/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Let \(D=\mathbb{N}-\\{1,2\\}\) a... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 6

Let \(D=\mathbb{N}-\\{1,2\\}\) and define \(d: D \rightarrow \mathbb{N} \cup\\{0\\}\) by \(d(n)=\) the number of diagonals of a convex polygon with \(n\) sides. In Preview Activity \(1,\) we showed that for values of \(n\) from 3 through 8 , $$ d(n)=\frac{n(n-3)}{2} $$ Use mathematical induction to prove that for all \(n \in D\), $$ d(n)=\frac{n(n-3)}{2} $$

Short Answer

Expert verified
To prove the formula \(d(n) = \frac{n(n-3)}{2}\) for all \(n \in D\), we can use mathematical induction. First, show that the formula holds for the base case: \(d(3) = 0\). Then, assume it holds true for \(n=k\): \(d(k) = \frac{k(k-3)}{2}\). Finally, show that it also holds for \(n=k+1\): \(d(k+1) = \frac{(k+1)((k+1)-3)}{2}\). Since we have demonstrated the validity of the formula for the base case and the inductive step, we can conclude that the formula is true for all \(n \in D\).

Step by step solution

01

Show the formula is true for n = 3

The base case for our domain is \(n = 3\). We must show that the formula holds true for this value. A convex polygon with 3 sides is a triangle, and since a triangle has no diagonals, we expect \(d(3)=0\). Now, we can plug \(n = 3\) into the formula: $$ d(3) = \frac{3(3 - 3)}{2} = \frac{3(0)}{2} = 0 $$ Since this result matches our expectation for a triangle, the formula holds true for the base case.
02

Inductive Hypothesis

Now, let's assume that the formula \(d(n) = \frac{n(n-3)}{2}\) holds true for some arbitrary \(n = k\) with \(k \in D\). So, for our inductive hypothesis, we have: $$ d(k) = \frac{k(k-3)}{2} $$
03

Prove the formula holds for n = k + 1

In this step, we must show that if the formula holds for \(n=k\), then it also holds for \(n=k+1\). In other words, we want to show that: $$ d(k + 1) = \frac{(k+1)((k+1)-3)}{2} $$ To do this, we will first find the difference between the diagonal count of polygons with \(k+1\) sides and \(k\) sides. When we add a new vertex to a convex polygon with \(k\) sides, we create \(k-2\) new diagonals (every new diagonal connects the new vertex with an old one excluding its adjacent vertices). Therefore, we have: $$ d(k + 1) = d(k) + (k-2) $$ Now, we can substitute the formula for \(d(k)\) from the inductive hypothesis and simplify: \begin{align*} d(k + 1) &= \frac{k(k-3)}{2} + (k-2) \\ &= \frac{k(k-3)}{2} + \frac{2(k-2)}{2} \\ &= \frac{k(k-3) + 2(k-2)}{2} \\ &= \frac{k^2 - 3k + 2k - 4}{2} \\ &= \frac{k^2 - k - 4}{2} \\ &= \frac{k(k-1) - 4}{2} \\ &= \frac{(k+1)((k+1) - 3)}{2} \end{align*} Thus, we have shown that if the formula holds for \(n=k\), it also holds for \(n=k+1\). Since we have demonstrated that the formula is true for the base case (\(n=3\)) and that the formula holds for \(n=k+1\) if it holds for \(n=k\), we can conclude that the formula \(d(n) = \frac{n(n-3)}{2}\) is true for all \(n \in D\) using mathematical induction.

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Key Concepts

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

Convex Polygon Diagonals
When we refer to the number of diagonals in a convex polygon, we're counting the line segments that connect non-adjacent vertices inside the polygon. In a convex shape, any two points within the polygon can be connected by a line that will not cross the polygon's sides.

Consider a triangle (a three-sided polygon), which has zero diagonals since every pair of vertices is connected by a side of the triangle. Now, as the number of sides increases, the number of diagonals grows quickly. This relationship between the number of sides (denoted as \(n\)) and the diagonals is given by the formula \(d(n) = \frac{n(n-3)}{2}\).

Why is this formula used? Think of it this way: every vertex can connect to \(n-3\) other vertices (excluding the one it's connected to and the two adjacent vertices), and since this applies to all \(n\) vertices, we initially get \(n(n-3)\) connections. However, we must divide by 2 because each diagonal is counted twice - once for each end vertex. This leads us to our formula for the number of diagonals in a convex polygon with \(n\) sides.
Inductive Hypothesis
The inductive hypothesis is a critical step in the process of proof by induction. Here's how it works: you assume that a statement or formula, in our case \(d(n) = \frac{n(n-3)}{2}\), is true for an arbitrary case of \(n=k\).

