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At what rate must the radius of a ball in a cusped domain shrink as it approaches the cusp?
Clash Royale CLAN TAG#URR8PPP
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For $pgeq 1$ let $D_p = ^1/p$. I would like some help to prove or disprove the following statement: There exists $epsilon>0$ such that for all $0< r< 1$, the containment $(x,y)in mathbb R^2: x^2 + (y - r)^2 < epsilon^2 r^2psubset D_p$ holds.
It is easy to see that the statement is true in the case $p = 1$ (for example, $epsilon = frac12sqrt 2$ works with room to spare) because one can easily compute the distance from the point $(0, r)$ to the boundary of $D_1$. The case $p>1$ is giving me trouble. In fact, I'd even be happy if I could gain some insight for the case $p = 2$.
real-analysis geometry analytic-geometry
asked Jul 29 at 3:46
BindersFull
498110
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up vote
0
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favorite
For $pgeq 1$ let $D_p = ^1/p$. I would like some help to prove or disprove the following statement: There exists $epsilon>0$ such that for all $0< r< 1$, the containment $(x,y)in mathbb R^2: x^2 + (y - r)^2 < epsilon^2 r^2psubset D_p$ holds.
It is easy to see that the statement is true in the case $p = 1$ (for example, $epsilon = frac12sqrt 2$ works with room to spare) because one can easily compute the distance from the point $(0, r)$ to the boundary of $D_1$. The case $p>1$ is giving me trouble. In fact, I'd even be happy if I could gain some insight for the case $p = 2$.
real-analysis geometry analytic-geometry
asked Jul 29 at 3:46
BindersFull
498110
1
Rewrite your set as $(x,y):$. Since $y^p$ is convex in $y$, the linearization $(x,y): $ is a subset for any $y_0$ and will give you a lower bound for the distance to the boundary. Take $y_0=r$, and I expect you'll obtain the desired result.
– Rahul
Jul 29 at 4:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For $pgeq 1$ let $D_p = ^1/p$. I would like some help to prove or disprove the following statement: There exists $epsilon>0$ such that for all $0< r< 1$, the containment $(x,y)in mathbb R^2: x^2 + (y - r)^2 < epsilon^2 r^2psubset D_p$ holds.
It is easy to see that the statement is true in the case $p = 1$ (for example, $epsilon = frac12sqrt 2$ works with room to spare) because one can easily compute the distance from the point $(0, r)$ to the boundary of $D_1$. The case $p>1$ is giving me trouble. In fact, I'd even be happy if I could gain some insight for the case $p = 2$.
real-analysis geometry analytic-geometry
asked Jul 29 at 3:46
BindersFull
498110
For $pgeq 1$ let $D_p = ^1/p$. I would like some help to prove or disprove the following statement: There exists $epsilon>0$ such that for all $0< r< 1$, the containment $(x,y)in mathbb R^2: x^2 + (y - r)^2 < epsilon^2 r^2psubset D_p$ holds.
It is easy to see that the statement is true in the case $p = 1$ (for example, $epsilon = frac12sqrt 2$ works with room to spare) because one can easily compute the distance from the point $(0, r)$ to the boundary of $D_1$. The case $p>1$ is giving me trouble. In fact, I'd even be happy if I could gain some insight for the case $p = 2$.
real-analysis geometry analytic-geometry
asked Jul 29 at 3:46
BindersFull
498110
asked Jul 29 at 3:46
BindersFull
498110
asked Jul 29 at 3:46
BindersFull
498110
asked Jul 29 at 3:46
BindersFull
498110
498110
1
Rewrite your set as $(x,y):$. Since $y^p$ is convex in $y$, the linearization $(x,y): $ is a subset for any $y_0$ and will give you a lower bound for the distance to the boundary. Take $y_0=r$, and I expect you'll obtain the desired result.
– Rahul
Jul 29 at 4:55
add a comment |Â
1
Rewrite your set as $(x,y):$. Since $y^p$ is convex in $y$, the linearization $(x,y): $ is a subset for any $y_0$ and will give you a lower bound for the distance to the boundary. Take $y_0=r$, and I expect you'll obtain the desired result.
– Rahul
Jul 29 at 4:55
1
1
Rewrite your set as $(x,y):$. Since $y^p$ is convex in $y$, the linearization $(x,y): $ is a subset for any $y_0$ and will give you a lower bound for the distance to the boundary. Take $y_0=r$, and I expect you'll obtain the desired result.
