2

Solve for $a_k$ in terms of $a_0$ and the other parameters in the following recurrence relation:

$a_k = ba_{k−1} + cr^k$, assuming $b \neq r$.

N. F. Taussig
  • 76,571
Valeria
  • 31
  • 3
    Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Dec 06 '20 at 21:30
  • $$a_k = ba_{k−1} + cr^k \tag{1}$$ $$a_{k+1} = ba_{k} + cr^{k+1} \tag{2}$$ Multiply $(1)$ with $r$ and subtract $(2)$ from the result. Solve what is left using Characteristic polynomial. – rtybase Dec 06 '20 at 22:49
  • Unfortunately the question is closed and I cannot provide you a details answer. But notice that if $b=0$ you already have the answer. Otherwise $$\dfrac{a_k}{b^k}-\dfrac{a_{k-1}}{b^{k-1}}=c\left(\dfrac{r}{b}\right)^k.$$ Now you can telescope this to compute a closed form for $a_k.$ – Bumblebee Dec 16 '20 at 21:52

3 Answers3

2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  • With $\ds{b \not= r,\quad a_{k} = ba_{k − 1} + cr^{k}}$.
  • \begin{align} \mbox{Lets}\quad d_{k} & \equiv {a_{k} \over r^{k}} \implies d_{k} = {b \over r}\,d_{k - 1} + c \\[2mm] \implies d_{k} - {rc \over r - b}& = {b \over r}\pars{d_{k - 1} - {rc \over r - b}} \end{align}
\begin{align} \implies & d_{k} - {rc \over r - b} = \pars{b \over r}^{2}\pars{d_{k - 2} - {rc \over r - b}} \\[2mm] & = \cdots = \pars{b \over r}^{k}\pars{d_{0} - {rc \over r - b}} \\[5mm] \implies & d_{k} = {rc \over r - b} + \pars{b \over r}^{k}\pars{d_{0} - {rc \over r - b}} \\[5mm] \implies & \bbx{a_{k} = {c \over r - b}\,r^{k + 1}\ +\ \pars{a_{0} - {rc \over r - b}}b^{k}} \\ & \end{align}
Felix Marin
  • 89,464
0

Welcome!! Generating function is the answer. Take $f(x)=\sum a_kx^k$, multiply by $x^k$ each sides and sum over $k$.. $$ \sum a_kx^k=bx\sum a_{k-1}x^{k-1}+c\sum r^kx^k $$ led to $$ f(x)=bxf(x)+\frac{c}{1-rx} $$ or $$ f(x)=\frac{c}{(1-bx)(1-rx)}. $$ Expand by partial fraction and you can find your $a_k$, or you can use the fact that $$ f(x)=c\sum b^kx^k\sum r^kx^k $$ can be written as $$ f(x)=c\sum_{n\geq0}\sum_{0\leq k\leq n}b^kr^{n-k}x^k $$ and so $$ f(x)=\sum_nc\frac{b^{n+1}-r^{n+1}}{b-r}x^n. $$

yngabl
  • 1,024
0

Hint:

$$a_k = ba_{k−1} + cr^k\\ a_{k+1} = ba_k+cr^{k+1}\\ \implies a_{k+1}-ba_k=r(a_k-ba_{k-1})\\ \implies a_{k+1}-ra_k=b(a_k-ra_{k-1}) $$

So both $a_{k+1}-ba_k$ and $a_{k+1}-ra_k$ are geometric sequences. Can you start from here?

If you still get stuck, take a look at this post.

Neat Math
  • 4,790