Lost in Proof for Recursive function

Question

Discrete math is fun but I still have a lot of difficulty with the algebra involved. I am attempting to prove, through induction, a recursive function. I am just starting my course in algorithm design and the assignment was to rewrite an iterative function into a recursive function and then prove it. I was able to implement the recursive function and I was able to test it using a brute force technique, but I am at a loss as to how to set up my proof. I don't think I am starting it correctly. I included my start with the proof. Thanks for any pointers you can give me.

edit #3 final proof completed thanks to help

f (k + 1) – f(k) = 
(k + 1) ^2 – ½ (k + 1) (k + 1 – 1) – k^2 – ½ (k (k -1)) =
k^2 + 2k + 1 – ½ (k^2 – k) – k^2 + ½ (k^2 - k) =  
2k + 1 - k =
k + 1

edit #2 This is my proof so far and I am sure I am way off.

Base Case, n = 1
When n is 1, 1 is returned        Line 1
 1^2-(1*(1-1))/2 = 1
Inductive Case, n > 1
Assume for k = n-1, show for n = k
triangular_recursive(k) =
triangular_recursive (k -1) + k =        Line 1
(k – 1) ^2 – ½(k-1) (k-1-1) + k =      Inductive Assumption
k^2 -2k +1 – ½ (k^2 -3k +2) + k =
k^2 – k + 1 – ½ (k^2 -3k + 2)
This doesn’t see, correct at all.

Below is my code:

    /*
 JLawley
 proof_of_correctness1.cpp
 This provides a brute force proof of my algorithm

 Originally, everything was integer type.
 I changed to double when I added pow.
 */

#include "stdafx.h"
#include <iostream>

// this is the original function
// we were to rewrite this as a recursive function
// so the proof would be simpler
double triangular(double n) {
    auto result = 0;
    for (auto i = 1; i <= n; i++) result += i;
    return result;
}

/*
 * This is my recursive implementation
 * It includes base case and post case
 */

// n > 0
double triangular_recursive(double n) {
    return  (n == 1) ? n : triangular_recursive(n - 1) + n;
}
// returns n^2 - (n(n-1)) / 2

// utility method to test my theory by brute force
double test_my_theory(double n)
{
    return pow(n, 2) - (n * (n - 1))/2;
}

int main(void)
{
    // at values much beyond 4000, this loop fails
   // edit added - the failure is due to values too large
   // the program crashes when this occurs
   //  this is a drawback of using recursive functions
    for (auto i = 1; i <= 4000; i++) 
        if (test_my_theory(i) != triangular_recursive(i) || test_my_theory(i) != triangular(i)) 
            std::cout << "\n!= at i = " << i;
    // I am not getting any "i ="'s so I assume a good brute force test
    return 0;
}

/*
 * My proof so far:
Base Case, n = 1
When n is 1, 1 is returned        Line 1
 1^2-(1*(1-1))/2 = 1
Inductive Case, n > 1
Assume for k = n-1, show for n = k
triangular_recursive(k) =
triangular_recursive (k -1) + k =        Line 1
(k – 1) ^2 – ½(k-1)(k-1-1) + k =      Inductive Assumption
 */

Answer 1

A recursive function typically has a form something like:

recursive(param) {
    if (base_case)
        return base_value;

    new_param = move_param_toward_base(param);
    return combine(present_value, recursive(new_param);
}

An inductive proof basically has two steps:

1. Prove some (usually trivial) base case.
2. Prove some way of extending it, so that if the base case is true, your extended version remains true for some larger set of input.
3. Prove that the extension can be applied more or less arbitrarily, so the result remains true for all inputs.

With a recursive function:

1. You show that you've detected and treated the base case correctly.
2. You show that your extension to some other value is correct.
3. You show that you're modifying the parameter in a way that will lead to the base case in a finite number of steps.

But, there are also some differences, including the one you seem to be running into here. In particular, in mathematics, a non-modular number can grow without limit--but on a computer, all numbers are modular; none of them can grow without limit.