python,math,tkinter,binomial-coefficients

The specific problem you mention is due to the fact that P0 is returning a string like "7.69%". You then try to use that value in the calculation that is failing. That string is what the error message is referring to when it says "can't multiply sequence...": the string is...

Your desired output isn't quite well defined, because the monomials you listed are not in the lexicographic order (which you used in the first line of your code). Anyway, using a double loop you can arrange coefficients in any specific way you want. Here is a natural way to do...

c++,long-integer,binomial-coefficients

Factorial grows really fast and even unsigned 64-bit integers overflow n! for n>20. The overflow free way to implement the binomial coefficient is to use this recursive definition: binom(n, k) = binom(n-1, k-1) + binom(n-1, k) This ensures that you get an overflow only when binom(n,k) is too large to...

python,python-2.7,math,binomial-coefficients

Guessing from what you are getting, you must be using py2. Here you will have to coerce to float that is you have to return float(Sum)/float(Pro). This is a small test to make you understand why >>> 1/30 0 >>> float(1/30) 0.0 >>> float(1)/float(30) 0.03333333333333333 ...

In the discrete case (i.e discrete n), choose(n,k) computes the number of distinct k-element subsets from a set of n elements, so if k > n, then you are counting subsets of a set which have more elements that the corresponding set. Since there are no such subsets, then the...

arrays,algorithm,pascals-triangle,binomial-coefficients

Here is a code which uses only one one dimensional array: int[] coefficients = new int[k + 1]; coefficients[0] = 1; for (int i = 1; i <= n; i++) { for (int j = k; j >= 1; j--) { coefficients[j] += coefficients[j - 1]; } } Why is...

c++,algorithm,math,binomial-coefficients

Just remove the outer loop: int main() { cout << "Enter a row number for Pascal's Triangle: "; cin >> n; int x = 1; for (int k = 0; k <= n; k++) { cout << x << '\t'; x = x * (n - k) / (k +...

haskell,recursion,tail-recursion,binomial-coefficients

Yes. There is a standard trick of using an accumulator to achieve tail recursion. In your case, you'll need two of them (or accumulate one rational number): binom :: Int -> Int -> Int binom = loop 1 1 where loop rn rd _ 0 = rn `div` rd loop...

Maybe this can get you started. I think it is what you expressed. However I think it would make more sense with an initialization like rnorm(nx,1,0.01) - i.e. something smoother. It might also make sense to have xsub interpolate between the generated points (i.e. make it a continuous function). It...