Recursion for Financial Maths

Arithmetic and geometric sequences can be applied in many areas of life, including simple and compound interest earnings, straight-line and unit depreciation, monthly rental accumulation and reducing balance loans.

Use this page to revise the following concepts within fecursion for financial maths:

Arithmetic recursion

When someone is saving money in equal instalments, the cumulative savings at each savings period form an arithmetic sequence. If the driver of a vehicle is travelling down a highway at a constant speed, the amount of petrol left in the tank, measured every minute of the trip, forms another arithmetic sequence. In fact, any time a quantity is changing by equal amounts at set time periods, the process can be considered as being arithmetic and therefore represented by an arithmetic sequence.

Worked Example

Tabitha starts with \(\$200\) in her piggy bank. At the end of the week she adds \(\$25\) and then continues to add \(\$25\) at the end of each successive week.

Find a rule to describe \(B_n\), the balance of Tabitha’s savings at the end of each week,  and find when her savings will reach \($450\).

Solution

The arithmetic sequence of savings generated is \($200, $225, $250, $275...\)  
This is arithmetic, so the starting value is a and the common difference is \($25\).

The balance of that beginning can be represented as \(B_0\). The end of the first week is represented by \(B_1\), the end of the second week by \(B_2\), and so on.

To find a general rule, use

\[B_n = a + nd \text{, where }\ B_0 = a\]

The first term gives \(a = 200\).
The common difference between terms gives us \(d = 25\).

Therefore, the formula for this arithmetic sequence is \(B_n = 200 + 25n\).

To find when the savings reach \(\$450\), substitute this into the general rule and solve for \(n\), which is the number of weeks:

\[\begin{align}450 &= 200 + 25n \\ 25n &= 250 \\ n &= 10\end{align}\]

Tabitha's savings will grown to \(\$450\) by the end of the 10th week.

Geometric Recursion

There are many applications of geometric sequences such as compound interest, exponential growth of bacteria, exponential decay of radioactive elements. If an amount is increasing or decreasing by a constant factor at set time periods, the process can be considered as being geometric.

The general form of this equation is:

\[u_n = ar^n\]

The recurrence rule for geometric recursion is given by:

\[u_{n + 1} = ru_{n}\text{, }u_{n} =a\]

The general form for geometric sequences is very similar to the compound interest formula:

\[A = P\left(1 + r\right)^{n}\]

The value of the investment after n time periods is defined by \(A\). The principal \(P\), which is the initial value of the investment, is equivalent to \(a\), the first term of a geometric sequence. The compound interest rate \(r\) is expressed as a percentage and added to \(1\) to give the common ratio of a geometric sequence.

Worked Example

The average rate of depreciation of the value of a Ferrari is \(14\%\) per year. A new Ferrari is bought for \(\$90,000\).

  1. What is the car worth after 1 year?

    2. Solution

    The car’s value will decrease by \(14\%\). This means that \(86\%\) will be remaining.

    Hence, \($90 000 \times 0.86 = $77,400\)

  2. Write a recurrence relation for \(V_n\), defining the value of the car after \(n\) years, with an initial condition \(V_0\).

    3. Solution

    When a vehicle depreciates by 14% every year, the value of the vehicle decreases to 86% of its value each year. This means the common ratio can be found as

    \[r = 1 - 0.14 = 0.86\]

    With \(r = 0.86\) with an initial value of \(V_0 = 90,000\), we can write the following recurrence relation:

    \[V_{n + 1} = 0.86 V_n, V_0 = 90,000\]

  3. Use your recurrence relation to determine a rule for the value of the Ferrari \(V_n\) after \(n\) years.

    Solution

    The general rule for a geometric sequence is given by \(u_{n} = ar^{n}\).

    Given that \(a=90,000\) and \(r=0.86\), the rule for \(V_n\) is
    \[V_n = 90,000 \times {0.86}^n\]

  4. What is the car worth after \(3\) years?

    5. Solution

    \(V_n = 90 000 \times {0.86}^n\)
    Solving for \(n = 3\) gives:

    \[\begin{align}V_{3} &= 90,000 \times 0.86^3 \\ &=\57,345.04\end{align}\]

Effective annual interest rate

When interest on an investment or loan is compounded, the amount of interest on that investment or loan will vary depending on how often it is compounded.

Consider two investment options for an investment of \(\$1,000\).

Investment Option 1 pays \(12\%\) interest compounded monthly.
Investment Option 2 pays \(12\%\) interest compounded semi-annually (twice a year).

Disregarding how often the investments are compounded, these appear to be equivalent investment options.
However if each is investigated further:

Investment Option 1 pays \(12\%\) p.a, compounded monthly, so \(1\%\) per month. If we look at it after one year, equivalent to \(12\) compounding periods,
\[\begin{align}V_{12} &= 1000 \times {1.01}^{12} \\ &= 1126.825\end{align}\]
So the investment would be worth \(\$1,126.83\) after one year.

Investment Option 2 pays 12% p.a, compounded semi-annually, so \(6\%\) per six months. If we look at it after one year, equivalent to 2 compounding periods,
\[\begin{align}V_2 &= 1000 \times{1.06}^2 \\ &=1123.60\end{align}\]
So the investment would be worth \($1,123.60\) after one year.

Here there is a difference of \(\$3.25\). While this is small over one year, this can become quite large for an investment or loan over a longer period of time.

For this reason, the effective annual interest rate is used to compare investments or loans that compound at different times.

\[r_{eff} = \left(\left(1 + \frac{r/n}{100}\right)^{n}-1\right) \times 100%\]

Where

  • \(r_{eff}\) is the effective annual interest rate
  • \(r\) is the nominal interest rate (annual interest rate)
  • \(n\) is the number of compounding periods in one year.

Worked Example

An investment of \($1,000\) is made for one year that pays \(24\%\)per annum.
Compare the effective annual interest rate for this investment if it compounded, annually, semi-annually, or monthly. Hence determine which compound structure would be best for this investment.

Solution

If it is compounded annually then \(n = 1\) as there is one compound per year.
Using the effective interest rate formula it can be found that:

\[\begin{align}r_{eff} &= \left(\left(1 + \frac{24/1}{100}\right)^{1} - 1\right) \times 100\% \\ &=24\% \end{align}\]

If it is compounded semi-annually then \(n = 2\) as there are two compounds per year.
Using the effective interest rate formula it can be found that:

\[\begin{align}r_{eff} &= \left(\left(1 + \frac{24/2}{100}\right)^{2} - 1\right) \times 100\% \\ &= 25.44\% \end{align}\]

If it is compounded monthly then \(n = 12\) as there are twelve compounds per year.
Using the effective interest rate formula it can be found that:

\[\begin{align}r_{eff} &= \left(\left(1 + \frac{24/12}{100}\right)^{12} - 1\right) \times 100\% \\ &= 26.83\% \end{align}\]

Hence it would be best to invest this \(\$1,000\) when it is compounded monthly as this gives the greatest effective interest rate.