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.
Renaming sb to just b. This is just a sum of a geometric progression
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.
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;
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:
(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:
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:
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.
Suppose we implement the following two methods to calculate the nth multiple of a real number x.
public static double multiply( double x, int n )
{
return x * n;
}
public static double iterativeAdd( double x, int n )
{
double a = 0.0;
for( int b = 0; b < n; b++ )
{
a += x;
}
return a;
}
Assume that n is a legal int and that both x and the exact mathematical product of n and x are no less in absolute value than Double.MIN_VALUE (unless both are 0.0) and no greater in absolute value than Double.MAX_VALUE. Here's what I'm wondering: In general, which is closer to the exact value of the product of x and n: the double returned by multiply( x, n ) or the double returned by iterativeAdd( x, n ) and how do you know?
According to my knowledge, the first method will produce more accurate result because in the second method after each addition probability that some of the digits will be truncated and rounded are more then a single multiplier operation as the result will be calculated once and then the digits will be truncated.
Generally, for every floating point operation you do, your epsilon increases. This happens because floating point numbers have a fixed size in memory, limiting their precision. Each operation is rounded to the nearest value that a float can represent. This rounding accumulates after a while.
Both numbers will get you very close to the answer, but if you run both methods on a large and varied set of numbers, you will see that on average iterativeAdd() has a greater distance from the actual value.
Additionally, multiply() will be significantly faster on any machine, so there's no benefit to ever using iterativeAdd().
Both will return approximately the same value, however there is greater chance that the iterativeAdd() will return more inapproximate value, but the difference will be negligible.
Any single float operation results in some precision loss however small.
In multiply() you make use of the float operation only once but in iterativeAdd() you use it n times.
In general we should avoid using any function like iterativeAdd() as it is will take up a lot of processor time with n floating point operations.
Here's my implementation of Fermat's little theorem. Does anyone know why it's not working?
Here are the rules I'm following:
Let n be the number to test for primality.
Pick any integer a between 2 and n-1.
compute a^n mod n.
check whether a^n = a mod n.
myCode:
int low = 2;
int high = n -1;
Random rand = new Random();
//Pick any integer a between 2 and n-1.
Double a = (double) (rand.nextInt(high-low) + low);
//compute:a^n = a mod n
Double val = Math.pow(a,n) % n;
//check whether a^n = a mod n
if(a.equals(val)){
return "True";
}else{
return "False";
}
This is a list of primes less than 100000. Whenever I input in any of these numbers, instead of getting 'true', I get 'false'.
The First 100,008 Primes
This is the reason why I believe the code isn't working.
In java, a double only has a limited precision of about 15 to 17 digits. This means that while you can compute the value of Math.pow(a,n), for very large numbers, you have no guarantee you'll get an exact result once the value has more than 15 digits.
With large values of a or n, your computation will exceed that limit. For example
Math.pow(3, 67) will have a value of 9.270946314789783e31 which means that any digit after the last 3 is lost. For this reason, after applying the modulo operation, you have no guarantee to get the right result (example).
This means that your code does not actually test what you think it does. This is inherent to the way floating point numbers work and you must change the way you hold your values to solve this problem. You could use long but then you would have problems with overflows (a long cannot hold a value greater than 2^64 - 1 so again, in the case of 3^67 you'd have another problem.
One solution is to use a class designed to hold arbitrary large numbers such as BigInteger which is part of the Java SE API.
As the others have noted, taking the power will quickly overflow. For example, if you are picking a number n to test for primality as small as say, 30, and the random number a is 20, 20^30 = about 10^39 which is something >> 2^90. (I took the ln of 10^39).
You want to use BigInteger, which even has the exact method you want:
public BigInteger modPow(BigInteger exponent, BigInteger m)
"Returns a BigInteger whose value is (this^exponent mod m)"
Also, I don't think that testing a single random number between 2 and n-1 will "prove" anything. You have to loop through all the integers between 2 and n-1.
