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Let $a$ be a real number, and define function $f$ on $\mathbb{N}$ $\cup$ {$0$} $\times \mathbb{N}$ $\cup$ {$0$} as follows:

$f(0,0) = 1$

$f(m,0)=f(0,n)=1$

$f(m,n)=af(m,n-1)+(1-a)f(m-1,n-1)$

where $m,n$ are positive integers.

$(a)$ Find all $a$ such that $|f(m,n)| \leq 1989, \forall m,n \in \mathbb{N}$.

$(b)$ Find the explicit expression of $f(m,n)$

dcxt
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    Firstly you need to state what you have tried ... even if you just say you have no idea where to start. Secondly I am getting the solution $f(m,n)=1$ ? – Donald Splutterwit Jul 04 '17 at 22:10
  • Do your statements in your definition hold for every $m,n \in \mathbb{N}$? Is $a$ a fixed constant or does it vary? – Sambo Jul 04 '17 at 22:10
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    Man, I wonder what year somebody came up with this problem. – Chris Jul 04 '17 at 22:11
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    For each $a$ there is only one such $f$ (induction). $f(m,n)\equiv1$ satisfies the recurrence and initial conditions. – Bettybel Jul 04 '17 at 22:28

1 Answers1

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

$\ds{\mrm{f}\pars{m,n} = a\,\mrm{f}\pars{m,n - 1} + \pars{1 - a}\,\mrm{f}\pars{m - 1,n - 1}\,,\quad \left\{\begin{array}{lclcl} \mrm{f}\pars{0,0} & \ds{=} & \ds{1}&& \\[1mm] \ds{\mrm{f}\pars{m,0}} & \ds{=} & \ds{\mrm{f}\pars{0,n}} & \ds{=} & \ds{1} \end{array}\right.}$

$\ds{m, n \in \mathbb{Z}_{\ \geq\ 0}\,}$.

$$ \begin{array}{|l|}\hline\mbox{}\\ \quad\mbox{As already pointed out, in an above comment, by}\ \color{#44f}{\texttt{@Bettybel}}; \quad \\[1mm] \quad\ds{\mrm{f}\pars{m,n} = 1\,,\ \forall\ m,n\ \in\ \mathbb{Z}_{\ \geq 0}}\ \mbox{satisfies the above}\ \ds{\mrm{f}\pars{m,n}}\ \mbox{conditions. However,} \quad \\[1mm] \quad\mbox{it would be interesting to illustrate a general procedure which, in the present case,}\quad \\[1mm] \quad\mbox{can be considered an}\ overkill.\quad \\ \mbox{}\\ \hline \end{array} $$ $$\bbx{% \mbox{Lets}\quad\mc{F}\pars{x,y} \equiv \sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty}\mrm{f}\pars{m,n}x^{m}\,y^{n}\quad \mbox{such that}\quad\bbx{\mrm{f}\pars{m,n} = \bracks{x^{m}\,y^{n}}\mc{F}\pars{x,y}}} $$ Then, \begin{align} &\!\!\!\!\!\!\!\!\!\!\!\!\! \sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}\mrm{f}\pars{m,n}x^{m}y^{n} = a\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}\mrm{f}\pars{m,n - 1}x^{m}y^{n} + \pars{1 - a}\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty} \mrm{f}\pars{m - 1,n - 1}x^{m}y^{n}\label{1}\tag{1} \end{align}

  • $\large\mathsf{Left\ Hand\ Side}$: \begin{align} &\sum_{m = 1}^{\infty}x^{m}\bracks{\sum_{n = 0}^{\infty}\mrm{f}\pars{m,n}y^{n} - \mrm{f}\pars{m,0}} = \sum_{n = 0}^{\infty}y^{n}\sum_{m = 1}^{\infty}\mrm{f}\pars{m,n}x^{m} - \sum_{m = 1}^{\infty}x^{m} \\[5mm] = &\ \sum_{n = 0}^{\infty}y^{n}\bracks{\sum_{m = 0}^{\infty}\mrm{f}\pars{m,n}x^{m} -\mrm{f}\pars{0,n}} - \sum_{m = 1}^{\infty}x^{m} = \bbx{\mc{F}\pars{x,y} - {x \over 1 - x} - {1 \over 1 - y}}\label{2}\tag{2} \end{align}
  • $\large\mathsf{Right\ Hand\ Side\ \underline{First\ Term}}$: \begin{align} &a\sum_{m = 1}^{\infty}\sum_{n = 0}^{\infty}\mrm{f}\pars{m,n}x^{m}y^{n + 1} = ay\sum_{n = 0}^{\infty}y^{n}\bracks{\sum_{m = 0}^{\infty}\mrm{f}\pars{m,n}x^{m} - \mrm{f}\pars{m,0}} \\[5mm] = &\ \bbx{ay\,\mc{F}\pars{x,y} - {ay \over 1 - y}} \end{align}
  • $\large\mathsf{Right\ Hand\ Side\ \underline{Second\ Term}}$: \begin{align} &\pars{1 - a}\sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty} \mrm{f}\pars{m,n}x^{m + 1}y^{n + 1} = \bbx{\pars{1- a}xy\,\mc{F}\pars{x,y}}\label{3}\tag{3} \end{align}
    With \eqref{1}, \eqref{2} and \eqref{3}: \begin{align} &\mc{F}\pars{x,y} - {x \over 1 - x} - {1 \over 1 - y} = \bracks{ay\,\mc{F}\pars{x,y} - {ay \over 1 - y}} + \bracks{\pars{1- a}xy\,\mc{F}\pars{x,y}} \\[5mm] & \implies \mc{F}\pars{x,y} = {1 \over 1 - x}\,{1 \over 1 - y} = \sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty}x^{m}y^{n} \implies \bbox[15px,#ffe,border:1px dotted navy]{\ds{\mrm{f}\pars{m,n} = 1\,,\ \forall\ m,n\ \in\ \mathbb{Z}_{\, \geq\ 0}}} \end{align}
Felix Marin
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