simple sup norm for a function on a unit circleHow to find the supremum of this?Geometric conditions equivalent to a set being the unit circle for some normWhy can we calculate the supremum of operator norm over unit circle?tangents at unit circle - parametrization leads to strange resultIs there a meaning to the notation “arg sup”?Show $sqrt2 = textsupA$ where $A = x in mathbb Q ;colon x^2 < 2$Supremum and infimum of a rational set bounded by irrational numbersMaximum and Supremum of IntervalsIntuition for $limsup$ and $liminf$For some $c geq 0$ $textsup c cdot f(x): textsome domain of $x$ $ = $c cdot textsup f(x): textsame domain of $x$ $
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simple sup norm for a function on a unit circle
How to find the supremum of this?Geometric conditions equivalent to a set being the unit circle for some normWhy can we calculate the supremum of operator norm over unit circle?tangents at unit circle - parametrization leads to strange resultIs there a meaning to the notation “arg sup”?Show $sqrt2 = textsupA$ where $A = x in mathbb Q ;colon x^2 < 2$Supremum and infimum of a rational set bounded by irrational numbersMaximum and Supremum of IntervalsIntuition for $limsup$ and $liminf$For some $c geq 0$ $textsup c cdot f(x): textsome domain of $x$ $ = $c cdot textsup f(x): textsame domain of $x$ $
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
From my understanding , the supremum norm would just be the maximum value
So if I have a point z that lies on the unit circle $|z|=1$
Question:
what would the $sup_ |2z-1|$ be ....
Is it just the maximum value of $(2z-1)$ for all points on the unit circle?
I attempted it in to ways, not sure if either is correct
First method
$ sup_ |2z-1| = 1$ As you simply just take the $|z|=1 $ so $ |2(1)-1|=1$ I dont think this is correct
Second method:
You just let $z = x+iy$ where $|z|=1$
$|2(x+iy)-1| $ $= |(2x-1)-2iy| = $ $sqrt4x^2-4x-1+4y^2$ so if z lies on the unit circle $x^2 + y^2 = 1 $ or $ x = cosθ $, $y = sinθ$
So the sup would just be the maximum value for $sqrt3-4cosθ = sqrt7? $
Not sure what the maximum value for $sup_ |2z-1|$ would be here.
real-analysis maxima-minima supremum-and-infimum
$endgroup$
add a comment |
$begingroup$
From my understanding , the supremum norm would just be the maximum value
So if I have a point z that lies on the unit circle $|z|=1$
Question:
what would the $sup_ |2z-1|$ be ....
Is it just the maximum value of $(2z-1)$ for all points on the unit circle?
I attempted it in to ways, not sure if either is correct
First method
$ sup_ |2z-1| = 1$ As you simply just take the $|z|=1 $ so $ |2(1)-1|=1$ I dont think this is correct
Second method:
You just let $z = x+iy$ where $|z|=1$
$|2(x+iy)-1| $ $= |(2x-1)-2iy| = $ $sqrt4x^2-4x-1+4y^2$ so if z lies on the unit circle $x^2 + y^2 = 1 $ or $ x = cosθ $, $y = sinθ$
So the sup would just be the maximum value for $sqrt3-4cosθ = sqrt7? $
Not sure what the maximum value for $sup_ |2z-1|$ would be here.
real-analysis maxima-minima supremum-and-infimum
$endgroup$
add a comment |
$begingroup$
From my understanding , the supremum norm would just be the maximum value
So if I have a point z that lies on the unit circle $|z|=1$
Question:
what would the $sup_ |2z-1|$ be ....
Is it just the maximum value of $(2z-1)$ for all points on the unit circle?
