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Chapter 2: Problem 77

Suppose that \(f\) is a function on the nonnegative integers such that \(f(0)=0\)and \(f(n)=n+f(n-1) .\) Prove that \(f(n)=n(n+1) / 2 .\) Notice that this gives athird proof that \(1+2+\cdots+n=n(n+1) / 2,\) because this sum satisfies the twoconditions for \(f\). (The sum has no terms and is thus 0 when \(n=0 .\) )

### Short Answer

Expert verified

\(f(n) = \frac{n(n+1)}{2}\) holds for all nonnegative integers by induction.

## Step by step solution

01

## Identify the Base Case

Given that the function is defined on nonnegative integers, check the value of the function at the base case.\(f(0) = 0\)

02

## Formulate the Recurrence Relation

Given the recurrence relation for \(f\): \(f(n) = n + f(n-1)\)

03

## Hypothesis for Induction

Assume for some \(k \geq 0\), \(f(k) = \frac{k(k+1)}{2}\).

04

## Inductive Step

Prove that assuming the hypothesis \(f(k) = \frac{k(k+1)}{2}\) holds for \(k\), it also holds for \(k+1\). That is, show \(f(k+1) = \frac{(k+1)(k+2)}{2}\). Start with the given recurrence relation: \[ f(k+1) = (k+1) + f(k) \] Substituting the inductive hypothesis \[ f(k+1) = (k+1) + \frac{k(k+1)}{2} \] Simplify: \[ f(k+1) = \frac{2(k+1) + k(k+1)}{2} = \frac{2(k+1)+k(k+1)}{2} = \frac{(k+1)(k+2)}{2} \]

05

## Conclusion

Since the base case and inductive step have been established and proven, by the principle of mathematical induction, \(f(n) = \frac{n(n+1)}{2}\) for all nonnegative integers \(n\).

## Key Concepts

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

###### recurrence relation

A recurrence relation is a way of defining a sequence of values by specifying how each term relates to the previous ones. In this problem, the recurrence relation is given by: \( f(n) = n + f(n-1) \). This means that the value of \( f \) at any nonnegative integer \( n \) is equal to \( n \) plus the value of \( f \) at \( n-1 \). Recurrence relations are crucial in understanding sequences because they show how each term is constructed from its predecessors.

Recurrence relations are useful because:

- They provide a way to recursively define complex sequences.
- They often make it simpler to write algorithms and compute terms.
- They can reveal patterns in the sequence that might not be immediately obvious.

###### base case

In mathematical induction, the base case is the initial step that shows the proposition holds for the smallest value, often \( n = 0 \) or \( n = 1 \). It serves as the foundation from which the inductive step builds. In our exercise, the base case is:

\( f(0) = 0 \)

This simply means that when \( n \) is 0, the function \( f \) returns 0. Proving the base case is critical because:

- It anchors the induction process.
- It ensures that the sequence or formula is correctly initialized.
- It provides a specific example where the proposition is true.

###### inductive step

The inductive step is where we prove that if a proposition holds for a specific case \( k \), then it also holds for the next case \( k + 1 \). This step ensures the proposition holds for all values starting from the base case onward. In this problem, we assume the hypothesis:

\( f(k) = \frac{k(k+1)}{2} \)

and use it to prove:

\( f(k+1) = \frac{(k+1)(k+2)}{2} \)

The calculation for the inductive step:

Start from the recurrence relation:

\( f(k+1) = (k+1) + f(k) \)

Substitute the hypothesis:

\( f(k+1) = (k+1) + \frac{k(k+1)}{2} \)

Combine terms:

\( f(k+1) = \frac{2(k+1) + k(k+1)}{2} \)

Simplify to get:

\( f(k+1) = \frac{(k+1)(k+2)}{2} \)

This step is essential in mathematical induction because:

- It shows that the proposition works for all subsequent values.
- It connects each case to the next, forming a chain of truth.
- It leverages the hypothesis to prove a broader truth.

###### nonnegative integers

Nonnegative integers are all the whole numbers starting from 0 and going upwards: \( 0, 1, 2, 3, \text{etc.} \). They do not include negative numbers. In our problem, the function \( f \) is defined on the set of nonnegative integers. This is important because:

- It sets the domain for which the function and the proposition are valid.
- It simplifies the problem by avoiding negative values that might complicate the recurrence relation.
- It ensures all operations within the recurrence relation are straightforward sums and counts.

Understanding nonnegative integers helps to contextualize where and how the recurrence relation and induction apply.

Mathematical induction, combined with recurrence relations for functions defined on nonnegative integers, offers a robust toolset for proving propositions in math.

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