Revision as of 17:50, 26 November 2023 by Admin (Created page with "'''Solution: D''' Fund A has value <math>(1.01)^m</math> after <math>m</math> months. Fund <math>\mathrm{B}</math> has value <math>e^{\int_0^t r / 6 d r}</math> after <math>t</math> years or <math>e^{\int_0^{m / 12} r / 6 d r}</math> after <math>m</math> months. Thus <math display="block"> 1.01^m=e^{r^2 /\left.12\right|_0 ^{m / 12}}=e^{m^2 / 12^3} \text { so } </math> <math>m \ln (1.01)=m^2 / 12^3</math> so <math>m=12^3 \ln (1.01)</math> months. Thus <math>T=m / 12=1...")
Exercise
ABy Admin
Nov 26'23
Answer
Solution: D
Fund A has value [math](1.01)^m[/math] after [math]m[/math] months. Fund [math]\mathrm{B}[/math] has value [math]e^{\int_0^t r / 6 d r}[/math] after [math]t[/math] years or [math]e^{\int_0^{m / 12} r / 6 d r}[/math] after [math]m[/math] months. Thus
[[math]]
1.01^m=e^{r^2 /\left.12\right|_0 ^{m / 12}}=e^{m^2 / 12^3} \text { so }
[[/math]]
[math]m \ln (1.01)=m^2 / 12^3[/math] so [math]m=12^3 \ln (1.01)[/math] months. Thus [math]T=m / 12=12^2 \ln (1.01)[/math].
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.