My first recursion method: the number e

So I was reading about binary numbers and I got to thinking I need to review on log and natural log. I was skimming through my notes from last year and I found out this:

Code:

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the number e = 1 + 1/1 + 1/(1*2) + 1/(1*2*3) + 1/(1*2*3*4) + ... 1/(n!)

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And just last night, I was reading about recursion methods. Instantaneously
I had an epiphany. The epiphany was simple: I'm gonna make a recursion method that returns the value of e. Here it is. :D

The MyMathClass class:

Code:

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public class MyMathClass {
       
        public static double e(int num) {
                if (num == 0)
                        return 1;
                else
                        return  ((double) 1 / factorial(num)) + e(num -1);
        }
       
        public static int factorial(int num) {
                if (num == 1)
                        return 1;
                else
                        return num * factorial(num -1);
        }
}

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The RecursionMethodsDriver class:

Code:

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public class RecursionMethods {
        public static void main(String[] args) {
                System.out.println(e(33));
        }
}

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I realize that there is a method that returns the value of e in the math class.
Unfortunately, if num > 33 in e(num), it returns "infinity." Is there anyway way I can get a more precise number?

p.s: so jaunty right now because I get recursion methods. ;) Although, there's a lot more to recursion, like mazes and stuff which I'm not getting. :(

Anyhow, feedbacks and comments appreciated.

I did some more research on binary numbers and I learned how to convert decimal numbers to binary numbers. Kudos to me! anyhow, I wanted to make a static method that would do this for me. First, I started off with a for loop:

Code:

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        public static int decimalToBinary(int num) {
                /*
                assume num = 52.
                2^x = 52
                multiply with log base of 10 on both sides
                log(2^x) = log(52)
                use the exponent log rule
                xlog(2) = log(52)
                x = log(52) / log(2)
                x = 5.7004 = 5
                */
                int pow = (int) (Math.log10(num) / Math.log10(2));
                String binaryNumber = "";
                int y;
                for (; pow >= 0; pow--) {
                        y = (int) Math.pow(2, pow);
                        if ((num-y) >= 0) {
                                binaryNumber += "1";
                                num -= y;
                        }
                        else
                                binaryNumber += "0";
                }
                return Integer.parseInt(binaryNumber);
        }

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Then I decided to use a recursion method. This required me two methods:

Code:

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        public static int decimalToBinary2(int num) {
                int pow = (int) (Math.log10(num) / Math.log10(2));
                return Integer.parseInt(decimaltoBinary(pow, num));
        }
        public static String decimaltoBinary(int pow, int num) {
               
                int y;
                if (pow == -1)
                        return "";
                else {
                        y = (int) Math.pow(2, pow);
                        if ((num-y) >= 0)
                                return "1" + decimaltoBinary(pow-1,num-y);
                        else
                                return "0" + decimaltoBinary(pow-1,num);
                }
                       
        }

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Question: is it possible to make the one recursion method (decimalToBinary(int pow, int num)) and one normal method (decimalToBinary2(int pow)) into a recursion method?

And finally, binaryToDecimal method (also recursive):

Code:

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        public static int binaryToDecimal(String num) {
                if (!num.matches("[01]+")) return -1;
                if (num.length() == 1)
                        return Integer.parseInt(num) * (int) Math.pow(2, 0);
                else
                        return (Integer.parseInt(num.substring(0,1)) * (int) Math.pow(2, num.length() -1)) + binaryToDecimal(num.substring(1));
        }

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I realize my decimalToBinary methods are not perfect (they do not handle any number that's less than or equal to 0).

Code:

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        public static void main(String[] args) {
                System.out.println("Decimal 55 to binary: " + MyMathClass.decimalToBinary(55));
                System.out.println("Binary 110111 to decimal: " +  MyMathClass.binaryToDecimal("110111"));

        }

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Output:

Code:

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Decimal 55 to binary: 110111
Binary 110111 to decimal: 55

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