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  • 1 Post By HallsofIvy

Math Help - solving for j

  1. #1
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    solving for j

    (1+\frac{i}{2})^2-1 = (1+\frac{j}{m})^m - 1

    (1+\frac{i}{2})^\frac{2}{m}= (1+\frac{j}{m})

    (1+\frac{i}{2})^\frac{2}{m}  -1=\frac{j}{m}

    j=m[(1+\frac{i}{2})^\frac{2}{m}  -1]

    j=m(1+\frac{i}{2})^\frac{2}{m}  -m

    but the correct answer is

    j=(1+\frac{i}{2})^\frac{2}{m}  -1

    what happened to the m? this is working out effective interests payments on a mortgage so perhaps there is some assumption I I'm unaware of about the periodic interest rate (maybe they put j but they mean j/m or something) , but mathematically shouldn't the m still be there?
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  2. #2
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    Re: solving for j

    Quote Originally Posted by kingsolomonsgrave View Post
    (1+\frac{i}{2})^2-1 = (1+\frac{j}{m})^m - 1

    (1+\frac{i}{2})^\frac{2}{m}= (1+\frac{j}{m})

    (1+\frac{i}{2})^\frac{2}{m}  -1=\frac{j}{m}

    j=m[(1+\frac{i}{2})^\frac{2}{m}  -1]

    j=m(1+\frac{i}{2})^\frac{2}{m}  -m
    This is the correct answer.

    but the correct answer is

    j=(1+\frac{i}{2})^\frac{2}{m}  -1
    No, this is not correct.

    what happened to the m? this is working out effective interests payments on a mortgage so perhaps there is some assumption I I'm unaware of about the periodic interest rate (maybe they put j but they mean j/m or something) , but mathematically shouldn't the m still be there?
    Thanks from kingsolomonsgrave
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