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Thread: Reversing A Formula

  1. #1
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    Reversing A Formula [solved]

    I'm sure there's an easy way to do this, but its been ages since I did algebra.

    I've got an arithmetic series. Each term increases by 10% of the original term. For instance: 40, 44, 48, 52, 56, etc.

    The formula I've come up with to calculate the nth term is:

    $\displaystyle v_n=v_0+n.v_0/10$

    Where $\displaystyle v_n$ is the nth term and $\displaystyle v_0$ is the initial term. This seems to work, although it may not be the most elegant expression of the series.

    Now I've decided that, given $\displaystyle n$ and $\displaystyle v_n$, I want to be able to calculate $\displaystyle v_0$. However solving my above equation for $\displaystyle v_0$ has left me going in circles.

    I'd just as soon NOT get the answer - a gentle hint in the right direction will be appreciated, helpful, and will hopefully leave my pride intact.
    Last edited by sapph; Aug 4th 2009 at 07:36 AM. Reason: Edited for formula clarity and solved tag
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by sapph View Post
    I'm sure there's an easy way to do this, but its been ages since I did algebra.

    I've got an arithmetic series. Each term increases by 10% of the original term. For instance: 40, 44, 48, 52, 56, etc.

    The formula I've come up with to calculate the nth term is:

    $\displaystyle v_{n}=v_{i}+(.1v_{i}(n-1))$

    Where $\displaystyle v_{n}$ is the nth value and $\displaystyle v_{i}$ is the initial value. This seems to work, although it may not be the most elegant expression of the series.

    Now I've decided that, given $\displaystyle n$ and $\displaystyle v_{n}$, I want to be able to calculate $\displaystyle v_{i}$. However solving my above equation for $\displaystyle v_{i}$ has left me going in circles.

    I'd just as soon NOT get the answer - a gentle hint in the right direction will be appreciated, helpful, and will hopefully leave my pride intact.
    Gentle hint: you can take out a common factor.

    Also, to ease your notation I would call your starting term $\displaystyle v_0$ instead of $\displaystyle v_i$ - this makes the fact that it is the starting term much clearer. Often you would want to find the "$\displaystyle i^{\text{th}}$" term in your series instead of the "$\displaystyle n^{\text{th}}$" term, so this will also stop that confusion. So in the series you gave you would have $\displaystyle v_0=40, v_1=44, v_2=48, \ldots$. This would mean your formula would be $\displaystyle v_n=v_0+n.v_0/10$.
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  3. #3
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    Hm. Maybe I need to let go of my pride. Here are the loops I am going through:

    $\displaystyle v_n=v_0+n.v_0/10$ to
    $\displaystyle 10v_n=10v_0+n.v_0$ to
    $\displaystyle 10v_n/n=10v_0/n+v_0$ to
    $\displaystyle 10v_n/n-10v_0/n=v_0$ to
    $\displaystyle (10v_n-10v_0)/n=v_0$ to
    $\displaystyle 10(v_n-v_0)/n=v_0$ Stuck.

    $\displaystyle v_n=v_0+n.v_0/10$ to
    $\displaystyle v_n-v_0=n.v_0/10$ to
    $\displaystyle 10(v_n-v_0)=n.v_0$ to
    $\displaystyle 10(v_n-v_0)/n=v_0$ Stuck.

    Because I essentially have an x+xy situation, I am having trouble isolating the x. I tried googling for "x+xy" to see if I could find common solutions. Unfortunately, I forgot that google tends to ignore punctuation in searches, so searching for XXY tends to turn up a lot about genetic disorders.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by sapph View Post
    Because I essentially have an x+xy situation, I am having trouble isolating the x. I tried googling for "x+xy" to see if I could find common solutions. Unfortunately, I forgot that google tends to ignore punctuation in searches, so searching for XXY tends to turn up a lot about genetic disorders.
    If I told you that $\displaystyle x+xy=x(1+y)$, would that help? That is what I was meaning when I said "common factor"...

    Also, wolfram alpha is good for helping with this sort of thing. Typing $\displaystyle x+xy$ into this and you will get that it is equivalent to $\displaystyle x(1+y)$.
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  5. #5
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    Aha!

    $\displaystyle v_n=v_0+v_0*0.1n$ to
    $\displaystyle v_n=v_0(1+0.1n)$ to
    $\displaystyle v_n/(1+0.1n)=v_0$

    Thanks for preserving the aha moment.
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