CODEWITHSUNDEEP
javamathcalculus
michaeladair
michaeladair

Reputation: 173

Summation equation in Java?

Summation Equation

I would like to know how I would go about writing this summation equation in java. But, the trick is, I need the summation to be equal to an amount.

x= Total Loss Streak amount
sb= Starting Bet
m= multiplier

The whole equation will equal to the current amount of currency in one's account. The amount of times the summation can complete itself while adding up needs to be less than or equal to the amount of currency in ones account.

Fyi, this is for a dicebot that work's on peerbet.org and I want to be able to show the user how many times he can loose in a row without wasting all his money.

If this question is bad, please do not answer it and let me delete it. Also, it thought the middle part was code, so I had to put it as such or it wouldn't let me post.

Upvotes: 3

Views: 10013

Answers (3)

Thomas
Thomas

Reputation: 17422

I'm not entirely clear I'm reading your formula correctly: are you summing up integers from 2 to x on the left-hand side of the equals sign, and you want that sum to be equal to the term on the right-hand side?

In that case, we could do the following transformation:

enter image description here

(Note that the first step might not be what you had in mind.)

We can now easily solve this using the quadratic formula to get:

enter image description here

Assuming that we're calculating in the reals, note that the root is only defined for non-negative arguments. The result of taking that root yields a non-negative number and substracting that non-negative number from -1 would give something <= -1, i.e., a negative number. Dividing it by 2 won't make it positive, either, but we've assumed from the get-go that our x must be >= 2, or else the very first sum wouldn't make any sense.

Therefore we can disregard the - case of the +/- in the formula altogether. Hence:

enter image description here

This should be straight-forward to translate into Java code, but note that the result is likely not to be an integer, so you will have to round if you're looking for an upper bound.

Upvotes: 2

Bathsheba
Bathsheba

Reputation: 234695

Renaming sb to just b. This is just a sum of a geometric progression

sum = b * (m<sup>2</sup> - m<sup>x + 1</sup>) / (1 - m)

In Java, you can write:

return b * (m * m - Math.pow(m, x + 1)) / (1 - m);

This will be considerably faster than using a loop, although you must check that m is not 1.

If you want to solve for x given a sum S then a rearrangement of the formula gives the following Java code:

double x = Math.log(m * m - S * (1 - m) / b) / log(m) - 1;

and truncate this result to get the integral value of x where the next integer bankrupts the player.

Upvotes: 5

Adam Yost
Adam Yost

Reputation: 3625

EDIT: apparently we are solving for x. still easily doable with a loop.

int sum = 0;
int x =2;
while(sum<=amount){
   sum+=sb*(Math.pow(m,x));
}    
return x;

A summation is really just an adding for loop right?

int sum = 0;
for(int i=2; i<x; i++){
    sum+=sb*(Math.pow(m,i));
}

return sum;

Upvotes: 4

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