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$f(f(x))=ax+bf(x)$

$f(f(f(x)))=f_3(x)=af(x)+b(ax+bf(x))=abx+(a+b^2)f(x)$

$f(f(f(f(x))))=f_4(x)=abf(x)+(a+b^2)(ax+bf(x))=(a^2+ab^2)x+(2ab+b^3)f(x)$

$f_n(x)=A_n(a,b)x+B_n(a,b)f(x)$

$f_n(f(x))=A_n(a,b)f(x)+B_n(a,b)(ax+bf(x))$

$f_n(f(x))=aB_n(a,b)x+(bB_n(a,b)+A_n(a,b))f(x))$

$f_{n+1}(x)=A_{n+1}(a,b)x+B_{n+1}(a,b)f(x)$


$A_2(a,b)=a$

$B_2(a,b)=b$

$A_3(a,b)=ab$

$B_3(a,b)=a+b^2$

$A_{n+1}(a,b)=aB_n(a,b)$

$B_{n+1}(a,b)=A_n(a,b)+bB_n(a,b)$


$A_{n+2}(a,b)=aB_{n+1}(a,b)=aA_n(a,b)+abB_n(a,b))=aA_n(a,b)+bA_{n+1}(a,b)$

$A_{n+2}(a,b)=aA_n(a,b)+bA_{n+1}(a,b)$

How can be found the closed form expression of $A_{n}(a,b)$?

Thanks a lot for answers

  • 0
    You want $A_2 = a$ and $B_2=b$. Otherwise your formular'd be wrong2012-10-10

2 Answers 2

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Rewrite it like:

  1. $\frac1aA_{n+1}(a,b)=B_n(a,b)$

  2. $B_{n+1}(a,b)=A_n(a,b)+\frac baA_{n+1}(a,b)$

  3. $A_{n+2}(a,b)=aB_{n+1}(a,b)=a(A_n(a,b)+\frac baA_{n+1}(a,b))=aA_n(a,b)+bA_{n+1}(a,b)$

Now $A_{n+2}(a,b)=bA_{n+1}(a,b)+aA_n(a,b)$ looks like a Lucas Sequence $U_n(b,-a)$. Equation (33) in the linked MathWorld page gives a closed form: $ A_n(a,b)=U_n(b,-a)=2^{1-n}\sum_{k=0}^{\lfloor(n-1)/2\rfloor} \binom{n}{2k+1}b^{n-2k-1}(b^2+4a)^k $

  • 0
    @math thanks for spotting...2012-10-10
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I would like to share my solution for the problem

$f(f(x))=ax+bf(x)$

$f(x)=rx$ is a solution of the equation.

$r^2x=(a+br)x$

$r^2-br-a=0$

$r_1=\frac{b+\sqrt{b^2+4a}}{2}$ $r_2=\frac{b-\sqrt{b^2+4a}}{2}$

$f_n(x)=r^nx$

$f_n(x)=A_n(a,b)x+B_n(a,b)f(x)=(A_n(a,b)+rB_n(a,b))x$

$A_n(a,b)+(\frac{b+\sqrt{b^2+4a}}{2})B_n(a,b)=(\frac{b+\sqrt{b^2+4a}}{2})^n$

$A_n(a,b)+(\frac{b-\sqrt{b^2+4a}}{2})B_n(a,b)=(\frac{b-\sqrt{b^2+4a}}{2})^n$

$B_n(a,b)=\cfrac{(\frac{b+\sqrt{b^2+4a}}{2})^n-(\frac{b-\sqrt{b^2+4a}}{2})^n}{\sqrt{b^2+4a}}$

$A_{n+1}(a,b)=aB_n(a,b)=a\cfrac{(\frac{b+\sqrt{b^2+4a}}{2})^n-(\frac{b-\sqrt{b^2+4a}}{2})^n}{\sqrt{b^2+4a}}$

$A_{n}(a,b)=a\cfrac{(\frac{b+\sqrt{b^2+4a}}{2})^{n-1}-(\frac{b-\sqrt{b^2+4a}}{2})^{n-1}}{\sqrt{b^2+4a}}$