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Taylor polynomial. System of equations for the coefficients.
#13
I misinterpreted what the Carleman matrix was. I tough that it contained the powers of the derivatives of a function (valued at zero), but it contains the derivatives of the powers of a function, so it actually haves the products of the aᵢ coefficients (of bᵢ in your notation).

________________

I tried to use this method to find the coefficients for exponentiation: bˣ=Σbᵢ.xⁿ

The condition is
b.(x+1)=b.Σbᵢ.xⁿ

which translates into
P.[bᵢ]=b.[bᵢ]

or
[P-b.I].[bᵢ]=0

The solution should be bᵢ=ln(b)ⁱ / i!

I found bᵢ=c. (ln(b)ⁱ/i!), where c is an arbitrary constant, because, obviously c.b⁽ˣ⁺¹⁾=b.(c.bˣ)

I was bugged for the fact that any equation for solving tetration I tried seems to have at least one degree of liberty. I think now that it should be explained by one (at least) arbitrary constant in the solution.

This looks analogous to constants found in the solution of differential equations, so I wonder if the evolvent of the curves generated by the constant is also a solution, and what is his meaning.
I have the result, but I do not yet know how to get it.
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Messages In This Thread
RE: Taylor polynomial. System of equations for the coefficients. - by marraco - 05/07/2015, 09:45 AM

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