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Chapter 4: Problem 2

For which natural numbers \(n\) is \(n^{2}<2^{n} ?\) Justify your conclusion.

Short Answer

Expert verified
The inequality \(n^2 < 2^n\) holds true for all natural numbers \(n\) where \(n \geq 3\). This is because the function \(n^2\) has a constant growth rate, whereas the function \(2^n\) has an increasing growth rate, causing \(2^n\) to outgrow \(n^2\) as values of \(n\) increase.

Step by step solution

01

Write down some values of n and their corresponding n^2 and 2^n values

We will try to find a pattern by listing down the values of \(n^2\) and \(2^n\) for few values of n. n | \(n^2\) | \(2^n\) --|-------|------ 1 | 1 | 2 2 | 4 | 4 3 | 9 | 8 4 | 16 | 16 5 | 25 | 32 6 | 36 | 64 7 | 49 | 128 8 | 64 | 256
02

Analyze the pattern for n, n^2 and 2^n

In the above table, we observe that when \(n=1\), the inequality \(n^2 < 2^n\) holds true (1 < 2). When \(n = 2\), \(n^2\) and \(2^n\) are equal (4 = 4), and when \(n\geq3\), the inequality \(n^2 < 2^n\) is true for our listed values. Now, we want to analyse further and provide a justification for our observation.
03

Checking for higher values of n

Looking at the pattern of the table, we infer that the inequality \(n^2 < 2^n\) might be true for all \(n \geq 3\). To justify this, let's compare the rate of growth of \(n^2\) and \(2^n\). The function \(n^2\) is a quadratic function with a constant growth rate, whereas the function \(2^n\) is an exponential function with an increasing growth rate. Therefore, with increasing values of n, the function \(2^n\) will eventually outgrow the function \(n^2\).
04

Conclusion

From our observations and analysis, we find that the inequality \(n^2 < 2^n\) holds true for all natural numbers \(n\) where \(n \geq 3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Numbers
Natural numbers are the simplest numbers we use daily, starting from 1 and going upwards (1, 2, 3, and so on). They are integers and positive, representing counts of objects or positions in a sequence.
Unlike whole numbers, they do not include zero, and unlike integers, they don鈥檛 include negative numbers.
  • Natural numbers are used in basic counting, like the number of apples in a basket.
  • They are the foundation of number systems and are often used in various mathematical operations.
  • In mathematical problems, natural numbers are usually the domain, representing quantities in real-world scenarios.
Understanding natural numbers is essential for grasping more complex mathematical concepts like inequalities and functions.
Exponential Growth
Exponential growth refers to the rapid increase of a quantity over time, characterized by a constant rate of growth proportional to the current value. In mathematics, this is often expressed as functions involving exponents, like the function \(2^n\) used in the inequality problem.
As the variable in the exponent increases, the value of the function grows much faster compared to linear or quadratic growth.
  • Exponential functions have a constant growth rate, meaning they double in value over consistent intervals.
  • This growth is evident in our comparison between \(n^2\) and \(2^n\), where \(2^n\) significantly increases compared to \(n^2\) as \(n\) grows larger.
  • This concept is foundational in understanding phenomena like compound interest, population growth, and nuclear chain reactions.
Recognizing the nature of exponential growth helps in predicting the behaviour of large numbers in problems involving inequalities.
Quadratic Functions
Quadratic functions are a type of polynomial function represented by the standard form \(ax^2 + bx + c\), and they graph as parabolas in coordinate geometry. In our problem, \(n^2\) is a simple quadratic function where \(a = 1\), and \(b\) and \(c\) are zero.
Quadratic functions grow at a rate proportional to the square of the variable, which is slower compared to exponential functions.
  • The graph of a quadratic function is symmetrical and has a single turning point called the vertex.
  • Quadratic growth means that the difference in outputs as the input increases is not constant; it grows larger like the outputs 1, 4, 9, 16, etc., for \(n^2\).
  • In the context of comparing growth rates, anything beyond a linear growth rate but slower than exponential can often involve quadratics.
Understanding how quadratic functions behave is important when comparing them with exponential functions to determine when one will surpass the other.

