Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 6 of 6
  1. MHB Apprentice
    Theia's Avatar
    Status
    Offline
    Join Date
    Mar 2016
    Posts
    48
    Thanks
    120 times
    Thanked
    111 times
    #1
    Hello!

    I started to study a differential equation, and turned the problem into a difference equation ($ \displaystyle a_0, a_1 \in \mathbb{R}, a_2 = 0, a_3 = \frac{a_0 + a_1}{6}$)

    $ \displaystyle a_{v+2} = \frac{v^2a_v + a_{v-1}}{(v+1)(v+2)}$, where $ \displaystyle v \ge 2$.

    The numbers $ \displaystyle a_v$ are coefficients of a serie solution to the original differential equation.

    But now the question: I'd like to prove if the serie converges or not (i.e. is it usefull at all). But how is this done in this case? As far as I can see, the difference equation has not err... a nice solution and so all I have is the difference equation. Sometimes I've seen, someone takes a limit of an equation and assumes $ \displaystyle a_{\infty} = a$ and then substituting this into the equation. But this method clearly fails in this case. Are there another methods?

    Thank you! ^^

  2. Indicium Physicus
    MHB Math Scholar
    MHB POTW Director
    MHB Ambassador

    Status
    Offline
    Join Date
    Jan 2012
    Location
    Raleigh, NC
    Posts
    3,620
    Thanks
    9,732 times
    Thanked
    8,825 times
    Thank/Post
    2.438
    Trophies
    1 Highscore
    Awards
    MHB Math Notes Award (2016)  

MHB Math Notes Award (2015)
    #2
    Hmm, well one thing you could say, if you were trying to take the limit as $v\to\infty$, is that
    \begin{align*}
    \lim_{v\to\infty}a_{v+2}&=\lim_{v\to\infty}\frac{v^2a_v+a_{v-1}}{v^2+3v+2} \\
    &=\lim_{v\to\infty}\frac{a_v+\frac{a_{v-1}}{v^2}}{1+\frac{3}{v}+\frac{2}{v^2}}.
    \end{align*}
    From this, we see that $a_{v+2}\to a_v$. This isn't a proof of convergence, but I think I'd be shocked if it didn't.

  3. MHB Seeker
    MHB Global Moderator
    MHB Math Scholar
    I like Serena's Avatar
    Status
    Offline
    Join Date
    Mar 2012
    Location
    Netherlands
    Posts
    6,029
    Thanks
    4,067 times
    Thanked
    11,384 times
    Thank/Post
    1.888
    Awards
    MHB Model Helper Award (2016)  

MHB Best Ideas (2016)  

MHB LaTeX Award (2016)  

MHB Calculus Award (2014)  

MHB Discrete Mathematics Award (Jul-Dec 2013)
    #3
    Quote Originally Posted by Ackbach View Post
    Hmm, well one thing you could say, if you were trying to take the limit as $v\to\infty$, is that
    \begin{align*}
    \lim_{v\to\infty}a_{v+2}&=\lim_{v\to\infty}\frac{v^2a_v+a_{v-1}}{v^2+3v+2} \\
    &=\lim_{v\to\infty}\frac{a_v+\frac{a_{v-1}}{v^2}}{1+\frac{3}{v}+\frac{2}{v^2}}.
    \end{align*}
    From this, we see that $a_{v+2}\to a_v$. This isn't a proof of convergence, but I think I'd be shocked if it didn't.
    I'm afraid that doesn't quite fly.
    Consider for instance $a_v = \ln v$ that satisfies the criterion $a_{v+2}\to a_v$. Successive terms get arbitrarily close to each other... but the sequence still does not converge.

    Anyway, if we pick for instance $a_0=a_1=0$, the sequence does converge.
    I'm not sure yet if it holds for the general case though.

  4. MHB Apprentice
    Theia's Avatar
    Status
    Offline
    Join Date
    Mar 2016
    Posts
    48
    Thanks
    120 times
    Thanked
    111 times
    #4 Thread Author
    *sigh* I've tried a ratio test to attack on this. Just to see that it is inconclusive... Also trying to write e.g. $ \displaystyle a_{v+4}$ in terms of lower indices seems not to work and all I got was that $ \displaystyle a_{v+4} \to a_{v+2}$, which we already know.

    To make a numerical study, I coded a program that computes the coefficients until one of them is smaller than $ \displaystyle 10^{-9}$. If I got everything right, it looks like that needs about million terms (plus some hundreds of thousands terms depending on initial values)... But still, it is not saying anything does it really converge or not... Maybe this needs quite a tricky way to be handled properly.

  5. MHB Apprentice
    Theia's Avatar
    Status
    Offline
    Join Date
    Mar 2016
    Posts
    48
    Thanks
    120 times
    Thanked
    111 times
    #5 Thread Author
    ... Nice. Post just disappeared. -.- Anyway, if one writes $ \displaystyle a_v = \alpha_va_0 + \beta_va_1$, then one obtains two sequences pure numbers, namely

    $ \displaystyle \alpha_0 = 1, \alpha_1 = \alpha_2 = 0, \alpha_3 = \frac{1}{6}, \ldots$ and $ \displaystyle \beta_0 = 0, \beta_1 = 1, \beta_2 = 0, \beta_3 = \frac{1}{6}, \ldots$.

    Then one can compute more easily the continuation (the same original equation applies to both sequences). But does this help or not, I have no idea.

  6. MHB Apprentice
    Theia's Avatar
    Status
    Offline
    Join Date
    Mar 2016
    Posts
    48
    Thanks
    120 times
    Thanked
    111 times
    #6 Thread Author
    *phew* After some err... months(?!) I finally got an answer whether the serie - mentioned in the first post - converges or not. The answer is that there is some sort of radius of convergence around 0 where the serie converges and it is the solution for the ode I was studying (well... I'm still).

    Shortly, I was/am studying ode $ \displaystyle y'' = y\sin x$. Near $ \displaystyle x = 0$ I derived another form of the ode, namely $ \displaystyle (1 - t^2)u'' -tu' - tu =0$ ($ \displaystyle \ ' = \tfrac{d}{dt}$), which clearly has singular points at $ \displaystyle t = \pm 1$ (with the fact that substitution $ \displaystyle t = \sin x$ was used), there is no point to try to search values other than $ \displaystyle |x| \le 1$. But some numerical computations I made, gave me also a hint that the serie will break earlier, about at $ \displaystyle x \approx \sin 1$. (Up to about 9 million terms used.)

    Being such, I think this question is answered and case closed. I'll see what I can and want to do with the ode. Thank you again all the helpers! ^^
    Last edited by Theia; December 23rd, 2016 at 16:48.

Similar Threads

  1. [SOLVED] Laplace transform of the integral of a difference equation
    By Roberto in forum Calculus
    Replies: 2
    Last Post: March 1st, 2015, 06:01
  2. The finite difference method for the heat equation-error
    By mathmari in forum Advanced Applied Mathematics
    Replies: 16
    Last Post: February 13th, 2014, 14:14
  3. Difference equation tutorial: draft of part II
    By chisigma in forum Discrete Mathematics, Set Theory, and Logic
    Replies: 2
    Last Post: March 2nd, 2013, 05:47
  4. [SOLVED] solving a difference equation
    By Poirot in forum Discrete Mathematics, Set Theory, and Logic
    Replies: 1
    Last Post: May 7th, 2012, 17:26
  5. Difference equation tutorial: draft of part I
    By chisigma in forum Discrete Mathematics, Set Theory, and Logic
    Replies: 3
    Last Post: March 6th, 2012, 03:02

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards