Asked by Lilly
                A particle P starts at the origin (x0,y0)=(0,0), of the plane. The particle moves in steps as follows.
1. P moves 1 unit to the right to the point (x1,y1)=(1,0), then
2. P moves 1/2 unit up to (x2,y2)=(1,1/2), then
3. P moves 1/4 unit to the left to (x3,y3)=(3/4,1/2), then
4. P moves 1/8 unit down to (x4,y4)=(3/4, 3/8), then
5. P moves 1/16 unit to the right to (x5,y5)= (13/16, 3/8)
6. and so on...
Find the limit point P-infinity= (x-infinity,y-infinity) of the particle P, where the limit to x-infinity= the limit of x sub k . as k approaches infinity and y-infinity= the limit of y sub k, as k approaches infinity.
I have no idea what I'm supposed to do or where to start. Any help will be much appreciated. Q.Q
            
        1. P moves 1 unit to the right to the point (x1,y1)=(1,0), then
2. P moves 1/2 unit up to (x2,y2)=(1,1/2), then
3. P moves 1/4 unit to the left to (x3,y3)=(3/4,1/2), then
4. P moves 1/8 unit down to (x4,y4)=(3/4, 3/8), then
5. P moves 1/16 unit to the right to (x5,y5)= (13/16, 3/8)
6. and so on...
Find the limit point P-infinity= (x-infinity,y-infinity) of the particle P, where the limit to x-infinity= the limit of x sub k . as k approaches infinity and y-infinity= the limit of y sub k, as k approaches infinity.
I have no idea what I'm supposed to do or where to start. Any help will be much appreciated. Q.Q
Answers
                    Answered by
            Reiny
            
    Take a look at the vertical and horizontal movements separately
vertical: 0 + 1/2 - 1/8 + 1/32 - .... ignoring the 0 , this is a GS where a = 1/2 and r = -1/4
sum (all terms) = a/(1-r) = 1/2 รท (5/4) = 2/5, so the y will approach 2/5
horizontal: 0 + 1 - 1/4 + 1/16 - .... again, let a = 1, r = -1/4
again sum of all the terms = a/(1-r) = 1/(5/4) = 4/5
the final position will be ((4/5, 2/5)
    
vertical: 0 + 1/2 - 1/8 + 1/32 - .... ignoring the 0 , this is a GS where a = 1/2 and r = -1/4
sum (all terms) = a/(1-r) = 1/2 รท (5/4) = 2/5, so the y will approach 2/5
horizontal: 0 + 1 - 1/4 + 1/16 - .... again, let a = 1, r = -1/4
again sum of all the terms = a/(1-r) = 1/(5/4) = 4/5
the final position will be ((4/5, 2/5)
                    Answered by
            Lilly
            
    I am confused as to why the sum of all the terms is a/(1-r). I thought it was (a-ar^n-1)/(1-r).
    
                    Answered by
            Reiny
            
    Sal Khan explains the formula very nicely here:
https://www.youtube.com/watch?v=b-7kCymoUpg
    
https://www.youtube.com/watch?v=b-7kCymoUpg
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