If I'm given the value of the following sum, Y, how do I explicitly solve for a?x + x*(x-a) + x*(x-a)*(x-2a) + x*(x-a)*(x-2a)*(x-3a) +x*(x-a)*(x-2a)*(x-3a)*(x-4a)......+x + x*(x-a)* ....*(x-na) = YX and a are known constants. a is small and n is possibly large. If I'm given Y is there a way I can get a closed-form solution for a?Thanks for any help.

Sorry, a is obviously not known. However I know x, n and y.

if you write f(a,0) = x, f(a,1) = x(x-a), f(a,2) = x(x-a)(x-2a) you can see that f(a,n+1) can be written f(a,n+1) = f(a,n) + (x-a(n+1))[f(a,n)-f(a,n-1)]Thus you can solve it algebraicly but i'd suggest using that recusion formula on a computer for a general N. shouldn't take more than 5 mins to code it in vba or whatever.

Public Function SolveForA(Y As Double, X As Double, N As Long) As Double Dim A As Double Dim tol As Double, shift As Double tol = 0.00001 shift = 0.01 Dim res As Double, resPrime As Double, deriv As Double A = X * N res = Y - FOfA(X, N, A) While Abs(res) > tol resPrime = Y - FOfA(X, N, A + shift) deriv = (resPrime - res) / shift If (Abs(deriv) < tol) Then GoTo out: End If A = A - res / deriv res = Y - FOfA(X, N, A) Wendout: SolveForA = AEnd FunctionPrivate Function FOfA(X As Double, N As Long, A As Double) As Double Dim temp As Double, sum As Double Dim i As Long temp = X sum = temp For i = 1 To N temp = temp * (X - i * A) sum = sum + temp Next i FOfA = sumEnd Functionpop these in a vba module and call SolveForA() from the worksheet.

well, maybe i wasn't too clear in my last post - writing f recursively allows us to compute f(a,k) relatively swiftly (order k). that allows for a numeric solver (like the newton-raphson below) to trawl through the derivatives without too much burden.i bet you can write a closed form for f(a,k) if you wish to sit through the tedious unpacking / repacking of products / sums, but with modern computing, a simple function like that is solved numerically in less time it takes to do the algebra.

Thanks very much for that Mutley, and also to you Outrun.

Explicit Solution.Construct the (n x n) diagonal matrix D with D(k,k) = k for k = 1 to n.Construct the (n x n) matrix A withA(k,k) = x for k = 1 to n,A(k,k+1) = -1 for k = 1 to n-1,A(n,1) = -Y/x,A(n,k) = A(n,k) + 1 for k = 1 to n,Let the matrix M = inv(D)*A. Then the solution to a are the eigenvalues of M, a = eig(M).There are exactly n solutions to a as the equation is an nth order polynomial in a.This solution is essentially re-writing the recurrence relations of f(a,n) in matrix form. The eigenvector appropriately normalised are the vectors [f(a,0);f(a,1);f(a,2),...,f(a,n-1)]. Edit: typo