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Chapter 10: Problem 4

The number of bacteria in a culture is 1000 (approximately), and this number increases \(250 \%\) every two hours. Use a recurrence relation to determine the number of bacteria present after one day.

Short Answer

Expert verified
Using an exponential growth recurrence relation, you would find that the number of bacteria in the culture would have increased to approximately 281,474,976,710,656 after one day.

Step by step solution

01

Understanding the Growth Rate

A growth of 250% is the same as multiplying the current number of bacteria by \(1 + \frac{250}{100} = 3.5\). This is because the initial number of bacteria is represented as 100% (or 1 in this case) and the growth is an additional 250%.
02

Calculate the Number of 2-hour Periods in a Day

There are \(24/2 = 12\) two-hour periods in a day. This number will be utilised in the recurrence relation.
03

Create a Recurrence Relation

A recurrence relation is a sequence where each successive term is defined as a function of the preceding ones. In this case, the relation will be defined as \(a_n = 3.5 \times a_{n-1}\) where \(a_0 = 1000\) and \(n\) is the number of 2-hour periods.
04

Use the Recurrence Relation

With this relation, calculate the value of the number of bacteria after each 2-hour period up to \(n = 12\), which is the total number of periods in a day. We keep multiplying the quantity of bacteria by 3.5 at each step.
05

Calculate the Total Number of Bacteria after One Day

At the end of the 12th 2-hour period (which would be one day), the number of bacteria in the culture would be obtained. This is the solution to the problem.

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Most popular questions from this chapter

Solve the following recurrence relations. a) \(a_{n+2}+3 a_{n+1}+2 a_{n}=3^{n}, \quad n \geq 0, a_{0}=0, a_{1}=1\) b) \(a_{n+2}+4 a_{n+1}+4 a_{n}=7, \quad n \geq 0, \quad a_{0}=1, a_{1}=2\) c) \(a_{n+2}-a_{n}=\sin (n \pi / 2), \quad n \geq 0, a_{0}=1, a_{1}=1\)

$$ \text { Use a recurrence relation to derive the formula for } \sum_{i=0}^{n} i^{2} . $$

In Exercise 14 of Section \(4.2\) we learned that \(F_{0}+F_{1}+F_{2}+\cdots+F_{n}=\sum_{i=0}^{n} F_{l}=F_{n+2}-1\). This is one of many such properties of the Fibonacci numbers that were discovered by the French mathematician Fran莽ois Lucas (1842-1891). Although we established the result by mathematical induction, we see that it is easy to develop this formula by adding the system of \(n+1\) equations $$ \begin{gathered} F_{0}=F_{2}-F_{1} \\ F_{1}=F_{3}-F_{2} \\ \cdots \\ F_{n-1}=F_{n+1}-F_{n} \\ F_{n}=F_{n+2}-F_{n+1} . \end{gathered} $$ Develop formulas for each of the following sums, and then check the general result by mathematical induction. a) \(F_{1}+F_{3}+F_{5}+\cdots+F_{2 n-1}\), where \(n \in \mathbf{Z}^{+}\) b) \(F_{0}+F_{2}+F_{4}+\cdots+F_{2 n}\), where \(n \in \mathbf{Z}^{+}\)

Solve the following recurrence relations. a) \(a_{n+2}^{2}-5 a_{n+1}^{2}+6 a_{n}^{2}=7 n, \quad n \geq 0, \quad a_{0}=a_{1}=1\) b) \(a_{n}+n a_{n-1}=n !, n \geq 1, a_{0}=1\) c) \(a_{n}^{2}-2 a_{n-1}=0, \quad n \geq 1, a_{0}=2\) (Let \(b_{n}=\log _{2} a_{n}, n \geq 0\).)

$$ \text { Solve the recurrence relation } a_{n+3}-3 a_{n+2}+3 a_{n+1}-a_{n}=3+5 n, n \geq 0 \text {. } $$
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