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

Re: Formula converting numbers to double the previous one

Quote:

Originally Posted by

**LadyJane** 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

$\displaystyle f(10n)=\frac{2^{10n}}{10}$

Re: Formula converting numbers to double the previous one

Hello, LadyJane!

Quote:

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

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

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

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

From the first two terms:

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

$\displaystyle \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 $

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

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

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