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    Hi, I'm really struggling to get the answer with these boundary locus methods. If anyone could help that would be great. The linear multistep method is

    y[n+2] - 4/3 y[n+1] +1/3 y[n] = 2/3 hf[n+2]

    I get the stability polynomial to be,

    pi(r,h) = r^2 - 4/3 r + 1/3 - 2/3 hr^2

    Then sub in r=e^(i*theta) and set the stability polynomial = 0, then find the endpoints of the interval such that Im(h(theta)) = 0 and end up with

    theta = 0 and theta = 9

    However after this I get really stuck, the answer is

    h is a member of (-infinity, 0) union (4, infinity)

    and I can't work out how to get the 4 at all.

    Thanks again for any help!
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    (Original post by Blue7195)
    I get the stability polynomial to be,

    pi(r,h) = r^2 - 4/3 r + 1/3 - 2/3 hr^2

    Then sub in r=e^(i*theta) and set the stability polynomial = 0, then find the endpoints of the interval such that Im(h(theta)) = 0 and end up with

    theta = 0 and theta = 9
    This really isn't my area - it looks like stability theory for linear multistep methods for the numerical solution of differential equations - but perhaps we can have a go if you tell us a bit more.

    I thought that the idea of the boundary locus method was that you attempt to find roots of the stability polynomial that are of absolute value equal to one (and hence the substitution of r=e^(i*theta)). But then you are looking for values of theta that are in [0, 2*Pi] aren't you? Where does the value 9 come from?
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    (Original post by Gregorius)
    This really isn't my area - it looks like stability theory for linear multistep methods for the numerical solution of differential equations - but perhaps we can have a go if you tell us a bit more.

    I thought that the idea of the boundary locus method was that you attempt to find roots of the stability polynomial that are of absolute value equal to one (and hence the substitution of r=e^(i*theta)). But then you are looking for values of theta that are in [0, 2*Pi] aren't you? Where does the value 9 come from?
    I was solving the imaginary part of h(theta) = 0, got the values sin(theta) = 0 and cos(theta) = 2/3 and subbed these back in to h(theta). But I really am clueless with how to do these questions so that may be wrong.
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    (Original post by Blue7195)
    I was solving the imaginary part of h(theta) = 0, got the values sin(theta) = 0 and cos(theta) = 2/3 and subbed these back in to h(theta). But I really am clueless with how to do these questions so that may be wrong.
    I'm puzzled as to what you are doing - any chance of a few more details? If I take

     \displaymode \pi(r,h) = (1 - 2h/3) r^2 -(4/3) r + 1/3

    and substitute in  r = e^{i \theta} to get g(\theta, h) and then solve this equal to zero, then I get roots for \theta that are functions of h - not the other way around - so I'm not sure what Im(h(theta))=0 means!
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    (Original post by Gregorius)
    I'm puzzled as to what you are doing - any chance of a few more details? If I take

     \displaymode \pi(r,h) = (1 - 2h/3) r^2 -(4/3) r + 1/3

    and substitute in  r = e^{i \theta} to get g(\theta, h) and then solve this equal to zero, then I get roots for \theta that are functions of h - not the other way around - so I'm not sure what Im(h(theta))=0 means!
    Im not really too sure of the method either, I'm following a lecture example. In the example they rearrange to get h(theta) in terms of cos(theta) and sin(theta). Then, they say they want the interval of absolute stability which are the endpoints of this interval at the points such that Im(h(theta)) = 0, where the region intersects the real axis. The solves to find theta then substitutes back into h(theta) to find it!
    I understood it at the time but when I try questions like this one it doesn't give me the right answer... Thanks for all your help btw!
 
 
 
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