This assumption does not mean that we believe the statement is true unconditionally; rather, it's a strategic move that allows us to show that if it holds for one case, it will hold for the next logical case (\(n=k+1\)).

The beauty of the inductive hypothesis lies in its power to create a domino effect: if you can prove that the statement is true for the lowest value in your range (the base case) and show that the truth carries over from one case \(k\) to the next \(k+1\), you can assert the truth of the statement for all values in the range. The hypothesis serves as the 'linking step' between these cases in the induction process.
Proof by Induction
Proof by induction is a technique used to prove a mathematical statement that is usually formulated in terms of \(n\), where \(n\) is a positive integer. The beauty of induction lies in its simplicity and power 鈥� it's like proving an infinite number of cases with just two steps.

To execute a proof by induction, you start with the base case. This is the lowest or first value of \(n\) for which the statement must be verified directly. For our diagonal counting formula, the base case is a triangle where \(n=3\), and indeed, a triangle has zero diagonals, matching our formula's prediction.

The second step involves applying the inductive hypothesis for some arbitrary case \(n=k\), and then proving that if the statement is true for this case, it must also be true for the next case, \(n=k+1\).

When both steps are successfully completed, the principle of mathematical induction allows us to leap from one true statement to a whole chain of true statements, thus confirming that our formula, \(d(n) = \frac{n(n-3)}{2}\), is valid for all positive integers in the domain of our problem.

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Most popular questions from this chapter

Let \(f: \mathbb{Z} \rightarrow \mathbb{Z}\) be defined by \(f(m)=3-m\). (a) Evaluate \(f(-7), f(-3), f(3),\) and \(f(7)\) (b) Determine the set of all of the preimages of 5 and the set of all of the preimages of 4 (c) Determine the range of the function \(f\). (d) This function can be considered a real function since \(\mathbb{Z} \subseteq \mathbb{R}\). Sketch a graph of this function. Note: The graph will be an infinite set of points that lie on a line. However, it will not be a line since its domain is not \(\mathbb{R}\) but is \(\mathbb{Z}\)

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In calculus, we learned that if \(f\) is real function that is continuous on the closed interval \([a, b],\) then the definite integral \(\int_{a}^{b} f(x) d x\) is a real number. In fact, one form of the Fundamental Theorem of Calculus states that $$ \int_{a}^{b} f(x) d x=F(b)-F(a) $$ where \(F\) is any antiderivative of \(f,\) that is, where \(F^{\prime}=f\) (a) Let \([a, b]\) be a closed interval of real numbers and let \(C[a, b]\) be the set of all real functions that are continuous on \([a, b]\). That is, $$ C[a, b]=\\{f:[a, b] \rightarrow \mathbb{R} \mid f \text { is continuous on }[a, b]\\} $$ i. Explain how the definite integral \(\int_{a}^{b} f(x) d x\) can be used to define a function \(I\) from \(C[a, b]\) to \(\mathbb{R}\). ii. Let \([a, b]=[0,2] .\) Calculate \(I(f),\) where \(f(x)=x^{2}+1\). iii. Let \([a, b]=[0,2] .\) Calculate \(I(g),\) where \(g(x)=\sin (\pi x)\). In calculus, we also learned how to determine the indefinite integral \(\int f(x) d x\) of a continuous function \(f\). (b) Let \(f(x)=x^{2}+1\) and \(g(x)=\cos (2 x)\). Determine \(\int f(x) d x\) and \(\int g(x) d x\) (c) Let \(f\) be a continuous function on the closed interval [0,1] and let \(T\) be the set of all real functions. Can the process of determining the indefinite integral of a continuous function be used to define a function from \(C[0,1]\) to \(T ?\) Explain. (d) Another form of the Fundamental Theorem of Calculus states that if \(f\) is continuous on the interval \([a, b]\) and if $$ g(x)=\int_{a}^{x} f(t) d t $$ for each \(x\) in \([a, b],\) then \(g^{\prime}(x)=f(x)\). That is, \(g\) is an antiderivative of \(f\). Explain how this theorem can be used to define a function from \(C[a, b]\) to \(T,\) where the output of the function is an antiderivative of the input. (Recall that \(T\) is the set of all real functions.)
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