Mathematics of Interest Rates and Finance

Abstract

A sequence is an ordered set of numbers. Consider the sequence of real numbers expressed as

Appendix A: Estimating Time \(t\) and New Interest Rate \(i'\) for Sect. 12.5

In the text we found ourselves faced with solving the following scary exponential equation

$$\begin{aligned} 118737.81(1.00417)^t- 90116.22(1.00333)^t - 371.21t -28620.96 =0 \end{aligned}$$

Solving this equation is equivalent to finding the zero(s) of the function

$$\begin{aligned}f(t)= 118737.81(1.00417)^t- 90116.22(1.00333)^t - 371.21t -28620.96 \end{aligned}$$

The graph of the above function indicates that an acceptable solution for $t$ is between 210 and 230 (Fig. 12.2).

Fig. 12.2

Graph of $118737.81(1.00417)^t- 90116.22(1.00333)^t - 371.21t -28620.96$

The Table 12.5 further narrows down the solution for $t$ between 215 and 220.

Table 12.5 Values of $f(t)$ for selected values of $t$

Using Taylor linear approximation, we expand the function around $t=220$.

$$\begin{aligned} f(220) = 96.20 \end{aligned}$$

$$\begin{aligned} f'(t)&= 118737.81(1.00417)^t\ln (1.00417) - 90116.22(1.00333)^t \ln (1.00333)\\&\quad -371.21\end{aligned}$$

$$\begin{aligned}f'(220)= 240.542\end{aligned}$$

Now we can write the linear approximation of the function as

$$\begin{aligned} f(t) = 96.20 +240.542(t-220)= -52822.99 +240.542 t\end{aligned}$$

The zero of this function is

$$\begin{aligned}f(t)=0 \;\;\; \longrightarrow \;\;\; -52822.99 +240.542 t = 0 \;\;\;\text {and} \;\;\; t = \dfrac{52822.99}{240.542} \approx 219.\end{aligned}$$

Next we encountered problem of determining $i'$ such that the it satisfies

$$\begin{aligned}\frac{(\frac{i'}{12})}{1-(1+\frac{i'}{12})^{-360}} \le 0.004873\end{aligned}$$

Assuming equality between the left and right hand sides, we simplify the expression to

$$\begin{aligned} \frac{i'}{12} = 0.004873 -0.004873(1+\frac{i'}{12})^{-360} \end{aligned}$$

$$\begin{aligned} 0.004873(1+\frac{i'}{12})^{-360} = 0.004873 - \frac{i'}{12}\end{aligned}$$

Taking the logarithm of both sides, we have

$$\begin{aligned}\ln (0.004873) -360\;\ln (1+\frac{i'}{12}) = \ln (0.004873-\frac{i'}{12})\end{aligned}$$

$$\begin{aligned}-5.32407 -360\;\ln (\frac{12+i'}{12}) = \ln (\frac{0.058475-i'}{12})\end{aligned}$$

$$\begin{aligned}-5.32407 -360\;\ln (12+i') + 360\;\ln (12)=\ln (0.058475-i') - \ln (12)\end{aligned}$$

and finally,

$$\begin{aligned} 360\;\ln (12+i') + \ln (0.058475 - i') -891.7272 = 0 \end{aligned}$$

(12.32)

Solution to this logarithmic function is equivalent of finding zero(s) of the function

$$\begin{aligned}f(i') = 360\;\ln (12+i') + \ln (0.058475 - i') -891.7272 \end{aligned}$$

$$\begin{aligned}f(i') = 0 \;\;\; \longrightarrow \;\;\; 360\;\ln (12+i') + \ln (0.058475 - i') -891.7272 = 0 \end{aligned}$$

Table 12.6 indicates that the solution must be in the neighborhood of $i' = 0.04$ (Fig. 12.3).

Table 12.6 Values of $f(i')$ and $f'(i')$ for selected values of $i'$

Fig. 12.3

Graph of $360\;\ln (12+i') + \ln (0.058475 - i') -891.7272$

We expand the function around this :

$$\begin{aligned} f(0.04)=0.45816\end{aligned}$$

$$\begin{aligned}f'(i')= \frac{360}{12+i'} -\frac{1}{0.058475 -i'}\end{aligned}$$

$$\begin{aligned}f'(0.04) = -24.2239\end{aligned}$$

The linear function is

$$\begin{aligned}f(i') = 0.045816-24.2239\;(i'-0.04)= 1.01489-24.2239\;i'\end{aligned}$$

The zero of this function is a linear approximation to $i'$:

$$\begin{aligned}f'(i') = 0 \;\; \longrightarrow \;\;\; 1.01489-24.2239\;i'= 0\;\;\; \text {leading to}\;\;i'= 0.04190 \approx 0.042\end{aligned}$$

Note that we can use WolframAlpha and solve the for $i'$ in Eq. (12.32). We enter (by using $x$ instead of $i'$)

$$\begin{aligned} solve[360\,log(12+x)+log(0.058475-x)-891.7272=0] \end{aligned}$$

There are two possible answers. The first one is a negative number that must be discarded. The second answer (to four decimal places) is $x=0.0417.$ WolframAlpha interpret $log$ as the natural logarithm. The symbol for $log$ at base 10 is $log10.$