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

= 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.   D b f i r s   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
 * $$0=\frac A R-\frac{1-(1+r)^{-n}}r$$
 * $$0=\frac A R-\frac {1-\sum_{i=0}^\infty \binom {-n}i r^i}r$$
 * $$0=\frac A R-\frac {-\sum_{i=1}^\infty \binom {-n}i r^{i}}r$$
 * $$0=\frac A R+\sum_{i=1}^\infty \binom {-n}{i} r^{i-1}$$
 * $$0=\frac A R+\sum_{i=0}^\infty \binom {-n}{i+1} r^{i}$$
 * $$0=\frac A R-n+\frac{n(n+1)}2 r-\frac{n(n+1)(n+2)}6 r^2 +\frac{n(n+1)(n+2)(n+3)}{24} r^3 -\cdots$$
 * $$0=\frac A R-n(1-\frac{n+1}2 r(1-\frac{n+2}3 r (1-\frac{n+3}{4} r (1-\cdots))))$$

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