^^ Sorry. I made a big mistake. We cannot substitute \( \\[25pt]
{b_i=f^i} \) of course.
Maybe \( \\[15pt]
{b_i=f^{k.i}} \) would work as an approximation, because we know that \( \\[15pt]
{b_i} \) tends very rapidly to a line on logarithmic scale. Anyways, it would be of little use.
We know that the \( \\[15pt]
{a_i} \) are the derivatives of \( \\[15pt]
{^xa|_0} \) , so a Fourier or Laplace transform would turn the derivatives into products. But that would mess with the rest of the equation.
{b_i=f^i} \) of course.
Maybe \( \\[15pt]
{b_i=f^{k.i}} \) would work as an approximation, because we know that \( \\[15pt]
{b_i} \) tends very rapidly to a line on logarithmic scale. Anyways, it would be of little use.
We know that the \( \\[15pt]
{a_i} \) are the derivatives of \( \\[15pt]
{^xa|_0} \) , so a Fourier or Laplace transform would turn the derivatives into products. But that would mess with the rest of the equation.
I have the result, but I do not yet know how to get it.