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Most popular questions from this chapter

The Future Value of an Ordinary Annuity. For an ordinary annuity, \(R\) dollars is deposited in an account at the end of each compounding period. It is assumed that the interest rate, \(i,\) per compounding period for the account remains constant. Let \(S_{t}\) represent the amount in the account at the end of the \(t\) th compounding period. \(S_{t}\) is frequently called the future value of the ordinary annuity. So \(S_{1}=R\). To determine the amount after two months, we first note that the amount after one month will gain interest and grow to \((1+i) S_{1} .\) In addition, a new deposit of \(R\) dollars will be made at the end of the second month. So $$ S_{2}=R+(1+i) S_{1} $$ (a) For each \(n \in \mathbb{N},\) use a similar argument to determine a recurrence relation for \(S_{n+1}\) in terms of \(R, i,\) and \(S_{n}\). (b) By recognizing this as a recursion formula for a geometric series, use Proposition 4.16 to determine a formula for \(S_{n}\) in terms of \(R, i,\) and \(n\) that does not use a summation. Then show that this formula can be written as $$ S_{n}=R\left(\frac{(1+i)^{n}-1}{i}\right) $$ (c) What is the future value of an ordinary annuity in 20 years if \(\$ 200\) dollars is deposited in an account at the end of each month where the interest rate for the account is \(6 \%\) per year compounded monthly? What is the amount of interest that has accumulated in this account during the 20 years?

In calculus, it can be shown that $$ \begin{array}{l} \int \sin ^{2} x d x=\frac{x}{2}-\frac{1}{2} \sin x \cos x+c \quad \text { and } \\ \int \cos ^{2} x d x=\frac{x}{2}+\frac{1}{2} \sin x \cos x+c \end{array} $$ Using integration by parts, it is also possible to prove that for each natural number \(n,\) $$ \begin{aligned} \int \sin ^{n} x d x &=-\frac{1}{n} \sin ^{n-1} x \cos x+\frac{n-1}{n} \int \sin ^{n-2} x d x \text { and } \\ \int \cos ^{n} x d x &=\frac{1}{n} \cos ^{n-1} x \sin x+\frac{n-1}{n} \int \cos ^{n-2} x d x \end{aligned} $$ (a) Determine the values of $$ \int_{0}^{\pi / 2} \sin ^{2} x d x $$ and $$ \int_{0}^{\pi / 2} \sin ^{4} x d x $$ (b) Use mathematical induction to prove that for each natural number \(n\), $$ \begin{aligned} \int_{0}^{\pi / 2} \sin ^{2 n} x d x &=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots(2 n)} \frac{\pi}{2} \text { and } \\ \int_{0}^{\pi / 2} \sin ^{2 n+1} x d x &=\frac{2 \cdot 4 \cdot 6 \cdots(2 n)}{1 \cdot 3 \cdot 5 \cdots(2 n+1)} \end{aligned} $$ These are known as the Wallis sine formulas. (c) Use mathematical induction to prove that $$ \begin{aligned} \int_{0}^{\pi / 2} \cos ^{2 n} x d x &=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots(2 n)} \frac{\pi}{2} \quad \text { and } \\ \int_{0}^{\pi / 2} \cos ^{2 n+1} x d x &=\frac{2 \cdot 4 \cdot 6 \cdots(2 n)}{1 \cdot 3 \cdot 5 \cdots(2 n+1)} \end{aligned} $$ These are known as the Wallis cosine formulas.

Which of the following sets are inductive sets? Explain. (a) \(\mathbb{Z}\) (c) \(\\{x \in \mathbb{Z} \mid x \leq 10\\}\) (b) \(\\{x \in \mathbb{N} \mid x \geq 4\\}\) (d) \(\\{1,2,3, \ldots, 500\\}\)

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1,\) and that for each natural number \(n\), $$ a_{n+1}=a_{n}+n \cdot n ! $$ (a) Compute \(n !\) for the first 10 natural numbers. (b) Compute \(a_{n}\) for the first 10 natural numbers. (c) Make a conjecture about a formula for \(a_{n}\) in terms of \(n\) that does not involve a summation or a recursion.

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=2\) and that for each \(n \in \mathbb{N}, a_{n+1}=a_{n}+5\) (a) Calculate \(a_{2}\) through \(a_{6}\). (b) Make a conjecture for a formula for \(a_{n}\) for each \(n \in \mathbb{N}\). (c) Prove that your conjecture in Exercise (9b) is correct.
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