RE: value of e

Question

sorry i may have no clarified what i was asking. i have to use an infinte series to determine the value of e. thats basically the basis of the program..but i did try that new equation you gave but it was giving me an error since y wasn't defined before it used itself in the equation. (i.e. e^(x)=1+(x/1!)+(x^(2)/2!)+.... ) using 20 terms. then the program calculates the error by subtracting that number from the defined value of e. any input would be much oblidged~ thanks again.

bn9 · Answer

Jiggles!! dear god you are a savior. thank you very much for all of your help.

jiggleslupiniv · Answer

This is a sample output 1 2 1.71828182845905 2 2.5 1.21828182845905 3 2.66666666666667 1.05161516179238 4 2.70833333333333 1.00994849512571 5 2.71666666666667 1.00161516179238 6 2.71805555555556 1.00022627290349 7 2.71825396825397 1.00002786020508 8 2.71827876984127 1.00000305861778 9 2.71828152557319 1.00000030288585 10 2.71828180114638 1.00000002731266 11 2.71828182619849 1.00000000226055 12 2.71828182828617 1.00000000017288 13 2.71828182844676 1.00000000001229 14 2.71828182845823 1.00000000000082 15 2.71828182845899 1.00000000000005 16 2.71828182845904 1 17 2.71828182845905 1 18 2.71828182845905 1 19 2.71828182845905 1 20 2.71828182845905 1

jiggleslupiniv · Answer

Try this Code If you not satisfied reply again Ok!. using System; namespace Infinite_Series { class Infinite_Series { static void Main(string[] args) { double y = 1; double x = Math.E; double [] e = new double[21]; double add = 1; e[0] = e[1] = 1; e[20] = 0; //Set this Value for cycle not to exceed and cause error. for (int n=2; n < 20 ;n++) { y = y * n; e[n]= 1/y; } for (int cycle = 1;cycle <= 20; cycle++) { add = add + e[cycle]; x = x - e[cycle]; Console.WriteLine ( {0} {1} {2}cycleaddx); } Console.ReadLine (); } } }