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Taylor polynomial. System of equations for the coefficients.
#11
(05/05/2015, 07:40 AM)Gottfried Wrote: P*A = A*Bb

I think that we are speaking of different things.

Obviously, there should be a way to demonstrate the equivalence of both, because they are trying to solve the same problem; looking for the same solution.

But as I understand, the Carleman matrix A only contains powers of a_i coefficients, yet if you look at the red side, it cannot be written as a matrix product A*Bb, because it needs to have products of a_i coefficients (like ). Maybe it is a power of A.Bb, or something like A^Bb?

The Pascal matrix on the blue side is the exponential of a much simpler matrix



Maybe the equation can be greatly simplified by taking a logarithm of both sides.
I have the result, but I do not yet know how to get it.
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Messages In This Thread
RE: Taylor polinomial. System of equations for the coefficients. - by marraco - 05/06/2015, 02:42 PM

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