11you mean they tend towards -0.693 but never reach there? Btw why have you marked there as -0.27?
I plotted my graph according to reply #11 [1]
30 Answers
62@sky no this is not the graph...
@varun.. This not a function! (so i think we should not use the word range!)
9coolehsrtol ur graph is correct if u replace -.027 by - .693 also there will be points on y axis cos x = 0 means y is integer which is very much possible
i donno wat aragorn trys to say
62that was not needed.. i mean rohit made a smalll mistake in plotting those points..
11
1hey the arrows start from the y axis............the arrow heads show..they wont exactly reach sumwhere..where i cannot exactly tell...........
11Sorry, Rohit[2]. I was blabering in my sleep[13][13][13][13][13]
1guys i guess mah was correct
simple explanations...
wen x=0.....ex=1 and so at all integral points on y axis where x=0....{y}=0........so (1/2)0=1...............hence it is valid at all integral points on y axis...........
now takin log on both sides.....we get x= - 0.3 .{y}..........where y =Integer part + fractioanal part.....so {y}=f
so we get x=-0.3 .f
where f is between 0 and 1....so the velue of 0.3*f <1 always....
so we get that when f=0.1.....x = - .03 and wen f=.9.....x=.27
1guys i guess mah was correct
simple explanations...
wen x=0.....ex=1 and so at all integral points on y axis where x=0....{y}=0........so (1/2)0=1...............hence it is valid at all integral points on y axis...........
now takin log on both sides.....we get x= - 0.3 .{y}..........where y =Integer part + fractioanal part.....so {y}=f
so we get x=-0.3 .f
where f is between 0 and 1....so the velue of 0.3*f <1 always....
so we get that when f=0.1.....x = - .03 and wen f=.9.....x=.27
11There won't be any points on the y axis as on yaxis x=0 and enonsense≠0. I don't think the graph which I posted and deleted is correct. But I'll put it anyway.[4]
POINTS ON THE Y AXIS ARE NOT INCLUDED
1this is very tuf..
seems at first look...
feelin toooooooooooo sleeeeeeeeeepppppppppyyyyyyyyyyy
zzzzzzzzzzzzzzzzzzzzzzzzzz
:(
1ex = 2^-{y}
range= [1/20 , 1/2)
so ex ε [1,1/2)
ex=1 => x=0
ex=1/2 =>x=ln(1/2) = -0.693
so domain=(-o.693,0]...
is it??
1taking log both sides -
x = -{y} ln 2.
x=-0.693{y}
hmm.. yes a portion where -0.693<x≤0 and y ε R
62varun look carefully.. if there is one point there will be infinite..
{y} is fractional part