5️⃣ Application of Financial Mathematics
Apply numerical sequences to finance through simple and compound interest: compute the interest and the acquired capital, convert interest rates across periods, and analyse simple financial situations.
- Compute simple interest and the acquired capital.
- Compute compound interest and the acquired capital.
- Convert an interest rate from one period to another.
- Recognise which regime (simple or compound) applies to a situation.
Simple interest and compound interest are two different ways of computing, for example, the return on invested capital or the cost of a loan. Each has its practical uses depending on the financial and economic context. They are the direct application of the arithmetic and geometric sequences studied in Module 4.
Simple Interest
With simple interest, interest is computed only on the initial capital, without reinvestment. The interest is the same for each period, so the acquired capital follows an arithmetic sequence.
Let \(C_0\) be the initial capital, \(i\) the interest rate per period, and \(n\) the number of periods. The interest of one period is \(C_0\,i\) (constant), so:
\[ \text{Interest over } n \text{ periods} = C_0\,i\,n, \qquad C_n = C_0\,(1 + i\,n). \]
- short-term loans;
- one-off investments without reinvestment;
- late-payment penalties and fees.
Worked example
We invest \(100{,}000\) € at a simple annual rate of \(4\%\) for \(4\) years. The yearly interest is constant: \(100{,}000 \times 0.04 = 4{,}000\) € per year. After \(4\) years the acquired capital is
\[ C_4 = 100{,}000\,(1 + 0.04 \times 4) = 100{,}000 \times 1.16 = 116{,}000 \text{ €}. \]
Converting Interest Rates
Sometimes the rate and the period are not expressed in the same unit of time. Under the simple (proportional) convention, we scale the rate proportionally to the period.
An annual rate \(i_{\text{year}}\) corresponds to a monthly rate \(i_{\text{month}} = \dfrac{i_{\text{year}}}{12}\), and conversely \(i_{\text{year}} = 12\,i_{\text{month}}\).
- Convert an annual rate of \(6\%\) to a monthly rate: \(\dfrac{6\%}{12} = 0.5\%\) per month.
- Convert a monthly rate of \(1\%\) to an annual rate: \(12 \times 1\% = 12\%\) per year.
- A capital of \(5{,}000\) € invested for \(4\) months earns \(50\) € of interest. Find the annual rate. The \(4\)-month rate is \(\dfrac{50}{5000} = 1\%\), so the monthly rate is \(0.25\%\) and the annual rate is \(12 \times 0.25\% = 3\%\).
Compound Interest
With compound interest, the interest earned each period is added to the capital and itself earns interest in the following periods (the “snowball” effect). The acquired capital follows a geometric sequence.
At the end of one period the capital is multiplied by \((1 + i)\): the capital \(C_0\) becomes \(C_1 = C_0 + C_0 i = C_0(1 + i)\). Repeating, \(C_2 = C_1(1 + i) = C_0(1 + i)^2\), and by induction
\[ C_n = C_0\,(1 + i)^{\,n}. \]
This is exactly a geometric sequence with first term \(C_0\) and common ratio \(q = 1 + i\). The interest earned over \(n\) periods is \(C_n - C_0 = C_0\big[(1 + i)^n - 1\big]\).
Worked example
We invest \(100{,}000\) € at a compound annual rate of \(5\%\) for \(10\) years:
\[ C_{10} = 100{,}000 \times 1.05^{10} \approx 162{,}889 \text{ €}. \]
For comparison, the same capital at a simple rate of \(6\%\) for \(10\) years gives \(100{,}000(1 + 0.06 \times 10) = 160{,}000\) €: over a long horizon the compounding effect can outweigh a higher simple rate — but not always, as the final-exam case in Module 8 shows.
Converting a compound rate between periods is not proportional. A monthly compound rate \(i_m\) corresponds to the annual rate \(i_{\text{year}} = (1 + i_m)^{12} - 1\). For \(i_m = 0.5\%\) this gives \((1.005)^{12} - 1 \approx 6.17\%\), not \(6\%\).
Average Placement Rate
The average placement rate is the single rate that would produce the same final result (the same acquired capital) as a set of distinct investments, each with its own rate, amount, and duration. It summarises the overall performance of several combined investments in one figure.
Practice
- \(8{,}000\) € at a simple annual rate of \(3\%\) for \(5\) years: acquired capital? \(8{,}000(1 + 0.03 \times 5) = 9{,}200\) €.
- \(8{,}000\) € at a compound annual rate of \(3\%\) for \(5\) years: acquired capital? \(8{,}000 \times 1.03^{5} \approx 9{,}274.19\) €.
- What annual compound rate doubles a capital in \(10\) years? Solve \((1 + i)^{10} = 2 \Rightarrow i = 2^{1/10} - 1 \approx 7.18\%\).