solving for j

**kingsolomonsgrave** · Oct 6th 2013, 04:49 PM

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

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

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

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

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

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?

**HallsofIvy** · Oct 6th 2013, 05:06 PM

Originally Posted by kingsolomonsgrave

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

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

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

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

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

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?

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