How to solve interest problem without using a crazy binomial expansion?

Question

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.

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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.

Answer 1

After $7.5$ years, Eric has $X(1+\frac i2)^{15}$ capital and thus makes an additional $X(1+\frac i2)^{15}\frac i2$ in the final six months. In the final year, Mike makes $2Xi$ from simple interest on his initial $2X$ investment. Thus in the final six months, he makes... well, he still just makes $2Xi$, delivered in December. However, if we assume that, we get the wrong answer, so it seems the question means for you to interpret "money made in the last half year" to mean "half the amount made in the full year", so we say he makes $Xi$ in the last six months. Thus we have:

$$X(1+\frac i2)^{15}\frac i2=Xi$$

this simplifies to

$$(1+\frac i2)^{15}=2$$

and so $i=2(\sqrt[15]2-1)$, as in the other answer.

Answer 2

$$\left(1+\frac{i}2\right)^{16}-\left(1+\frac{i}2\right)^{15}~=~i$$

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.