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Multivariate taylor

  1. Apr 26, 2008 #1

    gop

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    1. The problem statement, all variables and given/known data

    Calculate the taylor polynom of order 3 at (0,0,0) of the function with well-known series (that means I can't just take the derivatives)

    [tex]f(x,y,z)=\sqrt{e^{-x}+\sin y+z^{2}}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I wrote the functions within the square root as taylor polynomials and got

    [tex]f(x,y,z)=\sqrt{1+-x+\frac{1}{2}x^{2}-\frac{1}{6}x^{3}+y-\frac{1}{6}y^{3}+z^{2}}[/tex]

    But then I don't really know how to "remove" the square root. I already tried to just plug the term inside the square root in the taylor expansion of [tex]\sqrt{1+x}[/tex] but that didn't really work out very well.
     
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  3. Apr 27, 2008 #2

    HallsofIvy

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    Why did you do that? What not just write [itex]f(x,y,z)= (e^{-x}+ sin(y)+ z^2)^{1/2}[itex] and calculate the derivatives?
     
  4. Apr 27, 2008 #3

    benorin

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    Homework Helper

    Reference The formula for the Taylor series expansion of [itex]f(x,y,z)[/itex] about the point [itex](x_0,y_0,z_0)[/itex] is

    [tex]f(x,y,z)=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\sum_{j=0}^{\infty} \frac{\partial ^{n+k+j}f (x_0,y_0,z_0)}{\partial x^{n}\partial y^{k}\partial z^{j}} \cdot\frac{(x-x_0)^{n}}{n!} \cdot\frac{(y-y_0)^{k}}{k!}\cdot\frac{(z-z_0)^{j}}{j!} [/tex]​

    Use To compute the Tayor polynomial of order 3, only write out the terms for which [itex]n+k+j\le 3[/itex].
     
  5. Apr 27, 2008 #4

    gop

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    As stated I have to use "well-known series" to arrive at the taylor polynomial; thus, I'm not allowed to just take derivatives.
     
  6. Apr 27, 2008 #5

    HallsofIvy

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    Sorry, I missed reading that part!

    Okay, what is the Taylor's series for [iitex]\sqrt{x}[/itex]?
     
  7. Apr 27, 2008 #6

    gop

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    I can't really calculate the taylor series at x=0 because the derivatives is then of the form 1/0 and doesn't exist. I already tried sqrt(1+x) but that didn't produce a correct result.
     
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