Mixing
negative/inverse triangular inequality, asymptotic analysis
$begingroup$
For $r$ sufficiently large and $|z|ge r$ we have $$|z^2-4z+3|ge||z|^2-|4z-3||ge|z|^2-4|z|-3gefrac{|z|^2}{2}ge{r^2over2}$$
How do we get the second to last inequality?
Is it still true if we replace $2$ by any $alphain Bbb R_{>1}$?
inequality asymptotics
asked Jan 9 at 8:29
John CataldoJohn Cataldo
1,1841316
$endgroup$
add a comment |
$begingroup$
For $r$ sufficiently large and $|z|ge r$ we have $$|z^2-4z+3|ge||z|^2-|4z-3||ge|z|^2-4|z|-3gefrac{|z|^2}{2}ge{r^2over2}$$
How do we get the second to last inequality?
Is it still true if we replace $2$ by any $alphain Bbb R_{>1}$?
inequality asymptotics
asked Jan 9 at 8:29
John CataldoJohn Cataldo
1,1841316
$endgroup$
add a comment |
$begingroup$
For $r$ sufficiently large and $|z|ge r$ we have $$|z^2-4z+3|ge||z|^2-|4z-3||ge|z|^2-4|z|-3gefrac{|z|^2}{2}ge{r^2over2}$$
How do we get the second to last inequality?
Is it still true if we replace $2$ by any $alphain Bbb R_{>1}$?
inequality asymptotics
asked Jan 9 at 8:29
John CataldoJohn Cataldo
1,1841316
$endgroup$
For $r$ sufficiently large and $|z|ge r$ we have $$|z^2-4z+3|ge||z|^2-|4z-3||ge|z|^2-4|z|-3gefrac{|z|^2}{2}ge{r^2over2}$$
How do we get the second to last inequality?
Is it still true if we replace $2$ by any $alphain Bbb R_{>1}$?
inequality asymptotics
inequality asymptotics
asked Jan 9 at 8:29
John CataldoJohn Cataldo
1,1841316
asked Jan 9 at 8:29
John CataldoJohn Cataldo
1,1841316
edited Jan 9 at 8:37
John Cataldo
asked Jan 9 at 8:29
John CataldoJohn Cataldo
1,1841316
asked Jan 9 at 8:29
John CataldoJohn Cataldo
1,1841316
asked Jan 9 at 8:29
John CataldoJohn Cataldo
1,1841316
1,1841316
add a comment |
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
Parabolas grow faster than lines. That is, $1/2x^2>ax+b$ for $x$ large enough for all $a,b$. So if $|z|$ is sufficiently large, $1/2|z|^2>4|z|+3$.
answered Jan 9 at 8:35
RptoughsRptoughs
62
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$begingroup$
For $|z| ge r > 0$:
$$
|z|^2-4|z|-3 = |z|^2 left( 1 - frac{4}{|z|} -frac{3}{|z|^2} right)
ge |z|^2 left( 1 - frac{4}{r} -frac{3}{r^2} right)
$$
For $r > 0$ sufficiently large the second factor is $ge frac 12$.
answered Jan 9 at 8:35
Martin RMartin R
28k33355
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$begingroup$
Concerning your second question about replacing $2$ by $alpha > 1$:
Yes, the claim is still true for sufficiently large $r$.
Let $beta = 1-frac{1}{alpha}$, hence $beta in (0,1)$:
begin{eqnarray*}
|z|^2-4|z|-3 &ge & frac{|z|^2}{alpha }\
&Leftrightarrow & \
beta|z|^2 -4|z| &ge & 3\
&Leftrightarrow & \
|z|left( beta|z| -4right) &ge & 3\
end{eqnarray*}
The last inequality is surely satisfied for $boxed{|z| > maxleft( frac{7}{beta},3 right)}$.
answered Jan 9 at 9:39
trancelocationtrancelocation
10.8k1723
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Parabolas grow faster than lines. That is, $1/2x^2>ax+b$ for $x$ large enough for all $a,b$. So if $|z|$ is sufficiently large, $1/2|z|^2>4|z|+3$.
answered Jan 9 at 8:35
RptoughsRptoughs
62
$endgroup$
add a comment |
$begingroup$
Parabolas grow faster than lines. That is, $1/2x^2>ax+b$ for $x$ large enough for all $a,b$. So if $|z|$ is sufficiently large, $1/2|z|^2>4|z|+3$.
answered Jan 9 at 8:35
RptoughsRptoughs
62
$endgroup$
add a comment |
$begingroup$
Parabolas grow faster than lines. That is, $1/2x^2>ax+b$ for $x$ large enough for all $a,b$. So if $|z|$ is sufficiently large, $1/2|z|^2>4|z|+3$.
answered Jan 9 at 8:35
RptoughsRptoughs
62
$endgroup$
Parabolas grow faster than lines. That is, $1/2x^2>ax+b$ for $x$ large enough for all $a,b$. So if $|z|$ is sufficiently large, $1/2|z|^2>4|z|+3$.
answered Jan 9 at 8:35
RptoughsRptoughs
62
answered Jan 9 at 8:35
RptoughsRptoughs
62
answered Jan 9 at 8:35
RptoughsRptoughs
62
answered Jan 9 at 8:35
RptoughsRptoughs
62
62
add a comment |
add a comment |
$begingroup$
For $|z| ge r > 0$:
$$
|z|^2-4|z|-3 = |z|^2 left( 1 - frac{4}{|z|} -frac{3}{|z|^2} right)
ge |z|^2 left( 1 - frac{4}{r} -frac{3}{r^2} right)
$$
For $r > 0$ sufficiently large the second factor is $ge frac 12$.
