Math Is Fun Forum

He is inccorect:

the probablity is 1/3

He is misinterpreting/misquoting the Monty Hall / 3 Prisoners Problem of which his question is more related to the Monty Hall problem:

http://en.wikipedia.org/wiki/Monty_Hall_problem

In the Monty Hall problem, the player picks a door, and the host of the game, then picks one of the other two doors that ISN'T the prize, the probability of winning if you switch your door to the other available one is then increased from 1/3 to 2/3 as a result of the host's actions:

if the host does not give the player any information, or open one of the loosing doors, then the probablity is still 1/3 no matter which door you pick

each dimension is independant of the others in all but one calculation which calculates the length of the vector:

you have a vector representing the displacement of the body in 'n' dimensions, it's velocity in 'n' dimensons, you have a wind term, a vector in 'n' dimensions, and the gravity in 'n' dimensions, a force accumulator in 'n' dimensions etc, all of the operations are just addition and scaling by a scalar

the only calculation that is not independant of the dimension is the magnitude calculation, which is just sqrt(x^2 + y^2 + z^2 + w^2) etc.

Unless accuracy is absolutely needed, you can just simulate the missiles rather than using exact equations which would allow for much more varied scenarios, like wind that changes over time without any need for complex calculations:

You could simulate the missile simply as a particle, each frame you would calculate the forces being exerted on it, and the resultant acceleration to modify the velocity, and calculate the next position of the missile. something along the lines of:

Obviously, in application you wouldn't have wind etc stored as constants inside each missile, but in a singleton perhaps for the world variables.

package
{
   import flash.display.MovieClip;

   //treating missile as a straight line segement of length 'length'
   
   public class Missile extends MovieClip
   {
      private var px:Number, py:Number; //reason for using these rather than inherited x,y is for accuracy as x,y are rounded
      private var vx:Number, vy:Number; //velocity
      private var mass:Number, imass:Number; //mass, reciprocal of mass

      //assumes the missile graphic is centred, and horizontal at 0 rotation
      public function init(x:Number, y:Number, xv:Number, yv:Number, m:Number)
      {
          px = x; py = y;
          vx = xv; vy = yv;
          mass = m; imass = 1/m;
      }

      //h is the timestep in your simulation, at 60fps, it should be 1/60 etc.
      public function advance(h:Number):void
      {
         //force accumulators
         var fx:Number, fy:Number = fx = 0;
         //---compute forces
         //wind:
         fx -= 75;
         //gravity:
         fy += 220*mass;
         //air resistance:
         //relative velocity of missile with respect to the air
         var rvx:Number = vx+75;
         var rvy:Number = vy;
         if(rvx!=0||rvy!=0)
         {
	         var rvl:Number = 0.02*Math.sqrt(rvx*rvx+rvy*rvy); // 0.02 varied
	         fx -= rvx*rvl;
	         fy -= rvy*rvl;
		 }
         //---compute acceleration
         var ax:Number = fx*imass;
         var ay:Number = fy*imass;
         //---numerically integrate for new velocity and displacement
         vx += ax*h;
         vy += ay*h;
         px += vx*h;
         py += vy*h;
         //---update x,y,rotation based on these values
         x = px;
         y = py;
         //set rotation so that it faces direction of motion
         rotation = Math.atan2(vy,vx)*180/Math.PI;
      }
   }
}

the code in the .fla is then:

import Missile;

var missiles:Array = new Array();
function mdown(ev:MouseEvent):void
{
	var m:MissileClip = new MissileClip();
	m.init(0,400,mouseX*3,(mouseY-400)*3,5);
	
	addChild(m);
	missiles.push(m);
}
stage.addEventListener(MouseEvent.MOUSE_DOWN,mdown);

const h:Number = 1/60;
function main(ev:Event):void
{
	for(var i:int = 0; i<missiles.length; i++)
	{
		var m:MissileClip = missiles[i];
		m.advance(h);
		if(m.y>450)
		{
			removeChild(m);
			missiles.splice(i--,1);
		}
	}
}
stage.addEventListener(Event.ENTER_FRAME,main);

where MissileClip is the class name of the missile graphic clip, who's base class is Missile

Note how with the left wind of 45Newtons, when missiles begin to fall vertically, they actually fall a little to the left.
With the constant force of gravity and wind, the missiles reach a terminal velocity controlled by the drag constant, in this case 0.02 and their mass, and the values of gravity and wind.

http://spamtheweb.com/ul/upload/060409/ … issile.php

John + Core2Student, when computing the area between two curves, it matters not whether the curves themselves cross the x-axis, the only thing you have to worry about is that the area you are computing is such that one curve is always above the other one, if the intersect eachother within the area you are computing, then you have to split it into two integrals as you would if you were computing the area under a single curve that crossed the x-axis.

