Kishan Patel	9/26	HW4

1) 
a)	A will never catch up to T, as everytime the A reaches to the past position that T was at, T has travelled a further distance. Then A move to T's next position, however T has continued to move forward, this will continue as A will always catch up to T's past position, but never the actual position

b)	v_a*t=v_t*t+di
	200*t=50t+50
	150t=50
	t=1/3 hr

c)
1/4+1/16+1/64+1/256+1/1024
sum((1/4)^n)
=1/4*sum(x^n)
=1/(1-x)
=(1/4)*1/(1-.25)=1/3
d)
1/4+1/16+1/64=21/64
The next step is smaller than 1/64, so add 1/64
21/64+1/64=22/64 thus A passed T, then they passed between 21/64 and 22/64

2)
sum(x^n)=(1-x^(n+1))/(1-x)
n=0

sum(x^n)(x-1)=1-x^(n+1)
x^n*x^1-1=1-x^(n+1)
x^(n+1)-1=1-x^(n+1)

3)
a)
assume x=1/2
sum(x^n)=1/(1-x)
sum((1/2)^n)=1/(1-1/2)
2=1+1/2+1/4+1/8+1/16+1/32..
2=2

b)
A catches up in 1/1 seconds time
T runs 1*x distance
A catches up in 1/x time
T runs 1/x^2 distance more
A catches up in 1/x^2, T runs 1/x^4 time
where this continues forever, the sum of all of these creates
1/n(sum(x^n)) as A gets closer and x <1, they continue infinitely to a limit number.
thus the summation x^n = 1/(1-x)

v_a*t=x*t+1
1-x*t=1
t=1/(1-x)

4)
sum(1/2^n)
n=1	:=1/2=
n=2	:=1/2+1/4=3/4
n=3	:=1/2+1/4+1/8==7/8
n=4	:=1/2+1/4+1/8+1/16=15/16
the limit as n aproaches infinity, the sum(1/2^n)=1
this is because as n increases, the distance between the sum and 1 decreases so the limit aproaches 1