Mo logo [home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff] german flag
Mathematics-Online course: Prepcourse Mathematics - Analysis - Functions

Compound Interest

[previous page] [next page] [table of contents][page overview]
With an interest rate of $ p$ , a seed capital $ x$ results after $ n$ out-/incoming payments of $ r$ ($ r<0$ or . $ r>0$ respectively) in the final sum of

$\displaystyle y = (1+p)^nx + \frac{(1+p)^n - 1}{p} r\, . $

Here $ p=1$ corresponds to an interest rate of $ 100\%$ , speaking in annual rates.

With monthly interest calculation at an interest rate $ p_m$ , the annual percentage rate $ p_j$ calculates as

$\displaystyle p_j=(1+p_m)^{12}-1 \geq 12p_m\,. $

(Authors: Höllig/Hörner/Abele)
Nominal capital is already accounted for in the first period, while payments are only accounted for in the period following their occurrence. This yields for

\begin{displaymath} \begin{array}{ll} n=1: & y = x(1+p) + r, \\ n=2: & y = \big(x(1+p) + r\big)(1+p) + r. \end{array}\end{displaymath}

Generally it is

$\displaystyle y$ $\displaystyle =$ $\displaystyle \big(\cdots\big(x(1+p)+r\big)(1+p)+r\big)\cdots\big)$  
  $\displaystyle =$ $\displaystyle x(1+p)^n+\big(1+(1+p)+(1+p)^2+\cdots+(1+p)^{n-1}\big) r$  

The bracketed expression in front of $ r$ can be transformed by the formula given in the geometric series, which then yields the above result.
(Authors: Höllig/Hörner/Abele)
This content is only available in German at the moment.
[previous page] [next page] [table of contents][page overview]
  automatically generated 10/23/2009