# Change Function To Recurrence Relation

Usually when analyzing programs, we start with a recursive definition of the program and try to figure out a closed form or function for the recursive definition and then solve it’s time complexity. Here we are doing the opposite, we are starting with the closed form or function and changing it to a recursive definition.

Below we have our problem or function that we want to change into a recursive definition.

f(n) = 2^n + 1

The base case is the terminating case in recursion, that doesn’t use recursion to give an answer. The base case here is when n=0, to get f(0) = 2.

## Base Case: Try to solve for n=0

f(0) = 2⁰ + 1

= 1 + 1

= 2

The recursive case is the case when the function defines itself. Here we try to create a recursive case for our function f(n) = 2^n + 1. If f(n) = 2^(n)+1 then f(n+1) = 2^(n+1) + 1.

## Recursive Case: Try to solve f(n)

f(n+1) = 2^(n+1) + 1

= **2 * 2^n** + 1

= **(2 * 2^n )**+ 1

= **(2^n + 2^n )** + 1

= **2 ^n + 2^n** + 1

= 2 ^n + 2^n + 1 +** 0**

= 2 ^n + 2^n + 1 +** 1–1**

= 2 ^n **+ 1**+ 2^n + 1 **–1**

= **(2 ^n + 1)**+ **(2^n + 1)** –1

= **f(n)** + **f(n)** — 1

f(n+1) = f(n) + f(n) — 1

f(n+1 -1 ) = f(n)

f(n + 1–1) = f(n-1) + f(n-1) — 1

f(n) = f(n-1) + f(n-1) — 1

f(n) = 2f(n-1) — 1

**Conclusion:**

The recursive definition for f(n) = 2^n + 1 is below.

f(0) = 2

f(n) = 2f(n-1) — 1

This function time complexity is of course **O(2^n)**

If you would like to learn more about Algorithm Analysis , you can take my online course here. I also have a course on Udemy.com called Recurrence Relation Made Easy where I help students to understand how to solve recurrence relations and asymptotic terms such as Big-O, Big Omega, and Theta. You can check out my YouTube channel of videos where I solve recurrence relations and perform algorithm analysis on code that anyone can check out for free !

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