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Simple things made complicated … To be more expressive
No matter how many iterations you try, you can never get to a 2 by summing 1/(2^n) … So ur equation # 2 is invalid and hence, ur other equations are also invalid. A good approximation perhaps, but not spot-on. :nono:
Here n varies from 0 to enfinity … hence the serise will be something like this …
=1.0+ .5+0.25+0.125+0.0625+0.03125+0.015625+0.007813+0.003906+0.001953+0.000977=1.999023
the above serise is for first 11 numbers … so if u go on increasing the numbers u will finally get the required answer. Aftar @ 22 the value of 1/(2^n) becomes zero.
I agree with u. That’s what I said. A good approximation. But it will NEVER be exactly 2.0 … Could be 1.999999999999999999999999999999999999999999999, never 2 though. lol.
So here are the values
i 1/(2^n) SUM [1/(2^n)]
0 1.000000 1.000000
1 0.500000 1.500000
2 0.250000 1.750000
3 0.125000 1.875000
4 0.062500 1.937500
5 0.031250 1.968750
6 0.015625 1.984375
7 0.007813 1.992188
8 0.003906 1.996094
9 0.001953 1.998047
10 0.000977 1.999023
11 0.000488 1.999512
12 0.000244 1.999756
13 0.000122 1.999878
14 0.000061 1.999939
15 0.000031 1.999969
16 0.000015 1.999985
17 0.000008 1.999992
18 0.000004 1.999996
19 0.000002 1.999998
20 0.000001 1.999999
21 0.000000 2.000000
It’s a lost cause arguing with u. I give up !
HINT: That’s 7 significant figures only 
I rest my case !
dats why i hated finite and calculus
actually u cant express equations takign x,y,z or any other variables out of the blue....
[QUOTE]
*Originally posted by armughal: *
actually u cant express equations takign x,y,z or any other variables out of the blue....
[/QUOTE]
That's interesting... please elaborate !
u guys got too much time on ur hands to be doing that
This is what I’ll be doing for a living so …
Besides, it’s only outside of school that u give thought to such things