I attempted it in to ways, not sure if either is correct
First method
$ sup_ |2z-1| = 1$ As you simply just take the $|z|=1 $ so $ |2(1)-1|=1$ I dont think this is correct
Second method:
You just let $z = x+iy$ where $|z|=1$
$|2(x+iy)-1| $ $= |(2x-1)-2iy| = $ $sqrt4x^2-4x-1+4y^2$ so if z lies on the unit circle $x^2 + y^2 = 1 $ or $ x = cosθ $, $y = sinθ$
So the sup would just be the maximum value for $sqrt3-4cosθ = sqrt7? $
Not sure what the maximum value for $sup_ |2z-1|$ would be here.
real-analysis maxima-minima supremum-and-infimum
$endgroup$
From my understanding , the supremum norm would just be the maximum value
So if I have a point z that lies on the unit circle $|z|=1$
Question:
what would the $sup_ |2z-1|$ be ....
Is it just the maximum value of $(2z-1)$ for all points on the unit circle?
I attempted it in to ways, not sure if either is correct
First method
$ sup_ |2z-1| = 1$ As you simply just take the $|z|=1 $ so $ |2(1)-1|=1$ I dont think this is correct
Second method:
You just let $z = x+iy$ where $|z|=1$
$|2(x+iy)-1| $ $= |(2x-1)-2iy| = $ $sqrt4x^2-4x-1+4y^2$ so if z lies on the unit circle $x^2 + y^2 = 1 $ or $ x = cosθ $, $y = sinθ$
So the sup would just be the maximum value for $sqrt3-4cosθ = sqrt7? $
Not sure what the maximum value for $sup_ |2z-1|$ would be here.
real-analysis maxima-minima supremum-and-infimum
real-analysis maxima-minima supremum-and-infimum
edited Jun 30 at 8:42
Bernard
129k7 gold badges43 silver badges122 bronze badges
129k7 gold badges43 silver badges122 bronze badges
asked Jun 30 at 8:14
Swag34214Swag34214
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4 Answers
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$begingroup$
The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.
The second method is much better, however you seem to misplaced your $-$ signs. You should get
$$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
which has a maximum of $sqrt9 = 3$.
This could also be achieved through use of the triangle inequality. Consider
$$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.
$endgroup$
$begingroup$
Ahh yes thank you for the clarification.
$endgroup$
– Swag34214
Jun 30 at 8:38
add a comment |
$begingroup$
Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).
By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.
Thus, $$sup_ |2z - 1| = 3.$$
$endgroup$
add a comment |
$begingroup$
The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".
The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).
$endgroup$
add a comment |
$begingroup$
Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
$$sup_|2z-1|=2cdotfrac 32=colorred3.$$
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.
The second method is much better, however you seem to misplaced your $-$ signs. You should get
$$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
which has a maximum of $sqrt9 = 3$.
This could also be achieved through use of the triangle inequality. Consider
$$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.
$endgroup$
$begingroup$
Ahh yes thank you for the clarification.
$endgroup$
– Swag34214
Jun 30 at 8:38
add a comment |
$begingroup$
The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.
The second method is much better, however you seem to misplaced your $-$ signs. You should get
$$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
which has a maximum of $sqrt9 = 3$.
This could also be achieved through use of the triangle inequality. Consider
$$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.
$endgroup$
$begingroup$
Ahh yes thank you for the clarification.
$endgroup$
– Swag34214
Jun 30 at 8:38
add a comment |
$begingroup$
The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.
The second method is much better, however you seem to misplaced your $-$ signs. You should get
$$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
which has a maximum of $sqrt9 = 3$.
This could also be achieved through use of the triangle inequality. Consider
$$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.
$endgroup$
The first method doesn't work. The condition $|z| = 1$ doesn't imply $z = 1$, nor is $|2z - 1|$ the same as $2|z| - 1$.
The second method is much better, however you seem to misplaced your $-$ signs. You should get
$$|(2x - 1) colorred+ 2iy| = sqrt4x^2 - 4x colorred+ 1 + 4y^2 = sqrt5 - 4cos theta,$$
which has a maximum of $sqrt9 = 3$.
