Problem 21

Question

Using induction, prove that each is a solution to the corresponding recurrence relation, where $c$ is a constant and $f(n)$ a function of $n$. $$a_{n}=a_{0}+\sum_{i=1}^{n} f(i), a_{n}=a_{n-1}+f(n)$$

Step-by-Step Solution

Verified

Answer

We used induction to prove that $a_n = a_0 + \sum_{i=1}^{n} f(i)$ is a solution to the given recurrence relation $a_n = a_{n-1} + f(n)$ for all integers $n \ge 0$. The base case ($n=0$) holds true since $a_0 = a_0$. By assuming that the function holds true for some integer $k \ge 0$, we showed that it also holds true for $k + 1$ by plugging in the assumed true formula for $a_k$ and extending the sum. Thus, the function $a_n$ is a solution to the recurrence relation for all integers $n \ge 0$.

1Step 1: Base case

We start by verifying the base case when $n = 0$. For $n = 0$, we have $a_0 = a_{0-1} + f(0)$ Since $a_{n} = a_{0} + \sum_{i=1}^{n} f(i)$, we know that for $n = 0$, $a_0 = a_0 + \sum_{i=1}^{0} f(i)$. But the sum for an empty set is 0, so our equation becomes $a_0 = a_0$. The base case holds true.

2Step 2: Inductive hypothesis

Assume that the given function $a_n$ holds true for some integer $k \ge 0$, i.e. $a_k = a_0 + \sum_{i=1}^{k} f(i)$ Now, we need to show that the function holds true for the next integer $k + 1$.

3Step 3: Inductive step

To prove the inductive step, we need to show that the recurrence relation is true for $n = k + 1$. Using the recurrence relation formula for $n = k + 1$, we have $a_{k+1} = a_{k} + f(k+1)$ Now, from our inductive hypothesis, we plug in the assumed true formula for $a_k$: $a_{k+1} = (a_0 +\sum_{i=1}^{k} f(i)) + f(k+1)$ Next, we need to rewrite the above equation to match the form of the function $a_n$ for n = k + 1. This can be done by extending the sum as follows: $a_{k+1} = a_0 + \sum_{i=1}^{k+1} f(i)$

4Step 4: Conclusion

By the principle of mathematical induction, we have shown that if the given function $a_n$ holds true for $n = k$, then it also holds true for $n = k + 1$. Hence, the function $a_n = a_0 + \sum_{i=1}^{n} f(i)$ is a solution to the given recurrence relation $a_n = a_{n-1} + f(n)$ for all integers $n \ge 0$.

Key Concepts

Recurrence RelationsBase CaseInductive HypothesisInductive Step

Recurrence Relations

Recurrence relations are equations or inequalities that describe sequences by specifying how each element in the sequence relates to its predecessors. Instead of giving an explicit formula for each term, they define a chain of recurrence, highlighting the step-by-step evolution from one term to the next. In the context of this exercise, the recurrence relation describes how each term, $ a_n $, is generated by adding a function of $ n $, $ f(n) $, to the preceding term, $ a_{n-1} $. This is represented as:

$ a_{n} = a_{n-1} + f(n) $

This relation acts as the backbone of the sequence, guiding how expressions for future terms can be constructed by building on the previous ones. Understanding this progression is crucial for employing mathematical induction, as it frames the pattern we aim to prove across all numbers $ n $.

Base Case

The base case is the foundational step in mathematical induction. It sets the stage for proving that a statement holds true for an initial value, usually when $ n = 0 $ or $ n = 1 $. Think of it as laying the cornerstone for an inductive proof.

In the exercise, the base case examines the sequence when $ n = 0 $. Here, we confirm that:

$ a_0 = a_0 + \sum_{i=1}^{0} f(i) $

Since the sum over an empty set is zero, this simplifies to $ a_0 = a_0 $, ensuring the base condition is trivially satisfied. Establishing a true base case effectively anchors the inductive argument and permits extending the proof to higher values of $ n $.

Inductive Hypothesis

The inductive hypothesis is a crucial component of the induction process. It involves assuming the given statement or formula is true for an arbitrary, but specific integer $ k $. This assumption serves as a strategic stepping stone in proving the formula valid for the next integer.

In solving the exercise, the inductive hypothesis posits that:

$ a_k = a_0 + \sum_{i=1}^{k} f(i) $

Assuming the truth of this statement for $ n = k $ essentially sets the stage for demonstrating it for $ n = k + 1 $. It acts as a bridge, supporting the transition from one stage of the induction to the next.

Inductive Step

The inductive step is the final piece of the induction puzzle. It demonstrates that if a statement holds for an integer $ k $, it must also hold for $ k + 1 $. This logical leap is where the proof gains its momentum, cementing the argument across infinite instances.

For the given exercise, the inductive step requires proving $ a_{k+1} = a_0 + \sum_{i=1}^{k+1} f(i) $ assuming the correctness of:

$ a_k = a_0 + \sum_{i=1}^{k} f(i) $

Start with the recurrence relation for $ a_{k+1} $:

$ a_{k+1} = a_k + f(k+1) $

Substitute the known expression for $ a_k $ from the hypothesis:

$ a_{k+1} = (a_0 + \sum_{i=1}^{k} f(i)) + f(k+1) $

This simplifies to match the desired form, $ a_0 + \sum_{i=1}^{k+1} f(i) $, completing the pattern and ensuring continuity through the infinite sequence, proving the statement for all integers $ n \ge 0 $.

Problem 21

Other exercises in this chapter

Problem 21

Solve each LHRRWCC. $$\begin{array}{l}{a_{n}=-a_{n-1}+16 a_{n-2}+4 a_{n-3}-48 a_{n-4}, a_{0}=0, a_{1}=16, a_{2}=-2}{a_{3}=142}\end{array}$$

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Problem 21

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Problem 21

Solve each LHRRWCC. $$\begin{aligned} &a_{n}=-a_{n-1}+16 a_{n-2}+4 a_{n-3}-48 a_{n-4}, a_{0}=0, a_{1}=16, a_{2}=-2\\\ &a_{3}=142 \end{aligned}$$

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Problem 21

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