The trick, is rather than computing the areas of the two curves and the x-axis seperately, and then subtracting, you instead compute the area between the difference of the two curves themselves, and the x-axis, aka.

the fact that the value is negative simply means that the graph of y = 5-x is above the graph of y = x^2-9x+20 between x = 3 and x = 5

the area is then the absolute value of this:

you are correct; you just need to manipulate your equations a little further

i'd argue against your terminology 'calculating magnitude of OA and OB' as OA and OB refer to the vector between two points A, B and the origin, whereas A and B given are already vectors, and thus OA and OB don't make any sense.

But yes, you are correct; you would find the perimeter by calculating the magnitude of A and B, this is the length of the two sides, and thus the perimeter would be 2(|A| + |B|)

Actually, upon further investigation, mine is not quite the same as yours, in yours to form the cardioid you need to have the second circle having a speed 2x greater, whereas in mine it needs to have the same speed:

In yours the circles rotate in themselves, and it is the position of the circle that is rotating with the previous circle.

In mine, the circle itself rotates, but it is the entire circle (and all it's children) that is rotating with the previous circle.

Also the following code:

Matrix2x2 An(int n)
{
   Real tn = t[n];
   return Matrix2x2(cos(tn),-sin(tn),sin(tn),cos(tn));
}
Matrix2x2 Bn(int n)
{
   Real tn = t[n];
   return Matrix2x2(-sin(tn),-cos(tn),cos(tn),-sin(tn))*w[n];
}
Vector2 sn(int n)
{
   Real tn = t[n];
   return Vector2(cos(tn),sin(tn))*r[n];
}
Vector2 vn(int n)
{
   Real tn = t[n];
   return Vector2(-sin(tn),cos(tn))*r[n]*w[n];
}

//iterative methods:
Vector2 snr(int n, int k)
{
   if(n==k) return Vector2(0,0);
   return sn(k+1) + An(k+1)*snr(n,k+1);
}
Vector2 vnr(int n, int k)
{
   if(n==k) return Vector2(0,0);
   return vn(k+1) + An(k+1)*vnr(a,k+1) + Vn(k+1)*snr(a,k+1);
}
Vector2 sn0(int n)
{
   return snr(n,0);
}
Vector2 vn0(int n)
{
   return vnr(n,0);
}

//formula methods
Vector2 sn0eq(int n)
{
   if(n==0) return Vector2(0,0);
   
   Vector ret (0,0);
   for(int i = 1; i<=n; i++)
   {
      Matrix2x2 mat (1,0,0,1);
      for(int j = 0; j<=i-1; j++)
         mat *= An(j);
      ret += mat*sn(i);
   }
   
   return ret;
}
Vector2 vn0eq(int n)
{
   if(n==0) return Vector2(0,0);
   
   Vector ret (0,0);
   for(int i = 1; i<=n; i++)
   {
      Matrix2x2 mat (1,0,0,1);
      for(int j = 0; j<=i-1; j++)
         mat *= An(j);
      ret += mat*vn(i);
   }

   for(int i = 2; i<=n; i++)
   {
      Matrix2x2 mat (0,0,0,0);
      for(int j = 1; j<=i-1; j++)
      {
         Matrix2x2 matsub (1,0,0,1);
         for(int k = 1; k<=j-1; k++)
            matsub *= An(k);
         matsub *= Bn(j);
         for(int k = j+1; k<=i-1; k++)
            matsub *= An(k);
         
         mat += matsub;
      }
      ret += mat*sn(i);
   }
   
   return ret;
}

And once more, you insist on this 'growing' idea, which none of us have said, the 9's aren't growing, they are already there; we don't have to claim we can get them all, because we don't have anything to get; they already exist.

Wait, have I done something wrong here?

?

The method they present is the same as my own