Multiply 0,5?

M

Miqeh

New member
Joined
Jan 17, 2010
Messages
2
Hello :)

I would be happy if some one could help me with this little problem:

Each "day" the variable grows double so big as it was before, and it started out as 0,5
That means 0,5 goes 1 goes 2 goes 4 goes 8 and so on.
But i want a formula that can solve it, like i just put in how many days, and then the formula tells me what amount the variabel reaches at that amount of days.
Like day 4 Is 4
And day 7 is 32
It's hard to explain for me, but one of the variables is the "day" counter, and the other one is the "number" That has to be doubled each day, and that "number" starts on 0.5

Please tell me a easy way of doing that :)
 
It's hard to explain for me, but one of the variables is the "day" counter, and the other one is the "number" That has to be doubled each day, and that "number" starts on 0.5
 
\(\displaystyle Number\ is\ n.\)

\(\displaystyle After\ 1\ day,\ n\ has\ doubled\ to\ 2n=2^1n.\)

\(\displaystyle After\ 2\ days,\ 2n\ has\ doubled\ to\ 4n=2^2n.\)

\(\displaystyle After\ 3\ days,\ 4n\ has\ doubled\ to\ 8n=2^3n\)

The pattern is.....

the power of 2 is the number of elapsed days from the start.
 
I'm sticking with 2^(d-1)

d
0 : 1/2
1 : 1
2 : 2
3 : 4
4 : 8
5: 16
....
 
Given your value for n, Miqeh,
I'm backing up Denis.

The final answer and the number of days are variable.
The answer is variable because the number of days are variable.

However we stick with the initial value of 'n' to write the formula.

You gave an initial value of n=0.5, which is \(\displaystyle \frac{1}{2}=2^{-1}\)

If you plug that into the meat-grinder, you get \(\displaystyle 2^{-1}2^d=2^{d-1}\)

However, you can choose "n" to be other values and place it into \(\displaystyle n2^d\)
 
Miqeh said:
It's hard to explain for me, but one of the variables is the "day" counter, and the other one is the "number" That has to be doubled each day, and that "number" starts on 0.5
Click to expand...

You are putting INCOMPLETE information:

if you want - "number" That has to be doubled each day, and that "number" starts (meaning day 0 - not day 1) on 0.5 - then Sir Dennis is right.

However, if you want:

Like day 4 Is 4
And day 7 is 32
Click to expand...

Counting the beginning day as day 1 (not day 0) then

Then Sir Mark is right.

Sir Denis - can I please get out of the corner????
 
Miqeh said:
It's hard to explain for me, but one of the variables is the "day" counter, and the other one is the "number" That has to be doubled each day, and that "number" starts on 0.5
Click to expand...

Arise, Sir Denis

as time durations are endtime-startingtime.

I had to protect Subhotosh somehow.
 
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