# Formula converting numbers to double the previous one

• Jun 29th 2011, 07:44 AM
Formula converting numbers to double the previous one
Hi everyone,
I hope this is in the right forum.
I have a list of numbers and am looking for a formula that will convert these numbers in the following manner:
0.1 -> 0.1
0.2 -> 0.2
0.3 -> 0.4
0.4 -> 0.8
0.5 -> 1.6
0.6 -> 3.2
0.7 -> 6.4
0.8 -> 12.8
0.9 -> 25.6
1.0 -> 51.2

As you can see, numbers in the second column are always double the previous value.
I would like a formula (if one exists) that can be applied to each number in the first column to obtain the number in the second column. I would then use this formula to convert the in-between numbers (0.34, 0.567 etc...) in the same manner. Original number will always be in between 0 and 1 as they represent percentiles.

Thank-you
• Jun 29th 2011, 07:49 AM
Also sprach Zarathustra
Re: Formula converting numbers to double the previous one
Quote:

Hi everyone,
I hope this is in the right forum.
I have a list of numbers and am looking for a formula that will convert these numbers in the following manner:
0.1 -> 0.1
0.2 -> 0.2
0.3 -> 0.4
0.4 -> 0.8
0.5 -> 1.6
0.6 -> 3.2
0.7 -> 6.4
0.8 -> 12.8
0.9 -> 25.6
1.0 -> 51.2

As you can see, numbers in the second column are always double the previous value.
I would like a formula (if one exists) that can be applied to each number in the first column to obtain the number in the second column. I would then use this formula to convert the in-between numbers (0.34, 0.567 etc...) in the same manner. Original number will always be in between 0 and 1 as they represent percentiles.

Thank-you

$f(10n)=\frac{2^{10n}}{10}$
• Jun 29th 2011, 10:15 AM
Soroban
Re: Formula converting numbers to double the previous one

Quote:

I have a list of numbers and am looking for a formula for them.

. . $\begin{array}{c|c} x & f(x) \\ \hline 0.1 & 0.1 \\ 0.2 & 0.2 \\ 0.3 & 0.4 \\ 0.4 & 0.8 \\ 0.5 & 1.6 \\ \vdots & \vdots \end{array}$

As you can see, numbers in the second column are always double the previous value.

The formula is an exponential function of the form: . $f(x) \:=\:a\!\cdot\!2^{bx}$

From the first two terms:

. . $\begin{Bmatrix}f(0.1) =0.1\!: & a\cdot2^{0.1b} \:=\: 0.1 & [1] \\ \\[-3mm] f(0.2) =0.2\!: & a\cdot2^{0.2b} \:=\: 0.2 & [2] \end{Bmatrix}$

$\text{Divide [2] by [1]: }\:\frac{a\!\cdot\!2^{0.2b}}{a\!\cdot\!2^{0.1b}} \:=\:\frac{0.2}{0.1} \quad\Rightarrow\quad 2^{0.1b} \:=\:2$

$\text{Hence: }\:0.1b \:=\:1 \quad\Rightarrow\quad b \,=\,10$

$\text{Substitute into [1]: }\:a\!\cdot\!2^1 \:=\:0.1 \quad\Rightarrow\quad a \,=\,\tfrac{1}{20}$

$\text{Therefore: }\:f(x) \:=\:\tfrac{1}{20}\cdot 2^{10x}$