– Rahul
Jul 29 at 4:55
Rewrite your set as $(x,y):$. Since $y^p$ is convex in $y$, the linearization $(x,y): $ is a subset for any $y_0$ and will give you a lower bound for the distance to the boundary. Take $y_0=r$, and I expect you'll obtain the desired result.
– Rahul
Jul 29 at 4:55
add a comment |Â
1 Answer
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The case $p=1$ is simple, and needs special treatment. In the following I assume $p>1$ and reflect the figure on the line $x=y$. We therefore consider the curve
$$gamma:quad tmapstobf z(t):=(t,t^p)qquad(tgeq0)$$
with tangent vector $bf z'(t)=(1,pt^p-1)$.
Fix a $t>0$. The downwards normal $nu$ at $bf z(t)$ is given by
$$nu:quad smapsto bf n(s)=(t,t^p)+s(p t^p-1,-1)qquad(sgeq0) .$$
This normal intersects the $x$-axis when $s=t^p$ at the point $bf r(t)=bigl(r(t),0bigr)$, whereby
$$r(t)=t+p t^2p-1 .tag1$$
As $gamma$ is convex the function $tmapsto r(t)$ is a monotonically increasing function of $t$. The distance $rho(t):=|bf z(t)-bf r(t)|$ computes to
$$rho(t)=t^psqrt1+p^2 t^2p-2 .tag2$$
For your problem we have to express $t$ as a function of $r$, using $(1)$. One obtains
$$t=r-bigl(p+o(1)bigr)r^2p-1qquad(rto0+) .$$
Now you have to plug this into $(2)$ and obtain, after some calculations,
$$rho=r^pbigl(1+ (c+o(1)) r^alphabigr)qquad(rto0+)$$
whereby I leave the value of the $c$ and the $alpha>0$ to you.
answered Jul 29 at 12:49
Christian Blatter
163k7107306
Perfect! Thank you Christian!
– BindersFull
Jul 29 at 14:38
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The case $p=1$ is simple, and needs special treatment. In the following I assume $p>1$ and reflect the figure on the line $x=y$. We therefore consider the curve
$$gamma:quad tmapstobf z(t):=(t,t^p)qquad(tgeq0)$$
with tangent vector $bf z'(t)=(1,pt^p-1)$.
Fix a $t>0$. The downwards normal $nu$ at $bf z(t)$ is given by
$$nu:quad smapsto bf n(s)=(t,t^p)+s(p t^p-1,-1)qquad(sgeq0) .$$
This normal intersects the $x$-axis when $s=t^p$ at the point $bf r(t)=bigl(r(t),0bigr)$, whereby
$$r(t)=t+p t^2p-1 .tag1$$
As $gamma$ is convex the function $tmapsto r(t)$ is a monotonically increasing function of $t$. The distance $rho(t):=|bf z(t)-bf r(t)|$ computes to
$$rho(t)=t^psqrt1+p^2 t^2p-2 .tag2$$
For your problem we have to express $t$ as a function of $r$, using $(1)$. One obtains
$$t=r-bigl(p+o(1)bigr)r^2p-1qquad(rto0+) .$$
Now you have to plug this into $(2)$ and obtain, after some calculations,
$$rho=r^pbigl(1+ (c+o(1)) r^alphabigr)qquad(rto0+)$$
whereby I leave the value of the $c$ and the $alpha>0$ to you.
answered Jul 29 at 12:49
Christian Blatter
163k7107306
Perfect! Thank you Christian!
– BindersFull
Jul 29 at 14:38
add a comment |Â
up vote
1
down vote
accepted
The case $p=1$ is simple, and needs special treatment. In the following I assume $p>1$ and reflect the figure on the line $x=y$. We therefore consider the curve
$$gamma:quad tmapstobf z(t):=(t,t^p)qquad(tgeq0)$$
with tangent vector $bf z'(t)=(1,pt^p-1)$.
Fix a $t>0$. The downwards normal $nu$ at $bf z(t)$ is given by
$$nu:quad smapsto bf n(s)=(t,t^p)+s(p t^p-1,-1)qquad(sgeq0) .$$
This normal intersects the $x$-axis when $s=t^p$ at the point $bf r(t)=bigl(r(t),0bigr)$, whereby
$$r(t)=t+p t^2p-1 .tag1$$
As $gamma$ is convex the function $tmapsto r(t)$ is a monotonically increasing function of $t$. The distance $rho(t):=|bf z(t)-bf r(t)|$ computes to
$$rho(t)=t^psqrt1+p^2 t^2p-2 .tag2$$
For your problem we have to express $t$ as a function of $r$, using $(1)$. One obtains
$$t=r-bigl(p+o(1)bigr)r^2p-1qquad(rto0+) .$$
Now you have to plug this into $(2)$ and obtain, after some calculations,
$$rho=r^pbigl(1+ (c+o(1)) r^alphabigr)qquad(rto0+)$$
whereby I leave the value of the $c$ and the $alpha>0$ to you.
answered Jul 29 at 12:49
Christian Blatter
163k7107306
Perfect! Thank you Christian!
