Let f(x) be a function that adds the first x powers of 1/2:
f(x) = (1/2)^0 = 1
f(1) = (1/2)^0 + (1/2)^1 = 1+1/2 =3/2
f(2) = (1/2)^0 +(1/2)^1 + (1/2)^2 = 1+ 1/2+1/4=7/4

Question

Let f(x) be a function that adds the first x powers of 1/2:
f(x) = (1/2)^0 = 1
f(1) = (1/2)^0 + (1/2)^1 = 1+1/2 =3/2
f(2) = (1/2)^0 +(1/2)^1 + (1/2)^2 = 1+ 1/2+1/4=7/4

anonymous · Answer

@amistre64

anonymous · Answer

find a formula for f(x)

amistre64 · Answer

a sequence of numbers is generated by the partial sums .... you are in effect making a new sequence that will hopefully be able to have an explicit formula.

what are the next say, 3 partial sums?

anonymous · Answer

11/4 ?

amistre64 · Answer

$$\frac{1}{1},\frac32,\frac{7}{4},\frac{15}{8}$$

anonymous · Answer

oh yeah.

amistre64 · Answer

the next is double and add 1 ... 31/16

the next is double and add 1  63/64 .... seems to be the patter

anonymous · Answer

so how should I express that in terms of f(x) ?

amistre64 · Answer

well, a few ways, so a little more context is in order.  does it need to be a polynomial?  can it be a summation rule?  how do we need to define the rule?

amistre64 · Answer

$$f_{n}=f_{n-1}+\frac{1}{2^n}$$

amistre64 · Answer

we then determine the homogenous part, then work out the particular part

amistre64 · Answer

does that make sense?  or is that beyond the scope of your material?

anonymous · Answer

yes it is but your answer kind of makes sense, is that fx ? can't really see your subscript

amistre64 · Answer

its fn at the moment ...

anonymous · Answer

Oh can it be a simple f(x) = and with a formula ? lol. i waasn't sure because it goes up to infinity.

amistre64 · Answer

it can be, yes.

anonymous · Answer

hmm i think i figured it out, is it f(x) = 2 - (1/2)^x ? is that right ?

amistre64 · Answer

that actually does work :) http://www.wolframalpha.com/input/?i=table+%5B2+-+%281%2F2%29%5En%5D%2C+n%3D0..8

anonymous · Answer

thanks ;D

amistre64 · Answer

youre welcome :)

anonymous · Answer

okay so theres another part to this question lol

anonymous · Answer

use a graphing calculator to graph f(x) then make a conjecture about f(x) when x --> infinity

amistre64 · Answer

you got a graping calc?

anonymous · Answer

yes, it goes in a straight line at y=3

anonymous · Answer

right ?

amistre64 · Answer

2 - anything ... is not equal to 3.  we would say that 3 is an upper bound,  but it is not the smallest upper bound.  for that matter,  12 is an upper bound :)

amistre64 · Answer

(1/2)^x  approaches zero as x approaches infinity .. 2-0 = ??

anonymous · Answer

2 ?

amistre64 · Answer

yes,  so if we graph it ... the line it approaches will be 2.  we can conjecture that the limit of adding up this setup, for infinity ... will never exceed the value of 2.  that is the smallest upper bound we can place on it.