The number of pieces remaining after each day is a series: {25, 24, 23...}. How could you identify how many are left on any specific day of the month? Is there a general formula of calculating the candy remaining on any date?
Recursive Formulas
A sequence is an ordered list of objects. The simplest way to represent a sequence is by listing some of its terms.
The sequence of odd, positive integers is shown here:
1, 3, 5, 7 ... |
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In this lesson you will learn to represent a sequence recursively, which means that you need to know the previous term in order to find the next term in the sequence.
Consider the sequence shown above. What is the next term?
As long as you are familiar with the odd integers (i.e., you can count in 2’s) you can figure out that the next term is 9. If we want to describe this sequence in general, we can do so by stating what the first term is, and then by stating the relationship between successive terms. When we represent a sequence by describing the relationship between its successive terms, we are representing the sequence recursively.
The terms in a sequence are often denoted with a variable and a subscript. All of the terms in a given sequence are written with the same variable, and increasing subscripts. So we might list terms in a sequence as a_{1}, a_{2}, a_{3}, a_{4}, a_{5} ...
We can use this notation to represent the example above. This sequence is defined as follows:
a_{1} = 1 | |
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a_{n} = a_{n}_{-1} + 2 |
At first glance this notation may seem confusing. What is important to keep in mind is that the subscript of a term represents its “place in line.” So \begin{align*}a_n\end{align*} just means the n^{th} term in the sequence. The term \begin{align*}a_{n-1}\end{align*} just means the term before \begin{align*}a_n\end{align*}. In the sequence of odd numbers above, a_{1} = 1, a_{2} = 3, a_{3} = 5, a_{4} = 7, a_{5} = 9 and so on. If, for example, we wanted to find a_{10}, we would need to find the 9^{th} term in the sequence first. To find the 9^{th} term we need to find the 8^{th} term, and so on, back to a term that we know.
Examples
Example 1
Earlier, you were asked about a general formula for the candy remaining on any given date.
The remaining candy in an advent calendar is a standard arithmetic sequence, and can be described as a_{n} = a_{n}_{- 1} - 1.
Example 2
For the sequence of odd numbers, list a_{6}, a_{7}, a_{8}, a_{9}, and a_{10}.
Each term is two more than the previous term.
a_{6} = a_{5} + 2 = 9 + 2 = 11
a_{7} = a_{6} + 2 = 11 + 2 = 13
a_{8} = a_{7} + 2 = 13 + 2 = 15
a_{9} = a_{8} + 2 = 15 + 2 = 17
a_{10} = a_{9} + 2 = 17 + 2 = 19
The sequence of odd numbers is linear, and it is referred to as an arithmetic sequence. Every arithmetic sequence has a common difference, or a constant difference between each term. (The common difference is analogous to the slope of a line.) The sequence of odd numbers has a common difference of 2 because for all n, a_{n} - a_{n}_{- 1} = 2.
Finding terms in this sequence is relatively straightforward, as the pattern is familiar. However, this would clearly be tedious if you needed to find the 100^{th} term.
Example 3
Find the 5^{th} term for the sequence:
t_{1} = 3 | |
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t_{n} = 2t_{n}_{-1} |
t_{5} = 48
t_{2} = 2t_{1} = 2 × 3 = 6 | ||
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t_{1} = 3 | \begin{align*}\rightarrow\end{align*} | t_{3} = 2t_{2} = 2 × 6 = 12 |
t_{n} = 2 × t_{n}_{-1} | t_{4} = 2t_{3} = 2 × 12 = 24 | |
t_{5} = 2t_{4} = 2 × 24 = 48 |
This example is a geometric sequence. Every geometric sequence has a common ratio, which is 2 in this example, because for all n, \begin{align*}\frac{t_{n}}{t_{n-1}}=2\end{align*}. The terms of a geometric sequence follow an exponential pattern.
Example 4
Find the 4^{th} term for the sequence:
b_{1} = 2 | |
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b_{n} = (b_{n}_{-1})^{2} + 1 |
b_{4} = 677
b_{2} = (b_{1})^{2} + 1 = 2^{2} + 1 = 4 + 1 = 5 | ||
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b_{1} = 2 | \begin{align*}\rightarrow\end{align*} | b_{3} = (b_{2})^{2} + 1 = 5^{2} + 1 = 25 + 1 = 26 |
b_{n} = (b_{n}_{-1})^{2} + 1 | b_{4} = (b_{3})^{2} + 1 = 26^{2} + 1 = 676 + 1 = 677 |
This sequence is neither arithmetic nor geometric, though its values follow a cubic pattern. As you can see from just a few terms here, the terms in a sequence can grow quickly.