– BindersFull
Jul 29 at 14:38
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The case $p=1$ is simple, and needs special treatment. In the following I assume $p>1$ and reflect the figure on the line $x=y$. We therefore consider the curve
$$gamma:quad tmapstobf z(t):=(t,t^p)qquad(tgeq0)$$
with tangent vector $bf z'(t)=(1,pt^p-1)$.
Fix a $t>0$. The downwards normal $nu$ at $bf z(t)$ is given by
$$nu:quad smapsto bf n(s)=(t,t^p)+s(p t^p-1,-1)qquad(sgeq0) .$$
This normal intersects the $x$-axis when $s=t^p$ at the point $bf r(t)=bigl(r(t),0bigr)$, whereby
$$r(t)=t+p t^2p-1 .tag1$$
As $gamma$ is convex the function $tmapsto r(t)$ is a monotonically increasing function of $t$. The distance $rho(t):=|bf z(t)-bf r(t)|$ computes to
$$rho(t)=t^psqrt1+p^2 t^2p-2 .tag2$$
For your problem we have to express $t$ as a function of $r$, using $(1)$. One obtains
$$t=r-bigl(p+o(1)bigr)r^2p-1qquad(rto0+) .$$
Now you have to plug this into $(2)$ and obtain, after some calculations,
$$rho=r^pbigl(1+ (c+o(1)) r^alphabigr)qquad(rto0+)$$
whereby I leave the value of the $c$ and the $alpha>0$ to you.
answered Jul 29 at 12:49
Christian Blatter
163k7107306
The case $p=1$ is simple, and needs special treatment. In the following I assume $p>1$ and reflect the figure on the line $x=y$. We therefore consider the curve
$$gamma:quad tmapstobf z(t):=(t,t^p)qquad(tgeq0)$$
with tangent vector $bf z'(t)=(1,pt^p-1)$.
Fix a $t>0$. The downwards normal $nu$ at $bf z(t)$ is given by
$$nu:quad smapsto bf n(s)=(t,t^p)+s(p t^p-1,-1)qquad(sgeq0) .$$
This normal intersects the $x$-axis when $s=t^p$ at the point $bf r(t)=bigl(r(t),0bigr)$, whereby
$$r(t)=t+p t^2p-1 .tag1$$
As $gamma$ is convex the function $tmapsto r(t)$ is a monotonically increasing function of $t$. The distance $rho(t):=|bf z(t)-bf r(t)|$ computes to
$$rho(t)=t^psqrt1+p^2 t^2p-2 .tag2$$
For your problem we have to express $t$ as a function of $r$, using $(1)$. One obtains
$$t=r-bigl(p+o(1)bigr)r^2p-1qquad(rto0+) .$$
Now you have to plug this into $(2)$ and obtain, after some calculations,
$$rho=r^pbigl(1+ (c+o(1)) r^alphabigr)qquad(rto0+)$$
whereby I leave the value of the $c$ and the $alpha>0$ to you.
answered Jul 29 at 12:49
Christian Blatter
163k7107306
answered Jul 29 at 12:49
Christian Blatter
163k7107306
answered Jul 29 at 12:49
Christian Blatter
163k7107306
answered Jul 29 at 12:49
Christian Blatter
163k7107306
163k7107306
Perfect! Thank you Christian!
– BindersFull
Jul 29 at 14:38
add a comment |Â
Perfect! Thank you Christian!
– BindersFull
Jul 29 at 14:38
Perfect! Thank you Christian!
– BindersFull
Jul 29 at 14:38
Perfect! Thank you Christian!
– BindersFull
Jul 29 at 14:38
add a comment |Â
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1
Rewrite your set as $(x,y):$. Since $y^p$ is convex in $y$, the linearization $(x,y): $ is a subset for any $y_0$ and will give you a lower bound for the distance to the boundary. Take $y_0=r$, and I expect you'll obtain the desired result.
– Rahul
Jul 29 at 4:55