Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading Marcel B. Finan's A Basic Course in the Theory of Interest and Derivatives Markets: A Preparation for the Actuarial Exam FM/2 and have difficulty with two of his questions.

Problem 9.6 (page 73) is (paraphrased):

Given that $$1+\frac{i^{(n)}}n=\frac{1+\frac{i^{(4)}}4}{1+\frac{i^{(5)}}5}$$ find $n$.

(Note that $i^{(k)}$ denotes the nominal annual interest rate, which is convertible $k$ times per year for an effective annual interest rate $>i^{(k)}$ (when $k>1$). I assume that the three nominal annual interest rates in this question are the same, although the question doesn't so specify, since if they're not then there's obviously no way to figure out this problem.)

My only idea of how to solve this problem was that the right-hand side is the accumulation of $1+\frac{i^{(5)}}5$ over the next $\frac14-\frac15=\frac1{20}$ years, which would make $n=20$. But while that makes sense, I'm not sure it's correct and would welcome any feedback or ideas on how to solve this.

Problem 9.9 (page 74) is:

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 $i$.

I've got $$X\left((1+\frac i2)^{16}-(1+\frac i2)^{15}\right)=2X(1+\frac i2)$$ and thus $$i(1+\frac i2)^{14}=4$$ but have no idea how to proceed from there. Any help would be much appreciatred.

share|improve this question

1 Answer 1

For the second, Eric must have $2X$ at the end of $7.5$ years, as they will then each get six month's interest on the same amount. So $(1+\frac i2)^{15}=2$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.