Revision as of 17:51, 26 November 2023 by Admin (Created page with "'''Solution: B''' <math>\delta=\ln (1+i)</math> and <math>d=i /(1+i)=1-1 /(1+i)</math>. Thus <math display = "block"> \begin{aligned} \frac{d f}{d i}=\frac{d}{d i} \frac{(1+i) \ln (1+i)}{i} &=\frac{d}{d i}(1+1 / i) \ln (1+i) = \frac{-1}{i^2} \ln (1+i)+\left(1+\frac{1}{i}\right) \frac{1}{1+i} \\ &=\frac{-\ln (1+i)}{i^2}+\frac{1}{i}=\frac{-\delta}{i^2}+\frac{1}{i}=\frac{i-\delta}{i^2} \end{aligned} </math> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/...")
Exercise
ABy Admin
Nov 26'23
Answer
Solution: B
[math]\delta=\ln (1+i)[/math] and [math]d=i /(1+i)=1-1 /(1+i)[/math]. Thus
[[math]]
\begin{aligned}
\frac{d f}{d i}=\frac{d}{d i} \frac{(1+i) \ln (1+i)}{i} &=\frac{d}{d i}(1+1 / i) \ln (1+i) = \frac{-1}{i^2} \ln (1+i)+\left(1+\frac{1}{i}\right) \frac{1}{1+i} \\ &=\frac{-\ln (1+i)}{i^2}+\frac{1}{i}=\frac{-\delta}{i^2}+\frac{1}{i}=\frac{i-\delta}{i^2}
\end{aligned}
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.