This code inserts the number 4 into a vector's even indexes.vector<double> vecCoeffs;    // Put the coefficient 4 into each even index and 2 for each odd index    for(int i = 0; i <= 10; i++){        vecCoeffs[i];        if(i % 2 == 0 ){            vecCoeffs.push_back(4);        }else{            vecCoeffs.push_back(2);        }        // Starting and ending with coefficient 1        vecCoeffs[0] = 1;        vecCoeffs[10] = 1;
Populating a vector with alternating numbers
c++;vectors
If you know how large the vector needs to be, the specify the size in the constructor so that it doesn't need to guess.vecCoeffs[i]; is a useless statement.It would be clearer to pull the assignment of starting and ending coefficients out of the loop, and to avoid assigning vecCoeffs[10] twice.  You could assign the starting and ending coefficients in the same statement.The if-else would be better as a ternary conditional.vector<double> vecCoeffs(11);// Starting and ending coefficientsvecCoeffs[0] = vecCoeffs[vecCoeffs.size() - 1] = 1;// Put the coefficient 2 into each odd index and 4 into each even indexfor (int i = 1; i < vecCoeffs.size() - 1; i++) {    vecCoeffs[i] = (i % 2 ? 2 : 4);}But the modulo operator is relatively slow.  You would be better off with two loopsvector<double> vecCoeffs(11);// Starting and ending coefficientsvecCoeffs[0] = vecCoeffs[vecCoeffs.size() - 1] = 1;for (int i = 1; i < vecCoeffs.size() - 1; i += 2) {    vecCoeffs[i] = 2;                // Odd coefficients}for (int i = 2; i < vecCoeffs.size() - 1; i += 2) {    vecCoeffs[i] = 4;                // Even coefficients} or maybe just one, but getting the termination correct is trickier.  This version might eliminate an instruction or two from the loop, but I don't recommend it.vector<double> vecCoeffs(11);for (int i = 0; i < vecCoeffs.size() - 1; i += 2) {    vecCoeffs[i] = 4;                // Even coefficients    vecCoeffs[i + 1] = 2;            // Odd coefficients}// Starting and ending coefficientsvecCoeffs[0] = vecCoeffs[vecCoeffs.size() - 1] = 1;