Testwiki:Reference desk/Archives/Mathematics/2014 January 30

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January 30

ANNUITY

How can I make r as subject of the following formula? A=R(1-(1+r)^-n)/r Thank you.175.157.115.182 (talk) 16:13, 30 January 2014 (UTC)

In general (unless n is small) you cannot express r as a simple function of A, R and n. You can, however, solve for r using numerical methods - see internal rate of return. Gandalf61 (talk) 16:37, 30 January 2014 (UTC)

... and you can set that up very easily on a spreadsheet, then use goal seek. Dbfirs 17:37, 30 January 2014 (UTC)

MS Excel even has a built in function for this: IRR(). OldTimeNESter (talk) 20:08, 31 January 2014 (UTC)

The equation in terms of r is non-linear and there is no formula for it. However you can write

f(r) = A - R(1-(1+r)^-n)/r

And search for values of r that makes f(r) = 0 , probably the easiest is to ask the computer to plot the graph of f(r). 202.177.218.59 (talk) 23:15, 2 February 2014 (UTC)

Use the binomial theorem

$0=\frac{A}{R}-\frac{1-(1+r{)}^{-n}}{r}$

$0=\frac{A}{R}-\frac{1-\sum _{i=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{-n}{i})r^i}{r}$

$0=\frac{A}{R}-\frac{-\sum _{i=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{-n}{i})r^i}{r}$

$0=\frac{A}{R}+\sum _{i=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{-n}{i})r^{i-1}$

$0=\frac{A}{R}+\sum _{i=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{-n}{i+1})r^i$

$0=\frac{A}{R}-n+\frac{n(n+1)}{2}r-\frac{n(n+1)(n+2)}{6}{r}^2+\frac{n(n+1)(n+2)(n+3)}{24}{r}^3-\cdots$

$0=\frac{A}{R}-n(1-\frac{n+1}{2}r(1-\frac{n+2}{3}r(1-\frac{n+3}{4}r(1-\cdots ))))$

Truncate the series and solve the resulting algebraic equation numerically. Bo Jacoby (talk) 11:11, 3 February 2014 (UTC).