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Derive Multivariable Taylor Series

  1. Aug 17, 2010 #1
    Hello all,

    I am currently studying multivariable calculus, and I am interested in the Taylor series for two variable function.

    I am not sure where to begin; I cannot understand any of the proofs (which are apparently sparse) on the internet; they all just state it using a sigma sum; not very helpful to try to learn how it came to be :)

    I was wondering how the 2-variable Taylor series is derived? :)

  2. jcsd
  3. Aug 17, 2010 #2
    Basic idea:

    Let f(x,y) be a sufficiently differentiable real function around the point (x0, y0). Consider the one-variable function

    F(a) =f(x0 + a*nx, y0 + a*ny)

    where a is a real number and (nx, ny) a real 2-vector. Try to calculate the one-dimensional Taylor expansion for the function F(a), and at the end put

    nx = x - x0
    ny = y -y0

    You'll find the two-dimensional Taylor series for f(x,y).
  4. Aug 17, 2010 #3
    For a more direct approach, you may try this:

    [tex]f(x+\Delta x, y+\Delta y) = \sum_{k=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{d^k}{dx^k} f(x,y+\Delta y)[/tex]

    [tex]= \sum_{k=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{d^k}{dx^k}
    \sum_{j=0}^{\infty} \frac{(\Delta y)^j}{j!} \frac{d^j}{dy^j}

    [tex]= \sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{(\Delta y)^j}{j!} \frac{d^k}{dx^k} \frac{d^j}{dy^j} f(x,y)[/tex]

    [tex]= \sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \frac{(\Delta x)^k}{k!} \frac{(\Delta y)^j}{j!} f^{(k,j)}(x,y)[/tex]
    Last edited: Aug 17, 2010
  5. Aug 17, 2010 #4


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    Dearly Missed

    After ross' eminent start, you should try to switch the order of summation, in an intelligent manner, in order to gain a nice formula.
  6. Aug 18, 2010 #5
    Thanks everyone, I nearly understand. But why is it only the change in x and y raised to the power?

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