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Math Help - Interpolation functions

  1. #1
    Senior Member bugatti79's Avatar
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    Interpolation functions

    I am trying to establish what interpolation functions he used to obtain the 8 approximation functions (eqn 5.3.25a) shown in the attached excerpt from the book.


    He states that a cubic approximation can be used for w(x) and an interpendent quadratic approximation for \Psi(x) thus I write


    \Delta^e_1=w^e_1=c_1^e+c_2^ex_e+c_3^ex_e^2+c_4^ex_  e^3
    \Delta^e_2=S^e_1=c_1^e+c_2^ex_e+c_3^ex_e^2+0


    \displaystyle\Delta^e_3=w^e_2=c_1^e+c_2^ex_{e+1}+c  _3^ex_{e+1}^2+c_4^ex_{e+1}^3
    \displaystyle\Delta^e_4=S^e_2=c_1^e+c_2^ex_{e+1}+c  _3^ex_{e+1}^2+0


    Putting into matrix form and inverting to find the c values wolfram comes up with the matrix being singular (note I am using dummy variables).


    inverse[{{1,x,x^2,x^3},{1,x,x^2,0},{1,y,y^2,y^3},{1,y,y^2, 0}}] - Wolfram|Alpha

    Note: This has been submitted to PF 3 weeks ago with no response
    Interdependent Interpolation Element- Timoshenko Beam

    Regards
    bugatti
    Attached Thumbnails Attached Thumbnails Interpolation functions-interpolation-function.jpg  
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  2. #2
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    Re: Interpolation functions

    Hey bugatti79.

    Can you reduce it to reduced row-echelon form to show what is linearly dependent?
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  3. #3
    Senior Member bugatti79's Avatar
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    Re: Interpolation functions

    Hi Chiro,

    See link from Wolfram [{{1,x,x^2,x^3},{1,x,x^2,0},{1,y,y^2,y^3},{1,y,y^2, 0}}] row echelon form - Wolfram|Alpha

    This indicates the functions are linearly dependent, right?
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  4. #4
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    Re: Interpolation functions

    Yeah it's definitely linear dependent because the last row in the last column has a 0.

    The next step is to figure out exactly what other vectors the last is linearly dependent on: in other words write {1,y,y^2,0} in terms of a linear combination of the other basis vectors and this will highlight what other data it depends on.
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  5. #5
    Senior Member bugatti79's Avatar
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    Re: Interpolation functions

    Quote Originally Posted by chiro View Post
    Yeah it's definitely linear dependent because the last row in the last column has a 0.
    What is the significance of a_{44}=0 for linear dependence since we also have a_{24}=0 in the original matrix..?

    Quote Originally Posted by chiro View Post
    The next step is to figure out exactly what other vectors the last is linearly dependent on: in other words write {1,y,y^2,0} in terms of a linear combination of the other basis vectors and this will highlight what other data it depends on.
    Not sure what we are trying to establish here...? Is my interpretation of the matrix correct according to the book?

    thanks
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  6. #6
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    Re: Interpolation functions

    For the second question you need to consider how to write {1,y,y^2,0} = a*{1,x,x^2,x^3} + b*{1,x,x^2,0} + c*{1,y,y^2,y^3} for constants a, b, and c.

    For the first question you might want to calculate the symbolic determinant such that it is unequal to zero for a given value of a_24 and a_44 and see what values they have to be to get a proper inverse.
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  7. #7
    Senior Member bugatti79's Avatar
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    Re: Interpolation functions

    Quote Originally Posted by chiro View Post
    For the second question you need to consider how to write {1,y,y^2,0} = a*{1,x,x^2,x^3} + b*{1,x,x^2,0} + c*{1,y,y^2,y^3} for constants a, b, and c.

    For the first question you might want to calculate the symbolic determinant such that it is unequal to zero for a given value of a_24 and a_44 and see what values they have to be to get a proper inverse.
    Hmm, ok. Why do the vectors have to be linearly independent? Ie, does a matrix of linearly dependent vectors become singular upon inverse?
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  8. #8
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    Re: Interpolation functions

    You can't invert a matrix if the rows or columns are collinear: you get a singular matrix if this happens.

    To answer your question, yes: a matrix becomes singular when its rows or columns are linearly dependent in some way.
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