is there any way to get the sum of the product of euler's number and an integer, where the integer can range upto 10^6000?

Question

So I tried using BigDecimal which is giving me the value of Euler's number up to 50 decimal points. But since the constraint value is this high the number can range from 1 to 10^6000, and hence I am not getting any precision(making it optimal is another issue). Here is the block of code:

for(int i=1; i<n+1; i++){
 // i is the integer value and bd is the value of the euler's number, so below i am multiplying the value of i with the euler's number
 BigDecimal x = BigDecimal.valueOf(i).multiply(bd);
 //Here i am taking floor of the above calculated value
 x.setScale(0, RoundingMode.HALF_UP);
 //and here sum is floor[1*euler's no] + floor[2*euler's number] + ... + floor[n*euler's number] 
 sum += x.intValue();
}

What is `sum`? You might need to make that a BigDecimal as well. — lucasvw, Jun 08 '17 at 12:20
The code snippet suggests you want to sum up 10^6000 numbers: good luck with that. — Henry, Jun 08 '17 at 12:20
I first wanted to suggest using the formula `(n*(n+1) / 2)` and multiplying it with bd *once*, and then truncating, but that did not take into account that you need to add the the floor values, so 1*e would become 2, 2*e would become 5, giving a subtotal of 7, etc.while 3*e would become 8, giving an error of 1 already. You will have to loop, I'm afraid. — Rudy Velthuis, Jun 26 '17 at 16:08

Stephen C · Accepted Answer · 2017-06-26T22:37:14.633

I think you are asking how to compute Euler's number (i.e. e) to greater precision.

The answer would be to implement an algorithm that converges rapidly. With a little bit of research, I found this page on "Calculating the Value of e". At the bottom, it describes Brothers Formula:

e = Sum(n=0, n=∞, (2n+2) / (2n+1)! )

which converges very rapidly. You could code that in Java using BigDecimal to calculate to any number of digits precision that you need.

is there any way to get the sum of the product of euler's number and an integer, where the integer can range upto 10^6000?

1 Answers1