epsilon delta: limit test, such as x-> 2, for x ^ 2 = 4, with intervals (but without possible circular reasoning)

On page 200 (p. 208 in pdf) of Openstax Calculation, after solving the $ delta $, I wonder if it's a circular reasoning to use $ x ^ 2 $ in a test that involves $ x ^ 2 $ itself.

Although I know how to do it $ epsilon- delta $ limits with absolute values ​​(usually with $ lim_ {x to 2} x ^ 2 = 4 $we would have tied $ delta leq 1 $ and eventually $ delta = min {1, frac { epsilon} {5} } $) I also want to test it with intervals and see if I can find the proof that way.

I can get to the point where
$$ – (2- sqrt {4- epsilon}) <x-2 < sqrt {4+ epsilon} -2 $$
$$ – delta <x-a < delta $$
Where the leftmost expression is the $ delta $ (on the left) and the rightmost expression is the $ delta $ (on the right). Then I can find the $ delta $& # 39; s, but I'm not sure how to prove that these $ delta $indicate that $ | f (x) -L | < epsilon $ using intervals.

Link to book: https://d3bxy9euw4e147.cloudfront.net/oscms-prodcms/media/documents/CalculusVolume1-OP.pdf

geometric group theory: is there a circular packing theorem of a rapporteur?

Leave $ X_w $ be the presentation complex of a group of a rapporteur $ langle x_1, dotsc, x_n mid w rangle $, with $ w $ cyclically reduced, that is, $ X_w = R cup_w D $, with $ R $ the rose with $ n $ labeled petals $ x_1, dotsc, x_n $Y $ w colon partial D = S ^ 1 a R $ it's a dive that represents the word $ w $.

Leave $ f colon D a X_w $ be a reduced disk diagram in $ X_w $; $ D $ it is a complex flat plane (connected) simple connection whose edges (oriented) are labeled with $ x_i $ so that the word (cyclic) $ w $ (or $ w ^ 1 $) is read when crossing the limit of every two cells of $ D $ such that for any pair of two cells that are on a labeled edge $ x $, those occurrences of $ x $ they are in different places in $ w $. For each one $ D $, leave $ hat D $ be the complex of two planes that is obtained by forgetting the labels and eliminating all the two vertices of Valencia (except perhaps one in the case where $ D $ It has only one of two cells and is already essentially the disk).

A Weinbaum theorem says that there is no proper subword of $ w $ is in the normal closing of $ w $, which has the consequence that the closing of every two cells in $ D $ (and therefore $ hat D $) is the closed two cell. (Equivalently, the dual graph has no monogons.)

Consider now a compact (connected) of two complex flat planes simply connected $ E $ without valence-two vertices, so that the closing of every two cells is the closing of two cells. There is a word $ w $ and a reduced disk diagram $ f colon D a X_w $ such that $ E $ Y $ hat D $ Are they isomorphic as two complexes? Is Weinbaum's theorem the only restriction in the form of a reduced disk diagram in a group of a rapporteur?

Obviously the answer is no, but I can't prove anything.

Using Android phone sensors to track phone movement in a circular path

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Technique: guide path for circular time lapse

I am contemplating taking a picture of a subject from a different angle until I reach a complete circle. As for the frequency, let's assume something from the range of a photo every 1 degree (which would be an exhausting 360 photo) to something like every 10 degrees. I have seen photographers do this and the results are very good. The advice for this is to find a physical circular object outside to walk to the side.


Assuming we don't have a circular object simply lying in our backyard, what is another viable approach to get the most accurate circular path to walk while taking photos of your subject at more than 360 degrees?

circular file change is possible in the less command

I am thinking about how to access the buffers in vim. When it reaches the last buffer, it does not stop and emits a beep; Continue to the first buffer, which is smart and convenient.

Solving equations: how to solve the intersection points of two circular functions?

Solve can be used directly

pts = {x, y} /. 
   Solve({y^2 == 4 - 4 x^2, (1 - (x/2))^2 + (y - 1)^2 == 1}, {x, y}, 
    Reals) // FullSimplify

(* {{Root(144 - 160*#1 - 328*#1^2 + 
           120*#1^3 + 225*#1^4 & , 1, 
       0), Root(144 - 1280*#1 + 
           1688*#1^2 - 960*#1^3 + 
           225*#1^4 & , 2, 0)}, 
   {Root(144 - 160*#1 - 328*#1^2 + 
           120*#1^3 + 225*#1^4 & , 2, 
       0), Root(144 - 1280*#1 + 
           1688*#1^2 - 960*#1^3 + 
           225*#1^4 & , 1, 0)}} *)

Convert the Root objects to their numerical values

pts // N

(* {{0.539936, 1.68341}, {0.997732, 0.13463}} *)

   (y^2 == 4 - 4*x^2),
   ((1 - x/2)^2 + (y - 1)^2 == 1)},
  {x, -1.5, 4.5}, {y, -3, 3},
  Frame -> False,
  Axes -> True),
 Graphics({Red, AbsolutePointSize(4), Point(pts)}))

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How can I disassemble / repair a CPL (circular polarizing filter)?

One of my circular polarizing (rotating) filters has great hair between the two lenses. The hair moves and I always have to shake it aside until it is out of sight.

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Similar to this

How would I disassemble it and reassemble it (after getting rid of the little annoying hair)? I could not find any help / tutorial online and I am not sure how the pieces are welded.

ios – Animation line shaped like CAShapeLayer in a circular path

I am trying to animate my line in the form of CAShapeLayer along a circular path. I created an initial route of my line layer and the final route and added them to and from the CABasicAnimation values.

