Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

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

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

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

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

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


$A_2(a,b)=a$

$B_2(a,b)=b$

$A_3(a,b)=ab$

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

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

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


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

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

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

Thanks a lot for answers

share|improve this question
    
You want $A_2 = a$ and $B_2=b$. Otherwise your formular'd be wrong –  roman Oct 10 '12 at 9:47
add comment

2 Answers

up vote 1 down vote accepted

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: $$ A_n(a,b)=U_n(b,-a)=2^{1-n}\sum_{k=0}^{\lfloor(n-1)/2\rfloor} \binom{n}{2k+1}b^{n-2k-1}(b^2+4a)^k $$

share|improve this answer
    
Could you please check upper limit of summation? –  Mathlover Oct 10 '12 at 13:21
    
@math thanks for spotting... –  draks ... Oct 10 '12 at 15:16
add comment

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.

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

$$r^2-br-a=0$$

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

$$f_n(x)=r^nx$$

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

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

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

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

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.