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Math Help - [SOLVED] General form of a Taylor's polynomial approx. to a multivariable function

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] General form of a Taylor's polynomial approx. to a multivariable function

    The problem states : Find the Taylor's polynomial of degree 2n of the function f(x,y)=\frac{1}{1+xy} in the point (0,0).
    My attempt: I've calculated the polynomial of degree 0 and 2. ( I doubt my choice of degree 0 is a good one... but as there's no restriction over n, I can guess it is not a bad choice.)
    I got that P_0(0,0)=1 and P_2(0,0)=1.
    So P_{2n}=1, \forall n \geq 0.
    I can't believe my result. Calculating P_4(0,0) seems a really long work! Have I done what I've done right?
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    Quote Originally Posted by arbolis View Post
    The problem states : Find the Taylor's polynomial of degree 2n of the function f(x,y)=\frac{1}{1+xy} in the point (0,0).
    My attempt: I've calculated the polynomial of degree 0 and 2. ( I doubt my choice of degree 0 is a good one... but as there's no restriction over n, I can guess it is not a bad choice.)
    I got that P_0(0,0)=1 and P_2(0,0)=1.
    So P_{2n}=1, \forall n \geq 0.
    I can't believe my result. Calculating P_4(0,0) seems a really long work! Have I done what I've done right?
    Not really sure what your P_0(0,0) and P_2(0,0) is but you'll notice that the function is

    f = \frac{1}{1+u} where u = xy so you can use

     <br />
\frac{1}{1+u} = 1 - u + u^2 - u^3 \cdots<br />
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks for the quick response.
    Oh... I didn't realize that I could use \frac{1}{1+u} = 1 - u + u^2 - u^3 \cdots.

    A stupid question from my part : from your formula, is P_0(x,y)=1 and P_1(x,y)=1-xy, P_2(x,y)=1-xy+x^2y^2, etc...?
    If so then I was right by saying that P_{2n}(0,0)=1. (I don't think I was right since it would imply that P_n(0,0)=1, no matter what n is).
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    Quote Originally Posted by arbolis View Post
    Thanks for the quick response.
    Oh... I didn't realize that I could use \frac{1}{1+u} = 1 - u + u^2 - u^3 \cdots.

    A stupid question from my part : from your formula, is P_0(x,y)=1 and P_1(x,y)=1-xy, P_2(x,y)=1-xy+x^2y^2, etc...?
    If so then I was right by saying that P_{2n}(0,0)=1. (I don't think I was right since it would imply that P_n(0,0)=1, no matter what n is).
    OK, I see your form. The Taylor series in 2 spatial variables is

     <br />
z = f(a,b) + \frac{\partial f}{\partial x}(a,b)(x-a) + \frac{\partial f}{\partial y}(a,b)(y-b)

     + \frac{\partial^2 f}{\partial x^2}(a,b)\frac{(x-a)^2}{2!} + 2 \frac{\partial^2 f}{\partial x \partial y}(a,b)\frac{(x-a)(y-b)}{2!}+ \frac{\partial^2 f}{\partial^2 y}(a,b)\frac{(y-b) ^2}{2!} \cdots.
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by danny arrigo View Post
    OK, I see your form. The Taylor series in 2 spatial variables is

     <br />
z = f(a,b) + \frac{\partial f}{\partial x}(a,b)(x-a) + \frac{\partial f}{\partial y}(a,b)(y-b)

     + \frac{\partial^2 f}{\partial x^2}(a,b)\frac{(x-a)^2}{2!} + 2 \frac{\partial^2 f}{\partial x \partial y}(a,b)\frac{(x-a)(y-b)}{2!}+ \frac{\partial^2 f}{\partial^2 y}(a,b)\frac{(y-b) ^2}{2!} \cdots.
    This is exactly what I've done on my draft and I got P_2(0,0)=1 because all the terms were equal to 0, except the first (that is f(0,0) which is worth 1.)
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  6. #6
    MHF Contributor arbolis's Avatar
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    Oh! I just realized my error...
    P_2(x,y)=1-xy+x^2y^2 is right and P_{2n}=\sum _{i=1}^{n} (-1)^n (xy)^n thanks to you. It's obvious that any of these polynomials evaluated in (0,0) give 1!
    I don't know what I was thinking, I thought that the whole polynomial was worth 1 and that there were a polynomial for every values of x and y. Senseless.
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