#evthim Even if you have used the modPow function of the BigInteger class, you cannot get all the prime numbers in the range you selected correctly. To clarify the issue further, you will get all the prime numbers in the range, but some numbers you have are not prime. If you rearrange this code using the BigInteger class. When you try all 64-bit numbers, some non-prime numbers will also write. These numbers are as follows;
341, 561, 645, 1105, 1387, 1729, 1905, 2047, 2465, 2701, 2821, 3277, 4033, 4369, 4371, 4681, 5461, 6601, 7957, 8321, 8481, 8911, 10261, 10585, 11305, 12801, 13741, 13747, 13981, 14491, 15709, 15841, 16705, 18705, 18721, 19951, 23001, 23377, 25761, 29341, ...
https://oeis.org/a001567
161038, 215326, 2568226, 3020626, 7866046, 9115426, 49699666, 143742226, 161292286, 196116194, 209665666, 213388066, 293974066, 336408382, 376366, 666, 566, 566, 666 2001038066, 2138882626, 2952654706, 3220041826, ...
https://oeis.org/a006935
As a solution, make sure that the number you tested is not in this list by getting a list of these numbers from the link below.
http://www.cecm.sfu.ca/Pseudoprimes/index-2-to-64.html
The solution for C # is as follows.
public static bool IsPrime(ulong number)
{
return number == 2
? true
: (BigInterger.ModPow(2, number, number) == 2
? (number & 1 != 0 && BinarySearchInA001567(number) == false)
: false)
}
public static bool BinarySearchInA001567(ulong number)
{
// Is number in list?
// todo: Binary Search in A001567 (https://oeis.org/A001567) below 2 ^ 64
// Only 2.35 Gigabytes as a text file http://www.cecm.sfu.ca/Pseudoprimes/index-2-to-64.html
}
I have a requirement to calculate the average of a very large set of doubles (10^9 values). The sum of the values exceeds the upper bound of a double, so does anyone know any neat little tricks for calculating an average that doesn't require also calculating the sum?
I am using Java 1.5.
You can calculate the mean iteratively. This algorithm is simple, fast, you have to process each value just once, and the variables never get larger than the largest value in the set, so you won't get an overflow.
double mean(double[] ary) {
double avg = 0;
int t = 1;
for (double x : ary) {
avg += (x - avg) / t;
++t;
}
return avg;
}
Inside the loop avg always is the average value of all values processed so far. In other words, if all the values are finite you should not get an overflow.
The very first issue I'd like to ask you is this:
Do you know the number of values beforehand?
If not, then you have little choice but to sum, and count, and divide, to do the average. If Double isn't high enough precision to handle this, then tough luck, you can't use Double, you need to find a data type that can handle it.
If, on the other hand, you do know the number of values beforehand, you can look at what you're really doing and change how you do it, but keep the overall result.
The average of N values, stored in some collection A, is this:
A[0] A[1] A[2] A[3] A[N-1] A[N]
---- + ---- + ---- + ---- + .... + ------ + ----
N N N N N N
To calculate subsets of this result, you can split up the calculation into equally sized sets, so you can do this, for 3-valued sets (assuming the number of values is divisable by 3, otherwise you need a different divisor)
/ A[0] A[1] A[2] \ / A[3] A[4] A[5] \ // A[N-1] A[N] \
| ---- + ---- + ---- | | ---- + ---- + ---- | \\ + ------ + ---- |
\ 3 3 3 / \ 3 3 3 / // 3 3 /
--------------------- + -------------------- + \\ --------------
N N N
--- --- ---
3 3 3
Note that you need equally sized sets, otherwise numbers in the last set, which will not have enough values compared to all the sets before it, will have a higher impact on the final result.
