Cheesing a Math Problem with Code

It's definitely been a while since I posted things here! I'll use this blog space to share anything interesting that I feel like sharing here; don't expect a regular schedule. 😜 Also, I'm going to start using Prism for code syntax highlighting and KaTeX for math typesetting.

One more thing: It's Wikipedia's 20th birthday today (2020-01-15)!! Wikipedia is truly one of the greatest things we have; it's a vast, free, and open repository of human knowledge. Go donate to them if you can!

Back to the topic at hand: I am in a mathematics-related club and was presented with the following weekly problem.

I realized that I could solve it by repeatedly using the recursive function forwards or backwards until I got to my desired value. So, I created a .JS file, pulled up VS Code, started coding, and a few minutes later had my final code. (I've also included the comments I added after running the code.):

        
          /* PROBLEM:
           * Consider the function g such that g(x) + g(x − 1) = x² and g(17) = 153.
           * What is g(100)?
           * Bonus points if you can name the set formed by {g(1), g(2), g(3), g(4), …}
           * 
           * 
           * WORK:
           * Since g(x) = x² − g(x − 1),
           * g(x + 1) = (x + 1)² − g(x).
           * g(x − 1) = x² − g(x)
           */
          
          let resArr = [];
          /*Calculating values of g(x) for 0 ≤ x ≤ 17*/
          let currValue = 153; /*Value of g(17)*/
          for (let i = 16; i >= 0; i--) {
            currValue = (i + 1) ** 2 - currValue; /*x² − g(x); i is (x − 1), currValue is value of g(x) for previous x*/
            resArr.unshift(`g(${i}) = ${currValue}`); /*Insert values into array in reverse order since it's calculated from 16 to 0*/
          }
          resArr.push(`g(17) = 153 (given starting value)`);
          /*Calculating values of g(x) for 18 ≤ x ≤ 100*/
          currValue = 153;
          for (let i = 18; i <= 100; i++) {
            currValue = i ** 2 - currValue; /*(x + 1)² − g(x); i is (x + 1), currValue is value of g(x) for previous x*/
            resArr.push(`g(${i}) = ${currValue}`);
          }
          resArr.forEach(item => {
            console.log(item);
          });
          
          /* ANSWER (found from running program):
           * g(100) = 5050
           * {g(1), g(2), g(3), g(4), …} = {0, 1, 3, 6, …}
           * The above set is the set of triangular (or triangle) numbers.
           */
        
      

Running the code above gave me the following output:

        
          g(0) = 0
          g(1) = 1
          g(2) = 3
          g(3) = 6
          g(4) = 10
          g(5) = 15
          g(6) = 21
          g(7) = 28
          g(8) = 36
          g(9) = 45
          g(10) = 55
          g(11) = 66
          g(12) = 78
          g(13) = 91
          g(14) = 105
          g(15) = 120
          g(16) = 136
          g(17) = 153 (given starting value)
          g(18) = 171
          g(19) = 190
          g(20) = 210
          g(21) = 231
          g(22) = 253
          g(23) = 276
          g(24) = 300
          g(25) = 325
          g(26) = 351
          g(27) = 378
          g(28) = 406
          g(29) = 435
          g(30) = 465
          g(31) = 496
          g(32) = 528
          g(33) = 561
          g(34) = 595
          g(35) = 630
          g(36) = 666
          g(37) = 703
          g(38) = 741
          g(39) = 780
          g(40) = 820
          g(41) = 861
          g(42) = 903
          g(43) = 946
          g(44) = 990
          g(45) = 1035
          g(46) = 1081
          g(47) = 1128
          g(48) = 1176
          g(49) = 1225
          g(50) = 1275
          g(51) = 1326
          g(52) = 1378
          g(53) = 1431
          g(54) = 1485
          g(55) = 1540
          g(56) = 1596
          g(57) = 1653
          g(58) = 1711
          g(59) = 1770
          g(60) = 1830
          g(61) = 1891
          g(62) = 1953
          g(63) = 2016
          g(64) = 2080
          g(65) = 2145
          g(66) = 2211
          g(67) = 2278
          g(68) = 2346
          g(69) = 2415
          g(70) = 2485
          g(71) = 2556
          g(72) = 2628
          g(73) = 2701
          g(74) = 2775
          g(75) = 2850
          g(76) = 2926
          g(77) = 3003
          g(78) = 3081
          g(79) = 3160
          g(80) = 3240
          g(81) = 3321
          g(82) = 3403
          g(83) = 3486
          g(84) = 3570
          g(85) = 3655
          g(86) = 3741
          g(87) = 3828
          g(88) = 3916
          g(89) = 4005
          g(90) = 4095
          g(91) = 4186
          g(92) = 4278
          g(93) = 4371
          g(94) = 4465
          g(95) = 4560
          g(96) = 4656
          g(97) = 4753
          g(98) = 4851
          g(99) = 4950
          g(100) = 5050
        
      

Perfect! Now I knew that \(\displaystyle g(100) = 5050\) and the set \(\displaystyle \{ g(1), g(2), g(3), g(4), \ldots \} = \{ 1, 3, 6, 10, \ldots \} \) was the set of triangular numbers.

I still haven't endeavored to understand the actual method to solve it, but I'll get to that eventually. 😀