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Math Help - Issues with linear and quadratic multivariate Taylor series

  1. #1
    Newbie AmberLamps's Avatar
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    Question Issues with linear and quadratic multivariate Taylor series

    Hey guys

    I am having a few issues with my linear algebra course. Me and my friends are in a study group and we are working on this problem for one of our seminars:

    3. Find the linear and quadratic approximations to the funtion f(x,y)=e^(ax-by) around the point (0,0)

    We think we have found the linear approximation of this as:

    1 + xae^(ax-by) - ybe^(ax-by)

    Is this correct? How do we get the quadratic approximation?

    Hope you can help guys, it would be much appreciated.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by AmberLamps View Post
    Hey guys

    I am having a few issues with my linear algebra course. Me and my friends are in a study group and we are working on this problem for one of our seminars:

    3. Find the linear and quadratic approximations to the funtion f(x,y)=e^(ax-by) around the point (0,0)

    We think we have found the linear approximation of this as:

    1 + xae^(ax-by) - ybe^(ax-by)

    Is this correct? How do we get the quadratic approximation?

    Hope you can help guys, it would be much appreciated.
    The taylor series for
    g(x)=e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}

    This is valid for any real number. So define

    f(x,y)=g(ax-by)=\sum_{n=0}^{\infty}\frac{(ax-by)^n}{n!}

    Now just crank out as many terms as you need.

    So when n=1 you get the linear approx is

    L(x,y)=\frac{(ax-by)^0}{0!}+\frac{(ax-by)^1}{1!}=1+ax-by
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  3. #3
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    Quote Originally Posted by AmberLamps View Post
    Hey guys

    I am having a few issues with my linear algebra course. Me and my friends are in a study group and we are working on this problem for one of our seminars:

    3. Find the linear and quadratic approximations to the funtion f(x,y)=e^(ax-by) around the point (0,0)

    We think we have found the linear approximation of this as:

    1 + xae^(ax-by) - ybe^(ax-by)

    Is this correct? How do we get the quadratic approximation?

    Hope you can help guys, it would be much appreciated.
    Close- You have to take the value of the derivatives at (0,0):
    since e^(a0- b0)= 1, the linear approximation is 1+ ax- by.

    The quadratic approximation will be have the second derivatives and second powers added:

    1+ ax- by+ \frac{\partial^2 f}{\partial x^2}(0,0)x^2+ \frac{\partial^2 f}{\partial x\partial y}(0,0) xy+ \frac{\partial^2 f}{\partial y^2}(0,0) y^2
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  4. #4
    Newbie AmberLamps's Avatar
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    Thanks for the help guys! it is much appreciated. So following that for the quadratic you get the formula below when evaluated at (0,0), correct?

     1+ ax - by + a^2x^2 + b^2y^2 - 2baxy

    The reason I ask is because the way this material was presented in lectures was in vector/matrices form and following that method my answer is subtly different due to as it becomes


    1/2 [1+ ax - by + a^2x^2 + b^2y^2 - 2baxy]

    due to a  1/2! term.
    Last edited by AmberLamps; February 22nd 2010 at 03:51 PM. Reason: mistake
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  5. #5
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    You may have misunderstood that "1/2". If you have the quadratic form ax^2+ bxy+ cy^2, you can write that as the matrix multiplication
    \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}a & \frac{1}{2}b \\ \frac{1}{2}b & c \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}.

    Note that the "1/2" is only in the off diagonal terms. It is there so that when you multiply that out and add the two "xy" terms, it sums to bxy.
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  6. #6
    Newbie AmberLamps's Avatar
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    yeh apologies I meant to change my answer to:

     1+ax + by + 1/2a^2x^2+1/2b^2x^2+abxy

    So I understand how it comes to bxy in your example. But do not understand why the 1/2 is only in the off diagonal terms; because all the examples I am able to find online seem to have 1/2 for all terms? Thanks again for your help.
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