This could also be achieved through use of the triangle inequality. Consider
$$|2z - 1| le |2z| + |-1| = 2|z| + 1 = 3,$$
with equality if and only if $2z$ and $-1$ point in the same direction, i.e. when $z$ is a negative number. Substituting in $z = -1$ confirms that the value of $3$ is attained.
edited Jun 30 at 9:12
answered Jun 30 at 8:32
Theo BenditTheo Bendit
24.9k1 gold badge25 silver badges61 bronze badges
24.9k1 gold badge25 silver badges61 bronze badges
$begingroup$
Ahh yes thank you for the clarification.
$endgroup$
– Swag34214
Jun 30 at 8:38
add a comment |
$begingroup$
Ahh yes thank you for the clarification.
$endgroup$
– Swag34214
Jun 30 at 8:38
$begingroup$
Ahh yes thank you for the clarification.
$endgroup$
– Swag34214
Jun 30 at 8:38
$begingroup$
Ahh yes thank you for the clarification.
$endgroup$
– Swag34214
Jun 30 at 8:38
add a comment |
$begingroup$
Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).
By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.
Thus, $$sup_ |2z - 1| = 3.$$
$endgroup$
add a comment |
$begingroup$
Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).
By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.
Thus, $$sup_ |2z - 1| = 3.$$
$endgroup$
add a comment |
$begingroup$
Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).
By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.
Thus, $$sup_ |2z - 1| = 3.$$
$endgroup$
Geometrically, for $|z| = 1$, $|2z - 1|$ is the distance between a point on the circle of radius $2$ centered at the origin (in red), and the point $(1,0)$ (in blue).
By looking at the attached image, we can see that the maximum such distance is $3$. This can more easily be seen by considering a plot of the circle of radius $3$ centered at $(1,0)$ (in green). The point $(-2,0)$ is a distance of $3$ away from $(1,0)$, and each other point on the circle of radius $2$ is closer to $(1,0)$.
Thus, $$sup_ |2z - 1| = 3.$$
edited Jun 30 at 9:44
answered Jun 30 at 8:31
Eh WhaEh Wha
1415 bronze badges
1415 bronze badges
add a comment |
add a comment |
$begingroup$
The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".
The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).
$endgroup$
add a comment |
$begingroup$
The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".
The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).
$endgroup$
add a comment |
$begingroup$
The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".
The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).
$endgroup$
The first method is definitely wrong as you suggested. This is because $|z_1 + z_2| neq |z_1| + |z_2|$. In other words: you cannot just replace all variables inside $||$ with their "length".
The second method is correct since in every step you use a definition or some known fact. (Though you made a mistake when squaring $(2x-1)^2=4x^2-4x+1$. This gives you $sqrt5 - 4x)$. You do not need to represent $x$ using the angle $theta$. Just think about what $x$ would maximize this expression and can you find this value for $x$ on the unit circle. The conclusion should be that $x=-1$ which you can find on the unit circle (namely for $z=-1$).
answered Jun 30 at 8:34
araomisaraomis
8041 silver badge10 bronze badges
8041 silver badge10 bronze badges
add a comment |
add a comment |
$begingroup$
Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
$$sup_|2z-1|=2cdotfrac 32=colorred3.$$
$endgroup$
add a comment |
$begingroup$
Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
$$sup_|2z-1|=2cdotfrac 32=colorred3.$$
$endgroup$
add a comment |
$begingroup$
Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
$$sup_|2z-1|=2cdotfrac 32=colorred3.$$
$endgroup$
Note that $:|2z-1|=2Bigl|z-frac12Bigr|$, i.e. geometrically, twice the distance between the affix of $z$ and the point $;Abigl(frac12,0bigr)$ in the plane. It is attained at the intersection point of the diameter through $A$ (i.e. the $x$ axis) which has a negative abscissa: $(-1,0)$. Thus
$$sup_|2z-1|=2cdotfrac 32=colorred3.$$
answered Jun 30 at 8:53
BernardBernard
129k7 gold badges43 silver badges122 bronze badges
129k7 gold badges43 silver badges122 bronze badges
add a comment |
add a comment |
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