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Third degree Taylor polynomial in two variables

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the third degree Taylor polynomial about the origin of

    [itex]f(x,y) = \frac{\cos(x)}{1+xy}[/itex]

    2. Relevant equations

    3. The attempt at a solution

    From my ventures on the Internet, this is my attempt:

    I see that

    [itex]\cos(x) = 1 - \frac{1}{2}x^2 + \frac{1}{4!}x^4 - \cdots[/itex]

    [itex]\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots[/itex]

    and so

    [itex]\frac{1}{1+(xy)} = 1 - xy + x^2y^2 - x^3y^3 + \cdots[/itex]

    Therefore, in multiplying them out,

    [itex]f(x,y) = \frac{\cos(x)}{1+xy} = 1 - xy + x^2y^2 - x^3y^3 - x + x^2y - x^3y^2 + \cdots[/itex]

    And I suppose that would be my answer.

    Do I have the right idea?

    Thanks in advance.

    EDIT: Oops! I didn't substitute [itex](x,y) = (0,0)[/itex]. So, in doing that, I should get precisely [itex]1[/itex]. That is my answer.

    EDIT: But [itex]1[/itex] doesn't seem right...

    EDIT: I think I am getting confused. Pretty sure the product above would be the answer, D=
     
    Last edited: Oct 24, 2012
  2. jcsd
  3. Oct 24, 2012 #2

    Zondrina

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    Homework Helper

    Okay so, first things first, recall the taylor series of cos(x) around a = 0 ( Aka the maclaurin series ). As well as the one for your other fraction there which I believe you've already done.

    Now you want at most a polynomial of degree three, correct? So take all your terms from both series which are polynomials of degree three or less, so for example for cosx you would choose :

    [itex]1 - \frac{1}{2} x^2[/itex]

    Now do the same for your other series and multiply the two resulting equations together. What do you get?
     
  4. Oct 24, 2012 #3
    Thanks for the reply, Zondrina.

    Ohhh, so that is what is meant by "[itex]n[/itex]-th degree!"

    Alright, then I would have

    [itex]f(x,y) = (1 - \frac{1}{2}x^2) (1 - xy + x^2y^2 - x^3y^3)[/itex]

    so

    [itex]f(x,y) = 1 - xy + x^2y^2 -x^3y^3 - \frac{1}{2}x^2 + \frac{1}{2}x^3y - \frac{1}{2}x^4y^2 + \frac{1}{2}x^5y^3[/itex]

    However, as you said, I only want degree three or less, so I suppose I should eliminate all terms consisting of variables of degree greater than three, getting

    [itex]f(x,y) = 1 - xy + x^2y^2 -x^3y^3 - \frac{1}{2}x^2 + \frac{1}{2}x^3y[/itex]

    And, this would be my answer?
     
  5. Oct 24, 2012 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You are still mis-understanding what "nth degree" meas. [itex]x^2y^2[/itex] is of degree 4 and [itex]x^3y^3[/itex] is of degree 6.
     
  6. Oct 24, 2012 #5

    Thanks for the reply, HallsofIvy.

    Oh dear, I see. Hopefully this will be my last change of answer:

    [itex]f(x,y) = (1 - \frac{1}{2}x^2) (1 - xy) = 1 - xy - \frac{1}{2}x^2[/itex]

    Do I now have the correct idea? D=
     
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