For any of these sequences, as noted above, determining more than a few values by hand can be time consuming. In another lesson, we will introduce explicit formulas, which can be used to define a sequence in a way that makes finding the n^{th} term faster.
Example 5
Write the next 5 terms of the following sequence: \begin{align*}a_1 = -4 \text{, } a_2 = -4 \text{,}\end{align*} and \begin{align*}a_n = 2a_{n-1} + a_{n-2}\end{align*}.
\begin{align*}a_2 = 2(-4) + (-4) = -12\end{align*}
\begin{align*}a_3 = 2(-12) + (-4) = -28\end{align*}
\begin{align*}a_4 = 2(-28) + (-12) = -68\end{align*}
\begin{align*}a_5 = 2(-68) + (-28) = -164\end{align*}
\begin{align*}a_6 = 2(-164) +(-68) = -396\end{align*}
So our answer is: -12, -28, -68, -164, and -396.
Example 6
Write a recursive formula that fits the following sequence: 1, 5, 9, 13, 17.
In this problem we deduct that each term differs by the same amount. Once, we identify the difference of each term, +4 each time in this case, then we know that the sequence requires that we add the same amount (4) to each term.
We write that as: \begin{align*}a_n = a _{n-1} + 4\end{align*}.
Example 7
Given the following sequence, write a recursive formula, then find the next three numbers in the sequence: 3, -4, -1, -5, -6, -11, -17.
Getting to the next term is not going to be as easy as the previous example. We need to examine the number sequence more closely to solve this problem.
In this sequence, the Fibonacci Series was applied. What this means is that the two previous terms were added together to get the next term in the sequence.
It is written as: \begin{align*} a_n = a_{n-1} + a_{n - 2}\end{align*}.
Now that we know the formula, we can find the next three numbers in the sequence:
\begin{align*}a_8 = (-11) + (-17) = -28\end{align*}
\begin{align*}a_9 = (-17) + (-28) = -45\end{align*}
\begin{align*}a_{10} =(-28) + (-45) = -73\end{align*}
So the next three numbers in the sequence are: -28, -45, and -73.
Review
- A sequence in which you know the previous term in order to find the next term is:
- Why is the sequence of odd numbers linear?
- Which type of sequence has a common difference?
- A sequence that uses the same multiple to get from one term to another is:
- Find the value of a_{6} , given the sequence defined as: a_{1} = 4 a_{n} = 5a_{n}_{-1}
- Find the value of a_{5}, given the sequence defined as: a_{1} = 32 a_{n} = (1/2)a_{n}_{-1}
- Find the value of a_{n}_{-1} , given the sequence defined as: a_{1} = 1 a_{n} = 3a_{n}_{-1}-n
Using the given recursive formulas, identify the next 5 terms in the sequences that follow:
- \begin{align*}a_1 = -2a_2 = 1\end{align*} and \begin{align*}a_n = 3a_{n-1} -5a_{n-2}\end{align*}
- \begin{align*}a_1 = -2\end{align*} and \begin{align*}a_n = 3a_{n-1}\end{align*}
- \begin{align*}a_1 = 3a_2 = -2\end{align*} and \begin{align*}a_n = -5a_{n-1} + a_{n-2}\end{align*}
- \begin{align*}a_1 = 1\end{align*} and \begin{align*}a_n = 4a_{n-1}\end{align*}
- \begin{align*}a_1 = -4a_2 = 1\end{align*} and \begin{align*}a_n = -a_{n-1} + a_{n-2}\end{align*}
Given the following sequence of numbers find the recursive formula.
- 1, 5, 9, 13, 17
- -1, 3, 2, 5, 7, 12, 19
- -4, 16, -64, 256, -1024
Given the following sequence of numbers find the recursive formula and the next three numbers in the sequence.
- 1, -1, 1, -1, 1
- -5, -1, -6, -7, -13, -20, -33
- 1, - 3, 9, -27, 81
- -3, -4, -7, -11, -18, -29, -47
- -1, -5, -9, -13, -17
- Write the next three terms of the sequence: \begin{align*}a_n = (-1)^n \cdot 5a_{n-1}\end{align*}
- Given the formula: \begin{align*} a_n = 4n-1\end{align*}, is the number 27 a term in the sequence of numbers?
- Given the formula: \begin{align*} a_n = 4n -1\end{align*} is the number 97 a term in the sequence of numbers?
Review (Answers)
To see the Review answers, open this PDF file and look for section 7.1.