Question

Solve \(T(n)=T(n-1)+1\) for \(n \geq 1\) with \(T(0)=1\).

Short answer

The solution is \( T(n) = n + 1 \).

Step-by-step solution

Understanding the Recurrence Relation

The given recurrence relation is \( T(n) = T(n-1) + 1 \) with the initial condition \( T(0) = 1 \). This relation suggests that to find \( T(n) \), we add 1 to \( T(n-1) \).

Finding the Pattern

Let's compute the first few values to identify any pattern:- \( T(0) = 1 \)- \( T(1) = T(0) + 1 = 1 + 1 = 2 \)- \( T(2) = T(1) + 1 = 2 + 1 = 3 \)- \( T(3) = T(2) + 1 = 3 + 1 = 4 \)We observe a pattern: \( T(n) = n + 1 \).

Formulate the Hypothesis

Based on the identified pattern from Step 2, we hypothesize that \( T(n) = n + 1 \) for \( n \geq 0 \).

Verify the Hypothesis by Mathematical Induction

Perform induction to verify that \( T(n) = n + 1 \) holds for all \( n \geq 0 \):**Base case:** \( n = 0 \)\( T(0) = 1 \). Clearly, \( 0 + 1 = 1 \).**Inductive step:** Assume \( T(k) = k + 1 \) is true for \( n = k \).Show it holds for \( n = k + 1 \):\( T(k+1) = T(k) + 1 = (k + 1) + 1 = k + 2 \).Hence, \( T(k+1) = (k + 1) + 1 \).Therefore, by induction, \( T(n) = n + 1 \) is verified for all \( n \geq 0 \).

Concluding the Solution

The solution to the recurrence relation is \( T(n) = n + 1 \). This formula computes the value of \( T(n) \) for any non-negative integer \( n \).

Key concepts

Mathematical Induction

Mathematical induction is a powerful technique used to prove statements for all natural numbers. It works like this: first, we show the base case is true. Then, we assume it's true for a particular case, say \( n = k \), and show that it must then be true for the next case, \( n = k + 1 \).

Let's see how this applies to our given problem. We start by proving the base case for \( n = 0 \):

• Given \( T(0) = 1 \).
• We expect \( T(0) = 0 + 1 = 1 \), which matches the initial condition.

Once the base case is verified, we move on to the inductive step. Assume that the formula \( T(k) = k + 1 \) holds true for some integer \( k \). Then, we need to prove it for \( n = k + 1 \):

• Calculate \( T(k+1) = T(k) + 1 \).
• Substitute the hypothesis \( T(k) = k + 1 \), getting \( T(k+1) = (k + 1) + 1 = k + 2 \).

Once we've completed these steps, we conclude that since both the base case and the inductive step hold, by mathematical induction, \( T(n) = n + 1 \) for all \( n \geq 0 \).

This method provides a rigorous justification for our pattern discovery in recurrence relations.

Initial Conditions

Initial conditions in a problem involving recurrence relations are critical starting points that define how the sequence begins. They provide the first few terms, which are indispensable for both understanding the sequence and formulating further values.

In our exercise, the initial condition is given as \( T(0) = 1 \). This serves as a foundation because:

• It gives us the value at \( n = 0 \), which is crucial for beginning any calculations.
• As subsequent values depend on previous ones, without this starting point, we can't compute the rest.

This initial condition of \( T(0) = 1 \) directly determines \( T(1), T(2), \) and so on, by following the recurrence relation \( T(n) = T(n-1) + 1 \).

Therefore, initial conditions are not merely starting points but essential components that shape the entire sequence.

Pattern Recognition

Recognizing patterns in sequences can drastically simplify solving recurrence relations. By observing computed values, we often identify a trend or regularity that helps us predict subsequent terms.

For our problem, once we calculated a few terms:

• \( T(0) = 1 \)
• \( T(1) = 2 \)
• \( T(2) = 3 \)
• \( T(3) = 4 \)

We noticed that each term seems to be \( n+1 \) where \( n \) is the index of the term. This is a clear pattern, and recognizing it allows us to formulate a hypothesis about the general form \( T(n) = n + 1 \).

Pattern recognition complements the initial conditions because it uses the starting data points to construct an overall understanding of the sequence. Identifying such patterns doesn't just streamline computation but also strengthens our ability to solve recurrence relations.