tommy - sheldon method 1.1

Basicly the idea is to turn the disagreement fo the a_n by the method into agreement about the a_n.

f(x) is the half-iterate from the 2sinh method.

... + a_(n-1) x^(n-1) + a_n x^n + ... = f(x)

a_n x^n ( ... + a_(n+1)/a_n x + 1 + a_(n-1)/a_n 1/x + a_(n-2)/a_n /x^2 + ...) = f(x)

ln(a_n) + n ln(x) + ln( ... + 1 + a_(n-1)/a_n 1/x + ...) = ln(f(x))

ln(a_n) + n x + ln( ...+ a_(n+1)/a_n exp(x) + 1 + a_(n-1) / a_n exp(-x) + a_(n-2)/a_n exp(-2x) + ... ) = f(x)

a_n = MIN (f(x) - n x - ln( ... + 1 + a_(n-1) / a_n exp(-x) + a_(n-2)/a_n exp(-2x) + ... ))

And then use a system of n+1 MIN equations for the first n+1 values a_0 , a_1 , ... , a_n.

example n = 4

a_0 = MIN (f(x) - ln( a_4/a_0 exp(4x) + ... + 1 ))

a_1 = MIN (f(x) - x - ln( a_4/a_1 exp(3x) + ... + 1 + a_0 / a_1 exp(-x) ))

a_2 = MIN (f(x) - 2 x - ln( a_4/a_2 exp(2x) + ... + 1 + a_1 / a_2 exp(-x) + a_0/a_2 exp(-2x) + ... ))

a_3 = MIN (f(x) - 3 x - ln( a_4/a_3 exp(x) + 1 + a_2 / a_3 exp(-x) + a_1/a_3 exp(-2x) + a_0/a_3 exp(-3x) ))

a_4 = MIN (f(x) - 4 x - ln( 1 + a_3 / a_4 exp(-x) + a_2/a_4 exp(-2x) + ... a_0/a_4 exp(-4x)))

---

The urgent questions are uniqueness and existance for such systems of equations with a fixed n.

And then the convergence as n grows for each a_i.

But intuitively it seems correct.

How to solve these systems of MIN is another matter.

Probably by taking the derivative and/or Lagrange multipliers.

Then iterating ; going from a guess to a better one.

a_i(1) -> a_i(2).

regards

tommy1729

Basicly the idea is to turn the disagreement fo the a_n by the method into agreement about the a_n.

f(x) is the half-iterate from the 2sinh method.

... + a_(n-1) x^(n-1) + a_n x^n + ... = f(x)

a_n x^n ( ... + a_(n+1)/a_n x + 1 + a_(n-1)/a_n 1/x + a_(n-2)/a_n /x^2 + ...) = f(x)

ln(a_n) + n ln(x) + ln( ... + 1 + a_(n-1)/a_n 1/x + ...) = ln(f(x))

ln(a_n) + n x + ln( ...+ a_(n+1)/a_n exp(x) + 1 + a_(n-1) / a_n exp(-x) + a_(n-2)/a_n exp(-2x) + ... ) = f(x)

a_n = MIN (f(x) - n x - ln( ... + 1 + a_(n-1) / a_n exp(-x) + a_(n-2)/a_n exp(-2x) + ... ))

And then use a system of n+1 MIN equations for the first n+1 values a_0 , a_1 , ... , a_n.

example n = 4

a_0 = MIN (f(x) - ln( a_4/a_0 exp(4x) + ... + 1 ))

a_1 = MIN (f(x) - x - ln( a_4/a_1 exp(3x) + ... + 1 + a_0 / a_1 exp(-x) ))

a_2 = MIN (f(x) - 2 x - ln( a_4/a_2 exp(2x) + ... + 1 + a_1 / a_2 exp(-x) + a_0/a_2 exp(-2x) + ... ))

a_3 = MIN (f(x) - 3 x - ln( a_4/a_3 exp(x) + 1 + a_2 / a_3 exp(-x) + a_1/a_3 exp(-2x) + a_0/a_3 exp(-3x) ))

a_4 = MIN (f(x) - 4 x - ln( 1 + a_3 / a_4 exp(-x) + a_2/a_4 exp(-2x) + ... a_0/a_4 exp(-4x)))

---

The urgent questions are uniqueness and existance for such systems of equations with a fixed n.

And then the convergence as n grows for each a_i.

But intuitively it seems correct.

How to solve these systems of MIN is another matter.

Probably by taking the derivative and/or Lagrange multipliers.

Then iterating ; going from a guess to a better one.

a_i(1) -> a_i(2).

regards

tommy1729