13y=x+x2+x3+...
=>y=x(1+y)
=>y=x/(1-x)
[1]
my system has some problem for uploading image... sry... [2]
13 Answers
13
3for 0≤x≤1
the graph shud be
but the graph for x>1 is not defined since it tends to ∞
13bro... whats wrong in my function...!!!??? [11]
because x=2 => y=-2 from my function whereas if v substitute x=2 in d question, we get y→∞ ........... [7] [12]
33@
bro... whats wrong in my function...!!!???
because x=2 => y=-2 from my function whereas if v substitute x=2 in d question, we get y→∞ ...........
Whats wrong in his method.. why his method restricts domain [7]
62well abhishek... this might be a bit before time for u guys to know..
most ppl (engineers i mean) die without knowing the answer to this... ;)
bcos the method that we have used to find the sum does not have any restriction... and because it seems so strange.. doesnt it
The problem is that we have defined finite sums.. i mean sum of 1+2+3 or till n numbers..
but this has to do with sum of infinite sums...
There is something clalled convergence (it has to do with the existance of limits...)
Dont go into more details.. If u want i can post the whole thing.. but it will be not very useful for u
Have u ever wondered why we always use for a infinte GP that |r|<1 when in the proof nothing seems to go wrong!!
62it is like saying
Sn = t1+t2+t3+...tn
and this limit as n tends to infinity should exist... (for an infinite sum to exist..)
Here the limit does not exist...