A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

zmudz

  • one year ago

Let \(f(m,1) = f(1,n) = 1\) for \(m \geq 1, n \geq 1,\) and let \(f(m,n) = f(m-1,n) + f(m,n-1) + f(m-1,n-1)\) for \(m > 1\) and \(n > 1.\) Also, let \(S(n) = \sum_{a+b=n} f(a,b), \text{ for } a \geq 1, b \geq 1.\) Note: The summation notation means to sum over all positive integers \(a,b\) such that \(a+b=n.\) Given that \(S(n+2) = pS(n+1) + qS(n) \text{ for all } n \geq 2,\) for some constants p and q, find pq.

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I bet there's a neat solution that involves working through and possibly solving the first part with the double-index recurrence, but I'll admit I don't know how to work with those. Just working with the \(S(n)\) recurrence, I'm getting \(pq=2\).

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The first thing I did was to identify the first few terms of \(S(k)\). For \(f(m,n)\), you get this neat table: \[\begin{array}{cc|ccccc} &m&1&2&3&4&5&\cdots\\ n&\\ \hline 1&&\color{red}1&\color{green}1&1&1&1&\cdots\\ 2&&\color{green}1&\color{blue}3&5&7&9&\cdots\\ 3&&1&5&13&25&31&\cdots\\ 4&&1&7&25&63&129&\cdots\\ 5&&1&9&41&129&258&\cdots\\ \vdots&&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{array}\] with the pattern being \(\color{red}{\text{red}}+\color{green}{\text{green}}+\color{green}{\text{green}}=\color{blue}{\text{blue}}\). \(S(k)\) is the sequence given by the sum of the \(\color{green}{\text{greens}}\).

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So, we have the following recurrence: \[\begin{cases}S(2)=2\\S(3)=5\\S(k+2)=pS(k+1)+qS(k)&\text{for }k\ge2\end{cases}\] Denote the generating function of \(S(k)\) by \(\sigma(x)\), i.e. \(\sigma(x)=\sum\limits_{k\ge2}S(k)x^k\). You can solve for \(\sigma(x)\): \[\begin{align*} S(k+2)&=pS(k+1)+qS(k)\\[1ex] \sum_{k\ge2}S(k+2)x^k&=p\sum_{k\ge2}S(k+1)x^k+q\sum_{k\ge2}S(k)x^k\\[1ex] \frac{1}{x^2}\sum_{k\ge2}S(k+2)x^{k+2}&=\frac{p}{x}\sum_{k\ge2}S(k+1)x^{k+1}+q\sigma(x)\\[1ex] \frac{1}{x^2}\sum_{k\ge4}S(k)x^k&=\frac{p}{x}\sum_{k\ge3}S(k)x^k+q\sigma(x)\\[1ex] \frac{1}{x^2}\left(\sigma(x)-S(2)x^2-S(3)x^3\right)&=\frac{p}{x}\left(\sigma(x)-S(2)x^2\right)+q\sigma(x)\\[1ex] \sigma(x)&=\frac{(2p-5)x^3-2x^2}{qx^2+px-1} \end{align*}\]

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Finding the Taylor series for \(\sigma(x)\) by hand seemed excruciating to me, so I used Mathematica to find the first few terms (about \(x=0\)): \[\sigma(x)=2x^2+5x^3+(5p+2q)x^4+(5p^2+5q+2pq)x^5+\mathcal{O}(x^6)\] So, you have \[\begin{cases}S(4)=12\\S(5)=29\end{cases}~~\implies~~\begin{cases}5p+2q=12\\5p^2+5q+2pq=29\end{cases}~~\implies~~\begin{cases}5p+2q=12\\12p+5q=29\end{cases}\]

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Maybe this is the intended approach... As soon as you can establish the first few terms of \(S(k)\), the first half of the problem seems like fluff.

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Finding the Taylor series for \(\sigma(x)\) by hand seemed excruciating to me, so I used Mathematica to find the first few terms (about \(x=0\)): \[\sigma(x)=2x^2+5x^3+(5p+2q)x^4+(5p^2+5q+2pq)x^5+\mathcal{O}(x^6)\] So, you have \[\begin{cases}S(4)=12\\S(5)=29\end{cases}~~\implies~~\begin{cases}5p+2q=12\\5p^2+5q+2pq=29\end{cases}~~\implies~~\begin{cases}5p+2q=12\\12p+5q=29\end{cases}\]

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Also, using the GF is overkill. You can probably just solve this using the analogue of undetermined coefficients for difference equations.

  8. Zarkon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[S(n)=\frac{\sqrt{2}}{4}(1+\sqrt{2})^n-\frac{\sqrt{2}}{4}(1-\sqrt{2})n\]

  9. Zarkon
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    my \(n\) on the end fell down ;) \[\Large S(n)=\frac{\sqrt{2}}{4}(1+\sqrt{2})^n-\frac{\sqrt{2}}{4}(1-\sqrt{2})^n\]

  10. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.