Solve the recursion $a_n=\frac{1}{4}2^{n-1}-1+3a_{n-1}$

Question

I'd know how to solve it if it weren't for that -1. Because of it, I can't divide the particular equation with $2^{n-2}$ to solve it. What can be done here?

Oh, sorry. The starting conditions are $a_3=1$, and by "solve" I mean, find the explicit for of $a_n$ (the non recursive one)

What is the start condition? I mean, the value of $x_0$? – LeeNeverGup Nov 28 '13 at 20:25 — LeeNeverGup, Nov 28 '13 at 20:25

hhsaffar · Accepted Answer · 2013-11-28T20:35:04.360

2

Hint:

Consider $b_n=a_n-\frac{1}{2}$

Can you write a recursive equation for $b_n$?

edited Nov 28 '13 at 20:35

answered Nov 28 '13 at 20:29

hhsaffar

2,937

Hey! That works! Thanks for giving me a hint – Luka Horvat Nov 28 '13 at 20:43

score 0 · Answer 2 · answered Nov 28 '13 at 22:49

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $a_{n} - 1/2 = 2^{n - 1}/4 + 3\pars{a_{n -1} - 1/2}$. Set $a_{n} - 1/2 = 2^{n - 1}b_{n}/4\quad\iff\quad b_{n} = 2^{3 - n}\pars{a_{n} - 1/2}$ with $b_{3} = 1/2$. $$ {1 \over 4}\,2^{n - 1}b_{n} = {1 \over 4}\,2^{n - 1} + 3\,{1 \over 4}\,2^{n -2}b_{n - 1}\quad\imp\quad b_{n} = 1 + {3 \over 2}\,b_{n - 1} $$ $$ b_{n} + 2 = {3 \over 2}\pars{b_{n - 1} + 2} = \pars{3 \over 2}^{2}\pars{b_{n - 2} + 2} = \cdots = \pars{3 \over 2}^{n - 3}\bracks{b_{n - \pars{n - 3}} + 2} $$ $$ a_{n} = \half + 2^{n - 3}b_{n} = \half + 2^{n - 3}\bracks{-2 + \pars{3 \over 2}^{n - 3}\,{5 \over 2}} = \half - 2^{n - 2}+ {5 \over 2}\,3^{n - 3} $$ $$\color{#0000ff}{\large% a_{n} = \half - 2^{n - 2}+ {5 \over 2}\,3^{n - 3} } $$

Solve the recursion $a_n=\frac{1}{4}2^{n-1}-1+3a_{n-1}$

2 Answers2