answered Jan 9 at 8:35
Martin RMartin R
28k33355
$endgroup$
add a comment |
$begingroup$
For $|z| ge r > 0$:
$$
|z|^2-4|z|-3 = |z|^2 left( 1 - frac{4}{|z|} -frac{3}{|z|^2} right)
ge |z|^2 left( 1 - frac{4}{r} -frac{3}{r^2} right)
$$
For $r > 0$ sufficiently large the second factor is $ge frac 12$.
answered Jan 9 at 8:35
Martin RMartin R
28k33355
$endgroup$
add a comment |
$begingroup$
For $|z| ge r > 0$:
$$
|z|^2-4|z|-3 = |z|^2 left( 1 - frac{4}{|z|} -frac{3}{|z|^2} right)
ge |z|^2 left( 1 - frac{4}{r} -frac{3}{r^2} right)
$$
For $r > 0$ sufficiently large the second factor is $ge frac 12$.
answered Jan 9 at 8:35
Martin RMartin R
28k33355
$endgroup$
For $|z| ge r > 0$:
$$
|z|^2-4|z|-3 = |z|^2 left( 1 - frac{4}{|z|} -frac{3}{|z|^2} right)
ge |z|^2 left( 1 - frac{4}{r} -frac{3}{r^2} right)
$$
For $r > 0$ sufficiently large the second factor is $ge frac 12$.
answered Jan 9 at 8:35
Martin RMartin R
28k33355
edited Jan 9 at 8:47
answered Jan 9 at 8:35
Martin RMartin R
28k33355
answered Jan 9 at 8:35
Martin RMartin R
28k33355
answered Jan 9 at 8:35
Martin RMartin R
28k33355
28k33355
add a comment |
add a comment |
$begingroup$
Concerning your second question about replacing $2$ by $alpha > 1$:
Yes, the claim is still true for sufficiently large $r$.
Let $beta = 1-frac{1}{alpha}$, hence $beta in (0,1)$:
begin{eqnarray*}
|z|^2-4|z|-3 &ge & frac{|z|^2}{alpha }\
&Leftrightarrow & \
beta|z|^2 -4|z| &ge & 3\
&Leftrightarrow & \
|z|left( beta|z| -4right) &ge & 3\
end{eqnarray*}
The last inequality is surely satisfied for $boxed{|z| > maxleft( frac{7}{beta},3 right)}$.
answered Jan 9 at 9:39
trancelocationtrancelocation
10.8k1723
$endgroup$
add a comment |
$begingroup$
Concerning your second question about replacing $2$ by $alpha > 1$:
Yes, the claim is still true for sufficiently large $r$.
Let $beta = 1-frac{1}{alpha}$, hence $beta in (0,1)$:
begin{eqnarray*}
|z|^2-4|z|-3 &ge & frac{|z|^2}{alpha }\
&Leftrightarrow & \
beta|z|^2 -4|z| &ge & 3\
&Leftrightarrow & \
|z|left( beta|z| -4right) &ge & 3\
end{eqnarray*}
The last inequality is surely satisfied for $boxed{|z| > maxleft( frac{7}{beta},3 right)}$.
answered Jan 9 at 9:39
trancelocationtrancelocation
10.8k1723
$endgroup$
add a comment |
$begingroup$
Concerning your second question about replacing $2$ by $alpha > 1$:
Yes, the claim is still true for sufficiently large $r$.
Let $beta = 1-frac{1}{alpha}$, hence $beta in (0,1)$:
begin{eqnarray*}
|z|^2-4|z|-3 &ge & frac{|z|^2}{alpha }\
&Leftrightarrow & \
beta|z|^2 -4|z| &ge & 3\
&Leftrightarrow & \
|z|left( beta|z| -4right) &ge & 3\
end{eqnarray*}
The last inequality is surely satisfied for $boxed{|z| > maxleft( frac{7}{beta},3 right)}$.
answered Jan 9 at 9:39
trancelocationtrancelocation
10.8k1723
$endgroup$
Concerning your second question about replacing $2$ by $alpha > 1$:
Yes, the claim is still true for sufficiently large $r$.
Let $beta = 1-frac{1}{alpha}$, hence $beta in (0,1)$:
begin{eqnarray*}
|z|^2-4|z|-3 &ge & frac{|z|^2}{alpha }\
&Leftrightarrow & \
beta|z|^2 -4|z| &ge & 3\
&Leftrightarrow & \
|z|left( beta|z| -4right) &ge & 3\
end{eqnarray*}
The last inequality is surely satisfied for $boxed{|z| > maxleft( frac{7}{beta},3 right)}$.
answered Jan 9 at 9:39
trancelocationtrancelocation
10.8k1723
edited Jan 9 at 9:45
answered Jan 9 at 9:39
trancelocationtrancelocation
10.8k1723
answered Jan 9 at 9:39
trancelocationtrancelocation
10.8k1723
answered Jan 9 at 9:39
trancelocationtrancelocation
10.8k1723
10.8k1723
add a comment |
add a comment |
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