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so what i want is to find the coefficients which is $a,b,c,d$ in terms of $f(x)$ where $f(x)=(ax+b)/(cx+d)$ only in terms of $f(u)$,$f'(u)$ and $f''(u)$ where $u$ is some constant

i solved for a similar question but its instead $f(x)=1/(ax+b)+c$ which is

  • $a=-f'(0)(1/(f(0)-f(1))+1/(f'(0)))^2$
  • $b=1/(f(0)-f(1))+1/(f'(0))$
  • $c=f(0)-1/(1/(f(0)-f(1))+1/(f'(0)))$

and as you can tell from the solution above, i only used $f(0),f(1)$ and $f'(0)$, i never used the second derivative which makes me proud of the work, i tried doing $(ax+b)/(cx+d)$ but with no luck on a single coefficient, so i deciced to share it with you to try it for yourself. Maybe its only possible with thrid or higher derivatives but i hope its possible with only the second derivative.

Ashton Smith
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  • Note that $\frac{ax+b}{cx+d} = \frac {b-\frac {ad}c}{cx+d}+\frac ac$, which should allow you to proceed a bit further... Perhaps scaling to eliminate one of the coefficients would help. – abiessu Jan 22 '21 at 20:37
  • In other words a rational function of the form $$\frac{1}{ax+b} + c$$ is easily rewritten in the form you want. Such functions are generically called linear fractional transformations. – hardmath Jan 22 '21 at 20:45

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