Explain the following: Exam p le: For ever y n ∈ N 0we define the number a n∈ N 0b y means of

0 if n = 0 a n := { 2a n−1 + 1 if n > 0

e. g .

a 0 = 0 a 1 = 1 a 2 = 3 a 3 = 7 a 4 = 15 . . .

▷ Con j ecture: a n = 2 n − 1 for all n ∈ N 0 , i.e. { n ∈ N 0 | a n = 2 n − 1 } = N 0 .

▷ We have shown that X := { n ∈ N 0 | a n = 2 n − 1 } satisfies:

▷ 0 ∈ X (induction basis/base, base/initial case) ✓ ▷ For every n ∈ X ∩ N 0 also n + 1 ∈ X holds. (induction/inductive step) ✓

▷ B y virtue of the axiom of induction: N 0 ⊆ X, and thus X = N 0 .

▷ Hence, a n = 2 n − 1 for all n ∈ N 0 .

▷ Proofs b y induction ever y where, e. g . for correctness of recursive p ro g rams or p ro g rams with while-loo p s.

Let's break down the given explanation step by step and clarify the ideas presented in this recursive definition and the inductive proof.

### Recursive Definition of the Sequence \(\{a_n\}\):
We are given a recursively defined sequence \(\{a_n\}\) for every \(n \in \mathbb{N}_0\) (the set of non-negative integers), defined by the following rule:
\[
a_n = \begin{cases}
0 & \text{if } n = 0 \\
2a_{n-1} + 1 & \text{if } n > 0
\end{cases}
\]
In other words:
- The initial term \(a_0 = 0\),
- Each subsequent term \(a_n\) (for \(n > 0\)) is calculated as \(2a_{n-1} + 1\).

### Example Calculation:
Let’s compute the first few terms of the sequence to understand the pattern:

- \(a_0 = 0\) (given)
- \(a_1 = 2a_0 + 1 = 2(0) + 1 = 1\)
- \(a_2 = 2a_1 + 1 = 2(1) + 1 = 3\)
- \(a_3 = 2a_2 + 1 = 2(3) + 1 = 7\)
- \(a_4 = 2a_3 + 1 = 2(7) + 1 = 15\)

We observe the values: \(a_0 = 0, a_1 = 1, a_2 = 3, a_3 = 7, a_4 = 15, \dots\). This suggests a pattern.

### Conjecture:
The conjecture suggests that the general formula for \(a_n\) is:
\[
a_n = 2^n - 1 \quad \text{for all } n \in \mathbb{N}_0
\]
This means that for every non-negative integer \(n\), the sequence \(a_n\) follows the formula \(2^n - 1\).

### Inductive Proof:
We aim to prove this conjecture using **mathematical induction**.

#### Step 1: Base Case (Induction Basis)
We check the base case \(n = 0\):
\[
a_0 = 0 \quad \text{and} \quad 2^0 - 1 = 0
\]
Thus, the conjecture holds for \(n = 0\).

#### Step 2: Inductive Step
We assume that the conjecture holds for some \(n = k\), i.e., we assume:
\[
a_k = 2^k - 1 \quad \text{(inductive hypothesis)}
\]
Now, we need to prove that it also holds for \(n = k + 1\), i.e., we want to show:
\[
a_{k+1} = 2^{k+1} - 1
\]

Using the recursive definition for \(a_{k+1}\):
\[
a_{k+1} = 2a_k + 1
\]
By the inductive hypothesis, \(a_k = 2^k - 1\), so:
\[
a_{k+1} = 2(2^k - 1) + 1 = 2^{k+1} - 2 + 1 = 2^{k+1} - 1
\]
Thus, the conjecture holds for \(n = k + 1\).

#### Step 3: Conclusion
By the **principle of mathematical induction**, since the base case is true and the inductive step holds, we conclude that the conjecture is true for all \(n \in \mathbb{N}_0\):
\[
a_n = 2^n - 1 \quad \text{for all } n \in \mathbb{N}_0
\]

### Generalization and Application:
The proof by induction is a standard method to prove properties of recursively defined sequences, and it's commonly used in computer science to verify the correctness of algorithms, particularly those involving recursion or loops (e.g., programs with while-loops).

### Additional Questions:
1. What are the advantages of proving statements using mathematical induction?
2. Can you apply this method of induction to prove properties of other recursively defined sequences?
3. How does the recursive definition of a sequence relate to while-loops in programming?
4. How would the result change if we modified the recursion, for example, defining \(a_n = 3a_{n-1} + 2\)?
5. What is the role of the base case in induction, and what happens if it's missing?

### Tip:
In recursive problems, always identify the base case and ensure your inductive step covers the transition from \(n\) to \(n + 1\), ensuring completeness.

Explain the following: Example: For every n ∈ N0, we define the number an ∈ N0 by means of 0 if n = 0, an := { 2an−1 + 1 if n > 0. E.g. a0 = 0, a1 = 1, a2 = 3, a3 = 7, a4 = 15 ... Conjecture: an = 2n − 1 for all n ∈ N0, i.e. { n ∈ N0 | an = 2n − 1 } = N0. We have shown that X := { n ∈ N0 | an = 2n − 1 } satisfies: 0 ∈ X (induction basis/base, base/initial case) ✓ For every n ∈ X ∩ N0 also n + 1 ∈ X holds. (induction/inductive step) ✓ By virtue of the axiom of induction: N0 ⊆ X, and thus X = N0. Hence, an = 2n − 1 for all n ∈ N0. Proofs by induction everywhere, e.g. for correctness of recursive programs or programs with while-loops.

Explore the inductive proof of a recursively defined sequence an = 2^n - 1. Learn how recursion and the principle of mathematical induction work to confirm the correctness of recursive programs. This problem is ideal for students in advanced high school or university-level courses.

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