However, animation is not what I expected. I'm stuck here for the last 2 days.

func getAllLayers() -> (CALayer) 


    var layers = ()

    let pointerEndPoint = CGPoint(x: xPointOfPointer , y: yPointOfPointer)  // Final end point of pointer
    let radius = 5

    let pointerPath = UIBezierPath()
    pointerPath.addArc(withCenter: circleCenter, radius: radius, startAngle: 0, endAngle:2 * CGFloat.pi, clockwise: false)
    pointerPath.move(to: circleCenter)
    pointerPath.addLine(to: endPoint)

    let pointerPathLayer = CAShapeLayer()
    pointerPathLayer.path = pointerPath.cgPath
    pointerPathLayer.lineWidth = 2
    pointerPathLayer.fillColor = UIColor.black.cgColor
    pointerPathLayer.strokeColor = UIColor.black.cgColor

    let pointerInitialBezierPath = UIBezierPath()
    pointerInitialBezierPath.addArc(withCenter: circleCenter, radius: radius,startAngle: 0 , endAngle: 2 * CGFloat.pi , clockwise: false)
    pointerInitialBezierPath.move(to: circleCenter)
    pointerInitialBezierPath.addLine(to: CGPoint(x: circleCenter.x - 50 , y: centerY))

    let pointerInitialCGPath = pointerInitialBezierPath.cgPath
    let pointerFinalCGPath = pointerPathLayer.path

                    // Pointer Animation
    let pointerAnimation = CABasicAnimation(keyPath: #keyPath(CAShapeLayer.path))
     pointerAnimation.fromValue = pointerInitialCGPath
     pointerAnimation.toValue = pointerFinalCGPath
      pointerAnimation.duration = 0.5
      pointerAnimation.isCumulative = true
      pointerPathLayer.add(pointerAnimation, forKey: "Animation")

return layers

I looked for a solution in Stack over flow but I couldn't find my mistake. I want my animation to move like a clockwise.

My animation

Notice that the needle length is changing as the animation progresses. I want it to remain at a constant length and I just want the angle of the needle to change. I would greatly appreciate any help.

java – How to prevent the circular dependency from abstracting the navigation in the desktop application?

I'm trying to abstract some functionality from my desktop application (JavaFX) to change the views.

So, what I'm thinking is:

Browser interface with methods like goToAccountsOverview (), goToAccountDetails (account account) etc. The implementation will depend on a ViewFactory to create the views and a ViewingChanger to change the view that is displayed.

There will be multiple browser interfaces depending on the state of the application, starting with:

  • NotConnectedStateNavigator (goToConnectToDatabase (), goToCreateDatabase () etc)

  • ConnectedStateNavigator (goToAccountsOverview (), goToAccountDetails (account account) etc)

Therefore, each state will have its own browser and views factory and a common view changer. View factories and browsers will not share a common interface (createView (ViewEnum v) Y goToView (View View) respectively).

While implementing this, I run into a problem. An example of the problem is this:

At AccountsOverviewView There is a table with all the accounts. Each account has a button to view the details. Pressing that button should navigate to the AccountDetailsVista with the ViewModel of the account (I am using some type of MVVM).

To do that I would call connectedStateNavigator.goToAccountDetails (accountVM) what would i call viewChanger.changeView (connectedStateViewFactory.createAccountDetailsView (accountVM)).

So he AccountsOverviewView you need the ConnectedStateNavigator, who needs the ConnectedStateViewFactory, you need to be able to build the original AccountsOverviewView, who needs the ConnectedStateNavigator etc.

Some code to explain it better:

public class AccountsOverviewView {
public accounts Overview (ConnectedStateNavigator browser, ...) {
accountDetailsButton.onClick (e -> navigator.goToAccountDetails (account));

public class ConnectedStateNavigator {
ConnectedStateNavigator public (ConnectedStateViewFactory viewFactory, ...) {...}

public void goToAccountDetails (account account) {
viewChanger.changeView (viewFactory.createAccountDetailsView (account));

public class ConnectedStateViewFactory {
Public ConnectedStateViewFactory (ConnectedStateNavigator browser, ...) {...}

Public view createAccountsOverviewView () {
return new AccountsOverviewView (new AccountsOverviewViewModel (browser, ...));

Public view createAccountDetailsView (account account) {...}

How could I design this better to avoid this cyclical reference?
Is the only solution using a configurator or an asynchronous event system?

Complex Analysis – Circular Mapping Monday.

I want to map on circular Monday $ | z | <1, | z + i | <1 $ in the upper half plane

Since the points of intersection are $ z_1 = sqrt 3/2-i / 2 $ Y $ z_2 = – sqrt 3/2-i / 2 $ we have

$$ w_1 = frac {2z- sqrt 3 + i} {2z + sqrt 3 + i} $$

Now $ w_1 $ Monday's maps to the region between the lines. $ v = + sqrt 3 u $ Y $ v = – sqrt 3 u $ in the left half plane, that is

$$ 2 pi / 3 <arg w <4 pi / 3 $$

Now $ e ^ {- i2 pi / 3} w_1 $ maps for

$$ 0 <arg w <2 pi / 3 $$ and coming to power $ 3/2 $ We reach the upper half of the plane.

So my answer is $ e ^ {- i pi} w_1 ^ {3/2} $, but the answer at the end of my book is $ e ^ {i pi / 3} w_1 ^ 2 $. Could someone tell me if my answer is wrong or my book is wrong? Thank you!