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Math Help - Formula converting numbers to double the previous one

  1. #1
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    Question 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
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Formula converting numbers to double the previous one

    Quote Originally Posted by LadyJane View Post
    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}
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  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539

    Re: Formula converting numbers to double the previous one

    Hello, LadyJane!

    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}

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