Consider the numbers 1-7 in sequence, if you pick a set-size of 3, you'll get this result:
/ 1 2 3 \ / 4 5 6 \ / 7 \
| - + - + - | + | - + - + - | + | - |
\ 3 3 3 / \ 3 3 3 / \ 3 /
----------- ----------- ---
y y y
which gives:
2 5 7/3
- + - + ---
y y y
If y is 3 for all the sets, you get this:
2 5 7/3
- + - + ---
3 3 3
which gives:
2*3 5*3 7
--- + --- + ---
9 9 9
which is:
6 15 7
- + -- + -
9 9 9
which totals:
28
-- ~ 3,1111111111111111111111.........1111111.........
9
The average of 1-7, is 4. Obviously this won't work. Note that if you do the above exercise with the numbers 1, 2, 3, 4, 5, 6, 7, 0, 0 (note the two zeroes at the end there), then you'll get the above result.
In other words, if you can't split the number of values up into equally sized sets, the last set will be counted as though it has the same number of values as all the sets preceeding it, but it will be padded with zeroes for all the missing values.
So, you need equally sized sets. Tough luck if your original input set consists of a prime number of values.
What I'm worried about here though is loss of precision. I'm not entirely sure Double will give you good enough precision in such a case, if it initially cannot hold the entire sum of the values.
Apart from using the better approaches already suggested, you can use BigDecimal to make your calculations. (Bear in mind it is immutable)
IMHO, the most robust way of solving your problem is
sort your set
split in groups of elements whose sum wouldn't overflow - since they are sorted, this is fast and easy
do the sum in each group - and divide by the group size
do the sum of the group's sum's (possibly calling this same algorithm recursively) - be aware that if the groups will not be equally sized, you'll have to weight them by their size
One nice thing of this approach is that it scales nicely if you have a really large number of elements to sum - and a large number of processors/machines to use to do the math
Please clarify the potential ranges of the values.
Given that a double has a range ~= +/-10^308, and you're summing 10^9 values, the apparent range suggested in your question is values of the order of 10^299.
That seems somewhat, well, unlikely...
If your values really are that large, then with a normal double you've got only 17 significant decimal digits to play with, so you'll be throwing away about 280 digits worth of information before you can even think about averaging the values.
I would also note (since no-one else has) that for any set of numbers X:
mean(X) = sum(X[i] - c) + c
-------------
N
for any arbitrary constant c.
In this particular problem, setting c = min(X) might dramatically reduce the risk of overflow during the summation.
May I humbly suggest that the problem statement is incomplete...?
A double can be divided by a power of 2 without loss of precision. So if your only problem if the absolute size of the sum you could pre-scale your numbers before summing them. But with a dataset of this size, there is still the risk that you will hit a situation where you are adding small numbers to a large one, and the small numbers will end up being mostly (or completely) ignored.
for instance, when you add 2.2e-20 to 9.0e20 the result is 9.0e20 because once the scales are adjusted so that they numbers can be added together, the smaller number is 0. Doubles can only hold about 17 digits, and you would need more than 40 digits to add these two numbers together without loss.
So, depending on your data set and how many digits of precision you can afford to loose, you may need to do other things. Breaking the data into sets will help, but a better way to preserve precision might be to determine a rough average (you may already know this number). then subtract each value from the rough average before you sum it. That way you are summing the distances from the average, so your sum should never get very large.
Then you take the average delta, and add it to your rough sum to get the correct average. Keeping track of the min and max delta will also tell you how much precision you lost during the summing process. If you have lots of time and need a very accurate result, you can iterate.
You could take the average of averages of equal-sized subsets of numbers that don't exceed the limit.
divide all values by the set size and then sum it up
Option 1 is to use an arbitrary-precision library so you don't have an upper-bound.
Other options (which lose precision) are to sum in groups rather than all at once, or to divide before summing.
So I don't repeat myself so much, let me state that I am assuming that the list of numbers is normally distributed, and that you can sum many numbers before you overflow. The technique still works for non-normal distros, but somethings will not meet the expectations I describe below.
--
Sum up a sub-series, keeping track of how many numbers you eat, until you approach the overflow, then take the average. This will give you an average a0, and count n0. Repeat until you exhaust the list. Now you should have many ai, ni.
