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Math Help - Truncation and Taylor series

  1. #1
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    Truncation and Taylor series

    Hi all,

    If I expand the function

    y(t_{n+1}) about y(t_n) and get

    y(t_{n+1}) = y(t_n) + h y'(t_n) + h^2/2 y''(t_n) + ....

    what do I get if I expand the function

    y(t_{n+2}) = ?
    Last edited by CaptainBlack; October 16th 2009 at 01:32 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by davefulton View Post
    Hi all,

    If I expand the function

    y(t_{n+1}) about y(t_n) and get

    y(t_{n+1}) = y(t_n) + h y'(t_n) + h^2/2 y''(t_n) + ....

    what do I get if I expand the function

    y(t_{n+2}) = ?
    I think you ought to tell us what the actual question is.

    CB
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  3. #3
    Junior Member
    Joined
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    My question is this;

    Calculate the truncation error for the following scheme

     y_{n+2} - y_{n+1} = \frac{h}{2} (3f_{n+2}-f_{n})

    So I get

     y_{n+2}-y_{n+1}-\frac{h}{2}(3f_{n+2}-f_{n})=0

    I then assume  y_{n}=y(t_{n})

    I also know that  f(y_{n},t_{n})=y'

    This gives me

     y(t_{n+2})-y(t_{n+1})-\frac{h}{2}(3y'(t_{n+2})-y'(t_{n})\approx0

    Using a taylor series expansion about  t_{n}

    This is where I get confused.

    I have to get the equation all in terms of  t_{n}

    I know that  y(t_{n+1})

    is

     y(t_{n+1}) = y(t_{n}) + hy'(t_{n+1} + ...

    and that

     y(t_{n+2}) = y(t_{n}) + 2hy'(t_{n+1}) + ...

    since

     y(t_{n+2}) - y(t_{n}) = 2h

    but how do I expand the

     y'(t_{n+2})

    as a Taylor series or am I doing something wrong

    Thank you for your reply
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