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Math Help - Quadrature formula problem

  1. #1
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    Quadrature formula problem

    I am having some trouble with a math problem ( attached document). It is probably something easy, but I just can't see it. Thank you in advance!
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  2. #2
    Senior Member roninpro's Avatar
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    On part (a), did you try directly computing both sides and solving the resulting system of equations for \omega_0, \omega_1?
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  3. #3
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    I did. I get w0 + w1 =2. I was not sure that fully prove that w0=w1=1. Any ideas for b) ?
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  4. #4
    Senior Member roninpro's Avatar
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    You're right. Your single equation doesn't prove part (a) yet. Let me just review the situation for you:

    We want to compute \int_{-1}^1 f(x)\text{ dx}=\omega_0 f(-\alpha)+\omega_1 f(\alpha), where f(x)=mx+b, for m\ne 0. Taking the integral of the left hand side and evaluating the right hand side gives

    2b=\omega_0(-m\alpha+b)+\omega_1(m\alpha+b)

    Expanding the terms:

    2b=(\omega_0-\omega_1)(m\alpha)+(\omega_0+\omega_1)b

    Now, we have two systems of equations (by comparing the coefficients of each m\alpha and b on both sides):

    \omega_0-\omega_1=0
    \omega_0+\omega_1=2

    Solving the system shows that \omega_0=\omega_1=1.


    For part (b), I might apply the same strategy and see what happens.
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  5. #5
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    Any luck for b) ?
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  6. #6
    Senior Member roninpro's Avatar
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    Did you try it?
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  7. #7
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    I did. I applied the same strategy for b). However, I did no get the answer. To be honest, this happened probably because I am not exactly sure what I am supposed to get as an answer and whether I should put w0=w1=1 (the result from a) ). Any thoughts?
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  8. #8
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    I agree it is not very clear but to get a single value of \alpha in (b) you must use the values from (a): \omega_0= \omega_1= 1.

    For a quadratic function, a_0+ a_1x+ a_2x^2, the formula becomes \int_{-1}^1 a_0+ a_1x+ a_2x^2 dx= f(-\alpha)+ f(\alpha).

    It is easy to calculate that
    \int_{-1}^1 a_0+ a_1x+ a_2x^2 dx= 2a_0+ \frac{2}{3}a_2 while \omega_0f(-\alpha)+ \omega_1f(\alpha)=  a_0- a_1\alpha+ a_2\alpha^2+ a_0+ a_1\alpha+ a_2\alpha^2= 2a_0+ 2a_2\alpha^2.
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  9. #9
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    So, as a result we get that alpha equals +- the square root of 1/3. Is that correct?
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  10. #10
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    Yes, I believe it is.
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