Each ai and ni should be relatively close, with the possible exception of the last bite of the list. You can mitigate that by under-biting near the end of the list.
You can combine any subset of these ai, ni by picking any ni in the subset (call it np) and dividing all the ni in the subset by that value. The max size of the subsets to combine is the roughly constant value of the n's.
The ni/np should be close to one. Now sum ni/np * ai and multiple by np/(sum ni), keeping track of sum ni. This gives you a new ni, ai combination, if you need to repeat the procedure.
If you will need to repeat (i.e., the number of ai, ni pairs is much larger than the typical ni), try to keep relative n sizes constant by combining all the averages at one n level first, then combining at the next level, and so on.
First of all, make yourself familiar with the internal representation of double values. Wikipedia should be a good starting point.
Then, consider that doubles are expressed as "value plus exponent" where exponent is a power of two. The limit of the largest double value is an upper limit of the exponent, and not a limit of the value! So you may divide all large input numbers by a large enough power of two. This should be safe for all large enough numbers. You can re-multiply the result with the factor to check whether you lost precision with the multiplication.
Here we go with an algorithm
public static double sum(double[] numbers) {
double eachSum, tempSum;
double factor = Math.pow(2.0,30); // about as large as 10^9
for (double each: numbers) {
double temp = each / factor;
if (t * factor != each) {
eachSum += each;
else {
tempSum += temp;
}
}
return (tempSum / numbers.length) * factor + (eachSum / numbers.length);
}
and dont be worried by the additional division and multiplication. The FPU will optimize the hell out of them since they are done with a power of two (for comparison imagine adding and removing digits at the end of a decimal numbers).
PS: in addition, you may want to use Kahan summation to improve the precision. Kahan summation avoids loss of precision when very large and very small numbers are summed up.
I posted an answer to a question spawned from this one, realizing afterwards that my answer is better suited to this question than to that one. I've reproduced it below. I notice though, that my answer is similar to a combination of Bozho's and Anon.'s.
As the other question was tagged language-agnostic, I chose C# for the code sample I've included. Its relative ease of use and easy-to-follow syntax, along with its inclusion of a couple of features facilitating this routine (a DivRem function in the BCL, and support for iterator functions), as well as my own familiarity with it, made it a good choice for this problem. Since the OP here is interested in a Java solution, but I'm not Java-fluent enough to write it effectively, it might be nice if someone could add a translation of this code to Java.
Some of the mathematical solutions here are very good. Here's a simple technical solution.
Use a larger data type. This breaks down into two possibilities:
Use a high-precision floating point library. One who encounters a need to average a billion numbers probably has the resources to purchase, or the brain power to write, a 128-bit (or longer) floating point library.
I understand the drawbacks here. It would certainly be slower than using intrinsic types. You still might over/underflow if the number of values grows too high. Yada yada.
If your values are integers or can be easily scaled to integers, keep your sum in a list of integers. When you overflow, simply add another integer. This is essentially a simplified implementation of the first option. A simple (untested) example in C# follows
class BigMeanSet{
List<uint> list = new List<uint>();
public double GetAverage(IEnumerable<uint> values){
list.Clear();
list.Add(0);
uint count = 0;
foreach(uint value in values){
Add(0, value);
count++;
}
return DivideBy(count);
}
void Add(int listIndex, uint value){
if((list[listIndex] += value) < value){ // then overflow has ocurred
if(list.Count == listIndex + 1)
list.Add(0);
Add(listIndex + 1, 1);
}
}
double DivideBy(uint count){
const double shift = 4.0 * 1024 * 1024 * 1024;
double rtn = 0;
long remainder = 0;
for(int i = list.Count - 1; i >= 0; i--){
rtn *= shift;
remainder <<= 32;
rtn += Math.DivRem(remainder + list[i], count, out remainder);
}
rtn += remainder / (double)count;
return rtn;
}
}
Like I said, this is untested—I don't have a billion values I really want to average—so I've probably made a mistake or two, especially in the DivideBy function, but it should demonstrate the general idea.
