Give an estimate of the error, when the payment rate $x_m = r P_0 \frac{(1+r/m)^{mT}}{(1+r/m)^{mT}-1}$ (compounding and repayment m times per year) ist approximated bei $x_{\infty}=\frac{r P_0* e^{rt}}{e^{rt}-1}$.
Hello everyone,
$P_0$..initial loan
$t$...time when loan has to be repaid
$r$..nominal intrest rate
$x_m/m$.. repayment per period ($x_m$ is then repayment per year)
In lecture we derive for a loan of $P_0$ with runtime t,which is continous compounded with a nomial rate of r the formula $x_{\infty}=\frac{r P_0* e^{rt}}{e^{rt}-1}$.
It´s obviously that $ \lim_{m\rightarrow \infty} x_m=x_{\infty} $.
Should i estimate $|x_m-x_{\infty}|= |r P_0 \frac{(1+r/m)^{mT}}{(1+r/m)^{mT}-1} -\frac{r P_0 e^{rt}}{e^{rt}-1}|$= $r P_0|\frac{(1+r/m)^{mT}}{(1+r/m)^{mT}-1}-\frac{e^{rt}}{e^{rt}-1}|$ for fixed m? Or what should i do?
You need to notice that $T=t$ because both are the term of the loan. As you say $\lim_{m \to \infty} (1+\frac rm)^{mt}=e^{rt}$, which justifies your expression for $x_m-x_\infty$. You are asked to describe how close the left side is to the limit, presumably when $m$ is somewhat large. You can note that both sides are $t^{th}$ powers and compare $(1+\frac rm)^{m}$ and $e^{r}$ Now you can expand $e^r$ in a Taylor series and compare with the binomial expansion of $(1+\frac rm)^{m}$. You should be able to say something about how the approximation improves as $m$ gets larger