As $z neq 1$ by assumption, $frac2 leq 1$, so again, we're done.
Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.
answered Mar 26 at 11:04
KezerKezer
1,405621
$endgroup$
add a comment |
$begingroup$
The inequality is true. For $n=0,1$ the inequality is clear, hence let $n geq 2$. The case $z=1$ is also clear, so let $z neq 1$.
Note that $1+z+dots + z^n = fracz^n+1-1z-1$. So the given inequality is equivalent to
$$ left|z - frac1z^n right| geq |z-1| iff left|z - frac1z^n right|^2 geq |z-1|^2.$$
Now adhere to $|z|^2 = z overlinez$. So equivalently we have to prove
$$ z overlinez - left(fraczoverlinez^n + fracoverlinezz^n right) + frac1z^n overlinez^n geq zoverlinez - (z + overlinez) + 1.$$ This is equivalent to
$$2 operatornameRe(z) + frac1 - left(z^n+1 + overlinez^n+1 right) geq 1 iff 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 1. $$ But the estimation $$ 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 2 operatornameRe(z) + frac1 - frac2 overset?geq 1 $$ almost finishes. Note that $operatornameRe(z) geq 1$ by assumption. As $z neq 1$ by assumption, $frac2 leq 1$, so again, we're done.
Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.
answered Mar 26 at 11:04
KezerKezer
1,405621
$endgroup$
add a comment |
$begingroup$
The inequality is true. For $n=0,1$ the inequality is clear, hence let $n geq 2$. The case $z=1$ is also clear, so let $z neq 1$.
Note that $1+z+dots + z^n = fracz^n+1-1z-1$. So the given inequality is equivalent to
$$ left|z - frac1z^n right| geq |z-1| iff left|z - frac1z^n right|^2 geq |z-1|^2.$$
Now adhere to $|z|^2 = z overlinez$. So equivalently we have to prove
$$ z overlinez - left(fraczoverlinez^n + fracoverlinezz^n right) + frac1z^n overlinez^n geq zoverlinez - (z + overlinez) + 1.$$ This is equivalent to
$$2 operatornameRe(z) + frac1 - left(z^n+1 + overlinez^n+1 right) geq 1 iff 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 1. $$ But the estimation $$ 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 2 operatornameRe(z) + frac1 - frac2 overset?geq 1 $$ almost finishes. Note that $operatornameRe(z) geq 1$ by assumption. As $z neq 1$ by assumption, $frac2 leq 1$, so again, we're done.
Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.
answered Mar 26 at 11:04
KezerKezer
1,405621
$endgroup$
The inequality is true. For $n=0,1$ the inequality is clear, hence let $n geq 2$. The case $z=1$ is also clear, so let $z neq 1$.
Note that $1+z+dots + z^n = fracz^n+1-1z-1$. So the given inequality is equivalent to
$$ left|z - frac1z^n right| geq |z-1| iff left|z - frac1z^n right|^2 geq |z-1|^2.$$
Now adhere to $|z|^2 = z overlinez$. So equivalently we have to prove
$$ z overlinez - left(fraczoverlinez^n + fracoverlinezz^n right) + frac1z^n overlinez^n geq zoverlinez - (z + overlinez) + 1.$$ This is equivalent to
$$2 operatornameRe(z) + frac1 - left(z^n+1 + overlinez^n+1 right) geq 1 iff 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 1. $$ But the estimation $$ 2 operatornameRe(z) + frac1 - 2 operatornameReleft(z^n+1 right) geq 2 operatornameRe(z) + frac1 - frac2 overset?geq 1 $$ almost finishes. Note that $operatornameRe(z) geq 1$ by assumption. As $z neq 1$ by assumption, $frac2 leq 1$, so again, we're done.
Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.
answered Mar 26 at 11:04
KezerKezer
1,405621
edited Mar 26 at 11:26
answered Mar 26 at 11:04
KezerKezer
1,405621
answered Mar 26 at 11:04
KezerKezer
1,405621
answered Mar 26 at 11:04
KezerKezer
1,405621
1,405621
add a comment |
add a comment |
$begingroup$
Why does it seem to be true? For which integers did you verify it to be true? Do you have an example of equality?
$endgroup$
– uniquesolution
Mar 26 at 9:05
$begingroup$
I put several integers... That doesn't contradict this inequality
$endgroup$
– user561527
Mar 26 at 9:10
$begingroup$
It‘s definitely true for $n = 2$, I‘ve just verified that.
$endgroup$
– Kezer
Mar 26 at 10:28