This should provide as much accuracy as a double can represent and should work for any number of 32-bit elements, up to 232 - 1. If more elements are needed, then the count variable will need be expanded and the DivideBy function will increase in complexity, but I'll leave that as an exercise for the reader.
In terms of efficiency, it should be as fast or faster than any other technique here, as it only requires iterating through the list once, only performs one division operation (well, one set of them), and does most of its work with integers. I didn't optimize it, though, and I'm pretty certain it could be made slightly faster still if necessary. Ditching the recursive function call and list indexing would be a good start. Again, an exercise for the reader. The code is intended to be easy to understand.
If anybody more motivated than I am at the moment feels like verifying the correctness of the code, and fixing whatever problems there might be, please be my guest.
I've now tested this code, and made a couple of small corrections (a missing pair of parentheses in the List<uint> constructor call, and an incorrect divisor in the final division of the DivideBy function).
I tested it by first running it through 1000 sets of random length (ranging between 1 and 1000) filled with random integers (ranging between 0 and 232 - 1). These were sets for which I could easily and quickly verify accuracy by also running a canonical mean on them.
I then tested with 100* large series, with random length between 105 and 109. The lower and upper bounds of these series were also chosen at random, constrained so that the series would fit within the range of a 32-bit integer. For any series, the results are easily verifiable as (lowerbound + upperbound) / 2.
*Okay, that's a little white lie. I aborted the large-series test after about 20 or 30 successful runs. A series of length 109 takes just under a minute and a half to run on my machine, so half an hour or so of testing this routine was enough for my tastes.
For those interested, my test code is below:
static IEnumerable<uint> GetSeries(uint lowerbound, uint upperbound){
for(uint i = lowerbound; i <= upperbound; i++)
yield return i;
}
static void Test(){
Console.BufferHeight = 1200;
Random rnd = new Random();
for(int i = 0; i < 1000; i++){
uint[] numbers = new uint[rnd.Next(1, 1000)];
for(int j = 0; j < numbers.Length; j++)
numbers[j] = (uint)rnd.Next();
double sum = 0;
foreach(uint n in numbers)
sum += n;
double avg = sum / numbers.Length;
double ans = new BigMeanSet().GetAverage(numbers);
Console.WriteLine("{0}: {1} - {2} = {3}", numbers.Length, avg, ans, avg - ans);
if(avg != ans)
Debugger.Break();
}
for(int i = 0; i < 100; i++){
uint length = (uint)rnd.Next(100000, 1000000001);
uint lowerbound = (uint)rnd.Next(int.MaxValue - (int)length);
uint upperbound = lowerbound + length;
double avg = ((double)lowerbound + upperbound) / 2;
double ans = new BigMeanSet().GetAverage(GetSeries(lowerbound, upperbound));
Console.WriteLine("{0}: {1} - {2} = {3}", length, avg, ans, avg - ans);
if(avg != ans)
Debugger.Break();
}
}
A random sampling of a small set of the full dataset will often result in a 'good enough' solution. You obviously have to make this determination yourself based on system requirements. Sample size can be remarkably small and still obtain reasonably good answers. This can be adaptively computed by calculating the average of an increasing number of randomly chosen samples - the average will converge within some interval.
Sampling not only addresses the double overflow concern, but is much, much faster. Not applicable for all problems, but certainly useful for many problems.
