Problem 9.9
Eric deposits X into a savings account at time 0, which pays interest at
a nominal rate of i, compounded semiannually. Mike deposits 2X into a
different savings account at time 0, which pays simple interest at an annual
rate of i. Eric and Mike earn the same amount of interest during the last 6
months of the 8th year. Calculate i.
Hello everyone, I'm doing a practice problem from Finan's actuarial exam FM textbook, and there is an equation I came up with that comes up with the right answer, but I have to use a graphing calculator to solve it.
You can't use a graphing calculator on the actuary exams, so I have to find out a way to solve this without getting a headache with the binomial theorem.
First, I get this: $$A_E(8) – A_E(7.5) = A_M(8) – A_M(7.5)$$
Simplifying all that mess, I get this monster:
$$(1+ \frac i2)^{16} – (1+ \frac i2)^{15} = i$$
(I simplified all the i's because of simple interest)
Putting this beast in my calculator, I get the answer: $$i=.094588$$ ,which is the correct answer. Maybe you can figure out an easier way to solve this, but I would also like to know if there is a way to solve it this way without the use of a graphing calculator.
I'm thinking that this could possibly be solved by some sort of series expansion.
Best Answer
$$\left(1+\frac{i}2\right)^{16}-\left(1+\frac{i}2\right)^{15}~=~i$$
First in both terms of the L.H.S. contain the $15^{th}$ power of $\left(1+\frac{i}2\right)$. So lets rearrange this a little bit
$$\begin{align} \left(1+\frac{i}2\right)^{16}-\left(1+\frac{i}2\right)^{15}~&=~i\\ \left(1+\frac{i}2\right)^{15}\left(\left(1+\frac{i}2-1\right)\right)~&=~i\\ \left(1+\frac{i}2\right)^{15}\left(\frac{i}2\right)~&=~i\\ \end{align}$$ Now dividing both sides by $i$ and taking the $15^{th}$ root yields to
$$\begin{align} \left(1+\frac{i}2\right)^{15}\left(\frac12\right)~&=~1\\ \left(1+\frac{i}2\right)^{15}~&=~2\\ \left(1+\frac{i}2\right)~&=~\sqrt[15]{2}\\ i~&=2\cdot(\sqrt[15]{2}-1) \end{align}$$
which is approximately $i=0.094588245641$, your given solution.