Java Utililty Methods Number Sum

List of utility methods to do Number Sum

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Description

The list of methods to do Number Sum are organized into topic(s).

Method

doublesumAsDouble(Iterable values)
Calculate the total sum of the numbers in the specified collection. 
double total = 0;
for (Number value : values) {
    total += value.doubleValue();
return total;
floatSumDiffErr(float Term1Err, float Term2Err)
Calculate the error in a sum or difference, based on the error in the two terms 
return (float) Math.sqrt(Term1Err * Term1Err + Term2Err * Term2Err);
intsumDigits(String number)
Sums the digits of a natural number represented as a String. 
return number.chars().map(aChar -> Character.getNumericValue(aChar)).sum();
intsumDivisors(int n)
sum Divisors 
int sum = 0;
for (int i = 1; i < n; i++)
    if (n % i == 0)
        sum += i;
return sum;
doublesumError(long samples, long count, double m2, double mean)
Computes the standard deviation for the random variable S = sum(1 / p * X * Bern(p))

Derivation: 
 Var(S) = Var(sum(1 / p * X * Bern(p))) = sum(Var(1 / p * X * Bern(p)))                                                           [Bienayme formula] = n * Var(1 / p * X * Bern(p))                                                            [X * Bern(p) are iid] = n * 1 / p^2 * Var(X * Bern(p))                                                          [1 / p is constant] = n * 1 / p^2 * (Var(X) * Var(Bern(p)) + E(X)^2 * Var(Bern(p)) + Var(X) * E(Bern(p))^2    [Product of independent variables] = n * 1 / p^2 * (Var(X) * p(1 - p) + E(X)^2 * p(1 - p) + Var(X) * p^2)                    [Variance of a Bernoulli distribution] = n * 1 / p * (Var(X) + E(X)^2 * (1 - p)) = samples / p^2 * (Var(X) + E(X)^2 * (1 - p))                                             [samples = n * p, since it's only the observed rows]
Therefore Stddev(S) = 1 / p * sqrt(samples * (variance + mean^2 * (1 - p))) 
if (count == 0) {
    return Double.POSITIVE_INFINITY;
double p = samples / (double) count;
double variance = m2 / samples;
double error = 1 / p * Math.sqrt(samples * (variance + mean * mean * (1 - p)));
return conservativeError(error, p, samples);
longsumFirstIntegers(final long i)
Compute the sum of the i first integers. 
return ((i - 1L) * i) >> 1;
longsumFirstN(long n)
sum First N 
return (n * (n + 1)) / 2;
longsumFirstNDivisibleByM(long n, long m)
sum First N Divisible By M 
return sumFirstNDivisibleByM(n, m, true);
intsumFromNtoM(int n, int m)
Returns the sum from n to m (Ex: n = 1, m = 3 => 1 + 2 + 3 = 6) 
return Math.abs(m - n + 1) * (m + n) / 2;
booleansumHadOverflow(long a, long b, long sum)
sum Had Overflow 
return ((a ^ sum) & (b ^ sum)) < 0;