Consider this:
avg(n1) : n1 = a1
avg(n1, n2) : ((1/2)*n1)+((1/2)*n2) = ((1/2)*a1)+((1/2)*n2) = a2
avg(n1, n2, n3) : ((1/3)*n1)+((1/3)*n2)+((1/3)*n3) = ((2/3)*a2)+((1/3)*n3) = a3
So for any set of doubles of arbitrary size, you could do this (this is in C#, but I'm pretty sure it could be easily translated to Java):
static double GetAverage(IEnumerable<double> values) {
int i = 0;
double avg = 0.0;
foreach (double value in values) {
avg = (((double)i / (double)(i + 1)) * avg) + ((1.0 / (double)(i + 1)) * value);
i++;
}
return avg;
}
Actually, this simplifies nicely into (already provided by martinus):
static double GetAverage(IEnumerable<double> values) {
int i = 1;
double avg = 0.0;
foreach (double value in values) {
avg += (value - avg) / (i++);
}
return avg;
}
I wrote a quick test to try this function out against the more conventional method of summing up the values and dividing by the count (GetAverage_old). For my input I wrote this quick function to return as many random positive doubles as desired:
static IEnumerable<double> GetRandomDoubles(long numValues, double maxValue, int seed) {
Random r = new Random(seed);
for (long i = 0L; i < numValues; i++)
yield return r.NextDouble() * maxValue;
yield break;
}
And here are the results of a few test trials:
long N = 100L;
double max = double.MaxValue * 0.01;
IEnumerable<double> doubles = GetRandomDoubles(N, max, 0);
double oldWay = GetAverage_old(doubles); // 1.00535024998431E+306
double newWay = GetAverage(doubles); // 1.00535024998431E+306
doubles = GetRandomDoubles(N, max, 1);
oldWay = GetAverage_old(doubles); // 8.75142021696299E+305
newWay = GetAverage(doubles); // 8.75142021696299E+305
doubles = GetRandomDoubles(N, max, 2);
oldWay = GetAverage_old(doubles); // 8.70772312848651E+305
newWay = GetAverage(doubles); // 8.70772312848651E+305
OK, but what about for 10^9 values?
long N = 1000000000;
double max = 100.0; // we start small, to verify accuracy
IEnumerable<double> doubles = GetRandomDoubles(N, max, 0);
double oldWay = GetAverage_old(doubles); // 49.9994879713857
double newWay = GetAverage(doubles); // 49.9994879713868 -- pretty close
max = double.MaxValue * 0.001; // now let's try something enormous
doubles = GetRandomDoubles(N, max, 0);
oldWay = GetAverage_old(doubles); // Infinity
newWay = GetAverage(doubles); // 8.98837362725198E+305 -- no overflow
Naturally, how acceptable this solution is will depend on your accuracy requirements. But it's worth considering.
Check out the section for cummulative moving average
In order to keep logic simple, and keep performance not the best but acceptable, i recommend you to use BigDecimal together with the primitive type.
The concept is very simple, you use primitive type to sum values together, whenever the value will underflow or overflow, you move the calculate value to the BigDecimal, then reset it for the next sum calculation. One more thing you should aware is when you construct BigDecimal, you ought to always use String instead of double.
BigDecimal average(double[] values){
BigDecimal totalSum = BigDecimal.ZERO;
double tempSum = 0.00;
for (double value : values){
if (isOutOfRange(tempSum, value)) {
totalSum = sum(totalSum, tempSum);
tempSum = 0.00;
}
tempSum += value;
}
totalSum = sum(totalSum, tempSum);
BigDecimal count = new BigDecimal(values.length);
return totalSum.divide(count);
}
BigDecimal sum(BigDecimal val1, double val2){
BigDecimal val = new BigDecimal(String.valueOf(val2));
return val1.add(val);
}
boolean isOutOfRange(double sum, double value){
// because sum + value > max will be error if both sum and value are positive
// so I adapt the equation to be value > max - sum
if(sum >= 0.00 && value > Double.MAX - sum){
return true;
}
// because sum + value < min will be error if both sum and value are negative
// so I adapt the equation to be value < min - sum
if(sum < 0.00 && value < Double.MIN - sum){
return true;
}
return false;
}
From this concept, every time the result is underflow or overflow, we will keep that value into the bigger variable, this solution might a bit slowdown the performance due to the BigDecimal calculation, but it guarantee the runtime stability.
Why so many complicated long answers. Here is the simplest way to find the running average till now without any need to know how many elements or size etc..
long int i = 0;
double average = 0;
while(there are still elements)
{
average = average * (i / i+1) + X[i] / (i+1);
i++;
}
return average;