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Math Help - 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:

    v_n=v_0+n.v_0/10

    Where v_n is the nth term and 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 n and v_n, I want to be able to calculate v_0. However solving my above equation for 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; August 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:

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

    Where v_{n} is the nth value and 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 n and v_{n}, I want to be able to calculate v_{i}. However solving my above equation for 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 v_0 instead of v_i - this makes the fact that it is the starting term much clearer. Often you would want to find the " i^{\text{th}}" term in your series instead of the " n^{\text{th}}" term, so this will also stop that confusion. So in the series you gave you would have v_0=40, v_1=44, v_2=48, \ldots. This would mean your formula would be 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:

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

    v_n=v_0+n.v_0/10 to
    v_n-v_0=n.v_0/10 to
    10(v_n-v_0)=n.v_0 to
    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 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 x+xy into this and you will get that it is equivalent to x(1+y).
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  5. #5
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    Aha!

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

    Thanks for preserving the aha moment.
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